Sciences de l’ingénieur
2ndordre_a_inf_1_
Lycée Jacques Amyot
Auxerre 28/11/2003 Page 1 sur 3
Système du second ordre a<1
Recherche de la réponse à un échelon unitaire.
Michel Huguet Préambule :
La forme canonique d’un système du second ordre est de la forme :
2 2
1 1 2
) (
p a p
p K H
n n
⋅ +
⋅ ⋅ +
= ω ω
Le système étant sollicité par un échelon unitaire de type
e ( t ) = 1 ⋅ u ( t )
, l’image de cette entrée dans le domaine de Laplace est notée E(p) et vaut :p p
E 1
)
( =
. L’image de la réponse du système notée S(p) vaut alors :p p a p
p K S
p E p H p S
n n
1 1
1 2 ) (
) ( ) ( ) (
2 2
⋅
⋅ +
⋅ ⋅ +
=
⋅
=
ω ω
1) Décomposition de S(p) en éléments simples :
1.1)Recherche des racines de
2 1 0
1 + ⋅ a ⋅ p +
2⋅ p
2=
n
n
ω
ω
. Pour a<1 on trouve deux racines complexes :2 2
2 1
1 1
a j
a p
a j
a p
n n
n n
−
⋅
⋅
−
⋅
−
=
−
⋅
⋅ +
⋅
−
=
ω ω
ω ω
Une fois factorisée l’équation se met sous la forme :
1 ( ) ( ) 0
2 2
⋅ p − p
1⋅ p − p = ω
nOu
1
2⋅ ( p + a ⋅
n− j ⋅
n⋅ 1 − a
2) ⋅ ( p + a ⋅ n + j ⋅
n⋅ 1 − a
2) = 0
n
ω ω
ω ω ω
En développant on trouve :
( )
( ) ( 1 ) 0
1
0 1
1 2
2 2 2
2
2 2 2
2 2
2
⎟ =
⎠
⎜ ⎞
⎝
⎛ + ⋅ + ⋅ −
⋅
⎟ =
⎠
⎜ ⎞
⎝
⎛ + ⋅ ⋅ ⋅ + ⋅ + ⋅ −
⋅
a a
p
a a
p a p
n
n
n n
n n
n
ω ω ω
ω ω ω ω
finalement on met S(p) sous la forme :
( ) ( )
(
p a)
K(
a)
pp S
a p a
p p K
S
n n
n n n n
1 1
) (
1 1 1
) (
2 2 2
2
2 2 2
2
⋅
⎟⎠
⎜ ⎞
⎝
⎛ + ⋅ + ⋅ −
= ⋅
⋅
⎟⎠
⎜ ⎞
⎝
⎛ + ⋅ + ⋅ −
⋅
=
ω ω
ω
ω
ω ω
Sciences de l’ingénieur
2ndordre_a_inf_1_
Lycée Jacques Amyot
Auxerre 28/11/2003 Page 2 sur 3
Qui après mise en forme , s’écrit :
( p a ) ( a a ) p
a p K
S
n n
n
n
1
1 1 1
)
(
22 2
2
2
⋅
−
⋅ +
⋅ +
−
⋅ ⋅
−
= ⋅
ω ω
ω ω
Pour la suite on posera :
( )
1 )
(
12
S p
a p K
S
n⋅
−
= ⋅ ω
1.2)Décomposition de S1(p) :
( )
2(
2)
2( ( )
2( ) 2)
2
2 1
1 1
1 ) 1
(
a a
p
C a
p B p
A a p
a p p a S
n n
n
n n
n
−
⋅ +
⋅ +
+
⋅ + + ⋅
=
− ⋅
⋅ +
⋅ +
−
= ⋅
ω ω
ω ω
ω ω
• Recherche de A : On forme :
( )
2(
2)
2( ( )
2( ) 2)
2
2 1
1 1
) 1 (
a a
p
C a
p p B
p p A a
a p p a S p
n n
n
n n
n
−
⋅ +
⋅ +
+
⋅ +
⋅ ⋅ +
⋅
− =
⋅ +
⋅ +
−
= ⋅
⋅ ω ω
ω ω
ω ω
Et on pose p = 0 , il vient alors :
( a
n) ( n a a ) A a
nn n a a A
n
=
−
⋅ +
⋅
−
⇔ ⋅
− =
⋅ +
⋅
−
⋅
) 1 ( 1 1
1
2 2 2
2
2 2 2
2
2
ω ω
ω ω
ω ω
Finalement :
n
A a ω 1 −
2=
• Recherche de B et de C : On forme :
( ) (
1)
1( )2 2
2 a S p
a
p n n ⎟⋅
⎠
⎜ ⎞
⎝
⎛ + ⋅
ω
+ω
⋅ −( ) ( ) ( ) ( ) ( )
( )
2(
2)
22 2 2 2
2 2 2
1 1
1 1 :
a a
p
C a
p a B
a p p
a A a
p p a vient Il
n n
n n
n n
n n
−
⋅ +
⋅ +
+
⋅ +
⋅ ⋅
⎟⎠
⎜ ⎞
⎝
⎛ + ⋅ + ⋅ −
+
⎟⋅
⎠
⎜ ⎞
⎝
⎛ + ⋅ + ⋅ −
− =
⋅
ω ω ω ω
ω ω
ω ω
Et on pose p=p1 ou p=p2 :
(
2)
2 2
1 a j
a p
p p
n
+ ⋅ −
−
=
= ω
Il vient alors :
( )
B( (
a j a)
a)
Ca j a
a
n n
n
n = ⋅− ⋅ + ⋅ − + ⋅ +
−
⋅ +
⋅
−
−
⋅
ω ω
ω
ω
22 2
1 1
1
En multipliant le numérateur et le dénominateur due premier terme de l’égalité par le complexe conjugué, on peut écrire :
( )
( ) ( ( ) )
(
a j a)
j B a Ca
C a
a j a B
a j a a
n
n n
n n
n
+
−
⋅
⋅
⋅
−
=
−
⋅
−
⋅
−
−
+
⋅ +
−
⋅ +
⋅
−
⋅
=
−
⋅
−
⋅
−
− ⋅
⋅
2 2
2
2 2
2 2
1 1
1
1 1 1
ω
ω ω
ω ω ω
En identifiant les parties réelle et imaginaire, on obtient le système :
( )
( ) ( ( ) )
(
a j a)
j B a Ca
C a
a j a B
a j a a
n
n n
n n
n
+
−
⋅
⋅
⋅
−
=
−
⋅
−
⋅
−
−
+
⋅ +
−
⋅ +
⋅
−
⋅
=
−
⋅
−
⋅
−
− ⋅
⋅
2 2
2
2 2
2 2
1 1
1
1 1 1
ω
ω ω
ω ω
ω
Sciences de l’ingénieur
2ndordre_a_inf_1_
Lycée Jacques Amyot
Auxerre 28/11/2003 Page 3 sur 3
⎪⎩
⎪⎨
⎧
− =
⋅
− −
=
−
⋅
−
⎪⎩
⎪⎨
⎧
−
⋅
⋅
−
=
−
=
−
⋅
−
B a a
C a a a
B a
C a a
n
n 2
2 2
2 2
2
1 1
1 1
1
1
ω ω
Finalement on obtient :
⎪ ⎩
⎪ ⎨
⎧
− −
=
−
⋅
−
=
n
B a
a a
C
ω
2 2
1 1
Donc S(p) décomposée s’écrit :
( )
( ) ( ) ( ) ( ) ⎟ ⎟ ⎟ ⎟
⎟
⎠
⎞
⎜ ⎜
⎜ ⎜
⎜
⎝
⎛
−
⋅ +
⋅ +
−
− ⋅
−
⋅ +
⋅ +
⋅ +
− ⋅
−
−
− ⋅
− ⋅
= ⋅
22 2 2 2 2
2 2 2
2
1
1 1
1 1
1 1
) (
a a
p
a a
a a
p
a a p
p a a
p K S
n n
n n
n n
n n
ω ω
ω ω
ω ω ω
ω
Après simplification on obtient :
( )
( ) ( ) ( ) ( ) ⎟ ⎟ ⎠
⎞
⎜ ⎜
⎝
⎛
−
⋅ +
⋅ +
−
⋅
− −
−
⋅ +
⋅ +
⋅
− +
⋅
=
22 2
2 2 2
2 2
1 1 1 1
) 1 (
a a
p
a a
a a
a p
a p K p
p S
n n
n
n n
n
ω ω
ω ω
ω
ω
2) Transformée inverse :
En posant :
α = a ⋅ ω
net ω = ω
n⋅ 1 − a
2 :( )
( ) ( ) ( ( ) ) ( ( ) ) ( )
( p a ) ( a a ) ( p ) ( p ) e ( ) t
t p e
p p
p a
a p
a p
t
n n
n
t
n n
n
⋅
⋅
⎟⎟ =
⎠
⎜⎜ ⎞
⎝
⎛
+ + +
= +
−
⋅ +
⋅ +
−
⋅
⋅
⋅
⎟⎟ =
⎠
⎜⎜ ⎞
⎝
⎛
+ +
+ +
+
= +
−
⋅ +
⋅ +
⋅ +
⋅
−
⋅
−
ω ω α
ω ω
α ω ω
ω ω
ω ω α
α ω
α α ω
ω
ω
α α
sin 1
1
cos 1
2 2 2 2
2 2 2
2
2 2 2 2
2 2 2
1 -
1 -
L L
et
et
Finalement on obtient après transformation inverse :
( ) ( ) ⎟⎟
⎠
⎜⎜ ⎞
⎝
⎛ ⋅ ⋅ ⋅ − ⋅
− −
⋅
−
⋅
⋅
−
⋅
=
=
− ⋅ − ⋅e
− ⋅ − ⋅a t
a t a
a e
K t s p
S
n a t n n 1 a t n 22 2
1
sin 1
1 1
cos 1
) ( )) (
(
ω 2ω
ω 2ω
1
L-
( ) ( )
( ) ⎟ ⎟
⎠
⎞
⎜ ⎜
⎝
⎛ ⋅ − ⋅ ⋅ − ⋅ + ⋅ ⋅ − ⋅
− −
⋅
=
− ⋅ − ⋅a a t a a t
a K e
t
s
n nt
n a
2 2
2 2
1
1 sin
1 cos
1 1
1 )
(
2
ω
ω
ω
En posant :
cos ϕ = a
etsin ϕ = 1 − a
2 on peut écrire :( ) ( )
( ) ⎟ ⎟
⎠
⎞
⎜ ⎜
⎝
⎛ ⋅ ⋅ ⋅ − ⋅ + ⋅ ⋅ − ⋅
− −
⋅
=
− ⋅ − ⋅a t a t
a K e
t
s
n nt
n a
2 2
2 1
1 sin
cos 1
cos sin 1
1 )
(
2
ω ϕ ω
ω
ϕ
soit :
( ) ⎟ ⎟
⎠
⎞
⎜ ⎜
⎝
⎛ ⋅ ⋅ − ⋅ +
− −
⋅
=
−ω ⋅ − ⋅ω a t ϕ
a K e
t
s
nt
n a
2 2
1
1 sin
1 1 )
(
2
avec