Modern Control Engineering

352 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method

Hence, three root-locus branches meet at point s=–1.2.The angles of departures at point

s=–1.2of the root locus branches that approach the asymptotes are that is, 60° and

• 60°. (See Problem A–6–4.)
  Finally, we shall examine if root-locus branches cross the imaginary axis. By substituting s=jv
  into the characteristic equation, we have

or

This equation can be satisfied only if v=0, K=0.At point v=0,the root locus is tangent to

thejvaxis because of the presence of a double pole at the origin. There are no points where root-

locus branches cross the imaginary axis.

A sketch of the root loci for this system is shown in Figure 6–66(b).

A–6–4. Referring to Problem A–6–3, obtain the equations for the root-locus branches for the system

shown in Figure 6–66(a). Show that the root-locus branches cross the real axis at the breakaway

point at angles ;60°.

Solution.The equations for the root-locus branches can be obtained from the angle condition

which can be rewritten as

By substituting we obtain

or

By rearranging, we have

Taking tangents of both sides of this last equation, and noting that

we obtain

which can be simplified to

vs-v(s+0.4)

(s+0.4)s+v^2

=

v(s+3.6)+vs

s(s+3.6)-v^2

v

s+0.4

-

v

s

1 +

v

s+0.4

v

s

=

v

s

+

v

s+3.6

1 -

v

s

v

s+3.6

tan ctan-^1 a

v

s+3.6

b; 180 °(2k+1)d =

v

s+3.6

tan-^1 a

v

s+0.4

b-tan-^1 a

v

s

b=tan-^1 a

v

s

b+tan-^1 a

v

s+3.6

b; 180 °( 2 k+ 1 )

tan-^1 a

v

s+0.4

b- 2 tan-^1 a

v

s

b-tan-^1 a

v

s+3.6

b=; 180 °(2k+1)

/s+jv+0.4- 2 /s+jv- /s+jv+3.6=; 180 °(2k+1)

s=s+jv,

/s+0.4- 2 /s-/s+3.6=; 180 °(2k+1)

n

K(s+0.4)

s^2 (s+3.6)

=; 180 °(2k+1)

A0.4K-3.6v^2 B+jvAK-v^2 B= 0

(jv)^3 +3.6(jv)^2 +K(jv)+0.4K= 0

; 180 °3,

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