2006’ Automatic Control Class
Final Examination
학번:
성명:
점수
Problem 1 Sketch the root loci for the system shown in Figure 1.
Figure 1.
Solution A root locus exists on the real axis between points s=-1 and s=-3.6. The
asymptotes can be determined as follows:
 180  2k  1
 90  ,90 
Angles of asymptotes =
3 1
The intersection of the asymptotes and the real axis is found from
a 
(
0  0  3.6  1
  1 .3 )
3 1
Since the characteristic equation is
s 3  3.6s 2  K s  1  0
We have
s 3  3.6s 2
K 
s 1
The breakaway and break-in points are found from
dK
ds

3s
2



 7.2s s  1  s 3  3.6s 2
0
s  12
from which we
get
s  0,1.65  j 0.9367
Point s=0 corresponds to the actual breakaway point. But points s=-1.65±j0.9367 are
neither breakaway nor break-in points, because the corresponding gain values K
become complex quantities.
To check the points where root-locus branches may cross the imaginary axis,
substitute s=j into the characteristic equation.
1
2006’ Automatic Control Class
Final Examination
학번:
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K  3.6 2  j K   2  0




점수
Note that this equation can be satisfies only if =(0), K=0. Because of presence of a
double pole at the origin, the root locus is tangent to the j axis at =0. The root-locus
branches do not cross the j axis. Now draw the root-locus plot.
Problem 2 Sketch the root loci for the system shown in Figure 2, following the similar
procedure in Problem 1.
Figure 2.
Solution
2
2006’ Automatic Control Class
Final Examination
학번:
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점수
Problem 3 A control system with
G s  
H s   1
K
s s  1
2
is unstable for all positive values of gain K. Plot the root loci of the system. By using
this plot, show that this system can be stabilized by adding a zero on the negative real
axis or by modifying G(s) to G1(s), where
G s  
K s  a 
s 2 s  1
(0  a  1)
Solution A root-locus plot for the system with
G s  
H s   1
K
s s  1
2
is shown in Figure 3. Since two branches lie in the right half-plane, the system is
unstable for any values of K>0.
Addition of a zero to the transfer function G(s) bends
the right half-plane branches to the left and brings all root-locus branches to the left
half-plane, as shown in the root-locus plot in Figure 4.
G s  
K s  a 
s 2 s  1
H s   1 (0  a  1)
is stable for all K>0.
Figure 3.
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Thus the system with
2006’ Automatic Control Class
Final Examination
학번:
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점수
Figure 4.
Problem 4 Draw the Bode Diagram of the following transfer function:
G j  
e  jL
1  jT
where L=0.5 and T=1.
Solution
The log magnitude is
20 log G j   0  20 log
1
1  jT
The phase angle of G(j) is
G j   e  jL  
1
 L  tan 1 T
1  jT
The log magnitude and phase-angle curves for this transfer function with L=0.5 and
T=1 are shown in the figure below:
4
2006’ Automatic Control Class
Final Examination
학번:
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점수
Problem 5 Draw the Bode diagram of the following nonminimum-phase system:
C s 
 1  Ts
Rs 
Obtain the unit-ramp response of the system and plot c(t) versus t.
Solution
The Bode diagram of the system is shown in Figure 5.
Figure 5.
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2006’ Automatic Control Class
Final Examination
학번:
성명:
점수
For a unit ramp input,
C s  
1  Ts 1 T
 2 
s
s2
s
The inverse Laplace transform of C(s) gives
ct   t  T
fo r t  0
Figure 6 shows the response curve c(t) versus t.
Figure 6.
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