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Problem 4.25PP
A linear ODE model of the DC motor with negligible armature inductance {La = 0) and with a 
disturbance torque wwas given earlier in the chapter; it is restated here, in slightly different form.
= Mb + ^W,
Ki Ki
where dm is measured In radians. Dividing through by the coefficient o f’^ ,w e obtain 
Sm + 0\4m — ^Mfl + 
where
„ KtKe . K, ^ I
With rotating potentiometers, it is possible to measure the positioning error between 6 and the 
reference angle 0ror e = dre f- 0/n.With a tachometer we can measure the motor speed 6m-
With rotating potentiometers, it is possible to measure the positioning error between d and the 
reference angle drox e= dref - 0/n.With a tachometer we can measure the motor speed 
Consider using feedback of the error e and the motor speed ^ in the form
where K and TD are controller gains to be determined.
(a) Draw a block diagram of the resulting feedback system showing both dm and ^ as variables 
in the diagram representing the motor.
(b) Suppose the numbers work out so that a1 = 65. bO = 200, and cO = 10. If there is no load 
torque (w = 0), what speed {in rpm) results from va = 100 V?
(c) Using the parameter values given in part (b), let the control be D = kP + kDs and find kP and 
kD so that, using the results of Chapter 3, a step change in dref with zero load torque results in a 
transient that has an approximately 17% overshoot and that settles to within 5% of steady state 
in less than 0.05 sec.
(d) Derive an expression for the steady-state error to a reference angle input and compute its 
value for your design In part (c) assuming dref= 1 rad.
(e) Derive an expression for the steady-state error to a constant disturbance torque when dref= 0 
and compute Its value for your design in part (c) assuming w = 1.0.
Step-by-step solution
step 1 of 5
Sketch the block diagram of the resulting feedback sjrstem.
Thus, the required block diagram is sketched.
Step 2 of 5
(b)
I f Va = constant the system is in steady state
& = ^ K -
Find
6 which is given by
^ 200x100 60 rad s-’p = -
65 27T rpm
0= 2938 rpm
Thus, we get ̂ = 2938 rpm .
Step 3 of 5
(c)
Find —
0
0y
Conq>aring with standard second order equation
0 a j
0^
When = \1% we get
^ '= 0 .5 .
= 0.05 s 
Find .
= 0.05
k = 120
Thus, on comparing the coefficients, we get
i : = 72|
t; = 3 . 8 xlQT^I
step 4 of 5
(d)
Find the steady state error. 
£ ( s ) = 0 ^ -0
s ( s + a .T ^ IC b . \
s^ + s(a^ + 
1
For^,.« we get
5
s„ = lim s5 (s)
K = o |
Step 5 of 5
(e)
We know that
0
Q i s* + TjfKbft) + Kbfi
Find the response to torque.
0„ = lim s0{s)
" j-»o '■ '
0„ = lim s0{s) 
" j-*o '