Modern Control Engineering

Example Problems and Solutions 543

Solution.The angle of Gc(jv)is given by

At we have

Since

or

and also

we have

Thus, the angle of GcAjv 1 Bbecomes 0° at

A–7–22. Consider the control system shown in Figure 7–138. Determine the value of gain Ksuch that the

phase margin is 60°. What is the gain margin with this value of gain K?

Solution.The open-loop transfer function is

=

K(10s+1)

s^3 +1.5s^2 +0.5s

G(s)=K

s+0.1

s+0.5

10

s(s+1)

v=v 1 = 1  1 T 1 T 2.

/GcAjv 1 B= 0 °

tan-^1

1

bB

T 1

T 2

+tan-^1 b

B

T 2

T 1

= 90 °

tan-^1

B

T 1

T 2

+tan-^1

B

T 2

T 1

= 90 °

tan atan-^1

B

T 1

T 2

+tan-^1

B

T 2

T 1

b=

B

T 1

T 2

+

B

T 2

T 1

1 -

B

T 1

T 2 B

T 2

T 1

=q

/GcAjv 1 B=tan-^1

B

T 1

T 2

+tan-^1

B

T 2

T 1

• tan-^1

1

b

B

T 1

T 2

• tan-^1 b
  B

T 2

T 1

v=v 1 = 1  1 T 1 T 2 ,

=tan-^1 vT 1 +tan-^1 vT 2 - tan-^1 vT 1 b-tan-^1 vT 2 b

/Gc(jv)= njv+

1

T 1

+ njv+

1

T 2

• njv+

b

T 1

• njv+

1

bT 2

Kss++ 0.1 0.5

10

+– s(s+ 1)

Figure 7–138
Control system.