Modern Control Engineering

Example Problems and Solutions 553

From MATLAB Program 7–27 and Figure 7–148 it is clearly seen that the phase margin is ap-

proximately 50° and the gain margin is ±q dB. Since the static velocity error constant Kvis

20 sec–1, all the specifications are met. Before we conclude this problem, we need to check the

transient-response characteristics.

Unit-Step Response: We shall compare the unit-step response of the compensated system with

that of the original uncompensated system.

The closed-loop transfer function of the original uncompensated system is

The closed-loop transfer function of the compensated system is

MATLAB Program 7–28 produces the unit-step responses of the uncompensated and compen-

sated systems. The resulting response curves are shown in Figure 7–149. Clearly, the compensat-

ed system exhibits a satisfactory response. Note that the closed-loop zero and poles are located

as follows:

Zero at s=–3.0101

Poles at s=–5.2880;j5.6824, s=–4.7579

Unit-Ramp Response: It is worthwhile to check the unit-ramp response of the compensated

system. Since Kv=20sec–1, the steady-state error following the unit-ramp input will be

C(s)

R(s)

=

95.238s+286.6759

s^3 +15.3339s^2 +110.5719s+286.6759

C(s)

R(s)

=

10

s^2 +s+ 10

Frequency (rad/sec)

Bode Diagram of Gc(s)G(s)

− 200

− 100

− 150

− 50

− 100

Phase (deg); Magnitude (dB)

− 50

50

0

10 −^1100101102103

Figure 7–148
Bode diagram of
Gc(s)G(s).