Problem 1
Figure 1
The characteristic equation for the closed loop in figure 1 can be written as follows:
C
OL
2
2k (1 3 s)
1G (
s
)1
0.2s
0.4s 1
+
+=
+
++
(S1.1)
Setting (S1.1) equal to zero and substituting k
C
=2 yields:
32
0.2s
0.4s
5s 12
0
+
=
(S1.2)
Test 1: All the coefficients of (S1.2) are positive. Therefore, one can move to test 2.
Test 2: (S1.2) is a 3
rd
order polynomial. Thus, the RouthHurwitz array is composed of 4
rows.
Column 1
Column 2
Row 1
0.2
5
Row 2
0.4
12
Row 3
[(0.4*5) (12*0.2)] 0.4
1
−=
−
0
Row 4
( 1*12)
1 12
−−
=
0
As all the coefficients of the first column of the RouthHurwitz array are not positive, the
system is unstable. Moreover, as the sign changes two times (from row 2 to row 3 and
row 3 to row 4), there are two roots with positive real part.
Problem 2
Figure 2
For the proportional controller, the characteristic equation for the closed loop in figure 2
can be written as follows:
controller
valve
tank 1
Measuring
device
SP
y(
s
)
y(s)
C
G
1
3
1
(s 1)
+
1
controller
valve
tank 1
Measuring
device
SP
s
)
y(s)
C
k(
1 3s
)
+
1
2
1
0.2s
0.4s 1
+
+
2
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OL
3
k
1G (
s
)1
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 Spring '09
 Smith
 Complex number, Row, RouthHurwitz array, controller valve tank

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