Example on Root Locus Sketching and Control Design
MCE441  Spring 05
Dr. Richter
April 25, 2005
The following figure represents the system used for controlling the robotic manipulator of a Mars Rover.
We wish to:
K
(
s
2
+6
.
5
s
+12)
s
1
(
s
+1)(
s
+2)
U
(
s
)
Y
(
s
)
R
(
s
)
+

Figure 1: Control System
1. Provide a detailed handsketch of the root locus.
2. Find the gain
K
so that the system responds with an overshoot of 1 percent.
3. Find the gain
K
so that the system responds with an overshoot of less than 1 percent and a settling
time as fast as possible.
Root Locus Sketching
The poles of the openloop transfer function
G
(
s
)
K
(
s
) are given by
s
= 0
,
s
=

1
,
s
=

2
and there is a pair of complex zeros at
s
=

3
.
25
±
1
.
199
i
The order of the system (number of poles) is
n
= 3 and the number of zeros is
m
= 2. According to this,
m
= 2 poles go to the zeros as
K
→ ∞
and the remaining pole goes to infinity following an asymptote. The
angle of the asymptote is obviously 180
◦
, but in general, we find asymptote angles with the formula
φ
A
=
180
n

m
(2
k
+ 1)
where
k
= 0
,
1
, ..n

m

1. In our case,
n

m

1 = 0, so
φ
A
= 180
◦
Again, in this case it is unnecessary to find the center of the asymptotes, but in the general case it is obtained
by subtracting the sum of zeros from the sum of poles and dividing by
n

m
. In this case
σ
A
=
(

1

2)

(

3
.
25 + 1
.
199
i

3
.
25

1
.
199
i
)
1
= 3
.
5
With this information, we still have two possibilities for the shape of the root locus that comply with the
basic requirements, as shown in the figure below.
In order to determine which one of the possibilities is
1
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Possibility 1
Possibility 2
Figure 2: Two possibilities for the root locus
correct, we find the breakin and breakaway points. To do this, solve for
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 Fall '10
 WalterEversman
 Root Locus, 1 percent, Root Locus Sketching, 0.0642 1.5 1 K, 0.6094 radians

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