Modern Control Engineering

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262 Chapter 5 / Transient and Steady-State Response Analyses

If the input is a unit ramp, then the steady-state error is

Therefore, if kis chosen as

then the steady-state error for following a ramp input can be made equal to zero. Note that, if there

are any variations in the values of zand/orvndue to environmental changes or aging, then a

nonzero steady-state error for a ramp response may result.

A–5–26. Consider the stable unity-feedback control system with feedforward transfer function G(s).

Suppose that the closed-loop transfer function can be written

Show that

wheree(t)=r(t)-c(t)is the error in the unit-step response. Show also that

Solution.Let us define

and

Then

and

For a unit-step input,R(s)=1/sand

E(s)=

Q(s)-P(s)

sQ(s)

E(s)=

Q(s)-P(s)

Q(s)

R(s)

C(s)

R(s)

=

P(s)

Q(s)

AT 1 s+ 1 BAT 2 s+ 1 BpATn s+ 1 B=Q(s)

ATa s+ 1 BATb s+ 1 B p ATm s+ 1 B=P(s)

1

Kv

=

1

slimS 0 sG(s)

=AT 1 +T 2 +p+TnB-ATa+Tb+p+TmB

3

q

0

e(t)dt=AT 1 +T 2 +p+TnB-ATa+Tb+p+TmB

C(s)

R(s)

=

G(s)

1 +G(s)

=

ATa s+ 1 BATb s+ 1 BpATm s+ 1 B

AT 1 s+ 1 BAT 2 s+ 1 BpATn s+ 1 B

(mn)

k=

2 z

vn

=

2 zvn-v^2 n k

v^2 n

=slimS 0 s a

s^2 + 2 zvn s-v^2 n ks

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

b

1

s^2

e(q)=r(q)-c(q)

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