Transient Response of Second Order System
1
Transient Responce of Second Order Systems
Consider the transfer function G(s) =
2 Y (s) ωn = 2 2 U (s) s + 2ζωn s + ωn
(1.1)
Let U (s) = 1/s then Y (s) =
2 ωn 2 s (s2 + 2ζωn s + ωn )
(1.2)
In order to determine y(t) will perform partial fraction expansion and then take Lapalace inverse. The partial fraction expansion will depend on the roots of the second order polynomial in the denominator √ s = −ζωn ± ωn ζ 2 − 1 (1.3) We now have three cases depending on the value of ζ • For ζ > 1 the expression under the square root is positive and therefore the roots are real and distinct √ √ s = −ζωn + ωn ζ 2 − 1 s = −ζωn − ωn ζ 2 − 1 (1.4) • For ζ = 1 the expression under the square root is zero and therefore the roots are real and repeated s = −ωn (1.5)
• For ζ < 1 the expression under the square root is negative and therefore the roots are a complex conjugate pair √ s = −ζωn ± jωn 1 − ζ 2 (1.6) Consider first the case where ζ > 1, then Y (s) =
2 A B ωn C √ √ = + + 2 + 2ζω s + ω 2 ) s (s s n s + ζωn − ωn ζ 2 − 1 s + ζωn + ωn ζ 2 − 1 n
(1.7)
1
The constants A, B and C can be evaluated as follows A= B=
2 ωn 2 s2 + 2ζωn s + ωn
=1 s=0 (1.8) =( ζ 2 −1
) √ s s + ζωn + ωn ζ 2 − 1 1 ) √ 2 ζ2 − ζ ζ2 − 1 − 1 ( (
2 ωn
(
2 ωn
s=−ζωn +ωn
√
1 )( √ ) √ −ζ + ζ 2 − 1 2 ζ 2 − 1 (1.9)
=
C=
s s + ζωn − ωn (
√
) ζ2 − 1
s=−ζωn −ωn
√
=( ζ 2 −1
−ζ −
√
1 )( √ ) ζ 2 − 1 −2 ζ 2 − 1 (1.10)
= Therefore
1 ) √ 2 ζ2 + ζ ζ2 − 1 − 1 1 1 )( ) + ( √ √ s 2 ζ 2 − ζ ζ 2 − 1 − 1 s + ζω − ω ζ 2 − 1 n n 1 )( ) + ( √ √ 2+ζ 2−1−1 2−1 2 ζ ζ s + ζωn + ωn ζ
Y (s) =
(1.11)
and
e e )+ ( ) y(t) = 1 + ( √ √ 2 ζ2 − ζ ζ2 − 1 − 1 2 ζ2 + ζ ζ2 − 1 − 1
( ) √ −ζωn +ωn ζ 2 −1 t
) ( √ −ζωn −ωn ζ 2 −1 t
(1.12)
The solution has one input mode and two plant modes. The time constants for the two modes are τ1 = τ2 = The solution is of the form y(t) = 1 + Be−t/τ1 + Ce−t/τ2 (1.15) 1 −ζωn + ωn 1 −ζωn − ωn √ ζ2 − 1
Transient Responce of Second Order Systems
Consider the transfer function G(s) =
2 Y (s) ωn = 2 2 U (s) s + 2ζωn s + ωn
(1.1)
Let U (s) = 1/s then Y (s) =
2 ωn 2 s (s2 + 2ζωn s + ωn )
(1.2)
In order to determine y(t) will perform partial fraction expansion and then take Lapalace inverse. The partial fraction expansion will depend on the roots of the second order polynomial in the denominator √ s = −ζωn ± ωn ζ 2 − 1 (1.3) We now have three cases depending on the value of ζ • For ζ > 1 the expression under the square root is positive and therefore the roots are real and distinct √ √ s = −ζωn + ωn ζ 2 − 1 s = −ζωn − ωn ζ 2 − 1 (1.4) • For ζ = 1 the expression under the square root is zero and therefore the roots are real and repeated s = −ωn (1.5)
• For ζ < 1 the expression under the square root is negative and therefore the roots are a complex conjugate pair √ s = −ζωn ± jωn 1 − ζ 2 (1.6) Consider first the case where ζ > 1, then Y (s) =
2 A B ωn C √ √ = + + 2 + 2ζω s + ω 2 ) s (s s n s + ζωn − ωn ζ 2 − 1 s + ζωn + ωn ζ 2 − 1 n
(1.7)
1
The constants A, B and C can be evaluated as follows A= B=
2 ωn 2 s2 + 2ζωn s + ωn
=1 s=0 (1.8) =( ζ 2 −1
) √ s s + ζωn + ωn ζ 2 − 1 1 ) √ 2 ζ2 − ζ ζ2 − 1 − 1 ( (
2 ωn
(
2 ωn
s=−ζωn +ωn
√
1 )( √ ) √ −ζ + ζ 2 − 1 2 ζ 2 − 1 (1.9)
=
C=
s s + ζωn − ωn (
√
) ζ2 − 1
s=−ζωn −ωn
√
=( ζ 2 −1
−ζ −
√
1 )( √ ) ζ 2 − 1 −2 ζ 2 − 1 (1.10)
= Therefore
1 ) √ 2 ζ2 + ζ ζ2 − 1 − 1 1 1 )( ) + ( √ √ s 2 ζ 2 − ζ ζ 2 − 1 − 1 s + ζω − ω ζ 2 − 1 n n 1 )( ) + ( √ √ 2+ζ 2−1−1 2−1 2 ζ ζ s + ζωn + ωn ζ
Y (s) =
(1.11)
and
e e )+ ( ) y(t) = 1 + ( √ √ 2 ζ2 − ζ ζ2 − 1 − 1 2 ζ2 + ζ ζ2 − 1 − 1
( ) √ −ζωn +ωn ζ 2 −1 t
) ( √ −ζωn −ωn ζ 2 −1 t
(1.12)
The solution has one input mode and two plant modes. The time constants for the two modes are τ1 = τ2 = The solution is of the form y(t) = 1 + Be−t/τ1 + Ce−t/τ2 (1.15) 1 −ζωn + ωn 1 −ζωn − ωn √ ζ2 − 1