Student’s name:
Roll No.:
Section:
Final Exam, EE 250 (Control Systems Analysis), Spring 2013∗
D EPARTMENT OF E LECTRICAL E NGINEERING , IIT K ANPUR .
1) Write your name, roll number, and
section on every page.
2) If you find the space provided beneath each problem to be insufficient,
please continue only on the reverse
of the page on which the problem is
printed. Do not write on/behind any
other printed page.
3) Pencil is OK for trial and error, but
your final answer must use a pen.
4) Show all your assumptions.
5) You will need the following items:
1. A unity feedback control system
−
+
2
K
Plant
pen, pencil, ruler, eraser, calculator.
Borrowing is prohibited.
6) Unclear answers will lead to arbitrary
deduction of points. It is your responsibility to write legibly.
7) Giving or receiving help is prohibited.
has the following root locus for K ≥ 0.
locus
open loop poles
zeros
1.5
1
2
Imag. axis
0.5
0
-0.5
-1
-1.5
-2
-5
-4
-3
-2
-1
0
Root locus from u1 to y1 , gain=[0.000000,99.989999]: Real axis
1.1. [1 points] Write the plant transfer function and the characteristic equation corresponding to this root locus.
1.2. [1 points] Place arrows on the root locus to show its direction of evolution as K goes from 0 to ∞.
1.3. [2 points] Calculate graphically the value of K at the point marked 2.
∗ Instructor: Ramprasad Potluri, Western Labs 217 A, IIT Kanpur. Ph. (off): 259 · 6093, E-mail: potluri@iitk.ac.in.
EE250, Final exam
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1.4. [2 points] Calculate analytically the angle of departure (or, is it arrival?) at the open loop poles.
1.5. [2 points] Calculate analytically the roots of multiplicity greater than 1.
EE250, Final exam
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1.6. [4 points] For the K found in Problem 1.3, determine the unit step response of the closed-loop system.
EE250, Final exam
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Saturday, 27/04/2013, 9:00-12:00, L-7, ERES
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+
−
2. [2 points] The control system
Section:
G (s)
K
has the following Bode plot
6
G(jω), [deg]
20 log 10 |G(jω)|, [dB]
20
0
-20
-40
-60
-80
-100
-120
10−1
100
101
102
103
30
0
-30
-60
-90
-120
-150
-180
-210
-240
-270
-300
10−1
100
101
102
103
ω, [rad/s]
Determine the largest value of K > 0 for which the control system is stable.
G
(s)
OL
3. [2 points] Given GCL (s) = 1+G
. Evaluate GCL ( jω g ) for PM = 45◦ ; ω g is the gain crossover frequency.
(s)
OL
EE250, Final exam
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Saturday, 27/04/2013, 9:00-12:00, L-7, ERES
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d2 y
dy
4. [2 points] In the process of numerically integrating the equation dt2 + 3 dt + 4y = 2 du
dt + 7u, you have
arrived at the following diagram.
U (s)
u(t) +
−
s2 X ( s )
−
ẋ2
1
s
sX (s)
x2 = ẋ1
1
s
+
X (s)
+
x1
Y (s)
y(t)
Fill the appropriate numbers into the appropriate blank boxes of this diagram.
5. The loop-shaping method that I taught you works only approximately in the region around ω g . In this
problem, we will see how approximately. Consider the following minimum-phase ABMP with DD = 1 decade.
2
∆
PM/100
1.8
DD
−60 dB/dec
DD (in decades) & PM (in degrees)
1.6
1.4
1.2
1
0.8
0.6
−20 dB/dec
0.4
ωg
0.2
−60 dB/dec
0
0
10
20
30
40
50
60
Peak overshoot M (in %)
70
80
90
100
p
5.1. [1 points] Determine the phase margin that our loop-shaping technique attributes to this Bode plot.
sin Φmax
5.2. [1 points] Given that the DD-to-Φmax relationship for a lead compensator is DD = log10 11+
−sin Φmax , determine the Φmax of the lead compensator for DD = 1 decade.
5.3. [3 points] Now determine the true phase margin of the above Bode plot.
EE250, Final exam
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6. Consider the following proportional-integral control system for the armature current of a permanent magnet dc motor.
id +
−
s
+1 u +
ω0
K
−
s
1
sL + R
i
Kt
T+
1
Js + B
−
TL
ω
Kb
6.1. [1 points] Sketch the ABMP of the controller.
6.2. [1 points] We wish i to track step id with a small settling time. Assuming K = fixed, select the one option
that is most likely to result in this desired behavior, and justify your choice briefly.
6.2.1. Large ω0 .
6.2.2. Small ω0 .
6.3. [1 points] We wish i to track step id with a small overshoot. Assuming K = fixed, select the one option
that is most likely to result in this desired behavior, and justify your choice briefly.
6.3.1. Large ω0 .
6.3.2. Small ω0 .
6.4. [1 points] We wish i to track step id with a small settling time. Assuming ω0 = fixed, select the one option
that is most likely to result in this desired behavior, and justify your choice briefly.
6.4.1. Large K.
6.4.2. Small K.
6.5. [1 points] We wish i to track step id with a small steady-state error. Select the one option that is most likely
to result in this desired behavior, and justify your choice briefly.
6.5.1. Large K.
6.5.2. Small K.
6.5.3. Large ω0 .
6.5.4. Small ω0 .
6.5.5. K and ω0 do not affect the steady-state error.
EE250, Final exam
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7. The solution to the previous problem results in a well-regulated i.
The motor with well-regulated i can be represented
as
id
Kt
T +
TL
−
1
Js + B
On the other hand, the following scheme is a closedloop (CL) DOB scheme.
ω
TL
id
Here is an open-loop (OL) disturbance observer
bL of the load
(DOB) scheme to generate an estimate T
torque TL .
id
bL
T
Kt
1
τs + 1
TL
−
T +
+
1
Js + B
b
b
Js + B
bt (τs + 1)
− K
+
bi L
ω
+n
T+ −
1
Js + B
+
b
b
Js + B
+
bL
T
+
Kt
1
τs + 1
bt (τs + 1)
− K
ω
+
+n
bt
K
Here, the quantities with the hats are estimates of the
quantities without hats.
bt
K
bL for the OL DOB scheme.
7.1. [1 points] Write the transfer function (TF) from TL to T
bL for the CL DOB scheme.
7.2. [2 points] Determine the TF from TL to T
EE250, Final exam
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bt ≈ Kt , and that b
b are significantly different from J and B, for a unit step TL ,
7.3. [2 points] Assuming K
J and B
bL is closer to TL in steady state.
determine in which scheme T
Hint: Use the final value theorem y(t = ∞) = lims→0 sY (s).
7.4. [2 points] Assume that id = 1 A results in ω = 100 rad/s while TL = 0. If a non-zero TL appears,
determine which scheme rejects this disturbance better. That is, for a fixed id , in which scheme is ω less
affected in steady state by the appearance of a step TL ?
EE250, Final exam
Page 8 of 8
Saturday, 27/04/2013, 9:00-12:00, L-7, ERES