Modern Control Engineering

Example Problems and Solutions 539

Hence, we set

from which we obtain

Since the phase curve never crosses the –180° line, the gain margin is ±qdB. Noting that the

magnitude of G(jv)must be equal to 0 dB at v=2.3835, we have

from which we get

This Kvalue will give the phase margin of 50°.

A–7–17. For the standard second-order system

show that the bandwidth vbis given by

Note that vb/vnis a function only of z. Plot a curve of vb/vnversusz.

Solution.The bandwidth vbis determined from @CAjvbB/RAjvbB@=–3dB. Quite often, instead of

–3dB, we use –3.01dB, which is equal to 0.707. Thus,

Then

from which we get

v^4 n=0.5CAv^2 n-v^2 bB^2 + 4 z^2 v^2 n v^2 bD

v^2 n

3 Av^2 n-v^2 bB^2 +A 2 zvn vbB^2

=0.707

2

CAjvbB

RAjvbB

(^2) = 2 v
(^2) n
AjvbB^2 + 2 zvnAjvbB+v^2 n
(^2) =0.707
vb=vnA 1 - 2 z^2 + 24 z^4 - 4 z^2 + 2 B^1 ^2
C(s)
R(s)

=

v^2 n

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

K=

2.3835^2

222 +2.3835^2

=1.8259

2

K(jv+2)

(jv)^2

2

v=2.3835

= 1

vc=2.3835 radsec

tan-^1

vc

2

= 50 °

G(s)

+– K(s+ 2) s^12

Figure 7–133
Space-vehicle control
system.