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Problem 4.16PP
A compensated motor position control system is shown in Fig. Assume that the sensor dynamics 
are H(s) = 1.
Figure Control system
(a) Can the system track a step reference input rwith zero steady-state error? If yes. give the 
value of the velocity constant.
(a) Can the system track a step reference input rwith zero steady-state emor? If yes. give the 
value of the velocity constant.
(b) Can the system reject a step disturbance wwith zero steady-state en^or? If yes. give the value 
of the velocity constant.
(c) Compute the sensitivity of the closed-loop transfer function to changes in the plant pole at -2.
(d) In some instances there are dynamics in the sensor. Repeat parts (a) to (c) for |W(*) = 
and compare the corresponding velocity constants.
Step-by-step solution
step 1 of 8
Refer to Figure 4.29 in the textbook.
(a)
From the block diagram, it is clear that the system is Type 1 with
y ( 4 0 ( 4
R (s) l+ H (s )G (s )
y (« ) g ( 4
R (s) 1 + G ( j)
Write the expression for the error signal.
' ' i +a(s) '• ’
(2)
Step 2 of 8
, . 1 6 0 (j+ 4 ) , .
Substitute i for A^^jand —̂ — 30)^°'"
£ W = -
1 +
1 6 0 (j+ 4 )
s(i + 2)(*+30) 
(j+2)(i+30)
i ( i+ 2 ) ( i+ 3 0 ) + 1 6 0 ( i+ 4 )
Now. evaluate the steady-state system error. 
e„ - liin
= lin ii f -----^'-*• ^̂s(5 + 2)(5+30)+160(j+4)J
sO
Therefore, the system can track a step input with zero steady-state emor.
Step 3 of 8
Determine the value of velocity constant. 
= Hm j(7(5)
( ^ j( j+ 2 ) ( j+ 3 0 ) J
( 1 6 0 (j+ 4 ) ^™[(j -f2)(j +30) J
(»60)(4) 
- (2)(30)
=10.67
Therefore, the value of the velocity constant, is |1Q.67|-
Step 4 of 8
(b)
The system is Type 0 with respect to the disturbance and has the steady-state emor. Find the 
transfer function for 7 (4) •
1
y ( ^ ) = -
4(4 + 2 )
1 +
r i60(4+4)Y 1 ]
I IL«r«+2̂ l\ \ - — / ■ - / /
I
4(4 + 2)
4(4 + 2 ) (4 + 3 0 )+ I6 0 (4 + 4 ) 
4(4 + 2)(4 + 30)
4+30
4(4+2)(4+30)+160(4+4)
Determine the steady-state error to a disturbance input.
y .
*-*® ( 4̂(4 + 2)(4+30)+160(4+4)J
Hence, the system cannot reject a constant disturbance.
Step 5 of 8
(c)
In order to compute the sensitivity of the closed-loop transfer function to changes in the plant 
pole at _2. Now by determine the transfer function.
r ( ^ ) = -
l6 0 (^ + 4 )
«(« + >4)(s + 3 0 )+ I6 0 (2 + 4 )
Where. A was inserted for the pole at the nominal value of 2. 
Now evaluate the sensitivity of the system.
A ST
....... (3)TSA
Step 6 of 8
Apply partial differentiation to the transfer function with respect to A. 
160(5+4ST a ( 160(5+4) 'I
04 “ a4 [5 (5 + y ()(5 + 3 0 )+ I6 0 (5 + 4 )J(^5(5 + 4 )(5 + 30) +
160(5 + 4 )(5 )(5 + 30)
[5 (5 + 4 )(5 +30)+160 (5 + 4 ) ] ’
o r .Substitute — value in equation (3).
SA
4 [5 (5 + 4 )(5 + 3 0 )+ 1 6 0 (5 + 4 )][1 6 0 (5 + 4 )(5 )(5 + 3 0 )] 
" [1 6 0 (5 + 4 )][5 (5 + 4 )(5 + 30)+160(5 + 4 ) ] ‘
4 5 (5 + 3 0 )
5(5 + 4 )(5 +30)+160 (5 + 4)
25(5 + 3 0 )
(Since, 4 = 2)
5(5 + 2 )(5 + 30 )+ 160 (5+ 4 )
Therefore, the sensitivity of the closed-loop transfer function is
25(5 + 30)
5(5 + 2)(5 + 30)+160(5 + 4)
Step 7 of 8
((4
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