F-IRIS1-06

Transform´ee de Laplace F-IRIS1-06.tex
Transform´ee de Laplace Exercices Simples
1) Laplace
Calculer les transform´ees de Laplace suivantes :
a) Lht2+t−e−3tU(t)i
b) Lht+ 2U(t) + t+ 3U(t−2)i
c) Lht2+t+ 1e−2tU(t)i
2) Laplace inverse
Calculer les originaux suivants :
a) L−1p+ 2
(p+ 3)(p+ 4) 
b) L−13
(p+ 5)2
c) L−1p−1
(p2+ 2p+ 5) 
3) ´ Equations diff´erentielles
Utiliser la transform´ee de Laplace pour d´eterminer la solution particuli`ere de chacune des
´equations diff´erentielles suivantes :
a) x0(t) + x(t) = tU(t)−tU(t−1) condition initiale : x(0) = 0
b) x00(t) + x0(t) = U(t) conditions initiales : x(0) = 0
x0(0) = 0
c) x00(t)+4x(t)=2U(t) conditions initiales : x(0) = 0
x0(0) = 1
d) x00(t)+5x0(t)+4x(t) = e−2tU(t) conditions initiales : x(0) = 1
x0(0) = 0
e) x00(t)+2x0(t)+2x(t) = 0 conditions initiales : x(0) = 1
x0(0) = 1
♣♦
♥♠1 / 2 L
A
T
E
X 2ε

Transform´ee de Laplace F-IRIS1-06.tex
Transform´ee de Laplace Exercices d’entraˆınement
1) Calculer les transform´ees de Laplace suivantes :
a) L[cos(t)e−tU(t)] b) L[(5t)2e−5tU(t)]
c) L[(cos(2t)−sin(t)) e−3tU(t)] d) L[(t2+t+ 1)e−2tU(t)]
2) Calculer les originaux suivants :
a) L−13
p+ 2 −1 p3
b) L−1−2
(p+ 3)2
c) L−15
(p+ 3)(p2+ 3p+ 5)
d) L−1p
p2+ 4p+ 6
e) L−1p
(p+ 1)2
f) L−12p+ 3
2p2+ 4p+ 5
3) ´ Equations diff´erentielles
Utiliser la transform´ee de Laplace pour r´esoudre les ´equations suivantes :
a) x00(t)+3x0(t)+2x(t) = 0 avec : x(0) = 1 et x0(0) = 0
b) x00(t)+6x0(t)+9x(t) = e−2tU(t) avec : x(0) = 0 et x0(0) = 0
c) x00(t)−x(t) = (3e−2t+t2+ 1) U(t) avec : x(0) = 0 et x0(0) = 0
d) x00(t)−4x(t) = (3e−t−t2)U(t) avec : x(0) = 0 et x0(0) = 1
e) x00(t) + x(t) = etcos(t)U(t) avec : x(0) = 0 et x0(0) = 0
f) x00(t) + x(t) = U(t)−U(t−1) avec : x(0) = 2 et x0(0) = 0
♣♦
♥♠2 / 2 L
A
T
E
X 2ε

Transform´ee de Laplace F-IRIS1-06.tex
Exercices Simples (Solutions)
1) Laplace
Calculer les transform´ees de Laplace suivantes :
a) Lht2+t−e−3tU(t)i
f(t) = t2+t−e−3tU(t) = t2U(t) + tU(t)−e−3tU(t)
F(p) = 2
p3+1 p2−1 p+ 3
b) Lht+ 2U(t) + t+ 3U(t−2)i
f(t) = t+ 2U(t) + t+ 3U(t−2) = t+ 2U(t) + (t−2) + 5U(t−2)
Lt+ 5U(t)=1
p2+5 pLt+ 2U(t)=1 p2+2 p
L(t−2) + 5U(t−2)=1
p2+5 pe−2p=2p+ 1 p2
F(p) = 2p+ 1
p2+1 p2+5 pe−2p
c) Lht2+t+ 1e−2tU(t)if(t) = t2+t+ 1e−2tU(t)
Lht2+t+ 1U(t)i=2
p3+1 p2+1 p=p2+p+ 2 p3
Lht2+t+ 1e−2tU(t)i=(p+ 2)2+ (p+ 2) + 2
(p+ 2)3
F(t) = p2+ 5p+ 8
(p+ 2)3
2) Laplace inverse
Calculer les originaux suivants :
a) L−1p+ 2
(p+ 3)(p+ 4) 
F(p) = p+ 2
(p+ 3)(p+ 4) =2 p+ 4 +−1 p+ 3
f(t) = 2e−4t−e−3tU(t)
♣♦
♥♠3 / 10 L
A
T
E
X 2ε

Transform´ee de Laplace F-IRIS1-06.tex
b) L−13
(p+ 5)2
F(p) = 3
(p+ 5)2
L−13
p2= 3 tU(t)
f(t) = 3 t e−5tU(t)
c) L−1p−1
(p2+ 2p+ 5) 
F(p) = p−1
(p2+ 2p+ 5) =p+ 1 (p+ 1)2+ 22−2 (p+ 1)2+ 22
L−1p
p2+ 22−2 p2+ 22=cos(2t)−sin(2t)U(t)
f(t) = cos(2t)−sin(2t)e−tU(t)
3) ´ Equations diff´erentielles
a) x0(t) + x(t) = tU(t)−tU(t−1) condition initiale : x(0) = 0
x0(t) + x(t) = tU(t)−(t−1) + 1U(t−1)
(p X(p)−0) + X(p) = 1
p2−1 p2+1 pe−p
(p+ 1)X(p) = 1
p2−p+ 1 p2e−p
X(p) = 1
p2(p+ 1) −1 p2e−p
X(p) = 1
p2−1 p+1 p+ 1 −1 p2e−p
x(t) = t−1 + e−tU(t)−t−1U(t−1)
b) x00(t) + x0(t) = U(t) conditions initiales : x(0) = 0
x0(0) = 0
(p2X(p)−0−0) + (p X(p)−0) = 1
p
(p2+p)X(p) = 1
p
X(p) = 1
p(p2+p)=1 p2−1 p+1 p+ 1
x(t) = t−1 + e−tU(t)
♣♦
♥♠4 / 10 L
A
T
E
X 2ε

Transform´ee de Laplace F-IRIS1-06.tex
c) x00(t)+4x(t)=2U(t) conditions initiales : x(0) = 0
x0(0) = 1
(p2X(p)−0−1) + 4X(p) = 2
p
(p2+ 4)X(p) = 2
p+ 1
X(p) = p+ 2
p(p2+ 4) =
1
2
p+−1
2p+ 1
p2+ 4
=1
21 p−p p2+ 4 +2 p2+ 4
x(t) = 1
21−cos(2t) + sin(2t)U(t)
d) x00(t)+5x0(t)+4x(t) = e−2tU(t) conditions initiales : x(0) = 1
x0(0) = 0
(p2X(p)−p−0) + 5(p X(p)−1) + 4X(p) = 1
p+ 2
(p2+ 5p+ 4)X(p) = 1
p+ 2 +p+ 5
X(p) = p2+ 7p+ 11
(p+ 2)(p2+ 5p+ 4) =p2+ 7p+ 11 (p+ 2)(p+ 1)(p+ 4)
=5/3
p+ 1 +−1/2 p+ 2 +−1/6 p+ 4
x(t) = 5e−t
3−e−2t 2−e−4t 6U(t)
e) x00(t)+2x0(t)+2x(t) = 0 conditions initiales : x(0) = 1
x0(0) = 1
(p2X(p)−p−1) + 2(p X(p)−1) + 2X(p) = 0
(p2+ 2p+ 2)X(p) = p+ 3
X(p) = p+ 3
p2+ 2p+ 2
=p+ 1
(p+ 1)2+ 12+2 (p+ 1)2+ 12
x(t) = cos(t) + 2 sin(t)e−tU(t)
♣♦
♥♠5 / 10 L
A
T
E
X 2ε