4.4 The DC-motor speed control in Fig. 4.38 is described by the differential equation
y + 60 y = 600va − 1500 w,
where y is the motor speed, va is the armature voltage, and w is the load torque. Assume the
armature voltage is computed using the PI control law
(
t
)
va = k p e + k I ∫ edt ,
0
where e = r − y .
(a) Compute the transfer function from W to Y as a function of kp and kI .
(b) Compute values for kp and kI so that the characteristic equation of the closed-loop system
will have roots at −60 ± 60j.
Figure 4.38: Unity feedback system with preÞlter for Problem 4.4
Solution:
(a ) Transfer function: Set R ( s ) = 0, then E ( s ) = −Y ( s )
k ⎞
⎛
−600 ⎜ k p + I ⎟ Y ( s ) − 1500W ( s )
s ⎠
⎝
Y ( s) =
( s + 60)
Y ( s)
−1500s
= 2
W ( s ) s + 60(1 + 10k p ) s + 600k I
(b) For roots at s1,2 = −60 ± j 60,
⎧ s1 + s2 = −60(1 + 10k p ) = −120
⎨
⎩ s1s2 = 7200 = 600k I
⇒
k p = 0.1,
k I = 12
4.26 Consider the system shown in Fig. 4.49 with PI control.
(a) Determine the transfer function from R to Y .
(b) Determine the transfer function from W to Y .
(c) Use Routh's criteria to find the range of (k p , k I ) for which the system is stable.
(d*) Pick k p and k I so that the closed-loop system is stable. What is the steady state error
when r (t ) = 1(t ) and w(t ) = 1(t )?
1
Figure 4.49: Control system for Problem 4.26
Solution:
k p s + kI
10
10(k p s + k I )
Y ( s)
s
= s + s + 20
= 3 2
(a)
k p s + k I s + s + 10(2 + k p ) s + 10k I
10
R( s)
1+ 2
s + s + 20
s
10
2
Y (s)
10s
s + s + 20
=
= 3 2
(b)
k p s + k I s + s + 10(2 + k p ) s + 10k I
10
W ( s)
1+ 2
s + s + 20
s
(c) The characteristic equation is s 3 + s 2 + 10(2 + k p ) s + 10k I = 0. The Routh's array is
2
s 3:
1
10(2 + k p )
s2:
1
10k I
s : 10(2 + k p − k I )
1
s0:
10k I
For stability we must have k I > 0 and k p > k I − 2.
(d *) Choose k I = 1, k p = 1, The transfer functions become,
10 s
10s + 10
Y ( s)
Y ( s)
,
= 3 2
= 3 2
W ( s ) s + s + 30s + 10
R( s ) s + s + 30s + 10
Y (0)
Y ( s)
Since
= 1,
= 0, the steady state error is 0.
R(0)
W (s)
As a verification, let r (t ) = 1(t ) and w(t ) = 1(t ), using matlab, we can get the following response,
Response to r(t)=1(t), w(t)=1(t)
1.4
1.2
Response y(t)
1
0.8
0.6
0.4
0.2
0
0
5
10
15
time (sec)
20
2
25
30