Solution 4.6.4.9
The closed loop transfer function is
T
c
(s) =
K
(s +0:1)(s+4)(s+ 10)
1+
K
(s +0:1)(s+4)(s +10)
=
K
s
3
+14:1s
2
+41:4s+(K +4)
:
The characteristic equation is
s
3
+14:1s
2
+41:4s+(K +4)=0:
The MATLAB program
K=10
p=[1 14.1 41.1 4+K]
roots(p)
K=30.68
p=[1 14.1 41.1 4+K]
roots(p)
K=68
p=[1 14.1 41.1 4+K]
roots(p)
K=60
p=[1 14.1 41.1 4+K]
roots(p)
K=100
p=[1 14.1 41.1 4+K]
roots(p)
gh = zpk([],[-0.1 -4 -10],1)
K1 = [10 30.68 60 100]
[R,K] = rlocus(gh,K1)
plot(R,'kd')
print -deps rl4649.eps
produces the output
K=
10
1
p=
1.00000000000000 14.10000000000000 41.10000000000000 14.00000000000000
ans =
-10.20815647672320
-3.50000000000000
-0.39184352327679
K=
30.68000000000000
p=
1.00000000000000 14.10000000000000 41.10000000000000 34.68000000000000
ans =
-10.50039600358139
-1.79980199820931 + 0.25188360163417i
-1.79980199820931 - 0.25188360163417i
K=
68
p=
1.00000000000000 14.10000000000000 41.10000000000000 72.00000000000000
2
ans =
-10.94617029430882
-1.57691485284559 + 2.02261773677814i
-1.57691485284559 - 2.02261773677814i
K=
60
p=
1.00000000000000 14.10000000000000 41.10000000000000 64.00000000000000
ans =
-10.85749836877930
-1.62125081561034 + 1.80723258634298i
-1.62125081561034 - 1.80723258634298i
K=
100
p=
1.0e+02 *
0.01000000000000 0.14100000000000 0.41100000000000 1.04000000000000
ans =
-11.27239055897766
3
-1.41380472051116 + 2.68835232899338i
-1.41380472051116 - 2.68835232899338i
Zero/pole/gain:
1
--------------------
(s+0.1) (s+4) (s+10)
K1 =
1.0e+02 *
0.10000000000000 0.30680000000000 0.60000000000000 1.00000000000000
R=
Columns 1 through 2
-10.16131396237127 -10.45858281898964
-3.55065176580416 -1.82070859050518 - 0.03093642467982i
-0.38803427182457 -1.82070859050518 + 0.03093642467982i
Columns 3 through 4
-10.82056537115050 -11.23989884266038
-1.63971731442475 - 1.79610421381808i -1.43005057866981 - 2.68471742538630i
-1.63971731442475 + 1.79610421381808i -1.43005057866981 + 2.68471742538630i
K=
1.0e+02 *
0.10000000000000
0.30680000000000
0.60000000000000
1.00000000000000
4
-12 -10 -8 -6 -4 -2 0
-3
-2
-1
0
1
2
3
Figure 1: Plot of solutions
EDU>
The plot of the points is shown in Figure 1
For K = 10, the MATLAB dialogue
EDU>g = zpk([],[-0.1 -4 -10],10)
Zero/pole/gain:
10
--------------------
(s+0.1) (s+4) (s+10)
EDU>tc = feedback(g,1)
Zero/pole/gain:
10
-----------------------------
(s+10.16) (s+3.551) (s+0.388)
5
Time (sec.)
A
mp
li
tu
d
e
Step Response
0 5 10 15
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Figure 2: Step Response for K =30:68
EDU>step(tc)
EDU>print -deps sr4649a.eps
EDU>
produces the step response shown in Figure 2.
For K =30:68, the MATLAB dialogue
EDU>g = zpk([],[-0.1 -4 -10],30.68)
Zero/pole/gain:
30.68
--------------------
(s+0.1) (s+4) (s+10)
EDU>tc = feedback(g,1)
Zero/pole/gain:
30.68
6
Time (sec.)
A
mp
li
tu
d
e
Step Response
0 0.6 1.2 1.8 2.4 3 3.6
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Figure 3: Step Response for K =4:87
--------------------------------
(s+10.46) (s^2 + 3.641s + 3.316)
EDU>step(tc)
EDU>print -deps sr4649b.eps
EDU>
produces the step response shown in Figure 3.
For K = 60, the MATLAB dialogue
EDU>g = zpk([],[-0.1 -4 -10],60)
Zero/pole/gain:
60
--------------------
(s+0.1) (s+4) (s+10)
EDU>tc = feedback(g,1)
7
Time (sec.)
A
mp
li
tu
d
e
Step Response
0 0.6 1.2 1.8 2.4 3 3.6
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Figure 4: Step Response for K =60
Zero/pole/gain:
60
--------------------------------
(s+10.82) (s^2 + 3.279s + 5.915)
EDU>step(tc)
EDU>print -deps sr4649c.eps
EDU>
produces the step response shown in Figure 4.
For K = 100, the MATLAB dialogue
EDU>g = zpk([],[-0.1 -4 -10],100)
Zero/pole/gain:
100
--------------------
(s+0.1) (s+4) (s+10)
8
Time (sec.)
A
mp
li
tu
d
e
Step Response
0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Figure 5: Step Response for K =100
EDU>tc = feedback(g,1)
Zero/pole/gain:
100
-------------------------------
(s+11.24) (s^2 + 2.86s + 9.253)
EDU>step(tc)
EDU>print -deps sr4649d.eps
EDU>
produces the step response shown in Figure 5.
9