!

!

E7.10 (a) The characteristic equation is

1 +K(s+ 2)

s(s+ 1)= 0 .

Therefore,

K = −(s2 + s)

(s+ 2),

and

dK

ds= −

s2 + 4s + 2

(s+ 2)2= 0 .

Solving s2+4s+2 = 0 yields s = −0.586 and −3.414. Thus, the system

breakaway and entry points are at s = −0.586 and s = −3.414.

(b) The desired characteristic polynomial is

(s+ 2 + aj)(s + 2− aj) = s2 + 4s+ 4 + a2 = 0 ,

where a is not specified. The actual characteristic polynomial is

s2 + (1 +K)s+ 2K = 0 .

Equating coefficients and solving for K yields K = 3 and a =√

2.Thus, when K = 3, the roots are s1,2 = −2±

√

2j.

(c) The root locus is shown in Figure E7.10.

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1

xxo

Real Axis

Imag Axis

*

*

K=3, s=-2+1.414j

s=-0.58s=-3.41

FIGURE E7.10

Root locus for 1 +Ks+2

s(s+1)= 0.