Modern Control Engineering

Example Problems and Solutions 549

MATLAB Program 7–24 produces the Bode diagram shown in Figure 7–145. From this plot, the

phase margin is found to be 14°. The gain margin is ±qdB.

Frequency (rad/sec)

Bode Diagram of G 1 (s) = 20/[s(s + 1)]

− 200

− 100

− 150

− 50

− 100

Phase (deg); Magnitude (dB)

50

− 50

0

10 −^1100101102

Figure 7–145
Bode diagram of
G 1 (s).

MATLAB Program 7–24

num = [20];

den = [1 1 0];

w = logspace(-1,2,100);

bode(num,den,w)

title('Bode Diagram of G1(s) = 20/[s(s + 1)]')

Since the specification calls for a phase margin of 50°, the additional phase lead necessary to

satisfy the phase-margin requirement is 36°. A lead compensator can contribute this amount.

Noting that the addition of a lead compensator modifies the magnitude curve in the Bode di-

agram, we realize that the gain crossover frequency will be shifted to the right. We must offset the

increased phase lag of G 1 (jv)due to this increase in the gain crossover frequency. Taking the shift

of the gain crossover frequency into consideration, we may assume that fm,the maximum phase

lead required, is approximately 41°. (This means that approximately 5° has been added to com-

pensate for the shift in the gain crossover frequency.) Since

fm=41° corresponds to a=0.2077.Note that a=0.21corresponds to fm=40.76°. Whether

we choose fm=41° or fm=40.76° does not make much difference in the final solution. Hence,

let us choose a=0.21.

sinfm=

1 - a

1 +a