Modern Control Engineering

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178 Chapter 5 / Transient and Steady-State Response Analyses

1.0

0.8

0.6

0.4

0.2

0

–0.2

–0.4

–0.6

–0.8

–1.0

0 24681012

vnt

c(t)

vn

z = 0.1

z = 0.3

z = 0.5

z = 0.7

z = 1.0

Figure 5–14

Unit-impulse

response curves of

the system shown in

Figure 5–6.

Impulse Response of Second-Order Systems. For a unit-impulse input r(t), the

corresponding Laplace transform is unity, or R(s)=1. The unit-impulse response C(s)

of the second-order system shown in Figure 5-6 is

The inverse Laplace transform of this equation yields the time solution for the response

c(t)as follows:

For 0z<1,

fort 0 (5–26)

For z=1,

fort 0 (5–27)

For z>1,

fort 0 (5–28)

Note that without taking the inverse Laplace transform of C(s)we can also obtain

the time response c(t)by differentiating the corresponding unit-step response, since

the unit-impulse function is the time derivative of the unit-step function. A family of

unit-impulse response curves given by Equations (5–26) and (5–27) with various val-

ues of zis shown in Figure 5–14. The curves c(t)/vnare plotted against the dimen-

sionless variable vnt, and thus they are functions only of z. For the critically damped

and overdamped cases, the unit-impulse response is always positive or zero; that is,

c(t)0. This can be seen from Equations (5–27) and (5–28). For the underdamped

case, the unit-impulse response c(t)oscillates about zero and takes both positive and

negative values.

c(t)=

vn

22 z^2 - 1

e-Az-^2 z

(^2) - 1 Bvn t

-

vn

22 z^2 - 1

e-Az+^2 z

(^2) - 1 Bvn t

,

c(t)=v^2 n te-vn^ t,

c(t)=

vn

21 - z^2

e-zvn^ tsinvn 21 - z^2 t,

C(s)=

v^2 n

s^2 + 2 zvn s+v^2 n

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