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Transformées inverses de Laplace usuelles
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Intitulé : Transformées inverses de Laplace usuelles
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Outils mathématiques et informatiques pour la physique
.
Transformée de Laplace de la fonction
Fonction
1
{\displaystyle 1}
δ
(
t
)
{\displaystyle \delta (t)}
1
p
{\displaystyle {\frac {1}{p}}}
1
{\displaystyle 1}
1
p
2
{\displaystyle {\frac {1}{p^{2}}}}
t
{\displaystyle t}
1
p
n
{\displaystyle {\frac {1}{p^{n}}}}
t
n
−
1
(
n
−
1
)
!
∀
n
≥
1
{\displaystyle {\frac {t^{n-1}}{(n-1)!}}\qquad \forall n\geq 1}
1
p
{\displaystyle {\frac {1}{\sqrt {p}}}}
1
π
t
∀
t
∈
R
+
∗
{\displaystyle {\frac {1}{\sqrt {\pi t}}}\qquad \forall t\in R_{+}^{*}}
1
p
3
{\displaystyle {\frac {1}{\sqrt {p^{3}}}}}
2
t
π
{\displaystyle 2{\sqrt {\frac {t}{\pi }}}}
1
p
+
a
{\displaystyle {\frac {1}{p+a}}}
e
−
a
t
{\displaystyle e^{-at}}
1
p
(
p
+
a
)
{\displaystyle {\frac {1}{p(p+a)}}}
1
−
e
−
a
t
a
∀
a
≠
0
{\displaystyle {1-e^{-at} \over a}\qquad \forall a\neq 0}
1
p
2
(
p
+
a
)
{\displaystyle {\frac {1}{p^{2}(p+a)}}}
e
−
a
t
−
1
+
a
t
a
2
∀
a
≠
0
{\displaystyle {\frac {e^{-at}-1+at}{a^{2}}}\qquad \forall a\neq 0}
1
p
(
p
+
a
)
2
{\displaystyle {\frac {1}{p(p+a)^{2}}}}
1
−
e
−
a
t
(
1
+
a
t
)
a
2
∀
a
≠
0
{\displaystyle {1-e^{-at}(1+at) \over a^{2}}\qquad \forall a\neq 0}
1
(
p
+
a
)
(
p
+
b
)
{\displaystyle {\frac {1}{(p+a)(p+b)}}}
e
−
b
t
−
e
−
a
t
a
−
b
∀
a
≠
b
{\displaystyle {\frac {e^{-bt}-e^{-at}}{a-b}}\qquad \forall a\neq b}
p
(
p
+
a
)
(
p
+
b
)
{\displaystyle {\frac {p}{(p+a)(p+b)}}}
a
e
−
a
t
−
b
e
−
b
t
a
−
b
∀
a
≠
b
{\displaystyle {\frac {ae^{-at}-be^{-bt}}{a-b}}\qquad \forall a\neq b}
1
(
p
+
a
)
(
p
+
b
)
(
p
+
c
)
{\displaystyle {\frac {1}{(p+a)(p+b)(p+c)}}}
(
b
−
c
)
e
−
a
t
+
(
c
−
a
)
e
−
b
t
+
(
a
−
b
)
e
−
c
t
(
a
−
b
)
(
a
−
c
)
(
b
−
c
)
∀
a
≠
b
∨
a
≠
c
∨
b
≠
c
{\displaystyle {(b-c)e^{-at}+(c-a)e^{-bt}+(a-b)e^{-ct} \over (a-b)(a-c)(b-c)}\qquad \forall a\neq b\lor a\neq c\lor b\neq c}
1
(
p
+
a
)
2
{\displaystyle {\frac {1}{(p+a)^{2}}}}
t
e
−
a
t
{\displaystyle te^{-at}}
p
(
p
+
a
)
2
{\displaystyle {\frac {p}{(p+a)^{2}}}}
e
−
a
t
(
1
−
a
t
)
{\displaystyle e^{-at}(1-at)}
1
(
p
+
a
)
(
p
+
b
)
2
{\displaystyle {\frac {1}{(p+a)(p+b)^{2}}}}
e
−
a
t
+
[
(
a
−
b
)
t
−
1
]
e
−
b
t
(
a
−
b
)
2
∀
a
≠
b
{\displaystyle {\frac {e^{-at}+\left[(a-b)t-1\right]e^{-bt}}{(a-b)^{2}}}\qquad \forall a\neq b}
1
p
(
p
+
a
)
(
p
+
b
)
{\displaystyle {\frac {1}{p(p+a)(p+b)}}}
b
e
−
a
t
−
a
e
−
b
t
+
a
−
b
a
2
b
−
a
b
2
∀
a
≠
b
{\displaystyle {be^{-at}-ae^{-bt}+a-b \over \ a^{2}b-ab^{2}}\qquad \forall a\neq b}
p
+
c
p
(
p
+
a
)
(
p
+
b
)
{\displaystyle {\frac {p+c}{p(p+a)(p+b)}}}
b
(
c
−
a
)
e
−
a
t
+
a
(
b
−
c
)
e
−
b
t
+
c
(
a
−
b
)
a
b
(
a
−
b
)
∀
a
≠
b
{\displaystyle {b(c-a)e^{-at}+a(b-c)e^{-bt}+c(a-b) \over \ ab(a-b)}\qquad \forall a\neq b}
p
2
+
c
p
+
d
p
(
p
+
a
)
(
p
+
b
)
{\displaystyle {\frac {p^{2}+cp+d}{p(p+a)(p+b)}}}
b
(
a
2
−
a
c
+
d
)
e
−
a
t
−
a
(
b
2
−
b
c
+
d
)
e
−
b
t
+
d
(
a
−
b
)
a
b
(
a
−
b
)
∀
a
≠
b
{\displaystyle {b(a^{2}-ac+d)e^{-at}-a(b^{2}-bc+d)e^{-bt}+d(a-b) \over \ ab(a-b)}\qquad \forall a\neq b}
1
(
p
+
a
)
3
{\displaystyle {\frac {1}{(p+a)^{3}}}}
t
2
e
−
a
t
2
{\displaystyle {\frac {t^{2}e^{-at}}{2}}}
ln
(
p
+
a
p
+
b
)
{\displaystyle \ln \left({\frac {p+a}{p+b}}\right)}
e
−
b
t
−
e
−
a
t
t
∀
t
∈
R
+
∗
{\displaystyle {\frac {e^{-bt}-e^{-at}}{t}}\qquad \forall t\in R_{+}^{*}}
1
p
2
+
a
2
{\displaystyle {\frac {1}{p^{2}+a^{2}}}}
sin
(
a
t
)
a
∀
a
≠
0
{\displaystyle {\sin(at) \over a}\qquad \forall a\neq 0}
1
p
(
p
2
+
a
2
)
{\displaystyle {\frac {1}{p(p^{2}+a^{2})}}}
1
−
cos
(
a
t
)
a
2
∀
a
≠
0
{\displaystyle {1-\cos(at) \over a^{2}}\qquad \forall a\neq 0}
p
p
2
+
a
2
{\displaystyle {\frac {p}{p^{2}+a^{2}}}}
cos
(
a
t
)
{\displaystyle \cos(at)}
p
+
a
p
(
p
2
+
b
2
)
{\displaystyle {\frac {p+a}{p(p^{2}+b^{2})}}}
a
(
1
−
cos
(
b
t
)
)
+
b
sin
(
b
t
)
b
2
∀
b
≠
0
{\displaystyle {a(1-\cos(bt))+b\sin(bt) \over b^{2}}\qquad \forall b\neq 0}
p
2
+
c
p
+
d
p
(
p
2
+
b
2
)
{\displaystyle {\frac {p^{2}+cp+d}{p(p^{2}+b^{2})}}}
(
b
2
−
d
)
cos
(
b
t
)
+
b
c
sin
(
b
t
)
+
d
b
2
∀
b
≠
0
{\displaystyle {(b^{2}-d)\cos(bt)+bc\sin(bt)+d \over b^{2}}\qquad \forall b\neq 0}
1
p
2
−
a
2
{\displaystyle {\frac {1}{p^{2}-a^{2}}}}
sinh
(
a
t
)
a
∀
a
≠
0
{\displaystyle {\sinh(at) \over a}\qquad \forall a\neq 0}
p
p
2
−
a
2
{\displaystyle {\frac {p}{p^{2}-a^{2}}}}
cosh
(
a
t
)
{\displaystyle \cosh(at)}
1
(
p
+
b
)
2
+
a
2
{\displaystyle {\frac {1}{(p+b)^{2}+a^{2}}}}
e
−
b
t
sin
(
a
t
)
a
∀
a
≠
0
{\displaystyle {e^{-bt}\sin(at) \over a}\qquad \forall a\neq 0}
p
+
b
(
p
+
b
)
2
+
a
2
{\displaystyle {\frac {p+b}{(p+b)^{2}+a^{2}}}}
e
−
b
t
cos
(
a
t
)
{\displaystyle e^{-bt}\cos(at)}
1
(
p
2
+
a
2
)
2
{\displaystyle {\frac {1}{(p^{2}+a^{2})^{2}}}}
sin
(
a
t
)
−
a
t
cos
(
a
t
)
2
a
3
∀
a
≠
0
{\displaystyle {\sin(at)-at\cos(at) \over 2a^{3}}\qquad \forall a\neq 0}
p
(
p
2
+
a
2
)
2
{\displaystyle {\frac {p}{(p^{2}+a^{2})^{2}}}}
t
sin
(
a
t
)
2
a
∀
a
≠
0
{\displaystyle {t\sin(at) \over 2a}\qquad \forall a\neq 0}
p
2
(
p
2
+
a
2
)
2
{\displaystyle {\frac {p^{2}}{(p^{2}+a^{2})^{2}}}}
sin
(
a
t
)
+
a
t
cos
(
a
t
)
2
a
∀
a
≠
0
{\displaystyle {\sin(at)+at\cos(at) \over 2a}\qquad \forall a\neq 0}
1
p
3
+
a
3
{\displaystyle {\frac {1}{p^{3}+a^{3}}}}
e
−
a
t
−
e
a
t
2
(
cos
(
3
2
a
t
)
−
3
sin
(
3
2
a
t
)
)
3
a
2
∀
a
≠
0
{\displaystyle {e^{-at}-e^{\frac {at}{2}}\left(\cos \left({\frac {\sqrt {3}}{2}}at\right)-{\sqrt {3}}\sin \left({\frac {\sqrt {3}}{2}}at\right)\right) \over 3a^{2}}\qquad \forall a\neq 0}
p
p
3
+
a
3
{\displaystyle {\frac {p}{p^{3}+a^{3}}}}
e
a
t
2
(
3
sin
(
3
a
t
2
)
+
cos
(
3
a
t
2
)
)
−
e
−
a
t
3
a
∀
a
≠
0
{\displaystyle {e^{at \over 2}\left({\sqrt {3}}\sin \left({\frac {{\sqrt {3}}at}{2}}\right)+\cos \left({\frac {{\sqrt {3}}at}{2}}\right)\right)-e^{-at} \over 3a}\qquad \forall a\neq 0}
p
2
p
3
+
a
3
{\displaystyle {\frac {p^{2}}{p^{3}+a^{3}}}}
e
−
a
t
+
2
e
a
t
2
cos
(
3
a
t
2
)
3
{\displaystyle {e^{-at}+2e^{\frac {at}{2}}\cos \left({\frac {{\sqrt {3}}at}{2}}\right) \over 3}}
1
p
2
(
1
+
τ
1
p
)
(
1
+
τ
2
p
)
{\displaystyle {\frac {1}{p^{2}(1+\tau _{1}p)(1+\tau _{2}p)}}}
τ
1
2
e
−
t
/
τ
1
−
τ
2
2
e
−
t
/
τ
2
+
(
τ
1
−
τ
2
)
(
t
−
τ
1
−
τ
2
)
τ
1
−
τ
2
∀
τ
1
≠
τ
2
{\displaystyle {\tau _{1}^{2}e^{-t/\tau _{1}}-\tau _{2}^{2}e^{-t/\tau _{2}}+(\tau _{1}-\tau _{2})(t-\tau _{1}-\tau _{2}) \over \tau _{1}-\tau _{2}}\qquad \forall \tau _{1}\neq \tau _{2}}
![{\displaystyle {\frac {e^{-at}+\left[(a-b)t-1\right]e^{-bt}}{(a-b)^{2}}}\qquad \forall a\neq b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab15e1186fec8f1beed9aa9925af75fbab5fb6be)
