Solution 5.8.1.22
G
H
C
R
+
-
Figure 1:
For the system shown above
GH(s)=
K(s +2)
s(s +1)(s +40)
:
The rst step is to plot the poles and zeros of GH in the s-plane and
then nd the root locus on the real axis. The shaded regions of the real axis
in Figure 2 showwhere the root locus occurs. The rule is that root locus
occurs on the real axis to the left of an odd countofpoles and zeros. That
is, if you stand on the real axis and look to your right you must countan
odd numberofpoles and zeros.
The next step is to compute the asymptotes. The numberofasymptotes
is
p
ex
=Number of poles of GH ;Numberofnite zeros of GH:
j Im(s)
Re(s)
-40
-1
-2
Figure 2:
1
The asymptotes occur at the angles
`
=
1+2`
p
ex
180
` =0;;1;;:::p
ex
;1:
In the presentcase there are three poles and one nite zero so p
ex
=2.
Thus, there are two asymptotes at
0
=
1+(2)(0)
2
180
=90
1
=
1+(2)(1)
2
180
=270
The asymptotes intersect the real axis at
i
=
Sum of poles of GH ; Sum of zeros of GH
p
ex
=
[(;1)+ (;40)]; [(;2)]
2
= ;19:5
Since there is one nite zero and three poles, twoclosed loop poles will
migrate to zeros at innity,onesuchzeroatthe`end' of eachasymptote.
Twopossible root loci are shown in Figure 3 Of the two, that shown
in Figure 3 (a) is by far the more probable. Torule out case (b) requires
computing the gain along the real axis between s = ;40 and s = ;2Before
doing that, wenotethat there is also break-out pointbetween s = ;1and
s =0.Weknow that for lowgain the closed loop poles occur near the poles
of GH.Asthegain increases the closed loop poles migrate awayfrom the
poles of GH towards the zeros of GH.Inthis case all twozeros are zeros
at innity. The twoclosed loop poles migrating towards eachotheron the
interval [;1;;0] eventually meet and break-out of the real axis. The question
is where do they go after they break-out?
To answer that question which shows the more probable root locus and
a map of the gain between s = ;40 and s = ;2. The gain is calculated by
solving the equation
jGH(s)j =1
for points along the real axis between s = ;40 and s = ;2. That is
K =
jsjjs +1jjs +40j
js +2j
2
Im(s)
Re(s)
-40
-1
-2
j Im(s)
Re(s)
-40
-1
-2
(a)
(b)
Figure 3: Completed Root Locus
S -22 -21 -20 -19 -18
K 415.8 420 422.22 422.47 420.75
Table 1: SearchforMaximum in Gain
3
Table 1 summarizes the searchforpeaks and valleys.
From the table wesee that there is a maximum around s = ;19. Having
found the maximum, weknow the minimum will exist and so our earlier
supposition is veried, The correct root locus is that shown in Figure 3 (a).
4