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En raison de limitations techniques, la typographie souhaitable du titre, «
Outils mathématiques pour la physique (PCSI) : Transformées monolatérales de Laplace directes et inverses et leur utilisation Outils mathématiques pour la physique (PCSI)/Transformées monolatérales de Laplace directes et inverses et leur utilisation », n'a pu être restituée correctement ci-dessus.
Pierre-Simon Laplace [ 1] théorie des probabilités qu'il a initiée, mais elle fut découverte par Leonhard Euler [ 2]
Soit une fonction réelle
f
{\displaystyle \;f\;}
t
{\displaystyle \;t\;}
« les valeurs de la fonction sont nulles pour
t
<
0
{\displaystyle \;t<0\;}
[ 3]
(
{\displaystyle {\big (}}
[ 4]
)
{\displaystyle {\big )}}
« la fonction est continue par morceaux sur tout intervalle
]
0
;
t
0
]
{\displaystyle \;\left]0\,;\,t_{0}\right]\;}
[ 5] , [ 6]
« au voisinage de
t
=
0
{\displaystyle \;t=0}
∃
γ
∈
]
0
;
1
[
{\displaystyle \;\exists \;\gamma \in \left]0\,;1\right[\;}
lim
t
→
0
[
t
γ
|
f
(
t
)
|
]
=
0
{\displaystyle \;\lim \limits _{t\,\rightarrow \,0}\left[t^{\gamma }\;\vert f(t)\vert \right]=0\;}
[ 7]
« la fonction est “ d'ordre exponentiel
α
{\displaystyle \;\alpha \;}
α
∈
R
∪
{
−
∞
}
{\displaystyle \;\alpha \,\in \,\mathbb {R} \cup \left\lbrace -\infty \right\rbrace \;}
« la fonction est “ d'ordre expo
⇕
{\displaystyle \Updownarrow }
∃
M
>
0
{\displaystyle \;\exists \;M>0\;}
∀
β
⩾
α
,
|
exp
(
−
β
t
)
f
(
t
)
|
<
M
,
∀
t
>
τ
{\displaystyle \;\forall \,\beta \,\geqslant \,\alpha ,\;\;\vert \exp \!\left(-\beta \;t\right)\;f(t)\vert <M,\;\;\forall \,t>\tau \;}
∀
τ
>
0
{\displaystyle \;\forall \,\tau >0\;}
[ 8]
Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] [ 14] Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
d'une fonction à support positif étendu à gauche de
0
{\displaystyle \;0\;}
[
0
,
+
∞
[
∪
V
(
0
−
)
{\displaystyle \;\left[0\,,\,+\infty \right[\;\cup \;{\mathcal {V}}(0^{-})\;}
{
{\displaystyle \;{\big \{}}
V
(
0
−
)
{\displaystyle \;{\mathcal {V}}(0^{-})\;}
Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
0
{\displaystyle \;\color {transparent}{0}\;}
un voisinage ouvert à gauche de
0
{\displaystyle \;0}
Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
0
{\displaystyle \;\color {transparent}{0}\;}
telle que la restriction de la fonction à
[
0
,
+
∞
[
C
{\displaystyle \;\left[0\,,\,+\infty \right[^{\,C}\;}
[ 15] Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
0
{\displaystyle \;\color {transparent}{0}\;}
V
(
0
−
)
{\displaystyle \;{\mathcal {V}}(0^{-})\;}
[ 16] [ 17]
}
{\displaystyle {\big \}}\;}
Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
d'une distribution
(
{\displaystyle \;{\big (}}
[ 18] [ 1] [ 19]
)
{\displaystyle {\big )}}
« Fonction de Heaviside[ 20]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}}
Y
(
t
)
=
{
1
si
t
>
0
0
si
t
<
0
}
{\displaystyle \;Y(t)=\left\lbrace {\begin{array}{l}1\;{\text{si}}\;t>0\\0\;{\text{si}}\;t<0\end{array}}\right\rbrace \;}
⇒
{\displaystyle \Rightarrow }
L
{
Y
(
t
)
}
=
1
p
{\displaystyle \;{\mathcal {L}}\!\left\lbrace Y(t)\right\rbrace ={\dfrac {1}{p}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 21] « Fonction de Heaviside
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}}
Y
(
t
)
=
{
1
si
t
>
0
0
si
t
<
0
}
{\displaystyle \;\color {transparent}{Y(t)=\left\lbrace {\begin{array}{l}1\;{\text{si}}\;t>0\\0\;{\text{si}}\;t<0\end{array}}\right\rbrace }\;}
⇒
{\displaystyle \color {transparent}{\Rightarrow }}
«
L
{
Y
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
Y
(
t
)
d
t
=
∫
0
+
+
∞
exp
(
−
p
t
)
d
t
=
[
exp
(
−
p
t
)
−
p
]
0
+
+
∞
=
1
p
{\displaystyle \;{\mathcal {L}}\!\left\lbrace Y(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;Y(t)\;dt=\displaystyle \int _{0^{+}}^{+\infty }\exp(-p\,t)\;dt=\left[{\dfrac {\exp(-p\,t)}{-p}}\right]_{0^{+}}^{+\infty }={\dfrac {1}{p}}\;}
[ 13]
« Pic de Dirac [ 22]
δ
(
t
)
=
{
0
si
t
>
0
+
∞
si
t
=
0
0
si
t
<
0
}
=
Y
˙
(
t
)
{\displaystyle \;\delta (t)=\left\lbrace {\begin{array}{l}0\qquad {\text{si}}\;t>0\\+\infty \;\;\,{\text{si}}\;t=0\\0\qquad {\text{si}}\;t<0\end{array}}\right\rbrace ={\dot {Y}}(t)\;}
[ 23]
⇒
{\displaystyle \Rightarrow }
L
{
δ
(
t
)
}
=
1
∀
ℜ
(
p
)
∈
R
{\displaystyle \;{\mathcal {L}}\!\left\lbrace \delta (t)\right\rbrace =1\;\;\forall \;\Re (p)\in \mathbb {R} \;}
[ 24] « Pic de Dirac d'impulsion unité
δ
(
t
)
=
Y
˙
(
t
)
{\displaystyle \;\color {transparent}{\delta (t)={\dot {Y}}(t)}\;}
si
ℜ
(
p
)
{\displaystyle \;\Re (p)\;}
>
0
{\displaystyle \;>0}
L
{
δ
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
δ
(
t
)
d
t
=
∫
0
−
+
∞
exp
(
−
p
t
)
Y
˙
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\!\left\lbrace \delta (t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\delta (t)\;dt=\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;{\dot {Y}}(t)\;dt\;}
[ 13] « Pic de Dirac d'impulsion unité
δ
(
t
)
=
Y
˙
(
t
)
{\displaystyle \;\color {transparent}{\delta (t)={\dot {Y}}(t)}\;}
ℜ
(
p
)
{\displaystyle \;\color {transparent}{\Re (p)}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
«
L
{
δ
(
t
)
}
=
[
exp
(
−
p
t
)
Y
(
t
)
]
0
−
+
∞
−
∫
0
−
+
∞
−
p
exp
(
−
p
t
)
Y
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\!\left\lbrace \delta (t)\right\rbrace ={\cancel {\left[\exp(-p\,t)\;Y(t)\right]_{0^{-}}^{+\infty }}}-\displaystyle \int _{0^{-}}^{+\infty }-p\,\exp(-p\,t)\;Y(t)\;dt\;}
[ 13] [ 18] , [ 25] « Pic de Dirac d'impulsion unité
δ
(
t
)
=
Y
˙
(
t
)
{\displaystyle \;\color {transparent}{\delta (t)={\dot {Y}}(t)}\;}
ℜ
(
p
)
{\displaystyle \;\color {transparent}{\Re (p)}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
L
{
δ
(
t
)
}
=
p
∫
0
+
+
∞
exp
(
−
p
t
)
d
t
{\displaystyle \;{\mathcal {L}}\!\left\lbrace \delta (t)\right\rbrace =p\displaystyle \int _{0^{+}}^{+\infty }\exp(-p\,t)\;dt\;}
[ 13]
=
p
[
exp
(
−
p
t
)
−
p
]
0
+
+
∞
=
p
1
p
=
1
{\displaystyle \;=p\left[{\dfrac {\exp(-p\,t)}{-p}}\right]_{0^{+}}^{+\infty }=p\;{\dfrac {1}{p}}=1\;}
« Pic de Dirac d'impulsion unité
δ
(
t
)
=
Y
˙
(
t
)
{\displaystyle \;\color {transparent}{\delta (t)={\dot {Y}}(t)}\;}
si
ℜ
(
p
)
{\displaystyle \;\Re (p)\;}
⩽
0
{\displaystyle \;\leqslant 0}
[ 26] , [ 27]
« Fonction rampe
f
rampe
(
t
)
=
{
a
t
si
t
⩾
0
0
si
t
⩽
0
}
{\displaystyle \;f_{\text{rampe}}(t)=\left\lbrace {\begin{array}{l}a\;t\;{\text{si}}\;t\geqslant 0\\0\quad {\text{si}}\;t\leqslant 0\end{array}}\right\rbrace \;}
⇒
{\displaystyle \Rightarrow }
L
{
f
rampe
(
t
)
}
=
a
p
2
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{rampe}}(t)\right\rbrace ={\dfrac {a}{p^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 28] « Fonction rampe
f
rampe
(
t
)
=
{
a
t
si
t
⩾
0
0
si
t
⩽
0
}
{\displaystyle \;\color {transparent}{f_{\text{rampe}}(t)=\left\lbrace {\begin{array}{l}a\;t\;{\text{si}}\;t\geqslant 0\\0\quad {\text{si}}\;t\leqslant 0\end{array}}\right\rbrace }\;}
⇒
{\displaystyle \color {transparent}{\Rightarrow }}
«
L
{
f
rampe
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
rampe
(
t
)
d
t
=
∫
0
+
+
∞
exp
(
−
p
t
)
a
t
d
t
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{rampe}}(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f_{\text{rampe}}(t)\;dt=\displaystyle \int _{0^{+}}^{+\infty }\exp(-p\,t)\;a\;t\;dt\;}
[ 13] [ 25] « Fonction rampe
f
rampe
(
t
)
=
{
a
t
si
t
⩾
0
0
si
t
⩽
0
}
{\displaystyle \;\color {transparent}{f_{\text{rampe}}(t)=\left\lbrace {\begin{array}{l}a\;t\;{\text{si}}\;t\geqslant 0\\0\quad {\text{si}}\;t\leqslant 0\end{array}}\right\rbrace }\;}
⇒
{\displaystyle \color {transparent}{\Rightarrow }}
L
{
f
rampe
(
t
)
}
=
[
exp
(
−
p
t
)
−
p
a
t
]
0
−
+
∞
−
∫
0
−
+
∞
exp
(
−
p
t
)
−
p
a
d
t
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{rampe}}(t)\right\rbrace ={\cancel {\left[{\dfrac {\exp(-p\,t)}{-p}}\;a\;t\right]_{0^{-}}^{+\infty }}}-\displaystyle \int _{0^{-}}^{+\infty }{\dfrac {\exp(-p\,t)}{-p}}\;a\;dt\;}
[ 13]
=
a
p
[
exp
(
−
p
t
)
−
p
]
0
−
+
∞
=
a
p
2
{\displaystyle \;={\dfrac {a}{p}}\;\left[{\dfrac {\exp(-p\,t)}{-p}}\right]_{0^{-}}^{+\infty }={\dfrac {a}{p^{2}}}\;}
ℜ
(
p
)
{\displaystyle \;\Re (p)\;}
>
0
{\displaystyle \;>0\;}
Début d’un théorème
Théorème d'unicité (admis)
Fin du théorème
Soient «
f
1
(
t
)
{\displaystyle \;f_{1}(t)\;}
f
2
(
t
)
{\displaystyle \;f_{2}(t)\;}
(
{\displaystyle \;{\big (}}
distributions
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
1
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f_{1}(t)\right\rbrace \;}
L
{
f
2
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f_{2}(t)\right\rbrace \;}
Soient «
f
1
(
t
)
{\displaystyle \;\color {transparent}{f_{1}(t)}\;}
f
2
(
t
)
{\displaystyle \;\color {transparent}{f_{2}(t)}\;}
«
α
=
max
{
α
1
,
α
2
}
{\displaystyle \;\alpha =\max \left\lbrace \alpha _{1}\,,\,\alpha _{2}\right\rbrace \;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Soient «
a
{\displaystyle \;a\;}
{
{\displaystyle \;{\big \{}}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
}
{\displaystyle {\big \}}}
on démontre les propriétés suivantes «
L
{
f
1
(
t
)
+
f
2
(
t
)
}
=
L
{
f
1
(
t
)
}
+
L
{
f
2
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f_{1}(t)+f_{2}(t)\right\rbrace ={\mathcal {L}}\left\lbrace f_{1}(t)\right\rbrace +{\mathcal {L}}\left\lbrace f_{2}(t)\right\rbrace \;}
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
on démontre les propriétés suivantes «
L
{
a
f
1
(
t
)
}
=
a
L
{
f
1
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace a\;f_{1}(t)\right\rbrace =a\;{\mathcal {L}}\left\lbrace f_{1}(t)\right\rbrace \;}
ℜ
(
p
)
>
α
1
{\displaystyle \;\Re (p)>\alpha _{1}\;}
on démontre les propriétés suivantes «
L
{
a
f
2
(
t
)
}
=
a
L
{
f
2
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace a\;f_{2}(t)\right\rbrace =a\;{\mathcal {L}}\left\lbrace f_{2}(t)\right\rbrace \;}
ℜ
(
p
)
>
α
2
{\displaystyle \;\Re (p)>\alpha _{2}\;}
Soit
f
(
t
)
{\displaystyle \;f(t)\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t)\;dt\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13] Soit
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
(
t
)
}
=
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\exp(-p\,t)\;f(t)\;dt}\;}
«
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
Soit
f
˙
(
t
)
{\displaystyle \;{\dot {f}}(t)\;}
(
{\displaystyle \;{\big (}}
distribution [ 18]
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
˙
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
˙
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;{\dot {f}}(t)\;dt\;}
[ 29]
ℜ
(
p
)
>
α
′
{\displaystyle \;\Re (p)>\alpha '\;}
[ 13] Soit
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
avec «
α
′
{\displaystyle \;\alpha '\;}
L
{
f
˙
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace \;}
[ 30]
nous établissons ci-après la propriété suivante
Influence de la dérivation sur la transformée (monolatérale) de Laplace d'une fonction ou distribution
Démonstration : « pour calculer
L
{
f
˙
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
˙
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;{\dot {f}}(t)\;dt\;}
[ 13] , [ 35]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
Démonstration : « pour calculer
L
{
f
˙
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
˙
(
t
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;{\dot {f}}(t)\;dt}\;}
dans le cas de «
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
Démonstration : « pour calculer
L
{
f
˙
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace }\;}
dans l'hypothèse où «
f
˙
(
t
)
{\displaystyle \;{\dot {f}}(t)\;}
[ 36]
t
=
0
{\displaystyle \;t=0\;}
[ 35] , [ 25]
⇒
{\displaystyle \Rightarrow }
Démonstration : « pour calculer
L
{
f
˙
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace }\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
t
=
0
{\displaystyle \;\color {transparent}{t=0}\;}
«
L
{
f
˙
(
t
)
}
=
[
exp
(
−
p
t
)
f
(
t
)
]
0
−
∞
−
∫
0
−
+
∞
−
p
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace =\left[\exp(-p\,t)\;f(t)\right]_{0^{-}}^{\infty }-\displaystyle \int _{0^{-}}^{+\infty }-p\;\exp(-p\,t)\;f(t)\;dt\;}
[ 13] Démonstration : « pour calculer
L
{
f
˙
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace }\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
t
=
0
{\displaystyle \;\color {transparent}{t=0}\;}
L
{
f
˙
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace }}
=
−
f
(
0
−
)
+
p
L
{
f
(
t
)
}
{\displaystyle =-f(0^{-})+p\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
[ 37] Démonstration : « pour calculer
L
{
f
˙
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace }\;}
dans l'hypothèse où «
f
˙
(
t
)
{\displaystyle \;{\dot {f}}(t)\;}
[ 36]
t
=
0
{\displaystyle \;t=0\;}
⇒
{\displaystyle \Rightarrow }
∫
0
−
0
+
exp
(
−
p
t
)
f
˙
(
t
)
d
t
=
0
{\displaystyle \;\displaystyle \int _{0^{-}}^{0^{+}}\exp(-p\,t)\;{\dot {f}}(t)\;dt=0\;}
f
˙
(
t
)
{\displaystyle \;{\dot {f}}(t)\;}
ère espèce[ 38] Démonstration : « pour calculer
L
{
f
˙
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace }\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
t
=
0
{\displaystyle \;\color {transparent}{t=0}\;}
⇒
{\displaystyle \color {transparent}{\Rightarrow }}
exp
(
−
p
t
)
f
˙
(
t
)
d
t
=
0
{\displaystyle \;\color {transparent}{\exp(-p\,t)\;{\dot {f}}(t)\;dt=0}\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
continue en
t
=
0
{\displaystyle \;t=0\;}
[ 39] Démonstration : « pour calculer
L
{
f
˙
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace }\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
t
=
0
{\displaystyle \;\color {transparent}{t=0}\;}
«
L
{
f
˙
(
t
)
}
=
∫
0
−
0
+
exp
(
−
p
t
)
f
˙
(
t
)
d
t
+
∫
0
+
+
∞
exp
(
−
p
t
)
f
˙
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace ={\cancel {\displaystyle \int _{0^{-}}^{0^{+}}\exp(-p\,t)\;{\dot {f}}(t)\;dt\;+}}\,\displaystyle \int _{0^{+}}^{+\infty }\exp(-p\,t)\;{\dot {f}}(t)\;dt\;}
[ 13]
⇒
{\displaystyle \Rightarrow }
Démonstration : « pour calculer
L
{
f
˙
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace }\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
t
=
0
{\displaystyle \;\color {transparent}{t=0}\;}
«
L
{
f
˙
(
t
)
}
=
∫
0
+
+
∞
exp
(
−
p
t
)
f
˙
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace =\displaystyle \int _{0^{+}}^{+\infty }\exp(-p\,t)\;{\dot {f}}(t)\;dt\;}
[ 13] [ 25] Démonstration : « pour calculer
L
{
f
˙
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace }\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
t
=
0
{\displaystyle \;\color {transparent}{t=0}\;}
«
L
{
f
˙
(
t
)
}
=
[
exp
(
−
p
t
)
f
(
t
)
]
0
+
∞
−
∫
0
+
+
∞
−
p
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace =\left[\exp(-p\,t)\;f(t)\right]_{0^{+}}^{\infty }-\displaystyle \int _{0^{+}}^{+\infty }-p\;\exp(-p\,t)\;f(t)\;dt\;}
[ 13] Démonstration : « pour calculer
L
{
f
˙
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace }\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
t
=
0
{\displaystyle \;\color {transparent}{t=0}\;}
L
{
f
˙
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace }}
=
−
f
(
0
+
)
+
p
L
{
f
(
t
)
}
{\displaystyle =-f(0^{+})+p\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
[ 37] , [ 40]
Soit
f
(
t
)
{\displaystyle \;f(t)\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t)\;dt\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13] Soit
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
(
t
)
}
=
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\exp(-p\,t)\;f(t)\;dt}\;}
«
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
Soit
∫
0
t
f
(
t
′
)
d
t
′
{\displaystyle \;\displaystyle \int _{0}^{t}f(t')\;dt'\;}
(
{\displaystyle \;{\big (}}
distribution [ 18]
)
{\displaystyle {\big )}\;}
t
=
0
{\displaystyle \;t=0\;}
Soit
∫
0
t
f
(
t
′
)
d
t
′
{\displaystyle \;\color {transparent}{\displaystyle \int _{0}^{t}f(t')\;dt'}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
admettant pour transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Soit
∫
0
t
f
(
t
′
)
d
t
′
{\displaystyle \;\color {transparent}{\displaystyle \int _{0}^{t}f(t')\;dt'}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
«
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
{
∫
0
t
f
(
t
′
)
d
t
′
}
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace dt\;}
[ 29]
ℜ
(
p
)
>
α
″
{\displaystyle \;\Re (p)>\alpha ''\;}
[ 13] Soit
∫
0
t
f
(
t
′
)
d
t
′
{\displaystyle \;\color {transparent}{\displaystyle \int _{0}^{t}f(t')\;dt'}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
«
α
″
{\displaystyle \;\alpha ''\;}
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace \;}
[ 41]
nous établissons ci-après la propriété suivante
Influence de l'intégration sur la transformée (monolatérale) de Laplace d'une fonction ou distribution
Avec «
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
la transformée
(
{\displaystyle \;{\big (}}
monolatérale
)
{\displaystyle {\big )}\;}
de Laplace
[ 1] de la fonction
(
{\displaystyle \;{\big (}}
ou
distribution
)
{\displaystyle {\big )}}
f
(
t
)
{\displaystyle \;f(t)\;}
»
[ 31] Avec «
L
{
f
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace }\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
d'« abscisse de convergence
α
{\displaystyle \;\alpha \;}
» et
Avec «
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace \;}
celle de la primitive de la fonction
(
{\displaystyle \;{\big (}}
ou de la primitive de la
distribution [ 29]
)
{\displaystyle {\big )}}
∫
0
t
f
(
t
′
)
d
t
′
{\displaystyle \;\displaystyle \int _{0}^{t}f(t')\;dt'\;}
Avec «
L
{
f
(
t
′
)
d
t
′
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t')\;dt'\right\rbrace }\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}}
s'annulant en
t
=
0
{\displaystyle \;t=0\;}
»
Avec «
L
{
f
(
t
′
)
d
t
′
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t')\;dt'\right\rbrace }\;}
d'« abscisse de convergence
α
″
{\displaystyle \;\alpha ''\;}
[ 41] »,
ces deux transformées de Laplace
[ 1] sont liées par la relation suivante :
«
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
=
L
{
f
(
t
)
}
p
{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace ={\dfrac {{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{p}}\;}
ℜ
(
p
)
>
α
″
{\displaystyle \;\Re (p)>\alpha ''\;}
Démonstration : « pour calculer
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
{
∫
0
t
f
(
t
′
)
d
t
′
}
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace dt\;}
[ 13] , [ 35]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Démonstration : « pour calculer
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
{
∫
0
t
f
(
t
′
)
d
t
′
}
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace dt}\;}
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace \;}
Démonstration : « pour calculer
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
{
∫
0
t
f
(
t
′
)
d
t
′
}
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace dt}\;}
dans le cas de «
ℜ
(
p
)
>
α
″
{\displaystyle \;\Re (p)>\alpha ''\;}
α
″
{\displaystyle \;\alpha ''\;}
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace \;}
Démonstration : « pour calculer
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace }\;}
on procède en intégrant par parties[ 35] , [ 25]
⇒
{\displaystyle \Rightarrow }
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
=
[
exp
(
−
p
t
)
−
p
∫
0
t
f
(
t
′
)
d
t
′
]
0
−
∞
−
∫
0
−
+
∞
exp
(
−
p
t
)
−
p
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace =\left[{\dfrac {\exp(-p\,t)}{-p}}\;\displaystyle \int _{0}^{t}f(t')\;dt'\right]_{0^{-}}^{\infty }-\displaystyle \int _{0^{-}}^{+\infty }{\dfrac {\exp(-p\,t)}{-p}}\;f(t)\;dt\;}
[ 13] Démonstration : « pour calculer
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace }\;}
⇒
{\displaystyle \color {transparent}{\Rightarrow }}
L
{
∫
0
t
f
(
t
′
)
d
t
′
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace }}
=
L
{
f
(
t
)
}
p
{\displaystyle ={\dfrac {{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{p}}\;}
[ 42]
Soit
f
(
t
)
{\displaystyle \;f(t)\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t)\;dt\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13] Soit
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
(
t
)
}
=
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\exp(-p\,t)\;f(t)\;dt}\;}
«
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
Soit
f
′
(
t
)
=
f
(
t
′
)
{\displaystyle \;f'(t)=f(t')\;}
(
{\displaystyle \;{\big (}}
distribution [ 18]
)
{\displaystyle {\big )}\;}
t
′
=
t
−
t
0
{\displaystyle \;t'=t-t_{0}\;}
t
{\displaystyle \;t\;}
t
0
>
0
{\displaystyle \;t_{0}>0\;}
[ 43] Soit
f
′
(
t
)
=
f
(
t
′
)
{\displaystyle \;\color {transparent}{f'(t)=f(t')}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
admettant pour transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
′
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
′
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f'(t)\;dt\;}
[ 29] Soit
f
′
(
t
)
=
f
(
t
′
)
{\displaystyle \;\color {transparent}{f'(t)=f(t')}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
′
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace }}
=
∫
t
0
−
+
∞
exp
[
−
p
(
t
′
+
t
0
)
]
f
(
t
′
)
d
t
′
{\displaystyle =\displaystyle \int _{t_{0}^{-}}^{+\infty }\exp \!\left[-p\,(t'+t_{0})\right]\,f(t')\;dt'\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13] Soit
f
′
(
t
)
=
f
(
t
′
)
{\displaystyle \;\color {transparent}{f'(t)=f(t')}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
′
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace }}
[
{\displaystyle {\big [}}
f
(
t
)
{\displaystyle \;f(t)\;}
[ 44]
]
{\displaystyle {\big ]}}
nous établissons ci-après la propriété suivante connue sous le nom du « théorème du retard »
Théorème du retard
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
−
t
0
)
d
t
=
∫
−
t
0
−
+
∞
exp
[
−
p
(
u
+
t
0
)
]
f
(
u
)
d
u
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t-t_{0})\;dt=\displaystyle \int _{-t_{0}^{-}}^{+\infty }\exp \!\left[-p\,(u+t_{0})\right]\,f(u)\;du\;}
[ 13] , [ 35] Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
′
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \;}
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
[ 44] Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }\;}
« si
t
0
{\displaystyle \;t_{0}\;}
>
0
{\displaystyle \;>0\;}
[ 35]
[
−
t
0
−
,
+
∞
[
{\displaystyle \;\left[-t_{0}^{-}\,,\,+\infty \right[\;}
[
−
t
0
−
,
0
−
[
∪
[
0
−
,
+
∞
[
{\displaystyle \;\left[-t_{0}^{-}\,,\,0^{-}\right[\cup \left[0^{-}\,,\,+\infty \right[\;}
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }\;}
t
0
{\displaystyle \;\color {transparent}{t_{0}}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
«
L
{
f
′
(
t
)
}
=
∫
−
t
0
−
0
−
exp
[
−
p
(
u
+
t
0
)
]
f
(
u
)
d
u
+
∫
0
−
+
∞
exp
[
−
p
(
u
+
t
0
)
]
f
(
u
)
d
u
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\displaystyle \int _{-t_{0}^{-}}^{0^{-}}\exp \!\left[-p\,(u+t_{0})\right]\,f(u)\;du+\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-p\,(u+t_{0})\right]\,f(u)\;du\;}
[ 13] Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }\;}
t
0
{\displaystyle \;\color {transparent}{t_{0}}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
L
{
f
′
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace }}
=
exp
(
−
p
t
0
)
[
∫
t
0
−
0
−
exp
(
−
p
u
)
f
(
u
)
d
u
+
∫
0
−
+
∞
exp
(
−
p
u
)
f
(
u
)
d
u
]
{\displaystyle =\exp(-p\,t_{0})\left[\displaystyle \int _{t_{0}^{-}}^{0^{-}}\exp \!\left(-p\;u\right)\,f(u)\;du+\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left(-p\;u\right)\,f(u)\;du\right]\;}
[ 13] Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }\;}
t
0
{\displaystyle \;\color {transparent}{t_{0}}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
L
{
f
′
(
t
)
}
=
exp
(
−
p
t
0
)
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\exp(-p\,t_{0})}}
la 1ère intégrale est nulle
[
{\displaystyle \;{\big [}}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}}
f
(
t
)
{\displaystyle \;f(t)\;}
]
{\displaystyle {\big ]}\;}
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }\;}
t
0
{\displaystyle \;\color {transparent}{t_{0}}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
L
{
f
′
(
t
)
}
=
exp
(
−
p
t
0
)
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\exp(-p\,t_{0})}}
la 2nde est égale à
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }\;}
t
0
{\displaystyle \;\color {transparent}{t_{0}}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
«
L
{
f
′
(
t
)
}
=
exp
(
−
p
t
0
)
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\exp(-p\;t_{0})\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }\;}
« si
t
0
{\displaystyle \;t_{0}\;}
<
0
{\displaystyle \;<0\;}
[ 35]
[
−
t
0
−
,
+
∞
[
{\displaystyle \;\left[-t_{0}^{-}\,,\,+\infty \right[\;}
[
0
−
,
+
∞
[
{\displaystyle \;\left[0^{-}\,,\,+\infty \right[\;}
[
0
−
,
−
t
0
−
[
{\displaystyle \;\left[0^{-}\,,\,-t_{0}^{-}\right[\;}
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }\;}
t
0
{\displaystyle \;\color {transparent}{t_{0}}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
«
L
{
f
′
(
t
)
}
=
∫
0
−
+
∞
exp
[
−
p
(
u
+
t
0
)
]
f
(
u
)
d
u
−
∫
0
−
−
t
0
−
exp
[
−
p
(
u
+
t
0
)
]
f
(
u
)
d
u
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-p\,(u+t_{0})\right]\,f(u)\;du-\displaystyle \int _{0^{-}}^{-t_{0}^{-}}\exp \!\left[-p\,(u+t_{0})\right]\,f(u)\;du\;}
[ 13] Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }\;}
t
0
{\displaystyle \;\color {transparent}{t_{0}}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
L
{
f
′
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace }}
=
exp
(
−
p
t
0
)
[
∫
0
−
+
∞
exp
(
−
p
u
)
f
(
u
)
d
u
−
∫
0
−
t
0
−
exp
(
−
p
u
)
f
(
u
)
d
u
]
{\displaystyle =\exp(-p\,t_{0})\left[\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left(-p\;u\right)\,f(u)\;du-\displaystyle \int _{0^{-}}^{t_{0}^{-}}\exp \!\left(-p\;u\right)\,f(u)\;du\right]\;}
[ 13] Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }\;}
t
0
{\displaystyle \;\color {transparent}{t_{0}}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
L
{
f
′
(
t
)
}
=
exp
(
−
p
t
0
)
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\exp(-p\,t_{0})}}
la 1ère intégrale entre crochets est égale à
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }\;}
t
0
{\displaystyle \;\color {transparent}{t_{0}}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
L
{
f
′
(
t
)
}
=
exp
(
−
p
t
0
)
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\exp(-p\,t_{0})}}
la 2nde est non nulle
⇒
{\displaystyle \Rightarrow }
L
{
f
′
(
t
)
}
≠
exp
(
−
p
t
0
)
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \neq \exp(-p\;t_{0})\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
=
L
{
f
(
t
−
t
0
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace }}
d'où, pour vérifier «
L
{
f
′
(
t
)
}
=
exp
(
−
p
t
0
)
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\exp(-p\;t_{0})\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
t
0
>
0
{\displaystyle \;t_{0}>0}
Soit
f
(
t
)
{\displaystyle \;f(t)\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t)\;dt\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13] Soit
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
(
t
)
}
=
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\exp(-p\,t)\;f(t)\;dt}\;}
«
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
Soit
f
′
(
t
)
=
f
(
t
′
)
{\displaystyle \;f'(t)=f(t')\;}
(
{\displaystyle \;{\big (}}
distribution [ 18]
)
{\displaystyle {\big )}\;}
t
′
=
k
t
{\displaystyle \;t'=k\;t\;}
t
{\displaystyle \;t\;}
k
>
0
{\displaystyle \;k>0\;}
[ 45]
≠
1
{\displaystyle \;\neq 1\;}
Soit
f
′
(
t
)
=
f
(
t
′
)
{\displaystyle \;\color {transparent}{f'(t)=f(t')}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
admettant pour transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
′
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
′
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f'(t)\;dt\;}
[ 29] Soit
f
′
(
t
)
=
f
(
t
′
)
{\displaystyle \;\color {transparent}{f'(t)=f(t')}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
′
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace }}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
k
t
)
d
t
{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(k\;t)\;dt\;}
[ 29] , [ 13]
ℜ
(
p
)
>
k
α
{\displaystyle \;\Re (p)>k\;\alpha \;}
[ 46]
nous établissons ci-après la propriété suivante connue sous le nom du « règle de similitude »
Règle de similitude [ou influence d'un changement d'échelle de t sur la transformée (monolatérale) de Laplace d'une fonction ou distribution]
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
[
p
]
=
L
{
f
(
k
t
)
}
[
p
]
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace f(k\;t)\right\rbrace \!\left[p\right]\;}
[ 47]
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
k
t
)
d
t
=
∫
0
−
+
∞
exp
[
−
p
k
k
t
]
f
(
k
t
)
d
(
k
t
)
k
{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(k\;t)\;dt=\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-{\dfrac {p}{k}}\;k\;t\right]\,f(k\;t)\;{\dfrac {d(k\;t)}{k}}\;}
[ 13] , [ 35] Démonstration : « pour calculer
L
{
f
′
(
t
)
}
[
p
]
=
L
{
f
(
k
t
)
}
[
p
]
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace f(k\;t)\right\rbrace \!\left[p\right]}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
′
(
t
)
}
[
p
]
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]\;}
ℜ
(
p
)
>
k
α
{\displaystyle \;\Re (p)>k\;\alpha \;}
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
[ 46] Démonstration : « pour calculer
L
{
f
′
(
t
)
}
[
p
]
=
L
{
f
(
k
t
)
}
[
p
]
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace f(k\;t)\right\rbrace \!\left[p\right]}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
′
(
t
)
}
[
p
]
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]}\;}
ou encore, en remplaçant
k
t
{\displaystyle \;k\;t\;}
t
′
{\displaystyle \;t'}
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
[
p
]
=
L
{
f
(
t
′
)
}
[
p
]
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace f(t')\right\rbrace \!\left[p\right]\;}
[ 47]
=
∫
0
−
+
∞
exp
[
−
p
k
t
′
]
f
(
t
′
)
d
t
′
k
=
1
k
∫
0
−
+
∞
exp
[
−
p
k
t
′
]
f
(
t
′
)
d
t
′
{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-{\dfrac {p}{k}}\;t'\right]\,f(t')\;{\dfrac {dt'}{k}}={\dfrac {1}{k}}\,\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-{\dfrac {p}{k}}\;t'\right]\,f(t')\;dt'\;}
[ 13] , [ 35]
p
′
=
p
k
{\displaystyle \;p'={\dfrac {p}{k}}}
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
[
p
]
=
L
{
f
(
t
′
)
}
[
p
]
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace f(t')\right\rbrace \!\left[p\right]}\;}
=
∫
0
−
+
∞
exp
[
−
p
k
t
′
]
f
(
t
′
)
d
t
′
k
{\displaystyle \color {transparent}{=\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-{\dfrac {p}{k}}\;t'\right]\,f(t')\;{\dfrac {dt'}{k}}}}
=
1
k
∫
0
−
+
∞
exp
(
−
p
′
t
)
f
(
t
)
d
t
{\displaystyle ={\dfrac {1}{k}}\,\displaystyle \int _{0^{-}}^{+\infty }\exp(-p'\;t)\;f(t)\;dt\;}
[ 13] , [ 49]
=
1
k
L
{
f
(
t
)
}
[
p
′
]
{\displaystyle ={\dfrac {1}{k}}\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \!\left[p'\right]\;}
ℜ
(
p
′
)
>
α
{\displaystyle \;\Re (p')>\alpha \;}
⇔
{\displaystyle \Leftrightarrow }
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
[
p
]
=
L
{
f
(
t
′
)
}
[
p
]
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace f(t')\right\rbrace \!\left[p\right]}\;}
=
∫
0
−
+
∞
exp
[
−
p
k
t
′
]
f
(
t
′
)
d
t
′
k
{\displaystyle \color {transparent}{=\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-{\dfrac {p}{k}}\;t'\right]\,f(t')\;{\dfrac {dt'}{k}}}}
=
1
k
∫
0
−
+
∞
exp
(
−
p
′
t
)
f
(
t
)
d
t
{\displaystyle \color {transparent}{={\dfrac {1}{k}}\,\displaystyle \int _{0^{-}}^{+\infty }\exp(-p'\;t)\;f(t)\;dt}\;}
=
1
k
L
{
f
(
t
)
}
[
p
′
]
{\displaystyle \color {transparent}{={\dfrac {1}{k}}\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \!\left[p'\right]}\;}
ℜ
(
p
k
)
>
α
{\displaystyle \;\Re \left({\dfrac {p}{k}}\right)>\alpha \;}
ℜ
(
p
)
>
k
α
{\displaystyle \;\Re (p)>k\;\alpha }
Soit
f
(
t
)
{\displaystyle \;f(t)\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t)\;dt\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13] Soit
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
(
t
)
}
=
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\exp(-p\,t)\;f(t)\;dt}\;}
«
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
Soit
f
′
(
t
)
=
exp
(
−
a
t
)
f
(
t
)
{\displaystyle \;f'(t)=\exp(-a\;t)\;f(t)\;}
(
{\displaystyle \;{\big (}}
distribution [ 18]
)
{\displaystyle {\big )}\;}
f
(
t
)
{\displaystyle \;f(t)\;}
exp
(
−
a
t
)
{\displaystyle \;\exp(-a\;t)\;}
t
{\displaystyle \;t\;}
a
∈
R
∗
{\displaystyle \;a\in \mathbb {R} ^{*}\;}
[ 50] Soit
f
′
(
t
)
=
exp
(
−
a
t
)
f
(
t
)
{\displaystyle \;\color {transparent}{f'(t)=\exp(-a\;t)\;f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
admettant pour transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
′
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
′
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f'(t)\;dt\;}
[ 29] Soit
f
′
(
t
)
=
exp
(
−
a
t
)
f
(
t
)
{\displaystyle \;\color {transparent}{f'(t)=\exp(-a\;t)\;f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
′
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace }}
=
∫
0
−
+
∞
exp
[
−
(
p
+
a
)
t
]
f
(
t
)
d
t
{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-(p+a)\,t\right]\,f(t)\;dt\;}
[ 29] , [ 13] Soit
f
′
(
t
)
=
exp
(
−
a
t
)
f
(
t
)
{\displaystyle \;\color {transparent}{f'(t)=\exp(-a\;t)\;f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
′
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace }}
dans la mesure où
ℜ
(
p
)
>
α
−
a
{\displaystyle \;\Re (p)>\alpha -a\;}
[ 51]
nous établissons ci-après la propriété suivante connue sous le nom de « règle de translation en
p
{\displaystyle \;p\;}
Règle de translation en p [ou conséquence de la multiplication par une fonction exponentielle sur la transformée (monolatérale) de Laplace d'une fonction ou distribution]
Avec «
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
la transformée
(
{\displaystyle \;{\big (}}
monolatérale
)
{\displaystyle {\big )}\;}
de Laplace
[ 1] de la fonction
(
{\displaystyle \;{\big (}}
ou
distribution
)
{\displaystyle {\big )}}
f
(
t
)
{\displaystyle \;f(t)\;}
»
[ 31] Avec «
L
{
f
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace }\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
d'« abscisse de convergence
α
{\displaystyle \;\alpha \;}
» et
Avec «
L
{
f
′
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \;}
celle de la fonction
(
{\displaystyle \;{\big (}}
ou de la
distribution [ 29]
)
{\displaystyle {\big )}}
f
′
(
t
)
=
exp
(
−
a
t
)
f
(
t
)
{\displaystyle \;f'(t)=\exp(-a\;t)\;f(t)\;}
avec
a
∈
R
∗
{\displaystyle \;a\in \mathbb {R} ^{*}\;}
»
[ 50] Avec «
L
{
f
′
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace }\;}
L
{
f
′
(
t
)
}
=
L
{
exp
(
−
a
t
)
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace \exp(-a\;t)\;f(t)\right\rbrace \;}
d'« abscisse de convergence
α
′
=
α
−
a
{\displaystyle \;\alpha '=\alpha -a\;}
[ 51] »,
ces deux transformées de Laplace
[ 1] sont liées par la relation suivante :
«
L
{
f
′
(
t
)
}
[
p
]
=
L
{
exp
(
−
a
t
)
f
(
t
)
}
[
p
]
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace \exp(-a\;t)\;f(t)\right\rbrace \!\left[p\right]\;}
[ 47]
a
∈
R
∗
{\displaystyle \;a\in \mathbb {R} ^{*}\;}
=
L
{
f
(
t
)
}
[
p
+
a
]
{\displaystyle ={\mathcal {L}}\left\lbrace f(t)\right\rbrace \!\left[p+a\right]\;}
[ 52]
ℜ
(
p
)
>
α
−
a
{\displaystyle \;\Re (p)>\alpha -a\;}
[ 51]
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
[
p
]
=
L
{
exp
(
−
a
t
)
f
(
t
)
}
[
p
]
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace \exp(-a\;t)\;f(t)\right\rbrace \!\left[p\right]\;}
[ 47]
=
∫
0
−
+
∞
exp
[
−
(
p
+
a
)
t
]
f
(
t
)
d
t
{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-(p+a)\,t\right]\,f(t)\;dt\;}
[ 13] , [ 35]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Démonstration : « pour calculer
L
{
f
′
(
t
)
}
[
p
]
=
L
{
exp
(
−
a
t
)
f
(
t
)
}
[
p
]
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace \exp(-a\;t)\;f(t)\right\rbrace \!\left[p\right]}\;}
=
exp
[
−
(
p
+
a
)
t
]
f
(
t
)
d
t
{\displaystyle \color {transparent}{=\exp \!\left[-(p+a)\,t\right]\,f(t)\;dt}\;}
L
{
f
′
(
t
)
}
[
p
]
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]\;}
ℜ
(
p
)
>
α
−
a
{\displaystyle \;\Re (p)>\alpha -a\;}
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
[
p
]
=
L
{
exp
(
−
a
t
)
f
(
t
)
}
[
p
]
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace \exp(-a\;t)\;f(t)\right\rbrace \!\left[p\right]}\;}
=
exp
[
−
(
p
+
a
)
t
]
f
(
t
)
d
t
{\displaystyle \color {transparent}{=\exp \!\left[-(p+a)\,t\right]\,f(t)\;dt}\;}
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
[ 51] Démonstration : « pour calculer
L
{
f
′
(
t
)
}
[
p
]
=
L
{
exp
(
−
a
t
)
f
(
t
)
}
[
p
]
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace \exp(-a\;t)\;f(t)\right\rbrace \!\left[p\right]}\;}
=
exp
[
−
(
p
+
a
)
t
]
f
(
t
)
d
t
{\displaystyle \color {transparent}{=\exp \!\left[-(p+a)\,t\right]\,f(t)\;dt}\;}
en remplaçant
p
+
a
{\displaystyle \;p+a\;}
p
′
{\displaystyle \;p'}
Démonstration : « pour calculer
L
{
f
′
(
t
)
}
[
p
]
=
L
{
exp
(
−
a
t
)
f
(
t
)
}
[
p
]
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace \exp(-a\;t)\;f(t)\right\rbrace \!\left[p\right]\;}
[ 47]
=
∫
0
−
+
∞
exp
(
−
p
′
t
)
f
(
t
)
d
t
{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p'\;t)\;f(t)\;dt\;}
[ 13] , [ 35]
L
{
f
(
t
)
}
[
p
′
]
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \!\left[p'\right]\;}
[ 47]
ℜ
(
p
′
)
>
α
{\displaystyle \;\Re (p')>\alpha \;}
⇔
{\displaystyle \Leftrightarrow }
ℜ
(
p
)
>
α
−
a
{\displaystyle \;\Re (p)>\alpha -a\;}
[ 51]
Remarque 1 : « si
a
{\displaystyle \;a\;}
>
0
{\displaystyle \;>0\;}
exp
(
−
a
t
)
{\displaystyle \;\exp(-a\;t)\;}
↘
{\displaystyle \;\searrow \;}
t
{\displaystyle \;t\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}}
f
′
(
t
)
=
exp
(
−
a
t
)
f
(
t
)
{\displaystyle \;f'(t)=\exp(-a\;t)\;f(t)\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
α
′
=
α
−
a
<
α
{\displaystyle \;\alpha '=\alpha -a<\alpha \;}
Remarque 1 : « si
a
{\displaystyle \;a\;}
<
0
{\displaystyle \;<0\;}
exp
(
−
a
t
)
{\displaystyle \;\exp(-a\;t)\;}
↗
{\displaystyle \;\nearrow \;}
t
{\displaystyle \;t\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}}
f
′
(
t
)
=
exp
(
−
a
t
)
f
(
t
)
{\displaystyle \;f'(t)=\exp(-a\;t)\;f(t)\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
α
′
=
α
−
a
>
α
{\displaystyle \;\alpha '=\alpha -a>\alpha \;}
Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;{\underline {a}}\in \mathbb {C} ^{*}\;}
a
_
=
ℜ
(
a
_
)
+
i
ℑ
(
a
_
)
{\displaystyle \;{\underline {a}}=\Re ({\underline {a}})+i\,\Im ({\underline {a}})\;}
⇒
{\displaystyle \Rightarrow }
exp
(
−
a
_
t
)
=
exp
[
−
ℜ
(
a
_
)
t
]
exp
[
−
i
ℑ
(
a
_
)
t
]
{\displaystyle \;\exp \!\left(-{\underline {a}}\;t\right)=\exp \!\left[-\Re ({\underline {a}})\;t\right]\,\exp \!\left[-i\,\Im ({\underline {a}})\;t\right]\;}
Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
a
_
=
ℜ
(
a
_
)
+
i
ℑ
(
a
_
)
{\displaystyle \;\color {transparent}{{\underline {a}}=\Re ({\underline {a}})+i\,\Im ({\underline {a}})}\;}
⇒
{\displaystyle \color {transparent}{\Rightarrow }}
exp
(
−
a
_
t
)
=
exp
[
−
ℜ
(
a
_
)
t
]
exp
[
−
i
ℑ
(
a
_
)
t
]
{\displaystyle \;\color {transparent}{\exp \!\left(-{\underline {a}}\;t\right)=\exp \!\left[-\Re ({\underline {a}})\;t\right]\,\exp \!\left[-i\,\Im ({\underline {a}})\;t\right]}\;}
sont
{
ℜ
[
exp
(
−
a
_
t
)
]
=
exp
[
−
ℜ
(
a
_
)
t
]
cos
[
ℑ
(
a
_
)
t
]
ℑ
[
exp
(
−
a
_
t
)
]
=
−
exp
[
−
ℜ
(
a
_
)
t
]
sin
[
ℑ
(
a
_
)
t
]
}
{\displaystyle \;\left\lbrace {\begin{array}{r}\Re \left[\exp \!\left(-{\underline {a}}\;t\right)\right]\!\!&=&\!\!\exp \!\left[-\Re ({\underline {a}})\;t\right]\;\cos \!\left[\Im ({\underline {a}})\;t\right]\\\Im \left[\exp \!\left(-{\underline {a}}\;t\right)\right]\!\!&=&\!\!-\exp \!\left[-\Re ({\underline {a}})\;t\right]\;\sin \!\left[\Im ({\underline {a}})\;t\right]\end{array}}\right\rbrace }
Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
la « nouvelle fonction
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}}
f
′
_
(
t
)
=
exp
(
−
a
_
t
)
f
(
t
)
{\displaystyle \;{\underline {f'}}(t)=\exp \!\left(-{\underline {a}}\;t\right)\;f(t)\;}
{
{\displaystyle {\big \{}}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
f
(
t
)
{\displaystyle \;f(t)\;}
Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}}
f
′
_
(
t
)
=
exp
(
−
a
_
t
)
f
(
t
)
{\displaystyle \;\color {transparent}{{\underline {f'}}(t)=\exp \!\left(-{\underline {a}}\;t\right)\;f(t)}\;}
{
{\displaystyle \color {transparent}{\big \{}}
par l'exponentielle complexe
}
{\displaystyle {\big \}}\;}
Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
a pour « parties réelle et imaginaire
{
ℜ
[
f
′
_
(
t
)
]
=
ℜ
[
exp
(
−
a
_
t
)
f
(
t
)
]
=
exp
[
−
ℜ
(
a
_
)
t
]
cos
[
ℑ
(
a
_
)
t
]
f
(
t
)
ℑ
[
f
′
_
(
t
)
]
=
ℑ
[
exp
(
−
a
_
t
)
f
(
t
)
]
=
−
exp
[
−
ℜ
(
a
_
)
t
]
sin
[
ℑ
(
a
_
)
t
]
f
(
t
)
}
{\displaystyle \;\left\lbrace {\begin{array}{r}\Re \left[{\underline {f'}}(t)\right]=\Re \left[\exp \!\left(-{\underline {a}}\;t\right)\;f(t)\right]\!\!&=&\!\!\exp \!\left[-\Re ({\underline {a}})\;t\right]\;\cos \!\left[\Im ({\underline {a}})\;t\right]\;f(t)\\\Im \left[{\underline {f'}}(t)\right]=\Im \left[\exp \!\left(-{\underline {a}}\;t\right)\;f(t)\right]\!\!&=&\!\!-\exp \!\left[-\Re ({\underline {a}})\;t\right]\;\sin \!\left[\Im ({\underline {a}})\;t\right]\;f(t)\end{array}}\right\rbrace }
Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
lesquelles admettent toutes deux une transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
sous la même « condition de convergence
ℜ
(
p
)
>
α
′
{\displaystyle \;\Re (p)>\alpha '\;}
α
′
=
α
−
ℜ
(
a
_
)
{\displaystyle \;\alpha '=\alpha -\Re ({\underline {a}})\;}
[ 53] Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
on en déduit la « définition de la transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
_
{
}
{\displaystyle \;{\underline {\mathcal {L}}}\left\lbrace \right\rbrace \;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
f
′
_
(
t
)
{\displaystyle \;{\underline {f'}}(t)\;}
Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
à partir de celle des parties réelle et imaginaire de cette dernière selon
⟨
L
{
ℜ
[
f
′
_
(
t
)
]
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
ℜ
[
f
′
_
(
t
)
]
d
t
L
{
ℑ
[
f
′
_
(
t
)
]
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
ℑ
[
f
′
_
(
t
)
]
d
t
⟩
{\displaystyle \;{\Bigg \langle }{\begin{array}{c}{\mathcal {L}}\left\lbrace \Re \!\left[{\underline {f'}}(t)\right]\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;\Re \!\left[{\underline {f'}}(t)\right]\,dt\\{\mathcal {L}}\left\lbrace \Im \!\left[{\underline {f'}}(t)\right]\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;\Im \!\left[{\underline {f'}}(t)\right]\,dt\end{array}}{\Bigg \rangle }\;}
[ 54]
⇒
{\displaystyle \Rightarrow }
Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
«
L
_
{
f
′
_
(
t
)
}
=
L
{
ℜ
[
f
′
_
(
t
)
]
}
+
i
L
{
ℑ
[
f
′
_
(
t
)
]
}
{\displaystyle \;{\underline {\mathcal {L}}}\left\lbrace {\underline {f'}}(t)\right\rbrace ={\mathcal {L}}\left\lbrace \Re \!\left[{\underline {f'}}(t)\right]\right\rbrace +i\;{\mathcal {L}}\left\lbrace \Im \!\left[{\underline {f'}}(t)\right]\right\rbrace \;}
[ 55] Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
«
L
_
{
f
′
_
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
ℜ
[
f
′
_
(
t
)
]
d
t
+
i
∫
0
−
+
∞
exp
(
−
p
t
)
ℑ
[
f
′
_
(
t
)
]
d
t
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
′
_
(
t
)
d
t
{\displaystyle \;{\underline {\mathcal {L}}}\left\lbrace {\underline {f'}}(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;\Re \!\left[{\underline {f'}}(t)\right]\,dt+i\;\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;\Im \!\left[{\underline {f'}}(t)\right]\,dt=\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;{\underline {f'}}(t)\;dt\;}
[ 13] , [ 35] Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
la même définition que celle d'une fonction
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
il est alors aisé d'établir «
L
_
{
f
′
_
(
t
)
}
[
p
]
=
L
_
{
exp
(
−
a
_
t
)
f
(
t
)
}
[
p
]
=
L
{
f
(
t
)
}
[
p
+
a
_
]
{\displaystyle \;{\underline {\mathcal {L}}}\left\lbrace {\underline {f'}}(t)\right\rbrace \!\left[p\right]={\underline {\mathcal {L}}}\left\lbrace \exp(-{\underline {a}}\;t)\;f(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace f(t)\right\rbrace \!\left[p+{\underline {a}}\right]\;}
a
_
∈
C
∗
{\displaystyle \;{\underline {a}}\in \mathbb {C} ^{*}\;}
[ 56]
et Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
d'en déduire la « convergence de la transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
f
′
_
(
t
)
=
exp
(
−
a
_
t
)
f
(
t
)
{\displaystyle \;{\underline {f'}}(t)=\exp(-{\underline {a}}\;t)\;f(t)\;}
Remarque 2 : Envisageons maintenant
a
_
∈
C
∗
{\displaystyle \;\color {transparent}{{\underline {a}}\in \mathbb {C} ^{*}}\;}
ℜ
(
p
)
>
α
′
=
α
−
ℜ
(
a
_
)
{\displaystyle \;\Re (p)>\alpha '=\alpha -\Re ({\underline {a}})\;}
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
Fonction holomorphe en une valeur du domaine de définition
Une « fonction complexe
F
_
{\displaystyle \;{\underline {F}}\;}
z
_
{\displaystyle \;{\underline {z}}\;}
Ω
⊂
C
{\displaystyle \;\Omega \subset \mathbb {C} \;}
holomorphe en
z
0
_
∈
Ω
{\displaystyle \;{\underline {z_{0}}}\in \Omega \;}
si, « pour
z
_
∈
Ω
{\displaystyle \;{\underline {z}}\in \Omega }
lim
z
_
→
z
0
_
F
_
(
z
_
)
−
F
_
(
z
0
_
)
z
_
−
z
0
_
{\displaystyle \;\lim \limits _{{\underline {z}}\,\rightarrow \,{\underline {z_{0}}}}{\dfrac {{\underline {F}}({\underline {z}})-{\underline {F}}({\underline {z_{0}}})}{{\underline {z}}-{\underline {z_{0}}}}}\;}
z
0
_
{\displaystyle \;{\underline {z_{0}}}\;}
z
_
{\displaystyle \;{\underline {z}}\;}
F
_
(
z
_
)
{\displaystyle \;{\underline {F}}({\underline {z}})\;}
z
0
_
{\displaystyle \;{\underline {z_{0}}}\;}
F
_
′
(
z
0
_
)
{\displaystyle \;{\underline {F}}'({\underline {z_{0}}})\;}
D
F
_
(
z
0
_
)
{\displaystyle \;D{\underline {F}}({\underline {z_{0}}})\;}
[ 57]
en explicitant la direction d'approche,
[
h
_
{\displaystyle \;{\big [}{\underline {h}}\;}
z
0
_
{\displaystyle \;{\underline {z_{0}}}\;}
z
_
]
{\displaystyle \;{\underline {z}}{\big ]}}
lim
u
→
0
F
_
(
z
0
_
+
u
h
_
)
−
F
_
(
z
0
_
)
u
{\displaystyle \;\lim \limits _{u\,\rightarrow \,0}{\dfrac {{\underline {F}}({\underline {z_{0}}}+u\,{\underline {h}})-{\underline {F}}({\underline {z_{0}}})}{u}}\;}
∀
h
_
{\displaystyle \;\forall \,{\underline {h}}\;}
[ 58]
h
_
{\displaystyle \;{\underline {h}}\;}
z
0
_
{\displaystyle \;{\underline {z_{0}}}\;}
z
_
{\displaystyle \;{\underline {z}}\;}
F
_
(
z
_
)
{\displaystyle \;{\underline {F}}({\underline {z}})\;}
z
0
_
{\displaystyle \;{\underline {z_{0}}}\;}
D
F
_
(
z
0
_
)
{\displaystyle \;D{\underline {F}}({\underline {z_{0}}})\;}
[ 59]
Fonction holomorphe sur un ouvert du domaine de définition
Remarques : Dans la mesure où la fonction complexe
F
_
(
z
_
)
{\displaystyle \;{\underline {F}}({\underline {z}})\;}
holomorphe sur l'ouvert
Ω
{\displaystyle \;\Omega \;}
Remarques :
≻
{\displaystyle \succ \;}
F
_
′
(
z
_
)
=
D
F
_
(
z
_
)
=
d
F
_
d
z
_
(
z
_
)
{\displaystyle \;{\underline {F}}'({\underline {z}})=D{\underline {F}}({\underline {z}})={\dfrac {d{\underline {F}}}{d{\underline {z}}}}({\underline {z}})\;}
holomorphie et si celle-ci est établie,
Remarques :
≻
{\displaystyle \succ \;}
F
_
″
(
z
_
)
=
D
2
F
_
(
z
_
)
=
d
2
F
_
d
z
_
2
(
z
_
)
{\displaystyle \;{\underline {F}}''({\underline {z}})=D^{2}{\underline {F}}({\underline {z}})={\dfrac {d^{2}{\underline {F}}}{d{\underline {z}}^{2}}}({\underline {z}})\;}
holomorphie
Remarques :
≻
{\displaystyle \succ \;}
…
{\displaystyle \ldots }
Holomorphie de la transformée (monolatérale) de Laplace d'une fonction (ou d'une distribution)
Démonstration : Pour démontrer que la transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}}
f
(
t
)
{\displaystyle \;f(t)\;}
Démonstration : Pour démontrer que «
L
{
f
(
t
)
}
[
p
]
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \left[p\right]=\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;f(t)\;dt\;}
[ 13] , [ 35] holomorphe à n'importe quel ordre sur tout ouvert satisfaisant
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
Démonstration : Pour démontrer que «
L
{
f
(
t
)
}
[
p
]
=
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace \left[p\right]=\exp(-p\;t)\;f(t)\;dt}\;}
avec
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
[
p
]
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \left[p\right]\;}
Démonstration : Pour démontrer que «
L
{
f
(
t
)
}
[
p
]
=
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace \left[p\right]=\exp(-p\;t)\;f(t)\;dt}\;}
on vérifie d'abord la propriété pour
n
=
1
{\displaystyle \;n=1\;}
Démonstration : Pour démontrer
≻
{\displaystyle \succ \;}
n
=
1
{\displaystyle \;n=1\;}
d
L
{
f
(
t
)
}
d
p
=
d
[
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
)
d
t
]
d
p
{\displaystyle \;{\dfrac {d{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp}}={\dfrac {d\left[\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;f(t)\;dt\right]}{dp}}\;}
[ 13] , [ 35]
t
{\displaystyle \;t\;}
p
{\displaystyle \;p}
Démonstration : Pour démontrer
≻
{\displaystyle \color {transparent}{\succ }\;}
n
=
1
{\displaystyle \;\color {transparent}{n=1}\;}
d
L
{
f
(
t
)
}
d
p
{\displaystyle \;\color {transparent}{\dfrac {d{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp}}}
=
∫
0
−
+
∞
d
exp
(
−
p
t
)
d
p
f
(
t
)
d
t
=
∫
0
−
+
∞
−
t
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }{\dfrac {d\exp(-p\;t)}{dp}}\;f(t)\;dt=\displaystyle \int _{0^{-}}^{+\infty }-t\;\exp(-p\;t)\;f(t)\;dt\;}
[ 13] , [ 35]
d
L
{
f
(
t
)
}
d
p
=
−
L
{
t
f
(
t
)
}
{\displaystyle \;{\dfrac {d{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp}}=-{\mathcal {L}}\left\lbrace t\;f(t)\right\rbrace \;}
[ 64] Démonstration : Pour démontrer
≻
{\displaystyle \succ \;}
d
n
L
{
f
(
t
)
}
d
p
n
=
(
−
1
)
n
L
{
t
n
f
(
t
)
}
{\displaystyle \;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;{\mathcal {L}}\left\lbrace t^{n}\;f(t)\right\rbrace \;}
n
∈
N
∗
{\displaystyle \;n\in \mathbb {N} ^{*}\;}
Démonstration : Pour démontrer
≻
{\displaystyle \color {transparent}{\succ }\;}
on forme
d
(
n
+
1
)
L
{
f
(
t
)
}
d
p
(
n
+
1
)
=
d
[
(
−
1
)
n
∫
0
−
+
∞
exp
(
−
p
t
)
t
n
f
(
t
)
d
t
]
d
p
{\displaystyle \;{\dfrac {d^{(n+1)}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{(n+1)}}}={\dfrac {d\left[(-1)^{n}\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;t^{n}\;f(t)\;dt\right]}{dp}}\;}
[ 13] , [ 35]
t
{\displaystyle \;t\;}
p
{\displaystyle \;p}
Démonstration : Pour démontrer
≻
{\displaystyle \color {transparent}{\succ }\;}
d
(
n
+
1
)
L
{
f
(
t
)
}
d
p
(
n
+
1
)
{\displaystyle \;\color {transparent}{\dfrac {d^{(n+1)}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{(n+1)}}}}
=
(
−
1
)
n
∫
0
−
+
∞
d
exp
(
−
p
t
)
d
p
t
n
f
(
t
)
d
t
=
(
−
1
)
n
∫
0
−
+
∞
−
t
exp
(
−
p
t
)
t
n
f
(
t
)
d
t
{\displaystyle =(-1)^{n}\;\displaystyle \int _{0^{-}}^{+\infty }{\dfrac {d\exp(-p\;t)}{dp}}\;t^{n}\;f(t)\;dt=(-1)^{n}\;\displaystyle \int _{0^{-}}^{+\infty }-t\;\exp(-p\;t)\;t^{n}\;f(t)\;dt\;}
[ 13] , [ 35] Démonstration : Pour démontrer
≻
{\displaystyle \color {transparent}{\succ }\;}
«
d
(
n
+
1
)
L
{
f
(
t
)
}
d
p
(
n
+
1
)
=
(
−
1
)
(
n
+
1
)
L
{
t
(
n
+
1
)
f
(
t
)
}
{\displaystyle \;{\dfrac {d^{(n+1)}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{(n+1)}}}=(-1)^{(n+1)}\;{\mathcal {L}}\left\lbrace t^{(n+1)}\;f(t)\right\rbrace \;}
[ 64] Démonstration : Pour démontrer
≻
{\displaystyle \succ \;}
P
n
{\displaystyle \;{\mathcal {P}}_{n}\;}
d
n
L
{
f
(
t
)
}
d
p
n
=
(
−
1
)
n
L
{
t
n
f
(
t
)
}
{\displaystyle \;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;{\mathcal {L}}\left\lbrace t^{n}\;f(t)\right\rbrace \;}
n
=
1
{\displaystyle \;n=1\;}
P
n
{\displaystyle \;{\mathcal {P}}_{n}\;}
⇒
{\displaystyle \Rightarrow }
P
(
n
+
1
)
{\displaystyle \;{\mathcal {P}}_{(n+1)}\;}
n
∈
N
∗
{\displaystyle \;n\in \mathbb {N} ^{*}}
Soit
f
(
t
)
{\displaystyle \;f(t)\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t)\;dt\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13] Soit
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
(
t
)
}
=
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\exp(-p\,t)\;f(t)\;dt}\;}
«
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
Soit
f
′
(
t
)
=
t
n
f
(
t
)
{\displaystyle \;f'(t)=t^{n}\;f(t)\;}
(
{\displaystyle \;{\big (}}
distribution [ 18]
)
{\displaystyle {\big )}\;}
f
(
t
)
{\displaystyle \;f(t)\;}
t
n
{\displaystyle \;t^{n}\;}
ème de
t
{\displaystyle \;t\;}
n
∈
N
∗
{\displaystyle \;n\in \mathbb {N} ^{*}\;}
[ 65] Soit
f
′
(
t
)
=
t
n
f
(
t
)
{\displaystyle \;\color {transparent}{f'(t)=t^{n}\;f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
admettant pour transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
′
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
′
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f'(t)\;dt\;}
[ 29] Soit
f
′
(
t
)
=
t
n
f
(
t
)
{\displaystyle \;\color {transparent}{f'(t)=t^{n}\;f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
′
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace }}
=
∫
0
−
+
∞
exp
(
−
p
t
)
t
n
f
(
t
)
d
t
{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left(-p\;t\right)\;t^{n}\;f(t)\;dt\;}
[ 29] , [ 13] Soit
f
′
(
t
)
=
t
n
f
(
t
)
{\displaystyle \;\color {transparent}{f'(t)=t^{n}\;f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
′
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace }}
dans la mesure où
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 66]
de la relation «
d
n
L
{
f
(
t
)
}
d
p
n
=
(
−
1
)
n
L
{
t
n
f
(
t
)
}
{\displaystyle \;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;{\mathcal {L}}\left\lbrace t^{n}\;f(t)\right\rbrace \;}
n
∈
N
∗
{\displaystyle \;n\in \mathbb {N} ^{*}\;}
holomorphie de la transformée (monolatérale) de Laplace » plus haut dans ce chapitre, de la relation on en déduit «
L
{
f
′
(
t
)
}
=
L
{
t
n
f
(
t
)
}
=
(
−
1
)
n
d
n
L
{
f
(
t
)
}
d
p
n
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace t^{n}\;f(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{n}}}\;}
n
∈
N
∗
{\displaystyle \;n\in \mathbb {N} ^{*}\;}
▸
{\displaystyle \blacktriangleright \;}
n
=
1
{\displaystyle \;n=1}
L
{
t
f
(
t
)
}
=
−
d
L
{
f
(
t
)
}
d
p
{\displaystyle \;{\mathcal {L}}\left\lbrace t\;f(t)\right\rbrace =-{\dfrac {d{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp}}\;}
de la relation on en déduit «
L
{
f
′
(
t
)
}
=
L
{
t
n
f
(
t
)
}
=
(
−
1
)
n
d
n
L
{
f
(
t
)
}
d
p
n
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace t^{n}\;f(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{n}}}}\;}
n
∈
N
∗
{\displaystyle \;\color {transparent}{n\in \mathbb {N} ^{*}}\;}
▸
{\displaystyle \blacktriangleright \;}
n
=
2
{\displaystyle \;n=2}
L
{
t
2
f
(
t
)
}
=
d
2
L
{
f
(
t
)
}
d
p
2
{\displaystyle \;{\mathcal {L}}\left\lbrace t^{2}\;f(t)\right\rbrace ={\dfrac {d^{2}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{2}}}\;}
de la relation on en déduit «
L
{
f
′
(
t
)
}
=
L
{
t
n
f
(
t
)
}
=
(
−
1
)
n
d
n
L
{
f
(
t
)
}
d
p
n
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace t^{n}\;f(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{n}}}}\;}
n
∈
N
∗
{\displaystyle \;\color {transparent}{n\in \mathbb {N} ^{*}}\;}
▸
{\displaystyle \blacktriangleright \;}
n
=
3
{\displaystyle \;n=3}
L
{
t
3
f
(
t
)
}
=
−
d
3
L
{
f
(
t
)
}
d
p
3
{\displaystyle \;{\mathcal {L}}\left\lbrace t^{3}\;f(t)\right\rbrace =-{\dfrac {d^{3}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{3}}}\;}
de la relation on en déduit «
L
{
f
′
(
t
)
}
=
L
{
t
n
f
(
t
)
}
=
(
−
1
)
n
d
n
L
{
f
(
t
)
}
d
p
n
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace t^{n}\;f(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{n}}}}\;}
n
∈
N
∗
{\displaystyle \;\color {transparent}{n\in \mathbb {N} ^{*}}\;}
▸
{\displaystyle \blacktriangleright \;}
…
{\displaystyle \ldots }
Application : soit à « évaluer
L
{
t
n
Y
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace \;}
n
∈
N
∗
{\displaystyle \;n\in \mathbb {N} ^{*}\;}
Y
(
t
)
{\displaystyle \;Y(t)\;}
[ 20]
L
{
Y
(
t
)
}
=
1
p
{\displaystyle \;{\mathcal {L}}\left\lbrace Y(t)\right\rbrace ={\dfrac {1}{p}}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\Re e(p)>0\;}
[ 67] Application : soit à évaluer «
L
{
t
n
Y
(
t
)
}
=
(
−
1
)
n
d
n
L
{
Y
(
t
)
}
d
p
n
=
(
−
1
)
n
d
n
(
1
p
)
d
p
n
{\displaystyle \;{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace Y(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;{\dfrac {d^{n}\!\left({\dfrac {1}{p}}\right)}{dp^{n}}}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\Re e(p)>0\;}
∙
{\displaystyle \bullet \;}
L
{
t
Y
(
t
)
}
=
−
d
(
1
p
)
d
p
=
1
p
2
{\displaystyle \;{\mathcal {L}}\left\lbrace t\;Y(t)\right\rbrace =-{\dfrac {d\!\left({\dfrac {1}{p}}\right)}{dp}}={\dfrac {1}{p^{2}}}\;}
Application : soit à évaluer «
L
{
t
n
Y
(
t
)
}
=
(
−
1
)
n
d
n
L
{
Y
(
t
)
}
d
p
n
=
(
−
1
)
n
d
n
(
1
p
)
d
p
n
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace Y(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;{\dfrac {d^{n}\!\left({\dfrac {1}{p}}\right)}{dp^{n}}}}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\color {transparent}{\Re e(p)>0}\;}
∙
{\displaystyle \bullet \;}
L
{
t
2
Y
(
t
)
}
=
d
2
(
1
p
)
d
p
2
=
2
p
3
{\displaystyle \;{\mathcal {L}}\left\lbrace t^{2}\;Y(t)\right\rbrace ={\dfrac {d^{2}\!\left({\dfrac {1}{p}}\right)}{dp^{2}}}={\dfrac {2}{p^{3}}}\;}
[ 68] Application : soit à évaluer «
L
{
t
n
Y
(
t
)
}
=
(
−
1
)
n
d
n
L
{
Y
(
t
)
}
d
p
n
=
(
−
1
)
n
d
n
(
1
p
)
d
p
n
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace Y(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;{\dfrac {d^{n}\!\left({\dfrac {1}{p}}\right)}{dp^{n}}}}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\color {transparent}{\Re e(p)>0}\;}
∙
{\displaystyle \bullet \;}
L
{
t
3
Y
(
t
)
}
=
−
d
3
(
1
p
)
d
p
3
=
6
p
4
{\displaystyle \;{\mathcal {L}}\left\lbrace t^{3}\;Y(t)\right\rbrace =-{\dfrac {d^{3}\!\left({\dfrac {1}{p}}\right)}{dp^{3}}}={\dfrac {6}{p^{4}}}\;}
[ 69] Application : soit à évaluer «
L
{
t
n
Y
(
t
)
}
=
(
−
1
)
n
d
n
L
{
Y
(
t
)
}
d
p
n
=
(
−
1
)
n
d
n
(
1
p
)
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace Y(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;d^{n}\!\left({\dfrac {1}{p}}\right)}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\color {transparent}{\Re e(p)>0}\;}
∙
{\displaystyle \bullet \;}
L
{
t
n
Y
(
t
)
}
=
n
!
p
n
+
1
{\displaystyle \;{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace ={\dfrac {n!}{p^{n+1}}}\;}
Application : soit à évaluer «
L
{
t
n
Y
(
t
)
}
=
(
−
1
)
n
d
n
L
{
Y
(
t
)
}
d
p
n
=
(
−
1
)
n
d
n
(
1
p
)
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace Y(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;d^{n}\!\left({\dfrac {1}{p}}\right)}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\color {transparent}{\Re e(p)>0}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
(
{\displaystyle {\big (}}
n
=
1
)
{\displaystyle \;n=1{\big )}}
Application : soit à évaluer «
L
{
t
n
Y
(
t
)
}
=
(
−
1
)
n
d
n
L
{
Y
(
t
)
}
d
p
n
=
(
−
1
)
n
d
n
(
1
p
)
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace Y(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;d^{n}\!\left({\dfrac {1}{p}}\right)}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\color {transparent}{\Re e(p)>0}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
L
{
t
(
n
+
1
)
Y
(
t
)
}
=
L
{
t
[
t
n
Y
(
t
)
]
}
=
−
d
L
{
t
n
Y
(
t
)
}
d
p
{\displaystyle \;{\mathcal {L}}\left\lbrace t^{(n+1)}\;Y(t)\right\rbrace ={\mathcal {L}}\left\lbrace t\,\left[t^{n}\;Y(t)\right]\right\rbrace =-{\dfrac {d{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace }{dp}}}
Application : soit à évaluer «
L
{
t
n
Y
(
t
)
}
=
(
−
1
)
n
d
n
L
{
Y
(
t
)
}
d
p
n
=
(
−
1
)
n
d
n
(
1
p
)
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace Y(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;d^{n}\!\left({\dfrac {1}{p}}\right)}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\color {transparent}{\Re e(p)>0}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
L
{
t
(
n
+
1
)
Y
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace t^{(n+1)}\;Y(t)\right\rbrace }}
=
−
d
[
n
!
p
(
n
+
1
)
]
d
p
{\displaystyle =-{\dfrac {d\!\left[{\dfrac {n!}{p^{(n+1)}}}\right]}{dp}}}
Application : soit à évaluer «
L
{
t
n
Y
(
t
)
}
=
(
−
1
)
n
d
n
L
{
Y
(
t
)
}
d
p
n
=
(
−
1
)
n
d
n
(
1
p
)
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace Y(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;d^{n}\!\left({\dfrac {1}{p}}\right)}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\color {transparent}{\Re e(p)>0}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
L
{
t
(
n
+
1
)
Y
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace t^{(n+1)}\;Y(t)\right\rbrace }}
=
(
n
+
1
)
n
!
p
(
n
+
2
)
=
(
n
+
1
)
!
p
(
n
+
2
)
{\displaystyle =(n+1)\;{\dfrac {n!}{p^{(n+2)}}}={\dfrac {(n+1)!}{p^{(n+2)}}}\;}
Application : soit à évaluer «
L
{
t
n
Y
(
t
)
}
=
(
−
1
)
n
d
n
L
{
Y
(
t
)
}
d
p
n
=
(
−
1
)
n
d
n
(
1
p
)
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace Y(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;d^{n}\!\left({\dfrac {1}{p}}\right)}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\color {transparent}{\Re e(p)>0}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
ce qui valide l'hypothèse de récurrence d'où
Application : soit à évaluer «
L
{
t
n
Y
(
t
)
}
=
(
−
1
)
n
d
n
L
{
Y
(
t
)
}
d
p
n
=
(
−
1
)
n
d
n
(
1
p
)
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace =(-1)^{n}\;{\dfrac {d^{n}{\mathcal {L}}\left\lbrace Y(t)\right\rbrace }{dp^{n}}}=(-1)^{n}\;d^{n}\!\left({\dfrac {1}{p}}\right)}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\color {transparent}{\Re e(p)>0}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
L
{
t
n
Y
(
t
)
}
=
n
!
p
(
n
+
1
)
{\displaystyle \;{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace ={\dfrac {n!}{p^{(n+1)}}}\;}
n
∈
N
∗
{\displaystyle \;n\in \mathbb {N} ^{*}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\Re e(p)>0\;}
Soit
f
(
t
,
m
)
{\displaystyle \;f(t,\,m)\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
m
{\displaystyle \;m\;}
[ 29] Soit
f
(
t
,
m
)
{\displaystyle \;\color {transparent}{f(t,\,m)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
admettant pour transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
(
t
,
m
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
,
m
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t,\,m)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t,\,m)\;dt\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13]
α
{\displaystyle \;\alpha \;}
Soit
f
(
t
,
m
)
{\displaystyle \;\color {transparent}{f(t,\,m)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
(
t
,
m
)
}
=
exp
(
−
p
t
)
f
(
t
,
m
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t,\,m)\right\rbrace =\exp(-p\,t)\;f(t,\,m)\;dt}\;}
de convergence de
L
{
f
(
t
,
m
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t,\,m)\right\rbrace \;}
Soit
f
′
(
t
,
m
)
=
(
∂
f
∂
m
)
t
(
t
,
m
)
{\displaystyle \;f'(t,\,m)=\left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
m
{\displaystyle \;m\;}
[ 29] Soit
f
′
(
t
,
m
)
=
(
∂
f
∂
m
)
t
(
t
,
m
)
{\displaystyle \;\color {transparent}{f'(t,\,m)=\left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)}\;}
admettant pour transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
′
(
t
,
m
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
′
(
t
,
m
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t,\,m)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f'(t,\,m)\;dt\;}
[ 13] , [ 29] Soit
f
′
(
t
,
m
)
=
(
∂
f
∂
m
)
t
(
t
,
m
)
{\displaystyle \;\color {transparent}{f'(t,\,m)=\left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
′
(
t
,
m
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t,\,m)\right\rbrace }}
=
∫
0
−
+
∞
exp
(
−
p
t
)
(
∂
f
∂
m
)
t
(
t
,
m
)
d
t
{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\;dt\;}
[ 13] , [ 29] Soit
f
′
(
t
,
m
)
=
(
∂
f
∂
m
)
t
(
t
,
m
)
{\displaystyle \;\color {transparent}{f'(t,\,m)=\left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
cas général, la même condition de convergence que celle de
L
{
f
(
t
,
m
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t,\,m)\right\rbrace \;}
Soit
f
′
(
t
,
m
)
=
(
∂
f
)
t
(
t
,
m
)
{\displaystyle \;\color {transparent}{f'(t,\,m)=\left(\partial f\right)_{\!t}(t,\,m)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 70]
nous établissons ci-dessous la propriété de « permutation de la transformation
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
«
L
{
f
′
(
t
,
m
)
}
=
L
{
(
∂
f
∂
m
)
t
(
t
,
m
)
}
=
(
∂
L
{
f
(
t
,
m
)
}
∂
m
)
p
{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t,\,m)\right\rbrace ={\mathcal {L}}\left\lbrace \left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\right\rbrace =\left({\dfrac {\partial {\mathcal {L}}\left\lbrace f(t,\,m)\right\rbrace }{\partial m}}\right)_{\!p}\;}
Démonstration : partant de «
L
{
(
∂
f
∂
m
)
t
(
t
,
m
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
(
∂
f
∂
m
)
t
(
t
,
m
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace \left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\;dt\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13]
m
{\displaystyle \;m\;}
t
{\displaystyle \;t\;}
[ 71] Démonstration : partant de «
L
{
(
∂
f
∂
m
)
t
(
t
,
m
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
(
∂
f
∂
m
)
t
(
t
,
m
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\;dt}\;}
ℜ
(
p
)
>
α
{\displaystyle \;\color {transparent}{\Re (p)>\alpha }\;}
on obtient «
L
{
(
∂
f
∂
m
)
t
(
t
,
m
)
}
=
(
∂
[
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
,
m
)
d
t
]
∂
m
)
p
{\displaystyle \;{\mathcal {L}}\left\lbrace \left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\right\rbrace =\left({\dfrac {\partial \left[\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t,\,m)\;dt\right]}{\partial m}}\right)_{\!p}}
Démonstration : partant de «
L
{
(
∂
f
∂
m
)
t
(
t
,
m
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
(
∂
f
∂
m
)
t
(
t
,
m
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\;dt}\;}
ℜ
(
p
)
>
α
{\displaystyle \;\color {transparent}{\Re (p)>\alpha }\;}
L
{
(
∂
f
∂
m
)
t
(
t
,
m
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\right\rbrace }}
=
(
∂
L
{
f
(
t
,
m
)
}
∂
m
)
p
{\displaystyle =\left({\dfrac {\partial {\mathcal {L}}\left\lbrace f(t,\,m)\right\rbrace }{\partial m}}\right)_{\!p}\;}
Démonstration : partant de «
L
{
(
∂
f
∂
m
)
t
(
t
,
m
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
(
∂
f
∂
m
)
t
(
t
,
m
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\;dt}\;}
ℜ
(
p
)
>
α
{\displaystyle \;\color {transparent}{\Re (p)>\alpha }\;}
condition de convergence
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 70]
Soit
f
(
t
,
m
)
{\displaystyle \;f(t,\,m)\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
m
{\displaystyle \;m\;}
[ 29] Soit
f
(
t
,
m
)
{\displaystyle \;\color {transparent}{f(t,\,m)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
admettant pour transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
(
t
,
m
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
,
m
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t,\,m)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t,\,m)\;dt\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13]
α
{\displaystyle \;\alpha \;}
Soit
f
(
t
,
m
)
{\displaystyle \;\color {transparent}{f(t,\,m)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
(
t
,
m
)
}
=
exp
(
−
p
t
)
f
(
t
,
m
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t,\,m)\right\rbrace =\exp(-p\,t)\;f(t,\,m)\;dt}\;}
de convergence de
L
{
f
(
t
,
m
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t,\,m)\right\rbrace \;}
Soit
f
″
(
t
,
m
)
=
∫
0
m
f
(
t
,
m
′
)
d
m
′
{\displaystyle \;f''(t,\,m)=\displaystyle \int _{0}^{m}f(t,\,m')\;dm'\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
m
{\displaystyle \;m\;}
[ 29]
m
=
0
{\displaystyle \;m=0}
Soit
f
″
(
t
,
m
)
=
∫
0
m
f
(
t
,
m
′
)
d
m
′
{\displaystyle \;\color {transparent}{f''(t,\,m)=\displaystyle \int _{0}^{m}f(t,\,m')\;dm'}\;}
admettant pour transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
″
(
t
,
m
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
″
(
t
,
m
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f''(t,\,m)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f''(t,\,m)\;dt\;}
[ 13] , [ 29] Soit
f
″
(
t
,
m
)
=
∫
0
m
f
(
t
,
m
′
)
d
m
′
{\displaystyle \;\color {transparent}{f''(t,\,m)=\displaystyle \int _{0}^{m}f(t,\,m')\;dm'}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
″
(
t
,
m
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f''(t,\,m)\right\rbrace }}
=
∫
0
−
+
∞
exp
(
−
p
t
)
[
∫
0
m
f
(
t
,
m
′
)
d
m
′
]
d
t
{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left[\displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right]\;dt\;}
[ 13] , [ 29] Soit
f
″
(
t
,
m
)
=
∫
0
m
f
(
t
,
m
′
)
d
m
′
{\displaystyle \;\color {transparent}{f''(t,\,m)=\displaystyle \int _{0}^{m}f(t,\,m')\;dm'}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
avec, dans le cas général, la même condition de convergence que Soit
f
″
(
t
,
m
)
=
f
(
t
,
m
′
)
d
m
′
{\displaystyle \;\color {transparent}{f''(t,\,m)=f(t,\,m')\;dm'}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
celle de
L
{
f
(
t
,
m
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t,\,m)\right\rbrace \;}
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 72]
nous établissons ci-dessous la propriété de « permutation de la transformation
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
«
L
{
f
″
(
t
,
m
)
}
=
L
{
∫
0
m
f
(
t
,
m
′
)
d
m
′
}
=
∫
0
m
L
{
f
(
t
,
m
′
)
}
d
m
′
{\displaystyle \;{\mathcal {L}}\left\lbrace f''(t,\,m)\right\rbrace ={\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right\rbrace =\displaystyle \int _{0}^{m}{\mathcal {L}}\left\lbrace f(t,\,m')\right\rbrace \,dm'\;}
Démonstration : partant de «
L
{
∫
0
m
f
(
t
,
m
′
)
d
m
′
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
[
∫
0
m
f
(
t
,
m
′
)
d
m
′
]
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left[\displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right]\,dt\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13]
m
{\displaystyle \;m\;}
t
{\displaystyle \;t\;}
[ 73]
⇒
{\displaystyle \Rightarrow }
Démonstration : partant de «
L
{
∫
0
m
f
(
t
,
m
′
)
d
m
′
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
[
∫
0
m
f
(
t
,
m
′
)
d
m
′
]
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left[\displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right]\,dt}\;}
«
L
{
∫
0
m
f
(
t
,
m
′
)
d
m
′
}
=
∫
0
m
[
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
,
m
′
)
d
t
]
d
m
′
{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right\rbrace =\displaystyle \int _{0}^{m}\left[\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t,\,m')\;dt\right]\,dm'}
Démonstration : partant de «
L
{
∫
0
m
f
(
t
,
m
′
)
d
m
′
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
[
∫
0
m
f
(
t
,
m
′
)
d
m
′
]
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left[\displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right]\,dt}\;}
L
{
∫
0
m
f
(
t
,
m
′
)
d
m
′
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right\rbrace }}
=
∫
0
m
L
{
f
(
t
,
m
′
)
}
d
m
′
{\displaystyle =\displaystyle \int _{0}^{m}{\mathcal {L}}\left\lbrace f(t,\,m')\right\rbrace \,dm'\;}
Démonstration : partant de «
L
{
∫
0
m
f
(
t
,
m
′
)
d
m
′
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
[
∫
0
m
f
(
t
,
m
′
)
d
m
′
]
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left[\displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right]\,dt}\;}
L
{
∫
0
m
f
(
t
,
m
′
)
d
m
′
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right\rbrace }}
convergence
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 72]
Soit
f
(
t
)
{\displaystyle \;f(t)\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
T
{\displaystyle \;T\;}
f
(
t
+
T
)
=
f
(
t
)
{\displaystyle \;f(t+T)=f(t)\;}
Soit
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
admettant pour transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
f
(
t
,
m
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t,\,m)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t)\;dt\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13]
α
⩾
0
{\displaystyle \;\alpha \geqslant 0\;}
[ 74] Soit
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
{
f
(
t
)
}
=
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\exp(-p\,t)\;f(t)\;dt}\;}
de convergence de
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
nous établissons ci-dessous la propriété « traduisant la périodicité de la fonction
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
«
L
{
f
(
t
)
}
=
∫
0
T
exp
(
−
p
t
)
f
(
t
)
d
t
1
−
exp
(
−
p
T
)
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace ={\dfrac {\displaystyle \int _{0}^{T}\exp(-p\,t)\;f(t)\;dt}{1-\exp(-p\;T)}}\;}
[ 29]
Démonstration : partant de «
L
{
f
(
t
)
}
=
∫
0
−
+
∞
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t)\;dt\;}
[ 29]
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 13] Démonstration : partant de «
L
{
f
(
t
)
}
=
∫
0
T
exp
(
−
p
t
)
f
(
t
)
d
t
+
∫
T
2
T
exp
(
−
p
t
)
f
(
t
)
d
t
+
∫
2
T
3
T
exp
(
−
p
t
)
f
(
t
)
d
t
+
⋯
+
∫
n
T
(
n
+
1
)
T
exp
(
−
p
t
)
f
(
t
)
d
t
+
⋯
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\displaystyle \int _{0}^{T}\exp(-p\,t)\;f(t)\;dt+\displaystyle \int _{T}^{2\,T}\exp(-p\,t)\;f(t)\;dt+\displaystyle \int _{2\,T}^{3\,T}\exp(-p\,t)\;f(t)\;dt+\cdots +\displaystyle \int _{n\,T}^{(n+1)\,T}\exp(-p\,t)\;f(t)\;dt+\cdots \;}
[ 29] Démonstration : en faisant le « changement de variable
t
=
t
n
+
n
T
{\displaystyle \;t=t_{n}+n\;T\;}
∫
n
T
(
n
+
1
)
T
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;\displaystyle \int _{n\,T}^{(n+1)\,T}\exp(-p\,t)\;f(t)\;dt\;}
[
0
,
T
]
{\displaystyle \;\left[0\,,\,T\right]\;}
{
f
(
t
)
=
f
(
t
n
)
d
t
=
d
t
n
}
{\displaystyle \;\left\lbrace {\begin{array}{c}f(t)=f(t_{n})\\dt=dt_{n}\end{array}}\right\rbrace \;}
Démonstration : en faisant le « changement de variable
t
=
t
n
+
n
T
{\displaystyle \;\color {transparent}{t=t_{n}+n\;T}\;}
∫
n
T
(
n
+
1
)
T
exp
(
−
p
t
)
f
(
t
)
d
t
=
∫
0
T
exp
(
−
p
t
n
)
exp
(
−
p
n
T
)
f
(
t
n
)
d
t
n
{\displaystyle \;\displaystyle \int _{n\,T}^{(n+1)\,T}\exp(-p\,t)\;f(t)\;dt=\displaystyle \int _{0}^{T}\exp(-p\,t_{n})\;\exp(-p\,n\,T)\;f(t_{n})\;dt_{n}\;}
t
n
{\displaystyle \;t_{n}\;}
Démonstration : en faisant le « changement de variable
t
=
t
n
+
n
T
{\displaystyle \;\color {transparent}{t=t_{n}+n\;T}\;}
∫
n
T
(
n
+
1
)
T
exp
(
−
p
t
)
f
(
t
)
d
t
=
∫
0
T
exp
(
−
p
t
n
)
exp
(
−
p
n
T
)
f
(
t
n
)
d
t
n
{\displaystyle \;\color {transparent}{\displaystyle \int _{n\,T}^{(n+1)\,T}\exp(-p\,t)\;f(t)\;dt=\displaystyle \int _{0}^{T}\exp(-p\,t_{n})\;\exp(-p\,n\,T)\;f(t_{n})\;dt_{n}}\;}
exp
(
−
p
n
T
)
{\displaystyle \;\exp(-p\,n\,T)\;}
Démonstration : en faisant le « changement de variable
t
=
t
n
+
n
T
{\displaystyle \;\color {transparent}{t=t_{n}+n\;T}\;}
∫
n
T
(
n
+
1
)
T
exp
(
−
p
t
)
f
(
t
)
d
t
=
exp
(
−
p
n
T
)
∫
0
T
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;\displaystyle \int _{n\,T}^{(n+1)\,T}\exp(-p\,t)\;f(t)\;dt=\exp(-p\,n\,T)\;\displaystyle \int _{0}^{T}\exp(-p\,t)\;f(t)\;dt\;}
Démonstration : après factorisation par
∫
0
T
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;\displaystyle \int _{0}^{T}\exp(-p\,t)\;f(t)\;dt\;}
Démonstration : partant de «
L
{
f
(
t
)
}
=
[
1
+
∑
n
∈
N
∗
1
.
.
+
∞
exp
(
−
p
n
T
)
]
∫
0
T
exp
(
−
p
t
)
f
(
t
)
d
t
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\left[1+\sum \limits _{n\,\in \,\mathbb {N} ^{*}}^{1\,..\,+\infty }\exp(-p\,n\;T)\right]\displaystyle \int _{0}^{T}\exp(-p\,t)\;f(t)\;dt\;}
Démonstration : en reconnaissant dans les termes entre crochets la « somme infinie d'une progression géométrique de 1er terme
1
{\displaystyle \;1\;}
exp
(
−
p
T
)
{\displaystyle \;\exp(-p\,T)\;}
<
1
{\displaystyle \;<1\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 75] Démonstration :
[
{\displaystyle {\big [}}
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
⩾
0
{\displaystyle \;\geqslant 0\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
f
(
t
)
{\displaystyle \;f(t)}
T
{\displaystyle \;T}
]
{\displaystyle {\big ]}\;}
Démonstration : partant de «
L
{
f
(
t
)
}
=
∫
0
T
exp
(
−
p
t
)
f
(
t
)
d
t
1
−
exp
(
−
p
T
)
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace ={\dfrac {\displaystyle \int _{0}^{T}\exp(-p\,t)\;f(t)\;dt}{1-\exp(-p\;T)}}\;}
[ 29] [ 64]
Liste évidemment non exhaustive.
« Fonction de Heaviside[ 20]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}}
Y
(
t
)
=
{
1
si
t
>
0
0
si
t
<
0
}
{\displaystyle \;Y(t)=\left\lbrace {\begin{array}{l}1\;{\text{si}}\;t>0\\0\;{\text{si}}\;t<0\end{array}}\right\rbrace \;}
⇒
{\displaystyle \Rightarrow }
L
{
Y
(
t
)
}
=
1
p
{\displaystyle \;{\mathcal {L}}\!\left\lbrace Y(t)\right\rbrace ={\dfrac {1}{p}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 76]
« Fonction de Heaviside[ 20]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
Y
(
t
−
τ
)
=
{
1
si
t
>
τ
0
si
t
<
τ
}
{\displaystyle \;Y(t-\tau )=\left\lbrace {\begin{array}{l}1\;{\text{si}}\;t>\tau \\0\;{\text{si}}\;t<\tau \end{array}}\right\rbrace \;}
⇒
{\displaystyle \Rightarrow }
L
{
Y
(
t
−
τ
)
}
=
exp
(
−
p
τ
)
p
{\displaystyle \;{\mathcal {L}}\!\left\lbrace Y(t-\tau )\right\rbrace ={\dfrac {\exp(-p\,\tau )}{p}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 77]
« Fonction créneau unité sur
[
0
,
τ
]
{\displaystyle \;\left[0\,,\,\tau \right]}
C
[
0
,
τ
]
(
t
)
=
{
0
si
t
>
τ
1
si
0
<
t
<
τ
0
si
t
<
0
}
=
Y
(
t
)
−
Y
(
t
−
τ
)
{\displaystyle \;C_{\left[0\,,\,\tau \right]}(t)=\left\lbrace {\begin{array}{l}0\;\;{\text{si}}\;t>\tau \\1\;\;{\text{si}}\;0<t<\tau \\0\;\;{\text{si}}\;t<0\end{array}}\right\rbrace =Y(t)-Y(t-\tau )\;}
L
{
C
[
0
,
τ
]
(
t
)
}
=
L
{
Y
(
t
)
}
−
L
{
Y
(
t
−
τ
)
}
{\displaystyle \;{\mathcal {L}}\!\left\lbrace C_{\left[0\,,\,\tau \right]}(t)\right\rbrace ={\mathcal {L}}\!\left\lbrace Y(t)\right\rbrace -{\mathcal {L}}\!\left\lbrace Y(t-\tau )\right\rbrace \;}
[ 78] « Fonction créneau unité sur
[
0
,
τ
]
{\displaystyle \;\color {transparent}{\left[0\,,\,\tau \right]}}
C
[
0
,
τ
]
(
t
)
=
{
1
si
0
<
t
<
τ
}
=
Y
(
t
)
−
Y
(
t
−
τ
)
{\displaystyle \;\color {transparent}{C_{\left[0\,,\,\tau \right]}(t)=\left\lbrace 1\;\;{\text{si}}\;0<t<\tau \right\rbrace =Y(t)-Y(t-\tau )}\;}
«
L
{
C
[
0
,
τ
]
(
t
)
}
=
1
−
exp
(
−
p
τ
)
p
{\displaystyle \;{\mathcal {L}}\!\left\lbrace C_{\left[0\,,\,\tau \right]}(t)\right\rbrace ={\dfrac {1-\exp(-p\,\tau )}{p}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
« Pic de Dirac [ 22]
δ
(
t
)
=
{
0
si
t
>
0
+
∞
si
t
=
0
0
si
t
<
0
}
=
Y
˙
(
t
)
{\displaystyle \;\delta (t)=\left\lbrace {\begin{array}{l}0\qquad {\text{si}}\;t>0\\+\infty \;\;\,{\text{si}}\;t=0\\0\qquad {\text{si}}\;t<0\end{array}}\right\rbrace ={\dot {Y}}(t)\;}
[ 23]
⇒
{\displaystyle \Rightarrow }
L
{
δ
(
t
)
}
=
1
,
∀
ℜ
(
p
)
∈
R
{\displaystyle \;{\mathcal {L}}\!\left\lbrace \delta (t)\right\rbrace =1,\;\;\forall \;\Re (p)\in \mathbb {R} \;}
[ 76]
« Pic de Dirac [ 22]
δ
(
t
−
τ
)
=
{
0
si
t
>
τ
+
∞
si
t
=
τ
0
si
t
<
τ
}
=
Y
˙
(
t
−
τ
)
{\displaystyle \;\delta (t-\tau )=\left\lbrace {\begin{array}{l}0\qquad {\text{si}}\;t>\tau \\+\infty \;\;\,{\text{si}}\;t=\tau \\0\qquad {\text{si}}\;t<\tau \end{array}}\right\rbrace ={\dot {Y}}(t-\tau )\;}
[ 79]
⇒
{\displaystyle \Rightarrow }
L
{
δ
(
t
−
τ
)
}
=
exp
(
−
p
τ
)
,
∀
ℜ
(
p
)
∈
R
{\displaystyle \;{\mathcal {L}}\!\left\lbrace \delta (t-\tau )\right\rbrace =\exp(-p\,\tau ),\;\;\forall \;\Re (p)\in \mathbb {R} \;}
[ 77]
« Fonction rampe
f
rampe
(
t
)
=
{
a
t
si
t
⩾
0
0
si
t
⩽
0
}
{\displaystyle \;f_{\text{rampe}}(t)=\left\lbrace {\begin{array}{l}a\;t\;{\text{si}}\;t\geqslant 0\\0\quad {\text{si}}\;t\leqslant 0\end{array}}\right\rbrace \;}
⇒
{\displaystyle \Rightarrow }
L
{
f
rampe
(
t
)
}
=
a
p
2
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{rampe}}(t)\right\rbrace ={\dfrac {a}{p^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 76]
« Fonction rampe retardée
f
rampe
(
t
−
τ
)
=
{
a
t
si
t
⩾
τ
0
si
t
⩽
τ
}
{\displaystyle \;f_{\text{rampe}}(t-\tau )=\left\lbrace {\begin{array}{l}a\;t\;{\text{si}}\;t\geqslant \tau \\0\quad {\text{si}}\;t\leqslant \tau \end{array}}\right\rbrace \;}
⇒
{\displaystyle \Rightarrow }
L
{
f
rampe
(
t
−
τ
)
}
=
a
exp
(
−
p
τ
)
p
2
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{rampe}}(t-\tau )\right\rbrace ={\dfrac {a\;\exp(-p\,\tau )}{p^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 77]
« Fonction triangle symétrique sur
[
0
,
2
τ
]
{\displaystyle \;\left[0\,,\,2\,\tau \right]}
f
triang sym
(
t
)
=
{
0
si
t
⩾
2
τ
−
a
(
t
−
2
τ
)
si
t
∈
[
τ
,
2
τ
]
a
t
si
t
∈
[
0
,
τ
]
0
si
t
⩽
0
}
=
f
rampe
(
t
)
−
2
f
rampe
(
t
−
τ
)
+
f
rampe
(
t
−
2
τ
)
{\displaystyle \;f_{\text{triang sym}}(t)=\left\lbrace {\begin{array}{l}0\qquad \qquad \quad {\text{si}}\;t\geqslant 2\,\tau \\-a\;(t-2\,\tau )\;{\text{si}}\;t\in \left[\tau \,,\,2\,\tau \right]\\a\;t\qquad \qquad \;{\text{si}}\;t\in \left[0\,,\,\tau \right]\\0\qquad \qquad \quad {\text{si}}\;t\leqslant 0\end{array}}\right\rbrace =f_{\text{rampe}}(t)-2\,f_{\text{rampe}}(t-\tau )+f_{\text{rampe}}(t-2\,\tau )\;}
« Fonction triangle symétrique sur
[
0
,
2
τ
]
{\displaystyle \;\color {transparent}{\left[0\,,\,2\,\tau \right]}}
f
triang sym
(
t
)
=
{
−
a
(
t
−
2
τ
)
si
t
∈
[
τ
,
2
τ
]
}
=
{\displaystyle \;\color {transparent}{f_{\text{triang sym}}(t)=\left\lbrace -a\;(t-2\,\tau )\;{\text{si}}\;t\in \left[\tau \,,\,2\,\tau \right]\right\rbrace =}}
«
L
{
f
triang sym
(
t
)
}
=
a
[
1
−
2
exp
(
−
p
τ
)
+
exp
(
−
2
p
τ
)
]
p
2
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{triang sym}}(t)\right\rbrace ={\dfrac {a\,\left[1-2\,\exp(-p\,\tau )+\exp(-2\,p\,\tau )\right]}{p^{2}}}\;}
[ 78] , [ 77] « Fonction triangle symétrique sur
[
0
,
2
τ
]
{\displaystyle \;\color {transparent}{\left[0\,,\,2\,\tau \right]}}
f
triang sym
(
t
)
=
{
−
a
(
t
−
2
τ
)
si
t
∈
[
τ
,
2
τ
]
}
=
{\displaystyle \;\color {transparent}{f_{\text{triang sym}}(t)=\left\lbrace -a\;(t-2\,\tau )\;{\text{si}}\;t\in \left[\tau \,,\,2\,\tau \right]\right\rbrace =}}
L
{
f
triang sym
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\!\left\lbrace f_{\text{triang sym}}(t)\right\rbrace }}
=
a
[
1
−
exp
(
−
p
τ
)
]
2
p
2
{\displaystyle ={\dfrac {a\,\left[1-\exp(-p\,\tau )\right]^{2}}{p^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
« Fonction exponentielle réelle
f
exp
(
t
)
=
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;f_{\text{exp}}(t)=\exp(-a\,t)\;Y(t)\;}
a
∈
R
∗
{\displaystyle \;a\in \mathbb {R} ^{*}\;}
[ 80]
⇒
{\displaystyle \Rightarrow }
L
{
f
exp
(
t
)
}
=
1
p
+
a
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{exp}}(t)\right\rbrace ={\dfrac {1}{p+a}}\;}
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 81]
« Approche exponentielle réelle de l'échelon unité
f
app exp
(
t
)
=
[
1
−
exp
(
−
a
t
)
]
Y
(
t
)
=
Y
(
t
)
−
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;f_{\text{app exp}}(t)=\left[1-\exp(-a\,t)\right]\,Y(t)=Y(t)-\exp(-a\,t)\;Y(t)\;}
a
∈
R
+
∗
{\displaystyle \;a\in \mathbb {R} _{+}^{*}\;}
[ 82]
L
{
f
app exp
(
t
)
}
=
1
p
−
1
p
+
a
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{app exp}}(t)\right\rbrace ={\dfrac {1}{p}}-{\dfrac {1}{p+a}}\;}
[ 78] « Approche exponentielle réelle de l'échelon unité
f
app exp
(
t
)
=
[
1
−
exp
(
−
a
t
)
]
Y
(
t
)
=
Y
(
t
)
−
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{app exp}}(t)=\left[1-\exp(-a\,t)\right]\,Y(t)=Y(t)-\exp(-a\,t)\;Y(t)}\;}
a
∈
R
+
∗
{\displaystyle \;\color {transparent}{a\in \mathbb {R} _{+}^{*}}\;}
L
{
f
app exp
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\!\left\lbrace f_{\text{app exp}}(t)\right\rbrace }}
=
a
p
(
p
+
a
)
{\displaystyle ={\dfrac {a}{p\;(p+a)}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 83]
▸
{\displaystyle \blacktriangleright \;}
Fonction cosinusoïdale
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)\;}
ω
∈
R
+
∗
{\displaystyle \;\omega \in \mathbb {R} _{+}^{*}\;}
[ 82]
f
cos
(
t
)
=
ℜ
[
exp
(
i
ω
t
)
Y
(
t
)
]
{\displaystyle \;f_{\text{cos}}(t)=\Re \left[\exp(i\,\omega \,t)\;Y(t)\right]\;}
L
_
{
exp
(
i
ω
t
)
Y
(
t
)
}
=
1
p
−
i
ω
{\displaystyle \;{\underline {\mathcal {L}}}\!\left\lbrace \exp(i\,\omega \,t)\;Y(t)\right\rbrace ={\dfrac {1}{p-i\,\omega }}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 84] , [ 85]
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
p
{\displaystyle \;p}
ℜ
{
L
_
[
exp
(
i
ω
t
)
Y
(
t
)
]
}
=
L
{
ℜ
[
exp
(
i
ω
t
)
Y
(
t
)
]
}
{\displaystyle \;\Re \left\lbrace {\underline {\mathcal {L}}}\left[\exp(i\,\omega \,t)\;Y(t)\right]\right\rbrace ={\mathcal {L}}\left\lbrace \Re \left[\exp(i\,\omega \,t)\;Y(t)\right]\right\rbrace \;}
[ 55]
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
p
{\displaystyle \;\color {transparent}{p}}
L
{
f
cos
(
t
)
}
=
ℜ
[
1
p
−
i
ω
]
pour
p
=
ℜ
(
p
)
>
0
=
ℜ
[
p
+
i
ω
p
2
+
ω
2
]
pour
p
=
ℜ
(
p
)
>
0
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos}}(t)\right\rbrace =\Re \,\left[{\dfrac {1}{p-i\,\omega }}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,0}=\Re \,\left[{\dfrac {p+i\,\omega }{p^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
p
{\displaystyle \;\color {transparent}{p}}
L
{
f
cos
(
t
)
}
=
[
p
p
2
+
ω
2
]
pour
p
=
ℜ
(
p
)
>
0
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos}}(t)\right\rbrace =\left[{\dfrac {p}{p^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
p
{\displaystyle \;p\;}
[ 86]
L
{
cos
(
ω
t
)
Y
(
t
)
}
=
p
p
2
+
ω
2
{\displaystyle \;{\mathcal {L}}\!\left\lbrace \cos(\omega \,t)\;Y(t)\right\rbrace ={\dfrac {p}{p^{2}+\omega ^{2}}}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\Re e(p)>0\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
Autre démonstration
(
{\displaystyle \;{\big (}}
[ 87]
)
{\displaystyle {\big )}}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)\;}
T
{\displaystyle \;T}
T
=
2
π
ω
{\displaystyle \;T={\dfrac {2\;\pi }{\omega }}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
L
{
f
cos
(
t
)
}
=
∫
0
T
exp
(
−
p
t
)
f
cos
(
t
)
d
t
1
−
exp
(
−
p
T
)
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos}}(t)\right\rbrace ={\dfrac {\displaystyle \int _{0}^{T}\exp(-p\,t)\;f_{\text{cos}}(t)\;dt}{1-\exp(-p\,T)}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 88]
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
L
{
f
cos
(
t
)
}
=
∫
0
T
exp
(
−
p
t
)
cos
(
ω
t
)
d
t
1
−
exp
(
−
p
T
)
=
∫
0
T
exp
(
−
p
t
)
exp
(
i
ω
t
)
+
exp
(
−
i
ω
t
)
2
d
t
1
−
exp
(
−
p
T
)
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos}}(t)\right\rbrace ={\dfrac {\displaystyle \int _{0}^{T}\exp(-p\,t)\;\cos(\omega \,t)\;dt}{1-\exp(-p\,T)}}={\dfrac {\displaystyle \int _{0}^{T}\exp(-p\,t)\;{\dfrac {\exp(i\,\omega \,t)+\exp(-i\,\omega \,t)}{2}}\;dt}{1-\exp(-p\,T)}}\;}
[ 89]
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
L
{
f
cos
(
t
)
}
=
∫
0
T
{
exp
[
−
(
p
−
i
ω
)
t
]
+
exp
[
−
(
p
+
i
ω
)
t
]
}
d
t
2
[
1
−
exp
(
−
p
T
)
]
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos}}(t)\right\rbrace ={\dfrac {\displaystyle \int _{0}^{T}\left\lbrace \exp \!\left[-(p-i\;\omega )\;t\right]+\exp \!\left[-(p+i\;\omega )\;t\right]\right\rbrace \,dt}{2\,\left[1-\exp(-p\,T)\right]}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
nd membre se réécrivant en somme de deux intégrales :
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
≻
{\displaystyle \succ \;}
∫
0
T
exp
[
−
(
p
−
i
ω
)
t
]
d
t
=
[
−
exp
[
−
(
p
−
i
ω
)
t
]
p
−
i
ω
]
0
T
=
1
−
exp
(
−
p
T
)
exp
(
i
ω
T
)
p
−
i
ω
{\displaystyle \displaystyle \int _{0}^{T}\exp \!\left[-(p-i\;\omega )\;t\right]\;dt=\left[-{\dfrac {\exp \!\left[-(p-i\;\omega )\;t\right]}{p-i\;\omega }}\right]_{0}^{T}={\dfrac {1-\exp(-p\;T)\;\exp(i\;\omega \;T)}{p-i\;\omega }}}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
∫
0
T
exp
[
−
(
p
−
i
ω
)
t
]
d
t
{\displaystyle \color {transparent}{\displaystyle \int _{0}^{T}\exp \!\left[-(p-i\;\omega )\;t\right]\;dt}}
=
1
−
exp
(
−
p
T
)
p
−
i
ω
{\displaystyle ={\dfrac {1-\exp(-p\;T)}{p-i\;\omega }}\;}
exp
(
i
ω
T
)
=
exp
(
i
2
π
)
=
1
{\displaystyle \;\exp(i\;\omega \;T)=\exp(i\;2\;\pi )=1\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
≻
{\displaystyle \succ \;}
∫
0
T
exp
[
−
(
p
+
i
ω
)
t
]
d
t
=
[
−
exp
[
−
(
p
+
i
ω
)
t
]
p
+
i
ω
]
0
T
=
1
−
exp
(
−
p
T
)
exp
(
−
i
ω
T
)
p
+
i
ω
{\displaystyle \displaystyle \int _{0}^{T}\exp \!\left[-(p+i\;\omega )\;t\right]\;dt=\left[-{\dfrac {\exp \!\left[-(p+i\;\omega )\;t\right]}{p+i\;\omega }}\right]_{0}^{T}={\dfrac {1-\exp(-p\;T)\;\exp(-i\;\omega \;T)}{p+i\;\omega }}}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
∫
0
T
exp
[
−
(
p
+
i
ω
)
t
]
d
t
{\displaystyle \color {transparent}{\displaystyle \int _{0}^{T}\exp \!\left[-(p+i\;\omega )\;t\right]\;dt}}
=
1
−
exp
(
−
p
T
)
p
+
i
ω
{\displaystyle ={\dfrac {1-\exp(-p\;T)}{p+i\;\omega }}\;}
exp
(
−
i
ω
T
)
=
exp
(
−
i
2
π
)
=
1
{\displaystyle \;\exp(-i\;\omega \;T)=\exp(-i\;2\;\pi )=1\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
L
{
f
cos
(
t
)
}
=
1
2
[
1
p
−
i
ω
+
1
p
+
i
ω
]
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos}}(t)\right\rbrace ={\dfrac {1}{2}}\,\left[{\dfrac {1}{p-i\;\omega }}+{\dfrac {1}{p+i\;\omega }}\right]\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
L
{
cos
(
ω
t
)
Y
(
t
)
}
=
p
p
2
+
ω
2
{\displaystyle \;{\mathcal {L}}\!\left\lbrace \cos(\omega \,t)\;Y(t)\right\rbrace ={\dfrac {p}{p^{2}+\omega ^{2}}}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\Re e(p)>0\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos}}(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
L
{
cos
(
ω
t
)
Y
(
t
)
}
=
p
2
+
ω
2
{\displaystyle \;\color {transparent}{{\mathcal {L}}\!\left\lbrace \cos(\omega \,t)\;Y(t)\right\rbrace =p^{2}+\omega ^{2}}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\color {transparent}{\Re e(p)>0}\;}
[ 90]
▸
{\displaystyle \blacktriangleright \;}
Fonction sinusoïdale
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)\;}
ω
∈
R
+
∗
{\displaystyle \;\omega \in \mathbb {R} _{+}^{*}\;}
[ 82]
f
sin
(
t
)
=
ℑ
[
exp
(
i
ω
t
)
Y
(
t
)
]
{\displaystyle \;f_{\text{sin}}(t)=\Im \left[\exp(i\,\omega \,t)\;Y(t)\right]\;}
L
_
{
exp
(
i
ω
t
)
Y
(
t
)
}
=
1
p
−
i
ω
{\displaystyle \;{\underline {\mathcal {L}}}\!\left\lbrace \exp(i\,\omega \,t)\;Y(t)\right\rbrace ={\dfrac {1}{p-i\,\omega }}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 84] , [ 85]
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
p
{\displaystyle \;p}
ℑ
{
L
_
[
exp
(
i
ω
t
)
Y
(
t
)
]
}
=
L
{
ℑ
[
exp
(
i
ω
t
)
Y
(
t
)
]
}
{\displaystyle \;\Im \left\lbrace {\underline {\mathcal {L}}}\left[\exp(i\,\omega \,t)\;Y(t)\right]\right\rbrace ={\mathcal {L}}\left\lbrace \Im \left[\exp(i\,\omega \,t)\;Y(t)\right]\right\rbrace \;}
[ 55]
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
p
{\displaystyle \;\color {transparent}{p}}
L
{
f
sin
(
t
)
}
=
ℑ
[
1
p
−
i
ω
]
pour
p
=
ℜ
(
p
)
>
0
=
ℑ
[
p
+
i
ω
p
2
+
ω
2
]
pour
p
=
ℜ
(
p
)
>
0
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin}}(t)\right\rbrace =\Im \,\left[{\dfrac {1}{p-i\,\omega }}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,0}=\Im \,\left[{\dfrac {p+i\,\omega }{p^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
p
{\displaystyle \;\color {transparent}{p}}
L
{
f
sin
(
t
)
}
=
[
ω
p
2
+
ω
2
]
pour
p
=
ℜ
(
p
)
>
0
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin}}(t)\right\rbrace =\left[{\dfrac {\omega }{p^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
p
{\displaystyle \;p\;}
[ 86]
L
{
sin
(
ω
t
)
Y
(
t
)
}
=
ω
p
2
+
ω
2
{\displaystyle \;{\mathcal {L}}\!\left\lbrace \sin(\omega \,t)\;Y(t)\right\rbrace ={\dfrac {\omega }{p^{2}+\omega ^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
Autre démonstration
(
{\displaystyle \;{\big (}}
[ 87]
)
{\displaystyle {\big )}}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)\;}
T
{\displaystyle \;T}
T
=
2
π
ω
{\displaystyle \;T={\dfrac {2\;\pi }{\omega }}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
L
{
f
sin
(
t
)
}
=
∫
0
T
exp
(
−
p
t
)
f
sin
(
t
)
d
t
1
−
exp
(
−
p
T
)
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin}}(t)\right\rbrace ={\dfrac {\displaystyle \int _{0}^{T}\exp(-p\,t)\;f_{\text{sin}}(t)\;dt}{1-\exp(-p\,T)}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 88]
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
L
{
f
sin
(
t
)
}
=
∫
0
T
exp
(
−
p
t
)
sin
(
ω
t
)
d
t
1
−
exp
(
−
p
T
)
=
∫
0
T
exp
(
−
p
t
)
exp
(
i
ω
t
)
−
exp
(
−
i
ω
t
)
2
i
d
t
1
−
exp
(
−
p
T
)
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin}}(t)\right\rbrace ={\dfrac {\displaystyle \int _{0}^{T}\exp(-p\,t)\;\sin(\omega \,t)\;dt}{1-\exp(-p\,T)}}={\dfrac {\displaystyle \int _{0}^{T}\exp(-p\,t)\;{\dfrac {\exp(i\,\omega \,t)-\exp(-i\,\omega \,t)}{2\;i}}\;dt}{1-\exp(-p\,T)}}\;}
[ 91]
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
L
{
f
sin
(
t
)
}
=
∫
0
T
{
exp
[
−
(
p
−
i
ω
)
t
]
−
exp
[
−
(
p
+
i
ω
)
t
]
}
d
t
2
i
[
1
−
exp
(
−
p
T
)
]
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin}}(t)\right\rbrace ={\dfrac {\displaystyle \int _{0}^{T}\left\lbrace \exp \!\left[-(p-i\;\omega )\;t\right]-\exp \!\left[-(p+i\;\omega )\;t\right]\right\rbrace \,dt}{2\;i\,\left[1-\exp(-p\,T)\right]}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
nd membre se réécrivant en différence de deux intégrales :
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
≻
{\displaystyle \succ \;}
∫
0
T
exp
[
−
(
p
−
i
ω
)
t
]
d
t
=
…
1
−
exp
(
−
p
T
)
p
−
i
ω
{\displaystyle \displaystyle \int _{0}^{T}\exp \!\left[-(p-i\;\omega )\;t\right]\;dt{\overset {\ldots }{\;=\;}}\;{\dfrac {1-\exp(-p\;T)}{p-i\;\omega }}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
≻
{\displaystyle \succ \;}
∫
0
T
exp
[
−
(
p
+
i
ω
)
t
]
d
t
=
…
1
−
exp
(
−
p
T
)
p
+
i
ω
{\displaystyle \displaystyle \int _{0}^{T}\exp \!\left[-(p+i\;\omega )\;t\right]\;dt\;{\overset {\ldots }{\;=\;}}\;{\dfrac {1-\exp(-p\;T)}{p+i\;\omega }}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
L
{
f
sin
(
t
)
}
=
1
2
i
[
1
p
−
i
ω
−
1
p
+
i
ω
]
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin}}(t)\right\rbrace ={\dfrac {1}{2\;i}}\,\left[{\dfrac {1}{p-i\;\omega }}-{\dfrac {1}{p+i\;\omega }}\right]\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
L
{
sin
(
ω
t
)
Y
(
t
)
}
=
ω
p
2
+
ω
2
{\displaystyle \;{\mathcal {L}}\!\left\lbrace \sin(\omega \,t)\;Y(t)\right\rbrace ={\dfrac {\omega }{p^{2}+\omega ^{2}}}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\Re e(p)>0\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin}}(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
L
{
sin
(
ω
t
)
Y
(
t
)
}
=
p
2
+
ω
2
{\displaystyle \;\color {transparent}{{\mathcal {L}}\!\left\lbrace \sin(\omega \,t)\;Y(t)\right\rbrace =p^{2}+\omega ^{2}}\;}
ℜ
e
(
p
)
>
0
{\displaystyle \;\color {transparent}{\Re e(p)>0}\;}
[ 92]
▸
{\displaystyle \blacktriangleright \;}
Application : soit à déterminer la transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
f
(
t
)
=
cos
(
ω
t
+
φ
)
Y
(
t
)
{\displaystyle \;f(t)=\cos(\omega \,t+\varphi )\;Y(t)\;}
ω
∈
R
+
∗
{\displaystyle \;\omega \in \mathbb {R} _{+}^{*}\;}
[ 82]
φ
∈
]
−
π
,
+
π
]
{\displaystyle \;\varphi \in \left]-\pi \,,\,+\pi \right]\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
cos
(
ω
t
+
φ
)
=
cos
(
ω
t
)
cos
(
φ
)
−
sin
(
ω
t
)
sin
(
φ
)
{\displaystyle \;\cos(\omega \,t+\varphi )=\cos(\omega \,t)\;\cos(\varphi )-\sin(\omega \,t)\;\sin(\varphi )\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
f
(
t
)
=
cos
(
φ
)
cos
(
ω
t
)
Y
(
t
)
−
sin
(
φ
)
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;f(t)=\cos(\varphi )\;\cos(\omega \,t)\;Y(t)-\sin(\varphi )\;\sin(\omega \,t)\;Y(t)\;}
Application : soit à déterminer la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
«
L
{
cos
(
ω
t
+
φ
)
Y
(
t
)
}
=
cos
(
φ
)
L
{
cos
(
ω
t
)
Y
(
t
)
}
−
sin
(
φ
)
L
{
sin
(
ω
t
)
Y
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace \cos(\omega \,t+\varphi )\;Y(t)\right\rbrace =\cos(\varphi )\;{\mathcal {L}}\left\lbrace \cos(\omega \,t)\;Y(t)\right\rbrace -\sin(\varphi )\;{\mathcal {L}}\left\lbrace \sin(\omega \,t)\;Y(t)\right\rbrace \;}
[ 78]
⇒
{\displaystyle \Rightarrow }
Application : soit à déterminer la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
«
L
{
cos
(
ω
t
+
φ
)
Y
(
t
)
}
=
cos
(
φ
)
p
p
2
+
ω
2
−
sin
(
φ
)
ω
p
2
+
ω
2
{\displaystyle \;{\mathcal {L}}\left\lbrace \cos(\omega \,t+\varphi )\;Y(t)\right\rbrace =\cos(\varphi )\;{\dfrac {p}{p^{2}+\omega ^{2}}}-\sin(\varphi )\;{\dfrac {\omega }{p^{2}+\omega ^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
▸
{\displaystyle \blacktriangleright \;}
Fonction cosinusoïdale amortie exponentiellement
f
cos amort
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;f_{\text{cos amort}}(t)=\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)\;}
{
ω
∈
R
+
∗
a
∈
R
+
∗
}
{\displaystyle \;\left\lbrace \!{\begin{array}{c}\omega \in \mathbb {R} _{+}^{*}\\a\in \mathbb {R} _{+}^{*}\end{array}}\!\right\rbrace \;}
[ 82]
f
cos amort
(
t
)
=
ℜ
[
exp
(
−
a
t
)
exp
(
i
ω
t
)
Y
(
t
)
]
{\displaystyle \;f_{\text{cos amort}}(t)=\Re \left[\exp(-a\;t)\;\exp(i\,\omega \,t)\;Y(t)\right]\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos amort
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos amort}}(t)=\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)}\;}
L
_
{
exp
[
(
−
a
+
i
ω
)
t
]
Y
(
t
)
}
=
1
p
+
a
−
i
ω
{\displaystyle \;{\underline {\mathcal {L}}}\!\left\lbrace \exp \!\left[(-a+i\,\omega )\,t\right]\,Y(t)\right\rbrace ={\dfrac {1}{p+a-i\,\omega }}\;}
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 84] , [ 85]
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos amort
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos amort}}(t)}}
p
{\displaystyle \;p}
ℜ
⟨
L
_
{
exp
[
(
−
a
+
i
ω
)
t
]
Y
(
t
)
}
⟩
=
L
⟨
ℜ
{
exp
[
(
−
a
+
i
ω
)
t
]
Y
(
t
)
}
⟩
{\displaystyle \;\Re {\Big \langle }{\underline {\mathcal {L}}}\!\left\lbrace \exp \!\left[(-a+i\,\omega )\,t\right]\,Y(t)\right\rbrace \!{\Big \rangle }={\mathcal {L}}{\Big \langle }\Re \left\lbrace \exp \!\left[(-a+i\,\omega )\,t\right]\,Y(t)\right\rbrace \!{\Big \rangle }\;}
[ 55]
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos amort
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos amort}}(t)}}
p
{\displaystyle \;\color {transparent}{p}}
L
{
f
cos amort
(
t
)
}
=
ℜ
[
1
p
+
a
−
i
ω
]
pour
p
=
ℜ
(
p
)
>
−
a
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos amort}}(t)\right\rbrace =\Re \,\left[{\dfrac {1}{p+a-i\,\omega }}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,-a}}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos amort
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos amort}}(t)}}
p
{\displaystyle \;\color {transparent}{p}}
L
{
f
cos amort
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\!\left\lbrace f_{\text{cos amort}}(t)\right\rbrace }}
=
ℜ
[
p
+
a
+
i
ω
(
p
+
a
)
2
+
ω
2
]
pour
p
=
ℜ
(
p
)
>
−
a
{\displaystyle =\Re \,\left[{\dfrac {p+a+i\,\omega }{(p+a)^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,-a}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos amort
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos amort}}(t)}}
p
{\displaystyle \;\color {transparent}{p}}
L
{
f
cos amort
(
t
)
}
=
[
p
+
a
(
p
+
a
)
2
+
ω
2
]
pour
p
=
ℜ
(
p
)
>
−
a
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos amort}}(t)\right\rbrace =\left[{\dfrac {p+a}{(p+a)^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,-a}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
cos amort
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{cos amort}}(t)}}
p
{\displaystyle \;p\;}
[ 86]
L
{
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
}
=
p
+
a
(
p
+
a
)
2
+
ω
2
{\displaystyle \;{\mathcal {L}}\!\left\lbrace \exp(-a\,t)\;\cos(\omega \,t)\;Y(t)\right\rbrace ={\dfrac {p+a}{(p+a)^{2}+\omega ^{2}}}\;}
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
▸
{\displaystyle \blacktriangleright \;}
Fonction sinusoïdale amortie exponentiellement
f
sin amort
(
t
)
=
exp
(
−
a
t
)
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;f_{\text{sin amort}}(t)=\exp(-a\,t)\;\sin(\omega \,t)\;Y(t)\;}
{
ω
∈
R
+
∗
a
∈
R
+
∗
}
{\displaystyle \;\left\lbrace \!{\begin{array}{c}\omega \in \mathbb {R} _{+}^{*}\\a\in \mathbb {R} _{+}^{*}\end{array}}\!\right\rbrace \;}
[ 82]
f
sin amort
(
t
)
=
ℑ
[
exp
(
−
a
t
)
exp
(
i
ω
t
)
Y
(
t
)
]
{\displaystyle \;f_{\text{sin amort}}(t)=\Im \left[\exp(-a\;t)\;\exp(i\,\omega \,t)\;Y(t)\right]\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin amort
(
t
)
=
exp
(
−
a
t
)
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin amort}}(t)=\exp(-a\,t)\;\sin(\omega \,t)\;Y(t)}\;}
L
_
{
exp
[
(
−
a
+
i
ω
)
t
]
Y
(
t
)
}
==
1
p
+
a
−
i
ω
{\displaystyle \;{\underline {\mathcal {L}}}\!\left\lbrace \exp \!\left[(-a+i\,\omega )\,t\right]\,Y(t)\right\rbrace =={\dfrac {1}{p+a-i\,\omega }}\;}
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 84] , [ 85]
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin amort
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin amort}}(t)}}
p
{\displaystyle \;p}
ℑ
⟨
L
_
{
exp
[
(
−
a
+
i
ω
)
t
]
Y
(
t
)
}
⟩
=
L
⟨
ℑ
{
exp
[
(
−
a
+
i
ω
)
t
]
Y
(
t
)
}
⟩
{\displaystyle \;\Im {\Big \langle }{\underline {\mathcal {L}}}\!\left\lbrace \exp \!\left[(-a+i\,\omega )\,t\right]\,Y(t)\right\rbrace \!{\Big \rangle }={\mathcal {L}}{\Big \langle }\Im \left\lbrace \exp \!\left[(-a+i\,\omega )\,t\right]\,Y(t)\right\rbrace \!{\Big \rangle }\;}
[ 55]
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin amort
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin amort}}(t)}}
p
{\displaystyle \;\color {transparent}{p}}
L
{
f
sin amort
(
t
)
}
==
ℑ
[
1
p
+
a
−
i
ω
]
pour
p
=
ℜ
(
p
)
>
−
a
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin amort}}(t)\right\rbrace ==\Im \,\left[{\dfrac {1}{p+a-i\,\omega }}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,-a}}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin amort
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin amort}}(t)}}
p
{\displaystyle \;\color {transparent}{p}}
L
{
f
sin amort
(
t
)
}
{\displaystyle \;\color {transparent}{{\mathcal {L}}\!\left\lbrace f_{\text{sin amort}}(t)\right\rbrace }}
=
ℑ
[
p
+
a
+
i
ω
(
p
+
a
)
2
+
ω
2
]
pour
p
=
ℜ
(
p
)
>
−
a
{\displaystyle =\Im \,\left[{\dfrac {p+a+i\,\omega }{(p+a)^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,-a}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin amort
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin amort}}(t)}}
p
{\displaystyle \;\color {transparent}{p}}
L
{
f
sin amort
(
t
)
}
=
[
ω
(
p
+
a
)
2
+
ω
2
]
pour
p
=
ℜ
(
p
)
>
−
a
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin amort}}(t)\right\rbrace =\left[{\dfrac {\omega }{(p+a)^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,-a}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
f
sin amort
(
t
)
{\displaystyle \;\color {transparent}{f_{\text{sin amort}}(t)}}
p
{\displaystyle \;p\;}
[ 86]
L
{
exp
(
−
a
t
)
sin
(
ω
t
)
Y
(
t
)
}
=
ω
(
p
+
a
)
2
+
ω
2
{\displaystyle \;{\mathcal {L}}\!\left\lbrace \exp(-a\,t)\;\sin(\omega \,t)\;Y(t)\right\rbrace ={\dfrac {\omega }{(p+a)^{2}+\omega ^{2}}}\;}
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
▸
{\displaystyle \blacktriangleright \;}
Application : soit à déterminer la transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
f
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
+
φ
)
Y
(
t
)
{\displaystyle \;f(t)=\exp(-a\,t)\;\cos(\omega \,t+\varphi )\;Y(t)\;}
{
ω
∈
R
+
∗
a
∈
R
+
∗
}
{\displaystyle \;\left\lbrace \!{\begin{array}{c}\omega \in \mathbb {R} _{+}^{*}\\a\in \mathbb {R} _{+}^{*}\end{array}}\!\right\rbrace \;}
[ 82]
φ
∈
]
−
π
,
+
π
]
{\displaystyle \;\varphi \in \left]-\pi \,,\,+\pi \right]\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
exp
(
−
a
t
)
cos
(
ω
t
+
φ
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
cos
(
φ
)
−
exp
(
−
a
t
)
sin
(
ω
t
)
sin
(
φ
)
{\displaystyle \;\exp(-a\,t)\;\cos(\omega \,t+\varphi )=\exp(-a\,t)\;\cos(\omega \,t)\;\cos(\varphi )-\exp(-a\,t)\;\sin(\omega \,t)\;\sin(\varphi )\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
f
(
t
)
=
cos
(
φ
)
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
−
sin
(
φ
)
exp
(
−
a
t
)
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;f(t)=\cos(\varphi )\;\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)-\sin(\varphi )\;\exp(-a\,t)\;\sin(\omega \,t)\;Y(t)\;}
Application : soit à déterminer la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
«
L
{
exp
(
−
a
t
)
cos
(
ω
t
+
φ
)
Y
(
t
)
}
=
cos
(
φ
)
L
{
f
cos amort
(
t
)
}
−
sin
(
φ
)
L
{
f
sin amort
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace \exp(-a\,t)\;\cos(\omega \,t+\varphi )\;Y(t)\right\rbrace =\cos(\varphi )\;{\mathcal {L}}\left\lbrace f_{\text{cos amort}}(t)\right\rbrace -\sin(\varphi )\;{\mathcal {L}}\left\lbrace f_{\text{sin amort}}(t)\right\rbrace \;}
[ 78]
⇒
{\displaystyle \Rightarrow }
Application : soit à déterminer la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
«
L
{
exp
(
−
a
t
)
cos
(
ω
t
+
φ
)
Y
(
t
)
}
=
cos
(
φ
)
p
+
a
(
p
+
a
)
2
+
ω
2
−
sin
(
φ
)
ω
(
p
+
a
)
2
+
ω
2
{\displaystyle \;{\mathcal {L}}\left\lbrace \exp(-a\,t)\;\cos(\omega \,t+\varphi )\;Y(t)\right\rbrace =\cos(\varphi )\;{\dfrac {p+a}{(p+a)^{2}+\omega ^{2}}}-\sin(\varphi )\;{\dfrac {\omega }{(p+a)^{2}+\omega ^{2}}}\;}
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
Soit une fonction complexe
F
{\displaystyle \;F\;}
[ 93]
p
{\displaystyle \;p\;}
[ 93] holomorphe sur son domaine de définition.
Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] holomorphe dont l'originale est à support positif[ 14] Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
d'une fonction complexe holomorphe dont l'originale est à support positif étendu à gauche de
0
{\displaystyle \;0\;}
Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
support
[
0
,
+
∞
[
∪
V
(
0
−
)
{\displaystyle \;\left[0\,,\,+\infty \right[\;\cup \;{\mathcal {V}}(0^{-})\;}
{
{\displaystyle {\big \{}}
V
(
0
−
)
{\displaystyle \;{\mathcal {V}}(0^{-})\;}
Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
voisinage ouvert à gauche de
0
{\displaystyle \;0}
Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
telle que la restriction de la fonction à
[
0
,
+
∞
[
C
{\displaystyle \;\left[0\,,\,+\infty \right[^{\,C}\;}
[ 15] Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
dans
V
(
0
−
)
{\displaystyle \;{\mathcal {V}}(0^{-})\;}
[ 16] [ 17] Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
d'une fonction complexe holomorphe dont l'originale est une distribution
(
{\displaystyle \;{\big (}}
[ 18] Remarques : On prolonge la définition de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
de sa transformée de Laplace[ 1] [ 19]
)
{\displaystyle {\big )}}
Remarques : Dans le domaine d'holomorphie de
F
(
p
)
{\displaystyle \;F(p)\;}
p
{\displaystyle \;p\;}
F
(
p
)
{\displaystyle \;F(p)\;}
ère ou 2ème espèce, dans ce dernier cas Remarques : Dans le domaine d'holomorphie de
F
(
p
)
{\displaystyle \;\color {transparent}{F(p)}\;}
p
{\displaystyle \;\color {transparent}{p}\;}
F
(
p
)
{\displaystyle \;\color {transparent}{F(p)}\;}
ère ou
F
(
p
)
{\displaystyle F(p)\;}
distribution holomorphe [ 18]
Début d’un théorème
Théorème (admis) d'unicité de la transformée (monolatérale) inverse de Laplace d'une fonction complexe holomorphe
Fin du théorème
Soient «
F
1
(
p
)
{\displaystyle \;F_{1}(p)\;}
F
2
(
p
)
{\displaystyle \;F_{2}(p)\;}
(
{\displaystyle \;{\big (}}
distributions
)
{\displaystyle {\big )}\;}
holomorphes [ 95]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Soient «
F
1
(
p
)
{\displaystyle \;\color {transparent}{F_{1}(p)}\;}
F
2
(
p
)
{\displaystyle \;\color {transparent}{F_{2}(p)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
L
−
1
{
F
1
(
p
)
}
{\displaystyle \;{\mathcal {L}}^{-1}\!\left\lbrace F_{1}(p)\right\rbrace \;}
L
−
1
{
F
2
(
p
)
}
{\displaystyle \;{\mathcal {L}}^{-1}\!\left\lbrace F_{2}(p)\right\rbrace \;}
Soit avec «
α
=
max
{
α
1
,
α
2
}
{\displaystyle \;\alpha =\max \left\lbrace \alpha _{1}\,,\,\alpha _{2}\right\rbrace \;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Soient «
a
{\displaystyle \;a\;}
on démontre les propriétés suivantes
{
{\displaystyle \;{\big \{}}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
}
{\displaystyle {\big \}}}
on démontre les propriétés suivantes «
L
−
1
{
F
1
(
p
)
+
F
2
(
p
)
}
=
L
−
1
{
F
1
(
p
)
}
+
L
−
1
{
F
2
(
p
)
}
{\displaystyle \;{\mathcal {L}}^{-1}\!\left\lbrace F_{1}(p)+F_{2}(p)\right\rbrace ={\mathcal {L}}^{-1}\!\left\lbrace F_{1}(p)\right\rbrace +{\mathcal {L}}^{-1}\!\left\lbrace F_{2}(p)\right\rbrace \;}
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
on démontre les propriétés suivantes «
L
−
1
{
a
F
1
(
p
)
}
=
a
L
−
1
{
F
1
(
p
)
}
{\displaystyle \;{\mathcal {L}}^{-1}\!\left\lbrace a\;F_{1}(p)\right\rbrace =a\;{\mathcal {L}}^{-1}\!\left\lbrace F_{1}(p)\right\rbrace \;}
ℜ
(
p
)
>
α
1
{\displaystyle \;\Re (p)>\alpha _{1}\;}
on démontre les propriétés suivantes «
L
−
1
{
a
F
2
(
p
)
}
=
a
L
−
1
{
F
2
(
p
)
}
{\displaystyle \;{\mathcal {L}}^{-1}\!\left\lbrace a\;F_{2}(p)\right\rbrace =a\;{\mathcal {L}}^{-1}\!\left\lbrace F_{2}(p)\right\rbrace \;}
ℜ
(
p
)
>
α
2
{\displaystyle \;\Re (p)>\alpha _{2}\;}
[ modifier | modifier le wikicode ] Il n'existe pas de formule analytique générale permettant de calculer
f
(
t
)
{\displaystyle \;f(t)\;}
F
(
p
)
{\displaystyle \;F(p)\;}
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}}
F
(
p
)
{\displaystyle \;F(p)}
Il n'existe pas de formule analytique générale permettant de calculer la fonction
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}}
f
(
t
)
{\displaystyle \;f(t)\;}
Il n'existe pas de formule analytique générale permettant de calculer la fonction
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}}
sa définition peut nécessiter une forme intégrale
(
{\displaystyle \;{\big (}}
développement en série
)
{\displaystyle {\big )}}
…
{\displaystyle \;\ldots }
Une façon de définir l'originale de la fonction
(
{\displaystyle \;{\big (}}
distribution
)
{\displaystyle {\big )}}
F
(
p
)
{\displaystyle \;F(p)\;}
holomorphe sur son domaine de définition est «
f
(
t
)
=
L
−
1
{
F
(
p
)
}
=
1
2
π
i
∫
γ
−
i
∞
γ
+
i
∞
exp
(
p
t
)
F
(
p
)
d
p
{\displaystyle \;f(t)={\mathcal {L}}^{-1}\left\lbrace F(p)\right\rbrace ={\dfrac {1}{2\;\pi \;i}}\;\displaystyle \int _{\gamma \,-\,i\,\infty }^{\gamma \,+\,i\,\infty }\exp(p\,t)\;F(p)\;dp\;}
[ 96] Une façon de définir l'originale de la fonction
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}}
F
(
p
)
{\displaystyle \;\color {transparent}{F(p)}\;}
f
(
t
)
=
L
−
1
{
F
(
p
)
}
{\displaystyle \;\color {transparent}{f(t)={\mathcal {L}}^{-1}\left\lbrace F(p)\right\rbrace }}
γ
∈
R
{\displaystyle \;\gamma \in \mathbb {R} \;}
[ 97]
Préliminaire : une fonction
F
(
p
)
{\displaystyle \;F(p)\;}
p
{\displaystyle \;p\;}
rationnelle fonctions polynômes à valeurs dans
C
{\displaystyle \;\mathbb {C} }
Préliminaire : si
P
{\displaystyle \;P\;}
Q
{\displaystyle \;Q\;}
fonctions polynômes non identiquement nulles[ 98]
F
=
P
Q
{\displaystyle \;F={\dfrac {P}{Q}}\;}
p
{\displaystyle \;p\;}
Q
(
p
)
≠
0
{\displaystyle \;Q(p)\neq 0\;}
F
(
p
)
=
P
(
p
)
Q
(
p
)
{\displaystyle \;F(p)={\dfrac {P(p)}{Q(p)}}\;}
Soit « la fonction complexe rationnelle
F
(
p
)
=
P
(
p
)
Q
(
p
)
{\displaystyle \;F(p)={\dfrac {P(p)}{Q(p)}}\;}
polynômes
P
{\displaystyle \;P\;}
Q
{\displaystyle \;Q\;}
P
(
p
)
{\displaystyle \;P(p)\;}
<
{\displaystyle <\;}
Q
(
p
)
{\displaystyle \;Q(p)\;}
holomorphe sur
C
{\displaystyle \;\mathbb {C} \;}
p
{\displaystyle \;p\;}
Q
(
p
)
{\displaystyle \;Q(p)\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
f
(
t
)
=
{\displaystyle \;f(t)=}
L
−
1
{
F
(
p
}
{\displaystyle {\mathcal {L}}^{-1}\!\left\lbrace F(p\right\rbrace \;}
[ 99]
Soit «
F
(
p
)
=
P
(
p
)
Q
(
p
)
{\displaystyle \;F(p)={\dfrac {P(p)}{Q(p)}}\;}
[ 100]
P
{\displaystyle \;P\;}
Q
{\displaystyle \;Q\;}
d
e
g
[
P
]
<
d
e
g
[
Q
]
{\displaystyle \;\mathrm {deg} \left[P\right]<\mathrm {deg} \left[Q\right]\;}
Soit «
F
(
p
)
=
P
(
p
)
{\displaystyle \;\color {transparent}{F(p)=P(p)}\;}
la méthode pour déterminer
f
(
t
)
{\displaystyle \;f(t)}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
P
(
p
)
Q
(
p
)
{\displaystyle \;F(p)={\dfrac {P(p)}{Q(p)}}\;}
Soit «
F
(
p
)
=
P
(
p
)
{\displaystyle \;\color {transparent}{F(p)=P(p)}\;}
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}}
∙
{\displaystyle \bullet \;}
z
k
{\displaystyle \;z_{k}\;}
[
{\displaystyle \;{\big [}}
P
(
p
)
]
{\displaystyle \;P(p){\big ]}\;}
[ 101]
q
l
{\displaystyle \;q_{l}\;}
[
{\displaystyle \;{\big [}}
Q
(
p
)
]
{\displaystyle \;Q(p){\big ]}\;}
[ 101] Soit «
F
(
p
)
=
P
(
p
)
{\displaystyle \;\color {transparent}{F(p)=P(p)}\;}
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
P
(
p
)
=
A
∏
k
(
p
−
z
k
)
{\displaystyle \;P(p)=A\;\prod \limits _{k}\left(p-z_{k}\right)\;}
[ 102]
Q
(
p
)
=
∏
l
(
p
−
q
l
)
{\displaystyle \;Q(p)=\prod \limits _{l}\left(p-q_{l}\right)\;}
[ 103]
F
(
p
)
=
A
∏
k
(
p
−
z
k
)
∏
l
(
p
−
q
l
)
{\displaystyle \;F(p)=A\;{\dfrac {\prod \limits _{k}\left(p-z_{k}\right)}{\prod \limits _{l}\left(p-q_{l}\right)}}\;}
[ 102] , [ 103] Soit «
F
(
p
)
=
P
(
p
)
{\displaystyle \;\color {transparent}{F(p)=P(p)}\;}
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}}
∙
{\displaystyle \color {transparent}{\bullet }\;}
après avoir fait les simplifications éventuelles de façon à ce que la fonction rationnelle soit irréductible, puis Soit «
F
(
p
)
=
P
(
p
)
{\displaystyle \;\color {transparent}{F(p)=P(p)}\;}
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}}
∙
{\displaystyle \bullet \;}
fonction rationnelle irréductible en éléments simples soit, dans le cas où tous les pôles sont réels et simples, Soit «
F
(
p
)
=
P
(
p
)
{\displaystyle \;\color {transparent}{F(p)=P(p)}\;}
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
F
(
p
)
=
A
∑
l
ρ
l
p
−
q
l
{\displaystyle \;F(p)=A\;\sum \limits _{l}{\dfrac {\rho _{l}}{p-q_{l}}}\;}
ρ
l
∈
R
∗
{\displaystyle \;\rho _{l}\in \mathbb {R} ^{*}\;}
[ 104] , [ 105] Soit «
F
(
p
)
=
P
(
p
)
{\displaystyle \;\color {transparent}{F(p)=P(p)}\;}
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}}
∙
{\displaystyle \bullet \;}
Soit «
F
(
p
)
=
P
(
p
)
{\displaystyle \;\color {transparent}{F(p)=P(p)}\;}
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
L
−
1
{
ρ
l
p
−
q
l
}
=
ρ
l
exp
(
q
l
t
)
Y
(
t
)
{\displaystyle \;{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {\rho _{l}}{p-q_{l}}}\right\rbrace =\rho _{l}\;\exp(q_{l}\,t)\;Y(t)\;}
ℜ
(
p
)
>
q
l
{\displaystyle \;\Re (p)>q_{l}\;}
[ 106] Soit «
F
(
p
)
=
P
(
p
)
{\displaystyle \;\color {transparent}{F(p)=P(p)}\;}
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
L
−
1
{
F
(
p
)
}
=
A
[
∑
l
ρ
l
exp
(
q
l
t
)
]
Y
(
t
)
{\displaystyle \;{\mathcal {L}}^{-1}\!\left\lbrace F(p)\right\rbrace =A\left[\sum \limits _{l}\rho _{l}\;\exp(q_{l}\,t)\right]Y(t)\;}
ℜ
(
p
)
>
max
l
q
l
{\displaystyle \;\Re (p)>\max \limits _{l}q_{l}\;}
[ 107] , [ 106]
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;f(t)\;}
F
(
p
)
=
p
+
1
p
2
+
5
p
+
6
{\displaystyle \;F(p)={\dfrac {p+1}{p^{2}+5\,p+6}}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
on commence par
∙
{\displaystyle \bullet \;}
fonction rationnelle soit « un zéro
z
1
=
−
1
{\displaystyle \;z_{1}=-1\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
les pôles de cette dernière c'est-à-dire les solutions de
p
2
+
5
p
+
6
=
0
{\displaystyle \;p^{2}+5\,p+6=0\;}
Δ
=
5
2
−
4
×
6
=
1
>
0
{\displaystyle \;\Delta =5^{2}-4\times 6=1>0\;}
⇒
{\displaystyle \Rightarrow }
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
« deux pôles réels et simples
{
q
1
=
−
5
+
1
2
=
−
2
q
2
=
−
5
−
1
2
=
−
3
}
{\displaystyle \;\left\lbrace {\begin{array}{c}q_{1}={\dfrac {-5+{\sqrt {1}}}{2}}=-2\\q_{2}={\dfrac {-5-{\sqrt {1}}}{2}}=-3\end{array}}\right\rbrace \;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \bullet \;}
rationnelle holomorphe
F
(
p
)
=
p
+
1
(
p
+
2
)
(
p
+
3
)
{\displaystyle \;F(p)={\dfrac {p+1}{(p+2)\;(p+3)}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \bullet \;}
F
(
p
)
=
p
+
1
(
p
+
2
)
(
p
+
3
)
=
ρ
1
p
+
2
+
ρ
2
p
+
3
{\displaystyle \;F(p)={\dfrac {p+1}{(p+2)\;(p+3)}}={\dfrac {\rho _{1}}{p+2}}+{\dfrac {\rho _{2}}{p+3}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
[
{\displaystyle \;{\big [}}
ρ
1
{\displaystyle \;\rho _{1}\;}
ρ
2
{\displaystyle \;\rho _{2}\;}
[ 104]
]
{\displaystyle {\big ]}}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \succ \;}
ρ
1
{\displaystyle \;\rho _{1}\;}
p
+
2
{\displaystyle \;p+2\;}
p
=
−
2
{\displaystyle \;p=-2\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
[
p
+
1
p
+
3
]
p
=
−
2
=
ρ
1
+
[
ρ
2
(
p
+
2
)
p
+
3
]
p
=
−
2
{\displaystyle \left[{\dfrac {p+1}{p+3}}\right]_{\!p=-2}=\rho _{1}\;{\cancel {+\left[{\dfrac {\rho _{2}\;(p+2)}{p+3}}\right]_{\!p=-2}}}\;}
⇒
{\displaystyle \Rightarrow }
ρ
1
=
−
1
{\displaystyle \;\rho _{1}=-1\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \succ \;}
ρ
2
{\displaystyle \;\rho _{2}\;}
se déterminant en multipliant de part et d'autre par
p
+
3
{\displaystyle \;p+3\;}
p
=
−
3
{\displaystyle \;p=-3\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
[
p
+
1
p
+
2
]
p
=
−
3
=
[
ρ
1
(
p
+
3
)
p
+
2
]
p
=
−
3
+
ρ
2
{\displaystyle \left[{\dfrac {p+1}{p+2}}\right]_{\!p=-3}={\cancel {\left[{\dfrac {\rho _{1}\;(p+3)}{p+2}}\right]_{\!p=-3}+}}\;\rho _{2}\;}
⇒
{\displaystyle \Rightarrow }
ρ
2
=
2
{\displaystyle \;\rho _{2}=2\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
F
(
p
)
=
−
1
p
+
2
+
2
p
+
3
{\displaystyle \;F(p)=-{\dfrac {1}{p+2}}+{\dfrac {2}{p+3}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \bullet \;}
f
(
t
)
{\displaystyle \;f(t)\;}
F
(
p
)
=
p
+
1
p
2
+
5
p
+
6
{\displaystyle \;F(p)={\dfrac {p+1}{p^{2}+5\,p+6}}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
f
(
t
)
=
L
−
1
{
F
(
p
)
}
=
−
L
−
1
{
1
p
+
2
}
+
2
L
−
1
{
1
p
+
3
}
{\displaystyle \;f(t)={\mathcal {L}}^{-1}\!\left\lbrace F(p)\right\rbrace =-{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {1}{p+2}}\right\rbrace +2\;{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {1}{p+3}}\right\rbrace \;}
[ 107] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}}
=
L
−
1
{
p
+
1
p
2
+
5
p
+
6
}
=
[
−
exp
(
−
2
t
)
+
2
exp
(
−
3
t
)
]
Y
(
t
)
{\displaystyle ={\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {p+1}{p^{2}+5\,p+6}}\right\rbrace =\left[-\exp(-2\,t)+2\;\exp(-3\,t)\right]\,Y(t)\;}
[ 106]
ℜ
(
p
)
>
max
(
−
2
;
−
3
)
=
−
2
{\displaystyle \;\Re (p)>\max \left(-2\,;\,-3\right)=-2\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;f(t)\;}
F
(
p
)
=
p
+
1
p
(
p
2
+
4
p
+
4
)
{\displaystyle \;F(p)={\dfrac {p+1}{p\,(p^{2}+4\,p+4)}}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
on commence par
∙
{\displaystyle \bullet \;}
fonction rationnelle soit « un zéro
z
1
=
−
1
{\displaystyle \;z_{1}=-1\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
les pôles de cette dernière c'est-à-dire les solutions de
p
(
p
2
+
4
p
+
4
)
=
0
{\displaystyle \;p\,(p^{2}+4\,p+4)=0\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
« un 1er pôle réel simple
q
1
=
0
{\displaystyle \;q_{1}=0\;}
[ 108] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
d'autres pôles racines de
p
2
+
4
p
+
4
=
0
{\displaystyle \;p^{2}+4\,p+4=0\;}
ème degré Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
p
2
+
4
p
+
4
=
0
{\displaystyle \;\color {transparent}{p^{2}+4\,p+4=0}\;}
de discriminant réduit
Δ
′
=
2
2
−
4
=
0
{\displaystyle \;\Delta '=2^{2}-4=0\;}
⇒
{\displaystyle \Rightarrow }
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
« un pôle réel double
q
2
=
−
2
{\displaystyle \;q_{2}=-2\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \bullet \;}
rationnelle holomorphe
F
(
p
)
=
p
+
1
p
(
p
+
2
)
2
{\displaystyle \;F(p)={\dfrac {p+1}{p\;(p+2)^{2}}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \bullet \;}
F
(
p
)
=
p
+
1
p
(
p
+
2
)
2
=
ρ
1
p
+
ρ
2
(
p
+
2
)
2
+
ρ
′
2
p
+
2
{\displaystyle \;F(p)={\dfrac {p+1}{p\;(p+2)^{2}}}={\dfrac {\rho _{1}}{p}}+{\dfrac {\rho _{2}}{(p+2)^{2}}}+{\dfrac {{\rho '}_{\!2}}{p+2}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
[
{\displaystyle \;{\big [}}
ρ
1
{\displaystyle \;\rho _{1}}
ρ
2
{\displaystyle \;\rho _{2}\;}
ρ
′
2
{\displaystyle \;{\rho '}_{\!2}\;}
[ 104] , [ 109]
]
{\displaystyle {\big ]}}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \succ \;}
ρ
1
{\displaystyle \;\rho _{1}\;}
p
{\displaystyle \;p\;}
p
=
0
{\displaystyle \;p=0\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
[
p
+
1
(
p
+
2
)
2
]
p
=
0
=
ρ
1
+
[
ρ
2
p
(
p
+
2
)
2
]
p
=
0
+
[
ρ
′
2
p
p
+
2
]
p
=
0
{\displaystyle \left[{\dfrac {p+1}{(p+2)^{2}}}\right]_{\!p=0}=\rho _{1}\;{\cancel {+\left[{\dfrac {\rho _{2}\;p}{(p+2)^{2}}}\right]_{\!p=0}+\left[{\dfrac {{\rho '}_{2}\;p}{p+2}}\right]_{\!p=0}}}\;}
⇒
{\displaystyle \Rightarrow }
ρ
1
=
1
4
{\displaystyle \;\rho _{1}={\dfrac {1}{4}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \succ \;}
ρ
2
{\displaystyle \;\rho _{2}\;}
se déterminant en multipliant de part et d'autre par
(
p
+
2
)
2
{\displaystyle \;(p+2)^{2}\;}
p
=
−
2
{\displaystyle \;p=-2\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
[
p
+
1
p
]
p
=
−
2
=
[
ρ
1
(
p
+
2
)
2
p
]
p
=
−
2
+
ρ
2
+
[
ρ
′
2
(
p
+
2
)
]
p
=
−
2
{\displaystyle \left[{\dfrac {p+1}{p}}\right]_{\!p=-2}={\cancel {\left[{\dfrac {\rho _{1}\;(p+2)^{2}}{p}}\right]_{\!p=-2}+}}\;\rho _{2}\;{\cancel {+\left[{\rho '}_{2}\;(p+2)\right]_{p=-2}}}\;}
⇒
{\displaystyle \Rightarrow }
ρ
2
=
1
2
{\displaystyle \;\rho _{2}={\dfrac {1}{2}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \succ \;}
ρ
′
2
{\displaystyle \;{\rho '}_{2}\;}
se déterminant en multipliant de part et d'autre par
p
+
2
{\displaystyle \;p+2\;}
p
→
∞
{\displaystyle \;p\rightarrow \infty \;}
[ 110] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
[
p
+
1
p
(
p
+
2
)
]
p
→
∞
=
[
ρ
1
(
p
+
2
)
p
]
p
→
∞
+
[
ρ
2
p
+
2
]
p
→
∞
+
ρ
′
2
{\displaystyle {\cancel {\left[{\dfrac {p+1}{p\,(p+2)}}\right]_{\!p\rightarrow \infty }}}\;=\left[{\dfrac {\rho _{1}\;(p+2)}{p}}\right]_{\!p\rightarrow \infty }\;{\cancel {+\left[{\dfrac {\rho _{2}}{p+2}}\right]_{\!p\rightarrow \infty }+}}\;{\rho '}_{2}\;}
⇒
{\displaystyle \Rightarrow }
ρ
′
2
=
−
ρ
1
{\displaystyle \;{\rho '}_{2}=-\rho _{1}\;}
ρ
1
=
1
4
{\displaystyle \;\rho _{1}={\dfrac {1}{4}}}
ρ
′
2
=
−
1
4
{\displaystyle \;{\rho '}_{2}=-{\dfrac {1}{4}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
soit finalement «
F
(
p
)
=
1
4
p
+
1
2
(
p
+
2
)
2
−
1
4
(
p
+
2
)
{\displaystyle \;F(p)={\dfrac {1}{4\;p}}+{\dfrac {1}{2\,(p+2)^{2}}}-{\dfrac {1}{4\,(p+2)}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \bullet \;}
f
(
t
)
{\displaystyle \;f(t)\;}
F
(
p
)
=
p
+
1
p
(
p
2
+
4
p
+
4
)
{\displaystyle \;F(p)={\dfrac {p+1}{p\,(p^{2}+4\,p+4)}}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
f
(
t
)
=
L
−
1
{
F
(
p
)
}
=
1
4
L
−
1
{
1
p
}
+
1
2
L
−
1
{
1
(
p
+
2
)
2
}
−
1
4
L
−
1
{
1
p
+
2
}
{\displaystyle \;f(t)={\mathcal {L}}^{-1}\!\left\lbrace F(p)\right\rbrace ={\dfrac {1}{4}}\;{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {1}{p}}\right\rbrace +{\dfrac {1}{2}}\;{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {1}{(p+2)^{2}}}\right\rbrace -{\dfrac {1}{4}}\;{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {1}{p+2}}\right\rbrace \;}
[ 107] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}}
=
L
−
1
{
p
+
1
p
(
p
2
+
4
p
+
4
)
}
=
1
4
[
1
+
2
t
exp
(
−
2
t
)
−
exp
(
−
2
t
)
]
Y
(
t
)
{\displaystyle ={\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {p+1}{p\,(p^{2}+4\,p+4)}}\right\rbrace ={\dfrac {1}{4}}\left[1+2\;t\;\exp(-2\,t)-\exp(-2\,t)\right]\,Y(t)\;}
[ 106] , [ 111]
ℜ
(
p
)
>
max
(
0
;
−
2
)
{\displaystyle \;\Re (p)>\max \left(0\,;\,-2\right)\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}}
=
L
−
1
{
p
(
p
2
+
4
p
+
4
)
}
=
1
[
1
+
2
t
exp
(
−
2
t
)
−
exp
(
−
2
t
)
]
Y
(
t
)
{\displaystyle \color {transparent}{={\mathcal {L}}^{-1}\!\left\lbrace p\,(p^{2}+4\,p+4)\right\rbrace =1\left[1+2\;t\;\exp(-2\,t)-\exp(-2\,t)\right]\,Y(t)}\;}
ou
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;f(t)\;}
F
(
p
)
=
p
+
1
p
(
p
2
+
p
+
2
)
{\displaystyle \;F(p)={\dfrac {p+1}{p\,(p^{2}+p+2)}}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
on commence par
∙
{\displaystyle \bullet \;}
fonction rationnelle soit « un zéro
z
1
=
−
1
{\displaystyle \;z_{1}=-1\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
les pôles de cette dernière c'est-à-dire les solutions de
p
(
p
2
+
4
p
+
4
)
=
0
{\displaystyle \;p\,(p^{2}+4\,p+4)=0\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
« un 1er pôle réel simple
q
1
=
0
{\displaystyle \;q_{1}=0\;}
[ 112] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
d'autres pôles racines de
p
2
+
p
+
2
=
0
{\displaystyle \;p^{2}+p+2=0\;}
ème degré Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
p
2
+
p
+
2
=
0
{\displaystyle \;\color {transparent}{p^{2}+p+2=0}\;}
de discriminant
Δ
=
1
2
−
4
×
2
=
−
7
<
0
{\displaystyle \;\Delta =1^{2}-4\times 2=-7<0\;}
⇒
{\displaystyle \Rightarrow }
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
« deux pôles complexes conjugués qu'il n'est pas utiles de déterminer » puis Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \bullet \;}
rationnelle holomorphe
F
(
p
)
=
p
+
1
p
(
p
2
+
p
+
2
)
{\displaystyle \;F(p)={\dfrac {p+1}{p\;(p^{2}+p+2)}}\;}
[ 113] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \bullet \;}
F
(
p
)
=
p
+
1
p
(
p
2
+
p
+
2
)
=
ρ
1
p
+
ρ
2
p
+
ρ
′
2
p
2
+
p
+
2
{\displaystyle \;F(p)={\dfrac {p+1}{p\;(p^{2}+p+2)}}={\dfrac {\rho _{1}}{p}}+{\dfrac {\rho _{2}\;p+{\rho '}_{2}}{p^{2}+p+2}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
[
{\displaystyle \;{\big [}}
ρ
1
{\displaystyle \;\rho _{1}}
ρ
2
{\displaystyle \;\rho _{2}\;}
ρ
′
2
{\displaystyle \;{\rho '}_{\!2}\;}
[ 104] , [ 114]
]
{\displaystyle {\big ]}}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \succ \;}
ρ
1
{\displaystyle \;\rho _{1}\;}
p
{\displaystyle \;p\;}
p
=
0
{\displaystyle \;p=0\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
[
p
+
1
p
2
+
p
+
2
]
p
=
0
=
ρ
1
+
[
(
ρ
2
p
+
ρ
′
2
)
p
p
2
+
p
+
2
]
p
=
0
{\displaystyle \left[{\dfrac {p+1}{p^{2}+p+2}}\right]_{\!p=0}=\rho _{1}\;{\cancel {+\left[{\dfrac {\left(\rho _{2}\;p+{\rho '}_{2}\right)\,p}{p^{2}+p+2}}\right]_{\!p=0}}}\;}
⇒
{\displaystyle \Rightarrow }
ρ
1
=
1
2
{\displaystyle \;\rho _{1}={\dfrac {1}{2}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \succ \;}
ρ
2
{\displaystyle \;\rho _{2}\;}
se déterminant en multipliant de part et d'autre par
p
{\displaystyle \;p\;}
p
→
∞
{\displaystyle \;p\rightarrow \infty \;}
[ 115] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
[
p
+
1
p
2
+
p
+
2
]
p
→
∞
=
ρ
1
+
[
(
ρ
2
p
+
ρ
′
2
)
p
p
2
+
p
+
2
]
p
→
∞
{\displaystyle {\cancel {\left[{\dfrac {p+1}{p^{2}+p+2}}\right]_{\!p\rightarrow \infty }}}=\rho _{1}\;+\left[{\dfrac {\left(\rho _{2}\;p+{\rho '}_{2}\right)\,p}{p^{2}+p+2}}\right]_{\!p\rightarrow \infty }\;}
⇒
{\displaystyle \Rightarrow }
ρ
2
=
−
ρ
1
{\displaystyle \;\rho _{2}=-\rho _{1}\;}
⇒
{\displaystyle \Rightarrow }
ρ
2
=
−
1
2
{\displaystyle \;\rho _{2}=-{\dfrac {1}{2}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \succ \;}
ρ
′
2
{\displaystyle \;{\rho '}_{2}\;}
se déterminant en donnant une valeur particulière à
p
{\displaystyle \;p\;}
p
=
−
1
{\displaystyle \;p=-1\;}
[ 116] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
[
p
+
1
p
(
p
2
+
p
+
2
)
]
p
=
−
1
=
[
ρ
1
p
+
ρ
2
p
+
ρ
′
2
p
2
+
p
+
2
]
p
=
−
1
{\displaystyle {\cancel {\left[{\dfrac {p+1}{p\;(p^{2}+p+2)}}\right]_{\!p=-1}}}=\left[{\dfrac {\rho _{1}}{p}}+{\dfrac {\rho _{2}\;p+{\rho '}_{2}}{p^{2}+p+2}}\right]_{\!p=-1}\;}
⇒
{\displaystyle \Rightarrow }
ρ
′
2
2
=
ρ
1
+
ρ
2
2
=
ρ
1
2
{\displaystyle \;{\dfrac {{\rho '}_{2}}{2}}=\rho _{1}+{\dfrac {\rho _{2}}{2}}={\dfrac {\rho _{1}}{2}}\;}
[ 117]
⇒
{\displaystyle \Rightarrow }
ρ
′
2
=
ρ
1
{\displaystyle \;{\rho '}_{2}=\rho _{1}\;}
⇒
{\displaystyle \Rightarrow }
ρ
′
2
=
1
2
{\displaystyle \;{\rho '}_{2}={\dfrac {1}{2}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
soit finalement «
F
(
p
)
=
1
2
p
−
1
2
p
−
1
p
2
+
p
+
2
{\displaystyle \;F(p)={\dfrac {1}{2\;p}}-{\dfrac {1}{2}}\;{\dfrac {p-1}{p^{2}+p+2}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \bullet \;}
f
(
t
)
{\displaystyle \;f(t)\;}
F
(
p
)
=
p
+
1
p
(
p
2
+
p
+
2
)
{\displaystyle \;F(p)={\dfrac {p+1}{p\,(p^{2}+p+2)}}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
f
(
t
)
=
L
−
1
{
F
(
p
)
}
=
1
2
L
−
1
{
1
p
}
−
1
2
L
−
1
{
p
−
1
p
2
+
p
+
2
}
{\displaystyle \;f(t)={\mathcal {L}}^{-1}\!\left\lbrace F(p)\right\rbrace ={\dfrac {1}{2}}\;{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {1}{p}}\right\rbrace -{\dfrac {1}{2}}\;{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {p-1}{p^{2}+p+2}}\right\rbrace \;}
[ 107]
{
{\displaystyle \;{\Bigg \{}\!}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
2ème élément simple correspondant aux pôles complexes conjugués de façon à utiliser les transformées
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
inverses de Laplace[ 1]
[
L
−
1
{
p
+
a
(
p
+
a
)
2
+
ω
2
}
=
exp
(
−
a
t
)
cos
(
ω
t
)
L
−
1
{
ω
(
p
+
a
)
2
+
ω
2
}
=
exp
(
−
a
t
)
sin
(
ω
t
)
]
{\displaystyle \;\left[{\begin{array}{c}{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {p+a}{(p+a)^{2}+\omega ^{2}}}\right\rbrace =\exp(-a\,t)\;\cos(\omega \,t)\\{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {\omega }{(p+a)^{2}+\omega ^{2}}}\right\rbrace =\exp(-a\,t)\;\sin(\omega \,t)\end{array}}\right]\;}
[ 118]
soit, le dénominateur de ce 2ème Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
élément simple se réécrivant selon «
p
2
+
p
+
2
=
(
p
+
1
2
)
2
−
1
4
+
2
=
(
p
+
1
2
)
2
+
(
7
2
)
2
{\displaystyle \;p^{2}+p+2=\left(p+{\dfrac {1}{2}}\right)^{\!2}-{\dfrac {1}{4}}+2=\left(p+{\dfrac {1}{2}}\right)^{\!2}+\left({\dfrac {\sqrt {7}}{2}}\right)^{\!2}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
[
a
=
1
2
ω
=
7
2
]
{\displaystyle \left[{\begin{array}{c}a={\dfrac {1}{2}}\\\omega ={\dfrac {\sqrt {7}}{2}}\end{array}}\right]}
ème élément simple selon «
p
−
1
=
(
p
+
1
2
)
−
3
2
{\displaystyle \;p-1=\left(p+{\dfrac {1}{2}}\right)-{\dfrac {3}{2}}\;}
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
soit «
p
−
1
=
(
p
+
1
2
)
−
3
7
7
2
{\displaystyle \;p-1=\left(p+{\dfrac {1}{2}}\right)-{\dfrac {3}{\sqrt {7}}}\;{\dfrac {\sqrt {7}}{2}}\;}
ème élément simple selon Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
p
−
1
p
2
+
p
+
2
=
p
+
1
2
(
p
+
1
2
)
2
+
(
7
2
)
2
−
3
7
7
2
(
p
+
1
2
)
2
+
(
7
2
)
2
{\displaystyle \;{\dfrac {p-1}{p^{2}+p+2}}={\dfrac {p+{\dfrac {1}{2}}}{\left(p+{\dfrac {1}{2}}\right)^{\!2}+\left({\dfrac {\sqrt {7}}{2}}\right)^{\!2}}}-{\dfrac {3}{\sqrt {7}}}\;{\dfrac {\dfrac {\sqrt {7}}{2}}{\left(p+{\dfrac {1}{2}}\right)^{\!2}+\left({\dfrac {\sqrt {7}}{2}}\right)^{\!2}}}\;}
}
{\displaystyle {\Bigg \}}\;}
f
(
t
)
=
L
−
1
{
F
(
p
)
}
{\displaystyle \;f(t)={\mathcal {L}}^{-1}\!\left\lbrace F(p)\right\rbrace \;}
⇒
{\displaystyle \Rightarrow }
Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
f
(
t
)
=
1
2
[
L
−
1
{
1
p
}
−
L
−
1
{
p
+
1
2
(
p
+
1
2
)
2
+
(
7
2
)
2
}
+
3
7
L
−
1
{
7
2
(
p
+
1
2
)
2
+
(
7
2
)
2
}
]
{\displaystyle \;f(t)={\dfrac {1}{2}}\left[{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {1}{p}}\right\rbrace -{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {p+{\dfrac {1}{2}}}{\left(p+{\dfrac {1}{2}}\right)^{\!2}+\left({\dfrac {\sqrt {7}}{2}}\right)^{\!2}}}\right\rbrace +{\dfrac {3}{\sqrt {7}}}\;{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {\dfrac {\sqrt {7}}{2}}{\left(p+{\dfrac {1}{2}}\right)^{\!2}+\left({\dfrac {\sqrt {7}}{2}}\right)^{\!2}}}\right\rbrace \right]\;}
[ 107] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
f
(
t
)
=
L
−
1
{
F
(
p
)
}
=
1
2
[
1
−
exp
(
−
1
2
t
)
cos
(
7
2
t
)
+
3
7
exp
(
−
1
2
t
)
sin
(
7
2
t
)
]
Y
(
t
)
{\displaystyle \;f(t)={\mathcal {L}}^{-1}\!\left\lbrace F(p)\right\rbrace ={\dfrac {1}{2}}\left[1-\exp \!\left(-{\dfrac {1}{2}}\;t\right)\,\cos \!\left({\dfrac {\sqrt {7}}{2}}\;t\right)+{\dfrac {3}{\sqrt {7}}}\;\exp \!\left(-{\dfrac {1}{2}}\;t\right)\,\sin \!\left({\dfrac {\sqrt {7}}{2}}\;t\right)\right]\,Y(t)\;}
[ 106] , [ 118] Soit à « déterminer l'originale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
f
(
t
)
=
L
−
1
{
F
(
p
)
}
=
{\displaystyle \;\color {transparent}{f(t)={\mathcal {L}}^{-1}\!\left\lbrace F(p)\right\rbrace =}}
sous la condition
ℜ
(
p
)
>
max
(
0
;
−
1
2
)
=
0
{\displaystyle \;\Re (p)>\max \left(0\,;\,-{\dfrac {1}{2}}\right)=0\;}
Remarque : il est possible de mettre l'expression «
g
(
t
)
=
cos
(
7
2
t
)
−
3
7
sin
(
7
2
t
)
{\displaystyle \;g(t)=\cos \!\left({\dfrac {\sqrt {7}}{2}}\;t\right)-{\dfrac {3}{\sqrt {7}}}\;\sin \!\left({\dfrac {\sqrt {7}}{2}}\;t\right)\;}
g
(
t
)
=
C
cos
(
7
2
t
+
φ
)
{\displaystyle \;g(t)=C\;\cos \!\left({\dfrac {\sqrt {7}}{2}}\;t+\varphi \right)\;}
[ 119] Remarque : «
C
=
(
1
)
2
+
(
−
3
)
2
7
=
4
7
{\displaystyle \;C={\sqrt {(1)^{2}+{\dfrac {(-3)^{2}}{7}}}}={\dfrac {4}{\sqrt {7}}}\;}
⇒
{\displaystyle \Rightarrow }
g
(
t
)
=
4
7
[
7
4
cos
(
7
2
t
)
−
3
4
sin
(
7
2
t
)
]
{\displaystyle \;g(t)={\dfrac {4}{\sqrt {7}}}\left[{\dfrac {\sqrt {7}}{4}}\;\cos \!\left({\dfrac {\sqrt {7}}{2}}\;t\right)-{\dfrac {3}{4}}\;\sin \!\left({\dfrac {\sqrt {7}}{2}}\;t\right)\right]\;}
Remarque : «
{
cos
(
φ
)
=
7
4
sin
(
φ
)
=
3
4
}
{\displaystyle \left\lbrace {\begin{array}{c}\cos(\varphi )={\dfrac {\sqrt {7}}{4}}\\\sin(\varphi )={\dfrac {3}{4}}\end{array}}\right\rbrace \;}
φ
=
arctan
(
3
7
)
{\displaystyle \;\varphi =\arctan \!\left({\dfrac {3}{\sqrt {7}}}\right)\;}
[ 120]
⇒
{\displaystyle \Rightarrow }
g
(
t
)
=
4
7
cos
[
7
2
t
+
arctan
(
3
7
)
]
{\displaystyle \;g(t)={\dfrac {4}{\sqrt {7}}}\;\cos \!\left[{\dfrac {\sqrt {7}}{2}}\;t+\arctan \!\left({\dfrac {3}{\sqrt {7}}}\right)\right]\;}
Remarque : permettant de réécrire l'« originale de
F
(
p
)
=
p
+
1
p
(
p
2
+
p
+
2
)
{\displaystyle \;F(p)={\dfrac {p+1}{p\;(p^{2}+p+2)}}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Remarque : permettant de réécrire l'« originale «
f
(
t
)
=
L
−
1
{
p
+
1
p
(
p
2
+
p
+
2
)
}
=
1
2
{
1
−
4
7
exp
(
−
1
2
t
)
cos
[
7
2
t
+
arctan
(
3
7
)
]
}
Y
(
t
)
{\displaystyle \;f(t)={\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {p+1}{p\,(p^{2}+p+2)}}\right\rbrace ={\dfrac {1}{2}}\left\lbrace 1-{\dfrac {4}{\sqrt {7}}}\;\exp \!\left(-{\dfrac {1}{2}}\;t\right)\,\cos \!\left[{\dfrac {\sqrt {7}}{2}}\;t+\arctan \!\left({\dfrac {3}{\sqrt {7}}}\right)\right]\right\rbrace \,Y(t)\;}
[ 121]
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
Remarque : permettant de réécrire l'« originale «
f
(
t
)
≃
1
2
[
1
−
1
,
512
exp
(
−
0
,
5
t
)
cos
(
1
,
323
t
+
0
,
848
)
]
Y
(
t
)
{\displaystyle \;f(t)\simeq {\dfrac {1}{2}}\left[1-1,512\;\exp \!\left(-0,5\;t\right)\,\cos \!\left(1,323\;t+0,848\right)\right]\,Y(t)\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0}
Début d’un théorème
Théorème (admis) de la valeur initiale dans la transformation (monolatérale) de Laplace
Soit une « fonction réelle causale[ 4] , [ 9]
f
(
t
)
{\displaystyle \;f(t)\;}
t
{\displaystyle \;t\;}
[ 122] Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
convergeant en
t
=
0
{\displaystyle \;t=0\;}
[ 123] Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
telle qu'une transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
L
{
f
(
t
)
}
{\displaystyle \;F(p)={\mathcal {L}}\!\left\lbrace f(t)\right\rbrace \;}
Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
d'abscisse de convergence
α
∈
R
∪
−
∞
{\displaystyle \;\alpha \in \mathbb {R} \,\cup \,-\infty \;}
[ 124] [ 125] [ 126]
lim
t
→
0
+
[
f
(
t
)
]
=
lim
p
→
+
∞
[
p
F
(
p
)
]
p
∈
R
{\displaystyle \;\lim \limits _{t\,\rightarrow \,0^{+}}\!\left[f(t)\right]=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }\;}
Fin du théorème
À défaut de démonstration[ 126] valeur initiale » sur des exemples :
« fonction de Heaviside[ 20]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}}
f
(
t
)
=
{
1
si
t
>
0
0
si
t
<
0
}
{\displaystyle \;f(t)=\left\lbrace {\begin{array}{l}1\;{\text{si}}\;t>0\\0\;{\text{si}}\;t<0\end{array}}\right\rbrace \;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
1
p
{\displaystyle \;F(p)={\dfrac {1}{p}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 67] « fonction de Heaviside
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}}
f
(
t
)
=
{
1
si
t
>
0
0
si
t
<
0
}
{\displaystyle \;\color {transparent}{f(t)=\left\lbrace {\begin{array}{l}1\;{\text{si}}\;t>0\\0\;{\text{si}}\;t<0\end{array}}\right\rbrace }\;}
on vérifie que «
lim
p
→
+
∞
[
p
F
(
p
)
]
p
∈
R
=
1
=
f
(
0
+
)
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=1=f(0^{+})\;}
« fonction rampe
f
(
t
)
=
{
a
t
si
t
⩾
0
0
si
t
⩽
0
}
{\displaystyle \;f(t)=\left\lbrace {\begin{array}{l}a\;t\;{\text{si}}\;t\geqslant 0\\0\quad {\text{si}}\;t\leqslant 0\end{array}}\right\rbrace \;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
a
p
2
{\displaystyle \;F(p)={\dfrac {a}{p^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 67] « fonction rampe
f
(
t
)
=
{
a
t
si
t
⩾
0
0
si
t
⩽
0
}
{\displaystyle \;\color {transparent}{f(t)=\left\lbrace {\begin{array}{l}a\;t\;{\text{si}}\;t\geqslant 0\\0\quad {\text{si}}\;t\leqslant 0\end{array}}\right\rbrace }\;}
on vérifie que
lim
p
→
+
∞
[
p
F
(
p
)
]
p
∈
R
=
0
=
f
(
0
+
)
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=0=f(0^{+})\;}
« fonction exponentielle réelle
f
(
t
)
=
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;f(t)=\exp(-a\,t)\;Y(t)\;}
a
∈
R
∗
{\displaystyle \;a\in \mathbb {R} ^{*}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
1
p
+
a
{\displaystyle \;F(p)={\dfrac {1}{p+a}}\;}
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 106] « fonction exponentielle réelle
f
(
t
)
=
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;Y(t)}\;}
a
∈
R
∗
{\displaystyle \;\color {transparent}{a\in \mathbb {R} ^{*}}\;}
on vérifie que «
lim
p
→
+
∞
[
p
F
(
p
)
]
p
∈
R
=
1
=
f
(
0
+
)
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=1=f(0^{+})\;}
« fonction cosinusoïdale
f
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;f(t)=\cos(\omega \,t)\;Y(t)\;}
ω
∈
R
+
∗
{\displaystyle \;\omega \in \mathbb {R} _{+}^{*}\;}
[ 82]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
p
p
2
+
ω
2
{\displaystyle \;F(p)={\dfrac {p}{p^{2}+\omega ^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 127] « fonction cosinusoïdale
f
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\cos(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
on vérifie que «
lim
p
→
+
∞
[
p
F
(
p
)
]
p
∈
R
=
1
=
f
(
0
+
)
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=1=f(0^{+})\;}
« fonction sinusoïdale
f
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;f(t)=\sin(\omega \,t)\;Y(t)\;}
ω
∈
R
+
∗
{\displaystyle \;\omega \in \mathbb {R} _{+}^{*}\;}
[ 82]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
ω
p
2
+
ω
2
{\displaystyle \;F(p)={\dfrac {\omega }{p^{2}+\omega ^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 127] « fonction sinusoïdale
f
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\sin(\omega \,t)\;Y(t)}\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
on vérifie que «
lim
p
→
+
∞
[
p
F
(
p
)
]
p
∈
R
=
0
=
f
(
0
+
)
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=0=f(0^{+})\;}
« fonction cosinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;f(t)=\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)\;}
{
ω
∈
R
+
∗
a
∈
R
+
∗
}
{\displaystyle \;\left\lbrace {\begin{array}{c}\omega \in \mathbb {R} _{+}^{*}\\a\in \mathbb {R} _{+}^{*}\end{array}}\right\rbrace \;}
[ 82]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
p
+
a
(
p
+
a
)
2
+
ω
2
{\displaystyle \;F(p)={\dfrac {p+a}{(p+a)^{2}+\omega ^{2}}}\;}
« fonction cosinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)}\;}
{
ω
∈
R
+
∗
}
{\displaystyle \;\color {transparent}{\left\lbrace \omega \in \mathbb {R} _{+}^{*}\right\rbrace }\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
pour
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 128] « fonction cosinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)}\;}
{
ω
∈
R
+
∗
}
{\displaystyle \;\color {transparent}{\left\lbrace \omega \in \mathbb {R} _{+}^{*}\right\rbrace }\;}
on vérifie que «
lim
p
→
+
∞
[
p
F
(
p
)
]
p
∈
R
=
1
=
f
(
0
+
)
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=1=f(0^{+})\;}
« fonction sinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;f(t)=\exp(-a\,t)\;\sin(\omega \,t)\;Y(t)\;}
{
ω
∈
R
+
∗
a
∈
R
+
∗
}
{\displaystyle \;\left\lbrace {\begin{array}{c}\omega \in \mathbb {R} _{+}^{*}\\a\in \mathbb {R} _{+}^{*}\end{array}}\right\rbrace \;}
[ 82]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
ω
(
p
+
a
)
2
+
ω
2
{\displaystyle \;F(p)={\dfrac {\omega }{(p+a)^{2}+\omega ^{2}}}\;}
« fonction sinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;\sin(\omega \,t)\;Y(t)}\;}
{
ω
∈
R
+
∗
}
{\displaystyle \;\color {transparent}{\left\lbrace \omega \in \mathbb {R} _{+}^{*}\right\rbrace }\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
pour
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 128] « fonction sinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;\sin(\omega \,t)\;Y(t)}\;}
{
ω
∈
R
+
∗
}
{\displaystyle \;\color {transparent}{\left\lbrace \omega \in \mathbb {R} _{+}^{*}\right\rbrace }\;}
on vérifie que «
lim
p
→
+
∞
[
p
F
(
p
)
]
p
∈
R
=
0
=
f
(
0
+
)
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=0=f(0^{+})\;}
…
{\displaystyle \;\ldots }
Valeur initiale de la dérivée temporelle dans la transformation (monolatérale) de Laplace
Soit une « fonction réelle causale[ 4] , [ 9]
f
(
t
)
{\displaystyle \;f(t)\;}
[ 129]
f
˙
(
t
)
{\displaystyle \;{\dot {f}}(t)\;}
t
{\displaystyle \;t\;}
[ 122] Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
convergeant toutes deux en
t
=
0
{\displaystyle \;t=0\;}
[ 123] Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
telle qu'une transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
Laplace[ 1] [ 125] Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
∙
{\displaystyle \bullet \;}
L
{
f
(
t
)
}
=
F
(
p
)
{\displaystyle \;{\mathcal {L}}\!\left\lbrace f(t)\right\rbrace =F(p)\;}
Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
α
∈
R
∪
−
∞
{\displaystyle \;\alpha \in \mathbb {R} \,\cup \,-\infty \;}
[ 124] Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
∙
{\displaystyle \bullet \;}
L
{
f
˙
(
t
)
}
=
p
F
(
p
)
−
f
(
0
+
)
{\displaystyle \;{\mathcal {L}}\!\left\lbrace {\dot {f}}(t)\right\rbrace =p\;F(p)-f(0^{+})}
[ 130] Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
f
˙
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)}\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
de convergence
α
∈
R
∪
−
∞
{\displaystyle \;\alpha \in \mathbb {R} \,\cup \,-\infty \;}
[ 124] [ 131]
lim
t
→
0
+
[
f
˙
(
t
)
]
=
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
{\displaystyle \;\lim \limits _{t\,\rightarrow \,0^{+}}\!\left[{\dot {f}}(t)\right]=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }}
Reprenant les exemples des fonctions usuelles ayant servi à vérifier le théorème de la valeur initiale Reprenant les exemples des fonctions usuelles à l'exception de la fonction de Heaviside[ 20]
(
{\displaystyle \;{\big (}}
t
=
0
)
{\displaystyle \;t=0{\big )}\;}
Reprenant les exemples des fonctions usuelles pour justifier la validité de la détermination de la valeur initiale de la dérivée temporelle utilisant la transformation
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
« fonction rampe
f
(
t
)
=
{
a
t
si
t
⩾
0
0
si
t
⩽
0
}
{\displaystyle \;f(t)=\left\lbrace {\begin{array}{l}a\;t\;{\text{si}}\;t\geqslant 0\\0\quad {\text{si}}\;t\leqslant 0\end{array}}\right\rbrace \;}
[
{\displaystyle \;{\big [}}
f
(
0
+
)
=
0
{\displaystyle \;f(0^{+})=0\;}
]
{\displaystyle {\big ]}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
a
p
2
{\displaystyle \;F(p)={\dfrac {a}{p^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 67] [ 132]
f
˙
(
t
)
=
{
a
si
t
>
0
0
si
t
<
0
}
{\displaystyle \;{\dot {f}}(t)=\left\lbrace {\begin{array}{l}a\quad {\text{si}}\;t>0\\0\quad {\text{si}}\;t<0\end{array}}\right\rbrace \;}
(
{\displaystyle {\big (}}
ère espèce en
t
=
0
{\displaystyle \;t=0\;}
[ 38]
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
G
(
p
)
=
p
F
(
p
)
{\displaystyle \;G(p)=p\;F(p)\;}
[ 130]
=
a
p
{\displaystyle ={\dfrac {a}{p}}\;}
« dérivée temporelle
f
˙
(
t
)
=
{
a
si
t
>
0
}
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace a\quad {\text{si}}\;t>0\right\rbrace }\;}
(
{\displaystyle \color {transparent}{\big (}}
ère espèce en
t
=
0
{\displaystyle \;\color {transparent}{t=0}\;}
)
{\displaystyle \color {transparent}{\big )}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
G
(
p
)
=
p
F
(
p
)
{\displaystyle \;\color {transparent}{G(p)=p\;F(p)}\;}
pour
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 133] « dérivée temporelle
f
˙
(
t
)
=
{
a
si
t
>
0
0
si
t
<
0
}
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace {\begin{array}{l}a\quad {\text{si}}\;t>0\\0\quad {\text{si}}\;t<0\end{array}}\right\rbrace }\;}
(
{\displaystyle \color {transparent}{\big (}}
ère espèce en
t
=
0
{\displaystyle \;\color {transparent}{t=0}\;}
)
{\displaystyle \color {transparent}{\big )}\;}
on vérifie que «
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
=
a
=
f
˙
(
0
+
)
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)\;{\cancel {-p\;f(0^{+})}}\right]_{p\,\in \,\mathbb {R} }=a={\dot {f}}(0^{+})\;}
« fonction exponentielle réelle
f
(
t
)
=
exp
(
−
a
t
)
Y
(
t
)
=
{
exp
(
−
a
t
)
si
t
>
0
0
si
t
<
0
}
{\displaystyle \;f(t)=\exp(-a\,t)\;Y(t)=\left\lbrace {\begin{array}{c l}\exp(-a\,t)\!\!&{\text{si}}\;t>0\\0\!\!&{\text{si}}\;t<0\end{array}}\right\rbrace \;}
a
∈
R
∗
{\displaystyle \;a\in \mathbb {R} ^{*}\;}
[
{\displaystyle \;{\big [}}
f
(
0
+
)
=
1
{\displaystyle \;f(0^{+})=1\;}
]
{\displaystyle {\big ]}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
1
p
+
a
{\displaystyle \;F(p)={\dfrac {1}{p+a}}\;}
« fonction exponentielle réelle
f
(
t
)
=
exp
(
−
a
t
)
Y
(
t
)
=
{
exp
(
−
a
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;Y(t)=\left\lbrace \exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }\;}
a
∈
R
∗
{\displaystyle \;\color {transparent}{a\in \mathbb {R} ^{*}}\;}
[
{\displaystyle \;\color {transparent}{\big [}}
f
(
0
+
)
=
1
{\displaystyle \;\color {transparent}{f(0^{+})=1}\;}
]
{\displaystyle \color {transparent}{\big ]}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 106] [ 132]
f
˙
(
t
)
=
{
−
a
exp
(
−
a
t
)
si
t
>
0
0
si
t
<
0
}
=
−
a
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;{\dot {f}}(t)=\left\lbrace {\begin{array}{c l}-a\;\exp(-a\,t)\!\!&{\text{si}}\;t>0\\0\!\!&{\text{si}}\;t<0\end{array}}\right\rbrace =-a\;\exp(-a\,t)\;Y(t)\;}
(
{\displaystyle {\big (}}
ère espèce en
t
=
0
{\displaystyle \;t=0\;}
[ 38]
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
G
(
p
)
=
{\displaystyle \;G(p)=}
« dérivée temporelle
f
˙
(
t
)
=
{
−
a
exp
(
−
a
t
)
si
t
>
0
}
=
−
a
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace -a\;\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace =-a\;\exp(-a\,t)\;Y(t)}\;}
(
{\displaystyle \color {transparent}{\big (}}
ère espèce en
t
=
0
{\displaystyle \;\color {transparent}{t=0}\;}
)
{\displaystyle \color {transparent}{\big )}\;}
p
F
(
p
)
−
1
{\displaystyle p\;F(p)-1\;}
[ 130]
=
−
a
p
+
a
{\displaystyle ={\dfrac {-a}{p+a}}\;}
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 134] « dérivée temporelle
f
˙
(
t
)
=
{
−
a
exp
(
−
a
t
)
si
t
>
0
}
=
−
a
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace -a\;\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace =-a\;\exp(-a\,t)\;Y(t)}\;}
on vérifie que «
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
=
lim
p
→
+
∞
[
p
2
p
+
a
−
p
]
p
∈
R
=
lim
p
→
+
∞
[
−
p
a
)
p
+
a
]
p
∈
R
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {p^{2}}{p+a}}-p\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {-p\;a)}{p+a}}\right]_{p\,\in \,\mathbb {R} }}
« dérivée temporelle
f
˙
(
t
)
=
{
−
a
exp
(
−
a
t
)
si
t
>
0
}
=
−
a
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace -a\;\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace =-a\;\exp(-a\,t)\;Y(t)}\;}
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
{\displaystyle \;\color {transparent}{\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }}}
=
−
a
=
f
˙
(
0
+
)
{\displaystyle =-a={\dot {f}}(0^{+})\;}
« fonction cosinusoïdale
f
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
=
{
cos
(
ω
t
)
si
t
>
0
0
si
t
<
0
}
{\displaystyle \;f(t)=\cos(\omega \,t)\;Y(t)=\left\lbrace {\begin{array}{c l}\cos(\omega \,t)\!\!&{\text{si}}\;t>0\\0\!\!&{\text{si}}\;t<0\end{array}}\right\rbrace \;}
ω
∈
R
+
∗
{\displaystyle \;\omega \in \mathbb {R} _{+}^{*}\;}
[ 82]
[
{\displaystyle \;{\big [}}
f
(
0
+
)
=
1
{\displaystyle \;f(0^{+})=1\;}
]
{\displaystyle {\big ]}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
p
p
2
+
ω
2
{\displaystyle \;F(p)={\dfrac {p}{p^{2}+\omega ^{2}}}\;}
« fonction cosinusoïdale
f
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
=
{
cos
(
ω
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{f(t)=\cos(\omega \,t)\;Y(t)=\left\lbrace \cos(\omega \,t)\!\!{\text{si}}\;t>0\right\rbrace }\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
[
{\displaystyle \;\color {transparent}{\big [}}
f
(
0
+
)
=
1
{\displaystyle \;\color {transparent}{f(0^{+})=1}\;}
]
{\displaystyle \color {transparent}{\big ]}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
pour
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 127] [ 132]
f
˙
(
t
)
=
{
−
ω
sin
(
ω
t
)
si
t
>
0
0
si
t
<
0
}
=
−
ω
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;{\dot {f}}(t)=\left\lbrace {\begin{array}{c l}-\omega \;\sin(\omega \,t)\!\!&{\text{si}}\;t>0\\0\!\!&{\text{si}}\;t<0\end{array}}\right\rbrace =-\omega \;\sin(\omega \,t)\;Y(t)\;}
[ 135]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
G
(
p
)
=
p
F
(
p
)
−
1
{\displaystyle \;G(p)=p\;F(p)-1\;}
[ 130]
=
−
ω
2
p
2
+
ω
2
{\displaystyle ={\dfrac {-\omega ^{2}}{p^{2}+\omega ^{2}}}\;}
« dérivée temporelle
f
˙
(
t
)
=
{
−
ω
sin
(
ω
t
)
si
t
>
0
}
=
−
ω
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace -\omega \;\sin(\omega \,t)\!\!{\text{si}}\;t>0\right\rbrace =-\omega \;\sin(\omega \,t)\;Y(t)}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
G
(
p
)
=
p
F
(
p
)
−
1
{\displaystyle \;\color {transparent}{G(p)=p\;F(p)-1}\;}
pour
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 136] « dérivée temporelle
f
˙
(
t
)
=
{
−
ω
sin
(
ω
t
)
si
t
>
0
}
=
−
ω
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace -\omega \;\sin(\omega \,t)\!\!{\text{si}}\;t>0\right\rbrace =-\omega \;\sin(\omega \,t)\;Y(t)}\;}
on vérifie que «
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
=
lim
p
→
+
∞
[
p
3
p
2
+
ω
2
−
p
]
p
∈
R
=
lim
p
→
+
∞
[
−
p
ω
2
p
2
+
ω
2
]
p
∈
R
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {p^{3}}{p^{2}+\omega ^{2}}}-p\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {-p\;\omega ^{2}}{p^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }}
« dérivée temporelle
f
˙
(
t
)
=
{
−
ω
sin
(
ω
t
)
si
t
>
0
}
=
−
ω
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace -\omega \;\sin(\omega \,t)\!\!{\text{si}}\;t>0\right\rbrace =-\omega \;\sin(\omega \,t)\;Y(t)}\;}
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
{\displaystyle \;\color {transparent}{\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }}}
=
0
=
f
˙
(
0
+
)
{\displaystyle =0={\dot {f}}(0^{+})\;}
« fonction sinusoïdale
f
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
=
{
sin
(
ω
t
)
si
t
>
0
0
si
t
<
0
}
{\displaystyle \;f(t)=\sin(\omega \,t)\;Y(t)=\left\lbrace {\begin{array}{c l}\sin(\omega \,t)\!\!&{\text{si}}\;t>0\\0\!\!&{\text{si}}\;t<0\end{array}}\right\rbrace \;}
ω
∈
R
+
∗
{\displaystyle \;\omega \in \mathbb {R} _{+}^{*}\;}
[ 82]
[
{\displaystyle \;{\big [}}
f
(
0
+
)
=
0
{\displaystyle \;f(0^{+})=0\;}
]
{\displaystyle {\big ]}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
ω
p
2
+
ω
2
{\displaystyle \;F(p)={\dfrac {\omega }{p^{2}+\omega ^{2}}}\;}
« fonction sinusoïdale
f
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
=
{
sin
(
ω
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{f(t)=\sin(\omega \,t)\;Y(t)=\left\lbrace \sin(\omega \,t)\!\!{\text{si}}\;t>0\right\rbrace }\;}
ω
∈
R
+
∗
{\displaystyle \;\color {transparent}{\omega \in \mathbb {R} _{+}^{*}}\;}
[
{\displaystyle \;\color {transparent}{\big [}}
f
(
0
+
)
=
0
{\displaystyle \;\color {transparent}{f(0^{+})=0}\;}
]
{\displaystyle \color {transparent}{\big ]}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
pour
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 127] [ 132]
f
˙
(
t
)
=
{
ω
cos
(
ω
t
)
si
t
>
0
0
si
t
<
0
}
=
ω
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;{\dot {f}}(t)=\left\lbrace {\begin{array}{c l}\omega \;\cos(\omega \,t)\!\!&{\text{si}}\;t>0\\0\!\!&{\text{si}}\;t<0\end{array}}\right\rbrace =\omega \;\cos(\omega \,t)\;Y(t)\;}
{
{\displaystyle {\big \{}}
ère espèce[ 38]
}
{\displaystyle {\big \}}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
G
(
p
)
=
p
F
(
p
)
{\displaystyle \;G(p)=p\;F(p)\;}
[ 130] « dérivée temporelle
f
˙
(
t
)
=
{
ω
cos
(
ω
t
)
si
t
>
0
}
=
ω
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \omega \;\cos(\omega \,t)\!\!{\text{si}}\;t>0\right\rbrace =\omega \;\cos(\omega \,t)\;Y(t)}\;}
{
{\displaystyle \color {transparent}{\big \{}}
ère espèce
}
{\displaystyle \color {transparent}{\big \}}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
G
(
p
)
{\displaystyle \;\color {transparent}{G(p)}\;}
=
p
ω
p
2
+
ω
2
{\displaystyle ={\dfrac {p\;\omega }{p^{2}+\omega ^{2}}}\;}
« dérivée temporelle
f
˙
(
t
)
=
{
ω
cos
(
ω
t
)
si
t
>
0
}
=
ω
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \omega \;\cos(\omega \,t)\!\!{\text{si}}\;t>0\right\rbrace =\omega \;\cos(\omega \,t)\;Y(t)}\;}
{
{\displaystyle \color {transparent}{\big \{}}
ère espèce
}
{\displaystyle \color {transparent}{\big \}}\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
G
(
p
)
{\displaystyle \;\color {transparent}{G(p)}\;}
pour
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 137] « dérivée temporelle
f
˙
(
t
)
=
{
ω
cos
(
ω
t
)
si
t
>
0
}
=
ω
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \omega \;\cos(\omega \,t)\!\!{\text{si}}\;t>0\right\rbrace =\omega \;\cos(\omega \,t)\;Y(t)}\;}
on vérifie que «
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
=
lim
p
→
+
∞
[
p
2
ω
p
2
+
ω
2
]
p
∈
R
=
ω
=
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)\;{\cancel {-p\;f(0^{+})}}\;\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {p^{2}\;\omega }{p^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }=\omega =}
f
˙
(
0
+
)
{\displaystyle {\dot {f}}(0^{+})\;}
« fonction cosinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;f(t)=\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)\;}
{
ω
∈
R
+
∗
a
∈
R
+
∗
}
{\displaystyle \;\left\lbrace \!{\begin{array}{c}\omega \in \mathbb {R} _{+}^{*}\\a\in \mathbb {R} _{+}^{*}\end{array}}\!\right\rbrace \;}
[ 82]
[
{\displaystyle \;{\big [}}
f
(
0
+
)
=
1
{\displaystyle \;f(0^{+})=1\;}
]
{\displaystyle {\big ]}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] « fonction cosinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)}\;}
{
ω
∈
R
+
∗
}
{\displaystyle \;\color {transparent}{\left\lbrace \!\omega \in \mathbb {R} _{+}^{*}\!\right\rbrace }\;}
[
{\displaystyle \;\color {transparent}{\big [}}
f
(
0
+
)
=
1
{\displaystyle \;\color {transparent}{f(0^{+})=1}\;}
]
{\displaystyle \color {transparent}{\big ]}\;}
F
(
p
)
=
p
+
a
(
p
+
a
)
2
+
ω
2
{\displaystyle \;F(p)={\dfrac {p+a}{(p+a)^{2}+\omega ^{2}}}\;}
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 128] [ 132]
f
˙
(
t
)
=
{
[
−
a
cos
(
ω
t
)
−
ω
sin
(
ω
t
)
]
exp
(
−
a
t
)
si
t
>
0
0
si
t
<
0
}
=
[
−
a
cos
(
ω
t
)
−
ω
sin
(
ω
t
)
]
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;{\dot {f}}(t)=\left\lbrace {\begin{array}{c l}\left[-a\;\cos(\omega \,t)-\omega \;\sin(\omega \,t)\right]\,\exp(-a\,t)\!\!&{\text{si}}\;t>0\\0\!\!&{\text{si}}\;t<0\end{array}}\right\rbrace =\left[-a\;\cos(\omega \,t)-\omega \;\sin(\omega \,t)\right]\,\exp(-a\,t)\;Y(t)\;}
{
{\displaystyle {\big \{}}
ère espèce[ 38]
}
{\displaystyle {\big \}}\;}
« dérivée temporelle
f
˙
(
t
)
=
{
[
−
a
cos
(
ω
t
)
−
ω
sin
(
ω
t
)
]
exp
(
−
a
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \left[-a\;\cos(\omega \,t)-\omega \;\sin(\omega \,t)\right]\,\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }}
de « transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
G
(
p
)
=
p
F
(
p
)
−
f
(
0
+
)
{\displaystyle \;G(p)=p\;F(p)-f(0^{+})\;}
[ 130]
=
−
a
(
p
+
a
)
+
ω
2
(
p
+
a
)
2
+
ω
2
{\displaystyle =-{\dfrac {a\;(p+a)+\omega ^{2}}{(p+a)^{2}+\omega ^{2}}}\;}
« dérivée temporelle
f
˙
(
t
)
=
{
[
−
a
cos
(
ω
t
)
−
ω
sin
(
ω
t
)
]
exp
(
−
a
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \left[-a\;\cos(\omega \,t)-\omega \;\sin(\omega \,t)\right]\,\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
G
(
p
)
=
{\displaystyle \;\color {transparent}{G(p)=}}
pour
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 138] « dérivée temporelle
f
˙
(
t
)
=
{
[
−
a
cos
(
ω
t
)
−
ω
sin
(
ω
t
)
]
exp
(
−
a
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \left[-a\;\cos(\omega \,t)-\omega \;\sin(\omega \,t)\right]\,\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }}
on vérifie que «
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
=
lim
p
→
+
∞
[
p
2
(
p
+
a
)
(
p
+
a
)
2
+
ω
2
−
p
]
p
∈
R
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {p^{2}\,(p+a)}{(p+a)^{2}+\omega ^{2}}}-p\right]_{p\,\in \,\mathbb {R} }}
« dérivée temporelle
f
˙
(
t
)
=
{
[
−
a
cos
(
ω
t
)
−
ω
sin
(
ω
t
)
]
exp
(
−
a
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \left[-a\;\cos(\omega \,t)-\omega \;\sin(\omega \,t)\right]\,\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }}
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
{\displaystyle \;\color {transparent}{\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }}}
=
lim
p
→
+
∞
[
p
2
(
p
+
a
)
−
p
(
p
+
a
)
2
−
p
ω
2
(
p
+
a
)
2
+
ω
2
]
p
∈
R
{\displaystyle =\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {p^{2}\,(p+a)-p\,(p+a)^{2}-p\;\omega ^{2}}{(p+a)^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }}
« dérivée temporelle
f
˙
(
t
)
=
{
[
−
a
cos
(
ω
t
)
−
ω
sin
(
ω
t
)
]
exp
(
−
a
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \left[-a\;\cos(\omega \,t)-\omega \;\sin(\omega \,t)\right]\,\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }}
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
{\displaystyle \;\color {transparent}{\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }}}
=
lim
p
→
+
∞
[
−
p
2
a
−
p
(
a
2
+
ω
2
)
(
p
+
a
)
2
+
ω
2
]
p
∈
R
{\displaystyle =\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {-p^{2}\;a-p\,(a^{2}+\omega ^{2})}{(p+a)^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }}
« dérivée temporelle
f
˙
(
t
)
=
{
[
−
a
cos
(
ω
t
)
−
ω
sin
(
ω
t
)
]
exp
(
−
a
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \left[-a\;\cos(\omega \,t)-\omega \;\sin(\omega \,t)\right]\,\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }}
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
{\displaystyle \;\color {transparent}{\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }}}
=
−
a
=
f
˙
(
0
+
)
{\displaystyle =-a={\dot {f}}(0^{+})\;}
« fonction sinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;f(t)=\exp(-a\,t)\;\sin(\omega \,t)\;Y(t)\;}
{
ω
∈
R
+
∗
a
∈
R
+
∗
}
{\displaystyle \;\left\lbrace \!{\begin{array}{c}\omega \in \mathbb {R} _{+}^{*}\\a\in \mathbb {R} _{+}^{*}\end{array}}\!\right\rbrace \;}
[ 82]
[
{\displaystyle \;{\big [}}
f
(
0
+
)
=
0
{\displaystyle \;f(0^{+})=0\;}
]
{\displaystyle {\big ]}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] « fonction sinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;\sin(\omega \,t)\;Y(t)}\;}
{
ω
∈
R
+
∗
}
{\displaystyle \;\color {transparent}{\left\lbrace \!\omega \in \mathbb {R} _{+}^{*}\!\right\rbrace }\;}
[
{\displaystyle \;\color {transparent}{\big [}}
f
(
0
+
)
=
0
{\displaystyle \;\color {transparent}{f(0^{+})=0}\;}
]
{\displaystyle \color {transparent}{\big ]}\;}
F
(
p
)
=
ω
(
p
+
a
)
2
+
ω
2
{\displaystyle \;F(p)={\dfrac {\omega }{(p+a)^{2}+\omega ^{2}}}\;}
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 128] [ 132]
f
˙
(
t
)
=
{
[
−
a
sin
(
ω
t
)
+
ω
cos
(
ω
t
)
]
exp
(
−
a
t
)
si
t
>
0
0
si
t
<
0
}
=
[
−
a
sin
(
ω
t
)
+
ω
cos
(
ω
t
)
]
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;{\dot {f}}(t)=\left\lbrace {\begin{array}{c l}\left[-a\;\sin(\omega \,t)+\omega \;\cos(\omega \,t)\right]\,\exp(-a\,t)\!\!&{\text{si}}\;t>0\\0\!\!&{\text{si}}\;t<0\end{array}}\right\rbrace =\left[-a\;\sin(\omega \,t)+\omega \;\cos(\omega \,t)\right]\,\exp(-a\,t)\;Y(t)\;}
{
{\displaystyle {\big \{}}
ère espèce[ 38]
}
{\displaystyle {\big \}}\;}
« dérivée temporelle
f
˙
(
t
)
=
{
[
−
a
sin
(
ω
t
)
+
ω
cos
(
ω
t
)
]
exp
(
−
a
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \left[-a\;\sin(\omega \,t)+\omega \;\cos(\omega \,t)\right]\,\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }}
de « transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
G
(
p
)
=
p
F
(
p
)
{\displaystyle \;G(p)=p\;F(p)\;}
[ 130]
=
p
ω
(
p
+
a
)
2
+
ω
2
{\displaystyle ={\dfrac {p\;\omega }{(p+a)^{2}+\omega ^{2}}}\;}
« dérivée temporelle
f
˙
(
t
)
=
{
[
−
a
sin
(
ω
t
)
+
ω
cos
(
ω
t
)
]
exp
(
−
a
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \left[-a\;\sin(\omega \,t)+\omega \;\cos(\omega \,t)\right]\,\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
G
(
p
)
=
{\displaystyle \;\color {transparent}{G(p)=}}
pour
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 139] « dérivée temporelle
f
˙
(
t
)
=
{
[
−
a
sin
(
ω
t
)
+
ω
cos
(
ω
t
)
]
exp
(
−
a
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \left[-a\;\sin(\omega \,t)+\omega \;\cos(\omega \,t)\right]\,\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }}
on vérifie que «
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
=
lim
p
→
+
∞
[
p
2
ω
(
p
+
a
)
2
+
ω
2
]
p
∈
R
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {p^{2}\,\omega }{(p+a)^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }}
« dérivée temporelle
f
˙
(
t
)
=
{
[
−
a
sin
(
ω
t
)
+
ω
cos
(
ω
t
)
]
exp
(
−
a
t
)
si
t
>
0
}
{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \left[-a\;\sin(\omega \,t)+\omega \;\cos(\omega \,t)\right]\,\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }}
lim
p
→
+
∞
[
p
2
F
(
p
)
−
p
f
(
0
+
)
]
p
∈
R
{\displaystyle \;\color {transparent}{\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }}}
=
ω
=
f
˙
(
0
+
)
{\displaystyle =\omega ={\dot {f}}(0^{+})\;}
Début d’un théorème
Théorème (admis) de la valeur finale dans la transformation (monolatérale) de Laplace
Soit une « fonction réelle causale[ 4] , [ 9]
f
(
t
)
{\displaystyle \;f(t)\;}
t
{\displaystyle \;t\;}
[ 122] Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
convergeant pour
t
→
+
∞
{\displaystyle \;t\rightarrow +\infty \;}
[ 140] Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
localement intégrable au voisinage de
t
=
0
{\displaystyle \;t=0\;}
[ 141] Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
telle qu'une transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
L
{
f
(
t
)
}
{\displaystyle \;F(p)={\mathcal {L}}\!\left\lbrace f(t)\right\rbrace \;}
Soit une « fonction réelle causale
f
(
t
)
{\displaystyle \;\color {transparent}{f(t)}\;}
d'abscisse de convergence
α
∈
R
−
∪
−
∞
{\displaystyle \;\alpha \in \mathbb {R} ^{-}\,\cup \,-\infty \;}
[ 142] [ 125] [ 143]
lim
t
→
+
∞
[
f
(
t
)
]
=
lim
p
→
0
+
[
p
F
(
p
)
]
p
∈
R
{\displaystyle \;\lim \limits _{t\,\rightarrow \,+\infty }\!\left[f(t)\right]=\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }\;}
Fin du théorème
À défaut de démonstration[ 143] valeur finale » sur des exemples :
« fonction de Heaviside[ 20]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}}
f
(
t
)
=
{
1
si
t
>
0
0
si
t
<
0
}
{\displaystyle \;f(t)=\left\lbrace {\begin{array}{l}1\;{\text{si}}\;t>0\\0\;{\text{si}}\;t<0\end{array}}\right\rbrace \;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
1
p
{\displaystyle \;F(p)={\dfrac {1}{p}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 67] « fonction de Heaviside
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}}
f
(
t
)
=
{
1
si
t
>
0
0
si
t
<
0
}
{\displaystyle \;\color {transparent}{f(t)=\left\lbrace {\begin{array}{l}1\;{\text{si}}\;t>0\\0\;{\text{si}}\;t<0\end{array}}\right\rbrace }\;}
on vérifie que «
lim
p
→
0
+
[
p
F
(
p
)
]
p
∈
R
=
1
=
lim
t
→
+
∞
[
f
(
t
)
]
{\displaystyle \;\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=1=\lim \limits _{t\,\rightarrow \,+\infty }\!\left[f(t)\right]\;}
« fonction rampe
f
(
t
)
=
{
a
t
si
t
⩾
0
0
si
t
⩽
0
}
{\displaystyle \;f(t)=\left\lbrace {\begin{array}{l}a\;t\;{\text{si}}\;t\geqslant 0\\0\quad {\text{si}}\;t\leqslant 0\end{array}}\right\rbrace \;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
a
p
2
{\displaystyle \;F(p)={\dfrac {a}{p^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 67] on ne peut lui appliquer le théorème de la valeur finale , cette fonction divergeant à l'infini, néanmoins on vérifie que «
lim
p
→
0
+
[
p
F
(
p
)
]
p
∈
R
=
lim
p
→
0
+
[
p
a
p
2
]
p
∈
R
=
∞
{\displaystyle \;\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;{\dfrac {a}{p^{2}}}\right]_{p\,\in \,\mathbb {R} }=\infty \;}
« fonction exponentielle réelle
f
(
t
)
=
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;f(t)=\exp(-a\,t)\;Y(t)\;}
a
∈
R
∗
{\displaystyle \;a\in \mathbb {R} ^{*}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
1
p
+
a
{\displaystyle \;F(p)={\dfrac {1}{p+a}}\;}
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 106] , [ 144] « fonction exponentielle réelle
f
(
t
)
=
exp
(
−
a
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;Y(t)}\;}
a
∈
R
∗
{\displaystyle \;\color {transparent}{a\in \mathbb {R} ^{*}}\;}
on vérifie que «
lim
p
→
0
+
[
p
F
(
p
)
]
p
∈
R
=
lim
p
→
0
+
[
p
p
+
a
]
p
∈
R
=
0
=
lim
t
→
+
∞
[
f
(
t
)
]
{\displaystyle \;\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[{\dfrac {p}{p+a}}\right]_{p\,\in \,\mathbb {R} }=0=\lim \limits _{t\,\rightarrow \,+\infty }\!\left[f(t)\right]\;}
[ 145]
« fonction cosinusoïdale
f
(
t
)
=
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;f(t)=\cos(\omega \,t)\;Y(t)\;}
ω
∈
R
+
∗
{\displaystyle \;\omega \in \mathbb {R} _{+}^{*}\;}
[ 82]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
p
p
2
+
ω
2
{\displaystyle \;F(p)={\dfrac {p}{p^{2}+\omega ^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 127] on ne peut lui appliquer le théorème de la valeur finale , cette fonction n'ayant aucune limite à l'infini, ce qu'on vérifie car «
lim
p
→
0
+
[
p
F
(
p
)
]
p
∈
R
=
lim
p
→
0
+
[
p
p
p
2
+
ω
2
]
p
∈
R
=
0
{\displaystyle \;\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;{\dfrac {p}{p^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }=0\;}
on ne peut lui appliquer le théorème de la valeur finale, cette fonction n'ayant aucune limite à l'infini, ce qu'on vérifie car «
lim
t
→
+
∞
[
f
(
t
)
]
{\displaystyle \;\lim \limits _{t\,\rightarrow \,+\infty }\!\left[f(t)\right]\;}
« fonction sinusoïdale
f
(
t
)
=
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;f(t)=\sin(\omega \,t)\;Y(t)\;}
ω
∈
R
+
∗
{\displaystyle \;\omega \in \mathbb {R} _{+}^{*}\;}
[ 82]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
ω
p
2
+
ω
2
{\displaystyle \;F(p)={\dfrac {\omega }{p^{2}+\omega ^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 127] on ne peut lui appliquer le théorème de la valeur finale , cette fonction n'ayant aucune limite à l'infini, ce qu'on vérifie car «
lim
p
→
0
+
[
p
F
(
p
)
]
p
∈
R
=
lim
p
→
0
+
[
p
ω
p
2
+
ω
2
]
p
∈
R
=
0
{\displaystyle \;\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;{\dfrac {\omega }{p^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }=0\;}
on ne peut lui appliquer le théorème de la valeur finale, cette fonction n'ayant aucune limite à l'infini, ce qu'on vérifie car «
lim
t
→
+
∞
[
f
(
t
)
]
{\displaystyle \;\lim \limits _{t\,\rightarrow \,+\infty }\!\left[f(t)\right]\;}
« fonction cosinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;f(t)=\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)\;}
{
ω
∈
R
+
∗
a
∈
R
+
∗
}
{\displaystyle \;\left\lbrace \!{\begin{array}{c}\omega \in \mathbb {R} _{+}^{*}\\a\in \mathbb {R} _{+}^{*}\end{array}}\!\right\rbrace \;}
[ 82]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
p
+
a
(
p
+
a
)
2
+
ω
2
{\displaystyle \;F(p)={\dfrac {p+a}{(p+a)^{2}+\omega ^{2}}}\;}
« fonction cosinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)}\;}
{
ω
∈
R
+
∗
}
{\displaystyle \;\color {transparent}{\left\lbrace \!\omega \in \mathbb {R} _{+}^{*}\!\right\rbrace }\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
pour
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 128] , [ 144] « fonction cosinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)}\;}
{
ω
∈
R
+
∗
}
{\displaystyle \;\color {transparent}{\left\lbrace \!\omega \in \mathbb {R} _{+}^{*}\!\right\rbrace }\;}
on vérifie que «
lim
p
→
+
∞
[
p
F
(
p
)
]
p
∈
R
=
lim
p
→
+
∞
[
p
p
+
a
(
p
+
a
)
2
+
ω
2
]
p
∈
R
=
0
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;{\dfrac {p+a}{(p+a)^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }=0}
« fonction cosinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)}\;}
{
ω
∈
R
+
∗
}
{\displaystyle \;\color {transparent}{\left\lbrace \!\omega \in \mathbb {R} _{+}^{*}\!\right\rbrace }\;}
lim
p
→
+
∞
[
p
F
(
p
)
]
p
∈
R
{\displaystyle \;\color {transparent}{\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }}}
=
lim
t
→
+
∞
[
f
(
t
)
]
{\displaystyle =\lim \limits _{t\,\rightarrow \,+\infty }\!\left[f(t)\right]\;}
[ 146]
« fonction sinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;f(t)=\exp(-a\,t)\;\sin(\omega \,t)\;Y(t)\;}
{
ω
∈
R
+
∗
a
∈
R
+
∗
}
{\displaystyle \;\left\lbrace \!{\begin{array}{c}\omega \in \mathbb {R} _{+}^{*}\\a\in \mathbb {R} _{+}^{*}\end{array}}\!\right\rbrace \;}
[ 82]
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
F
(
p
)
=
ω
(
p
+
a
)
2
+
ω
2
{\displaystyle \;F(p)={\dfrac {\omega }{(p+a)^{2}+\omega ^{2}}}\;}
« fonction sinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;\sin(\omega \,t)\;Y(t)}\;}
{
ω
∈
R
+
∗
}
{\displaystyle \;\color {transparent}{\left\lbrace \!\omega \in \mathbb {R} _{+}^{*}\!\right\rbrace }\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
pour
ℜ
(
p
)
>
−
a
{\displaystyle \;\Re (p)>-a\;}
[ 128] , [ 144] « fonction sinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
cos
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;\cos(\omega \,t)\;Y(t)}\;}
{
ω
∈
R
+
∗
}
{\displaystyle \;\color {transparent}{\left\lbrace \!\omega \in \mathbb {R} _{+}^{*}\!\right\rbrace }\;}
on vérifie que «
lim
p
→
+
∞
[
p
F
(
p
)
]
p
∈
R
=
lim
p
→
+
∞
[
p
ω
(
p
+
a
)
2
+
ω
2
]
p
∈
R
=
0
{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;{\dfrac {\omega }{(p+a)^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }=0}
« fonction sinusoïdale amortie exponentiellement
f
(
t
)
=
exp
(
−
a
t
)
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{f(t)=\exp(-a\,t)\;\sin(\omega \,t)\;Y(t)}\;}
{
ω
∈
R
+
∗
}
{\displaystyle \;\color {transparent}{\left\lbrace \!\omega \in \mathbb {R} _{+}^{*}\!\right\rbrace }\;}
lim
p
→
+
∞
[
p
F
(
p
)
]
p
∈
R
{\displaystyle \;\color {transparent}{\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }}}
=
lim
t
→
+
∞
[
f
(
t
)
]
{\displaystyle =\lim \limits _{t\,\rightarrow \,+\infty }\!\left[f(t)\right]\;}
[ 147]
…
{\displaystyle \;\ldots }
Nous cherchons des « solutions réelles causales »[ 4] , [ 9] Nous cherchons des « solutions réelles causales » d'« par utilisation de la transformation
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
[
{\displaystyle \;{\big [}}
er et 2nd membre de l'équation devant être pourvus Nous cherchons des « solutions réelles causales » d'« par utilisation de la transformation
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
[
{\displaystyle \;\color {transparent}{\big [}}
d'une transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
⇒
{\displaystyle \Rightarrow }
nd Nous cherchons des « solutions réelles causales » d'« par utilisation de la transformation
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
[
{\displaystyle \;\color {transparent}{\big [}}
membre de l'équation doit être continue, discontinue de 1ère ou 2ème Nous cherchons des « solutions réelles causales » d'« par utilisation de la transformation
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
[
{\displaystyle \;\color {transparent}{\big [}}
espèce en
t
=
0
{\displaystyle \;t=0\;}
[ 38] , [ 148] Nous cherchons des « solutions réelles causales » d'« par utilisation de la transformation
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
[
{\displaystyle \;\color {transparent}{\big [}}
en aucun cas “ discontinue de 3ème espèce ”[ 149] , [ 150]
]
{\displaystyle {\big ]}}
Nous cherchons des « solutions pouvant être continues ou discontinues de 1ère espèce en
t
=
0
{\displaystyle \;t=0\;}
[ 38] er ordre[ 151] Nous cherchons des « solutions nécessairement continues en
t
=
0
{\displaystyle \;t=0\;}
ème ordre ou plus[ 152]
Remarque : Si cette méthode par transformation
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] [ 153] Remarque : Si cette méthode par transformation
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
elle ne doit pas être employée en physique par un étudiant de P.C.S.I. car Remarque : Si cette méthode par transformation
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
en physique il est exigé que ce soit par recherche de solutions libre et forcée Remarque : Si cette méthode par transformation
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
suivie d'utilisation des C.I[ 154] [ 155]
« Linéarité de la transformation
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
L
{
λ
f
(
t
)
+
μ
g
(
t
)
}
=
λ
L
{
f
(
t
)
}
+
μ
L
{
g
(
t
)
}
,
{\displaystyle \;{\mathcal {L}}\left\lbrace \lambda \;f(t)+\mu \;g(t)\right\rbrace =\lambda \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace +\mu \;{\mathcal {L}}\left\lbrace g(t)\right\rbrace ,\;}
[ 107] « Linéarité de la transformation
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
∀
(
λ
,
μ
)
∈
R
2
{\displaystyle \forall \left(\lambda \,,\,\mu \right)\in \mathbb {R} ^{2}\;}
∀
{
f
(
t
)
,
g
(
t
)
}
{\displaystyle \;\forall \left\lbrace f(t)\,,\,g(t)\right\rbrace \;}
(
{\displaystyle \;{\big (}}
distributions
)
{\displaystyle {\big )}\;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] « Linéarité de la transformation
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
∀
(
λ
,
μ
)
∈
R
2
{\displaystyle \;\color {Transparent}{\forall \left(\lambda \,,\,\mu \right)\in \mathbb {R} ^{2}}\;}
∀
{
f
(
t
)
,
g
(
t
)
}
{\displaystyle \;\color {transparent}{\forall \left\lbrace f(t)\,,\,g(t)\right\rbrace }\;}
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
d'abscisses de convergence finies ou égales à
−
∞
{\displaystyle \;-\infty \;}
« Transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] [ 20]
L
{
Y
(
t
)
}
=
1
p
{\displaystyle \;{\mathcal {L}}\left\lbrace Y(t)\right\rbrace ={\dfrac {1}{p}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 67]
« Transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] pic de Dirac [ 22]
L
{
δ
(
t
)
}
=
1
{\displaystyle \;{\mathcal {L}}\left\lbrace \delta (t)\right\rbrace =1\;}
∀
ℜ
(
p
)
∈
R
{\displaystyle \;\forall \;\Re (p)\,\in \,\mathbb {R} \;}
[ 67]
« Transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] ère d'une fonction
f
(
t
)
{\displaystyle \;f(t)\;}
[ 14]
f
(
t
)
{\displaystyle \;f(t)\;}
t
=
0
{\displaystyle \;t=0\;}
« Transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
ère «
L
{
f
˙
(
t
)
}
=
{\displaystyle \;{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace =}
p
L
{
f
(
t
)
}
−
f
(
0
+
)
{\displaystyle p\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace -f(0^{+})\;}
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
[ 130]
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
« Transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] nde d'une fonction
f
(
t
)
{\displaystyle \;f(t)\;}
[ 14]
f
˙
(
t
)
{\displaystyle \;{\dot {f}}(t)\;}
t
=
0
{\displaystyle \;t=0\;}
« Transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
nde «
L
{
f
¨
(
t
)
}
=
p
L
{
f
˙
(
t
)
}
−
f
˙
(
0
+
)
{\displaystyle \;{\mathcal {L}}\left\lbrace {\ddot {f}}(t)\right\rbrace =p\;{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace -{\dot {f}}(0^{+})\;}
ℜ
(
p
)
>
α
′
{\displaystyle \;\Re (p)>\alpha '\;}
[ 130]
α
′
{\displaystyle \;\alpha '\;}
L
{
f
˙
}
{\displaystyle \;{\mathcal {L}}\left\lbrace {\dot {f}}\right\rbrace \;}
« Transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
nde «
L
{
f
¨
(
t
)
}
=
p
2
L
{
f
(
t
)
}
−
p
f
(
0
+
)
−
f
˙
(
0
+
)
{\displaystyle \;{\mathcal {L}}\left\lbrace {\ddot {f}}(t)\right\rbrace =p^{2}\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace -p\;f(0^{+})-{\dot {f}}(0^{+})\;}
ℜ
(
p
)
>
α
{\displaystyle \;\Re (p)>\alpha \;}
α
{\displaystyle \;\alpha \;}
L
{
f
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \;}
« Transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
nde «
L
{
f
¨
(
t
)
}
=
p
2
L
{
f
(
t
)
}
−
p
f
(
0
+
)
−
f
˙
(
0
+
)
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace {\ddot {f}}(t)\right\rbrace =p^{2}\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace -p\;f(0^{+})-{\dot {f}}(0^{+})}\;}
avec
α
⩾
α
′
{\displaystyle \;\alpha \geqslant \alpha '\;}
[ 30]
er exemple : équation différentielle linéaire à cœfficients réels constants du 2ème ordre en y(t) hétérogène, sans terme du 1er ordre, à excitation sinusoïdale[ modifier | modifier le wikicode ]
Soit à résoudre, dans le domaine des fonctions à support positif[ 14]
t
=
0
{\displaystyle \;t=0}
Soit à résoudre, l'équation différentielle linéaire à cœfficients réels constants du 2ème ordre en
y
(
t
)
{\displaystyle \;y(t)\;}
er ordre, Soit à résoudre, l'équation différentielle linéaire à « excitation sinusoïdale
e
(
t
)
=
E
2
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;e(t)=E\;{\sqrt {2}}\;\sin(\omega \,t)\;Y(t)}
(
E
,
ω
)
∈
(
R
+
∗
)
2
{\displaystyle \;\left(E\,,\,\omega \right)\in \left(\mathbb {R} _{+}^{*}\right)^{2}\;}
[ 82] , [ 156] Soit à résoudre, l'équation différentielle linéaire avec les « C.I[ 154]
y
(
0
+
)
{\displaystyle \;y(0^{+})}
=
y
0
{\displaystyle =y_{0}\;}
y
˙
(
0
+
)
=
y
˙
0
{\displaystyle \;{\dot {y}}(0^{+})={\dot {y}}_{0}\;}
Soit à résoudre, l'équation différentielle linéaire à cœfficients réels constants du 2ème ordre en
y
(
t
)
{\displaystyle \;y(t)\;}
er ordre, s'écrivant sous forme normalisée
«
y
¨
(
t
)
+
c
y
(
t
)
=
e
(
t
)
=
E
2
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;{\ddot {y}}(t)+c\;y(t)=e(t)=E\;{\sqrt {2}}\;\sin(\omega \,t)\;Y(t)\;}
c
∈
R
∗
{\displaystyle \;c\in \mathbb {R} ^{*}\;}
[ 156]
« La transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
e
(
t
)
=
E
2
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;e(t)=E\;{\sqrt {2}}\;\sin(\omega \,t)\;Y(t)\;}
L
{
e
(
t
)
}
=
E
2
L
{
sin
(
ω
t
)
Y
(
t
)
}
=
E
2
ω
p
2
+
ω
2
{\displaystyle \;{\mathcal {L}}\left\lbrace e(t)\right\rbrace =E\;{\sqrt {2}}\;{\mathcal {L}}\left\lbrace \sin(\omega \,t)\;Y(t)\right\rbrace =E\;{\sqrt {2}}\;{\dfrac {\omega }{p^{2}+\omega ^{2}}}\;}
« La transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
e
(
t
)
=
E
2
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;\color {transparent}{e(t)=E\;{\sqrt {2}}\;\sin(\omega \,t)\;Y(t)}\;}
L
{
e
(
t
)
}
=
E
2
L
{
sin
(
ω
t
)
Y
(
t
)
}
=
{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace e(t)\right\rbrace =E\;{\sqrt {2}}\;{\mathcal {L}}\left\lbrace \sin(\omega \,t)\;Y(t)\right\rbrace =}}
pour
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 127] rappel des propriétés utilisées énoncé plus haut dans ce chapitre et en notant «
L
{
y
(
t
)
}
=
Y
_
(
p
)
{\displaystyle \;{\mathcal {L}}\left\lbrace y(t)\right\rbrace ={\underline {Y}}(p)\;}
[ 157]
«
[
p
2
Y
_
(
p
)
−
p
y
0
−
y
˙
0
]
+
c
Y
_
(
p
)
=
E
2
ω
p
2
+
ω
2
{\displaystyle \;\left[p^{2}\;{\underline {Y}}(p)-p\;y_{0}-{\dot {y}}_{0}\right]+c\;{\underline {Y}}(p)=E\;{\sqrt {2}}\;{\dfrac {\omega }{p^{2}+\omega ^{2}}}\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
De l'équation précédente on tire aisément «
Y
_
(
p
)
=
E
2
ω
(
p
2
+
ω
2
)
(
p
2
+
c
)
+
p
p
2
+
c
y
0
+
1
p
2
+
c
y
˙
0
{\displaystyle \;{\underline {Y}}(p)=E\;{\sqrt {2}}\;{\dfrac {\omega }{(p^{2}+\omega ^{2})\;(p^{2}+c)}}+{\dfrac {p}{p^{2}+c}}\;y_{0}+{\dfrac {1}{p^{2}+c}}\;{\dot {y}}_{0}\;}
[ 157]
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
p
2
+
c
{\displaystyle \;p^{2}+c\;}
≠
0
{\displaystyle \;\neq 0\;}
Il reste donc à déterminer l'« originale de
Y
_
(
p
)
{\displaystyle \;{\underline {Y}}(p)\;}
[ 157]
p
2
+
c
≠
0
{\displaystyle \;p^{2}+c\neq 0\;}
[
{\displaystyle {\big [}}
c
{\displaystyle \;c\;}
>
0
{\displaystyle \;>0\;}
[ 158] Il reste donc à déterminer l'« originale de
Y
_
(
p
)
{\displaystyle \;\color {transparent}{{\underline {Y}}(p)}\;}
p
2
+
c
≠
0
{\displaystyle \;\color {transparent}{p^{2}+c\neq 0}\;}
[
{\displaystyle \color {transparent}{\big [}}
mais qui nécessite une restriction de domaine de
p
{\displaystyle \;p\;}
c
{\displaystyle \;c\;}
<
0
{\displaystyle \;<0\;}
[ 159]
]
{\displaystyle {\big ]}}
Si
c
{\displaystyle \;c\;}
>
0
{\displaystyle \;>0}
fonction rationnelle
A
(
p
)
=
1
(
p
2
+
ω
2
)
(
p
2
+
c
)
{\displaystyle \;A(p)={\dfrac {1}{(p^{2}+\omega ^{2})\;(p^{2}+c)}}\;}
Si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
≻
{\displaystyle \succ \;}
c
≠
ω
2
{\displaystyle \;c\neq \omega ^{2}\;}
A
(
p
)
=
ρ
1
p
+
ρ
′
1
p
2
+
ω
2
+
ρ
2
p
+
ρ
′
2
p
2
+
c
{\displaystyle \;A(p)={\dfrac {\rho _{1}\,p+{\rho '}_{1}}{p^{2}+\omega ^{2}}}+{\dfrac {\rho _{2}\,p+{\rho '}_{2}}{p^{2}+c}}\;}
Si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
≻
{\displaystyle \color {transparent}{\succ }\;}
c
≠
ω
2
{\displaystyle \;\color {transparent}{c\neq \omega ^{2}}\;}
«
(
ρ
1
,
ρ
′
1
)
{\displaystyle \;\left(\rho _{1}\,,\,{\rho '}_{1}\right)\;}
p
2
+
ω
2
{\displaystyle \;p^{2}+\omega ^{2}\;}
p
=
±
i
ω
{\displaystyle \;p=\pm i\;\omega \;}
[ 104]
1
c
−
ω
2
=
{
ρ
1
i
ω
+
ρ
′
1
pour
p
=
i
ω
−
ρ
1
i
ω
+
ρ
′
1
pour
p
=
−
i
ω
}
{\displaystyle \;{\dfrac {1}{c-\omega ^{2}}}=\left\lbrace {\begin{array}{c}\rho _{1}\,i\,\omega +{\rho '}_{1}\;{\text{pour}}\;p=i\,\omega \\-\rho _{1}\,i\,\omega +{\rho '}_{1}\;{\text{pour}}\;p=-i\,\omega \end{array}}\right\rbrace \;}
Si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
≻
{\displaystyle \color {transparent}{\succ }\;}
c
≠
ω
2
{\displaystyle \;\color {transparent}{c\neq \omega ^{2}}\;}
(
ρ
1
,
ρ
′
1
)
{\displaystyle \;\color {transparent}{\left(\rho _{1}\,,\,{\rho '}_{1}\right)}\;}
p
2
+
ω
2
{\displaystyle \;\color {transparent}{p^{2}+\omega ^{2}}\;}
p
=
±
i
ω
{\displaystyle \;\color {transparent}{p=\pm i\;\omega }\;}
d'où «
ρ
1
=
0
{\displaystyle \;\rho _{1}=0\;}
ρ
′
1
=
1
c
−
ω
2
{\displaystyle \;{\rho '}_{1}={\dfrac {1}{c-\omega ^{2}}}\;}
Si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
≻
{\displaystyle \color {transparent}{\succ }\;}
c
≠
ω
2
{\displaystyle \;\color {transparent}{c\neq \omega ^{2}}\;}
«
(
ρ
2
,
ρ
′
2
)
{\displaystyle \;\left(\rho _{2}\,,\,{\rho '}_{2}\right)\;}
p
2
+
c
{\displaystyle \;p^{2}+c\;}
p
=
±
i
c
{\displaystyle \;p=\pm i\;{\sqrt {c}}\;}
[ 104]
1
−
c
+
ω
2
=
{
ρ
2
i
c
+
ρ
′
2
pour
p
=
i
c
−
ρ
2
i
c
+
ρ
′
2
pour
p
=
−
i
c
}
{\displaystyle \;{\dfrac {1}{-c+\omega ^{2}}}=\left\lbrace {\begin{array}{c}\rho _{2}\,i\,{\sqrt {c}}+{\rho '}_{2}\;{\text{pour}}\;p=i\,{\sqrt {c}}\\-\rho _{2}\,i\,{\sqrt {c}}+{\rho '}_{2}\;{\text{pour}}\;p=-i\,{\sqrt {c}}\end{array}}\right\rbrace \;}
Si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
≻
{\displaystyle \color {transparent}{\succ }\;}
c
≠
ω
2
{\displaystyle \;\color {transparent}{c\neq \omega ^{2}}\;}
(
ρ
2
,
ρ
′
2
)
{\displaystyle \;\color {transparent}{\left(\rho _{2}\,,\,{\rho '}_{2}\right)}\;}
p
2
+
c
{\displaystyle \;\color {transparent}{p^{2}+c}\;}
p
=
±
i
c
{\displaystyle \;\color {transparent}{p=\pm i\;{\sqrt {c}}}\;}
d'où «
ρ
2
=
0
{\displaystyle \;\rho _{2}=0\;}
ρ
′
2
=
−
1
c
−
ω
2
{\displaystyle \;{\rho '}_{2}={\dfrac {-1}{c-\omega ^{2}}}\;}
Si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
≻
{\displaystyle \color {transparent}{\succ }\;}
c
≠
ω
2
{\displaystyle \;\color {transparent}{c\neq \omega ^{2}}\;}
soit finalement «
A
(
p
)
=
1
(
p
2
+
ω
2
)
(
p
2
+
c
)
=
1
c
−
ω
2
[
1
p
2
+
ω
2
−
1
p
2
+
c
]
{\displaystyle \;A(p)={\dfrac {1}{(p^{2}+\omega ^{2})\;(p^{2}+c)}}={\dfrac {1}{c-\omega ^{2}}}\,\left[{\dfrac {1}{p^{2}+\omega ^{2}}}-{\dfrac {1}{p^{2}+c}}\right]\;}
Si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
≻
{\displaystyle \succ \;}
c
=
ω
2
{\displaystyle \;c=\omega ^{2}\;}
A
(
p
)
=
1
(
p
2
+
ω
2
)
(
p
2
+
c
)
{\displaystyle \;A(p)={\dfrac {1}{(p^{2}+\omega ^{2})\;(p^{2}+c)}}\;}
A
(
p
)
=
1
(
p
2
+
ω
2
)
2
=
1
(
p
2
+
c
)
2
{\displaystyle \;A(p)={\dfrac {1}{\left(p^{2}+\omega ^{2}\right)^{2}}}={\dfrac {1}{\left(p^{2}+c\right)^{2}}}\;}
Il reste alors à chercher l'« originale de la transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
Y
_
(
p
)
=
E
2
ω
(
p
2
+
ω
2
)
(
p
2
+
c
)
+
p
p
2
+
c
y
0
+
1
p
2
+
c
y
˙
0
{\displaystyle \;{\underline {Y}}(p)=E\;{\sqrt {2}}\;{\dfrac {\omega }{(p^{2}+\omega ^{2})\;(p^{2}+c)}}+{\dfrac {p}{p^{2}+c}}\;y_{0}+{\dfrac {1}{p^{2}+c}}\;{\dot {y}}_{0}\;}
[ 157] Il reste alors à chercher l'« originale de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
Y
_
(
p
)
{\displaystyle \;\color {transparent}{{\underline {Y}}(p)}}
pour
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
p
2
+
c
≠
0
{\displaystyle \;p^{2}+c\neq 0\;}
[ 160] Il reste alors à chercher l'« originale
≻
{\displaystyle \succ \;}
c
>
0
{\displaystyle \;c>0\;}
≠
ω
2
{\displaystyle \;\neq \omega ^{2}\;}
Y
_
(
p
)
=
E
2
c
−
ω
2
[
ω
p
2
+
ω
2
−
ω
c
c
p
2
+
c
]
+
p
p
2
+
c
y
0
+
c
p
2
+
c
y
˙
0
c
{\displaystyle \;{\underline {Y}}(p)={\dfrac {E\;{\sqrt {2}}}{c-\omega ^{2}}}\,\left[{\dfrac {\omega }{p^{2}+\omega ^{2}}}-{\dfrac {\omega }{\sqrt {c}}}\;{\dfrac {\sqrt {c}}{p^{2}+c}}\right]+{\dfrac {p}{p^{2}+c}}\;y_{0}+{\dfrac {\sqrt {c}}{p^{2}+c}}\;{\dfrac {{\dot {y}}_{0}}{\sqrt {c}}}\;}
[ 157] , [ 161] Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
≠
ω
2
{\displaystyle \;\color {transparent}{\neq \omega ^{2}}\;}
des originales de transformées
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
≠
ω
2
{\displaystyle \;\color {transparent}{\neq \omega ^{2}}\;}
«
y
(
t
)
=
{
E
2
c
−
ω
2
[
sin
(
ω
t
)
−
ω
c
sin
(
c
t
)
]
+
y
0
cos
(
c
t
)
+
y
˙
0
c
sin
(
c
t
)
}
Y
(
t
)
{\displaystyle \;y(t)=\left\lbrace {\dfrac {E\;{\sqrt {2}}}{c-\omega ^{2}}}\,\left[\sin(\omega \;t)-{\dfrac {\omega }{\sqrt {c}}}\;\sin({\sqrt {c}}\;t)\right]+y_{0}\;\cos({\sqrt {c}}\;t)+{\dfrac {{\dot {y}}_{0}}{\sqrt {c}}}\;\sin({\sqrt {c}}\;t)\right\rbrace Y(t)\;}
[ 127] Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
≠
ω
2
{\displaystyle \;\color {transparent}{\neq \omega ^{2}}\;}
«
y
(
t
)
=
{
y
0
cos
(
c
t
)
+
[
y
˙
0
c
−
E
2
c
−
ω
2
ω
c
]
sin
(
c
t
)
+
E
2
c
−
ω
2
sin
(
ω
t
)
}
Y
(
t
)
{\displaystyle \;y(t)=\left\lbrace y_{0}\,\cos({\sqrt {c}}\;t)+\left[{\dfrac {{\dot {y}}_{0}}{\sqrt {c}}}-{\dfrac {E\;{\sqrt {2}}}{c-\omega ^{2}}}\;{\dfrac {\omega }{\sqrt {c}}}\right]\sin({\sqrt {c}}\;t)+{\dfrac {E\;{\sqrt {2}}}{c-\omega ^{2}}}\;\sin(\omega \,t)\right\rbrace Y(t)\;}
Il reste alors à chercher l'« originale
≻
{\displaystyle \succ \;}
c
>
0
{\displaystyle \;c>0\;}
=
ω
2
{\displaystyle \;=\omega ^{2}\;}
Y
_
(
p
)
=
E
2
ω
(
p
2
+
ω
2
)
2
+
p
p
2
+
ω
2
y
0
+
ω
p
2
+
ω
2
y
˙
0
ω
{\displaystyle \;{\underline {Y}}(p)=E\;{\sqrt {2}}\;{\dfrac {\omega }{\left(p^{2}+\omega ^{2}\right)^{2}}}+{\dfrac {p}{p^{2}+\omega ^{2}}}\;y_{0}+{\dfrac {\omega }{p^{2}+\omega ^{2}}}\;{\dfrac {{\dot {y}}_{0}}{\omega }}\;}
[ 157] , [ 162] Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
transformées
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] er terme du 2nd membre
∝
{\displaystyle \;\propto \;}
ω
(
p
2
+
ω
2
)
2
{\displaystyle \;{\dfrac {\omega }{\left(p^{2}+\omega ^{2}\right)^{2}}}\;}
Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
«
y
(
t
)
=
E
2
L
−
1
{
ω
(
p
2
+
ω
2
)
2
}
+
{
y
0
cos
(
ω
t
)
+
y
˙
0
ω
sin
(
ω
t
)
}
Y
(
t
)
{\displaystyle \;y(t)=E\;{\sqrt {2}}\;{\mathcal {L}}^{-1}\!\!\left\lbrace {\dfrac {\omega }{\left(p^{2}+\omega ^{2}\right)^{2}}}\right\rbrace +\left\lbrace y_{0}\;\cos(\omega \;t)+{\dfrac {{\dot {y}}_{0}}{\omega }}\;\sin(\omega \;t)\right\rbrace Y(t)\;}
[ 127] Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
pour déterminer l'originale de
B
(
p
)
=
ω
(
p
2
+
ω
2
)
2
{\displaystyle \;B(p)={\dfrac {\omega }{\left(p^{2}+\omega ^{2}\right)^{2}}}\;}
L
−
1
{
B
(
p
)
}
=
L
−
1
{
ω
(
p
2
+
ω
2
)
2
}
{\displaystyle \;{\mathcal {L}}^{-1}\!\!\left\lbrace B(p)\right\rbrace ={\mathcal {L}}^{-1}\!\!\left\lbrace {\dfrac {\omega }{\left(p^{2}+\omega ^{2}\right)^{2}}}\right\rbrace \;}
Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
d
[
p
p
2
+
ω
2
]
d
p
=
(
p
2
+
ω
2
)
−
p
2
p
(
p
2
+
ω
2
)
2
=
ω
2
−
p
2
(
p
2
+
ω
2
)
2
=
−
ω
2
+
p
2
(
p
2
+
ω
2
)
2
+
2
ω
2
(
p
2
+
ω
2
)
2
{\displaystyle \;{\dfrac {d\left[{\dfrac {p}{p^{2}+\omega ^{2}}}\right]}{dp}}={\dfrac {\left(p^{2}+\omega ^{2}\right)-p\;2\;p}{\left(p^{2}+\omega ^{2}\right)^{2}}}={\dfrac {\omega ^{2}-p^{2}}{\left(p^{2}+\omega ^{2}\right)^{2}}}=-{\dfrac {\omega ^{2}+p^{2}}{\left(p^{2}+\omega ^{2}\right)^{2}}}+{\dfrac {2\;\omega ^{2}}{\left(p^{2}+\omega ^{2}\right)^{2}}}}
Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
d
[
p
2
+
ω
2
]
{\displaystyle \;\color {transparent}{d\left[p^{2}+\omega ^{2}\right]}}
=
−
1
p
2
+
ω
2
+
2
ω
2
(
p
2
+
ω
2
)
2
{\displaystyle =-{\dfrac {1}{p^{2}+\omega ^{2}}}+{\dfrac {2\;\omega ^{2}}{\left(p^{2}+\omega ^{2}\right)^{2}}}\;}
Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
«
B
(
p
)
=
ω
(
p
2
+
ω
2
)
2
=
1
2
ω
d
[
p
p
2
+
ω
2
]
d
p
+
1
2
ω
2
ω
p
2
+
ω
2
{\displaystyle \;B(p)={\dfrac {\omega }{\left(p^{2}+\omega ^{2}\right)^{2}}}={\dfrac {1}{2\;\omega }}\;{\dfrac {d\left[{\dfrac {p}{p^{2}+\omega ^{2}}}\right]}{dp}}+{\dfrac {1}{2\;\omega ^{2}}}\;{\dfrac {\omega }{p^{2}+\omega ^{2}}}\;}
Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
{
L
[
cos
(
ω
t
)
Y
(
t
)
]
=
p
p
2
+
ω
2
d
L
[
f
(
t
)
]
d
p
=
−
L
[
t
f
(
t
)
]
}
{\displaystyle \;\left\lbrace {\begin{array}{c c c}{\mathcal {L}}\!\left[\cos(\omega \,t)\;Y(t)\right]\!\!&=&\!\!{\dfrac {p}{p^{2}+\omega ^{2}}}\\{\dfrac {d{\mathcal {L}}\!\left[f(t)\right]}{dp}}\!\!&=&\!\!-{\mathcal {L}}\!\left[t\;f(t)\right]\end{array}}\right\rbrace \;}
[ 127] , [ 163] Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
«
B
(
p
)
=
ω
(
p
2
+
ω
2
)
2
=
−
1
2
ω
L
{
t
cos
(
ω
t
)
Y
(
t
)
}
+
1
2
ω
2
L
{
sin
(
ω
t
)
Y
(
t
)
}
{\displaystyle \;B(p)={\dfrac {\omega }{\left(p^{2}+\omega ^{2}\right)^{2}}}=-{\dfrac {1}{2\;\omega }}\;{\mathcal {L}}\!\left\lbrace t\;\cos(\omega \,t)\;Y(t)\right\rbrace +{\dfrac {1}{2\;\omega ^{2}}}\;{\mathcal {L}}\!\left\lbrace \sin(\omega \,t)\;Y(t)\right\rbrace \;}
[ 127] Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
B
(
p
)
{\displaystyle \;\color {transparent}{B(p)}}
=
1
2
ω
2
L
{
[
sin
(
ω
t
)
−
ω
t
cos
(
ω
t
)
]
Y
(
t
)
}
{\displaystyle ={\dfrac {1}{2\;\omega ^{2}}}\;{\mathcal {L}}\!\left\lbrace \left[\sin(\omega \;t)-\omega \;t\;\cos(\omega \;t)\right]\,Y(t)\right\rbrace \;}
B
(
p
)
=
ω
(
p
2
+
ω
2
)
2
{\displaystyle \;B(p)={\dfrac {\omega }{\left(p^{2}+\omega ^{2}\right)^{2}}}\;}
Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
«
L
−
1
{
ω
(
p
2
+
ω
2
)
2
}
=
1
2
ω
2
[
sin
(
ω
t
)
−
ω
t
cos
(
ω
t
)
]
Y
(
t
)
{\displaystyle \;{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {\omega }{\left(p^{2}+\omega ^{2}\right)^{2}}}\right\rbrace ={\dfrac {1}{2\;\omega ^{2}}}\,\left[\sin(\omega \;t)-\omega \;t\;\cos(\omega \;t)\right]\,Y(t)\;}
Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
«
y
(
t
)
=
{
E
2
2
ω
2
[
sin
(
ω
t
)
−
ω
t
cos
(
ω
t
)
]
+
y
0
cos
(
ω
t
)
+
y
˙
0
ω
sin
(
ω
t
)
}
Y
(
t
)
{\displaystyle \;y(t)=\left\lbrace {\dfrac {E\;{\sqrt {2}}}{2\;\omega ^{2}}}\,\left[\sin(\omega \,t)-\omega \;t\;\cos(\omega \,t)\right]+y_{0}\;\cos(\omega \;t)+{\dfrac {{\dot {y}}_{0}}{\omega }}\;\sin(\omega \;t)\right\rbrace Y(t)\;}
Il reste alors à chercher l'« originale
≻
{\displaystyle \color {transparent}{\succ }\;}
c
>
0
{\displaystyle \;\color {transparent}{c>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
«
y
(
t
)
=
{
y
0
cos
(
ω
t
)
+
[
y
˙
0
ω
+
E
2
ω
2
]
sin
(
ω
t
)
−
E
2
ω
t
cos
(
ω
t
)
}
Y
(
t
)
{\displaystyle \;y(t)=\left\lbrace y_{0}\,\cos(\omega \;t)+\left[{\dfrac {{\dot {y}}_{0}}{\omega }}+{\dfrac {E}{{\sqrt {2}}\;\omega ^{2}}}\right]\sin(\omega \;t)-{\dfrac {E}{{\sqrt {2}}\;\omega }}\;t\;\cos(\omega \,t)\right\rbrace Y(t)\;}
Rappel : D'après la note « 159
p
2
+
c
{\displaystyle \;p^{2}+c\;}
≠
0
{\displaystyle \;\neq 0\;}
c
<
0
{\displaystyle \;c<0}
p
{\displaystyle \;p\;}
ℜ
(
p
)
>
−
c
{\displaystyle \;\Re (p)>{\sqrt {-c}}\;}
Rappel : D'après la note « 159 » plus haut dans ce chapitre, pour que
p
2
+
c
{\displaystyle \;\color {transparent}{p^{2}+c}\;}
≠
0
{\displaystyle \;\color {transparent}{\neq 0}\;}
c
<
0
{\displaystyle \;\color {transparent}{c<0}}
«
Y
_
(
p
)
=
E
2
ω
(
p
2
+
ω
2
)
(
p
2
+
c
)
+
p
p
2
+
c
y
0
+
1
p
2
+
c
y
˙
0
{\displaystyle \;{\underline {Y}}(p)=E\;{\sqrt {2}}\;{\dfrac {\omega }{(p^{2}+\omega ^{2})\;(p^{2}+c)}}+{\dfrac {p}{p^{2}+c}}\;y_{0}+{\dfrac {1}{p^{2}+c}}\;{\dot {y}}_{0}\;}
[ 157]
ℜ
(
p
)
>
−
c
{\displaystyle \;\Re (p)>{\sqrt {-c}}\;}
«
c
{\displaystyle \;c\;}
<
0
{\displaystyle \;<0\;}
▸
{\displaystyle \blacktriangleright \;}
fonction rationnelle
A
′
(
p
)
=
1
(
p
2
+
ω
2
)
(
p
2
+
c
)
=
1
(
p
2
+
ω
2
)
(
p
+
−
c
)
(
p
−
−
c
)
{\displaystyle \;A'(p)={\dfrac {1}{(p^{2}+\omega ^{2})\;(p^{2}+c)}}={\dfrac {1}{(p^{2}+\omega ^{2})\;(p+{\sqrt {-c}})\;(p-{\sqrt {-c}})}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
A
′
(
p
)
=
(
p
2
+
ω
2
)
(
p
2
+
c
)
=
(
p
2
+
ω
2
)
(
p
+
−
c
)
(
p
−
−
c
)
{\displaystyle \;\color {transparent}{A'(p)=(p^{2}+\omega ^{2})\;(p^{2}+c)=(p^{2}+\omega ^{2})\;(p+{\sqrt {-c}})\;(p-{\sqrt {-c}})}\;}
réels simples opposés d'autre part » «
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
sa décomposition en éléments irréductibles simples est «
A
′
(
p
)
=
1
(
p
2
+
ω
2
)
(
p
+
−
c
)
(
p
−
−
c
)
=
ρ
1
p
+
ρ
′
1
p
2
+
ω
2
+
ρ
2
p
+
−
c
+
ρ
3
p
−
−
c
{\displaystyle \;A'(p)={\dfrac {1}{(p^{2}+\omega ^{2})\;(p+{\sqrt {-c}})\;(p-{\sqrt {-c}})}}={\dfrac {\rho _{1}\,p+{\rho '}_{1}}{p^{2}+\omega ^{2}}}+{\dfrac {\rho _{2}}{p+{\sqrt {-c}}}}+{\dfrac {\rho _{3}}{p-{\sqrt {-c}}}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
∙
{\displaystyle \bullet \;}
(
ρ
1
,
ρ
′
1
)
{\displaystyle \;\left(\rho _{1}\,,\,{\rho '}_{1}\right)\;}
p
2
+
ω
2
{\displaystyle \;p^{2}+\omega ^{2}}
p
=
±
i
ω
{\displaystyle \;p=\pm i\;\omega \;}
[ 104]
1
c
−
ω
2
=
{
ρ
1
i
ω
+
ρ
′
1
pour
p
=
i
ω
−
ρ
1
i
ω
+
ρ
′
1
pour
p
=
−
i
ω
}
{\displaystyle \;{\dfrac {1}{c-\omega ^{2}}}=\left\lbrace {\begin{array}{c}\rho _{1}\,i\,\omega +{\rho '}_{1}\;{\text{pour}}\;p=i\,\omega \\-\rho _{1}\,i\,\omega +{\rho '}_{1}\;{\text{pour}}\;p=-i\,\omega \end{array}}\right\rbrace \;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
∙
{\displaystyle \color {transparent}{\bullet }\;}
(
ρ
1
,
ρ
′
1
)
{\displaystyle \;\color {transparent}{\left(\rho _{1}\,,\,{\rho '}_{1}\right)}\;}
p
2
+
ω
2
{\displaystyle \;\color {transparent}{p^{2}+\omega ^{2}}\;}
p
=
±
i
ω
{\displaystyle \;\color {transparent}{p=\pm i\;\omega }\;}
d'où «
ρ
1
=
0
{\displaystyle \;\rho _{1}=0\;}
ρ
′
1
=
−
1
ω
2
−
c
{\displaystyle \;{\rho '}_{1}=-{\dfrac {1}{\omega ^{2}-c}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
∙
{\displaystyle \bullet \;}
ρ
2
{\displaystyle \;\rho _{2}\;}
p
+
−
c
{\displaystyle \;p+{\sqrt {-c}}}
p
=
−
−
c
{\displaystyle \;p=-\;{\sqrt {-c}}\;}
1
(
−
c
+
ω
2
)
(
−
2
−
c
)
=
ρ
2
{\displaystyle \;{\dfrac {1}{(-c+\omega ^{2})\,(-2\;{\sqrt {-c}})}}=\rho _{2}\;}
ρ
2
=
−
1
2
(
ω
2
−
c
)
−
c
)
{\displaystyle \;\rho _{2}=-{\dfrac {1}{2\;(\omega ^{2}-c)\;{\sqrt {-c}})}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
∙
{\displaystyle \bullet \;}
ρ
3
{\displaystyle \;\rho _{3}\;}
p
−
−
c
{\displaystyle \;p-{\sqrt {-c}}}
p
=
−
c
{\displaystyle \;p=\;{\sqrt {-c}}\;}
1
(
−
c
+
ω
2
)
(
2
−
c
)
=
ρ
3
{\displaystyle \;{\dfrac {1}{(-c+\omega ^{2})\,(2\;{\sqrt {-c}})}}=\rho _{3}\;}
ρ
3
=
1
2
(
ω
2
−
c
)
−
c
)
{\displaystyle \;\rho _{3}={\dfrac {1}{2\;(\omega ^{2}-c)\;{\sqrt {-c}})}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
∙
{\displaystyle \bullet \;}
A
′
(
p
)
=
1
(
p
2
+
ω
2
)
(
p
2
+
c
)
=
1
ω
2
−
c
{
−
1
p
2
+
ω
2
+
1
2
−
c
[
1
p
−
−
c
−
1
p
+
−
c
]
}
{\displaystyle \;A'(p)={\dfrac {1}{(p^{2}+\omega ^{2})\;(p^{2}+c)}}={\dfrac {1}{\omega ^{2}-c}}\,\left\lbrace -{\dfrac {1}{p^{2}+\omega ^{2}}}+{\dfrac {1}{2\;{\sqrt {-c}}}}\,\left[{\dfrac {1}{p-{\sqrt {-c}}}}-{\dfrac {1}{p+{\sqrt {-c}}}}\right]\right\rbrace \;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \blacktriangleright \;}
fonction rationnelle
B
′
(
p
)
=
p
p
2
+
c
=
p
(
p
+
−
c
)
(
p
−
−
c
)
{\displaystyle \;B'(p)={\dfrac {p}{p^{2}+c}}={\dfrac {p}{(p+{\sqrt {-c}})\;(p-{\sqrt {-c}})}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
sa décomposition en éléments irréductibles simples s'écrit est «
B
′
(
p
)
=
p
(
p
+
−
c
)
(
p
−
−
c
)
=
ζ
1
p
+
−
c
+
ζ
2
p
−
−
c
{\displaystyle \;B'(p)={\dfrac {p}{(p+{\sqrt {-c}})\;(p-{\sqrt {-c}})}}={\dfrac {\zeta _{1}}{p+{\sqrt {-c}}}}+{\dfrac {\zeta _{2}}{p-{\sqrt {-c}}}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
∙
{\displaystyle \bullet \;}
ζ
1
{\displaystyle \;\zeta _{1}\;}
p
+
−
c
{\displaystyle \;p+{\sqrt {-c}}}
p
=
−
−
c
{\displaystyle \;p=-\;{\sqrt {-c}}\;}
[ 104]
−
−
c
−
2
−
c
=
ζ
1
{\displaystyle \;{\dfrac {-\;{\sqrt {-c}}}{-2\;{\sqrt {-c}}}}=\zeta _{1}\;}
ζ
1
=
1
2
{\displaystyle \;\zeta _{1}={\dfrac {1}{2}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
∙
{\displaystyle \bullet \;}
ζ
2
{\displaystyle \;\zeta _{2}\;}
p
−
−
c
{\displaystyle \;p-{\sqrt {-c}}}
p
=
−
c
{\displaystyle \;p=\;{\sqrt {-c}}\;}
[ 104]
−
c
2
−
c
)
=
ζ
2
{\displaystyle \;{\dfrac {\sqrt {-c}}{2\;{\sqrt {-c}})}}=\zeta _{2}\;}
ζ
2
=
1
2
{\displaystyle \;\zeta _{2}={\dfrac {1}{2}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
∙
{\displaystyle \bullet \;}
B
′
(
p
)
=
p
p
2
+
c
=
1
2
[
1
p
+
−
c
+
1
p
−
−
c
]
{\displaystyle \;B'(p)={\dfrac {p}{p^{2}+c}}={\dfrac {1}{2}}\,\left[{\dfrac {1}{p+{\sqrt {-c}}}}+{\dfrac {1}{p-{\sqrt {-c}}}}\right]\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \blacktriangleright \;}
fonction rationnelle
C
′
(
p
)
=
1
p
2
+
c
=
1
(
p
+
−
c
)
(
p
−
−
c
)
{\displaystyle \;C'(p)={\dfrac {1}{p^{2}+c}}={\dfrac {1}{(p+{\sqrt {-c}})\;(p-{\sqrt {-c}})}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
sa décomposition en éléments irréductibles simples est «
C
′
(
p
)
=
1
(
p
+
−
c
)
(
p
−
−
c
)
=
ζ
3
p
+
−
c
+
ζ
4
p
−
−
c
{\displaystyle \;C'(p)={\dfrac {1}{(p+{\sqrt {-c}})\;(p-{\sqrt {-c}})}}={\dfrac {\zeta _{3}}{p+{\sqrt {-c}}}}+{\dfrac {\zeta _{4}}{p-{\sqrt {-c}}}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
∙
{\displaystyle \bullet \;}
ζ
3
{\displaystyle \;\zeta _{3}\;}
p
+
−
c
{\displaystyle \;p+{\sqrt {-c}}}
p
=
−
−
c
{\displaystyle \;p=-\;{\sqrt {-c}}\;}
[ 104]
1
−
2
−
c
=
ζ
3
{\displaystyle \;{\dfrac {1}{-2\;{\sqrt {-c}}}}=\zeta _{3}\;}
ζ
3
=
−
1
2
−
c
{\displaystyle \;\zeta _{3}=-{\dfrac {1}{2\;{\sqrt {-c}}}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
∙
{\displaystyle \bullet \;}
ζ
4
{\displaystyle \;\zeta _{4}\;}
p
−
−
c
{\displaystyle \;p-{\sqrt {-c}}}
p
=
−
c
{\displaystyle \;p=\;{\sqrt {-c}}\;}
[ 104]
1
2
−
c
=
ζ
4
{\displaystyle \;{\dfrac {1}{2\;{\sqrt {-c}}}}=\zeta _{4}\;}
ζ
4
=
1
2
−
c
{\displaystyle \;\zeta _{4}={\dfrac {1}{2\;{\sqrt {-c}}}}\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \color {transparent}{\blacktriangleright }\;}
∙
{\displaystyle \bullet \;}
C
′
(
p
)
=
1
p
2
+
c
=
1
2
−
c
[
1
p
−
−
c
−
1
p
+
−
c
]
{\displaystyle \;C'(p)={\dfrac {1}{p^{2}+c}}={\dfrac {1}{2\;{\sqrt {-c}}}}\,\left[{\dfrac {1}{p-{\sqrt {-c}}}}-{\dfrac {1}{p+{\sqrt {-c}}}}\right]\;}
«
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
▸
{\displaystyle \blacktriangleright \;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
c
{\displaystyle \;c\;}
<
0
{\displaystyle \;<0\;}
«
Y
_
(
p
)
=
E
2
ω
2
−
c
{
−
ω
p
2
+
ω
2
+
ω
2
−
c
[
1
p
−
−
c
−
1
p
+
−
c
]
}
+
y
0
2
[
1
p
+
−
c
+
1
p
−
−
c
]
+
y
˙
0
2
−
c
[
1
p
−
−
c
−
1
p
+
−
c
]
{\displaystyle \;{\underline {Y}}(p)={\dfrac {E\;{\sqrt {2}}}{\omega ^{2}-c}}\,\left\lbrace -{\dfrac {\omega }{p^{2}+\omega ^{2}}}+{\dfrac {\omega }{2\;{\sqrt {-c}}}}\,\left[{\dfrac {1}{p-{\sqrt {-c}}}}-{\dfrac {1}{p+{\sqrt {-c}}}}\right]\right\rbrace +{\dfrac {y_{0}}{2}}\,\left[{\dfrac {1}{p+{\sqrt {-c}}}}+{\dfrac {1}{p-{\sqrt {-c}}}}\right]+{\dfrac {{\dot {y}}_{0}}{2\;{\sqrt {-c}}}}\,\left[{\dfrac {1}{p-{\sqrt {-c}}}}-{\dfrac {1}{p+{\sqrt {-c}}}}\right]\;}
[ 157]
ℜ
(
p
)
>
−
c
{\displaystyle \;\Re (p)>{\sqrt {-c}}\;}
Y
_
(
p
)
=
−
E
2
ω
2
−
c
ω
p
2
+
ω
2
+
[
E
ω
(
ω
2
−
c
)
−
2
c
+
y
˙
0
+
y
0
−
c
2
−
c
]
1
p
−
−
c
−
[
E
ω
(
ω
2
−
c
)
−
2
c
+
y
˙
0
−
y
0
−
c
2
−
c
]
1
p
+
−
c
{\displaystyle \;{\underline {Y}}(p)=-{\dfrac {E\;{\sqrt {2}}}{\omega ^{2}-c}}\;{\dfrac {\omega }{p^{2}+\omega ^{2}}}+\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\,{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}+y_{0}\;{\sqrt {-c}}}{2\;{\sqrt {-c}}}}\right]\,{\dfrac {1}{p-{\sqrt {-c}}}}-\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\,{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}-y_{0}\;{\sqrt {-c}}}{2\;{\sqrt {-c}}}}\right]\,{\dfrac {1}{p+{\sqrt {-c}}}}\;}
[ 157]
ℜ
(
p
)
>
−
c
{\displaystyle \;\Re (p)>{\sqrt {-c}}\;}
Il reste alors à chercher l'« originale de la transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
Y
_
(
p
)
{\displaystyle \;{\underline {Y}}(p)\;}
[ 157]
c
<
0
{\displaystyle \;c<0}
ℜ
(
p
)
>
−
c
{\displaystyle \;\Re (p)>{\sqrt {-c}}}
Il reste alors à chercher l'« originale de
Y
_
(
p
)
=
−
E
2
ω
2
−
c
ω
p
2
+
ω
2
+
[
E
ω
(
ω
2
−
c
)
−
2
c
+
y
˙
0
+
y
0
−
c
2
−
c
]
1
p
−
−
c
−
[
E
ω
(
ω
2
−
c
)
−
2
c
+
y
˙
0
−
y
0
−
c
2
−
c
]
1
p
+
−
c
{\displaystyle {\underline {Y}}(p)=-{\dfrac {E\;{\sqrt {2}}}{\omega ^{2}-c}}\;{\dfrac {\omega }{p^{2}+\omega ^{2}}}+\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\,{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}+y_{0}\;{\sqrt {-c}}}{2\;{\sqrt {-c}}}}\right]\,{\dfrac {1}{p-{\sqrt {-c}}}}-\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\,{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}-y_{0}\;{\sqrt {-c}}}{2\;{\sqrt {-c}}}}\right]\,{\dfrac {1}{p+{\sqrt {-c}}}}\;}
[ 157] Il reste alors à chercher l'« originale de la transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
Y
_
(
p
)
{\displaystyle \;\color {transparent}{{\underline {Y}}(p)}\;}
c
<
0
{\displaystyle \;\color {transparent}{c<0}}
en reconnaissant des originales de transformées
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
d'où «
y
(
t
)
=
{
−
E
2
ω
2
−
c
sin
(
ω
t
)
+
[
E
ω
(
ω
2
−
c
)
−
2
c
+
y
˙
0
+
y
0
−
c
2
−
c
]
exp
(
−
c
t
)
−
[
E
ω
(
ω
2
−
c
)
−
2
c
+
y
˙
0
−
y
0
−
c
2
−
c
]
exp
(
−
−
c
t
)
}
Y
(
t
)
{\displaystyle \;y(t)=\left\lbrace -{\dfrac {E\;{\sqrt {2}}}{\omega ^{2}-c}}\;\sin(\omega \;t)+\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\,{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}+y_{0}\;{\sqrt {-c}}}{2\;{\sqrt {-c}}}}\right]\,\exp \!\left({\sqrt {-c}}\,t\right)-\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\,{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}-y_{0}\;{\sqrt {-c}}}{2\;{\sqrt {-c}}}}\right]\,\exp \!\left(-{\sqrt {-c}}\,t\right)\right\rbrace \;Y(t)\;}
[ 127] , [ 106]
Nous rappelons que la méthode de résolution d'une équation différentielle linéaire à cœfficients réels constants homogène ou hétérogène Nous rappelons que la méthode de résolution par transformation
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] [ 164] Nous rappelons que seule la méthode de résolution par solutions libre et forcée suivie de l'utilisation des C.I[ 154]
nous allons donc vérifier, dans l'exemple de «
y
¨
(
t
)
+
c
y
(
t
)
=
e
(
t
)
=
E
2
sin
(
ω
t
)
Y
(
t
)
{\displaystyle \;{\ddot {y}}(t)+c\;y(t)=e(t)=E\;{\sqrt {2}}\;\sin(\omega \,t)\;Y(t)\;}
c
∈
R
∗
{\displaystyle \;c\in \mathbb {R} ^{*}\;}
(
E
,
ω
)
∈
R
+
∗
2
{\displaystyle \;\left(E\,,\,\omega \right)\in {\mathbb {R} ^{+\;*}}^{2}\;}
[ 156] nous allons donc vérifier, que nous trouvons le même résultat dans l'ensemble des fonctions réelles « causales »[ 4] , [ 9] [ 154]
y
(
0
+
)
=
y
0
{\displaystyle \;y(0^{+})=y_{0}\;}
y
˙
(
0
+
)
=
y
˙
0
{\displaystyle \;{\dot {y}}(0^{+})={\dot {y}}_{0}\;}
nous allons donc vérifier, en reprenant la résolution par méthode classique de recherche des solutions libre et forcée suivie de l'utilisation des C.I[ 154]
Soit à résoudre «
y
l
¨
(
t
)
+
c
y
l
(
t
)
=
0
{\displaystyle \;{\ddot {y_{l}}}(t)+c\;y_{l}(t)=0\;}
[ 165]
s
2
+
c
=
0
{\displaystyle \;s^{2}+c=0\;}
[ 166]
c
{\displaystyle \;c}
« si
c
{\displaystyle \;c\;}
>
0
{\displaystyle \;>0\;}
[ 165]
s
±
_
=
±
i
c
{\displaystyle \;{\underline {s_{\pm }}}=\pm i\,{\sqrt {c}}\;}
⇒
{\displaystyle \Rightarrow }
y
l
(
t
)
=
A
cos
(
c
t
+
φ
)
{\displaystyle \;y_{l}(t)=A\,\cos({\sqrt {c}}\;t+\varphi )\;}
(
A
,
φ
)
∈
R
2
{\displaystyle \;\left(A\,,\,\varphi \right)\in \mathbb {R} ^{2}\;}
[ 167]
« si
c
{\displaystyle \;c\;}
<
0
{\displaystyle \;<0\;}
s
±
=
±
−
c
{\displaystyle \;s_{\pm }=\pm {\sqrt {-c}}\;}
⇒
{\displaystyle \Rightarrow }
y
l
(
t
)
=
A
+
exp
(
−
c
t
)
+
A
−
exp
(
−
−
c
t
)
{\displaystyle \;y_{l}(t)=A_{+}\,\exp({\sqrt {-c}}\;t)+A_{-}\,\exp(-{\sqrt {-c}}\;t)\;}
(
A
+
,
A
−
)
∈
R
2
{\displaystyle \;\left(A_{+}\,,\,A_{-}\right)\in \mathbb {R} ^{2}\;}
Soit à déterminer la « solution forcée
y
f
(
t
)
{\displaystyle \;y_{f}(t)\;}
y
f
¨
(
t
)
+
c
y
f
(
t
)
=
E
2
sin
(
ω
t
)
pour
t
>
0
{\displaystyle \;{\ddot {y_{f}}}(t)+c\;y_{f}(t)=E\;{\sqrt {2}}\;\sin(\omega \,t)\;\;{\text{pour}}\;t>0\;}
[ 168]
e
(
t
)
=
E
2
sin
(
ω
t
)
pour
t
>
0
{\displaystyle \;e(t)=E\;{\sqrt {2}}\;\sin(\omega \,t)\;\;{\text{pour}}\;t>0\;}
Soit à déterminer la « solution forcée c'est-à-dire sous forme générale «
y
f
(
t
)
=
Y
2
sin
(
ω
t
+
φ
y
)
pour
t
>
0
{\displaystyle \;y_{f}(t)=Y\;{\sqrt {2}}\;\sin(\omega \,t+\varphi _{y})\;\;{\text{pour}}\;t>0\;}
[ 169]
(
Y
,
φ
y
)
∈
R
{\displaystyle \;\left(Y\,,\,\varphi _{y}\right)\in \mathbb {R} \;}
Soit à déterminer la « solution forcée
y
f
(
t
)
{\displaystyle \;\color {transparent}{y_{f}(t)}\;}
sous forme particulière «
y
f
(
t
)
=
Y
2
sin
(
ω
t
)
pour
t
>
0
{\displaystyle \;y_{f}(t)=Y\;{\sqrt {2}}\;\sin(\omega \,t)\;\;{\text{pour}}\;t>0\;}
[
{\displaystyle {\big [}}
er ordre dans l'équation différentielle
]
{\displaystyle {\big ]}}
Soit à déterminer la « solution forcée
y
f
(
t
)
{\displaystyle \;\color {transparent}{y_{f}(t)}\;}
y
f
(
t
)
=
Y
2
sin
(
ω
t
)
pour
t
>
0
{\displaystyle \;\color {transparent}{y_{f}(t)=Y\;{\sqrt {2}}\;\sin(\omega \,t)\;\;{\text{pour}}\;t>0}\;}
le choix de
φ
y
=
0
{\displaystyle \;\varphi _{y}=0\;}
« si
c
{\displaystyle \;c\;}
>
0
{\displaystyle \;>0\;}
y
f
(
t
)
=
Y
2
sin
(
ω
t
)
{\displaystyle \;y_{f}(t)=Y\;{\sqrt {2}}\;\sin(\omega \,t)\;}
(
−
ω
2
+
c
)
Y
2
sin
(
ω
t
)
=
E
2
sin
(
ω
t
)
∀
t
>
0
{\displaystyle \;\left(-\omega ^{2}+c\right)\,Y\;{\sqrt {2}}\;\sin(\omega \,t)=E\;{\sqrt {2}}\;\sin(\omega \,t)\;\;\forall \;t>0\;}
« si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
≻
{\displaystyle \succ \;}
c
≠
ω
2
{\displaystyle \;c\neq \omega ^{2}\;}
Y
=
E
c
−
ω
2
{\displaystyle \;Y={\dfrac {E}{c-\omega ^{2}}}\;}
y
f
(
t
)
=
E
c
−
ω
2
2
sin
(
ω
t
)
pour
t
>
0
{\displaystyle \;y_{f}(t)={\dfrac {E}{c-\omega ^{2}}}\;{\sqrt {2}}\;\sin(\omega \,t)\;\;{\text{pour}}\;t>0\;}
« si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
≻
{\displaystyle \succ \;}
c
=
ω
2
{\displaystyle \;c=\omega ^{2}\;}
Y
{\displaystyle \;Y\;}
« si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
c
=
ω
2
{\displaystyle \;\color {transparent}{c=\omega ^{2}}\;}
on cherche alors une « solution particulière de la forme[ 170]
y
f
(
t
)
=
Y
′
2
t
sin
(
ω
t
+
φ
y
′
)
{\displaystyle \;y_{f}(t)=Y'\;{\sqrt {2}}\;t\;\sin(\omega \,t+\varphi _{y'})\;}
[ 171]
⇒
{\displaystyle \Rightarrow }
« si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
c
=
ω
2
{\displaystyle \;\color {transparent}{c=\omega ^{2}}\;}
«
y
f
˙
(
t
)
=
Y
′
2
sin
(
ω
t
+
φ
y
′
)
+
ω
Y
′
2
t
cos
(
ω
t
+
φ
y
′
)
{\displaystyle \;{\dot {y_{f}}}(t)=Y'\;{\sqrt {2}}\;\sin(\omega \,t+\varphi _{y'})+\omega \;Y'\;{\sqrt {2}}\;t\;\cos(\omega \,t+\varphi _{y'})\;}
⇒
{\displaystyle \Rightarrow }
y
f
¨
(
t
)
=
2
ω
Y
′
2
cos
(
ω
t
+
φ
y
′
)
−
ω
2
Y
′
2
t
sin
(
ω
t
+
φ
y
′
)
{\displaystyle \;{\ddot {y_{f}}}(t)=2\;\omega \;Y'\;{\sqrt {2}}\;\cos(\omega \,t+\varphi _{y'})-\omega ^{2}\;Y'\;{\sqrt {2}}\;t\;\sin(\omega \,t+\varphi _{y'})\;}
« si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
c
=
ω
2
{\displaystyle \;\color {transparent}{c=\omega ^{2}}\;}
en reportant dans l'équation différentielle et en tenant compte de «
c
=
ω
2
{\displaystyle \;c=\omega ^{2}\;}
(
Y
′
,
φ
y
′
)
{\displaystyle \;\left(Y'\,,\,\varphi _{y'}\right)\;}
« si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
c
=
ω
2
{\displaystyle \;\color {transparent}{c=\omega ^{2}}\;}
«
2
ω
Y
′
2
cos
(
ω
t
+
φ
y
′
)
−
ω
2
Y
′
2
t
sin
(
ω
t
+
φ
y
′
)
+
c
Y
′
2
t
sin
(
ω
t
+
φ
y
′
)
=
E
2
sin
(
ω
t
)
∀
t
>
0
{\displaystyle \;2\;\omega \;Y'\;{\sqrt {2}}\;\cos(\omega \,t+\varphi _{y'})\;{\cancel {-\;\omega ^{2}\;Y'\;{\sqrt {2}}\;t\;\sin(\omega \,t+\varphi _{y'})+c\;Y'\;{\sqrt {2}}\;t\;\sin(\omega \,t+\varphi _{y'})}}\;=E\;{\sqrt {2}}\;\sin(\omega \,t)\;\;\forall \;t>0\;}
« si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
≻
{\displaystyle \color {transparent}{\succ }\;}
c
=
ω
2
{\displaystyle \;\color {transparent}{c=\omega ^{2}}\;}
«
φ
y
′
=
−
π
2
{\displaystyle \;\varphi _{y'}=-{\dfrac {\pi }{2}}\;}
[ 172]
Y
′
=
E
2
ω
=
E
2
c
{\displaystyle \;Y'={\dfrac {E}{2\,\omega }}={\dfrac {E}{2\,{\sqrt {c}}}}\;}
y
f
(
t
)
=
{\displaystyle \;y_{f}(t)=}
E
2
ω
2
t
sin
(
ω
t
−
π
2
)
=
−
E
2
ω
2
t
cos
(
ω
t
)
pour
t
>
0
{\displaystyle {\dfrac {E}{2\,\omega }}\;{\sqrt {2}}\;t\;\sin \!\left(\omega \,t-{\dfrac {\pi }{2}}\right)=-{\dfrac {E}{2\,\omega }}\;{\sqrt {2}}\;t\;\cos(\omega \,t)\;\;{\text{pour}}\;t>0\;}
[ 173]
« si
c
{\displaystyle \;c\;}
<
0
{\displaystyle \;<0\;}
y
f
(
t
)
=
Y
2
sin
(
ω
t
)
{\displaystyle \;y_{f}(t)=Y\;{\sqrt {2}}\;\sin(\omega \,t)\;}
(
−
ω
2
+
c
)
Y
2
sin
(
ω
t
)
=
E
2
sin
(
ω
t
)
∀
t
>
0
{\displaystyle \;\left(-\omega ^{2}+c\right)\,Y\;{\sqrt {2}}\;\sin(\omega \,t)=E\;{\sqrt {2}}\;\sin(\omega \,t)\;\;\forall \;t>0\;}
⇒
{\displaystyle \Rightarrow }
Y
=
−
E
ω
2
−
c
{\displaystyle \;Y=-{\dfrac {E}{\omega ^{2}-c}}\;}
« si
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
«
y
f
(
t
)
=
{\displaystyle \;y_{f}(t)=}
−
E
2
ω
2
−
c
sin
(
ω
t
)
pour
t
>
0
{\displaystyle -{\dfrac {E\;{\sqrt {2}}}{\omega ^{2}-c}}\;\sin(\omega \,t)\;\;{\text{pour}}\;t>0\;}
La solution complète avant utilisation des C.I[ 154]
y
(
t
)
=
y
l
(
t
)
+
y
f
(
t
)
{\displaystyle \;y(t)=y_{l}(t)+y_{f}(t)\;}
y
l
(
t
)
{\displaystyle \;y_{l}(t)\;}
[ 174]
y
f
(
t
)
{\displaystyle \;y_{f}(t)\;}
[ 175]
« Si
c
{\displaystyle \;c\;}
>
0
{\displaystyle \;>0\;}
≠
ω
2
{\displaystyle \;\neq \omega ^{2}\;}
y
(
t
)
=
y
l
(
t
)
+
y
f
(
t
)
=
A
cos
(
c
t
+
φ
)
+
E
c
−
ω
2
2
sin
(
ω
t
)
pour
t
>
0
{\displaystyle \;y(t)=y_{l}(t)+y_{f}(t)=A\,\cos({\sqrt {c}}\;t+\varphi )+{\dfrac {E}{c-\omega ^{2}}}\;{\sqrt {2}}\;\sin(\omega \,t)\;\;{\text{pour}}\;t>0\;}
(
A
,
φ
)
∈
R
2
{\displaystyle \;\left(A\,,\,\varphi \right)\in \mathbb {R} ^{2}\;}
[ 154]
« si
c
{\displaystyle \;c\;}
>
0
{\displaystyle \;>0\;}
=
ω
2
{\displaystyle \;=\omega ^{2}\;}
y
(
t
)
=
y
l
(
t
)
+
y
f
(
t
)
=
A
cos
(
ω
t
+
φ
)
−
E
2
ω
2
t
cos
(
ω
t
)
pour
t
>
0
{\displaystyle \;y(t)=y_{l}(t)+y_{f}(t)=A\,\cos(\omega \;t+\varphi )-{\dfrac {E}{2\,\omega }}\;{\sqrt {2}}\;t\;\cos(\omega \,t)\;\;{\text{pour}}\;t>0\;}
[ 176]
(
A
,
φ
)
∈
R
2
{\displaystyle \;\left(A\,,\,\varphi \right)\in \mathbb {R} ^{2}\;}
[ 154]
« si
c
{\displaystyle \;c\;}
<
0
{\displaystyle \;<0\;}
y
(
t
)
=
y
l
(
t
)
+
y
f
(
t
)
=
A
+
exp
(
−
c
t
)
+
A
−
exp
(
−
−
c
t
)
−
E
ω
2
−
c
2
sin
(
ω
t
)
pour
t
>
0
{\displaystyle \;y(t)=y_{l}(t)+y_{f}(t)=A_{+}\,\exp({\sqrt {-c}}\;t)+A_{-}\,\exp(-{\sqrt {-c}}\;t)-{\dfrac {E}{\omega ^{2}-c}}\;{\sqrt {2}}\;\sin(\omega \,t)\;\;{\text{pour}}\;t>0\;}
(
A
+
,
A
−
)
{\displaystyle \;\left(A_{+}\,,\,A_{-}\right)}
∈
R
2
{\displaystyle \in \mathbb {R} ^{2}\;}
[ 154] Il faut donc écrire dans les expressions précédentes «
{
y
(
0
+
)
=
y
0
y
˙
(
0
+
)
=
y
˙
0
}
{\displaystyle \;\left\lbrace {\begin{array}{c}y(0^{+})=y_{0}\\{\dot {y}}(0^{+})={\dot {y}}_{0}\end{array}}\right\rbrace \;}
« si
c
{\displaystyle \;c\;}
>
0
{\displaystyle \;>0\;}
≠
ω
2
{\displaystyle \;\neq \omega ^{2}\;}
{
y
(
0
+
)
=
A
cos
(
φ
)
=
y
0
y
˙
(
0
+
)
=
−
A
c
sin
(
φ
)
+
E
c
−
ω
2
2
ω
=
y
˙
0
}
{\displaystyle \;\left\lbrace {\begin{array}{l}y(0^{+})=A\,\cos(\varphi )\qquad \qquad \qquad \qquad \quad =y_{0}\\{\dot {y}}(0^{+})=-A\,{\sqrt {c}}\,\sin(\varphi )+{\dfrac {E}{c-\omega ^{2}}}\;{\sqrt {2}}\;\omega ={\dot {y}}_{0}\end{array}}\right\rbrace \;}
{
A
cos
(
φ
)
=
y
0
A
sin
(
φ
)
=
E
c
−
ω
2
2
ω
c
−
y
˙
0
c
}
{\displaystyle \;\left\lbrace {\begin{array}{l}A\,\cos(\varphi )=y_{0}\\A\,\sin(\varphi )={\dfrac {E}{c-\omega ^{2}}}\;{\sqrt {2}}\;{\dfrac {\omega }{\sqrt {c}}}-{\dfrac {{\dot {y}}_{0}}{\sqrt {c}}}\end{array}}\right\rbrace \;}
« si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
≠
ω
2
{\displaystyle \;\color {transparent}{\neq \omega ^{2}}\;}
en utilisant
A
cos
(
c
t
+
φ
)
=
A
cos
(
φ
)
cos
(
c
t
)
−
A
sin
(
φ
)
sin
(
c
t
)
{\displaystyle \;A\,\cos({\sqrt {c}}\;t+\varphi )=A\,\cos(\varphi )\,\cos({\sqrt {c}}\;t)-A\,\sin(\varphi )\,\sin({\sqrt {c}}\;t)}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] «
y
(
t
)
=
y
0
cos
(
c
t
)
+
[
y
˙
0
c
−
E
2
c
−
ω
2
ω
c
]
sin
(
c
t
)
+
E
2
c
−
ω
2
sin
(
ω
t
)
pour
t
>
0
{\displaystyle \;y(t)=y_{0}\,\cos({\sqrt {c}}\;t)+\left[{\dfrac {{\dot {y}}_{0}}{\sqrt {c}}}-{\dfrac {E\;{\sqrt {2}}}{c-\omega ^{2}}}\;{\dfrac {\omega }{\sqrt {c}}}\right]\sin({\sqrt {c}}\;t)+{\dfrac {E\;{\sqrt {2}}}{c-\omega ^{2}}}\;\sin(\omega \,t)\;\;{\text{pour}}\;t>0\;}
« si
c
{\displaystyle \;c\;}
>
0
{\displaystyle \;>0\;}
=
ω
2
{\displaystyle \;=\omega ^{2}\;}
{
y
(
0
+
)
=
A
cos
(
φ
)
=
y
0
y
˙
(
0
+
)
=
−
A
ω
sin
(
φ
)
−
E
2
ω
2
=
y
˙
0
}
{\displaystyle \;\left\lbrace {\begin{array}{l}y(0^{+})=A\,\cos(\varphi )\qquad \qquad \quad \quad =y_{0}\\{\dot {y}}(0^{+})=-A\,\omega \,\sin(\varphi )-{\dfrac {E}{2\,\omega }}\;{\sqrt {2}}={\dot {y}}_{0}\end{array}}\right\rbrace \;}
{
A
cos
(
φ
)
=
y
0
A
sin
(
φ
)
=
−
E
2
ω
2
2
−
y
˙
0
ω
}
{\displaystyle \;\left\lbrace {\begin{array}{l}A\,\cos(\varphi )=y_{0}\\A\,\sin(\varphi )=-{\dfrac {E}{2\,\omega ^{2}}}\;{\sqrt {2}}-{\dfrac {{\dot {y}}_{0}}{\omega }}\end{array}}\right\rbrace \;}
« si
c
{\displaystyle \;\color {transparent}{c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}\;}
=
ω
2
{\displaystyle \;\color {transparent}{=\omega ^{2}}\;}
en utilisant
A
cos
(
ω
t
+
φ
)
=
A
cos
(
φ
)
cos
(
ω
t
)
−
A
sin
(
φ
)
sin
(
ω
t
)
{\displaystyle \;A\,\cos(\omega \;t+\varphi )=A\,\cos(\varphi )\,\cos(\omega \;t)-A\,\sin(\varphi )\,\sin(\omega \;t)}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] «
y
(
t
)
=
y
0
cos
(
ω
t
)
+
[
y
˙
0
ω
+
E
2
ω
2
]
sin
(
ω
t
)
−
E
2
ω
t
cos
(
ω
t
)
pour
t
>
0
{\displaystyle \;y(t)=y_{0}\,\cos(\omega \;t)+\left[{\dfrac {{\dot {y}}_{0}}{\omega }}+{\dfrac {E}{{\sqrt {2}}\;\omega ^{2}}}\right]\sin(\omega \;t)-{\dfrac {E}{{\sqrt {2}}\;\omega }}\;t\;\cos(\omega \,t)\;\;{\text{pour}}\;t>0\;}
« si
c
{\displaystyle \;c\;}
<
0
{\displaystyle \;<0\;}
{
y
(
0
+
)
=
A
+
+
A
−
=
y
0
y
˙
(
0
+
)
=
A
+
−
c
−
A
−
−
c
−
E
ω
2
−
c
2
ω
=
y
˙
0
}
{\displaystyle \;\left\lbrace {\begin{array}{l}y(0^{+})=A_{+}+A_{-}\qquad \qquad \qquad \qquad \qquad \quad \;\;\,=y_{0}\\{\dot {y}}(0^{+})=A_{+}\,{\sqrt {-c}}-A_{-}\,{\sqrt {-c}}-{\dfrac {E}{\omega ^{2}-c}}\;{\sqrt {2}}\;\omega ={\dot {y}}_{0}\end{array}}\right\rbrace \;}
{
A
+
+
A
−
=
y
0
A
+
−
A
−
=
E
ω
2
−
c
2
ω
−
c
+
y
˙
0
−
c
}
{\displaystyle \;\left\lbrace {\begin{array}{l}A_{+}+A_{-}=y_{0}\\A_{+}-A_{-}={\dfrac {E}{\omega ^{2}-c}}\;{\sqrt {2}}\;{\dfrac {\omega }{\sqrt {-c}}}+{\dfrac {{\dot {y}}_{0}}{\sqrt {-c}}}\end{array}}\right\rbrace \;}
« si
c
{\displaystyle \;\color {transparent}{c}\;}
<
0
{\displaystyle \;\color {transparent}{<0}\;}
«
{
A
+
=
1
2
[
y
0
+
E
ω
2
−
c
2
ω
−
c
+
y
˙
0
−
c
]
A
−
=
1
2
[
y
0
−
E
ω
2
−
c
2
ω
−
c
−
y
˙
0
−
c
]
}
{\displaystyle \;\left\lbrace {\begin{array}{l}A_{+}={\dfrac {1}{2}}\left[y_{0}+{\dfrac {E}{\omega ^{2}-c}}\;{\sqrt {2}}\;{\dfrac {\omega }{\sqrt {-c}}}+{\dfrac {{\dot {y}}_{0}}{\sqrt {-c}}}\right]\\A_{-}={\dfrac {1}{2}}\left[y_{0}-{\dfrac {E}{\omega ^{2}-c}}\;{\sqrt {2}}\;{\dfrac {\omega }{\sqrt {-c}}}-{\dfrac {{\dot {y}}_{0}}{\sqrt {-c}}}\right]\end{array}}\right\rbrace \;}
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1] «
y
(
t
)
=
[
E
ω
(
ω
2
−
c
)
−
2
c
+
y
˙
0
+
y
0
−
c
2
−
c
]
exp
(
−
c
t
)
−
[
E
ω
(
ω
2
−
c
)
−
2
c
+
y
˙
0
−
y
0
−
c
2
−
c
]
exp
(
−
−
c
t
)
−
E
2
ω
2
−
c
sin
(
ω
t
)
pour
t
>
0
{\displaystyle \;y(t)=\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\;{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}+y_{0}\,{\sqrt {-c}}}{2\,{\sqrt {-c}}}}\right]\exp \!\left({\sqrt {-c}}\,t\right)-\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\;{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}-y_{0}\,{\sqrt {-c}}}{2\,{\sqrt {-c}}}}\right]\exp \!\left(-{\sqrt {-c}}\,t\right)-{\dfrac {E\,{\sqrt {2}}}{\omega ^{2}-c}}\,\sin(\omega \,t)\;\;{\text{pour}}\;t>0\;}
ème exemple : équation différentielle linéaire à cœfficients réels constants du 2ème ordre en y(t) hétérogène à excitation constante discontinue de 2ème espèce[ modifier | modifier le wikicode ]
Soit à résoudre, dans le domaine des fonctions à support positif[ 14] Soit à résoudre, l'équation différentielle linéaire à cœfficients réels constants du 2ème ordre en
y
(
t
)
{\displaystyle \;y(t)\;}
Soit à résoudre, l'équation différentielle linéaire à « excitation
e
(
t
)
=
E
Y
(
t
)
+
E
′
δ
(
t
)
{\displaystyle \;e(t)=E\;Y(t)+E'\;\delta (t)}
(
E
,
E
′
)
∈
R
∗
2
{\displaystyle \;\left(E\,,\,E'\right)\in {\mathbb {R} ^{*}}^{2}\;}
[ 156] ème espèce en
t
=
0
{\displaystyle \;t=0\;}
[ 148] Soit à résoudre, l'équation différentielle linéaire avec les « C.I[ 154]
y
(
0
−
)
=
0
{\displaystyle \;y(0^{-})=0\;}
y
˙
(
0
−
)
=
0
{\displaystyle \;{\dot {y}}(0^{-})=0\;}
[ 177] Soit à résoudre, l'équation différentielle linéaire à cœfficients réels constants du 2ème ordre en
y
(
t
)
{\displaystyle \;y(t)\;}
ème espèce en
t
=
0
{\displaystyle \;t=0\;}
[ 148]
«
y
¨
(
t
)
+
b
y
˙
(
t
)
+
c
y
(
t
)
=
e
(
t
)
=
E
Y
(
t
)
+
E
′
δ
(
t
)
{\displaystyle \;{\ddot {y}}(t)+b\;{\dot {y}}(t)+c\;y(t)=e(t)=E\;Y(t)+E'\;\delta (t)\;}
[ 178]
(
b
,
c
)
∈
(
R
+
∗
)
2
{\displaystyle \;\left(b\,,\,c\right)\in \left(\mathbb {R} _{+}^{*}\right)^{2}\;}
[ 82] , [ 156]
c
=
ω
0
2
{\displaystyle \;c=\omega _{0}^{2}\;}
ω
0
∈
R
+
∗
{\displaystyle \;\omega _{0}\in \mathbb {R} _{+}^{*}\;}
[ 82]
b
=
2
σ
ω
0
{\displaystyle \;b=2\;\sigma \;\omega _{0}\;}
σ
∈
R
+
∗
{\displaystyle \;\sigma \in \mathbb {R} _{+}^{*}\;}
[ 82]
y
¨
(
t
)
+
2
σ
ω
0
y
˙
(
t
)
+
ω
0
2
y
(
t
)
=
e
(
t
)
=
E
Y
(
t
)
+
E
′
δ
(
t
)
{\displaystyle \;{\ddot {y}}(t)+2\;\sigma \;\omega _{0}\;{\dot {y}}(t)+\omega _{0}^{2}\;y(t)=e(t)=E\;Y(t)+E'\;\delta (t)\;}
La « transformée
(
{\displaystyle \;{\big (}}
)
{\displaystyle {\big )}\;}
[ 1]
e
(
t
)
=
E
Y
(
t
)
+
E
′
δ
(
t
)
{\displaystyle \;e(t)=E\;Y(t)+E'\;\delta (t)\;}
L
{
e
(
t
)
}
=
E
L
{
Y
(
t
)
}
+
E
′
L
{
δ
(
t
)
}
{\displaystyle \;{\mathcal {L}}\left\lbrace e(t)\right\rbrace =E\;{\mathcal {L}}\left\lbrace Y(t)\right\rbrace +E'\;{\mathcal {L}}\left\lbrace \delta (t)\right\rbrace \;}
[ 107]
=
E
p
+
E
′
{\displaystyle ={\dfrac {E}{p}}+E'\;}
[ 67]
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 179] La « transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
nous obtenons, en utilisant le « rappel des propriétés utilisées » énoncé plus haut dans ce chapitre ainsi que La « transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
le caractère divergent en
t
=
0
{\displaystyle \;t=0\;}
y
¨
(
t
)
{\displaystyle \;{\ddot {y}}(t)\;}
[ 178] La « transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
en notant «
L
{
y
(
t
)
}
=
Y
_
(
p
)
{\displaystyle \;{\mathcal {L}}\left\lbrace y(t)\right\rbrace ={\underline {Y}}(p)\;}
[ 157] La « transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
«
[
p
2
Y
_
(
p
)
−
p
y
(
0
−
)
−
y
˙
(
0
−
)
]
+
b
[
p
Y
_
(
p
)
−
y
(
0
−
)
]
+
c
Y
_
(
p
)
=
E
p
+
E
′
{\displaystyle \;\left[p^{2}\;{\underline {Y}}(p){\cancel {-p\;y(0^{-})-{\dot {y}}(0^{-})}}\right]+b\,\left[p\;{\underline {Y}}(p){\cancel {-y(0^{-})}}\right]+c\;{\underline {Y}}(p)={\dfrac {E}{p}}+E'\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 157] , [ 180] La « transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
«
p
2
Y
_
(
p
)
+
b
p
Y
_
(
p
)
+
c
Y
_
(
p
)
=
E
p
+
E
′
{\displaystyle \;p^{2}\;{\underline {Y}}(p)+b\;p\;{\underline {Y}}(p)+c\;{\underline {Y}}(p)={\dfrac {E}{p}}+E'\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
La « transformée
(
{\displaystyle \;\color {transparent}{\big (}}
)
{\displaystyle \color {transparent}{\big )}\;}
«
p
2
Y
_
(
p
)
+
2
σ
ω
0
p
Y
_
(
p
)
+
ω
0
2
Y
_
(
p
)
=
E
p
+
E
′
{\displaystyle \;p^{2}\;{\underline {Y}}(p)+2\;\sigma \;\omega _{0}\;p\;{\underline {Y}}(p)+\omega _{0}^{2}\;{\underline {Y}}(p)={\dfrac {E}{p}}+E'\;}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
De l'équation précédente on tire aisément «
Y
_
(
p
)
=
E
p
(
p
2
+
b
p
+
c
)
+
E
′
p
2
+
b
p
+
c
{\displaystyle \;{\underline {Y}}(p)={\dfrac {E}{p\;(p^{2}+b\,p+c)}}+{\dfrac {E'}{p^{2}+b\,p+c}}\;}
[ 157]
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
p
2
+
b
p
+
c
{\displaystyle \;p^{2}+b\,p+c\;}
≠
0
{\displaystyle \;\neq 0\;}
De l'équation précédente on tire aisément «
Y
_
(
p
)
=
E
p
(
p
2
+
2
σ
ω
0
p
+
ω
0
2
)
+
E
′
p
2
+
2
σ
ω
0
p
+
ω
0
2
{\displaystyle \;{\underline {Y}}(p)={\dfrac {E}{p\;(p^{2}+2\,\sigma \,\omega _{0}\,p+\omega _{0}^{2})}}+{\dfrac {E'}{p^{2}+2\,\sigma \,\omega _{0}\,p+\omega _{0}^{2}}}\;}
[ 157]
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
p
2
+
2
σ
ω
0
p
+
ω
0
2
{\displaystyle \;p^{2}+2\,\sigma \,\omega _{0}\,p+\omega _{0}^{2}\;}
≠
0
{\displaystyle \;\neq 0\;}
Il reste donc à déterminer l'« originale de
Y
_
(
p
)
{\displaystyle \;{\underline {Y}}(p)\;}
[ 157]
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
p
2
+
b
p
+
c
≠
0
{\displaystyle \;p^{2}+b\,p+c\neq 0\;}
[
{\displaystyle \;{\big [}}
ℜ
(
p
)
>
0
{\displaystyle \;\Re (p)>0\;}
[ 181]
]
{\displaystyle {\big ]}}
ère fonction rationnelle en éléments irréductibles simples dans le cas où la grandeur b2 - 4 c est > 0[ modifier | modifier le wikicode ] Si
b
2
−
4
c
{\displaystyle \;b^{2}-4\;c\;}
>
0
{\displaystyle \;>0}
p
2
+
b
p
+
c
{\displaystyle \;p^{2}+b\;p+c\;}
q
(
±
)
=
−
b
±
b
2
−
4
c
2
{\displaystyle \;q_{(\pm )}={\dfrac {-b\pm {\sqrt {b^{2}-4\;c}}}{2}}\;}
Si
b
2
−
4
c
{\displaystyle \;\color {transparent}{b^{2}-4\;c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
la décomposition de
A
(
p
)
=
1
p
(
p
2
+
b
p
+
c
)
=
1
p
(
p
−
q
(
+
)
)
(
p
−
q
(
−
)
)
{\displaystyle \;A(p)={\dfrac {1}{p\,(p^{2}+b\,p+c)}}={\dfrac {1}{p\;(p-q_{(+)})\;(p-q_{(-)})}}\;}
A
(
p
)
=
ρ
1
p
+
ρ
(
+
)
p
−
q
(
+
)
+
ρ
(
−
)
p
−
q
(
−
)
{\displaystyle \;A(p)={\dfrac {\rho _{1}}{p}}+{\dfrac {\rho _{(+)}}{p-q_{(+)}}}+{\dfrac {\rho _{(-)}}{p-q_{(-)}}}\;}
Si
b
2
−
4
c
{\displaystyle \;\color {transparent}{b^{2}-4\;c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
on détermine
∙
{\displaystyle \bullet \;}
ρ
1
{\displaystyle \;\rho _{1}\;}
p
{\displaystyle \;p\;}
p
=
0
{\displaystyle \;p=0\;}
[ 104]
1
c
=
ρ
1
{\displaystyle \;{\dfrac {1}{c}}=\rho _{1}\;}
ρ
1
=
1
c
{\displaystyle \;\rho _{1}={\dfrac {1}{c}}\;}
Si
b
2
−
4
c
{\displaystyle \;\color {transparent}{b^{2}-4\;c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
∙
{\displaystyle \bullet \;}
ρ
(
+
)
{\displaystyle \;\rho _{(+)}\;}
p
−
q
(
+
)
{\displaystyle \;p-q_{(+)}\;}
p
=
q
(
+
)
{\displaystyle \;p=q_{(+)}\;}
[ 104]
1
q
(
+
)
[
q
(
+
)
−
q
(
−
)
]
=
ρ
(
+
)
{\displaystyle \;{\dfrac {1}{q_{(+)}\,[q_{(+)}-q_{(-)}]}}=\rho _{(+)}\;}
1
q
(
+
)
b
2
−
4
c
=
ρ
(
+
)
{\displaystyle \;{\dfrac {1}{q_{(+)}\;{\sqrt {b^{2}-4\;c}}}}=\rho _{(+)}\;}
[ 182] Si
b
2
−
4
c
{\displaystyle \;\color {transparent}{b^{2}-4\;c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
ρ
(
+
)
=
1
b
2
−
4
c
1
q
(
+
)
{\displaystyle \;\rho _{(+)}={\dfrac {1}{\sqrt {b^{2}-4\;c}}}\;{\dfrac {1}{q_{(+)}}}\;}
q
(
+
)
=
−
b
+
b
2
−
4
c
2
{\displaystyle \;q_{(+)}={\dfrac {-b+{\sqrt {b^{2}-4\;c}}}{2}}\;}
⇒
{\displaystyle \Rightarrow }
ρ
(
+
)
=
2
b
2
−
4
c
(
−
b
+
b
2
−
4
c
)
{\displaystyle \;\rho _{(+)}={\dfrac {2}{{\sqrt {b^{2}-4\;c}}\left(-b+{\sqrt {b^{2}-4\;c}}\right)}}\;}
Si
b
2
−
4
c
{\displaystyle \;\color {transparent}{b^{2}-4\;c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
∙
{\displaystyle \bullet \;}
ρ
(
−
)
{\displaystyle \;\rho _{(-)}\;}
p
−
q
(
−
)
{\displaystyle \;p-q_{(-)}\;}
p
=
q
(
−
)
{\displaystyle \;p=q_{(-)}\;}
[ 104]
1
q
(
−
)
[
q
(
−
)
−
q
(
+
)
]
=
ρ
(
−
)
{\displaystyle \;{\dfrac {1}{q_{(-)}\,[q_{(-)}-q_{(+)}]}}=\rho _{(-)}\;}
−
1
q
(
−
)
b
2
−
4
c
=
ρ
(
−
)
{\displaystyle \;{\dfrac {-1}{q_{(-)}\;{\sqrt {b^{2}-4\;c}}}}=\rho _{(-)}\;}
[ 182] Si
b
2
−
4
c
{\displaystyle \;\color {transparent}{b^{2}-4\;c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
∙
{\displaystyle \color {transparent}{\bullet }\;}
«
ρ
(
−
)
=
1
b
2
−
4
c
−
1
q
(
−
)
{\displaystyle \;\rho _{(-)}={\dfrac {1}{\sqrt {b^{2}-4\;c}}}\;{\dfrac {-1}{q_{(-)}}}\;}
q
(
−
)
=
−
b
−
b
2
−
4
c
2
{\displaystyle \;q_{(-)}={\dfrac {-b-{\sqrt {b^{2}-4\;c}}}{2}}\;}
⇒
{\displaystyle \Rightarrow }
ρ
(
−
)
=
2
b
2
−
4
c
(
b
+
b
2
−
4
c
)
{\displaystyle \;\rho _{(-)}={\dfrac {2}{{\sqrt {b^{2}-4\;c}}\left(b+{\sqrt {b^{2}-4\;c}}\right)}}}
Si
b
2
−
4
c
{\displaystyle \;\color {transparent}{b^{2}-4\;c}\;}
>
0
{\displaystyle \;\color {transparent}{>0}}
soit finalement la décomposition de
A
(
p
)
{\displaystyle \;A(p)\;}
b
2
−
4
c
{\displaystyle \;b^{2}-4\;c\;}
>
0
{\displaystyle \;>0\;}
![{\displaystyle \;\left]0\,;\,t_{0}\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65b2d44f989e9f1160dd0718b3c67f66047cb7a7)
![{\displaystyle \;\exists \;\gamma \in \left]0\,;1\right[\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e0d11ddac42795119352f596e2b554337606479)
![{\displaystyle \;\lim \limits _{t\,\rightarrow \,0}\left[t^{\gamma }\;\vert f(t)\vert \right]=0\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a27dc0eb564a89e617b8c7b0f0f7aec36cded4c)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace Y(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;Y(t)\;dt=\displaystyle \int _{0^{+}}^{+\infty }\exp(-p\,t)\;dt=\left[{\dfrac {\exp(-p\,t)}{-p}}\right]_{0^{+}}^{+\infty }={\dfrac {1}{p}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4cffc30cc6ed3a08e581dcfd815ee436a8eb60a6)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace \delta (t)\right\rbrace ={\cancel {\left[\exp(-p\,t)\;Y(t)\right]_{0^{-}}^{+\infty }}}-\displaystyle \int _{0^{-}}^{+\infty }-p\,\exp(-p\,t)\;Y(t)\;dt\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9da40158a8b22ec33d19f824d0484f7117ab6354)
![{\displaystyle \;=p\left[{\dfrac {\exp(-p\,t)}{-p}}\right]_{0^{+}}^{+\infty }=p\;{\dfrac {1}{p}}=1\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4a824a5d79119317e53dba9cbb40becc8b6765b)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{rampe}}(t)\right\rbrace ={\cancel {\left[{\dfrac {\exp(-p\,t)}{-p}}\;a\;t\right]_{0^{-}}^{+\infty }}}-\displaystyle \int _{0^{-}}^{+\infty }{\dfrac {\exp(-p\,t)}{-p}}\;a\;dt\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc392370e4b1c8e05336696ad99e54612ee52e55)
![{\displaystyle \;={\dfrac {a}{p}}\;\left[{\dfrac {\exp(-p\,t)}{-p}}\right]_{0^{-}}^{+\infty }={\dfrac {a}{p^{2}}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df55663dcdab0f4a2d0fd8fa8f02032bb73f133a)
![{\displaystyle \;{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace =\left[\exp(-p\,t)\;f(t)\right]_{0^{-}}^{\infty }-\displaystyle \int _{0^{-}}^{+\infty }-p\;\exp(-p\,t)\;f(t)\;dt\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/37a31194fac4ac2786f4d9269c1b6a3ce9d217d0)
![{\displaystyle \;{\mathcal {L}}\left\lbrace {\dot {f}}(t)\right\rbrace =\left[\exp(-p\,t)\;f(t)\right]_{0^{+}}^{\infty }-\displaystyle \int _{0^{+}}^{+\infty }-p\;\exp(-p\,t)\;f(t)\;dt\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/463b88a19411e8b6877ded06e2046f3c17c79e6a)
![{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{t}f(t')\;dt'\right\rbrace =\left[{\dfrac {\exp(-p\,t)}{-p}}\;\displaystyle \int _{0}^{t}f(t')\;dt'\right]_{0^{-}}^{\infty }-\displaystyle \int _{0^{-}}^{+\infty }{\dfrac {\exp(-p\,t)}{-p}}\;f(t)\;dt\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3952763e0ce88ca35f6cd89a0bdd91db75725bf9)
![{\displaystyle =\displaystyle \int _{t_{0}^{-}}^{+\infty }\exp \!\left[-p\,(t'+t_{0})\right]\,f(t')\;dt'\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7896983e2e8a2ad648f61412958d8c0adaaa6f5d)
![{\displaystyle {\big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31d63350d68291e3f56604a089e0e0b5a86c2472)
![{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace ={\mathcal {L}}\left\lbrace f(t-t_{0})\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t-t_{0})\;dt=\displaystyle \int _{-t_{0}^{-}}^{+\infty }\exp \!\left[-p\,(u+t_{0})\right]\,f(u)\;du\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5fdb35c1bbdb3e07601c06f0f9e3c749eadae631)
![{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\displaystyle \int _{-t_{0}^{-}}^{0^{-}}\exp \!\left[-p\,(u+t_{0})\right]\,f(u)\;du+\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-p\,(u+t_{0})\right]\,f(u)\;du\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd3f80f5b12b64d36c7e1de45bbdedddab6b1ba1)
![{\displaystyle =\exp(-p\,t_{0})\left[\displaystyle \int _{t_{0}^{-}}^{0^{-}}\exp \!\left(-p\;u\right)\,f(u)\;du+\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left(-p\;u\right)\,f(u)\;du\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bce2dbc6862681edef69aa2583ec59593c5e4d4b)
![{\displaystyle {\big ]}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6409e9848c1422eef16462fbeae301d8fa0f831)
![{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-p\,(u+t_{0})\right]\,f(u)\;du-\displaystyle \int _{0^{-}}^{-t_{0}^{-}}\exp \!\left[-p\,(u+t_{0})\right]\,f(u)\;du\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c94a60dc938f37a4ac14baadec9cfb55e4bdeff)
![{\displaystyle =\exp(-p\,t_{0})\left[\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left(-p\;u\right)\,f(u)\;du-\displaystyle \int _{0^{-}}^{t_{0}^{-}}\exp \!\left(-p\;u\right)\,f(u)\;du\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/857409e534769a7a4cd910985495ba01a3597724)
![{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace f(k\;t)\right\rbrace \!\left[p\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aef1550de3409eb2878b6b237dc31f472e6781ca)
![{\displaystyle ={\dfrac {1}{k}}\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \!\left[{\dfrac {p}{k}}\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/914c5f12e3f77e5f94d4e5f7efcc152894bac95a)
![{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(k\;t)\;dt=\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-{\dfrac {p}{k}}\;k\;t\right]\,f(k\;t)\;{\dfrac {d(k\;t)}{k}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cced8d0b204f405d9811abf9c21fdfa76cfd5810)
![{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace f(k\;t)\right\rbrace \!\left[p\right]}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09d70169480e24d86ca68c171cb54b891d944530)
![{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/957b5fc091364244677497166128c953de5cb96a)
![{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/119b13847635faeb2c727682a24303005e0f4e1f)
![{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace f(t')\right\rbrace \!\left[p\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d5d1206e156f758279925086f6bc798d958ff95)
![{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-{\dfrac {p}{k}}\;t'\right]\,f(t')\;{\dfrac {dt'}{k}}={\dfrac {1}{k}}\,\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-{\dfrac {p}{k}}\;t'\right]\,f(t')\;dt'\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48f83c2e00b838daf8a387f2398a9ae185f09345)
![{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace f(t')\right\rbrace \!\left[p\right]}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c715ad70bbe4a841d5c56466b2f7a9ee8a67aa1c)
![{\displaystyle \color {transparent}{=\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-{\dfrac {p}{k}}\;t'\right]\,f(t')\;{\dfrac {dt'}{k}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1811e32595755ae92a06421447a4c2bab8293fa)
![{\displaystyle ={\dfrac {1}{k}}\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \!\left[p'\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c2e7a2a8df13acea3d4decb81170cf26e30ab4b)
![{\displaystyle \color {transparent}{={\dfrac {1}{k}}\;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \!\left[p'\right]}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f412a083e39adc82b0a8c36bb572be91e3053701)
![{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp \!\left[-(p+a)\,t\right]\,f(t)\;dt\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3aa9553f59ab1a775e361ca79f9e06c6afcbf57)
![{\displaystyle \;{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace \exp(-a\;t)\;f(t)\right\rbrace \!\left[p\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af7e105ec73f809ee7d3c899f170fa5791363c0f)
![{\displaystyle ={\mathcal {L}}\left\lbrace f(t)\right\rbrace \!\left[p+a\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/208d7641599bc92464ffb670cfc1c6255e30b4dd)
![{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f'(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace \exp(-a\;t)\;f(t)\right\rbrace \!\left[p\right]}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bd8d68889836703304a465da58dea271856d8b3)
![{\displaystyle \color {transparent}{=\exp \!\left[-(p+a)\,t\right]\,f(t)\;dt}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0929c6ac84db5fd2972f9bdbf931758da7de18a4)
![{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \!\left[p'\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd12f708bfda058d5015d3a5f2d610bdec4423fa)
![{\displaystyle \;\exp \!\left(-{\underline {a}}\;t\right)=\exp \!\left[-\Re ({\underline {a}})\;t\right]\,\exp \!\left[-i\,\Im ({\underline {a}})\;t\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bdf5815659e283b597e77344f3c43589af07fbe)
![{\displaystyle \;\color {transparent}{\exp \!\left(-{\underline {a}}\;t\right)=\exp \!\left[-\Re ({\underline {a}})\;t\right]\,\exp \!\left[-i\,\Im ({\underline {a}})\;t\right]}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d7497c55e11f6c0219214c783071119a12b22b0)
![{\displaystyle \;\left\lbrace {\begin{array}{r}\Re \left[\exp \!\left(-{\underline {a}}\;t\right)\right]\!\!&=&\!\!\exp \!\left[-\Re ({\underline {a}})\;t\right]\;\cos \!\left[\Im ({\underline {a}})\;t\right]\\\Im \left[\exp \!\left(-{\underline {a}}\;t\right)\right]\!\!&=&\!\!-\exp \!\left[-\Re ({\underline {a}})\;t\right]\;\sin \!\left[\Im ({\underline {a}})\;t\right]\end{array}}\right\rbrace }](https://wikimedia.org/api/rest_v1/media/math/render/svg/7bc8f04c53cb3c53d9438789a04e0298b634a290)
![{\displaystyle \;\left\lbrace {\begin{array}{r}\Re \left[{\underline {f'}}(t)\right]=\Re \left[\exp \!\left(-{\underline {a}}\;t\right)\;f(t)\right]\!\!&=&\!\!\exp \!\left[-\Re ({\underline {a}})\;t\right]\;\cos \!\left[\Im ({\underline {a}})\;t\right]\;f(t)\\\Im \left[{\underline {f'}}(t)\right]=\Im \left[\exp \!\left(-{\underline {a}}\;t\right)\;f(t)\right]\!\!&=&\!\!-\exp \!\left[-\Re ({\underline {a}})\;t\right]\;\sin \!\left[\Im ({\underline {a}})\;t\right]\;f(t)\end{array}}\right\rbrace }](https://wikimedia.org/api/rest_v1/media/math/render/svg/27c7aea4051ae0dd2e9f40d8c0446c3a2fe7f93f)
![{\displaystyle \;{\Bigg \langle }{\begin{array}{c}{\mathcal {L}}\left\lbrace \Re \!\left[{\underline {f'}}(t)\right]\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;\Re \!\left[{\underline {f'}}(t)\right]\,dt\\{\mathcal {L}}\left\lbrace \Im \!\left[{\underline {f'}}(t)\right]\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;\Im \!\left[{\underline {f'}}(t)\right]\,dt\end{array}}{\Bigg \rangle }\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e15a6ded1f4b62675405214136923de42fb127e)
![{\displaystyle \;{\underline {\mathcal {L}}}\left\lbrace {\underline {f'}}(t)\right\rbrace ={\mathcal {L}}\left\lbrace \Re \!\left[{\underline {f'}}(t)\right]\right\rbrace +i\;{\mathcal {L}}\left\lbrace \Im \!\left[{\underline {f'}}(t)\right]\right\rbrace \;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ed3c6acdf060150cc3942ff95a0c4e332c8d7dc)
![{\displaystyle \;{\underline {\mathcal {L}}}\left\lbrace {\underline {f'}}(t)\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;\Re \!\left[{\underline {f'}}(t)\right]\,dt+i\;\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;\Im \!\left[{\underline {f'}}(t)\right]\,dt=\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;{\underline {f'}}(t)\;dt\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d367cac85e1615610a50883cf2f412abdd0de8fe)
![{\displaystyle \;{\underline {\mathcal {L}}}\left\lbrace {\underline {f'}}(t)\right\rbrace \!\left[p\right]={\underline {\mathcal {L}}}\left\lbrace \exp(-{\underline {a}}\;t)\;f(t)\right\rbrace \!\left[p\right]={\mathcal {L}}\left\lbrace f(t)\right\rbrace \!\left[p+{\underline {a}}\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e5178dd7b101184b94feb64fe56f7e04e234f08)
![{\displaystyle \;{\underline {z}}{\big ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/66aaf0bc112fa9d1439107a80bc4743173323f4c)
![{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \left[p\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2cfaf6e89ee2a8d4fac51f110ce6b6ad4ff6dfb9)
![{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace \left[p\right]}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/465e53ef0ca93820d1c2c30e98b5ebd7b93267c3)
![{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace \left[p\right]=\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;f(t)\;dt\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/95e3021e2fd7f97f7bb09f441bb46942eb5c7f00)
![{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace f(t)\right\rbrace \left[p\right]=\exp(-p\;t)\;f(t)\;dt}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b1de7e2f7492582b63f918bbf37ac40149bff5a)
![{\displaystyle \;{\dfrac {d{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp}}={\dfrac {d\left[\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;f(t)\;dt\right]}{dp}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9e0777e96b1a72ef28fc627182704fb067de944)
![{\displaystyle \;{\dfrac {d^{(n+1)}{\mathcal {L}}\left\lbrace f(t)\right\rbrace }{dp^{(n+1)}}}={\dfrac {d\left[(-1)^{n}\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\;t)\;t^{n}\;f(t)\;dt\right]}{dp}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e406bacf93cf8d4336bb10ef1bd493c948ef01f)
![{\displaystyle \;{\mathcal {L}}\left\lbrace t^{(n+1)}\;Y(t)\right\rbrace ={\mathcal {L}}\left\lbrace t\,\left[t^{n}\;Y(t)\right]\right\rbrace =-{\dfrac {d{\mathcal {L}}\left\lbrace t^{n}\;Y(t)\right\rbrace }{dp}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25e5a67d4a56ac7de70b1775101bb2dbeb91cf00)
![{\displaystyle =-{\dfrac {d\!\left[{\dfrac {n!}{p^{(n+1)}}}\right]}{dp}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44527acc32a4e1920046fada986d0bd7282304d6)
![{\displaystyle \;{\mathcal {L}}\left\lbrace \left({\dfrac {\partial f}{\partial m}}\right)_{\!t}(t,\,m)\right\rbrace =\left({\dfrac {\partial \left[\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t,\,m)\;dt\right]}{\partial m}}\right)_{\!p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd806f72ce1e54a3021beb45053c526a0bb03bd5)
![{\displaystyle =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left[\displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right]\;dt\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1423d6ccde4888f63fdf929a532ac643730ca7c)
![{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left[\displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right]\,dt\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86ca5fe6180c2dcc9c3406e85e407f6ef8829449)
![{\displaystyle \;\color {transparent}{{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right\rbrace =\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;\left[\displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right]\,dt}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa865becc4a47005c49406c6b0c1fb567a49a3b3)
![{\displaystyle \;{\mathcal {L}}\left\lbrace \displaystyle \int _{0}^{m}f(t,\,m')\;dm'\right\rbrace =\displaystyle \int _{0}^{m}\left[\displaystyle \int _{0^{-}}^{+\infty }\exp(-p\,t)\;f(t,\,m')\;dt\right]\,dm'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/332592e08c6a238b9d7aa40fa082172961570459)
![{\displaystyle \;\left[0\,,\,T\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e72ca9d8af3f23fc21a660ab248da207e3ad3cf6)
![{\displaystyle \;{\mathcal {L}}\left\lbrace f(t)\right\rbrace =\left[1+\sum \limits _{n\,\in \,\mathbb {N} ^{*}}^{1\,..\,+\infty }\exp(-p\,n\;T)\right]\displaystyle \int _{0}^{T}\exp(-p\,t)\;f(t)\;dt\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94e7c64eeb39e1b74316fce2f2f42f5b2d89eff3)
![{\displaystyle \;\left[0\,,\,\tau \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/46630ecbef65fdc0b7ab22c7a6b2f5e98c3edb62)
![{\displaystyle \;C_{\left[0\,,\,\tau \right]}(t)=\left\lbrace {\begin{array}{l}0\;\;{\text{si}}\;t>\tau \\1\;\;{\text{si}}\;0<t<\tau \\0\;\;{\text{si}}\;t<0\end{array}}\right\rbrace =Y(t)-Y(t-\tau )\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69e5e7382197613e0dea5942ea638740982cfa79)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace C_{\left[0\,,\,\tau \right]}(t)\right\rbrace ={\mathcal {L}}\!\left\lbrace Y(t)\right\rbrace -{\mathcal {L}}\!\left\lbrace Y(t-\tau )\right\rbrace \;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0de18ba93c9455ac99977e53c5f4ced7657bf16)
![{\displaystyle \;\color {transparent}{\left[0\,,\,\tau \right]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc24fb4dd6571206f7b4c2a2a1027006b885a892)
![{\displaystyle \;\color {transparent}{C_{\left[0\,,\,\tau \right]}(t)=\left\lbrace 1\;\;{\text{si}}\;0<t<\tau \right\rbrace =Y(t)-Y(t-\tau )}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d96acb828018950c58ad9fa8377e06ed582e189)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace C_{\left[0\,,\,\tau \right]}(t)\right\rbrace ={\dfrac {1-\exp(-p\,\tau )}{p}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7cc848a6c850cc2099a6268caddcfcf2b8f8046b)
![{\displaystyle \;\left[0\,,\,2\,\tau \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b5b9a4cdd43877e53afc23d8c2fe5223fb918fa)
![{\displaystyle \;f_{\text{triang sym}}(t)=\left\lbrace {\begin{array}{l}0\qquad \qquad \quad {\text{si}}\;t\geqslant 2\,\tau \\-a\;(t-2\,\tau )\;{\text{si}}\;t\in \left[\tau \,,\,2\,\tau \right]\\a\;t\qquad \qquad \;{\text{si}}\;t\in \left[0\,,\,\tau \right]\\0\qquad \qquad \quad {\text{si}}\;t\leqslant 0\end{array}}\right\rbrace =f_{\text{rampe}}(t)-2\,f_{\text{rampe}}(t-\tau )+f_{\text{rampe}}(t-2\,\tau )\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c33fc44a34eb6811f91d63a7739469e4b01a252)
![{\displaystyle \;\color {transparent}{\left[0\,,\,2\,\tau \right]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/215d274fa2bb4091b9edca074828d3a77355612d)
![{\displaystyle \;\color {transparent}{f_{\text{triang sym}}(t)=\left\lbrace -a\;(t-2\,\tau )\;{\text{si}}\;t\in \left[\tau \,,\,2\,\tau \right]\right\rbrace =}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f58780f2a874c3a3291231299709f4e7cb3ac23)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{triang sym}}(t)\right\rbrace ={\dfrac {a\,\left[1-2\,\exp(-p\,\tau )+\exp(-2\,p\,\tau )\right]}{p^{2}}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77fcefdbc34003214cad53143dc030a25587c442)
![{\displaystyle ={\dfrac {a\,\left[1-\exp(-p\,\tau )\right]^{2}}{p^{2}}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5764412338200a01326f2c255a46c10914ce7757)
![{\displaystyle \;f_{\text{app exp}}(t)=\left[1-\exp(-a\,t)\right]\,Y(t)=Y(t)-\exp(-a\,t)\;Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16f736c5f54c49c2673cac4c0c29bf8e7bd16aa8)
![{\displaystyle \;\color {transparent}{f_{\text{app exp}}(t)=\left[1-\exp(-a\,t)\right]\,Y(t)=Y(t)-\exp(-a\,t)\;Y(t)}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e39274ab9c560a648b7489eeec9303909c67267f)
![{\displaystyle \;f_{\text{cos}}(t)=\Re \left[\exp(i\,\omega \,t)\;Y(t)\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3ca78a419c34140e0b21d0721049d5cddcb9995)
![{\displaystyle \;\Re \left\lbrace {\underline {\mathcal {L}}}\left[\exp(i\,\omega \,t)\;Y(t)\right]\right\rbrace ={\mathcal {L}}\left\lbrace \Re \left[\exp(i\,\omega \,t)\;Y(t)\right]\right\rbrace \;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/500afc76069329658d795b1d2d6a29391794f43a)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos}}(t)\right\rbrace =\Re \,\left[{\dfrac {1}{p-i\,\omega }}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,0}=\Re \,\left[{\dfrac {p+i\,\omega }{p^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,0}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eee337284e4289f06c0a68ed3a073939b6268179)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos}}(t)\right\rbrace =\left[{\dfrac {p}{p^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,0}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0f6adcafba1df9ff0932e22a7ceb4d4eb396784)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos}}(t)\right\rbrace ={\dfrac {\displaystyle \int _{0}^{T}\left\lbrace \exp \!\left[-(p-i\;\omega )\;t\right]+\exp \!\left[-(p+i\;\omega )\;t\right]\right\rbrace \,dt}{2\,\left[1-\exp(-p\,T)\right]}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c410f1d5d4f1cd014b924715fc69e62e28f9073)
![{\displaystyle \displaystyle \int _{0}^{T}\exp \!\left[-(p-i\;\omega )\;t\right]\;dt=\left[-{\dfrac {\exp \!\left[-(p-i\;\omega )\;t\right]}{p-i\;\omega }}\right]_{0}^{T}={\dfrac {1-\exp(-p\;T)\;\exp(i\;\omega \;T)}{p-i\;\omega }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4ed74a7f094e21bcd79d24d88866b158d9f4fd8)
![{\displaystyle \color {transparent}{\displaystyle \int _{0}^{T}\exp \!\left[-(p-i\;\omega )\;t\right]\;dt}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d2ea990e7f0f594c34b401fae72099bca5eee53)
![{\displaystyle \displaystyle \int _{0}^{T}\exp \!\left[-(p+i\;\omega )\;t\right]\;dt=\left[-{\dfrac {\exp \!\left[-(p+i\;\omega )\;t\right]}{p+i\;\omega }}\right]_{0}^{T}={\dfrac {1-\exp(-p\;T)\;\exp(-i\;\omega \;T)}{p+i\;\omega }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/589f6b768f166a63a57a1355325caefc73d6c1bd)
![{\displaystyle \color {transparent}{\displaystyle \int _{0}^{T}\exp \!\left[-(p+i\;\omega )\;t\right]\;dt}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad79524a99ac36f6af37497d904d2a9b8381178a)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos}}(t)\right\rbrace ={\dfrac {1}{2}}\,\left[{\dfrac {1}{p-i\;\omega }}+{\dfrac {1}{p+i\;\omega }}\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b27deac3f67111e3e5d7221806acfdca20746174)
![{\displaystyle \;f_{\text{sin}}(t)=\Im \left[\exp(i\,\omega \,t)\;Y(t)\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42d9a5106150ce5a5303001b6f998c52854bc808)
![{\displaystyle \;\Im \left\lbrace {\underline {\mathcal {L}}}\left[\exp(i\,\omega \,t)\;Y(t)\right]\right\rbrace ={\mathcal {L}}\left\lbrace \Im \left[\exp(i\,\omega \,t)\;Y(t)\right]\right\rbrace \;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f705620575b959caa7a64282eb521c754631a4ea)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin}}(t)\right\rbrace =\Im \,\left[{\dfrac {1}{p-i\,\omega }}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,0}=\Im \,\left[{\dfrac {p+i\,\omega }{p^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,0}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6efee43fa64c4e8e5e5e91e34526190c17ba8a93)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin}}(t)\right\rbrace =\left[{\dfrac {\omega }{p^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,0}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8dc6016916bbb245b88e0bae91a621d459cdb737)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin}}(t)\right\rbrace ={\dfrac {\displaystyle \int _{0}^{T}\left\lbrace \exp \!\left[-(p-i\;\omega )\;t\right]-\exp \!\left[-(p+i\;\omega )\;t\right]\right\rbrace \,dt}{2\;i\,\left[1-\exp(-p\,T)\right]}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3821fe03691f2d716f3ffac7fe7b529e6f611e6)
![{\displaystyle \displaystyle \int _{0}^{T}\exp \!\left[-(p-i\;\omega )\;t\right]\;dt{\overset {\ldots }{\;=\;}}\;{\dfrac {1-\exp(-p\;T)}{p-i\;\omega }}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d76ae8350ef0cf76e1be43e60cacd645a884487)
![{\displaystyle \displaystyle \int _{0}^{T}\exp \!\left[-(p+i\;\omega )\;t\right]\;dt\;{\overset {\ldots }{\;=\;}}\;{\dfrac {1-\exp(-p\;T)}{p+i\;\omega }}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a2e9192a239a223a43b26e2d8d0bb459612029f)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin}}(t)\right\rbrace ={\dfrac {1}{2\;i}}\,\left[{\dfrac {1}{p-i\;\omega }}-{\dfrac {1}{p+i\;\omega }}\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08b79b67421c6953c742e8d229bfd1cb4103799f)
![{\displaystyle \;\varphi \in \left]-\pi \,,\,+\pi \right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2dd4581a2a72689a584e4d7a3534106d965848cf)
![{\displaystyle \;f_{\text{cos amort}}(t)=\Re \left[\exp(-a\;t)\;\exp(i\,\omega \,t)\;Y(t)\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8cff1118b9a07ed4693895584b1727361c54228)
![{\displaystyle \;{\underline {\mathcal {L}}}\!\left\lbrace \exp \!\left[(-a+i\,\omega )\,t\right]\,Y(t)\right\rbrace ={\dfrac {1}{p+a-i\,\omega }}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d85fde15c496b2418ba3c5c755216d3d4af6d4db)
![{\displaystyle \;\Re {\Big \langle }{\underline {\mathcal {L}}}\!\left\lbrace \exp \!\left[(-a+i\,\omega )\,t\right]\,Y(t)\right\rbrace \!{\Big \rangle }={\mathcal {L}}{\Big \langle }\Re \left\lbrace \exp \!\left[(-a+i\,\omega )\,t\right]\,Y(t)\right\rbrace \!{\Big \rangle }\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3175998d8e1f6c93457c3f1720d6d3190d97f5b)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos amort}}(t)\right\rbrace =\Re \,\left[{\dfrac {1}{p+a-i\,\omega }}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,-a}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b21efb4a601e7fbd6b8dc88bbaf6b2b05b6db360)
![{\displaystyle =\Re \,\left[{\dfrac {p+a+i\,\omega }{(p+a)^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,-a}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/705860fae9cb0a24e7f0335666f6d7f7261787a8)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{cos amort}}(t)\right\rbrace =\left[{\dfrac {p+a}{(p+a)^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,-a}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aeaf045b3ac2214994ad0caef111a298b4b11d7e)
![{\displaystyle \;f_{\text{sin amort}}(t)=\Im \left[\exp(-a\;t)\;\exp(i\,\omega \,t)\;Y(t)\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/711e88b51f4e11db43f88435fbc6cfedd9a8f4b0)
![{\displaystyle \;{\underline {\mathcal {L}}}\!\left\lbrace \exp \!\left[(-a+i\,\omega )\,t\right]\,Y(t)\right\rbrace =={\dfrac {1}{p+a-i\,\omega }}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/003987eb5f77ba4acfcfea4f8374c155058cfece)
![{\displaystyle \;\Im {\Big \langle }{\underline {\mathcal {L}}}\!\left\lbrace \exp \!\left[(-a+i\,\omega )\,t\right]\,Y(t)\right\rbrace \!{\Big \rangle }={\mathcal {L}}{\Big \langle }\Im \left\lbrace \exp \!\left[(-a+i\,\omega )\,t\right]\,Y(t)\right\rbrace \!{\Big \rangle }\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8467b11e2bd57b1cb6b1c599ee3789b4ca8744a1)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin amort}}(t)\right\rbrace ==\Im \,\left[{\dfrac {1}{p+a-i\,\omega }}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,-a}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e87804317eeae68e22e2cce3e6152e8a33785a3e)
![{\displaystyle =\Im \,\left[{\dfrac {p+a+i\,\omega }{(p+a)^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,-a}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3802e596b3bda2fd95df4ed894c691ceb6878491)
![{\displaystyle \;{\mathcal {L}}\!\left\lbrace f_{\text{sin amort}}(t)\right\rbrace =\left[{\dfrac {\omega }{(p+a)^{2}+\omega ^{2}}}\right]_{{\text{pour }}p\,=\,\Re (p)\,>\,-a}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0607a0ffd34e3ffdd2e2ece410a3b8d25fc29efa)
![{\displaystyle \;\mathrm {deg} \left[P\right]<\mathrm {deg} \left[Q\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a24a5fa9520c5435fdc78ffa0b8fae38f571508c)
![{\displaystyle \;P(p){\big ]}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da3f94b38715be027ce070815d6c3d31e68821ba)
![{\displaystyle \;Q(p){\big ]}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd1c83d0c7d9ddfd64ba2bd2e3ea07947d321261)
![{\displaystyle \;{\mathcal {L}}^{-1}\!\left\lbrace F(p)\right\rbrace =A\left[\sum \limits _{l}\rho _{l}\;\exp(q_{l}\,t)\right]Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d9571b5e7ba917dd028fb7ce3544413f36bc451)
![{\displaystyle \left[{\dfrac {p+1}{p+3}}\right]_{\!p=-2}=\rho _{1}\;{\cancel {+\left[{\dfrac {\rho _{2}\;(p+2)}{p+3}}\right]_{\!p=-2}}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84f58a2c7ab794a025e5047249228aac2fc61a3a)
![{\displaystyle \left[{\dfrac {p+1}{p+2}}\right]_{\!p=-3}={\cancel {\left[{\dfrac {\rho _{1}\;(p+3)}{p+2}}\right]_{\!p=-3}+}}\;\rho _{2}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d466ae90ed98d28cac432374983f06590a1ee2db)
![{\displaystyle ={\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {p+1}{p^{2}+5\,p+6}}\right\rbrace =\left[-\exp(-2\,t)+2\;\exp(-3\,t)\right]\,Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43692ce8513cd33a27df84340b0195ec59985c6a)
![{\displaystyle \left[{\dfrac {p+1}{(p+2)^{2}}}\right]_{\!p=0}=\rho _{1}\;{\cancel {+\left[{\dfrac {\rho _{2}\;p}{(p+2)^{2}}}\right]_{\!p=0}+\left[{\dfrac {{\rho '}_{2}\;p}{p+2}}\right]_{\!p=0}}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/faaf78274997c1586ac6b08577ba1c2e355d15fa)
![{\displaystyle \left[{\dfrac {p+1}{p}}\right]_{\!p=-2}={\cancel {\left[{\dfrac {\rho _{1}\;(p+2)^{2}}{p}}\right]_{\!p=-2}+}}\;\rho _{2}\;{\cancel {+\left[{\rho '}_{2}\;(p+2)\right]_{p=-2}}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a6d8ee84d1834dfb9786ccf06511ff23ca925ec)
![{\displaystyle {\cancel {\left[{\dfrac {p+1}{p\,(p+2)}}\right]_{\!p\rightarrow \infty }}}\;=\left[{\dfrac {\rho _{1}\;(p+2)}{p}}\right]_{\!p\rightarrow \infty }\;{\cancel {+\left[{\dfrac {\rho _{2}}{p+2}}\right]_{\!p\rightarrow \infty }+}}\;{\rho '}_{2}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3af53a1e4056588079470c73b66b87cfdb858bd7)
![{\displaystyle ={\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {p+1}{p\,(p^{2}+4\,p+4)}}\right\rbrace ={\dfrac {1}{4}}\left[1+2\;t\;\exp(-2\,t)-\exp(-2\,t)\right]\,Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b7d4f4f9405f67fa4be22e3945974157e741a2d)
![{\displaystyle \color {transparent}{={\mathcal {L}}^{-1}\!\left\lbrace p\,(p^{2}+4\,p+4)\right\rbrace =1\left[1+2\;t\;\exp(-2\,t)-\exp(-2\,t)\right]\,Y(t)}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e171afc1c78ff3446a675f925fdb41ee3377ddec)
![{\displaystyle \left[{\dfrac {p+1}{p^{2}+p+2}}\right]_{\!p=0}=\rho _{1}\;{\cancel {+\left[{\dfrac {\left(\rho _{2}\;p+{\rho '}_{2}\right)\,p}{p^{2}+p+2}}\right]_{\!p=0}}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3a1e94a6141ab57a4d15e96eb0fea1e496b4612)
![{\displaystyle {\cancel {\left[{\dfrac {p+1}{p^{2}+p+2}}\right]_{\!p\rightarrow \infty }}}=\rho _{1}\;+\left[{\dfrac {\left(\rho _{2}\;p+{\rho '}_{2}\right)\,p}{p^{2}+p+2}}\right]_{\!p\rightarrow \infty }\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9e187880c95fb6aec9488b71960212ee29eaeb6)
![{\displaystyle {\cancel {\left[{\dfrac {p+1}{p\;(p^{2}+p+2)}}\right]_{\!p=-1}}}=\left[{\dfrac {\rho _{1}}{p}}+{\dfrac {\rho _{2}\;p+{\rho '}_{2}}{p^{2}+p+2}}\right]_{\!p=-1}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c6102760f34bc155c1ba39f0f2429fb3f853aa0)
![{\displaystyle \;\left[{\begin{array}{c}{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {p+a}{(p+a)^{2}+\omega ^{2}}}\right\rbrace =\exp(-a\,t)\;\cos(\omega \,t)\\{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {\omega }{(p+a)^{2}+\omega ^{2}}}\right\rbrace =\exp(-a\,t)\;\sin(\omega \,t)\end{array}}\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e075d1d77e946c96017576b04d5d90f31adb6a4b)
![{\displaystyle \left[{\begin{array}{c}a={\dfrac {1}{2}}\\\omega ={\dfrac {\sqrt {7}}{2}}\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3582223f810d76d781e1ae7e4e24e41c04cfe2b4)
![{\displaystyle \;f(t)={\dfrac {1}{2}}\left[{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {1}{p}}\right\rbrace -{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {p+{\dfrac {1}{2}}}{\left(p+{\dfrac {1}{2}}\right)^{\!2}+\left({\dfrac {\sqrt {7}}{2}}\right)^{\!2}}}\right\rbrace +{\dfrac {3}{\sqrt {7}}}\;{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {\dfrac {\sqrt {7}}{2}}{\left(p+{\dfrac {1}{2}}\right)^{\!2}+\left({\dfrac {\sqrt {7}}{2}}\right)^{\!2}}}\right\rbrace \right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5513181b1dc08da3c809ae9f0ab4fa8fd15a1927)
![{\displaystyle \;f(t)={\mathcal {L}}^{-1}\!\left\lbrace F(p)\right\rbrace ={\dfrac {1}{2}}\left[1-\exp \!\left(-{\dfrac {1}{2}}\;t\right)\,\cos \!\left({\dfrac {\sqrt {7}}{2}}\;t\right)+{\dfrac {3}{\sqrt {7}}}\;\exp \!\left(-{\dfrac {1}{2}}\;t\right)\,\sin \!\left({\dfrac {\sqrt {7}}{2}}\;t\right)\right]\,Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a10459bc2628c151b07c73fe53f15d4636a5c858)
![{\displaystyle \;g(t)={\dfrac {4}{\sqrt {7}}}\left[{\dfrac {\sqrt {7}}{4}}\;\cos \!\left({\dfrac {\sqrt {7}}{2}}\;t\right)-{\dfrac {3}{4}}\;\sin \!\left({\dfrac {\sqrt {7}}{2}}\;t\right)\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05d198f903c6353ed557d573be2534d8ad68b363)
![{\displaystyle \;g(t)={\dfrac {4}{\sqrt {7}}}\;\cos \!\left[{\dfrac {\sqrt {7}}{2}}\;t+\arctan \!\left({\dfrac {3}{\sqrt {7}}}\right)\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c0c339a1df9308530329d4e107523b4163472e1)
![{\displaystyle \;f(t)={\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {p+1}{p\,(p^{2}+p+2)}}\right\rbrace ={\dfrac {1}{2}}\left\lbrace 1-{\dfrac {4}{\sqrt {7}}}\;\exp \!\left(-{\dfrac {1}{2}}\;t\right)\,\cos \!\left[{\dfrac {\sqrt {7}}{2}}\;t+\arctan \!\left({\dfrac {3}{\sqrt {7}}}\right)\right]\right\rbrace \,Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bd8f4e19b95e3fbd8f127fb596d82b8c5675fb9)
![{\displaystyle \;f(t)\simeq {\dfrac {1}{2}}\left[1-1,512\;\exp \!\left(-0,5\;t\right)\,\cos \!\left(1,323\;t+0,848\right)\right]\,Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf8a0f1210b5419e52fe9e88900179da24beb114)
![{\displaystyle \;\lim \limits _{t\,\rightarrow \,0^{+}}\!\left[f(t)\right]=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8093fa889bac463ffc6ea4cb25b785248290864e)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=1=f(0^{+})\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f8c9d967623f7341114f75a4bde13f4ad003647)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=0=f(0^{+})\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/070ef58eb8cd14528c4566ac97dd18deca169688)
![{\displaystyle \;\lim \limits _{t\,\rightarrow \,0^{+}}\!\left[{\dot {f}}(t)\right]=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26f2c2518396f1090c75d1fcfa16359051fd4efc)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)\;{\cancel {-p\;f(0^{+})}}\right]_{p\,\in \,\mathbb {R} }=a={\dot {f}}(0^{+})\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f289867453900b0a57a7cb4a8255ca49a9aef67)
![{\displaystyle \color {transparent}{\big ]}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a93d1d51146785f3ced70f6522b1ea5046be592)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {p^{2}}{p+a}}-p\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {-p\;a)}{p+a}}\right]_{p\,\in \,\mathbb {R} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b225499f65c813cf68bc729cb452804fd4a2d487)
![{\displaystyle \;\color {transparent}{\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d07f417db0b180f9dfb44cf8022234758305c5f1)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {p^{3}}{p^{2}+\omega ^{2}}}-p\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {-p\;\omega ^{2}}{p^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/938ff1c960d5b4192abe80c38c8a52c09eac39aa)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)\;{\cancel {-p\;f(0^{+})}}\;\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {p^{2}\;\omega }{p^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }=\omega =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f41625022edcf6f56930965ba0a5da092db648f0)
![{\displaystyle \;{\dot {f}}(t)=\left\lbrace {\begin{array}{c l}\left[-a\;\cos(\omega \,t)-\omega \;\sin(\omega \,t)\right]\,\exp(-a\,t)\!\!&{\text{si}}\;t>0\\0\!\!&{\text{si}}\;t<0\end{array}}\right\rbrace =\left[-a\;\cos(\omega \,t)-\omega \;\sin(\omega \,t)\right]\,\exp(-a\,t)\;Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbf182ff69bef269bb4c16aedf8c5318c1023fe9)
![{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \left[-a\;\cos(\omega \,t)-\omega \;\sin(\omega \,t)\right]\,\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2de369e4b42fbb7ea0acdddc3415ae57422db3c)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {p^{2}\,(p+a)}{(p+a)^{2}+\omega ^{2}}}-p\right]_{p\,\in \,\mathbb {R} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f9460a3861ffda7b3f1901a3707534c8b223f9b)
![{\displaystyle =\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {p^{2}\,(p+a)-p\,(p+a)^{2}-p\;\omega ^{2}}{(p+a)^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d673cbe3e735ac0898219ef51578d849ef15c766)
![{\displaystyle =\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {-p^{2}\;a-p\,(a^{2}+\omega ^{2})}{(p+a)^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be4fb1e0d0d761247b45d6738052fa36279939fd)
![{\displaystyle \;{\dot {f}}(t)=\left\lbrace {\begin{array}{c l}\left[-a\;\sin(\omega \,t)+\omega \;\cos(\omega \,t)\right]\,\exp(-a\,t)\!\!&{\text{si}}\;t>0\\0\!\!&{\text{si}}\;t<0\end{array}}\right\rbrace =\left[-a\;\sin(\omega \,t)+\omega \;\cos(\omega \,t)\right]\,\exp(-a\,t)\;Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6071d1d28f520031e04d9af6ca6f2fade2c545cf)
![{\displaystyle \;\color {transparent}{{\dot {f}}(t)=\left\lbrace \left[-a\;\sin(\omega \,t)+\omega \;\cos(\omega \,t)\right]\,\exp(-a\,t)\!\!{\text{si}}\;t>0\right\rbrace }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7cccfc56d317c7379b1f9db445d3fc41ddb5c269)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p^{2}\;F(p)-p\;f(0^{+})\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[{\dfrac {p^{2}\,\omega }{(p+a)^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b22905ad25a50f16a0a5d179f7e7cc07c997cd2)
![{\displaystyle \;\lim \limits _{t\,\rightarrow \,+\infty }\!\left[f(t)\right]=\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/696ea58b00fc5334f933a2e3f3d9b57b7572bea4)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=1=\lim \limits _{t\,\rightarrow \,+\infty }\!\left[f(t)\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e11088d8807e41ab026cc6f26015b4402e10ca8)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;{\dfrac {a}{p^{2}}}\right]_{p\,\in \,\mathbb {R} }=\infty \;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b9fc3285c51aff5bba27b8273748ff5b5b72a92)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[{\dfrac {p}{p+a}}\right]_{p\,\in \,\mathbb {R} }=0=\lim \limits _{t\,\rightarrow \,+\infty }\!\left[f(t)\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a70826fa768737df999fa2131be510a94c5fd7f)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;{\dfrac {p}{p^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }=0\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac8414a798cb10bc48601d70b3103916c4130835)
![{\displaystyle \;\lim \limits _{t\,\rightarrow \,+\infty }\!\left[f(t)\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b6845461c622d2305cab2970821d16f65b97deb)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,0^{+}}\!\left[p\;{\dfrac {\omega }{p^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }=0\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da0906f4cabb31897c13f62bd94f4e9088573ce9)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;{\dfrac {p+a}{(p+a)^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44657a3dd49f8693a8a6e70da340c07c23892d0e)
![{\displaystyle \;\color {transparent}{\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a9fc2028d2be4a33c12e1f6a68c26d9e6acf35e)
![{\displaystyle =\lim \limits _{t\,\rightarrow \,+\infty }\!\left[f(t)\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cd03922e879829abe98f09b51bbb9900eab474a)
![{\displaystyle \;\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;F(p)\right]_{p\,\in \,\mathbb {R} }=\lim \limits _{p\,\rightarrow \,+\infty }\!\left[p\;{\dfrac {\omega }{(p+a)^{2}+\omega ^{2}}}\right]_{p\,\in \,\mathbb {R} }=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6eed7c64ec0dd5520827408ea830a0a7da8130b8)
![{\displaystyle \;\left[p^{2}\;{\underline {Y}}(p)-p\;y_{0}-{\dot {y}}_{0}\right]+c\;{\underline {Y}}(p)=E\;{\sqrt {2}}\;{\dfrac {\omega }{p^{2}+\omega ^{2}}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/efe0af8c591ee663c22650366a78fb66ba7ccb2f)
![{\displaystyle \;A(p)={\dfrac {1}{(p^{2}+\omega ^{2})\;(p^{2}+c)}}={\dfrac {1}{c-\omega ^{2}}}\,\left[{\dfrac {1}{p^{2}+\omega ^{2}}}-{\dfrac {1}{p^{2}+c}}\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b1f31ce6db670790395f9d67e83ad38a6e5a9b4)
![{\displaystyle \;{\underline {Y}}(p)={\dfrac {E\;{\sqrt {2}}}{c-\omega ^{2}}}\,\left[{\dfrac {\omega }{p^{2}+\omega ^{2}}}-{\dfrac {\omega }{\sqrt {c}}}\;{\dfrac {\sqrt {c}}{p^{2}+c}}\right]+{\dfrac {p}{p^{2}+c}}\;y_{0}+{\dfrac {\sqrt {c}}{p^{2}+c}}\;{\dfrac {{\dot {y}}_{0}}{\sqrt {c}}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4450706b939b606cc5beaf0be886439df9e4bcc0)
![{\displaystyle \;y(t)=\left\lbrace {\dfrac {E\;{\sqrt {2}}}{c-\omega ^{2}}}\,\left[\sin(\omega \;t)-{\dfrac {\omega }{\sqrt {c}}}\;\sin({\sqrt {c}}\;t)\right]+y_{0}\;\cos({\sqrt {c}}\;t)+{\dfrac {{\dot {y}}_{0}}{\sqrt {c}}}\;\sin({\sqrt {c}}\;t)\right\rbrace Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df531ece8fd8c69e9275b57b062d80b705046834)
![{\displaystyle \;y(t)=\left\lbrace y_{0}\,\cos({\sqrt {c}}\;t)+\left[{\dfrac {{\dot {y}}_{0}}{\sqrt {c}}}-{\dfrac {E\;{\sqrt {2}}}{c-\omega ^{2}}}\;{\dfrac {\omega }{\sqrt {c}}}\right]\sin({\sqrt {c}}\;t)+{\dfrac {E\;{\sqrt {2}}}{c-\omega ^{2}}}\;\sin(\omega \,t)\right\rbrace Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/54302346a9989c1cc66f4c1bc3c564fff8922d7f)
![{\displaystyle \;{\dfrac {d\left[{\dfrac {p}{p^{2}+\omega ^{2}}}\right]}{dp}}={\dfrac {\left(p^{2}+\omega ^{2}\right)-p\;2\;p}{\left(p^{2}+\omega ^{2}\right)^{2}}}={\dfrac {\omega ^{2}-p^{2}}{\left(p^{2}+\omega ^{2}\right)^{2}}}=-{\dfrac {\omega ^{2}+p^{2}}{\left(p^{2}+\omega ^{2}\right)^{2}}}+{\dfrac {2\;\omega ^{2}}{\left(p^{2}+\omega ^{2}\right)^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b650097307271eaa6ab80879af83a35812a6eb38)
![{\displaystyle \;\color {transparent}{d\left[p^{2}+\omega ^{2}\right]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23758ca7f650f81f3c06a6b0f377ac3676f61fce)
![{\displaystyle \;B(p)={\dfrac {\omega }{\left(p^{2}+\omega ^{2}\right)^{2}}}={\dfrac {1}{2\;\omega }}\;{\dfrac {d\left[{\dfrac {p}{p^{2}+\omega ^{2}}}\right]}{dp}}+{\dfrac {1}{2\;\omega ^{2}}}\;{\dfrac {\omega }{p^{2}+\omega ^{2}}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6d2e09b7b9958f85121aa0592e48800253c812b)
![{\displaystyle \;\left\lbrace {\begin{array}{c c c}{\mathcal {L}}\!\left[\cos(\omega \,t)\;Y(t)\right]\!\!&=&\!\!{\dfrac {p}{p^{2}+\omega ^{2}}}\\{\dfrac {d{\mathcal {L}}\!\left[f(t)\right]}{dp}}\!\!&=&\!\!-{\mathcal {L}}\!\left[t\;f(t)\right]\end{array}}\right\rbrace \;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d1f145ce49d20a1aa7f30f94017eda5e9742aa2)
![{\displaystyle ={\dfrac {1}{2\;\omega ^{2}}}\;{\mathcal {L}}\!\left\lbrace \left[\sin(\omega \;t)-\omega \;t\;\cos(\omega \;t)\right]\,Y(t)\right\rbrace \;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f035e6814b9dd6f971a4264e717b80654603f75a)
![{\displaystyle \;{\mathcal {L}}^{-1}\!\left\lbrace {\dfrac {\omega }{\left(p^{2}+\omega ^{2}\right)^{2}}}\right\rbrace ={\dfrac {1}{2\;\omega ^{2}}}\,\left[\sin(\omega \;t)-\omega \;t\;\cos(\omega \;t)\right]\,Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f897f113f50d6b2c44880920bf7ecb95d531b8f)
![{\displaystyle \;y(t)=\left\lbrace {\dfrac {E\;{\sqrt {2}}}{2\;\omega ^{2}}}\,\left[\sin(\omega \,t)-\omega \;t\;\cos(\omega \,t)\right]+y_{0}\;\cos(\omega \;t)+{\dfrac {{\dot {y}}_{0}}{\omega }}\;\sin(\omega \;t)\right\rbrace Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b769c0a3f7882e4d30affb0e9fa52ef23b9d99b0)
![{\displaystyle \;y(t)=\left\lbrace y_{0}\,\cos(\omega \;t)+\left[{\dfrac {{\dot {y}}_{0}}{\omega }}+{\dfrac {E}{{\sqrt {2}}\;\omega ^{2}}}\right]\sin(\omega \;t)-{\dfrac {E}{{\sqrt {2}}\;\omega }}\;t\;\cos(\omega \,t)\right\rbrace Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/840840fb0db43a77320e40541394bb5e3d632d96)
![{\displaystyle \;A'(p)={\dfrac {1}{(p^{2}+\omega ^{2})\;(p^{2}+c)}}={\dfrac {1}{\omega ^{2}-c}}\,\left\lbrace -{\dfrac {1}{p^{2}+\omega ^{2}}}+{\dfrac {1}{2\;{\sqrt {-c}}}}\,\left[{\dfrac {1}{p-{\sqrt {-c}}}}-{\dfrac {1}{p+{\sqrt {-c}}}}\right]\right\rbrace \;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7df3f67d5de59abc82b0be3eb9a13b3376acdb2a)
![{\displaystyle \;B'(p)={\dfrac {p}{p^{2}+c}}={\dfrac {1}{2}}\,\left[{\dfrac {1}{p+{\sqrt {-c}}}}+{\dfrac {1}{p-{\sqrt {-c}}}}\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2730b0dfc7e8f66505ac722cbf7b671defc50b1c)
![{\displaystyle \;C'(p)={\dfrac {1}{p^{2}+c}}={\dfrac {1}{2\;{\sqrt {-c}}}}\,\left[{\dfrac {1}{p-{\sqrt {-c}}}}-{\dfrac {1}{p+{\sqrt {-c}}}}\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f607f5e43bbc99343f71f22c092979116e88eb83)
![{\displaystyle \;{\underline {Y}}(p)={\dfrac {E\;{\sqrt {2}}}{\omega ^{2}-c}}\,\left\lbrace -{\dfrac {\omega }{p^{2}+\omega ^{2}}}+{\dfrac {\omega }{2\;{\sqrt {-c}}}}\,\left[{\dfrac {1}{p-{\sqrt {-c}}}}-{\dfrac {1}{p+{\sqrt {-c}}}}\right]\right\rbrace +{\dfrac {y_{0}}{2}}\,\left[{\dfrac {1}{p+{\sqrt {-c}}}}+{\dfrac {1}{p-{\sqrt {-c}}}}\right]+{\dfrac {{\dot {y}}_{0}}{2\;{\sqrt {-c}}}}\,\left[{\dfrac {1}{p-{\sqrt {-c}}}}-{\dfrac {1}{p+{\sqrt {-c}}}}\right]\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/939b8641d51df587c7544960648f4f5437fb1690)
![{\displaystyle \;{\underline {Y}}(p)=-{\dfrac {E\;{\sqrt {2}}}{\omega ^{2}-c}}\;{\dfrac {\omega }{p^{2}+\omega ^{2}}}+\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\,{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}+y_{0}\;{\sqrt {-c}}}{2\;{\sqrt {-c}}}}\right]\,{\dfrac {1}{p-{\sqrt {-c}}}}-\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\,{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}-y_{0}\;{\sqrt {-c}}}{2\;{\sqrt {-c}}}}\right]\,{\dfrac {1}{p+{\sqrt {-c}}}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bea902d927e8a39be68df31219120efec7eeb7d)
![{\displaystyle {\underline {Y}}(p)=-{\dfrac {E\;{\sqrt {2}}}{\omega ^{2}-c}}\;{\dfrac {\omega }{p^{2}+\omega ^{2}}}+\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\,{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}+y_{0}\;{\sqrt {-c}}}{2\;{\sqrt {-c}}}}\right]\,{\dfrac {1}{p-{\sqrt {-c}}}}-\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\,{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}-y_{0}\;{\sqrt {-c}}}{2\;{\sqrt {-c}}}}\right]\,{\dfrac {1}{p+{\sqrt {-c}}}}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e9c84dd1efc87a484b3dfc8d17b2f6f2989de5a)
![{\displaystyle \;y(t)=\left\lbrace -{\dfrac {E\;{\sqrt {2}}}{\omega ^{2}-c}}\;\sin(\omega \;t)+\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\,{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}+y_{0}\;{\sqrt {-c}}}{2\;{\sqrt {-c}}}}\right]\,\exp \!\left({\sqrt {-c}}\,t\right)-\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\,{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}-y_{0}\;{\sqrt {-c}}}{2\;{\sqrt {-c}}}}\right]\,\exp \!\left(-{\sqrt {-c}}\,t\right)\right\rbrace \;Y(t)\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34b93b1b7b7e68744d194dfecebdf0deb89fd395)
![{\displaystyle \;y(t)=y_{0}\,\cos({\sqrt {c}}\;t)+\left[{\dfrac {{\dot {y}}_{0}}{\sqrt {c}}}-{\dfrac {E\;{\sqrt {2}}}{c-\omega ^{2}}}\;{\dfrac {\omega }{\sqrt {c}}}\right]\sin({\sqrt {c}}\;t)+{\dfrac {E\;{\sqrt {2}}}{c-\omega ^{2}}}\;\sin(\omega \,t)\;\;{\text{pour}}\;t>0\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9577d82e39ce6ed418ccc7dc0d2cbfb79236edfa)
![{\displaystyle \;y(t)=y_{0}\,\cos(\omega \;t)+\left[{\dfrac {{\dot {y}}_{0}}{\omega }}+{\dfrac {E}{{\sqrt {2}}\;\omega ^{2}}}\right]\sin(\omega \;t)-{\dfrac {E}{{\sqrt {2}}\;\omega }}\;t\;\cos(\omega \,t)\;\;{\text{pour}}\;t>0\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/643e1bf3455846b1ae74ac8692fb556a34e2c003)
![{\displaystyle \;\left\lbrace {\begin{array}{l}A_{+}={\dfrac {1}{2}}\left[y_{0}+{\dfrac {E}{\omega ^{2}-c}}\;{\sqrt {2}}\;{\dfrac {\omega }{\sqrt {-c}}}+{\dfrac {{\dot {y}}_{0}}{\sqrt {-c}}}\right]\\A_{-}={\dfrac {1}{2}}\left[y_{0}-{\dfrac {E}{\omega ^{2}-c}}\;{\sqrt {2}}\;{\dfrac {\omega }{\sqrt {-c}}}-{\dfrac {{\dot {y}}_{0}}{\sqrt {-c}}}\right]\end{array}}\right\rbrace \;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8815d5c6cf0d43d0a12573c282265eeac93adbe2)
![{\displaystyle \;y(t)=\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\;{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}+y_{0}\,{\sqrt {-c}}}{2\,{\sqrt {-c}}}}\right]\exp \!\left({\sqrt {-c}}\,t\right)-\left[{\dfrac {E\;\omega }{(\omega ^{2}-c)\;{\sqrt {-2\;c}}}}+{\dfrac {{\dot {y}}_{0}-y_{0}\,{\sqrt {-c}}}{2\,{\sqrt {-c}}}}\right]\exp \!\left(-{\sqrt {-c}}\,t\right)-{\dfrac {E\,{\sqrt {2}}}{\omega ^{2}-c}}\,\sin(\omega \,t)\;\;{\text{pour}}\;t>0\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2acc10fa530d22b2258f806cd43179515a12fec)
![{\displaystyle \;\left[p^{2}\;{\underline {Y}}(p){\cancel {-p\;y(0^{-})-{\dot {y}}(0^{-})}}\right]+b\,\left[p\;{\underline {Y}}(p){\cancel {-y(0^{-})}}\right]+c\;{\underline {Y}}(p)={\dfrac {E}{p}}+E'\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d7804151ed575742eac16afcfbb43909be9398d)
![{\displaystyle \;{\dfrac {1}{q_{(+)}\,[q_{(+)}-q_{(-)}]}}=\rho _{(+)}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d719321de45aae21b0f43eeb448afd608566487)
![{\displaystyle \;{\dfrac {1}{q_{(-)}\,[q_{(-)}-q_{(+)}]}}=\rho _{(-)}\;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ccdda3b2fe2a83559788f43886fb3e5352913ce7)
