Modern Control Engineering

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Problems 263

Since the system is stable, converges to a constant value. Noting that

we have

Since

we have

For a unit-step input r(t), since

we have

Note that zeros in the left half-plane (that is, positive ) will improve Kv. Poles close

to the origin cause low velocity-error constants unless there are zeros nearby.

Problems

Ta ,Tb ,p,Tm

1

Kv

=

1

slimS 0 sG(s)

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

=slimS 0

1

1 +G(s)

1

s

=

1

limsS 0 sG(s)

=

1

3 Kv

q

0

e(t)dt=limsS 0 E(s)=slimS 0

1

1 +G(s)

R(s)

3

q

0

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

slimS 0 Q¿(s)=T^1 +T^2 +p+Tn

slimS 0 P¿(s)=Ta+Tb+p+Tm

=limsS 0 CQ¿(s)-P¿(s)D

=limsS 0

Q¿(s)-P¿(s)

Q(s)+sQ¿(s)

3

q

0

e(t)dt=limsS 0

Q(s)-P(s)

sQ(s)

3

q

0

e(t)dt=limsS 0 s

E(s)

s

=limsS 0 E(s)

1

q

0 e(t)dt

B–5–1.A thermometer requires 1 min to indicate 98%of
the response to a step input. Assuming the thermometer to
be a first-order system, find the time constant.
If the thermometer is placed in a bath, the temperature
of which is changing linearly at a rate of 10°min, how much
error does the thermometer show?

B–5–2.Consider the unit-step response of a unity-feedback
control system whose open-loop transfer function is

G(s)=

1

s(s+1)

Obtain the rise time, peak time, maximum overshoot, and

settling time.

B–5–3.Consider the closed-loop system given by

Determine the values of zandvnso that the system

responds to a step input with approximately 5%overshoot

and with a settling time of 2 sec. (Use the 2%criterion.)

C(s)

R(s)

=

v^2 n

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