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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