# Transient Response of Second Order System

Pages: 7 (1158 words) Published: February 19, 2012
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 (1.13) (1.14)

√ ζ2 − 1

The settling time for each mode is 4τ i.e the time required for the exponential to decay to e−4 = 0.0183. The settling time of the system depends on the slower mode (the one with the larger time constant). Now for ζ = 1 Y (s) = or 2 ωn = A (s + ωn )2 + Bs (s + ωn ) + Cs ) ( ) ( 2 = A s2 + ωn + 2sωn + B s2 + sωn + Cs ( 2) = s2 (A + B) + s (2Aωn + Bωn + C) + s0 Aωn 2 B C A ωn + + 2 = s (s + ωn ) (s + ωn )2 s (s + ωn )

(1.16)

(1.17)

2

Comparing the like powers of s we have A = 1, B = −1 and C = −ωn . Therefore y(t) = 1 − eωn t − ωn teωn t Finally for ζ < 1 Y (s) = 2 A B C ωn √ √ = + + 2 + 2ζω s + ω 2 ) 2 s (s s n s + ζωn − jωn 1 − ζ s + ζωn + jωn 1 − ζ 2 n

(1.18)

(1.19)

It follows that A= B=
2 ωn 2 s2 + 2ζωn s + ωn

=1
s=0

(1.20) =(
1−ζ 2

) √ s s + ζωn + jωn 1 − ζ 2 2 ζ 2 − 1 − jζ ( ( 1 ) √ 1 − ζ2 √ )

(

2 ωn

s=−ζωn +jωn

1 )( √ ) √ −ζ + j 1 − ζ 2 2j 1 − ζ 2 (1.21)

=

C=

2 ωn

s s + ζωn − jωn 2 ζ 2 − 1 + jζ ( 1 √

1 − ζ2 )

s=−ζωn −jωn

=(
1−ζ 2

−ζ − j

1 − ζ2

1 )(

−2j

1 − ζ2

)

=

1 − ζ2

(1.22)

Y (s) =

1 1 )( ) + ( √ √ s 2 ζ 2 − 1 − jζ 1 − ζ 2 s + ζω − jω ζ 2 − 1 n n 1 )( ) + ( √ √ 2 ζ 2 − 1 + jζ 1 − ζ 2 s + ζωn + jωn ζ 2 − 1 (1.23)

and e e )+ ( ) y(t) = 1 + ( √ √ 2 ζ 2 − 1 − jζ 1 − ζ 2 2 ζ 2 − 1 + jζ 1 − ζ 2 Deﬁne β = △ ( ) √ −ζωn +jωn ζ 2 −1 t ( ) √ −ζωn −jωn ζ 2 −1 t

(1.24)

1 − ζ 2 . Then [ ] e−ζωn t e−jβωn t ejβωn t y(t) = 1 + + 2 −β 2 − jζβ −β 2 + jζβ [ ] e−ζωn t (−β 2 + jζβ) ejβωn t + (−β 2 − jζβ) e−jβωn t =1+ 2 β 2 (β 2 + ζ 2 ) [ ( ) ( )] ζ ejβωn t − e−jβωn t e−ζωn t − ejβωn t + e−jβωn t +j =1+ 2 β + ζ2 2 β 2 [( ) ( jβω t )] ejβωn t + e−jβωn t e−ζωn t ζ e n − e−jβωn t =1− 2 + β + ζ2 2 β 2j 3

(1.25)

but β 2 + ζ 2 = 1, and so y(t) = 1 − e
−ζωn t

[ ] ζ cos βωn t + sin βωn t β (1.26)

e−ζωn t =1− [(sin βωn t) ζ + (cos βωn t) β] β Now using the trignometric identity sin (A + B) = sin A cos B + cos A sin B...