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Problem 10.17PP
(Contributed by Prof. L. Swindlehurst) The feedback control system shown in Fig. is proposed as 
a position control system. A key component of this system is an armature-controlled DC motor. 
The input potentiometer produces a voltage Ei that is proportional to the desired shaft position; Ei 
= Kpdi. Similarly, the output potentiometer produces a voltage EO that is proportional to the actual 
shaft position: £0 = Kpd. Note that we have assumed that both potentiometers have the same 
proportionality constant. The error signal Ei - EO drives a compensator, which in turn produces 
an armature voltage that drives the motor. The motor has an armature resistance Ra, an 
armature inductance La, a torque constant Kt, and a back emf constant Ke. The moment of 
inertia of the motor shaft is Jm, and the rotational damping due to bearing friction is Bm. Finaiiy, 
the gear ratio is A/; 1, the moment of inertia of the load is JL, and the load damping is BL.
(a) Write the differential equations that describe the operation of this feedback system.
(aj Wnife tne ainefenfiai dquaiio'nS tnar'deschOS thfe operation bt inis reeaPaciCsystem.
(b) Find the transfer function relating 00 and 6i(s) for a general compensator Dc(s).
(c) The open-loop frequency-response data shown in Table were taken using the armature 
voltage va of the motor as an input and the output potentiometer voltage EO as the output. 
Assuming that the motor is linear and minimum-phase, make an estimate of the transfer function 
of the motor.
0m(s)G(s) =
where dm is the angular position of the motor shaft.
(d) Determine a set of perfonnance specifications that are appropriate for a position control 
system and will yield good performance. Design Dc(s) to meet these specifications.
(e) Verify your design through analysis and simulation using Matlab.
Figure A servomechanism with gears on the motor shaft and potentiometer sensore
Frequency-Response Data
Frequency (rad/sec) Frequency (rad/sec)
0.1 60.0 10.0 14.0
0.2 54.0 20.0 2.0
0.3 50.0 40.0 -10.0
0.5 46.0 60.0 -20.0
0.8 42.0 65.0 -21.0
1.0 40.0 80.0 -24.0
2.0 34.0 100.0 -30.0
3.0 30.5 200.0 -48.0
4.0 27.0 300.0 -59.0
5.0 23.0 500.0 -72.0
7.0 19.5
Step-by-step solution
step 1 of 6
Step 2 of 6
Using KIRCHHOFF’S voltage laws we can write
The torque of the motor. T is proportional to the angular speed, hence
dt
Using NEWTON Law OF Motion, we have 
J A = T - P r .
J
Where r̂ , fee radius of the gear is connected to the motor shaft and is the radius of 
fee gear connected to the ou^ut shaft.
We have
r, = M-. => e. = -N%
Step 3 of 6
First we will find the transfer function from ^ to Sg and then we will find the 
closed loop transfer function.
Nk,
[ i f i J . + J , )
Next
0 1 0 So 0
e. = 0 - a -P e. + 0
4 . 0 K .4 . 1
^ A.. .A..
And so the closed loop transfer function is
M £ 1 _ k ,a {s )D ,{S )
step 4 of 6
O(S’) . k
S(£r+5)(S+70)
step 5 of 6
1 p
^ * — * 0-3 sec
" 6
Step 6 of 6
 ̂ ' ( S + 2 0 0 f