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GATE EC BY RK Kanodia
MULTIPLE CHOICE QUESTION
Electronics & Communication Engineering
Fifth Edition
R. K. Kanodia
B.Tech.
NODIA & COMAPNY
JAIPUR
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GATE EC BY RK Kanodia
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GATE EC BY RK Kanodia
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GATE EC BY RK Kanodia
CHAPTER
1.1
BASIC CONCEPTS
1. A solid copper sphere, 10 cm in diameter is deprived
of 10
20
electrons by a charging scheme. The charge on
the sphere is
(A) 160.2 C
(B) -160.2 C
(C) 16.02 C
(D) -16.02 C
(A) 1 A
(B) 2 A
(C) 3 A
(D) 4 A
6. In the circuit of fig P1.1.6 a charge of 600 C is
delivered to the 100 V source in a 1 minute. The value
of v1 must be
v1
2. A lightning bolt carrying 15,000 A lasts for 100 ms. If
the lightning strikes an airplane flying at 2 km, the
charge deposited on the plane is
(A) 13.33 mC
(B) 75 C
(C) 1500 mC
(D) 1.5 C
60 V
20 W
100 V
3. If 120 C of charge passes through an electric
Fig. P.1.1.6
conductor in 60 sec, the current in the conductor is
(A) 0.5 A
(B) 2 A
(C) 3.33 mA
(D) 0.3 mA
(A) 240 V
(B) 120 V
(C) 60 V
(D) 30 V
4. The energy required to move 120 coulomb through
7. In the circuit of the fig P1.1.7, the value of the
3 V is
(B) 360 J
(C) 40 J
(D) 2.78 mJ
voltage source E is
0V
+
+
(A) 25 mJ
–
–
2V
5. i = ?
+
–
+
5V
–
–
4V
i
E
+
1A
1V
10 V
2A
5A
Fig. P.1.1.7
3A
4A
Fig. P.1.1.5
(A) -16 V
(b) 4 V
(C) -6 V
(D) 16 V
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GATE EC BY RK Kanodia
UNIT 1
8. Consider the circuit graph shown in fig. P1.1.8. Each
Networks
12. v1 = ?
branch of circuit graph represent a circuit element. The
v1 –
+
value of voltage v1 is
1 kW
2
7V
+ 105 V –
– 15 V +
+
+
55 V
+ v1 –
–
8V
35 V
–
–
5V
3
6V
kW
+
–
100 V
– 10 V +
+
65 V
kW
V
30
V
30
4 kW
+
–
Fig. P1.1.12
Fig. P.1.1.8
(A) -30 V
(B) 25 V
(C) -20 V
(D) 15 V
9. For the circuit shown in fig P.1.1.9 the value of
(A) -11 V
(B) 5 V
(C) 8 V
(D) 18 V
13. The voltage vo in fig. P1.1.11 is always equal to
1A
voltage vo is
5W
+
vo
4W
+
vo
–
1A
15 V
5V
Fig. P1.1.11
–
Fig. P.1.1.9
(A) 10 V
(B) 15 V
(C) 20 V
(D) None of the above
(A) 1 V
(B) 5 V
(C) 9 V
(D) None of the above
14. Req = ?
5W
10 W
10 W
10 W
10. R1 = ?
60 W
Req
10 W
+
10 W
10 W
up to ¥
R1
100 V
R2
+
20 V
–
70 V
Fig. P1.1.14
–
Fig. P.1.1.10
(A) 11.86 W
(B) 10 W
(C) 25 W
(D) 11.18 W
15. vs = ?
(A) 25 W
(B) 50 W
(C) 100 W
(D) 2000 W
180 W
+
60 W
11. Twelve 6 W resistor are used as edge to form a cube.
vs
The resistance between two diagonally opposite corner
of the cube is
5
(A) W
6
(C) 5 W
Page
4
(B)
90 W
6
W
5
(D) 6 W
40 W
20 V
–
Fig. P.1.1.15
(A) 320 V
(B) 280 V
(C) 240 V
(D) 200 V
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GATE EC BY RK Kanodia
UNIT 1
24. Let i( t) = 3te-100 t A and v( t) = 0.6(0.01 - t) e -100 t V for
Networks
28. vab = ?
a
the network of fig. P.1.1.24. The power being absorbed
2
8
–
b
0.3i1
A
2
W
W
N
0.2i1
i1
A
6
+
W
R
i
v
2
by the network element at t = 5 ms is
Fig. P.1.1.24
(A) 18.4 mW
(B) 9.2 mW
(C) 16.6 mW
(D) 8.3 mW
Fig. P.1.1.28
25. In the circuit of fig. P.1.1.25 bulb A uses 36 W when
lit, bulb B uses 24 W when lit, and bulb C uses 14.4 W
(A) 15.4 V
(B) 2.6 V
(C) -2.6 V
(D) 15.4 V
29. In the circuit of fig. P.1.1.29 power is delivered by
when lit. The additional A bulbs in parallel to this
500 W
circuit, that would be required to blow the fuse is
400 W
ix
20 A
200 W
2ix
40 V
12 V
C
B
A
Fig. P.1.1.29
(A) dependent source of 192 W
Fig. P.1.1.25
(A) 4
(B) 5
(B) dependent source of 368 W
(C) 6
(D) 7
(C) independent source of 16 W
26. In the circuit of fig. P.1.1.26, the power absorbed by
(D) independent source of 40 W
the load RL is
30. The dependent source in fig. P.1.1.30
i1
5W
RL = 2 W
2i1
1W
1V
v1
20 V
v1
5
5W
Fig. P.1.1.26
(A) 2 W
(B) 4 W
(C) 6 W
(D) 8 W
Fig. P.1.1.30
27. vo = ?
(A) delivers 80 W
(B) delivers 40 W
(C) absorbs 40 W
(D) absorbs 80 W
31. In the circuit of fig. P.1.1.31 dependent source
0.2 A
5W
+
v1 0.3v1
–
8W
+
v2
–
5v2
18 W
+
vo
–
+ 8V –
ix
Page
6
(A) 6 V
(B) -6 V
(C) -12 V
(D) 12 V
2ix
4A
Fig. P.1.1.27
Fig. P.1.1.31
(A) supplies 16 W
(B) absorbs 16 W
(C) supplies 32 W
(D) absorbs 32 W
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GATE EC BY RK Kanodia
Basic Concepts
Chap 1.1
32. A capacitor is charged by a constant current of 2 mA
36. The waveform for the current in a 200 mF capacitor
and results in a voltage increase of 12 V in a 10 sec
is shown in fig. P.1.1.36 The waveform for the capacitor
interval. The value of capacitance is
voltage is
(A) 0.75 mF
(B) 1.33 mF
(C) 0.6 mF
(D) 1.67 mF
i(mA)
5
33. The energy required to charge a 10 mF capacitor to
100 V is
Fig. P. 1.1.36
(A) 0.10 J
(C) 5 ´ 10
t(ms)
4
(B) 0.05 J
-9
v
(D) 10 ´ 10
J
-9
J
v
50m
5
34. The current in a 100 mF capacitor is shown in fig.
t(ms)
4
P.1.1.34. If capacitor is initially uncharged, then the
4
(A)
(B)
v
waveform for the voltage across it is
t(ms)
v
50m
250m
i(mA)
t(ms)
4
2
4
(C)
t(ms)
t(ms)
(D)
Fig. P. 1.1.34
v
37. Ceq = ?
v
10
10
2
4
t(ms)
2
(A)
v
2.5 mF
4
t(ms)
1.5 mF
(B)
v
2
4
1 mF
Ceq
0.2
0.2
t(ms)
2
(C)
4
2 mF
t(ms)
(D)
Fig. P.1.1.37
35. The voltage across a 100 mF capacitor is shown in
fig. P.1.1.35. The waveform for the current in the
(A) 3.5 mF
(B) 1.2 mF
capacitor is
(C) 2.4 mF
(D) 2.6 mF
v
6
38. In the circuit shown in fig. P.1.1.38
1
2
3
t(ms)
iin ( t) = 300 sin 20 t mA, for t ³ 0.
Fig. P.1.1.35
i(mA)
6
C2
C2
C2
C2
+
i(mA)
600
vin C1
iin
1
2
3
t(ms)
1
2
3
C1
C1
C1
60 mF
t(ms)
–
(A)
(B)
i(mA)
6
Fig. P. 1.1.38
i(mA)
600
2
1
(C)
3
t(ms)
2
1
3
t(ms)
Let C1 = 40 mF and C2 = 30 mF. All capacitors are
initially uncharged. The vin ( t) would be
(D)
(A) -0.25cos 20t V
(B) 0.25cos 20t V
(C) -36cos 20t mV
(D) 36cos 20t mV
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GATE EC BY RK Kanodia
Basic Concepts
Chap 1.1
v3 - 30 = v2
SOLUTIONS
Þ
v3 = 65 V
105 + v4 - v3 - 65 = 0 Þ
v4 = 25 V
v4 + 15 - 55 + v1 = 0 Þ v1 = 15 V
1. (C) n = 10 20 , Q = ne = e10 20 = 16.02 C
Charge on sphere will be positive.
9. (B) Voltage is constant because of 15 V source.
2. (D) DQ = i ´ D t = 15000 ´ 100m = 15
. C
10. (C) Voltage across 60 W resistor = 30 V
30
Current =
= 0.5 A
60
3. (B) i =
dQ 120
=
=2 A
dt
60
4. (B) W = Qv = 360 J
Voltage across R1 is = 70 - 20 = 50 V
50
R1 =
= 100 W
0.5
6. (A)
11. (C) The current i will be distributed in the cube
branches symmetrically
1A
i=1A
a
5A
2A
3A
4A
i
3
i
i
3
6A
1A
i
3
2A
i
6
Fig. S 1.1.5
6. (A) In order for 600 C charge to be delivered to the
100 V source, the current must be anticlockwise.
dQ 600
i=
=
= 10 A
dt
60
Applying KVL we get
0V
Fig. S. 1.1.11
+
+
1V
–
–
+
–
12. (C) If we go from +side of 1 kW through 7 V, 6 V and
5 V, we get v1 = 7 + 6 - 5 = 8 V
E
13. (D) It is not possible to determine the voltage across
–
–
+
+
5V
1 A source.
10 V
Fig. S 1.1.7
14. (D) Req = 5 +
10 + 5 + E + 1 = 0 or E = -16 V
8. (D) 100 = 65 + v2 Þ
v2 = 35 V
+ 105 V –
+
30
+
Req
Fig. S 1.1.14
–
–
10 W
–
V
v2
Req
55 V
+ v1 –
5W
– 10 V +
30
V
–
+ v3 –
10 + 5 + Req
5W
+
v4
10 ( Req + 5)
+
–
+
– 15 V +
+
65 V
100 V
b
6i 6i 6i
+
+
= 5 i,
3
6
3
v
Req = ab = 5 W
i
7. (A) Going from 10 V to 0 V
4V
i
vab =
v1 + 60 - 100 = 10 ´ 20 or v1 = 240 V
2V
i
3
Þ Req2 + 15 Req = 5 Req + 75 + 10 Req + 50
Fig. S 1.1.8
Þ
Req = 125 = 1118
. W
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GATE EC BY RK Kanodia
Basic Concepts
vo - 20 vo 20
+
=
5
5
5
Power is P = vo ´
Þ
We can say Cd = 20 mF, Ceq = 20 + 40 = 60 mF
1
1 æ 300
ö
cos 20 t ÷ ´ 10 -3 = - 0.25 cos 20 t V
vC = ò idt=
çC
60m è 20
ø
vo = 20 V
20
v1
= 20 ´
= 80 W
5
5
39. (C) iC1 =
31. (D) Power P = vi = 2 ix ´ ix = 2 ix2
ix = 4 A, P = 32 W (absorb)
32. (D) vt 2 - vt1 =
Þ
1
C
t2
ò idt
Þ
iin C1
= 0.8 sin 600 t mA
C1 + C2
At t = 2 ms, iC1 = 0.75 mA
Þ
12 =
t1
12 C = 2m ´ 10
Chap 1.1
1
2m ( t2 - t1 )
C
40. (B) vC1 =
vin C2
4 vin
=
C1 + C2 6 + 4
Þ
vc1
= 0.4
vin
C = 1.67 mF
41. (D) V = 2 + 3 + 5 = 10, Q = 1 C, C =
1
33. (B) E = Cv 2 = 5 ´ 10 -6 ´ 100 2 = 0.05 J
2
42. (A) vL = L
di
Þ
dt
This 0.2 V increases linearly from 0 to 0.2 V. Then
43. (B) vL = L
di
= 0.01 ´ 2( 377 cos 377 t) V
dt
current is zero. So capacitor hold this voltage.
= 7.54 cos 377 t V
1
34. (D) vc =
c
35. (D) i = C
-3
10 ´ 10
-3
ò0 idt = 100 ´ 10 -6 (2 ´ 10 ) = 0.2 V
2m
dv
dt
For 0 < t < 1 , C
dv
6 -0
= 100 ´ 10 -6 ´
= 600 mA
dt
10 -3 - 0
1
1
12000
vdt =
120 cos 3t dt =
sin 377 t
Lò
0.01 ò
377
12000 ´ 120
P = vi =
(sin 377 t)(cos 377 t)
377
45. (D) vL = L
1
1
idt =
ò
C
200 ´ 10 -6
5m
ò 4m tdt = 3125 t
vC = 3vL Þ
iC = 3 LC
46. (B) vL = L
It will be parabolic path. at t = 0 t-axis will be tangent.
combination
is
in
series
with
1.5
mF.
15
. (2 + 1)
C1 =
= 1mF, C1 is in parallel with 2.5 mF
15
. +2+1
. mF
Ceq = 1 + 2.5 = 35
diL
dt
æ -100 - 0 ö
vL = (0.05)ç
÷ = - 2.5 V
2
è
ø
For 4 < t £ 8,
æ 100 + 100 ö
vL = (0.05)ç
÷ = 2.5 V
4
è
ø
For 8 < t £ 10,
æ 0 - 100 ö
vL = (0.05)ç
÷ = - 2.5 V
2
è
ø
Thus (B) is correct option.
30 ´ 60
30(20 + 40)
38. (A) Ca =
= 20 mF, Cb =
= 20 mF
30 + 60
30 + 20 + 0
C2
Cc
C2
Cb
C2
Ca
47. (C) Algebraic sum of the current entering or leaving
a cutset is equal to 0.
6 16
i2 + i4 + i3 = 0
Þ
+
+ i3 = 0
2
4
C2
+
i3 = - 7 A,
C1
d 2 iL
= - 9.6 sin 4 t A
dt
For 2 < t £ 4,
37. (A) 2 mF is in parallel with 1 mF and this
vin
diL
dv
, iC = C C
dt
dt
2
At t = 4 ms, vc = 0.05 V
iin
L = 2 mH
= 1910 sin 754 t W
36. (B) For 0 £ t £ 4,
Cd
Þ
44. (A) i =
For 1 ms < t < 2 ms,
dv
0-6
C
= 100 ´ 10 -6 ´
= - 600 mA
( 3 - 2)m
dt
vC =
100m = L
200m
4m
Q
= 0.1 F
V
C1
C1
C1
v3 = - 7 ´ 3 = - 21 V
60 mF
–
Fig. S 1.1.38
*********
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GATE EC BY RK Kanodia
CHAPTER
1.2
GRAPH THEORY
1. Consider the following circuits :
Non-planner graphs are
(A) 1 and 3
(B) 4 only
(C) 3 only
(D) 3 and 4
3. A graph of an electrical network has 4 nodes and 7
branches. The number of links l, with respect to the
(1)
(2)
chosen tree, would be
(A) 2
(B) 3
(C) 4
(D) 5
4. For the graph shown in fig. P.1.1.4 correct set is
(3)
(4)
Fig. P.1.1.4
The planner circuits are
(A) 1 and 2
(B) 2 and 3
(C) 3 and 4
(D) 4 and 1
2. Consider the following graphs
Node
Branch
Twigs
Link
(A)
4
6
4
2
(B)
4
6
3
3
(C)
5
6
4
2
(D)
5
5
4
1
5. A tree of the graph shown in fig. P.1.2.5 is
c
(1)
(2)
b
3
g
e
d
4
f
2
a
1
5
h
Fig. P.1.2.5
(3)
Page
12
(4)
(A) a d e h
(B) a c f h
(C) a f h g
(D) a e f g
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GATE EC BY RK Kanodia
UNIT 1
é 1 -1 0 ù
(A) ê-1 0
1ú
ê
ú
1 -1úû
êë 0
é-1 -1 0 ù
(C) ê 0
1 1ú
ê
ú
êë 1 0 -1ûú
é 1 0 -1ù
(B) ê-1 -1 0 ú
ê
ú
1 1úû
êë 0
1ù
é-1 0
ê
(D) 0
1 -1ú
ê
ú
êë 1 -1 0 úû
2
2
4
1
3
4
1
5
13. The incidence matrix of a graph is as given below
3
5
(C)
é-1 1 1 0 0 0 ù
ê 0 0 -1 1 1 0 ú
ú
A =ê
ê 0 -1 0 -1 0 -1ú
ê 1 0 0 0 -1 -1ú
ë
û
(D)
15. The incidence matrix of a graph is as given below
é-1 1 1 0 0 0 ù
ê 0 0 -1 1 1 0 ú
ú
A =ê
ê 0 -1 0 -1 0 0 ú
ê 1 0 0 0 -1 -1ú
ë
û
The graph is
2
Networks
2
The graph is
4
4
3
1
3
1
(A)
1
2
3
1
2
3
(B)
2
2
4
4
(A)
4
(B)
4
3
1
3
1
(C)
1
2
3
1
2
3
(D)
14. The incidence matrix of a graph is as given below
é- 1 0 0 - 1 1 0 0 ù
ê 0 0 0
1 0
1 1ú
ê
ú
A =ê 0 0
1 0 0 0 -1ú
ê 0
1 0 0 -1 -1 0 ú
ê
ú
êë 1 -1 -1 0 0 0 0 úû
4
4
(C)
(D)
16. The graph of a network is shown in fig. P.1.1.16. The
number of possible tree are
The graph is
2
2
4
1
3
4
1
3
Fig. P.1.1.16
5
(A)
Page
14
5
(A) 8
(B) 12
(C) 16
(D) 20
(B)
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GATE EC BY RK Kanodia
UNIT 1
Networks
22. The fundamental cut-set matrix of a graph is
é1 -1 0 0 0
ê0 -1 1 0
1
QF = ê
0
0
0
0
1
ê
ê0 0 0 1 -1
ë
0ù
0ú
ú
1ú
0 úû
The oriented graph of the network is
2
1
(C)
24. A graph is shown in fig. P.1.2.24 in which twigs are
2
1
3
3
4
solid line and links are dotted line. For this tree
4
5
5
6
fundamental loop matrix is given as below
é1 1 1 0 ù
BF = ê
ú
ë1 0 1 1û
6
(A)
(D)
(B)
1
2
1
2
1
3
3
2
4
3
4
4
5
5
6
6
Fig. P.1.2.24
(C)
(D)
The oriented graph will be
23. A graph is shown in fig. P.1.2.23 in which twigs are
solid line and links are dotted line. For this chosen tree
fundamental set matrix is given below.
é 1 1
BF = ê 0 -1
ê
êë 0 0
0
0
1
1
1
0
0
1
1
0ù
0ú
ú
1úû
(A)
(B)
3
2
1
4
5
6
(C)
Fig. P. 1.2.23
The oriented graph will be
(D)
25. Consider the graph shown in fig. P.1.2.25 in which
twigs are solid line and links are dotted line.
1
5
6
(A)
2
4
3
(B)
Fig. P. 1.2.25
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16
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GATE EC BY RK Kanodia
Graph Theory
A fundamental loop matrix for this tree is given as
below
é 1
BF = ê 0
ê
êë 0
1
1ù
0ú
ú
1 -1úû
0
0
1
0 -1 -1
0
1
0
0
é 1
ê 0
(D) ê
ê 0
ê-1
ë
Chap 1.2
1
0
0
0
0
0
1
1 -1
0 0
1
1
0
1
0
1
1
0
0
0
0
0
1
0
0
0
1
0ù
0ú
ú
0ú
1úû
27. Branch current and loop current relation are
The oriented graph will be
expressed in matrix form as
(A)
é i1 ù é 0
êi ú ê 0
ê 2ú ê
ê i3 ú ê 1
êi ú ê-1
ê 4ú = ê
ê i5 ú ê 1
êi ú ê 0
6
ê ú ê
i
ê 7ú ê 0
êë i8 úû êë 0
(B)
1 -1
0 -1
0
0
1
0
0
0
1
0
0
1
0
0
0ù
1ú
ú
-1ú é I1 ù
0 ú êI 2 ú
úê ú
0 ú êI 3 ú
0 ú êëI 4 úû
ú
0ú
1úû
where i j represent branch current and I k loop
current. The number of independent node equation are
(C)
(D)
(A) 4
(B) 5
(C) 6
(D) 7
28. If the number of branch in a network is b, the
26. In the graph shown in fig. P.1.2.26 solid lines are
twigs and dotted line are link. The fundamental loop
number of nodes is n and the number of dependent loop
is l, then the number of independent node equations
will be
matrix is
i
c
a
e
(A) n + l - 1
(B) b - 1
(C) b - n + 1
(D) n - 1
Statement for Q.29–30:
h
b
Branch current and loop current relation are
f
d
expressed in matrix form as
1 0ù
é i1 ù é 0 0
êi ú ê-1 -1 -1 0 ú
ê 2ú ê
ú
1 0 0 ú é I1 ù
ê i3 ú ê 0
êi ú ê 1 0 0 0 ú ê I ú
ê 4ú = ê
ú ê 2ú
i
0
0
1
1
ê 5ú ê
ú êI 3 ú
ê i ú ê 1 1 0 -1ú êI ú
ë 4û
6
ê ú ê
ú
ê i7 ú ê 1 0 0 0 ú
êë i8 úû êë 0 0 0
1úû
g
Fig. P.1.2.26
é 1
ê 0
(A) ê
ê 0
ê-1
ë
1
0
0
0
0
1 -1 -1
1
0
0
1
0
0
1 -1
1
0
0
0 -1
0 -1
0
0
0
0
0
0
1
0
0
1
0
1 -1
0
0
0
é-1 1 0 0
ê 0 -1 -1 -1
(B) ê
ê 0 0 0 -1
ê 1 0
1 0
ë
é
ê
(C) ê
ê
ê
ë
0
1
0
0
0
1
0
1
1
0
1
0
1
0
0
0
0
0 -1
1 0
0
1
0
0
0
1
1 -1
0 -1
1
0
0
0
0
0
0ù
0ú
ú
0ú
1úû
0ù
0ú
ú
0ú
1úû
0ù
0ú
ú
0ú
1úû
where i j represent branch current and I k loop
current.
29. The rank of incidence matrix is
(A) 4
(B) 5
(C) 6
(D) 8
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UNIT 1
30. The directed graph will be
Networks
33. The oriented graph for this network is
8
8
5
5
6
2
3
1
2
7
1
6
3
1
4
3
2
7
4
1
5
8
5
6
3
2
4
1
2
6
5
5
(C)
2
3
1
4
3
4
5
8
5
3
4
1
2
7
3
1
4
(D)
7
4
(C)
************
(D)
31. A network has 8 nodes and 5 independent loops.
The number of branches in the network is
(A) 11
(B) 12
(C) 8
(D) 6
32. A branch has 6 node and 9 branch. The independent
loops are
(A) 3
(B) 4
(C) 5
(D) 6
Statement for Q.33–34:
For a network branch voltage and node voltage
relation are expressed in matrix form as follows:
1ù
é v1 ù é 1 0 0
êv ú ê 0
1 0 0ú
ê 2ú ê
ú
1 0 ú é V1 ù
ê v3 ú ê 0 0
êv ú ê 0 0 0
1ú êV2 ú
ê 4ú = ê
úê ú
ê v5 ú ê 1 -1 0 0 ú ê V3 ú
êv ú ê 0
1 -1 0 ú êëV4 úû
6
ê ú ê
ú
1 -1ú
ê v7 ú ê 0 0
êë v8 úû êë 1 0 -1 0 úû
where vi is the branch voltage and Vk is the node
voltage with respect to datum node.
33. The independent mesh equation for this network
are
(A) 4
(B) 5
(C) 6
(D 7
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18
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Graph Theory
Chap 1.2
7. (D) D is not a tree
SOLUTIONS
1. (A) The circuit 1 and 2 are redrawn as below. 3 and 4 can
not be redrawn on a plane without crossing other branch.
(A)
(1)
(B)
(2)
Fig. S1.2.1
2. (B) Other three circuits can be drawn on plane
without crossing
(C)
(1)
(D)
(2)
Fig. S .1.2.7
8. (D) it is obvious from the following figure that 1, 3,
and 4 are tree
2
2
(3)
a
Fig. S1.2.1
c
1
3. (C) l = b - ( n - 1) = 4.
a
b
e
(1)
2
a
c
1
g
e
a
b
f
a
(2)
2
c
d
4
4
5. (C) From fig. it can be seen that a f h g is a tree of
b
f
f
l=b-n+1=3
given graph
3
e
d
4. (B) There are 4 node and 6 branches.
t = n - 1 = 3,
c
1
3
d
b
c
1
3
e
d
b
3
e
d
f
f
4
4
h
(3)
Fig. S 1.2.5
(4)
2
6. (B) From fig. it can be seen that a d f is a tree.
c
b
a
a
e
d
b
c
1
e
d
f
3
f
4
(5)
Fig. S. 1.2.6
Fig. S. 1.2.8
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UNIT 1
So independent mesh equation =Number of link.
i
l1
a
34. (D) We know that [ vb ] = ArT [ Vn ]
e
c
So reduced
é1 0
ê0 1
Ar = ê
ê0 0
ê0 0
ë
l4
l2
h
b
l3
Networks
d
f
g
Fig. S 1.2.26
incidence matrix is
0 0
1 0 0
1ù
0 0 -1 1 0 0 ú
ú
1 0 0 -1 1 -1ú
0 1 0 0 -1 0 úû
At node-1, three branch leaves so the only option is (D).
This in similar to matrix in (A). Only place of rows has
been changed.
27. (A) Number of branch =8
***********
Number of link =4
Number of twigs =8 - 4 = 4
Number of twigs =number of independent node
equation.
28. (D) The number of independent node equation are
n - 1.
29. (A) Number of branch b = 8
Number of link l = 4
Number of twigs t = b - l = 4
rank of matrix = n - 1 = t = 4
30. (B) We know the branch current and loop current
are related as
[ ib ] = [ B T ] [ I L ]
So fundamental loop matrix is
1 1 0ù
é0 -1 0 1 0
ê0 -1 1 0 0
1 0 0ú
ú
Bf = ê
ê1 -1 0 0 -1 0 0 0 ú
ê0 0 0 0 -1 -1 0 1ú
ë
û
f-loop 1 include branch (2, 4, 6, 7) and direction of
branch–2 is opposite to other (B only).
31. (B) Independent loops =link
l = b - ( n - 1)
Þ
5 = b - 7, b = 12
32. (B) Independent loop =link
l = b - ( n - 1) = 4
33. (A) There are 8 branches and 4 + 1 = 5 node
Number of link = 8 - 5 + 1 = 4
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CHAPTER
1.3
METHODS OF ANALYSIS
1. v1 = ?
6R
6R
4vs
+
v1
–
(B) -120 V
(D) -90 V
(A) 120 V
(C) 90 V
3R
vs
4. va = ?
10 W
4W
12 V
10 V
Fig. P1.3.1
(A) 0.4vs
(B) 1.5vs
(C) 0.67vs
(D) 2.5vs
va
4A
1W
2. va = ?
2W
Fig. P1.3.4
3A
(A) 4.33 V
(C) 8.67 V
2W
5. v2 = ?
va
3W
(B) 4.09 V
(D) 8.18 V
20 W
1A
+
30 W
60 W
v2 –
10 V
0.5 A
30 W
Fig. P1.3.2
(A) -11 V
(B) 11 V
(C) 3 V
(D) -3 V
Fig. P1.3.5
(A) 0.5 V
(C) 1.5 V
3. v1 = ?
10 W
6. ib = ?
30 V
3A
(B) 1.0 V
(D) 2.0 V
64 W
30 W
37 W
ib
20 W
0.5 A
10 V
60 W
69 W
36 W
–
v1
+
9A
6A
60 W
Fig. P1.3.6
Fig. P1.3.3
(A) 0.6 A
(C) 0.4 A
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(B) 0.5 A
(D) 0.3 A
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UNIT 1
7. i1 = ?
6A
8W
(A) 20 mA
(B) 15 mA
(C) 10 mA
(D) 5 mA
11. i1 = ?
2W
i1
Networks
50 W
10 W
5W
3W
4W
75 V
100 W
i1
6.6 V
Fig. P1.3.7
(A) 3.3 A
(B) 2.1 A
(C) 1.7 A
(D) 1.1 A
0.1 A
40 W
0.06 A
60W
8. i1 = ?
0.1A
i2
90 kW
7.5mA
i1
10 kW
Fig. P1.3.11
10 kW
75 V
(A) 0.01 A
(B) -0.01 A
(C) 0.03 A
(D) 0.02 A
90 kW
12. The value of the current measured by the ammeter
in Fig. P1.3.12 is
Fig. P1.3.8
(A) 1 mA
(B) 1.5 mA
(C) 2 mA
(D) 2.5 mA
2A
Ammeter
4W
7W
9. i1 = ?
2A
6W
3A
5W
3V
4W
2W
Fig. P1.3.12
3W
4A
2W
i1
(B) 3 A
(C) 6 A
(D) 5 A
2
A
3
(C) -
Fig. P1.3.9
(A) 4 A
(A)
5
A
6
(B)
5
A
3
(D)
2
A
9
13. i1 = ?
200 W
10. i1 = ?
2 kW
45 V
i1
40 mA
500 W
Fig. P1.3.10
Page
24
15 mA
100 W
50 W
i1
Fig. 1.3.13
(A) 10 mA
(B) -10 mA
(C) 0.4 mA
(D) -0.4 mA
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Methods of Analysis
Chap 1.3
5W
8W
4W
14. The values of node voltage are va = 12 V, vb = 9.88 V
and vc = 5.29 V. The power supplied by the voltage
i1
12 V
source is
2W
i3
2W
i2
20 V
8V
6W
va
4W
3W
vb
12 V
1A
vc
Fig. 1.3.17
0 ù éi1 ù é12 ù
é 4 -2
(A) ê-2
8 -2 ú êi2 ú = ê-8 ú
ê
úê ú ê ú
5 úû êëi3 úû êë20 úû
êë 0 -2
2W
-2
0 ù éi1 ù é12 ù
é6
(B) ê2 -12
2 ú êi2 ú = ê 8 ú
ê
úê ú ê ú
2 -7 úû êëi3 úû êë20 úû
êë0
Fig. 1.3.14
(A) 19.8 W
(B) 27.3 W
(C) 46.9 W
(D) 54.6 W
0 ù éi1 ù é12 ù
é 6 -2
(C) ê-2 12 -2 ú êi2 ú = ê 8 ú
ê
úê ú ê ú
7 úû êëi3 úû êë20 úû
êë 0 -2
15. i1 , i2 , i3 = ?
2W
3W
15 V
9W
i1
6W
i2
0 ù éi1 ù é12 ù
é 4 -2
ê
(D) 2 -8
2 ú êi2 ú = ê 8 ú
ê
úê ú ê ú
2 -5 úû êëi3 úû êë20 úû
êë 0
18. For the circuit shown in Fig. P1.3.18 the mesh
21 V
i3
equation are
6 kW
Fig. P1.3.15
6 kW
(A) 3 A, 2 A, and 4 A
(B) 3 A, 3 A, and 8 A
(C) 1 A, 3 A, and 4 A
(D) 1 A, 2 A, and 8 A
i3
6 kW
i1
6V
i2
5 mA
6 kW
16. vo = ?
Fig. 1.3.18
4 mA
2 kW
2 mA
é 6 k -12 k -12 k ù éi1 ù
é-6 ù
ê
ú
ê
ú
(A) -6 k
6 k -18 k i2 = ê 0 ú
ê
úê ú
ê ú
-1k
0 k úû ëêi3 úû
êë -1k
êë 5 úû
1 kW
1 kW
+
1 kW
2 kW
1 mA
vo
é 6 k 12 k -12 k ù éi1 ù
é-6 ù
(B) ê-6 k -6 k
18 k ú êi2 ú = ê 0 ú
ê
úê ú
ê ú
1k
0 k úû êëi3 úû
êë -1k
êë 5 úû
–
Fig. P1.3.16
(A)
6
V
5
(B)
8
V
5
(C)
6
V
7
(D)
5
V
7
é-6 k -12 k 12 k ù éi1 ù
é-6 ù
(C) ê 6 k
-6 k 18 k ú êi2 ú = ê 0 ú
ê
úê ú
ê ú
1k 0 k úû êëi3 úû
êë 1k
êë 5 úû
17. The mesh current equation for the circuit in Fig.
P1.3.17 are
é-6 k 12 k -12 k ù éi1 ù
é-6 ù
(D) ê-6 k 6 k -18 k ú êi2 ú = ê 0 ú
ê
úê ú
ê ú
1k
0 k úû êëi3 úû
êë -1k
êë 5 úû
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Methods of Analysis
R1
i3
R2
Chap 1.3
(A) 66.67 mA
(B) 46.24 mA
(C) 23.12 mA
(D) 33.33 mA
R3
29. va = ?
i1
R4
i2
10 W
v2
v1
25i2
4A
50 W
va
40 W
Fig. P1.3.25
200 W
10 A
The value of R4 is
(A) 40
(B) 15
(C) 5
(D) 20
2.5 kW
10 kW
20 V
10 kW
(A) 342 V
(B) 171 V
(C) 198 V
(D) 396 V
30. ia = ?
va
5 kW
20 A
Fig. P1.3.29
26. va = ?
10 kW
20 W
100 W
5A
50 W
150 W
4 mA
225 W
100 W
200 W
ia
Fig. P1.3.26
2V
(A) 26 V
(B) 19 V
(C) 13 V
(D) 18 V
75 W
2A
50 W
(A) 14 mA
(B) -6.5 mA
(C) 7 mA
(D) -21 mA
31. v2 = ?
20 W
v
8V
Fig. P1.3.30
27. v = ?
10 W
4V
50 W
– v2 +
4A
100 W
10 V
15 W
0.04v2
5W
Fig. P.3.1.27
Fig. P1.3.31
(A) 60 V
(B) -60 V
(A) 5 V
(B) 75 V
(C) 30 V
(D) -30 V
(C) 3 V
(D) 10 V
32. i1 = ?
28. i1 = ?
2W
300 W
40 V
i1
500 W
8V
0.4i1
0.5i1
4A
4W
6V
i1
Fig. P1.3.32
Fig. P1.3.28
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UNIT 1
(A) -1.636 A
(B) -3.273 A
(C) -2.314 A
(D) -4.628 A
Networks
37. va = ?
0.8va
16 A
33. vx = ?
+
100 W
1.6 A
2.5 W
5W
10 A
vx
50 W
0.02vx
2W
va
–
Fig. P1.3.37
Fig. P1.3.33
(A) 32 V
(B) -32 V
(C) 12 V
(D) -12 V
(A) 25.91 V
(B) -25.91 V
(C) 51.82 V
(D) -51.82 V
38. For the circuit of Fig. P1.3.38 the value of vs , that
34. ib = ?
will result in v1 = 0, is
1 kW
va
3A
3 kW
2A
ib
0.1v1
6V
4va
2 kW
10 W
20 W
+
vs
40 W
Fig. P1.3.34
(A) 4 mA
(B) -4 mA
(C) 12 mA
(D) -12 mA
4 kW
vb
(A) 28 V
(B) -28 V
(C) 14 V
(D) -14 V
39. i1 , i2 = ?
2ix
4W
2V
48 V
Fig. P1.3.38
35. vb = ?
ia
v1
–
2W
5ia
2 kW
ix
15 V
6W
i1
i2
18 V
Fig. P1.3.35
(A) 1 V
(B) 1.5 V
(C) 4 V
(D) 6 V
Fig. P1.3.39
36. vx = ?
(A) 2.6 A, 1.4 A
(B) 2.6 A, -1.4 A
(C) 1.6 A, 1.35 A
(D) 1.2 A, -1.35 A
40. v1 = ?
50 W
3W
iy
+
2A
100 W
25iy
50 W
vx
+
vy
–
0.2vx
–
3W
2A
6W
14 V
+
v1
–
Fig. P1.3.36
Page
28
(A) -3 V
(B) 3 V
(C) 10 V
(D) -10 V
7A
Fig. P1.3.40
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Methods of Analysis
(A) 10 V
(B) -10 V
(C) 7 V
(D) -7 V
SOLUTIONS
1. (B) Applying the nodal analysis
v
4 vs
+ s
R
R
6
3
v1 =
= 15
. vs
1
1
1
+
+
6 R 3R 6 R
41. vx = ?
– vx +
500 W
0.5vx
500 W
900 W
600 W
0.6 A
Chap 1.3
2. (C) va = 2( 3 + 1) + 3 (1) = 11 V
0.3 A
3. (D) -
v1
-v
+ 1 + 6 =9
60 60
Þ
v1 = - 90 V
Fig. P1.3.41
(A) 9 V
(B) -9 V
(C) 10 V
(D) -10 V
42. The power being dissipated in the 2 W resistor in the
4. (C)
va - 10 va
+
=4
4
2
5. (D)
v2 v2 + 10
+
= 0.5
20
30
Þ
va = 8.67 V
Þ
v2 = 2 V
circuit of Fig. P1.3.42 is
5W
6. (B) Using Thevenin equivalent and source transform
ia
3W
8W
3
2W
i1
2W
10 W
va
60 V
2A
2.5 A
4W
6ia
3W
25 V
30 V
Fig. P1.3.42
Fig. S.1.3.6
(A) 76.4 W
(B) 305.6 W
(C) 52.5 W
(D) 210.0 W
43. i1 = ?
500 W
+ vx –
100 W
180 V
400 W
5W
25
60
+
+ 2 15
va =
= 15.23 V
3
1
1
+ +
14 3 15
25 - 15.23
i1 =
= 2.09 A
14
3
8
3
0.6 A
+
vy
–
7. (A) ib =
0.001vy
100 W
10
+ 0.5 = 0.6 A
64 + 36
8. (B) 75 = 90 ki1 + 10 k( i1 - 7.5m)
150 = 100 ki1
Þ
i1 = 15
. mA
9. (B) 3 = 2 i1 + 3( i1 - 4)
0.005vy
Þ
i1 = 3 A
Fig. P1.3.43
10. (B) 45 = 2 ki1 + 500 ( i1 + 15m)
(A) 0.12 A
(B) 0.24 A
(C) 0.36 A
(D) 0.48 A
Þ
i1 = 15 mA
11. (D)
6.6 = 50 i1 + 100( i1 + 0.1) + 40( i1 - 0.06) + 60( i1 - 0.1)
*****************
i1 = 0.02 A
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UNIT 1
38. (D) If v1 = 0, the dependent source is a short circuit
v1
v - vs v1 - 48
+ 1
+
=2 - 3 Þ
40
10
20
48
v
- s = - 1 Þ vs = - 14 V
10 20
3A
v1 = 0
ia =
Networks
30 + 7.5 + 2
= 329
. A
12
Þ
6 ia = 19.75 V
voltage across 2 W resistor
30 - 19.75 = 10.25 V,
P=
2A
(10.25) 2
= 52.53 W
2
43. (A) vx = 500 i1
10 W
40 W
vs
v y = 400( i1 - 0.001vx ) = 400 ( i1 - 0.5 i1 ) = 200 i1
20 W
+
v1
–
180 = 500 i1 + 100( i1 - 0.6) + 200 i1 + 100( i1 + 0.005 v y)
48 V
180 = 900 i1 - 60 + 100 ´ 0.005 ´ 200 i1
i1 = 0.12 A
Fig. S1.3.38
************
39. (D) ix = i1 - i2
15 = 4 i1 - 2( i1 - i2 ) + 6( i1 - i2 )
Þ
8 i1 - 4 i2 = 15
K(i)
-18 = 2 i2 + 6( i2 - i1 )
Þ
3i1 - 4 i2 = 9
K(ii)
i1 = 12
. A, i2 = -1.35 A
40. (B) 14 = 3i1 + v y + 6( i1 - 2 - 7) + 2 v y + 2( i1 - 7)
v y = 3( i1 - 2)
14 = 3i1 + 9( i1 - 2) + 6( i1 - 9) + 2( i1 - 7)
14 = 20 i1 - 18 - 54 - 14
Þ
i1 = 5 A
v1 = 6(5 - 2 - 7) + 2 ´ 3(5 - 2) + 2(5 - 7) = -10 V
3W
+
vy
3W
–
2A
6W
14 V
i1
+
v1
–
2vy
7A
2W
Fig. S1.3.40
41. (D) Let i1 and i2 be two loop current
0.5 vx = 500 i1 + 500( i1 - i2 ),
vx = -500 i1
Þ
5 i1 - 2 i2 = 0
K(i)
500( i2 - i1 ) + 900( i2 + 0.3) + 600( i2 - 0.6) = 0
-5 i1 + 20 i2 = 0.9
K(ii)
i1 = 20 mA, vx = -500 ´ 20m = -10 V
42. (C) 30 = 5 ia + 3( ia - 2.5) + 4( ia - 2.5 + 2)
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CHAPTER
1.4
NETWORKS THEOREM
1. vTH , RTH = ?
3W
4. A simple equivalent circuit of the 2 terminal network
2W
6W
6V
shown in fig. P1.4.4 is
vTH, RTH
R
i
Fig. P.1.4.1
(A) 2 V, 4 W
(B) 4 V, 4 W
(C) 4 V, 5 W
(D) 2 V, 5 W
2. i N , R N = ?
2W
v
Fig. P.1.4.4
R
2W
R
v
4W
15 V
iN, RN
(A)
(B)
Fig. P.1.4.2
10
W
3
(A) 3 A,
(B) 10 A, 4 W
(C) 1,5 A, 6 W
2W
3W
i
i
(D) 1.5 A, 4 W
3. vTH , RTH = ?
2A
R
R
(C)
(D)
5. i N , R N = ?
1W
2W
vTH, RTH
6A
4W
3W
iN RN
Fig. P.1.4.3
(A) -2 V,
6
W
5
5
(C) 1 V,
W
6
(B) 2 V,
5
W
6
6
(D) -1 V,
W
5
Fig. P.1.4.5
(A) 4 A, 3 W
(B) 2 A, 6 W
(C) 2 A, 9 W
(D) 4 A, 2 W
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UNIT 1
6. vTH , RTH = ?
Networks
The value of the parameter are
25 W
30 W
20 W
vTH, RTH
5V
5A
vTH
RTH
iN
RN
(A)
4 V
2 W
2 A
2 W
(B)
4 V
2 W
2 A
3 W
(C)
8 V
1.2 W
30
3
A
1.2 W
8 V
5 W
8
5
A
5 W
(D)
Fig. P.1.4.6
10. v1 = ?
(A) -100 V, 75 W
(B) 155 V, 55 W
(C) 155 V, 37 W
(D) 145 V, 75 W
2W
3W
1W
6W
2W
8V
7. RTH = ?
1W
+
v1
6W
18 V
–
6W
Fig. P.1.4.10
6W
2A
RTH
5V
Fig. P.1.4.7
(A) 6 V
(B) 7 V
(C) 8 V
(D) 10 V
11. i1 = ?
(A) 3 W
(B) 12 W
(C) 6 W
(D) ¥
4 kW
8. The Thevenin impedance across the terminals ab of
12 V
i1
20 V
4 kW
6 kW
24 V
3 kW
4 kW
the network shown in fig. P.1.4.8 is
a
3W
6W
2A
Fig. P.1.4.11
2V
8W
(A) 3 A
(B) 0.75 mA
(C) 2 mA
(D) 1.75 mA
8W
Statement for Q.12–13:
b
Fig. P.1.4.8
(A) 2 W
(B) 6 W
(C) 6.16 W
4
(D) W
3
A circuit is given in fig. P.1.4.12–13. Find the
Thevenin equivalent as given in question..
10 W
16 W
x
y
40 W
5V
8W
9. For In the the circuit shown in fig. P.1.4.9 a network
and its Thevenin and Norton equivalent are given
2W
x’
3W
y’
Fig. P.1.4.12–13
RTH
4V
iN
2A
vTH
RN
12. As viewed from terminal x and x¢ is
(A) 8 V, 6 W
(B) 5 V, 6 W
(C) 5 V, 32 W
(D) 8 V, 32 W
Fig. P.1.4.9
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GATE EC BY RK Kanodia
Network Theorems
13. As viewed from terminal y and y¢ is
(A) 8 V, 32 W
(B) 4 V, 32 W
(C) 5 V, 6 W
(D) 7 V, 6 W
Chap 1.4
19. vTH , RTH = ?
6W
3i1
14. A practical DC current source provide 20 kW to a
i1
iN,
4W
RN
50 W load and 20 kW to a 200 W load. The maximum
power, that can drawn from it, is
Fig. P1.4.19
(A) 22.5 kW
(B) 45 kW
(C) 30.3 kW
(D) 40 kW
Statement for Q.15–16:
In the circuit of fig. P.1.4.15–16 when R = 0 W , the
(A) 0 W
(B) 1.2 W
(C) 2.4 W
(D) 3.6 W
20. vTH , RTH = ?
current iR equals 10 A.
2W
4W
4V
2W
+
E
4W
2W
R
5W
0.1v1
4A
vTH RTH
v1
iR
–
Fig. P.1.4.20
Fig. P.1.4.15–16.
15. The value of R, for which it absorbs maximum
(A) 8 V, 5 W
(B) 8 V, 10 W
power, is
(C) 4 V, 5 W
(D) 4 V, 10 W
(A) 4 W
(B) 3 W
(C) 2 W
(D) None of the above
21. RTH = ?
3W
2W
+
16. The maximum power will be
(A) 50 W
(B) 100 W
(C) 200 W
(D) value of E is required
vx
4
4V
vx
RTH
–
17. Consider a 24 V battery of internal resistance
Fig. P.1.4.21
r = 4 W connected to a variable resistance RL . The rate
of heat dissipated in the resistor is maximum when the
current drawn from the battery is i . The current drawn
form the battery will be i 2 when RL is equal to
(A) 3 W
(B) 1.2 W
(C) 5 W
(D) 10 W
(A) 2 W
(B) 4 W
22. In the circuit shown in fig. P.1.4.22 the effective
(C) 8 W
(D) 12 W
resistance faced by the voltage source is
18. i N , R N = ?
4W
5W
10 W
i1
20i1
iN,
30 W
vs
RN
i
4
i
Fig. P.1.4.22
Fig. P.1.4.18
(A) 2 A, 20 W
(B) 2 A, -20 W
(A) 4 W
(B) 3 W
(C) 0 A, 20 W
(D) 0 A, -20 W
(C) 2 W
(D) 1 W
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UNIT 1
23. In the circuit of fig. P1.4.23 the value of RTH at
Networks
26. The value of RL will be
terminal ab is
ix
0.75va
16 V
–
a
Fig. P.1.4.26–27
va
+
4W
b
Fig. P.1.4.23
(A) -3 W
9
W
8
(B)
8
W
3
(C) -
RL
2W
8W
9V
3W
0.9 A
(A) 2 W
(B) 3 W
(C) 1 W
(D) None of the above
27. The maximum power is
(D) None of the above
(A) 0.75 W
(B) 1.5 W
(C) 2.25 W
(D) 1.125 W
28. RTH = ?
24. RTH = ?
-2ix
200 W
–
va
100
va
+
100 W
50 W
RTH
100 W
300 W
Fig. P.1.4.24
(A) ¥
(C)
(D)
125
W
3
maximum power if RL is equal to
(B) 136.4 W
(C) 200 W
(D) 272.8 W
29. Consider the circuits shown in fig. P.1.4.29
ia
100 W
200 W
2W
6W
6W
2W
2W
3i
–
(A) 100 W
i
6V
RTH
Fig. P.1.4.28
25. In the circuit of fig. P.1.4.25, the RL will absorb
40 W
+
vx
800 W
ix
(B) 0
3
W
125
100 W
0.01vx
RL
12 V
12 V
8V
Fig. P.1.4.25
6W
(A)
400
W
3
(B)
2
kW
9
(C)
800
W
3
(D)
4
kW
9
ib
2W
6W
6W
2W
Statement for Q.26–27:
18 V
2W
6W
3A
In the circuit shown in fig. P1.4.26–27 the
maximum power transfer condition is met for the load
RL .
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GATE EC BY RK Kanodia
Network Theorems
Chap 1.4
33. If vs1 = 6 V and vs 2 = - 6 V then the value of va is
The relation between ia and ib is
(A) ib = ia + 6
(B) ib = ia + 2
(A) 4 V
(B) -4 V
(C) ib = 15
. ia
(D) ib = ia
(C) 6 V
(D) -6 V
34. A network N feeds a resistance R as shown in fig.
30. Req = ?
12 W
P1.4.34. Let the power consumed by R be P. If an
4W
identical network is added as shown in figure, the
Req
6W
power consumed by R will be
2W
18 W
6W
9W
R
N
N
N
R
Fig. P.1.4.30
Fig. P.1.4.34
72
(B)
W
13
(A) 18 W
36
(C)
W
13
(D) 9 W
31. In the lattice network the value of RL for the
maximum power transfer to it is
(A) equal to P
(B) less than P
(C) between P and 4P
(D) more than 4P
35. A certain network consists of a large number of
ideal linear resistors, one of which is R and two
constant ideal source. The power consumed by R is P1
7W
when only the first source is active, and P2 when only
6
W
the second source is active. If both sources are active
RL
5
W
simultaneously, then the power consumed by R is
9W
(A) P1 ± P2
(B)
P1 ± P2
(C) ( P1 ± P2 ) 2
(D) ( P1 ± P2 ) 2
Fig. P.1.4.31
(A) 6.67 W
(B) 9 W
36. A battery has a short-circuit current of 30 A and an
(C) 6.52 W
(D) 8 W
open circuit voltage of 24 V. If the battery is connected
to an electric bulb of resistance 2 W, the power
dissipated by the bulb is
Statement for Q.32–33:
A circuit is shown in fig. P.1.4.32–33.
12 W
1W
3W
3W
1W
(A) 80 W
(B) 1800 W
(C) 112.5 W
(D) 228 W
37.
+
The
following
results
were
obtained
from
measurements taken between the two terminal of a
vs1
1W
va
vs2
resistive network
–
Terminal voltage
12 V
0V
Terminal current
0A
1.5 A
Fig. P.1.4.32–33
32. If vs1 = vs 2 = 6 V then the value of va is
(A) 3 V
(B) 4 V
(C) 6 V
(D) 5 V
The Thevenin resistance of the network is
(A) 16 W
(B) 8 W
(C) 0
(D) ¥
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UNIT 1
SOLUTIONS
38. A DC voltmeter with a sensitivity of 20 kW/V is used
to find the Thevenin equivalent of a linear network.
Reading on two scales are as follows
1. (B) vTH =
(a) 0 - 10 V scale : 4 V
The Thevenin voltage and the Thevenin resistance
(C) 18 V,
32
V,
3
1
MW
15
(D) 36 V,
200
kW
3
(B)
2
MW
15
( 6)( 6)
= 4 V,
3+ 6
RTH = ( 3||6) + 2 = 4 W
(b) 0 -15 V scale : 5 V
of the network is
16
200
(A)
V,
kW
3
3
Networks
2. (A)
2W
isc
Fig. S.1.4.2
R N = 2 ||4 + 2 =
+
RL
4W
15 V
39. Consider the network shown in fig. P.1.4.39.
Linear
Network
2W
v1
10
W,
3
15
2
v1 =
=6 V
1 1 1
+ +
2 2 4
v
isc = i N = 1 = 3 A
2
vab
–
Fig. P.1.4.39
The power absorbed by load resistance RL is
shown in table :
(2)( 3)(1)
= 1 V,
3+ 3
5
= 1||5 = W
6
3. (C) vTH =
RL
10 kW
30 kW
P
3.6 MW
4.8 MW
RTH
4. (B) After killing all source equivalent resistance is R
The value of RL , that would absorb maximum
power, is
(A) 60 kW
(B) 100 W
(C) 300 W
(D) 30 kW
Open circuit voltage = v1
5. (D) The short circuit current across the terminal is
2W
isc
40. Measurement made on terminal ab of a circuit of
6A
4W
3W
fig.P.1.4.40 yield the current-voltage characteristics
shown in fig. P.1.4.40. The Thevenin resistance is
Fig. S1.4.5
i(mA)
+
30
Resistive
Network
20
10
-4
-3 -2
-1
0
vab
–
1
2
v
Fig. P.1.4.40
a
b
isc =
6´ 4
= 4 A = iN ,
4+2
R N = 6 ||3 = 2 W
6. (B) For the calculation of RTH if we kill the sources
then 20 W resistance is inactive because 5 A source will
(A) 300 W
(B) -300 W
be open circuit
(C) 100 W
(D) -100 W
RTH = 30 + 25 = 55 W,
vTH = 5 + 5 ´ 30 = 155 V
***********
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Network Theorems
Chap 1.4
12. (B) We Thevenized the left side of xx¢ and source
7. (C) After killing the source, RTH = 6 W
transformed right side of yy¢
6W
8W
16 W
x
y
8W
6W
RTH
4V
8V
x’
Fig. S.1.4.7
y’
Fig. S1.4.12
8. (B) After killing all source,
RTH = 3||6 + 8 ||8 = 6 W
4
8
+
= 8 24 = 5 V,
1
1
+
8 24
vxx ¢ = vTH
9. (D) voc = 2 ´ 2 + 4 = 8 V = vTH
RTH = 2 + 3 = 5 W = R N ,
iN =
RTH = 8 ||(16 + 8) = 6 W
vTH 8
= A
RTH 5
13. (D) Thevenin equivalent seen from terminal yy¢ is
4
8
+
24
8 = 7 V,
=
1
1
+
24 8
10. (A) If we solve this circuit direct, we have to deal
with three variable. But by simple manipulation
v yy¢ = vTH
variable can be reduced to one. By changing the LHS
and RHS in Thevenin equivalent
1W
1W
1W
RTH = ( 8 + 16)||8 = 6 W
2W
14. (A)
+
6W
4V
12 V
v1
–
i
RL
r
Fig. S1.4.10
Fig. S1.4.14
4
12
+
v1 = 1 + 1 1 + 2 = 6 V
1
1
1
+ +
1+1 6 1+2
2
æ ir ö
ç
÷ 50 = 20 k,
è r + 50 ø
( r + 200) 2 = 4( r + 50) 2
11. (B) If we solve this circuit direct, we have to deal
with three variable. But by simple manipulation
Þ
r = 100 W
i = 30 A,
Pmax =
variable can be reduced to one. By changing the LHS
and RHS in Thevenin equivalent
2 kW
i1
4 kW
2
æ ir ö
ç
÷ 200 = 20 k
è r + 200 ø
20 V
6V
( 30) 2 ´ 100
= 22.5 kW
4
15. (C) Thevenized the circuit across R, RTH = 2 W
4W
2 kW
2W
2W
4W
8V
2W
Fig. S1.4.15
Fig. S1.4.11
i1 =
20 - 6 - 8
= 0.75 mA
2k + 4k + 2k
16. (A) isc = 10 A, RTH = 2 W,
2
æ 10 ö
Pmax = ç
÷ ´ 2 = 50 W
è 2 ø
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UNIT 1
Now in this circuit all straight-through connection
Networks
i 2 R = ( P1 ± P2 ) 2
have been cut as shown in fig. S1.4.32b
6W
36. (C) r =
1W
3W
2W
+
P=
voc
= 1. 2 W
isc
24 2
´ 2 = 112.5 W
(1. 2 + 2) 2
6V
va
37. (B) RTH =
–
voc 12
=
=8W
isc 15
.
Fig. S.1.4.32b
va =
6 ´ (2 + 3)
=5 V
2 + 3+1
38. (A) Let
33. (B) Since both source have opposite polarity, hence
short circuit the all straight-through connection as
shown in fig. S.1.4.33
6W
2W
For 0 -10 V scale Rm = 10 ´ 20 k = 200 kW
For 0 -50 V scale Rm = 50 ´ 20 k = 1 MW
4
For 4 V reading i =
´ 50 = 20 mA
10
vTH = 20mRTH + 20m ´ 200 k = 4 + 20mRTH
5
For 5 V reading i =
´ 50m = 5 mA
50
1W
3W
1
1
=
= 50 mA
sensitivity
20 k
+
vTH = 5m ´ RTH + 5m ´ 1M = 5 + 5mRTH
6V
va
Solving (i) and (ii)
16
200
V, RTH =
vTH =
kW
3
3
–
Fig. S1.4.33
39. (D) v10 k = 10 k ´ 3.6m = 6
6 ´ ( 6 ||3)
va = = -4 V
2 +1
v30 k = 30 k ´ 4.8m = 12 V
34. (C) Let Thevenin equivalent of both network
RTH
RTH
vTH
R
vTH
RTH
R
vTH
6 =
10
vTH
10 + RTH
12 =
30 vTH
30 + RTH
Þ
Þ
10 vTH = 6 RTH + 60
5 vTH = 2 RTH + 60
RTH = 30 kW
40. (D) At v = 0 , isc = 30 mA
Fig. S1.4.34
At i = 0, voc = - 3 V
v
-3
RTH = oc =
= - 100 W
isc 30m
2
æ VTH ö
÷÷ R
P = çç
è RTH + R ø
æ
ç V
TH
P¢ = ç
çç R + RTH
2
è
2
ö
÷
æ
VTH
÷ R = 4çç
÷÷
è 2 R + RTH
ø
2
ö
÷÷ R
ø
************
Thus P < P ¢ < 4 P
35. (C) i1 =
P1
P2
and i2 =
R
R
using superposition i = i1 + i2 =
Page
42
P1
±
R
P2
R
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GATE EC BY RK Kanodia
CHAPTER
1.6
THE RLC CIRCUITS
1. The natural response of an RLC circuit is described
(A) iL¢¢ ( t) + 1100 i¢¢L ( t) + 11 ´ 108 iL ( t) = 108 is ( t)
by the differential equation
(B) i¢¢L ( t) + 1100 i¢¢L ( t) + 11 ´ 108 iL ( t) = 108 is ( t)
d 2v
dv
dv(0)
+2
+ v = 0, v(0) = 10,
= 0.
dt 2
dt
dt
The v( t) is
(A) 10(1 + t) e - t V
(B) 10(1 - t) e - t V
(C) 10e - t V
(D) 10te - t V
vs
2W
1 mH
. i¢¢L ( t)
iL¢¢ ( t) 11
. iL ( t) = is ( t)
+
+ 11
108
10 4
(D)
i¢¢L ( t) 11i¢L ( t)
+
+ 11iL ( t) = is ( t)
108
10 4
4. In the circuit of fig. P.1.6.4 vs = 0 for t > 0. The initial
2. The differential equation for the circuit shown in fig.
P1.6.2. is
(C)
condition are v(0) = 6 V and dv(0) dt = -3000 V s. The
v( t) for t > 0 is
v
1H
10 mF
100 W
vs
25 mF
80 W
+
vC
–
Fig. P1.6.2
(A) v¢¢( t) + 3000 v¢( t) + 102
. ´ 108 v( t) = 108 vs ( t)
Fig. P1.6.4
(B) v¢¢( t) + 1000 v¢ ( t) + 102
. ´ 10 v( t) = 10 vs ( t)
8
8
(A) -2 e -100 t + 8 e -400 t V
(B) 6 e -100 t + 8 e -400 t V
(D) None of the above
(C)
2 v¢ ( t)
v¢¢( t)
. v( t) = vs ( t)
+
+ 102
108
10 5
(C) 6 e -100 t - 8 e -400 t V
(D)
2 v¢ ( t)
v¢¢( t)
. v( t) = vs ( t)
+
+ 198
8
10
10 5
5. The circuit shown in fig. P1.6.5 has been open for a
long time before closing at t = 0. The initial condition is
3. The differential equation for the circuit shown in fig.
v(0) = 2 V. The v( t) for t > is
P1.6.3 is
t=0
1H
10 W
is
3
4W
1
3F
+
vC
–
10 mF
100 W
Fig. P.1.6.5
iL
Fig. P.1.6.3
Page
54
-t
(A) 5 e - 7 e
-3t
V
(C) - e - t + 3e -3t V
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The RLC Circuits
Statement for Q.6–7:
Chap 1.6
10. The switch of the circuit shown in fig. P1.6.10 is
Circuit is shown in fig. P.1.6. Initial conditions are
opened at t = 0 after long time. The v( t) , for t > 0 is
i1 (0) = i2 (0) = 11 A
t=0
3W
i1
i2
2H
1W
1
2H
1W
6V
+
vC
–
1
4F
2W
3H
Fig. P1.6.10
Fig. P1.6.6–7
6. i1 (1 s) = ?
(A) 4 e -2 t sin 2 t V
(B) -4 e -2 t sin 2 t V
(C) 4 e -2 t cos 2 t V
(D) -4 e -2 t cos 2 t V
(A) 0.78 A
(B) 1.46 A
11. In the circuit of fig. P1.6.23 the switch is opened at
(C) 2.56 A
(D) 3.62 A
t = 0 after long time. The current iL ( t) for t > 0 is
4H
iL
7. i2 (1 s) = ?
(A) 0.78 A
(B) 1.46 A
(C) 2.56 A
(D) 3.62 A
t=0
2W
1
4F
8W
8. vC ( t) = ? for t > 0
4W
7A
25 mH
Fig. P1.6.11
10 mF
100 W
30u(-t) mA
+
vC
–
(A) e -2 t (2 cos t + 4 sin t) A
(B) e -2 t ( 3 sin t - 4 cos t) A
(C) e -2 t ( -4 sin t + 2 cos t) A
(D)e -2 t (2 sin t - 4 cos t) A
Statement for Q.12–14:
Fig. P1.6.8
In the circuit shown in fig. P1.6.12–14 all initial
(A) 4 e -1000 t - e -2000 t V
(B) ( 3 + 6000 t) e -2000 t V
(C) 2 e -1000 t + e -2000 t V
(D) ( 3 - 6000 t) e -2000 t V
condition are zero.
iL
9. The circuit shown in fig. P1.6.9 is in steady state
with switch open. At t = 0
10 mH
1 mF
+
vL
–
the switch is closed. The
output voltage vC ( t) for t > 0 is
Fig. P1.5.12-14
12. If is ( t) = 1 A, then the inductor current iL ( t) is
0.8 H
250 W
100
W
65
isu(t) A
500 W
t=0
9V
5 mF
+
vC
–
(A) 1 A
(B) t A
(C) t + 1 A
(D) 0 A
13. If is ( t) = 0.5 t A,
Fig. P1.6.9
A
(B) 2 t - 3250 A
-3
A
(D) 2 t + 3250 A
(A) 0.5 t + 3. 25 ´ 10
(C) 0.5 t - 0. 25 ´ 10
then iL ( t) is
-3
(B) e
-400 t
[ 3 cos 300 t + 4 sin 300 t ]
14. If is ( t) = 2 e -250 t A then iL ( t) is
4000 -250 t
4000 -250 t
(A)
A
(B)
A
te
e
3
3
(C) e
-300 t
[ 3 cos 400 t + 4 sin 300 t ]
(C)
(A) -9 e
-400 t
+ 12 e
-300 t
(D) e -300 t [ 3 cos 400 t + 2. 25 sin 300 t ]
200 -250 t
A
e
7
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(D)
200 -250 t
A
te
7
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55
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UNIT 1
15. The forced response for the capacitor voltage v f ( t) is
Networks
19. In the circuit shown in fig. P 1.5.19 v( t) for t > 0 is
100 W
2u(-t) A
iL
– vx +
avx
50 W
1H
20 mH
0.04 F
+ vC –
2W
4W
Fig. P1.6.15
(A) 0. 2 t + 117
. ´ 10 -3 V
(B) 0. 2 t - 117
. ´ 10 -3 V
(C) 117
. ´ 10 -3 t - 0. 2 V
(D) 117
. ´ 10 -3 t + 0. 2 V
50u(t) V
Fig. P1.6.19
16. For a RLC series circuit R = 20 W , L = 0.6 H, the
value of C will be
(B) 50 + ( 46.5 sin 3t + 62 cos 3t) e -4 t V
[CD =critically damped, OD =over damped,
(C) 50 + ( 62 cos 4 t + 46.5 sin 4 t) e -3t V
UD =under damped].
CD
OD
UD
(A) C = 6 mF
C >6 mF
C <6 mF
(B) C = 6 mF
C < 6 mF
C > 6 mF
(C) C >6 mF
C = 6 mF
C < 6 mF
(D) C < 6 mF
C =6 mF
C > 6 mF
(A) 50 - ( 46.5 sin 3t + 62 cos 3t) e -4 t V
(D) 50 - ( 62 cos 4 t + 46.5 sin 4 t) e -3t V
20. In the circuit of fig. P1.6.20 the switch is closed at
t = 0 after long time. The current i( t) for t > 0 is
1
16 F
+ vC –
17. The circuit shown in fig. P1.6.17 is critically
damped. The value of R is
20 V
t=0
1
4H
Fig. P1.6.20
120 W
R
5W
iL
10 mF
4H
(A) -10 sin 8 t A
(B) 10 sin 8 t A
(C) -10 cos 8 t A
(D) 10 cos 8 t A
21. In the circuit of fig. P1.6.21 switch is moved from 8
Fig. P1.6.17
(A) 40 W
(B) 60 W
(C) 120 W
(D) 180 W
V to 12 V at t = 0. The voltage v( t) for t > 0 is
2W
t=0
18. The step response of an RLC series circuit is given
8V
12 V
by
1H
d 2 i( t) 2 di( t)
di(0 + )
= 4.
+
+ 5 i( t) = 10, i(0 + ) = 2,
dt
dt
dt
Fig. P1.6.21
(A) 12 - ( 4 cos 2 t + 2 sin 2 t) e - t V
The i( t) is
(A) 1 + e - t cos 4 t A
(B) 4 - 2 e - t cos 4 t A
(B) 12 - ( 4 cos 2 t + 8 sin 2 t) e - t V
(C) 2 + e - t sin 4 t A
(D) 10 + e - t sin 4 t A
(C) 12 + ( 4 cos 2 t + 8 sin 2 t) e - t V
(D) 12 + ( 4 cos 2 t + 2 sin 2 t) e - t V
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56
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1
F
6
+
vC
–
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The RLC Circuits
22. In the circuit of fig. P1.5.22 the voltage v( t) is
5W
3u(t) A
25. In the circuit shown in fig. P1.6.25 a steady state
has been established before switch closed. The i( t) for
1W
5H
t > 0 is
5W
+
vC
–
0.2 F
20 V
Chap 1.6
20 W
1H
2W
i
t=0
1
F
25
5W
100 V
Fig. P1.6.22
(A) 40 - (20 cos 0.6 t + 15 sin 0.6 t) e -0 .8 t V
(B) 35 + (15 cos 0.6 t + 20 sin 0.6 t) e
-0 .8 t
(C) 35 - (15 cos 0.6 t + 20 sin 0.6 t) e
(D) 35 - 15 cos 0.6 t e
-0 .8 t
-0 .8 t
Fig. P1.6.25
V
(A) 0.73e -2 t sin 4.58 t A
V
(B) 0.89 e -2 t sin 6.38 t A
V
(C) 0.73e -4 t sin 4.58 t A
23. In the circuit of fig. P1.6.23 the switch is opened at
t = 0 after long time. The current i( t) for t > 0 is
(D) 0.89 e -4 t sin 6.38 t A
26. The switch is closed after long time in the circuit of
fig. P1.6.26. The v( t) for t > 0 is
2A
2A
3
4H
10 W
1
3F
t=0
1W
1H
6W
5W
10 W
4V
t=0
1
25 F
+
vC
–
Fig. P1.6.23
(A) e-2 .306 t + e-0 . 869t A
Fig. P1.6.26
(B) -e -2 .306 t + 2 e -0 . 869t A
(A) -8 + 6 e -3t sin 4 t V
(C) e -4 .431 t + e -0 .903t A
(B) -12 + 4 e -3t cos 4 t V
(D) 2 e -4 .431 t - e -0 .903t A
(C) -12 + ( 4 cos 4 t + 3 sin 4 t) e -3t V
24. In the circuit of fig. P1.6.24 switch is moved from
position a to b at t = 0. The iL ( t) for t > 0 is
0.02 F
(D) -12 + ( 4 cos 4 t + 6 sin 4 t) e -3t V
27. i( t) = ?
2 kW
14 W
b
i
2 H t=0
2W
12 V
5 mF
12u(t) V
8 mH
a
iL
6W
Fig. P1.6.27
(A) 6 - ( 6 cos 500 t + 6 sin 5000 t) e -50 t mA
(B) 8 - ( 8 cos 500 t + 0.06 sin 5000 t) e -50 t mA
4A
Fig. P1.6.24
(A) ( 4 - 6 t) e 4 t A
(B) ( 3 - 6 t) e -4 t A
(C) ( 3 - 9 t) e -5t A
(D) ( 3 - 8 t) e -5t A
(C) 6 - ( 6 cos 5000 t + 0.06 sin 5000 t) e -50 t mA
(D) 6 e -50 t sin 5000 t mA
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UNIT 1
Networks
SOLUTIONS
28. In the circuit of fig. P1.6.28 i(0) = 1 A and v(0) = 0.
The current i( t) for t > 0 is
i
4u(t) A
1. (A) s 2 + 2 s + 1 = 0
2W
1H
s = -1, - 1,
v( t) = ( A1 + A2 t) e
dv(0)
v(0) = 10 V,
= 0 = - 1 A1 + A2
dt
0.5 F
A1 = A2 = 10
Fig. P1.6.28
(A) 4 + 6.38 e -0 .5t A
Þ
-t
(B) 4 - 6.38 e -0 .5t A
(C) 4 + ( 3 cos 1.32 t + 113
. sin 1.32 t) e -0 .5t A
2. (A) iL =
(D) 4 - ( 3 cos 1.32 t + 113
. sin 1.32 t) e -0 .5t A
v
dv
+ 10 ´ 10 -6
100
dt
iL
29. In the circuit of fig. P1.6.29 a steady state has been
2W
1 mH
v
established before switch closed. The vo( t) for t > 0 is
10 W
vs
t=0
5W
3A
+
10 mF
1H
Fig. S1.6.2
vo
–
vs = 2 iL + 10 -3
Fig. P1.6.29
(A) 100 te -10 t V
(B) 200 te -10 t V
(C) 400 te -50 t V
(D) 800 te -50 t V
108 vs ( t) = v¢¢( t) + 3000 v¢( t) + 102
. v( t)
t > 0 is
established before switch closed. The i( t) for
i
1H
1
4F
2W
t=0
(B) -e
sin 2 t A
(C) -2(1 - t) e
-2 t
-2 t
sin 2 t A
(D) 2(1 - t) e -2 t A
A
4. (A)
31. In the circuit of fig. P1.6.31 a steady state has been
established. The i( t) for t > 0 is
i
3A
10 W
6u(t) A
diL
di
d 2 iL
+ iL + 10 -4 L + 10 -8
dt
dt
dt 2
.
i¢¢L ( t) 11
. iL ( t) = is ( t)
+
iL¢ ( t) + 11
108
10 4
Þ
Fig. P1.6.30
(A) 2 e
vC
dvC
+ iL + 10m
100
dt
di
vC = 10 iL + 10 -3 L
dt
di
d
di
is = 0.1iL + 10 -5 L + iL + 10 -5 (10 iL + 10 -3 L )
dt
dt
dt
3. (C) is =
= 0.1iL + 10 -5
6V
-2 t
diL
+v
dt
æ 1 dv
dv ö
d2 v ö
æ v
÷+ v
÷ + 10 -3 çç
= 2ç
+ 10 -6 ´ 10 - t
+ 10 ´ 10 -6
dt ø
dt 2 ÷ø
è 100
è 100 dt
30. In the circuit of fig. P1.6.30 a steady state has been
1W
10 mF
100 W
40 W
10 mF
4H
v
dv
+ 25m
+ ( v - vs ) dt = 0
80
dt ò
d 2v
dv
+ 500
+ 40000 = 0
2
dt
dt
Þ
s 2 + 500 s + 40000 = 0
Þ
s = -100, - 400,
v( t) = Ae -100 t + Be -400 t
A + B = 6, -100 A - 400 B = -3000
Fig. P1.6.31
(A) 9 + 2 e
-10 t
- 8e
-2 .5t
(B) 9 - 8 e
10 t
A
(C) 9 + (2 cos 10 t + sin 10 t) e
-2 .5t
A
(D) 9 + (cos 10 t + 2 sin 10 t) e -2 .5t A
***************
Page
58
+ 2e
-2 .5t
Þ
B = 8, A = -2
A
5. (C) The characteristic equation is s 2 +
After putting the values,
v( t) = Ae - t + Be -3t ,
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s2 + 4 s + 3 = 0
1
s
+
=0
RC LC
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v(0 + ) = 2 V
Þ
iL (0 + ) = 0Þ
iR (0) =
-C
9. (B) vC (0 + ) = 3 V , iL (0 + ) = -12 mA
A + B =2
dv(0 + ) 8
=
dt
3
vC
dvC
+ iL + 5 ´ 10 -6
=0
250
dt
2
8
= ,
34 3
3
dvC (0 + )
dvC (0 + )
- 12m + 5 ´ 10 -6
=0 Þ
=0
250
dt
dt
1
s
+
=0
s2 +
-6
250 ´ 5 ´ 10
0.8 ´ 5 ´ 10 -6
dv(0 + )
= - 8,
dt
Þ
- A - 3B = -8, B = 3, A = -1
Þ
di1
di
- 3 2 = 0,
dt
dt
3di2
di
2 i2 +
- 3 1 =0
dt
dt
6. (D) i1 + 5
Þ
i1 = A e
+ Be
-2 t
10. (B) v(0 + ) = 0,
s2 + 4 s + 8 = 0
A1 = 0,
, i(0) = A + B = 11
-
1
6
i2 = - e
2W
1
4F
-
t
6
+ De
+ 12 e
di2 (0) -143
C
=
= - - 2D
dt
6
6
A, i2 (1 s) = e
+
vC
–
8W
diL (0 + )
diL (0 + )
= 8 - ( -4) ´ 8 Þ
= 10
dt
dt
1
s
vC + vC + iL = 0, vC = 4 siL + 8 iL
4
2
4
-2 t
s 2 iL + 4 siL + 5 = 0, s = -2 ± j
and D = 12
-2 t
4H
Fig. S1.6.11
-
1
6
+ 12 e
-2
= 0.78 A
iL ( t) = e-2 t ( A1 cos t + A2 sin t)
A1 = -4,
8. (B) vC (0 + ) = 30m ´ 100 = 3 V
dvC (0 - )
dvC (0 + )
= iL (0 - ) = 0 = iL (0 + ) = C
dt
dt
100
1
s2 +
s+
25 ´ 10 -3
25 ´ 10 -3 ´ 10 ´ 10 -6
C
Þ
vC (0 + ) = 8 V
iL
+ 8 e -2 = 3.62 A
i2 (0) = 11 = C + D,
t
6
11. (D) iL (0 + ) = -4,
+ 8e ,
7. (A) i2 = Ce
-
dvC (0 + )
= -8 = -2 (0 + 0) + (0 + 2 A2 ), A2 = -4
dt
-2 t
i1 (1 s) = 3e
C = -1
s = - 2 ± j2
vC ( t) = e ( A1 cos 2 t + A2 sin 2 t)
di1 (0 + )
33 di2 (0 + )
143
=,
=dt
2
dt
6
A
33
- 2 B = - , Þ A = 3, B = 8,
6
2
i1 = 3e
Þ
1 dvC (0 + )
= -2
4 xdt
iL (0 + ) = 2 A,
-2 t
In differential equation putting t = 0 and solving
t
6
s = -400 ± j 300
vC ( t) = e-400 t ( A1 cos 300 t + A2 sin 300 t)
dvC (0)
A1 = 3,
= -400 A1 + 300 A2 , A2 = 4
dt
6 s 2 + 13s + 2 = 0
1
s = - , -2
6
1
- t
6
s 2 + 800 s + 25 ´ 10 4 = 0
Þ
(1 + 5 s) i1 - 3si2 = 0, -3si1 + (2 + 3s) i2 = 0
( 3s)( 3s) i1
(1 + 5 s) i1 =0
2 + 3s
Þ
Chap 1.6
s = -2000, -2000
diL (0 + )
= 10 = -2 ( A1 + 0) + A2 , A2 = 2
dt
12. (A) is =
is =
v
dv
di
+ 10 -3
+ iL , v = 10 ´ 10 -3 L
100 65
dt
dt
65
di
d 2 iL
(10 ´ 10 -3) L + 10 -3(10 ´ 10 -3)
+ iL = 0
100
dt
dt
d 2 iL
di
+ 650 L + 10 5 iL = 10 5 is
dt
dt
vC ( t) = ( A1 + A2 t) e -2000 t
dvC ( t)
= A2 e -2000 t + ( A1 + A2 t) e -2000 t ( -2000)
dt
dvC (0)
= A2 - 2000 ´ 3 = 0
vC (0 + ) = A1 = 3,
dt
0 + 0 + 10 5 B = 10 5,
Þ
0 + 650 A + ( At + B)10 5 = 10 5(0.5 t), A = 0.5
A2 = 6000
Trying iL ( t) = B
B = 1,
iL = 1 A
13. (A) Trying iL ( t) = At + B,
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The RLC Circuits
Chap 1.6
di(0 + ) -16
=
= -4.431 A - 0.903B
dt
3
28.(D) i(0 + ) = 1 A, v(0 + ) =
A = 1, B = 1
a=
Ldi(0 + )
dt
1
= 0.5,
2 ´ 2 ´ 0.5
4´ 6
=3
6 +2
dv (0)
dvC (0)
0.02 C
= iL (0) = 3 Þ
= 150
dt
dt
6 + 14
1
a=
= 5,
wo =
=5
2´2
2 ´ 0.02
1 = 4 + A,
Þ A = -3
di(0)
= 0 = 0.5 A + 1.32 B,
dt
a = wo critically damped
29. (B) Vo(0 + ) = 0 ,iL (0 + ) = 1 A
24. (C) vC (0) = 0,
v( t) = 12 + ( A + Bt) e
0 = 12 + A,
v( t) = 12 + (90 t - 12) e
A = -12,
iL ( t) = 0.02( -5) e (90 t - 12) + 0.02(90) e
-5t
= ( 3 - 9 t) e
-5t
diL (0 + )
= v1 (0) = 0
dt
1
a=
= 10,
2 ´ 5 ´ 0.01
B = -113
.
Wo =
1
1 ´ 0.01
= 10
a = Wo, so critically damped response
100 ´ 5
50
,
25. (A) v(0 ) =
=
5 + 5 + 20
3
+
+
iL (0 ) = 0
s = -10, - 10
i( t) = 3( A + Bt) e -10 t ,
if = 0 A
i(0) = 1 = 3 + A
+
di(0 )
= -10 A + B
dt
diL (0 + )
50 10
= 20 =
dt
3
3
4
1
a=
= 2, wo =
=5
2´1
1
1´
25
iL ( t) = 3 - (2 + 20 t) e -10 t ,
i( t) = ( A cos 4.58 t + B sin 4.58 t) e -2 t
26.(A) iL (0 + ) = 0, vL (0 + ) = 4 - 12 = -8
+
LdiL ( t)
= 200 te -10 t
dt
-6
di(0 + )
= -2 A,
vc (0 + ) = 2 ´ 1 = 2 =
1+2
dt
1
1
1
a=
=
= 2,
Wo =
=2
2 RC 2 ´ 1 ´ 0.25
LC
a = Wo, critically damped response
s = -2 , -2
1 dvL (0 )
= iL (0 + ) = 0
25
dt
6
1
a = = 3, Wo =
=5
2
1 ´ 1 / 25
A = -2
i( t) = ( A + Bt) e -2 t ,
di( t)
= ( -2 + Bt) e 2 t ( -2) + (0 + B) e -2 t
dt
b = -3 ± 9 - 25 = -3 ± j 4
v1 ( t) = -12 + ( A cos 4 t + B sin 4 t) e
vo =
30. (C) i(0 + ) =
s = -2 ± 4 - 25 = -2 ± j 4.58
At t = 0, Þ B = -2
-3t
vL (0) = -8 = 12 + A,
Þ A =4
dvL (0)
= 0 = -3 A + 4 B,
Þ B=3
dt
1
1
=
= 50
2 RC 2 ´ 2 k ´ 54
1
1
=
= 5000
LC
8 m ´ 5m
27. (C) a =
Wo =
B = 90
-5t
-5t
= 2
i( t) = 4 + ( A cos 1.32 t + B sin 1.32 t) e -0 .5t
-5t
Þ
1 - 0.5
s = -0.5 ± 0.5 2 - 2 = 0.5 ± j1.323
iL (0) =
150 = -5 A + B
1
Wo =
31. (A) i(0 + ) = 3 A, vC (0 + ) = 0 V =
is = 9 A, R = 10||40 = 8 W
1
1
a=
=
= 6.25
2 RC 2 ´ 8 ´ 0.01
1
1
Wo =
=
=5
LC
4 ´ 0.01
a > Wo, so overdamped response
a < Wo, underdamped response.
s = -6.25 ± 6.25 2 - 25 = -10, -2.5
s = -50 ± 50 2 - 5000 2 = -50 ± j5000
i( t) = 9 + Ae -10 t + Be -2 .5t
i( t) = 6 + ( A cos 5000 t + B sin 5000 t) e -50 t mA
3 = 9 + A + B,
i(0) = 6 = 6 + A,
Þ A = -6
di(0)
= -50 A + 5000 B = 0,
B = -0.06
dt
4 di(0 + )
dt
0 = -10 A - 2.5 B
On solving, A = 2, B = -8
************
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CHAPTER
1.7
SINUSOIDAL STEADY STATE ANALYSIS
1. i( t) = ?
20cos 300t V
(A)
3W
i
(C)
~
25 mH
1
2
1
2
cos (2 t - 45 ° ) V
(B)
sin (2 t - 45 ° ) V
(D)
1
2
1
2
cos (2 t + 45 ° ) V
sin (2 t + 45 ° ) V
4. vC ( t) = ?
Fig. P1.7.1
3H
(A) 20 cos ( 300 t + 68. 2 ° ) A
(B) 20 cos( 300 t - 68. 2 ° ) A
8cos 5t V
(C) 2.48 cos( 300 t + 68. 2 ° ) A
~
+
vC
–
50 mF
9W
(D) 2.48 cos( 300 t - 68. 2 ° ) A
Fig. P1.7.4
2. vC ( t) = ?
(A) 2. 25 cos (5 t + 150 ° ) V
3
cos 10 t A
~
2W
+
vC
–
1 mF
(B) 2. 25 cos (5 t - 150 ° ) V
(C) 2. 25 cos (5 t + 140.71° ) V
(D) 2. 25 cos (5 t - 140.71° ) V
Fig. P1.7.2
(A) 0.89 cos (10 3 t - 63.43° ) V
5. i( t) = ?
(B) 0.89 cos (10 3 t + 63.43° ) V
1W
4W
(C) 0.45 cos (10 t + 26.57 ° ) V
3
i
(D) 0.45 cos (10 t - 26.57 ° ) V
3
10cos 2t V
~
0.25 F
4H
3. vC ( t) = ?
5W
cos 2t V
~
0.1 F
Fig. P1.7.5
+
vC
–
(A) 2 sin (2 t + 5.77 ° ) A
(B) cos (2 t - 84. 23° ) A
(C) 2 sin (2 t - 5.77 ° ) A
(D) cos (2 t + 84. 23° ) A
Fig. P1.7.3
Page
62
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UNIT 1
13. In the bridge shown in fig. P1.7.13, Z1 = 300 W,
Z 2 = ( 300 - j 600) W, Z 3 = (200 + j 100)W. The Z 4
Networks
Statement for Q.17-18:
at
The circuit is as shown in fig. P1.7.17-18
balance is
1W
1W
1H
Z
Z
3
1
1H
1W
~
i1
o
10cos (4t-30 ) V
1F
Z
Z
~
i2
4
2
5cos 4t V
Fig. P1.7.17–18
~
17. i1 ( t) = ?
(A) 2.36 cos ( 4 t - 4107
. °) A
Fig. P1.7.13
(A) 400 + j 300 W
(B) 400 - j 300 W
(C) j100 W
(D) - j900 W
(B) 2.36 cos ( 4 t + 4107
. °) A
(C) 1.37 cos ( 4 t - 4107
. °) A
(D) 2.36 cos ( 4 t + 4107
. °) A
14. In a two element series circuit, the applied
voltage
and
the
resulting
current
are
v( t) = 60 + 66 sin (10 3 t) V, i( t) = 2.3 sin (10 3 t + 68.3° ) A.
The nature of the elements would be
(A) R - C
(B) L - C
(C) R - L
(D) R - R
18. i2 ( t) = ?
(A) 2.04 sin ( 4 t + 92.13° ) A
(B) -2.04 sin ( 4 t + 2.13° ) A
(C) 2.04 cos ( 4 t + 2.13° ) A
(D) -2.04 cos ( 4 t + 92.13° ) A
19. I x = ?
15. Vo = ?
40 W
j20
~
120Ð-15o V
~
50 W
-j30
Ix
6Ð30 A
10Ð30 V
o
o
~
Fig. P1.7.15
-j2 W
(B) 223Ð56 ° V
(C) 124 Ð - 154 ° V
(D) 124 Ð154 ° V
16. vo( t) = ?
(A) 394
. Ð46. 28 ° A
(B) 4.62 Ð97.38 ° A
(C) 7.42 Ð92.49 ° A
(D) 6.78 Ð49. 27 ° A
20. Vx = ?
3W
j10 W
20 W
o
10sin (t+30 ) V
~
1F
+
vo
–
j3 W
Fig. P1.7.19
(A) 223Ð - 56 ° V
1H
0.5Ix
4W
Vo
~
o
20cos (t-45 ) V
4Vx
3Ð0o A
~
20 W
+
Vx
–
Fig. P1.7.16
Fig. P1.7.20
(A) 315
. cos ( t + 112 ° ) V
(B) 43. 2 cos ( t + 23° ) V
(A) 29.11Ð166 ° V
(B) 29.11Ð - 166 ° V
(C) 315
. cos ( t - 112 ° ) V
(C) 43. 24 Ð124 ° V
(D) 43. 24 Ð - 124 ° V
(D) 43. 2 cos ( t - 23° ) V
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64
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UNIT 1
Statement for Q.27–32:
Networks
35. In the circuit shown in fig. P1.7.35 power factor is
Determine the complex power for hte given values
-j2
4W
in question.
27. P = 269 W, Q = 150 VAR (capacitive)
(A) 150 - j269 VA
(B) 150 + j269 VA
(C) 269 - j150 VA
(D) 269 + j150 VA
-j2
j5
Fig. P1.7.35
28. Q = 2000 VAR, pf = 0.9 (leading)
(A) 4129.8 + j2000 VA
(B) 2000 + j 4129.8 VA
(C) 2000 - j 4129
. .8 VA
(D) 4129.8 - j2000 VA
(A) 56.31 (leading)
(B) 56.31 (lagging)
(C) 0.555 (lagging)
(D) 0.555 (leading)
36. The power factor seen by the voltage source is
29. S = 60 VA, Q = 45 VAR (inductive)
(A) 39.69 + j 45 VA
(B) 39.69 - j 45 VA
(C) 45 + j 39.69 VA
(D) 45 - j 39.69 VA
4W
1W
+ v1 –
10cos 2t V
30. Vrms = 220 V, P = 1 kW, |Z |= 40 W (inductive)
(A) 1000 - j 68125
. VA
(B) 1000 + j 68125
. VA
(C) 68125
. + j1000 VA
(D) 68125
. - j1000 VA
31. Vrms = 21Ð20 ° V, Vrms = 21Ð20 ° V, I rms = 8.5 Ð - 50 ° A
(A) 154.6 + j 89.3 VA
(B) 154.6 - j 89.3 VA
(C) 61 + j167.7 VA
(D) 61 - j167.7 VA
~
1
3F
3v
4 1
Fig. P1.7.36
(A) 0.8 (leading)
(B) 0.8 (lagging)
(C) 36.9 (leading)
(D) 39.6 (lagging)
37. The average power supplied by the dependent
source is
32. Vrms = 120 Ð30 ° V, Z = 40 + j 80 W
Ix
(A) 72 + j144 VA
(B) 72 - j144 VA
(C) 144 + j72 VA
(D) 144 - j72 VA
~
2Ð90 A
o
j1.92
4.8 W
8W
1.6Ix
33. Vo = ?
Fig. P1.7.37
+
6Ð0o A
~
16 kW
0.9 pf lagging
VO
20 kW
0.8 pf lagging
(A) 96 W
(B) -96 W
(C) 92 W
(D) -192 W
–
38. In the circuit of fig. P1.7.38 the maximum power
Fig. P1.7.33
absorbed by Z L is
(A) 7.1Ð32. 29 ° kV
(B) 42.59 Ð32.29 ° kV
(C) 38.49 Ð24.39 ° kV
(D) 38.49 Ð32. 29 ° kV
10 W
120Ð0 V
o
34. A relay coil is connected to a 210 V, 50 Hz supply. If
~
j15
-j10
it has resistance of 30 W and an inductance of 0.5 H, the
Fig. P1.7.38
apparent power is
Page
66
(A) 30 VA
(B) 275.6 VA
(C) 157 VA
(D) 187 VA
(A) 180 W
(B) 90 W
(C) 140 W
(D) 700 W
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Sinusoidal Steady State Analysis
Chap 1.7
39. The value of the load impedance, that would
The line impedance connecting the source to load is
absorbs the maximum average power is
0.3 + j0.2 W. If the current in a phase of load 1 is
j100
3Ð20o A
~
I = 10 Ð20 ° A rms , the current in source in ab branch is
-j40
80 W
ZL
(A) 15 Ð - 122 ° A rms
(B) 8.67 Ð - 122 ° A rms
(C) 15 Ð27.9 ° A rms
(D) 8.67 Ð - 57.9 ° A rms
45.
Fig. P1.7.39
(A) 12.8 - j 49.6 W
(B) 12.8 + j 49.6 W
(C) 339
. - j 86.3 W
(D) 339
. + j 86.3 W
An
abc
phase
sequence
3-phase
balanced
Y-connected source supplies power to a balanced
D –connected load. The impedance per phase in the load
is 10 + j 8 W. If the line current in a phase is
I aA = 28.10 Ð - 28.66 ° A rms and the line impedance is
zero, the load voltage V AB is
Statement for Q.40–41:
In a balanced Y–connected three phase generator
Vab = 400 Vrms
40. If phase sequence is abc then phase voltage
Va , Vb , and Vc are respectively
(A) 207.8 Ð - 140 ° Vrms
(B) 148.3Ð40 ° Vrms
(C) 148.3Ð - 40 ° Vrms
(D) 207.8 Ð40 ° Vrms
46. The magnitude of the complex power supplied by a
3-phase balanced Y-Y system is 3600 VA. The line
voltage is 208 Vrms . If the line impedance is negligible
(A) 231Ð0 ° , 231Ð120 ° , 231Ð240 °
and the power factor angle of the load is 25°, the load
(B) 231Ð - 30 ° , 231Ð - 150 ° , 231Ð90 °
impedance is
(C) 231Ð30 ° , 231Ð150 ° , 231Ð - 90 °
(A) 5.07 + j10.88 W
(B) 10.88 + j5.07 W
(D) 231Ð60 ° , 231Ð180 ° , 231Ð - 60 °
(C) 432
. + j14.6 W
(D) 14.6 + j 432
. W
41. If phase sequence is acb then phase voltage are
(A) 231Ð0 ° , 231Ð120 ° , 231Ð240 °
(B) 231Ð - 30 ° , 231Ð - 150 ° , 231Ð90 °
***********
(C) 231Ð30 ° , 231Ð150 ° , 231Ð - 90 °
(D) 231Ð60 ° , 231Ð180 ° , 231Ð - 60 °
42. A balanced three-phase Y-connected load has one
phase voltage Vc = 277 Ð45 ° V. The phase sequence is
abc. The line to line voltage V AB is
(A) 480 Ð45 ° V
(B) 480 Ð - 45 ° V
(C) 339 Ð45 ° V
(D) 339 Ð - 45 ° V
43. A three-phase circuit has two parallel balanced D
loads, one of the 6 W resistor and one of 12 W resistors.
The magnitude of the total line current, when the
line-to-line voltage is 480 Vrms , is
(A) 120 A rms
(B) 360 A rms
(C) 208 A rms
(D) 470 A rms
44. In a balanced three-phase system, the source has an
abc phase sequence and is connected in delta. There are
two parallel Y-connected load. The phase impedance of
load 1 and load 2 is 4 + j 4 W and 10 + j 4 W respectively.
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UNIT 1
SOLUTIONS
æ -j ö
7. (C) Z = çç
÷÷||( 6 + j(27m) w)
è w(22m ) ø
1. (D) Z = 3 + j(25m)( 300) = 3 + j7.5 W = 8.08 Ð68. 2 °
I=
20 Ð0
= 2.48 Ð - 68. 2 ° A
8.08 Ð68. 2 °
i( t) = 2.48 cos ( 300 t - 68. 2 ° ) A
1
+ j(1m)(10 3) = 0.5 + j = 112
. Ð63.43°
2
(1Ð0)
VC =
= 0.89 Ð - 63.43° V
112
. Ð63.43°
2. (A) Y =
vC ( t) = 0.89 cos (10 3 t - 63.43° ) V
(1Ð0) (5 Ð - 90 ° )
vC ( t) =
5 2 Ð - 45 °
1
2
1
=
2
27 ´ 10 3 j 6 ´ 106
- j106
( 6 + j27 ´ 10 -3 w)
22
22 w
= 22 w
=
6
10
æ
106 ö
6 + j(27mw )
÷
6 + jwçç 27m 22 w
22 w2 ÷ø
è
j27 ´ 10 3
- j 36 ´ 106
22
w22
æ
106
wçç 27m 22 w2
è
Þ w = 1278
w
1278
Hz =
f =
= 203 Hz
2p
2p
V1 = Vs - V2 = 7.68 Ð47° - 7.51Ð35° = 159
. Ð125°
9. (B) vin = 32 + (14 - 10) 2 = 5
Ð - 45 ° V
10. (C) I1 = 744 Ð - 118 ° mA,
cos (2 t - 45 ° ) V
I 2 = 540 Ð100 ° mA
4. (D) Z = 9 + j( 3)(5) +
I = I1 + I 2 = 744 Ð - 118 ° + 540.5 Ð100 °
-j
= 9 + j11
(50m) (5)
= 460 Ð - 164 °
i( t) = 460 cos ( 3t - 164 ° ) mA
Þ
Z = 14.21Ð50.71° W
( 8 Ð0)( 4 Ð - 90 ° )
VC =
= 2. 25 Ð140.71° V
14. 21Ð50.71°
11. (A)
2 Ð45° =
VC
V - 20 Ð0
+ C
- j4
j5 + 10
vC ( t) = 2. 25 cos (5 t - 140.71° ) V
j50
10 Ð0
10 Ð0
1
5. (B) Va =
V
=
1
1
1
105
. + j0.4
+
+
1 - j2 4 + j 8
I=
ö
÷÷ = 0
ø
8. (C) Vs = 7.68 Ð47° V, V2 = 7.51Ð35°
-j
3. (A) Z = 5 +
= 5 - j5 = 5 5 Ð - 45 °
(0.1)(2)
VC =
Networks
Ö2Ð45o A
~
+
-j4
Va
10 Ð0
=
= 1Ð - 84. 23 A
4 + j 8 1 + j10
1W
Va
10Ð0 V
~
–
~
20Ð0o V
Fig. S1.7.11
4W
I
o
VC
10 W
-j2
j8
(1 + j)( - j 4)(10 + j5) = VC (10 + j5 - j 4) + j 8
Þ
60 - j100 = VC (10 + j)
Þ
VC = 11.6 Ð - 64.7°
12. (D) X = X L + X C = 0
Fig. S1.7.5
i( t) = cos (2 t - 84.23° ) A
6. (D) w = 2 p ´ 10 ´ 10 3 = 2 p ´ 10 4
-j
1
Y = j(1m )(2 p ´ 10 4 ) +
+
4
(160m )(2 p ´ 10 ) 36
= 0.0278 - j0.0366 S
1
Z=
= 1316
. + j17.33 W
Y
Page
68
So reactive power drawn from the source is zero.
13. (B) Z1 Z 4 = Z 3Z 2
300 Z 4 = ( 300 - j 600)(200 + j100)
Þ Z 4 = 400 - j 300
14. (A) R - C causes a positive phase shift in voltage
Z =|Z |Ðq , -90 ° < q < 0 ,
I=
V
V
=
Ð-q
Z |Z |
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Sinusoidal Steady State Analysis
120 Ð15 °
- 6 Ð30 °
40 + j20
15. (C) Vo =
= 124 Ð - 154 °
1
1
1
+
+
40 + j20 - j 30 50
20. (B) Let Vo be the voltage across current source
Vo - 4 Vx Vo - Vx
+
=3
20
j10
Vo(20 + j10) - (20 + j 40) Vx = j 600
Vo(20)
V
Vx =
Þ
Vo = x (2 + j)
20 + j10
2
16. (C) 10 sin ( t + 30 ° ) = 10 cos ( t - 60 ° )
10 Ð - 60 ° 20 Ð - 45 °
+
j
3
Vo =
1
1
1
+
+
j -j 3
æ (2 + j)(20 + j10)
ö
Vx = ç
- 20(1 + j2) ÷ = j 600
2
è
ø
j 600
Vx =
= 29. 22 Ð - 166 °
-5 - j20
= 30 Ð - 150 °+20 Ð - 45 °
. Ð - 112 ° V
Vo = 315
10Ð-60 V
æ j ö V - V2
21. (A) I1 = V3ç ÷ + 3
= j0.1V2 + j0.4 V3
j10
è2 ø
3W
j1
o
+
~
-j1 W
~
Vo
–
= (0.1Ð90 ° )(0.757 Ð66.7 ° ) + (0.4 Ð90 ° )(0.606 Ð - 69.8 ° )
Þ
20Ð-45 V
o
I1 = 0.196 Ð35.6 °
22. (A)
Fig. S.1.7.16
1W
5Ð0o V
~
1W
10Ð-30o V
12Ð0 V
o
-j0.25 W
j4
+
~
2W
I1
Vo
I2
–
2W
Þ
( 8 + j15) I1 - ( 4 - j) I 2 = 20 Ð0
j
j
-10 Ð - 30 ° = I 2 (1 + j 4 + 1 - ) - I1 (1 - )
4
4
( 4 - j) I1 - ( 8 + j15) I 2 = 40 Ð - 30 °
~
...(i)
2Ð0o A
Fig. P1.7.23
...(ii)
12 Ð0 ° = I1 ( - j 3 + 2 + 2) + 8 Ð90 °-4 Ð0 °
= (20 Ð0)( 8 + j15) - ( 40 Ð - 30 ° )( 4 - j)
Þ
I1 ( -176 + j248) = 41.43 + j 414.64
Vo = 2( 352
. + j0.64 + j 4) = 11.65 Ð52.82 ° V
Þ
2
I1 = 103
. - j0.9 = 1.37 Ð - 4107
.
I1 = 352
. + j0.64
24. (D) I 2 = 3Ð0 ° A , I 4 - I 3 = 6 Ð0 ° A
( 8 + j15)(103
. - j0.9) - 20 Ð0 °
18. (B) I 2 =
4-j
= -0.076 + j2.04
Þ
o
2W
I1 [( 8 + j15) - ( 4 - j) ]
2
4Ð90 V
~
Fig. S.1.7.17
Þ
Vo = 5.65 Ð - 75 °
-j3
~
I2
Þ
23. (D) I 2 = 4 Ð90 ° , I 3 = 2 Ð0 °
j4
1W
I1
Vo Vo - 3Vo
+
= 4 Ð - 30 °
2
j4
Vo(0.5 + j0.5) = 3.46 - j2
jö
jö
æ
æ
17. (C) 5 Ð0 ° = I1 ç j 4 + 1 + 1 - ÷ - I 2 ç 1 - ÷
4ø
4ø
è
è
j4
Chap 1.7
Io
I 2 = 2.04 Ð92.13°
15Ð90o V
~
2W
I2
-j4
j2
~
3Ð0o A
19. (B) 10 Ð30 ° = 4 I1 - 0.5 I x + ( - j2) I x
( - j2) I x = ( I1 - I x ) j 3, I1 =
æ4
ö
10 Ð30 ° = ç - 0.5 - j2 ÷I x
è3
ø
Ix
3
1W
Þ
Ix =
10 Ð30 °
2.17 Ð - 67.38 °
I3
~
6Ð0o A
I4
1W
Fig. S.1.7.24
I 3(1) + ( I 3 - I o)( - j 4) + ( I 4 + I 2 )( j2) + I 4 = 0
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UNIT 1
I 3 + ( I 3 - I o)( - j 4) + ( I 3 + 6 Ð0 °+3Ð0 ° )( j2) + I 3 + 60 ° = 0
I 3(2 - j2) + I o( j 4) = -18 j - 6
-I ( j2) - 3 - 9 j
I3 = o
(1 - j)
S2 = 20 + j
15 Ð90 ° = ( I o + 3Ð0 ° )(2) + ( I o - I 3)( - j 4)
Þ
VTH =
|V |2
(120) 2
=
= 72 + j144 VA
Z*
40 - j 80
20
sin (cos -1 (0.8)) = 20 + j15
0.8
S = VoI * = 6 Vo
( j10)( 8 - j5)
=
= 9 + j 4.4
8 + j10 - j5
Þ
S=
|V |2
(210) 2
=
*
Z
30 - j157
...(i)
Apparent power =|S |=
...(ii)
35. (D) Z = 4 +
300 I 2 = 3V1 , V1 = ( - j 300)( I1 - I 2 )
I 2 = - j 3( I1 - I 2 )
Þ
3I1 = ( 3 + j) I 2
Solving (i) and (ii) I 2 = 12.36 Ð - 16 ° mA
36. (A)
S=
Þ
1W
~
3V
4 1
Q
2000
=
= 4588.6 VA
sin q sin 25.84 °
-j1.5
Fig. S.1.7.36
I1 = 1Ð36.9 °
(1Ð36.9 ° )(10 Ð0 ° )
S=
= 5 Ð - 36.9 °
2
S = 4129.8 - j2000
29. (A) Q = S sin q
V1 = 4 Ð36.9 °,
+ V1 –
I1
q = 25.84 °
P = S cos q = 4129.8,
Þ
4W
10Ð0o V
Þ
= 275.6 VA
( - j2)( j5 - j2)
- j2 + j5 - j2
10 - V1
V1 3
+ V1 =
4
4
1 - j15
.
27. (C) S = P - jQ = 269 - j150 VA
Q = S sin q
30 2 + 152 2
pf = cos 56.31° = 0.555 leading
-2 V1 - V1 = 0 Þ V1 = 0
9 Ð0 °
Þ I sc =
= 15 Ð0 ° mA
600
V
311
. Ð - 16 °
Z TH = oc =
= 247 Ð - 16 ° W
I sc 15 Ð0 °´10 -3
Þ
(210) 2
= 4 - j 6 = 7.21Ð - 56.31°,
. Ð - 16 °
Voc = 300 I 2 = 371
28. (D) pf = cos q = 0.9
Vo = 7.1Ð32.29 °
34. (B) Z = 30 + j(0.5)(2 p)(50) = 30 + j157,
( 32 Ð0 ° )( j10)
= 339
. Ð58 ° V
8 + j10 - j5
26 (D) ( 600 - j 300) I1 + j 300 I 2 = 9
Þ
16
sin (cos -1 (0.9)) = 16 + j7.75
0.9
S = S1 + S2 = 36 + j2.75 = 42.59 Ð32.29 °
j15 = 2 I o + 6 + ( j 4)( 3 - j 6)
25. (A) Z TH
32. (A) S =
33. (A) S1 = 16 + j
I3 = Io + 3 - j6
Þ
Networks
Þ
sin q =
Q 45
or
=
S 60
q = 48.59 °,
pf = cos 36.9 ° = 0.8 leading
37. (A) (2 Ð - 90 ° ) 4.8 = - I x ( 4.8 + j192
. ) + 0.6 I x ( 8)
P = S cos q = 39.69,
Ix
j1.92
Va
S = 39.69 + j 45 VA
4.8 W
|V |2 (220 2 )
30. (B) S = rms =
= 1210
|Z |
40
1.6Ix
(2Ð90 )4.8 V
o
P 1000
cos q = =
= 0.8264 or q = 34.26 °,
S 1210
Fig. S.1.7.37
Q = S sin q = 68125,
.
S = 1000 + j 68125
. VA
*
31. (C) S = Vrms I rms
= (21Ð20 ° )( 8.5 Ð50 ° )
I x = 5 Ð0 °,
Va = 0.6 ´ 5 ´ 8 = 24 Ð0 °,
1
Pave = ´ 24 ´ 1.6 ´ 5 = 96
2
= 61 + j167.7 VA
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70
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Sinusoidal Steady State Analysis
38. (A) Z TH =
VTH =
( - j10)(10 + j15)
= 8 - j14 W
10 + j15 - j10
120( - j10)
= 107.3Ð - 116.6 ° V
10 + j5
40. (B) Va =
3
I aA
3
Ð( q + 30 ° ) = 16.22 Ð1.34 ° A rms
= 207.8 Ð40 ° Vrms
46. (B) |S |= 3VL I L
ZY =
( - j 40)( 80 + j100)
= 12.8 - j 49.6 W
80 + j 60
400
45. (D) I AB =
V AB = I AB × Z D = (16. 22 Ð1.340 ° )(10 + j 8)
107.3Ð - 116.6 °
IL =
= 6.7 Ð - 116.6 °
16
1
PLmax = ( 6.7) 2 ´ 8 = 180 W
2
39. (B) Z TH =
Chap 1.7
208
10 3
Þ
IL =
3600
208 3
= 10 A rms
Ð25 ° = 12 Ð25 ° = 10.88 + j5.07 W
********
Ð - 30 ° = 231Ð - 30 ° V
Vb = 231Ð - 150 ° V, Vc = 231Ð - 270 ° V
41. (C) For the acb sequence
Vab = Va - Vb = Vp Ð0 ° - Vp Ð120 °
æ
1
3
400 = Vp çç 1 + - j
2
2
è
400
Þ Vp =
Ð30 °
3
ö
÷ = Vp 3Ð - 30 °
÷
ø
Va = Vp Ð0 ° = 231Ð30 ° V,
Vb = Vp Ð120 ° = 231Ð150 ° V
Vc = Vp Ð240 ° = 231Ð - 90 ° V
42. (B) V A = 277 Ð( 45 °-120 ° ) = 277 Ð - 75 ° V
VB = 277 Ð( 45 ° + 120 ° ) = 277 Ð165 ° V
V AB = V A - VB = 480 Ð - 45 ° V
43. (C) Z A = 6 ||12 = 4,
IP =
480
= 120 A rms
4
I L = 3I P = 208 A rms
44 (B) I =
I aA (10 + j 4)
= 10 Ð20 °
(10 + j 4) + ( 4 + j 4)
IaA
a
Iac
c
Iab
Ibc
IbB
b
icC
Fig. S.1.7.44
I aA = 15 Ð - 27.9 ° A rms
|I |
I ab = - aA Ð( q + 30 ° ) = 8.67 Ð - 122.1° A rms
3
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CHAPTER
1.8
CIRCUIT ANALYSIS IN THE S-DOMAIN
s2 + 1
s2 + 2 s + 1
2 s2 + 1
(C) 2
s + 2s + 2
1. Z ( s) = ?
2( s 2 + 1)
( s + 1) 2
s2 + 1
(D)
3s + 2
(A)
1F
2H
Z(s)
1W
1W
(B)
4. Z ( s) = ?
1W
Z(s)
1H
1W
2
s 2 + 3s + 1
s( s + 1)
2 s 2 + 3s + 1
(D)
2 s( s + 1)
(B)
W
Fig. P1.8.1
s 2 + 15
. s+1
(A)
s( s + 1)
2 s 2 + 3s + 2
(C)
s( s + 1)
0.5 F
Fig. P1.8.4
2. Z ( s) = ?
3s 2 + 8 s + 7
s(5 s + 6)
3s 2 + 7 s + 6
(C)
s(5 s + 6)
s(5 s + 6)
3s 2 + 8 s + 7
s(5 s + 6)
(D)
3s 2 + 7 s + 6
(B)
(A)
1F
Z(s)
1W
1H
1W
5. The s-domain equivalent of the circuit of Fig.P1.8.5. is
t=0
Fig. P1.8.2
s + s+1
s( s + 1)
s( s + 1)
(C)
2 s2 + s + 1
2s + s + 1
s( s + 1)
s( s + 1)
(D) 2
s + s+1
2
3W
2
(A)
(B)
6V
3F
+
vC
–
Fig. P1.8.5
3. Z ( s) = ?
3W
3W
+
1
3s
1H
Z(s)
2W
1F
Fig. P1.8.3
6 V
s
(A)
(C) Both A and B
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72
1
3s
VC(s)
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+
VC(s)
-
-
(B)
(D) None of these
2A
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GATE EC BY RK Kanodia
Circuit Analysis in the s-Domain
Chap 1.8
6. The s–domain equivalent of the circuit shown in Fig.
9. For the network shown in fig. P1.8.9 voltage ratio
P1.8.6 is
transfer function G12 is
t=0
+
12 W
2A
2H
1H
1H
1F
1F
vL
v1
–
1F
1F
1F
+
v2
-
Fig. P1.8.6
Fig. P1.8.9
(A)
(B)
+
+
(A)
( s 2 + 2)
5 s + 5 s2 + 1
(B)
s2 + 1
5 s + 5 s2 + 1
(C)
( s 2 + 2) 2
5 s4 + 5 s2 + 1
(D)
( s 2 + 1) 2
5 s4 + 5 s2 + 1
2s
12 W
12 W
VL
2A
s
VL
2s
4
4
4V
-
-
(C) Both A and B
10. For the network shown in fig. P1.8.10, the
admittance transfer function is
(D) None of these
Y12 =
Statement for Q.7-8:
K ( s + 1)
( s + 2)( s + 4)
3
W
2
The circuit is as shown in fig. P1.8.7–8. Solve the
i1
problem and choose correct option.
i2
1W
+
+
is
1W
1H
io
v1
+
vs
1F
1F
1W
2F
2F
3
-
Fig. P1.8.10
The value of K is
(A) -3
V ( s)
7. H1 ( s) = o
=?
Vs ( s)
(C)
(A) s( s3 + 2 s2 + 3s + 1) -1
(B) 3
1
3
(D) -
1
3
11. In the circuit of fig. P1.8.11 the switch is in position
(B) ( s 3 + 3s 2 + 2 s + 1) -1
1 for a long time and thrown to position 2 at t = 0. The
(C) ( s 3 + 2 s 2 + 3s + 2) -1
equation for the loop currents I1 ( s) and I 2 ( s) are
(D) s( s 3 + 3s 2 + 2 s2 + 2) -1
1F
1
I o( s)
=?
Vs ( s)
2
t=0
12 V
(A)
-s
( s + 3s + 2 s + 1)
(B) -( s 3 + 3s 2 + 2 s + 1) -1
(C)
-s
3
2
( s + 2 s + 3s + 1)
(D) ( s 3 + 2 s 2 + 3s + 2) -1
2
v2
-
Fig. P1.8.7–8
3
6
vo
–
8. H 2 ( s) =
1W
i1
3H
2W
i2
1F
Fig. P1.8.11
1
é
ê2 + 3s + s
(A) ê
ê -3s
ë
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ù
- 3s ú
1 ú
2+ ú
s û
é12 ù
é I1 ( s) ù ê s ú
êI ( s) ú = ê ú
ë 2 û ê0 ú
ë û
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GATE EC BY RK Kanodia
UNIT 1
1
é
ê2 + 3s + s
(B) ê
ê -3s
ë
ù
- 3s ú
1 ú
2+ ú
s û
é 12 ù
é I1 ( s) ù ê- s ú
ú
êI ( s) ú = ê
ë 2 û ê 0 ú
ë
û
1
é
-3s
ê2 + 3s + s
(C) ê
2 + 3s +
ê -3s
ë
é 12 ù
ù
ú é I1 ( s) ù ê- s ú
=
ú
1 ú êëI 2 ( s) úû ê
ê 0 ú
ú
ë
û
sû
1
é
-3s
ê2 + 3s + s
(D) ê
2 + 3s +
ê -3s
ë
ù
é12 ù
ú é I1 ( s) ù ê s ú
=
1 ú êëI 2 ( s) úû ê ú
ú
ê0 ú
sû
ë û
Networks
3F
S1
-
4F
Va
+
5V
+
10 V
2F
1V
S2
-
+
6V
5V
-
Fig. P1.8.14
(A)
9
t
(B) 9e - t V
V
(C) 9 V
(D) 0 V
15. A unit step current of 1 A is applied to a network
whose driving point impedance is
12. In the circuit of fig. P1.8.12 at terminal ab
1
The steady state and initial values of the voltage
s
a
4 A
+
Vo(s)
-
2W
(s+1)
developed across the source would be respectively
2Vo(s)
b
(A)
3
4
(C)
3
4
V, I V
1
4
V,
3
4
V
(D) 1 V,
3
4
V
(B)
V, 0 V
16. In the circuit of Fig. P1.8.16 i(0) = 1 A, vC (0) = 8
Fig. P1.8.12
4
V and v1 = 2 e -2 ´10 t u( t). The i( t) is
(A) VTH ( s) =
-8( s + 2)
-(2 s + 1)
, Z TH ( s) =
3s( s + 1)
3s
(B) VTH ( s) =
8( s + 2)
(2 s + 1)
, Z TH ( s) =
3s( s + 1)
3s
(C) VTH ( s) =
4( s + 3)
(2 s + 1)
, Z TH ( s) =
3s( s + 1)
6s
50 W
i
1m H
2.5 mF
v1
+
vC
–
Fig. P1.8.16
4
-4( s + 3)
-(2 s + 1)
(D) VTH ( s) =
, Z TH ( s) =
3s( s + 1)
6s
(A)
1
15
[10 e-10
(B)
1
15
[ -10 e -10
13. In the circuit of fig. P1.8.13 just before the closing of
(C) 13 [10 e-10
switch at t = 0, the initial conditions are known to be
(D) 13 [ -10 e -10
-
V ( s) ( s + 3)
=
I ( s) ( s + 2) 2
Z ( s) =
Thevenin equivalent is
-
vC1 (0 ) = 1 V, vC2 (0 ) = 0. The voltage vC1 ( t) is
4
- 3e-2 ´10
t
4
t
4
t
4
- 22 e-4 ´10 t ]u( t) A
t
+ 3e -2 ´10
+ 3e-2 ´10
t
4
4
4
4
+ 22 e -4 ´10 t ]u( t) A
4
+ 22 e-4 ´10 t ]u( t) A
t
+ 3e -2 ´10
t
4
t
4
- 22 e -4 ´10 t ]u( t) A
17. In the circuit shown in Fig. P1.8.18 v(0 - ) = 8 V and
+
vC1
-
iin ( t) = 4d( t). The vC ( t) for t ³ 0 is
t=0
1F
1F
+
vC2
-
iin
50 W
20 mF
Fig. P1.8.13
(A) u( t) V
(C) 0.5 e
-t
(B) 0.5 u( t) V
V
Fig. P1.8.17
(D) e - t V
-
14. The initial condition at t = 0 of a switched capacitor
(A) 164e- t V
(B) 208e- t V
(C) 208(1 - e -3t ) V
(D) 164 e -3t V
circuit are shown in Fig. P1.8.14. Switch S1 and S2 are
closed at t = 0. The voltage va ( t) for t > 0 is
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74
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vC
–
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Circuit Analysis in the s-Domain
18. The driving point impedance Z ( s) of a network has
Chap 1.8
The steady state voltage across capacitor is
the pole zero location as shown in Fig. P1.8.18. If
(A) 6 V
(B) 0 V
Z(0) = 3, the Z ( s) is
(C) ¥
(D) 2 V
jw
23.
1
-3
transformed
VC ( s) =
-1
4( s + 3)
s2 + s + 1
(B)
2( s + 3)
(C) 2
s + 2s + 2
2( s + 3)
2
s + 2s + 2
4( s + 3)
(D) 2
s + s+2
20 s + 6
(10 s + 3)( s + 4)
(B) - 0.12 mA
(C) 0.48 mA
(D) - 0.48 mA
24. The current through an 4 H inductor is given by
I L ( s) =
The circuit is as shown in the fig. P1.8.19–21. All
initial conditions are zero.
io
1H
60 mF
the
(A) 0.12 mA
Statement for Q.19-21:
iin
across
The initial current through capacitor is
Fig. P1.8.18
(A)
voltage
capacitor is given by
s
-1
The
1W
1F
10
s( s + 2)
The initial voltage across inductor is
(A) 40 V
(B) 20 V
(C) 10 V
(D) 5 V
25. The amplifier network shown in fig. P1.8.25
1W
is stable if
4W
1F
1H
Fig. P1.8.19–21
19.
I o( s)
=?
I in ( s)
(A)
( s + 1)
2s
+
+
v1 Amplifier v
2
gain=K
-
2W
(B) 2 s( s + 1) -1
(C) ( s + 1) s -1
Fig.P1.8.25
(D) s( s + 1) -1
(A) K £ 3
20. If iin ( t) = 4d( t) then io( t) will be
(C) K £
(A) 4d( t) - e- t u( t) A
(B) 4 d( t) - 4 e- t u( t) A
(B) K ³ 3
1
3
(D) K ³
1
3
26. The network shown in fig. P1.8.26 is stable if
(C) 4 e - t u( t) - 4 d( t) A
2F
1W
(D) e - t u( t) - d( t) A
+
21. If iin ( t) = tu( t) then io( t) will be
(A) e - t u( t) A
(B) (1 - e - t ) u( t) A
(C) u( t) A
(D) (2 - e - t ) u( t) A
Kv2
1W
1F
v2
-
Fig.P1.8.26
22. The voltage across 200 mF capacitor is given by
VC ( s) =
2s + 6
s( s + 3)
5
2
2
(C) K ³
5
(A) K ³
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5
2
2
(D) K £
5
(B) K £
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GATE EC BY RK Kanodia
UNIT 1
27. A circuit has a transfer function with a pole s = - 4
and a zero which may be adjusted in position as s = - a
The response of this system to a step input has a term
of form Ke -4 t . The K will be
aö
æ
(A) H ç 1 - ÷
4ø
è
(H= scale factor)
aö
æ
(B) H ç 1 + ÷
4ø
è
aö
æ
(C) H ç 4 - ÷
4ø
è
aö
æ
(D) H ç 4 + ÷
4ø
è
io( t) = 2 sin 2 t u( t) A. The circuit had no internal stored
energy at t = 0. The admittance transfer function is
2
s
(A)
(B)
s
2
(D)
31. The current ratio transfer function
Io
is
Is
(A)
s( s + 4)
s + 3s + 4
(B)
s( s + 4)
( s + 1)( s + 3)
(C)
s 2 + 3s + 4
s( s + 4)
(D)
( s + 1)( s + 3)
s( s + 4)
2
32. The response is
28. A circuit has input vin ( t) = cos 2 t u( t) V and output
(C) s
Networks
1
s
(A) Over damped
(B) Under damped
(C) Critically damped
(D) can’t be determined
33. If input is is 2u( t) A, the output current io is
(A) (2 e - t - 3te -3t ) u( t) A
(B) ( 3te - t - e -3t ) u( t) A
(C) ( 3e - t - e -3t ) u( t) A
(D) ( e -3t - 3e - t ) u( t) A
34. In the network of Fig. P1.8.34, all initial condition
are zero. The damping exhibited by the network is
29. A two terminal network consists of a coil having an
inductance L and resistance R shunted by a capacitor
1F
4
C. The poles of the driving point impedance function Z
of this network are at - 12 ± j
3
2
and zero at -1. If
+
Z(0) = 1 the value of R, L, C are
1
(A) 3 W, 3 H, F
3
2H
vs
2W
vo
-
1
(B) 2 W, 2 H, F
2
Fig. P1.8.34
1
(C) 1 W, 2 H, F
2
(D) 1 W, 1 H, 1 F
(A) Over damped
(B) Under damped
30. The current response of a network to a unit step
input is
(C) Critically damped
(D) value of voltage is requires
10( s + 2)
Io = 2
s ( s + 11s + 30)
35. The voltage response of a network to a unit step
input is
The response is
(A) Under damped
(B) Over damped
(C) Critically damped
(D) None of the above
Vo( s) =
10
s( s 2 + 8 s + 16)
The response is
Statement for Q.31-33:
(A) under damped
(B) over damped
(C) critically damped
(D) can’t be determined
The circuit is shown in fig. P1.8.31-33.
36. The response of an initially relaxed circuit to a
1H
io
(
signal vs is e -2 t u( t). If the signal is changed to vs +
, the response would be
1F
is
3
4W
(A) 5 e -2 t u( t)
(B) -3e -2 t u( t)
(C) 4 e -2 t u( t)
(D) -4 e -2 t u( t)
Fig. P1.8.31-33
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dt
)
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Circuit Analysis in the s-Domain
37. Consider the following statements in the circuit
shown in fig. P1.8.37
i
4W
2H
1W
10 V
1F
2
+
vC
–
42. The network function
(B) RL admittance
(C) LC impedance function
(D) LC admittance
43. A valid immittance function is
( s + 4)( s + 8)
s( s + 1)
(A)
(B)
( s + 2)( s - 5)
( s + 2)( s + 5)
s( s + 2)( s + 3)
( s + 1)( s + 4)
(C)
1. It is a first order circuit with steady state value of
10
5
, i= A
vC =
3
3
2. It is a second order circuit with steady state of
vC = 2 V , i = 2 A
V ( s)
has one pole.
I ( s)
4. The network function
V ( s)
has two poles.
I ( s)
(D)
44. The network function
(B) 1 and 4
(C) 2 and 3
(D) 2 and 4
38. The network function
(B) RC admittance
(C) LC admittance
(D) Above all
Z ( s) =
3W
s 2 + 10 s + 24
represent a
s 2 + 8 s + 15
1F
(D) None of the above
(C) LC impedance
(D) None of these
3W
8F
1F
(A)
(B)
1F
1W
3
1H
1F
8
s( 3s + 8)
40. The network function
represents an
( s + 1)( s + 3)
1W
3W
3
(C)
(A) RL admittance
(B) RC impedance
(C) RC admittance
(D) None of the above
41. The network function
1W
3
8
s( s + 4)
represents
( s + 1)( s + 2)( s + 3)
(B) RL impedance
1W
3
1F
(C) LC impedance
(A) RC impedance
3( s + 2)( s + 4)
s( s + 3)
The network for this function is
(B) RL impedance
an
s 2 + 8 s + 15
is a
s2 + 6 s + 8
45. A impedance function is given as
(A) RC admittance
39. The network function
s( s + 2)( s + 6)
( s + 1)( s + 4)
(A) RLadmittance
The true statements are
(A) 1 and 3
s2 + 7 s + 6
is a
s+2
(A) RL impedance function
Fig. P1.8.37
3. The network function
Chap 1.8
( s + 1)( s + 4)
is a
s( s + 2)( s + 5)
1H
8
3W
(D)
************
(A) RL impedance function
(B) RC impedance function
(C) LC impedance function
(D) Above all
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UNIT 1
18. (B) Z ( s) =
K ( s + 3)
K ( s + 3)
= 2
( s - ( -1 + j))( s - (1 - j)) s + 2 s + 2
3K
Z (0) =
=3
2
Þ
I ( s)
19. (D) o
=
I in ( s)
26. (B) Let v1 be the node voltage of middle node
V1 ( s) =
K =2
s
s+1
s
1
+
s+1 s+1
=
4s
4
=4s+1
s+1
Þ
s
s+1
Þ (2 s + 1) V2 ( s) = 2 sV1 ( s)
Þ ( 3s + 1)(2 s + 1) = 2 s(2 s + K )
2 s 2 + (5 - 2 K ) s + 1 = 0,
5
5 - 2 K > 0, K <
2
io( t) = 4 d( t) - 4 e - t u( t)
H ( s + a)
s+4
27. (A) H ( s) =
1
,
s2
1
1
1
I o( s) =
= s( s + 1) s s + 1
21. (B) I in ( s) =
aö
æ
Hç 1 - ÷
H ( s + a) Ha
4ø
R( s) =
=
+ è
s( s + 4)
4s
s+4
io( t) = u( t) - e - t u( t) = (1 - e - t ) u( t)
22. (D) vC ( ¥) = lim sVC ( s) = lim
s ®0
KV2 ( s) + 2 sV2 ( s)
1 + 2s + s
Þ ( 3s + 1) V1 ( s) = (2 s + K ) V2 ( s)
2 sV1 ( s)
Þ V2 ( s) =
2s + 1
20. (B) I in ( s) = 4
I o( s) =
Networks
s ®0
r ( t) =
2s + 6
=2 V
s+3
Ha
aö
æ
u( t) + H ç 1 - ÷e - 4 t
4
4ø
è
28. (A) Vin ( s) =
23. (D) vC (0 + ) = lim sVC ( s)=
s ®¥
CdvC
iC =
dt
Þ
s(20 s + 6)
=2 V
(10 s + 3)( s + 4)
I C ( s) = C[ sVC ( s) - vC (0 )]
æ s(20 s + 6)
ö -480 ´ 10 -6 (10 s + 3)
= 60 ´ 10 -6 çç
- 2 ÷÷ =
10 s 2 + 43s + 12
è (10 s + 3)( s + 4)
ø
s ®¥
Rö
1æ
1
çs + ÷
C
Lø
è
sC =
29. (D) Z ( s) =
R
1
1
s2 +
sL + R +
+
L LC
sC
K ( s + 1)
K ( s + 1)
Z ( s) =
= 2
(
s
+ s + 1)
æ
1
3 öæ
1
3ö
çs + + j
֍ s + - j
÷
ç
÷
ç
÷
2
2 øè
2
2 ø
è
( sL + R)
+
iC (0 + ) = lim sI C ( s) = - 480 ´ 10 -6 = - 0.48 mA
s
2
I o( s) 2
, I o( s) = 2
,
=
s2 + 1
s + 1 Vin ( s) s
d iL
Þ VL ( s) = L [ sI L ( s) - iL (0 + )]
dt
10
iL (0 + ) = lim sI L ( s) =
=0
s ®¥
s+2
24. (A) vL = L
VL ( s) =
s ®¥
s 40
= 40
s+2
25. (A) V2 ( s) = KV1 ( s)
V1 ( s) V1 ( s) - KV1 ( s)
+
=0
1
2
4+s+
s
1
4 + s + + 2 - 2K = 0
s
Þ
s2 + ( 6 - 2 K ) s + 1 = 0
(6 - 2K ) > 0
Page
80
Þ
1
Cs
R
40 s
40
=
s( s + 2) s + 2
vL (0 + ) = lim sVL ( s) =
Þ
sL
Z(s)
Fig. S1.8.29
Since Z (0) = 1, thus K = 1
1
R
1
= 1,
= 1,
=1
C
L
LC
Þ
C = 1, L = 1, R = 1
30. (B) The characteristic equation is
s 2 ( s 2 + 11s + 30) = 0
Þ
s 2 ( s + 6) ( s + 5) =0
s = -6, - 5, Being real and unequal, it is overdamped.
K <3
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GATE EC BY RK Kanodia
Circuit Analysis in the s-Domain
31. (B)
Io
s+4
s( s + 4)
=
=
3
Is s + 4 +
( s + 1)( s + 3)
s
39. (D) Poles and zero does not interlace on negative
real axis so it is not a immittance function.
32. (A) The characteristic equation is ( s + 1) ( s + 3) = 0.
Being real and unequal root, it is overdamped response.
33. (C) is = 2 u( t)
I o ( s) =
40. (C) The singularity nearest to origin is a zero. So it
may be RL impedance or RC
admittance function.
Because of (D) option it is required to check that it is a
valid RC
2
I s ( s) =
s
Þ
Chap 1.8
admittance function. The poles and zeros
interlace along the negative real axis. The residues of
YRC ( s)
are real and positive.
s
2( s + 4)
3
1
=
( s + 1)( s + 3)
s+1 s+ 3
io = ( 3e- t - e-3t ) u( t) A
41. (B) The singularity nearest to origin is a pole. So it
may be RC impedance or RL admittance function.
V ( s)
2
1
34. (B) o
=
= 2
4
Vs ( s)
+ 2s + 2 s + s + 2
s
42. (A)
The roots are imaginary so network is underdamped.
s 2 + 7 s + 6 ( s + 1)( s + 6)
=
s+2
( s + 2)
The singularity nearest to origin is at zero. So it may be
35. (C) The characteristic equation is
RC admittance or RL impedance function.
s( s 2 + 8 s + 16) = 0, ( s + 4) 2 = 0, s = -4, - 4
43. (D)
Being real and repeated root, it is critically damped.
(A) pole lie on positive real axis
36. (B) vo = e -2 t u( t)
v¢s = vs +
2 dvs
dt
Þ
Þ
Vo( s) = H ( s) Vs ( s) =
(B) poles and zero does not interlace on axis.
1
s+2
(C) poles and zero does not interlace on axis.
(D) is a valid immittance function.
Vs¢( s) = (1 + 2 s) Vs ( s)
44. (A)
Vo¢( s) = H ( s) Vs¢( s) = (1 + 2 s) Vs ( s) H ( s)
Vo¢( s) =
1 + 2s
3
=2 s+2
s+2
Þ
v¢o = 2 d( s) - 3e -2 t u( t)
The singularity nearest to origin is a pole. So it may be
a RL admittance or RC impedance function.
37. (C) It is a second order circuit. In steady state
i=
10
=2 A , v =2 ´ 1 =2 V
4+1
I ( s) =
10
2s + 4 +
V ( s) =
1
1
1+ s
2
=
10
1
1+ s
2
1
(2 s + 4) +
1
1+ s
2
s 2 + 8 s + 15 ( s + 3) ( s + 5)
=
s 2 + 6 s + 8 ( s + 2) ( s + 4)
45. (A) The singularity nearest to origin is a pole. So
this is RC impedance function.
8
1
8
13
Z ( s) = 3 + +
=3+ +
s s+3
s 1+ s
3
5( s + 2)
( s + 2) 2 + 1
**************
=
10
( s + 2) 2 + 1
V ( s)
2
, It has one pole at s = -2
=
I ( s) s + 2
38. (D)
s 2 + 10 s + 24 ( s + 4)( s + 6)
=
s 2 + 8 s + 15 ( s + 3)( s + 5)
The singularity near to origin is pole. So it may be RC
impedance or RL admittance function.
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GATE EC BY RK Kanodia
CHAPTER
1.9
MAGNETICALLY COUPLED CIRCUITS
4. If i1 = e -2 t V and i2 = 0, the voltage v2 is
Statement for Q.1-2:
In the circuit of fig. P1.9.1-2 i1 = 4 sin 2 t A, and
i2 = 0.
i2
i1
1H
+
2H
v1
+
(A) -6 e -2 t V
(B) 6 e -2 t V
(C) 15
. e -2 t V
(D) -15
. e -2 t V
Statement for Q.5-6:
Consider the circuit shown in fig. P19.5-6
v2
1H
i2
i1
-
-
2H
+
+
Fig. P1.9.1-2
v1
2H
3H
v2
1. v1 = ?
(A) -16 cos 2 t V
(B) 16 cos 2 t V
(C) 4 cos 2 t V
(D) -4 cos 2 t V
Fig. P1.9.5-6
5. If current i1 = 3 cos 4 t A and i2 = 0, then voltage v1 and
2. v2 = ?
(A) 2 cos 2 t V
(B) -2 cos 2 t V
v2 are
(C) 8 cos 2 t V
(D) -8 cos 2 t V
(A) v1 = -24 sin 4 t V,
v2 = -24 sin 4 t V
(B) v1 = 24 sin 4 t V,
v2 = -36 sin 4 t V
(C) v1 = 15
. sin 4 t V,
v2 = sin 4 t V
(D) v1 = -15
. sin 4 t V,
v2 = -sin 4 t V
Statement for Q.3-4:
Consider the circuit shown in Fig. P1.9.3-4
i2
i1
3H
+
v1
3H
+
4H
v2
-
-
Fig. P1.9.5-6
3. If i1 = 0 and i2 = 2 sin 4 t A, the voltage v1 is
Page
82
-
-
(A) 24 cos 4 t V
(B) -24 cos 4 t V
(C) 15
. cos 4 t V
(D) -15
. cos 4 t V
6. If current i1 = 0 and i2 = 4 sin 3t A, then voltage v1 and
v2 are
(A) v1 = 24 cos 3t V,
v2 = 36 cos 3t V
(B) v1 = 24 cos 3t V,
v2 = -36 cos 3t V
(C) v1 = -24 cos 3t V,
v2 = 36 cos 3t V
(D) v1 = -24 cos 3t V,
v2 = -36 cos 3t V
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Magnetically Coupled Circuits
Statement for Q.7-8:
Chap 1.9
12. Leq = ?
In the circuit shown in fig. P1.9.7-8, i1 = 3 cos 3t A
3.6 H
and i2 = 4 sin 3t A.
Leq
1H
+
v1
1H
i2
i1
2H
1.4 H
+
v2
2H
Fig. P1.9.12
-
-
(A) 4 H
(B) 6 H
(C) 7 H
(D) 0 H
Fig. P1.9.7-8
7. v1 = ?
(A) 6( -2 cos t + 3 sin t) V
(B) 6( 2 cos t + 3 sin t) V
(C) -6(2 cos t + 3 sin t) V
(D) 6(2 cos t - 3 sin t) V
13. Leq = ?
4H
Leq
2H
8. v2 = ?
2H
(A) 3( 8 cos 3t - 3 sin t) V
(B) 6(2 cos t + 3 sin t) V
(C) 3( 8 cos 3t + 3 sin 3t) V
(D) 6(2 cos t - 3 sin t) V
Statement for Q.9-10:
In the circuit shown in fig. P1.9.9-10, i1 = 5 sin 3t A
and i2 = 3 cos 3t A
3H
+
(A) 2 H
(B) 4 H
(C) 6 H
(D) 8 H
14. Leq = ?
i2
i1
Fig. P1.9.13
4H
+
Leq
v1
3H
v2
4H
6H
4H
-
-
Fig. P1.9.14
Fig. P1.9.9-10
9. v1 =?
(A) 9(5 cos 3t + 3 sin 3t) V
(B) 9(5 cos 3t - 3 sin 3t) V
(C) 9( 4 cos 3t + 5 sin 3t) V
(D) 9(5 cos 3t - 3 sin 3t) V
(A) 8 H
(B) 6 H
(C) 4 H
(D) 2 H
15. Leq = ?
2H
10. v2 = ?
(A) 9( -4 sin 3t + 5 cos 3t) V
(B) 9( 4 sin 3t - 5 cos 3t) V
(C) 9( -4 sin 3t - 5 cos 3t) V
(D) 9( 4 sin 3t + 5 cos 3t) V
Leq
11. In the circuit shown in fig. P1.9.11 if current
Fig. P1.9.15
i1 = 5 cos (500 t - 20 ° ) mA and i2 = 20 cos (500 t - 20 ° ) mA,
the total energy stored in system at t = 0 is
i2
i1
k=0.6
+
2H
4H
(A) 0.4 H
(B) 2 H
(C) 1.2 H
(D) 6 H
+
16. The equivalent inductance of a pair of a coupled
v1
2.5 H
0.4 H
-
v2
inductor in various configuration are
-
(a) 7 H after series adding connection
Fig. P1.9.11
(A) 151.14 mJ
(B) 45.24 mJ
(C) 249.44 mJ
(D) 143.46 mJ
(b) 1.8 H after series opposing connection
(c) 0.5 H after parallel connection with dotted
terminal connected together.
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GATE EC BY RK Kanodia
UNIT 1
Networks
20. Leq = ?
The value of L1 , L2 and M are
(A) 3 H, 1.6 H, 1.2 H
(B) 1.6 H, 3 H, 1.4 H
(C) 3.7 H, 0.7 H, 1.3 H
(D) 2 H, 3 H, 3 H
1H
1H
Leq
2H
17. Leq = ?
2H
2H
3H
Leq
2H
4H
Fig. P1.9.20
Fig. P1.9.17
(A) 0.2 H
(B) 1 H
(C) 0.4 H
(D) 2 H
(A) 1 H
(B) 2 H
(C) 3 H
(D) 4 H
21. Leq = ?
4H
18. Leq = ?
Leq
3H
Leq
3H
3H
2H
2H
3H
5H
Fig. P1.9.21
Fig. P1.9.18
41
(A)
H
5
(B)
49
H
5
51
H
5
(D)
39
H
5
(C)
(A) 1 H
(B) 2 H
(C) 3 H
(D) 4 H
Statement for Q.22-24:
19. In the network of fig. P1.9.19 following terminal are
Consider the circuit shown in fig. P1.9.22–24.
connected together
A
(i) none
(ii) A to B
(iii) B to C
(iv) A to C
B
2H
a
A
2H
3H
6t A
20 H
C
5H
4H
3H
B
D
5H
Fig. P1.9.22–24
1H
b
22. The voltage V AG of terminal AD is
C
Fig. P1.9.19
(A) 60 V
(B) -60 V
(C) 180 V
(D) 240 V
The correct match for equivalent induction seen at
terminal a - b is
Page
84
4H
6H
15t A
23. The voltage vBG of terminal BD is
(i)
(ii)
(iii)
(iv)
(A) 45 V
(B) 33 V
(A)
1 H
0.875 H
0.6 H
0.75 H
(C) 69 V
(D) 105 V
(B)
13 H
0.875 H
0.6 H
0.75 H
(C)
13 H
7.375 H
6.6 H
2.4375 H
(A) 30 V
(B) 0 V
(D)
1 H
7.375 H
6.6 H
2.4375 H
(C) -36 V
(D) 36 V
24. The voltage vCG of terminal CD is
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GATE EC BY RK Kanodia
UNIT 1
Networks
33. In the circuit of fig. P1.9.33 the w = 2 rad s. The
37. In the circuit of fig. P1.9.37 the maximum power
resonance occurs when C is
delivered to RL is
C
10 W
2H
Zin
1 : 4
4W
2H
2H
100 Vrms
~
RL
Fig. P1.9.33
(A) 1 F
(C)
1
3
F
Fig. P1.9.37
(B)
1
2
F
(D)
1
6
F
34. In the circuit of fig. P1.9.34, the voltage gain is zero
at w = 333.33 rad s. The value of C is
20 W
vin
0.12 H
~
(B) 200 W
(C) 150 W
(D) 100 W
38. The average power delivered to the 8 W load in the
circuit of fig. P1.9.38 is
40 W
0.09 H
(A) 250 W
300 W
0.27 H
2F
I1
I2
+
+
vout
-
50 Vrms
C
~
+
8W
V1
-
-0.04V2
Fig. P1.9.34
5 : 1
Fig. P1.9.38
(A) 100 mF
(B) 75 mF
(A) 8 W
(B) 1.25 kW
(C) 50 mF
(D) 25 mF
(C) 625 kW
(D) 2.50 kW
35. In the circuit of fig. P1.9.35 at w = 333.33 rad s, the
39. In the circuit of fig. P1.9.39 the ideal source supplies
voltage gain vout vin is zero. The value of C is
1000 W, half of which is delivered to the 100 W load. The
C
value of a and b are
20 W
4W
40 W
k=0.5
vin
~
0.12 H
25 W
1 : a
+
0.27 H
20 W
100 Vrms
vout
1 : b
~
100 W
-
Fig. P1.9.39
Fig. P1.9.35
(A) 3.33 mF
(B) 33.33 mF
(C) 3.33 mF
(D) 33.33 mF
36. The Thevenin equivalent at terminal ab for the
(A) 6, 0.47
(B) 5, 0.89
(C) 0.89, 5
(D) 0.47, 6
40. I 2 = ?
25 W
network shown in fig. P1.9.36 is
60 W
a
I2
2W
3 : 1
1 : 4
50 Vrms
4 : 3
~
3W
20Ix
20 W
Fig. P1.9.40
Ix
b
Fig. P1.9.36
Page
86
V2
-
(A) 6 V, 10 W
(B) 6 V, 4 W
(C) 0 V, 4 W
(D) 0 V, 10 W
(A) 1.65 A rms
(B) 0.18 A rms
(C) 0.66 A rms
(D) 5.90 A rms
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Magnetically Coupled Circuits
41. V2 = ?
Chap 1.9
SOLUTIONS
50 W
40 W
1. (B) v1 = 2
5 : 2
+
~
80 Vrms
10 W
V2
2. (C) v2 = (1)
-
Fig. P1.9.41
(A) -12.31 V
(B) 12.31 V
(C) -9.231 V
(D) 9.231 V
42. The power being dissipated in 400 W resistor is
1W
~
di2
di
di
+ (1) 1 = 1 = 8 cos 2 t V
dt
dt
dt
3. (B) v1 = 3
di1
di
di
- 3 2 = - 3 2 = - 24 cos 4 t V
dt
dt
dt
4. (C) v2 = 4
di2
di
di
- 3 1 = - 3 1 = 6 e -2 t V
dt
dt
dt
di1
di
di
- 2 2 = 2 1 = -24 sin 4 t V
dt
dt
dt
di2
di1
di1
v2 = -3
+2
=2
= -24 sin 4 t V
dt
dt
dt
5. (A) v1 = 2
4W
1 : 2
10 Vrms
di1
di
di
+ 1 2 = 2 1 = 16 cos 2 t V
dt
dt
dt
1 : 5
48 W
400 W
di1
di
di
- 2 2 = - 2 2 = - 24 cos 3t V
dt
dt
dt
di
di
di
v2 = 3 2 + 2 1 = -3 2 = -36 cos 3t V
dt
dt
dt
6. (D) v1 = 2
Fig. P1.9.42
(A) 3 W
(B) 6 W
(C) 9 W
(D) 12 W
7. (A) v1 = 2
43. I x = ?
8W
di1
di
+1 2
dt
dt
= -18 sin t + 12 cos t = 6 (2 cos t - 3 sin t) V
10 W
2 : 1
100Ð0 V
o
8. (A) v2 = 2
~
j6
= 24 cos 3t - 9 sin 3t = 3 ( 8 cos 3t - 3 sin 3t) V
-j4
Ix
9. (A) v1 = 3
Fig. P1.9.43
(A) 1921
.
Ð57.4 ° A
(B) 2.931Ð59.4 ° A
(C) 1.68 Ð43.6 ° A
(D) 179
. Ð43.6 ° A
di1
di
-3 2
dt
dt
= 45 cos 3t + 27 sin 3t = 9 (5 cos 3t + 3 sin 3t) V
10. (D) v2 = -4
44. Z in = ?
j16
di2
di
+1 1
dt
dt
6W
1 : 5
= 36 sin 3t + 45 cos 3t = 9 ( 4 sin 3t + 5 cos 3t) V
6W
24 W
4 : 1
Zin
di2
di
+3 1
dt
dt
11. (A) W =
-j10
1
1
L1 i12 + L2 i22 + Mi1 i2
2
2
At t = 0, i1 = 4 cos ( -20 ° ) = 4.7 mA
i2 = 20 cos ( -20 ° ) = 18.8 mA ,
Fig. P1.9.44
(A) 46.3 + j 6.8 W
(B) 432.1 + j0.96 W
(C) 10.8 + j9.6 W
(D) 615.4 + j0.38 W
********************
M = 0.6 2.5 ´ 0.4 = 0.6
W =
1
1
(2.5)( 4.7) 2 + (0.4)(18.8) 2 + 0.6( 4.7)(18.8)
2
2
= 151.3 mJ
12. (C) Leq = L1 + L2 + 2 M = 7 H
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UNIT 1
13. (A) Leq = L1 + L2 - 2 M = 4 + 2 - 2 ´ 2 = 2 H
Networks
22. (C) v AG = 20
14. (C) Leq =
L1 L2 - M 2
24 - 16
=
=4 H
L1 + L2 - 2 M 6 + 4 - 8
23. (B) vBG = 3
15. (A) Leq =
L1 L2 - M 2
8 -4
=
= 0.4 H
L1 + L2 + 2 M 6 + 4
24. (C) vCG = -6
16. (C) L1 + L2 + 2 M = 7, L1 + L2 - 2 M = 1.8
Þ
d( 6 t)
d(15 t)
+4
= 180 V
dt
dt
d(15 t)
d 6( t)
d ( 6 t)
+4
-6
= 33 V
dt
dt
dt
d( 6 t)
= -36 V
dt
25. (B) Z = Z11 +
L1 + L2 = 4.4, M = 1.3
w2 M 2
Z 22
2
L1 L2 - M 2
= 0.5, L1 L2 - 1.32 = 0.5 ´ 1.8
L1 + L2 - 2 M
æ1ö
(50) 2 ç ÷
æ 1 ö
è5 ø
= 4 + j (50) ç
÷+
è 10 ø 5 + j (50) 1
2
L1 L2 = 2.59, ( L1 - L2 ) 2 = 4.4 2 - 4 ´ 2.59 = 9
L1 - L2 = 3, L1 = 37
. , L2 = 0.7
= 4.77 + j 115
. W
2
17. (D) Leq = L1 -
M
4
= 4 - =2 H
L2
2
26. (B) Vs = j (0.8)10(12
. Ð0) - j (0.2)(10)(2 Ð0)
+[ 3 + j(0.5)(10)] (12
. Ð0 + 2 Ð0)
18. (B) Leq = L1 -
= 9.6 + j21.6 = 26.64 Ð66.04 ° V
2
M
9
=5 - =2 H
L2
3
27. (A) [ j(100 p) (2) + 10 ]I 2 + j(100 p)(0.4) (2 Ð0) = 0
19. (A)
Þ
2H
-1 H
= 4 Ð - 179.1°
3H
Þ
2H
5H
I 2 = -0.4 - j0.0064,
Vo = 10 I 2 = -4 - j0.064
vo = 4 cos (100 pt - 179.1° ) V
28. (B) 30 Ð30 ° = I ( - j 6 + j 8 - j 4 + j12 - j 4 + 10)
Fig. S.1.9.19
Þ
20. (D) VL1 = 1sI + 1sI = 2 sI
30 Ð30 °
= 2.57 - j0.043
(10 + j 6)
Vo = I ( j12 - j 4 + 10)
VL 2 = 2 sI + 1sI - 2 sI = sI ,
= (2.57 - j0.043)(10 + 8 j)
VL 3 = 3sI - 2 sI = sI
VL = VL1 + VL 2 + VL 3 = 4 sI
I=
Þ
= 26.067 + j20.14 = 32.9 Ð37.7 ° V
Leq = 4 H
21. (B) Let I1 be the current through 4 H inductor and
29. (A)
-j
2W
I 2 and I 3 be the current through 3 H, and 2 H inductor
- Vx +
respectively
I1 = I 2 + I 3 , V2 = V3
3Ð-90o A
3sI 2 + 3sI1 = 2 sI 3 + 2 sI1
Þ
Þ
3I 2 + I1 = 2 I 3
Þ
4
I1
I 2 = , I 3 = I1
5
5
~
j4
j4
I1
j
4I2 = I3
V = 4 sI1 + 3sI 2 + 2 sI 3 + 3sI 2 + 3s I1
6s
2 ´ 4s
= 7 sI1 +
I1 +
I3
5
5
49
49
H
V =
sI1 , Leq =
5
5
Page
88
2W
Fig. S1.9.29
( - j + 2 + j 4) I1 - jI 2 = - j 3
( j 4 + 2) I 2 - jI1 = -12 Ð30 ° V
I1 = -1.45 - j0.56,
.
Vx = -2 I1 = 2.9 + j112
= 311
. Ð2112
. °V
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I2
~
12Ð30 V
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GATE EC BY RK Kanodia
Magnetically Coupled Circuits
Chap 1.9
20 W
30. (D)
j8
j10
0.03 H
j10
j20
j10
2F
0.09 H
35. (D) The p equivalent circuit of coupled coil is shown
in fig. S1.9.35
= 112
. + j112
. W
M
31. (C) Z in = ( - j 6)||( Z o)
(12) 2
Z o = j20 +
= 0.52 + j15.7
( j 30 + j5 - j2 + 4)
L1
L1L2 - M 2
L1 L2 - M 2
=
M
w2 M 2
Z L + jwL2
= j250 ´ 10 3 ´ 2 ´ 10 -6 +
L1L2 - M 2
L1 - M
L2 - M
Fig. S1.9.35
M 2 = 160 ´ 10 -12
32. (D) M = k L1 L2 ,
M
L1L2 - M 2
L2
( - j 6)(0.52 + j15.7)
= 0.20 - j9.7 W
( - j 6 + 0.52 + j15.7)
Z in = jwL1 +
Vout
-
Fig. P1.9.34
( j14) ( j10 + 2 - j 6)
= 10 + j 8 +
j14 + j10 + 2 - j 6
Z in =
j30
-j3
1000C
Fig. S.1.9.30
Z eq
0.18 H
+
~
Vin
j18
40 W
L1 L2 (1 - k2 )
0.12 ´ 0.27 (10
. .5 2 )
=
= 0.27
k
0.5
Output is zero if
(250 ´ 10 3)2 ´ 160 ´ 10 -12
2 + j10 + j ´ 250 ´ 10 3 ´ 80 ´ 10 -6
Z in = 0.02 + j0.17 W
C=
-j
+ jCw = 0
0.27 w
1
= 33.33 mF
0.27 w2
36. (C) Applying 1 V test source at ab terminal,
33. (D)
-j
2C
60 W
j2 2
V1
~
I1
j4
1 : 4
j4
I2
4W
20 W
1V
20Ix
Ix
Fig. S.1.9.33
V1 = -
jI1
+ j 4 I1 + j2 2 I 2
2C
0 = ( 4 + 4 j) I 2 + j2 2 I1
Þ
I2 =
- j 2 I1
2 (1 + j)
2
V1 - j
- j + j 8 C + 2 C - j2 C
=
+ j4 +
=
1+ j
I1 2 C
2C
Z in =
- j + j8C + 2C - j 2C
2C
Im ( Z in ) = 0
Þ
C=
Þ
- j + j 8 C - j2 C = 0
1
6
34. (A) j 30 -
3j
= 0, C = 100 mF
1000 C
Fig. S1.9.36
Vab = 1 V, I x =
1
= 0.05 A, V2 = 4 V ,
20
4 = 60 I 2 + 20 ´ 0.05
Þ
I 2 = 0.05 A
I in = I x + I1 = I x + 4 I 2 = 0.25 A
1
RTH =
=4W ,
VTH = 0
I in
37. (A) Impedance seen by RL = 10 ´ 4 2 = 160 W
For maximum power RL = 160 W, Z o = 10 W
2
æ 100 ö
PLmax = ç
÷ ´ 10 = 250 W
è 10 + 10 ø
38. (B) I 2 =
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V2
I
V
, I1 = 2 = 2 , V1 = 5 V2
8
5 40
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CHAPTER
1.10
TWO PORT NETWORK
Statement for Q.1-4:
The circuit is given in fig. P.1.10.1–4
2W
I1
2W
I2
+
+
3W
1W
V1
V2
-
-
Fig. P.1.10.1–4
1. [ z ] = ?
é 1
ê(A) ê 2
ê- 17
ë 6
é 17
ê(C) ê 6
ê-1
ë 6
3ù
2ú
ú
1ú
2û
é1
ê
(B) ê 2
ê17
ë6
3ù
2ú
ú
1ú
2û
1ù
2ú
ú
3ú
2û
é17
ê
(D) ê 6
ê1
ë6
1ù
2ú
ú
3ú
2û
-
-
3. [ h] = ?
3ù
é6
ê17
17 ú
(A) ê
ú
24 ú
ê3
ë17
17 û
é 8
ê
(B) ê 3
ê- 1
ë 3
1ù
3ú
ú
2ú
3û
é 6
ê
(C) ê 17
ê- 3
ë 17
é8
ê
(D) ê 3
ê1
ë3
1ù
- ú
3
ú
2ú
3û
3ù
17 ú
ú
24 ú
17 û
4. [ T ] = ?
é17
ù
8ú
(A) ê 3
ê
ú
3 û
ë2
é 17
(B) ê 3
ê
ë-2
é 17
ù
- 8ú
(C) ê 3
ê
ú
ë 2 - 3û
é17
(D) ê 3
ê
ë 2
5. [ z ] = ?
2W
I1
ù
- 8ú
ú
3û
ù
- 8ú
ú
- 3û
2W
I2
+
1W
V1
2. [ y ] = ?
1ù
é3
ê8
8ú
(A) ê
ú
1
17
ê
ú
ë8
24 û
é 3
ê
(B) ê 8
ê- 1
ë 8
é17
ê
(C) ê 6
ê1
ë2
é 17
ê
(D) ê 6
ê- 1
ë 8
1ù
2ú
ú
3ú
2û
+
1ù
8ú
ú
17 ú
24 û
-
1ù
- ú
2
ú
3ú
2û
3W
2W
-
V2
-
Fig. P.1.10.5
é
ê
(A) ê
ê
ë
21
16
1
8
é 21
ê
(C) ê 16
ê- 1
ë 8
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1 ù
8 ú
ú
7 ú
12 û
é
ê
(B) ê
ê
ë
7
9
1
6
1ù
6ú
ú
7ú
4û
1ù
8ú
ú
7 ú
12 û
é
ê
(D) ê
ê
ë
7
9
1
3
1ù
3ú
ú
7ú
4û
-
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UNIT 1
6. [ y ] = ?
Networks
11. [ y ] = ?
2W
1W
I1
3W
+
+
V2
V1
-
-
+
2W
V1
1W
1W
I1
I2
-
+
2W
1W
2 ù
41 ú
ú
19 ú
41 û
é 11
ê
(B) ê 41
ê- 2
ë 41
2 ù
41 ú
ú
19 ú
41 û
é19
ê
(C) ê 41
ê2
ë 41
2 ù
41 ú
ú
11 ú
41 û
é 19
ê
(D) ê 41
ê- 2
ë 41
2 ù
41 ú
ú
11 ú
41 û
é
ê
(A) ê
ê
ë
ù
1ú
ú
-1ú
û
1
2
3
2
é 1
ê
(C) ê 2
ê- 1
ë 4
é
ê
(B) ê
ê
ë
1ù
2ú
ú
3ú
4û
3
2
1
2
ù
-1ú
ú
1ú
û
é 1
ê(D) ê 4
ê 1
ë 2
3ù
4ú
ú
1ú
2û
12. [ z ] = ?
Statement for Q.7-10:
2W
I1
is
V2
Fig. P.1.10.11
é 11
ê
(A) ê 41
ê2
ë 41
port
I1
-
Fig. P.1.10.6
A two
I2
described
by
V1 = I1 + 2 V2 ,
2V1
I2
+
+
I 2 = - 2 I1 + 0.4 V2
1W
V1
7. [ z ] = ?
2W
V2
-
é 11
(A) ê
ë-5
-5 ù
2.5 úû
é 11 5 ù
(B) ê
ú
ë 5 2.5 û
é1
(C) ê
ë5
-2ù
0.4 úû
é 1
(D) ê
ë-2
-
Fig. P.1.10.12
2ù
0.4 úû
é 4
ê
(A) ê 3
ê- 2
ë 3
2ù
3ú
ú
2ú
3û
é
ê
(B) ê
ê
ë
1
2
1
2
1ù
- ú
2
ú
1ú
û
é 2
ê(C) ê 3
ê 4
ë 3
2ù
3ú
ú
2ú
3û
é
ê
(D) ê
ê
ë
1
2
1
2
ù
1ú
8. [ y ] = ?
é11 5 ù
(A) ê
ú
ë 5 2.5 û
é 1
(B) ê
ë-2
-2 ù
4.4 úû
é-2 4.4 ù
(C) ê
ú
ë 4 -2 û
é 11
(D) ê
ë- 5
-5 ù
2.5 úû
ú
1
- ú
2û
13. [ y ] = ?
2W
2W
I1
I2
+
+
9. [ h] = ?
é3
(A) ê
ë4
- 6ù
- 4 úû
é 1
(C) ê
ë-2
é 4
(B) ê
ë-2
-2 ù
4.4 úû
2ù
0.4 úû
é 11
(D) ê
ë 5
5ù
2.5 úû
0.5 ù
0.5 úû
é 2. 2 - 0.5 ù
(B) ê
ú
ë 0. 2 - 0.5 û
10. [ T ] = ?
é 2. 2
(A) ê
ë 0. 2
é 1
(C) ê
ë-2
Page
92
2ù
0.4 úû
-2 ù
é 1
(D) ê
ú
ë-2 - 0.4 û
2W
V1
2V1
1W
V2
-
-
Fig. P.1.10.13
é
ê
(A) ê
ê
ë
7
4
1
2
é
ê
(C) ê
ê
ë
10
19
6
19
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1ù
- ú
4
ú
5ú
4û
2 ù
19 ú
ú
14 ú
19 û
é
ê
(B) ê
ê
ë
7
4
3
4
é
ê
(D) ê
ê
ë
6
19
10
19
-
1ù
4ú
ú
5ú
4û
14 ù
19 ú
ú
2 ú
19 û
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Two Port Networks
14. [ z ] = ?
Chap 1.10
17. [ z ] = ?
4V3
I1
2W
I1
2W
I2
+
+
I1
I2
+
2W
2I2
+
1W
V1
V3
2W
V1
+
1W
-
Fig. P.1.10.17
-
é3
(A) ê
ê6
ë
Fig. P.1.10.14
é2
(A) ê
ë3
3ù
3úû
é-3
(B) ê
ë 3
- 2ù
3úû
é3
(C) ê
ë3
3ù
2 úû
é 3
(D) ê
ë-3
3ù
- 2 úû
é
ê
(C) ê
ê
ë
15. [ z ] = ?
2V1
2W
I1
2ù
1ú
ú
7û
ù
1ú
ú
3ú
û
7
4
1
2
+
2W
V1
2W
3
3V
2 2
I1
Fig. P.1.1.15
2ù
ú
2ú
û
é
2
(C) ê
ê
ë2
3ù
2ú
ú
2û
é
ê
(D) ê
ê
ë
ù
3ú
ú
1ú
û
1
2
7
4
1V
5 1
4W
I2
+
1V
10 2
V1
-
é2
(A) ê 3
ê
ë2
1ù
7ú
ú
2û
+
V2
-
é
6
(B) ê
ê
ë3
18. [ T ] = ?
I2
+
V2
-
V2
-
2W
4W
V2
-
é
-2
(B) ê
ê
ë 2
3ù
2ú
ú
-2 û
é 2 -2 ù
ú
(D) ê 3
ê2ú
ë 2
û
-
Fig. P.1.10.18
é 0.35 - 1 ù
(A) ê
ú
ë 2 - 3.33û
- 3.33ù
é 2
(B) ê
0
.
35
- 1 úû
ë
é 2
(C) ê
ë0.35
é 0.35
(D) ê
ë 2
3.33ù
1 úû
1 ù
3.33úû
19. [ h] = ?
16. [ y ] = ?
I1
1W
2W
I1
I2
+
V2
2W
I2
+
+
+
3W
V1
1
2 I2
V1
V2
4W
V2
-
-
V2
-
Fig. P.1.10.19
-
1ù
é-1
(A) ê
ú
ë-1 - 2 û
é 1 - 1ù
(B) ê
ú
ë 1 - 2û
é
ê 4
(A) ê
ê-2
ë
1ù
é 2
ê- 3
3ú
(C) ê
ú
ê- 1 - 1 ú
ë 3
3û
1ù
é 2
ê- 3 - 3 ú
(D) ê
ú
ê 1 - 1ú
ë 3
3û
3ù
é
ê 4 - 2ú
(C) ê
ú
1ú
ê2
ë
2û
Fig. P.1.10.16
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3ù
2ú
ú
1ú
2û
é
ê-2
(B) ê
ê 4
ë
1ù
2ú
ú
3ú
2û
é
ê2
(D) ê
ê4
ë
1ù
2ú
ú
3
- ú
2û
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Two Port Networks
Z ab ù
éZ + Z ab
(A) ê a
Z b + Z ab úû
ë Z ab
éZ - Z ab
(B) ê a
ë Z ab
éZ + Z ab
(C) ê a
ë - Z ab
éZ - Z a
(D) ê ab
ë Z ab
- Z ab
ù
Z b + Z ab úû
ù
Z b - Z ab úû
Z ab
ù
- Z b úû
Z ab
Z ab
Chap 1.10
The value of
Vo
is
Vs
(A)
3
32
(B)
1
16
(C)
2
33
(D)
1
17
27. [ y ] = ?
30. The T-parameters of a 2-port network are
Yab
Ya
é2
[T ] = ê
ë1
Yb
1ù
.
1 úû
If such two 2-port network are cascaded, the
Fig. P.1.10.27
éY + Yab
(A) ê a
ë - Yab
- Yab ù
Yb + Yab úû
éY - Yab
(B) ê a
ë Yab
Yab ù
Yb - Yab úû
éY - Ya
(C) ê ab
ë Yab
Yab ù
Yab - Ya úû
éY - Yab
(D) ê a
ë - Yab
- Yab ù
Yb - Yab úû
28. The y-parameters of a 2-port network are
é5
[ y] = ê
ë1
3ù
S
2 úû
z –parameter for the cascaded network is
1ù
é 5
- ú
é 2 - 2ù
ê 3
3
ú
(A) ê 1
(B) ê
ú
1
2
ê1ú
ê
ú
ë 2
û
ë 3
3û
é
ê
(C) ê
ê
ë
5
3
1
3
1ù
3ú
ú
2ú
3û
é2
(D) ê 1
ê
ë2
2ù
ú
1ú
û
31. [ y ] = ?
2W
A resistor of 1 ohm is connected across as shown in
1W
fig. P.1.10.2 8. The new y –parameter would be
1W
1W
2W
1W
2
1W
[ y] = 5 3
1 2
Fig. P.1.10.31
é 19
ê
(A) ê 10
ê- 9
ë 10
Fig. P.1.10.28
é6
(A) ê
ë2
4ù
S
3 úû
é6
(B) ê
ë0
2ù
S
3 úû
é5
(C) ê
ë2
4ù
S
2 úû
é4
(D) ê
ë2
4ù
S
1 úû
é
ê
(C) ê
ê
ë
é2 0 ù
29. For the 2-port of fig. P.1.10.29, [ ya ] = ê
ú mS
ë0 10 û
19
10
9
10
-9 ù
10 ú
ú
31 ú
10 û
9 ù
10 ú
ú
31 ú
10 û
é 19
ê
(B) ê 10
ê- 7
ë 10
-7 ù
10 ú
ú
31 ú
10 û
é
ê
(D) ê
ê
ë
7 ù
10 ú
ú
31 ú
10 û
19
10
7
10
32. [ y ] = ?
2F
I1(s)
60 W
I2(s)
+
+
[ ya]
Vs
100 W
+
Vo
-
300 W
V1(s)
1W
3
2F
2V1(s)
-
1W
4
2F
V2(s)
-
Fig. P.1.10.32
Fig. P.1.10.29
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UNIT 1
és+3
(A) ê
ë2 s + 2
é s+3
(B) ê
ë-2 s - 2
2 sù
4 úû
é s + 3 - 2 sù
(C) ê
ú
ë-2 s - 2 4 û
- 2s ù
4 s + 4 úû
Networks
37. The circuit shown in fig. P.1.10.37 is reciprocal if a
is
é 3s + 3 - 2 s ù
(D) ê
ú
ë- 2 s - 2 4 s + 4 û
I1
0.5V1
1W
I2
+
+
2W
33. h21 = ?
V1
I1
R
R
+
V1
V2
I2
aI1
+
R
-
-
Fig. P.1.10.37
V2
-
-
Fig. P.1.10.33
(A) -
3
2
(B)
1
2
(C) -
1
2
(D)
3
2
(A) 2
(B) -2
(C) 1
(D) -1
38. Z in = ?
Zin
I2
I1
1 kW
+
+
34. In the circuit shown in fig. P.1.10.34, when the
Vs
[ y] = 4 -0.1 mS
50 1
V1
voltage V1 is 10 V, the current I is 1 A. If the applied
V2
1 kW
-
-
voltage at port-2 is 100 V, the short circuit current
Fig. P.1.10.38
flowing through at port 1 will be
V1
Linear
Resistive
Network
I
(A) 86.4 W
(B) 64.3 W
(C) 153.8 W
(D) 94.3 W
39. V1 , V2 = ?
25 W
Fig. P.1.10.34
(A) 0.1 A
(B) 1 A
(C) 10 A
(D) 100 A
+
+
[ y] = 10 -5 mS
50 20
V1
100 V
V2
100 W
-
-
Fig. P.1.10.39
35. For a 2-port symmetrical bilateral network, if
transmission parameters A = 3 and B = 1 W, the value of
parameter C is
(A) -68.6 V,
114.3 V
(B) 68.6 V, - 114.3 V
(C) 114.3 V,
- 68.6 V
(D) -114.3 V,
68.6 V
(A) 3
(B) 8 S
(C) 8 W
40. A 2-port network is driven by a source Vs = 100 V in
(D) 9
series with 5 W, and terminated in a 25 W resistor. The
impedance parameters are
36. A 2-port resistive network satisfy the condition
3
4
A = D = B = C. The z11 of the network is
2
3
4
3
(A)
(B)
3
4
2
(C)
3
Page
96
3
(D)
2
é 20
[z ] = ê
ë 40
2 ù
W
10 úû
The Thevenin equivalent circuit presented to the
25 W resistor is
(A) 80 V, 2.8 W
(B) 160 V, 6.8 W
(C) 100 V, 2.4 W
(D) 120 V, 6.4 W
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Two Port Networks
é
ê
[z ] = ê
ê
ë
7
4
1
2
Chap 1.10
Z ¢R = n2 4 = 36
ù
1ú
ú
3ú
û
Z’R
I1=0
+
1 : 3
+
V1
18. (D) Let I 3 be the clockwise loop current in center
I2
4W
+
V’2
9W
V’1
-
-
V2
-
loop
I1 =
V2
+ I 3,
10
Þ
I1 = 0.35 V2 - I 2
V2 = 4( I 2 + I 3)
Þ
I 3 = 0.25 V2 - I 2
...(i)
V1 = 4 I1 - 0.2 V1 + V2
Þ V1 = 2 V2 - 3.33I 2
...(ii)
I ö
1
æ
19. (A) V2 = 4ç I 2 + I1 - 2 ÷Þ I 2 = -2 I1 + V2
2 ø
2
è
( V - V2 ) - V2
I
V
V
= -I1 + 2 + 1 - V2
I1 = 2 + 1
2
2
4
2
3
Þ V1 = 4 I1 + V2
2
...(ii)
+
+
+
V’1
-
-
V’2
9 W V2
-
Fig. S1.10.23a
æ4 ö
V2 = V2¢ = nV1¢ = 3ç I1 ÷
è5 ø
z11 =
Þ
V1
= 0.8
I1
z 21 =
V2
+ 0( -I 2 ),
5
...(ii)
I1 = (0) V2 + 5( -I 2 )
...(i)
27. (A) I1 = ( V1 - V2 ) Yab + V1 Ya
...(ii)
I2=0
1 : 3
Þ
V2 = 2 sI1 + 3sI 2
...(ii)
...(i)
V2 = sI1 + (2 + 2 s) I 2
ZR
4
I1
5
Þ
V2 = ( Z a + Z ab ) I 2 + Z ab I1 = Z ab I1 + ( Z a + Z ab ) I 2
...(ii)
9 9
= =1
n2 9
V1 = ( 4||1) I1 =
V2 = 3sI 2 + 2 sI1
...(i)
...(i)
...(i)
I1
æ1
ö
+ sI1 + sI 2 = ç + s ÷I1 + sI 2
s
ès
ø
4W
V1 = 6 sI1 + 2 sI 2
I1 = V1 ( Ya + Yab ) - V2 Yab
V2
= 2.4,
I1
....(ii)
...(i)
I 2 = ( V2 - V1 ) Yab + V2 Yb = -V1 Yab + V2 ( Yb + Yab )
I1 = (2 + j2) V1 - jV2
V2
+ V1 + j( V2 - V1 ) = (1 - j) V1 + (1 + j) V2
I2 =
1
V1
Þ
Þ
Þ
I1
V2
= 7.2,
I2
26. (A) V1 = ( Z a + Z ab ) I1 + Z ab I 2
21. (D) I1 = 2 V1 + jV1 + j( V1 - V2 )
23. (D) Z R =
z 22 =
...(i)
V1 V1 - V2 3
3
+
= V1 - V2
1
2
2
2
V
V - V1 3
3
I 2 = 2 V1 + 2 + 2
= V1 + V2
1
2
2
2
Þ
Þ
24. (C) V1 = 3sI1 + 3sI1 - 3sI1 + 3sI1 + 2 sI 2
25. (C) V1 =
20. (B) I1 = -V2 +
V2 = 2 I 2 + 2 s I 2 + sI1
V2 = ( 36 || 9) I 2 = 7.2 I 2
z12 = z 21 = 2.4
12
. V1 = 4(0.35 V2 - I 2 ) + V2 = 2.4 V2 - 4 I 2
22. (B) V1 =
Fig. S1.10.23b
...(ii)
28. (B) y-parameter of 1 W resistor network are
é 1
ê-1
ë
- 1ù
1úû
é5
New y-parameter = ê
ë1
3ù é 1 - 1 ù
é6
+
= ê
2 úû êë-1
1úû
ë0
2ù
.
3 úû
-1
0 ù
é5000 0 ù
é2 mS
29. (A) [ z a ] = ê
ú =ê 0
100 úû
mS
0
10
ë
ë
û
é5000
[z ] = ê
ë 0
0 ù é100
+
100 úû êë100
100 ù é600
=
100 úû êë100
100 ù
200 úû
V1 = 600 I1 + 100 I 2 , V2 = 100 I1 + 200 I 2
Vs = 60 I1 + V1 = 660 I1 + 100 I 2 , V2 = Vo = -300 I 2
2
V
Vo = 100 I1 - Vo Þ
I1 = o
60
3
V
Vo
3
Vs = 11Vo - o Þ
=
V5 32
3
é2
30. (C) [ TN ] = ê
ë1
1ù
1 úû
é2
ê1
ë
1ù é5
=
1 úû êë3
3ù
2 úû
V1 = 5 V2 - 3I 2 , I1 = 3V2 - 2 I 2
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UNIT 1
3V1 - 5 I1 = I 2
V2 =
Þ
V1 =
5
1
I1 + I 2
3
3
...(i)
1
2
I1 + I 2
3
3
31. (B)
...(ii)
Networks
I ¢2
= 0.1, I 2¢ = 100 ´ 0.1 = 10
V1¢
Interchanging the port
35. (B) For symmetrical network A = D = 3
For bilateral AD - BC = 1, 9 - C = 1,
1W
2W
C=8 S
2W
36. (A) z11 =
1W
1W
2
1W
A 4
=
C 3
37. (A) V1 = 0.5 V1 + I1 + 2( I1 + I 2 ) + aI1
Fig. S.1.10.31a & b
é3
[ za ] = ê
ë1
1ù
é 2
é 3
ê 5 - 5ú
ê 2
1ù
, [ ya ] = ê
ú , [ yb ] = ê
ú
2û
3ú
ê- 1
ê- 1
ë 5
ë 2
5û
é 19
ê
[ y ] = [ ya ] + [ yb ] = ê 10
ê- 7
ë 10
-
7 ù
10 ú
ú
31 ú
10 û
Þ
V1 = ( 6 + 2 a) I1 + 4 I 2
V2 = 2( I1 + I 2 ) + aI1
...(i)
Þ
V2 = (2 + a) I1 + 2 I 2
...(ii)
For reciprocal network
z12 = z 21 , 4 = 2 + a
Þ
a =2
38. (C) I1 = 4 ´ 10 3 V1 - 0.1 ´ 10 -3 V2
I 2 = 50 ´ 10 -3 V1 + 10 -3 V2 , V2 = -10 3 I 2
-10 -3 V2 = 50 ´ 10 -3 V1 + 10 -3 V2 , V2 = -25 V1
- 2s ù
é 3
, [ yb ] = ê
ú
4 sû
ë-2
é 3s
32. (D) [ ya ] = ê
ë-2 s
1ù
- ú
2
ú
5ú
2û
0ù
4 úû
2F
10 3 I1 = 4 V1 + 2.5 V1 ,
V1 10 3
=
= 153.8
I1
6.5
39. (B) I1 = 10 ´ 10 -3 V1 - 5 ´ 10 -3 V2 ,
100 = 25 I1 + V1
2F
2F
100 - V1 = 0. 25 V1 - 0.125 V2
-3
Þ
800 = 10 V1 - V2
...(i)
-3
I 2 = 50 ´ 10 V1 + 20 ´ 10 V2 , V2 = -100 I 2
V2 = -5 V1 - 2 V2
Fig. S.1.10.32a
1W
3
40. (B) 100 = 5 I1 + V1 ,
Þ
I2
I1
I1
, -I 2 =
V2 = 0
- 2s ù
4 s + 4 úû
R
Þ
I2
R
-
Fig. S.1.10.33
34. (C)
I2
V1
= y21 =
V2 = 0
1
= 0.1
10
V2 z 21
=
V1 z11
42. (B) I 2 = y21 V1 + y22 V2 , I 2 = -V2 YL
y21 V1 + ( y22 + YL ) V2 = 0 ,
+
V1
V2 = 160 + 6.8 I 2
41. (B) V1 = z11 I1 , V2 = z 21 I1 ,
I
1
I1 R
, 2 =R + R I1
2
R
V1 = 20 I1 + 2 I 2
VTH = 160 V, RTH = 6.8 W
Fig. S.1.10.32b
33. (C) h21 =
V2 = -114.3 V.
100 = 25 I1 + 2 I 2 , V2 = 40 I1 + 10 I 2
800 - 5 V2 = -34 I 2
é 3s + 3
[ y ] = [ ya ] = [ yb ] = ê
ë-2 s - 2
3V2 + 5 V1 = 0
From (i) and (ii) V1 = 68.6 V,
1W
4
2V1
Þ
V2
- y21
=
V1 ( y22 + yL )
43. (A) V2 = z 21 I1 + z 22 I 2 ,
V2 = -Z L I 2
æ V ö
V2 = z 21 I1 + z 22 çç - 2 ÷÷
è ZL ø
V2 ( Z L + z 22 ) = z 21 Z L I1 ,
V2
z Z
= 21 L
I1 z 22 + Z L
**********
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100
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GATE EC BY RK Kanodia
CHAPTER
1.11
FREQUENCY RESPONSE
Statement for Q.1-3:
7. A parallel circuit has R = 1 kW , C = 50 mF and L = 10
A parallel resonant circuit has a resistance of 2 kW
and half power frequencies of 86 kHz and 90 kHz.
1. The value of capacitor is
(A) 6 mF
(C) 2 nF
BW = 10 3 rad/s. The value of R and C will be
(A) 100 mF, 10 W
(C) 100 pF, 10 MW
(B) 43 mH
(D) 1.6 mH
The required R for the BW 15.9 Hz is
(A) 22
(C) 48
(A) 0.1 W
(C) 15.9 mW
(B) 100
(D) 200
Statement for Q.4-5:
(B) 0.2 W
(D) 500 W
10. For the RLC parallel resonant circuit when
resonant
admittance of 25 ´ 10
(B) 100 pF, 10 W
(D) 100 mF, 10 MW
9. A series resonant circuit has L = 1 mH and C = 10 mF.
3. The quality factor is
-3
(B) 90.86
(D) None of the above
The resonant frequency wo = 106 rad s and bandwidth is
2. The value of inductor is
A parallel
(A) 100
(C) 70.7
8. A series resonant circuit has an inductor L = 10 mH.
(B) 20 nF
(D) 60 mF
(A) 4.3 mH
(C) 0.16 mH
mH. The quality factor at resonance is
circuit
has
a
midband
S, quality factor of 80 and a
resonant frequency of 200 krad s.
R = 8 kW, L = 40 mH and C = 0. 25 mF, the quality factor
Q is
(A) 40
(B) 20
(C) 30
(D) 10
4. The value of R is
(A) 40 W
(C) 80 W
11. The maximum voltage across capacitor would be
(B) 56.57 W
(D) 28. 28 W
0.105v1
5. The value of C is
(A) 2 mF
(C) 10 mF
+
(B) 28.1 mF
(D) 14.14 mF
25 W
3V
6. A parallel RLC circuit has R = 1 kW and C = 1 mF. The
quality factor at resonance is 200. The value of inductor is
(A) 35.4 mH
(B) 25 mH
(C) 17.7 mH
(D) 50 mH
v1
–
4H
10 W
~
1
mF
4
+
vC
–
Fig. P1.11.11
(A) 3200 V
(C) -3 V
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(B) 3 V
(D) 1600 V
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GATE EC BY RK Kanodia
UNIT 1
12. For the circuit shown in fig. P1.1.11 resonant
frequency fo is
10 W
Networks
16. H ( w) =
Vo
=?
Vi
+
v1 -
+
40 W
1 mH
Zin
vs
v1
2
50 nF
~
10 W
0.5 F
vO
–
Fig. P1.11.16
Fig. P1.11.12
(A) 346 kHz
(C) 196 kHz
(B) 55 kHz
(D) 286 kHz
(A) (5 + j20 w) -1
(B) (5 + j 4 w) -1
(C) (5 + j 30 w) -1
(D) 5(1 + j 6 w) -1
13. For the circuit shown in fig. P1.11.13 the resonant
17. The value of input frequency is required to cause a
frequency fo is
gain equal to 1.5. The value is
2 kW
10 mH
600 pF
vs
22 kW
~
60 mF
+
vO
–
1.8 W
Fig. P1.11.17
Fig. P1.11.13
(A) 12.9 kHz
(B) 12.9 MHz
(C) 2.05 MHz
(D) 2.05 kHz
14.
The
network
function
of
circuit
shown
fig.P1.11.14 is
in
(A) 20 rad s
(B) 20 Hz
(C) 10 rad s
(D) No such value exists.
18. In the circuit of fig. P1.11.18 phase shift equal to
-45 ° is required at frequency w = 20 rad s . The value of
R is
4
V
H ( w) = o =
V1 1 + j0.01w
2 kW
10 W
15 kW
Vs
+
vi
~
+
vC
–
C
~
1 mF
+
vO
–
vo
AvC
–
Fig. P1.11.18
Fig. P1.11.14
The value of the C and A is
(A) 10 mF, 6
(B) 5 mF, 10
(C) 5 mF, 6
(D) 10 mF, 10
15. H ( w) =
ia
Vo
=?
Vi
(A) 200 kW
(B) 150 kW
(C) 100 kW
(D) 50 kW
19. For the circuit of fig. P1.11.19 the input frequency is
adjusted until the gain is equal to 0.6. The value of the
frequency is
20 W
2H
+
vi
~
3ia
4H
0.25 F
vO
vs
~
30 W
+
vO
–
–
Fig. P1.11.15
0.6
(A)
jw(1 + j0. 2 w)
3
(C)
jw(1 + jw)
Page
102
0.6
jw(5 + jw)
3
(D)
jw(20 + j 4 w)
Fig. P1.11.19
(B)
(A) 20 rad s
(B) 20 Hz
(C) 40 rad s
(D) 40 Hz
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UNIT 1
Networks
31. Bode diagram of the network function Vo Vs for the
SOLUTIONS
circuit of fig. P1.11.30 is
4W
1. (B) BW = w2 - w1 = 2 p(90 - 86)k = 8 p krad s
+
vs
2W
~
BW=
vo
30 mF
Fig.P1.11.30
2. (C) wo =
dB
c.
5.56 16.7
dB
/d
e
0
20
log w
5.56 16.7
=
5.56 16.7
.
0 dB
1
1
=
RBW
8 p ´ 10 3 ´ 2 ´ 10 3
( w1 + w2 ) 2 p(90 + 86)k
=
= 176p krad s
2
2
1
Þ L= 2 C
wo
1
= 0.16 mH
(176 p ´ 10 ) (20 ´ 10 -9)
log w
wo 176 pk
=
= 22
B
8 pk
-40
-2
dB
B
0d
/de
/d
ec
log w
C =
3 2
3. (A) Q =
c.
5.56 16.7
log w
LC
(B)
(A)
0 dB
1
wo =
c.
dB
/de
40
0
Þ
= 19.89 nF
–
dB
1
RC
4. (A) At mid-band frequency Z = R , Y =
R=
(C)
(D)
1
= 40 W
25 ´ 10 -3
5. (C) Q = wo RC
Q
80
Þ C=
=
= 10 mF
wo R 200 ´ 10 3 ´ 40
***************
C
L
6. (B) Qo = R
Þ
8. (B) wo =
C=
BW =
1
LC
1
-3
10 ´ 10 ´ (106 ) 2
R
L
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= 100 pF
R = 10 ´ 10 -3 ´ 10 3 = 10
Þ
R
L
R
= 15.9 ´ 2 p = 0.1 W
1 ´ 10 -3
10. (B) Q = R
Page
104
10 -6
L
C
50 ´ 10 -6
= 10 3
= 70.7
L
10 ´ 10 -3
9. (A) BW =
Þ
200 = 10 3
L = 25 mH
7. (C) Qo = R
Þ
Þ
C
L
1
R
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Frequency Response
= 8 ´ 10 3
0. 25 ´ 10 -6
= 20
40 ´ 10 -3
15. (A) I a =
11. (A) Thevenin equivalent seen by L-C combination
æ v - 0.105 v1 ö
3 = v1 + 10ç 1
÷
125
è
ø
I sc =
Þ
v1 = 100
1100
= 0.8 V
125
Open Circuit : v1 = 0, voc = 3 V
RTH =
Chap 1.11
3
= 375
. W, wo =
0.8
1
LC
= 1000
w L 1000 ´ 4
Qo = o =
= 1066.67
R
375
.
|vC|max = Qo vTH = 1066.67 ´ 3 = 3200 V
12. (B) Applying 1 A at input port V1 = 10 V
Vtest = 10 + jw10
j
(5 + 1)
w50 ´ 10 -9
Z in = Vtest
wo 10 -3 =
6
wo 50 ´ 10 -9
Þ
wo = 346 kHz
fo = 55 kHz
10
10
jw(0.51)
16. (A) Z1 =
=
1
+ 10 1 + j 3w
jw0.5
Vo
Z1
=
=
Vi 40 + Z1
=
= jw6 ´ 10 -10 + 45.45 +
1
1
+
2 ´ 10 3 1.8 + jw10 -5
1.8 - jw10 -5
3. 24 + w2 10 -50
At resonance Im { Y } = 0
wo 6 ´ 10 -10 ( 324
. + w2o 10 -10 ) - wo 10 -5 = 0
3. 24 + w 10
2
o
-10
17. (D) H ( w) =
Vo = AVc
Þ
Þ
18. (D) H ( w) =
= - tan -1 wCR = - 450 °
=
Vi
1 + j2 ´ 10 3 Cw
(15 k)
2 AVc
2 AVi
=
=
16 k + 30 k
3
3(1 + j2 ´ 10 3 Cw)
Þ
C = 5 mF
20 ´ 1 ´ 10 -6 R = 1
19. (A) H ( w) =
gain =
A = 6,
Þ
R = 50 kW.
Vo
R
=
Vs
1 + jwL
R
R +w L
2
2
2
=
30
900 + 4 w2 + 0.6
2
50 2 - 30 2
= 20 rad s
2
20. (A) H ( w) =
1
2 ´ 10 3 +
jCw
Vo
1
=
Vs 1 + jwCR
wCR = 1,
w=
2A
Vo
3
=
Vi 1 + j2 p ´ 10 3 Cw
2A
=4
3
(1 + w2 RC)
For any value of w, R, C gain £ 1.
= 16.67 ´ 10 wo = 12.9 Mrad s
w
fo = o = 2.05 MHz
2p
14. (C) VC =
Vo
1
=
Vi 1 + jwRC
1
gain =
3
Vi
jCw
10
1 + j5 w
10
+ 40
1 + j5 w
10
= (5 + j20 w) -1
50 + j200 w
phase shift
13. (C) Y = jw600 ´ 10 -12 +
3I a
0. 25 jw
Thus (D) is correct option.
At resonance Im { Z in } = 0
Þ
Vo =
Vo
3
0.6
=
=
V1
jw(5 + jw) jw(1 + j0.2 w)
voltage across 1 A source
-3
Vi
,
20 + j 4w
Vo
1
1
=
=
Vs 1 + jwCR 1 + j
Phase shift = - tan -1 wCR = - 45 °
gain =
1
1
=
= 0.707
| j + 1|
2
21. (B) BW= w2 - w1 = 2 p( 456 - 434) = 44 p
wo = 2 pfo = QBW = 20 ´ 44 p
fo = 440 Hz
2 ´ 10 3 C = 0.01
22. (C) fo =
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2 p LC
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GATE EC BY RK Kanodia
UNIT 1
=
1
2 p 360 ´ 10
fo =
-12
´ 240 ´ 10
-6
1
2 p 50 ´ 10
-12
´ 240 ´ 10 -6
Þ
= 541 kHz
= 1.45 MHz
R=
Networks
1
. kW
= 106
2 p ´ 15 ´ 10 ´ 10 -6
30. (B) 20 log H = 20 log
1
R2
- 2
LC L
1
= -40 log w
w2
R
400
10 7
=
=
-6
L 240 ´ 10
6
1
jw
1+
Vo
jw30 ´ 10 -3
.67
16
31. (D)
=
=
jw
1
Vs 6 +
1+
.
jw30 ´ 10 -3
356
1
1
1016
=
=
LC 240 ´ 10 -6 ´ 120 ´ 10 -12 288
20 dB/decade line starting from w = 16.67 rad s
23. (B) fo =
1
2p
2+
-20 dB/decade line starting from w = 5.56 rad s
Hence -20 dB/decade line for 5.56 < w < 16.67
R
1
1
<
, fo =
= 938 kHz
L LC
2 p LC
parallel to w axis to w > 16.67
1
, R and C should be as small as possible.
RC
(1.8)
R = ( 3.3)
= 1165
.
kW
3.3.+1.8
( 30)
C = (10)
= 7.5 pF
(10 + 30)
24. (B) wo =
w=
1
= 114.5 ´ 106 rad s
.
1165
´ 7.5 ´ 10 -9
25. (D) R¢ = K m R = 800 ´ 12 ´ 10 3 = 9.6 MW
L¢ =
Km
800
=
40 ´ 10 -6 = 32 mF
K f L 1000
C¢ =
C
10 -9
K F = 30 ´
´ 1000 = 0.375 pF
Km
80
26. (A) L¢C¢ =
LC
K 2f
Þ
K 2f =
4 ´ 20 ´ 10 -3 ´ 10 -6
1´ 6
Þ K f = 2 ´ 10 -4
(1)(20 ´ 10 -6 )
L¢ L 2
Þ K m2 =
= Km
(2) ( 4 ´ 10 -3)
C¢ C
Þ
K m = 0.05
1
RC
1
= 15.9 W
R=
2 p ´ 20 ´ 10 3 ´ 0.5 ´ 10 -6
27. (D) wc = 2 pfc =
Þ
28. (A) RTH across the capacitor is
RTH = (1k + 4 k)||5 k = 2.5 kW
1
. kHz
fc =
= 106
2 p ´ 2.5 ´ 10 3 ´ 40 ´ 10 -9
29. (B) wc = 2 pfc =
Page
106
1
RC
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GATE EC BY RK Kanodia
CHAPTER
2.1
SEMICONDUCTOR PHYSICS
In the problems assume the parameter given in
3. Two semiconductor material have exactly the same
following table. Use the temperature T = 300 K unless
properties except that material A has a bandgap of 1.0
otherwise stated.
eV and material B has a bandgap energy of 1.2 eV. The
ratio of intrinsic concentration of material A to that of
Property
Si
Ga As
Ge
material B is
Bandgap Energy
1.12
1.42
0.66
(A) 2016
(B) 47.5
Dielectric Constant
11.7
13.1
16.0
(C) 58.23
(D) 1048
Effective density of
2.8 ´ 1019
states in conduction
band N c (cm -3)
4.7 ´ 1017
104
. ´ 1019
Effective density of
104
. ´ 1019
states in valence
band N v(cm -3)
7.0 ´ 1018
Intrinsic carrier
concertration
ni (cm -3)
Mobility
Electron
Hole
15
. ´ 1010
4. In silicon at T = 300 K the thermal-equilibrium
concentration of electron is n0 = 5 ´ 10 4 cm -3. The hole
concentration is
1.8 ´ 106
6.0 ´ 1018
2.4 ´ 1013
(A) 4.5 ´ 1015 cm -3
(B) 4.5 ´ 1015 m -3
(C) 0.3 ´ 10 -6 cm -3
(D) 0.3 ´ 10 -6 m -3
5. In silicon at T = 300 K if the Fermi energy is 0.22 eV
above the valence band energy, the value of p0 is
1350
480
8500
400
3900
1900
(A) 2 ´ 1015 cm -3
(B) 1015 cm -3
(C) 3 ´ 1015 cm -3
(D) 4 ´ 1015 cm -3
6. The thermal-equilibrium concentration of hole p0 in
1. In germanium semiconductor material at T = 400 K
silicon at T = 300 K is 1015 cm -3. The value of n0 is
the intrinsic concentration is
(A) 3.8 ´ 108 cm -3
(B) 4.4 ´ 10 4 cm -3
(C) 2.6 ´ 10 4 cm -3
(D) 4.3 ´ 108 cm -3
(A) 26.8 ´ 1014 cm -3
(B) 18.4 ´ 1014 cm -3
(C) 8.5 ´ 1014 cm -3
(D) 3.6 ´ 1014 cm -3
7. In germanium semiconductor at T = 300 K, the
2. The intrinsic carrier concentration in silicon is to be
acceptor concentrations is N a = 1013 cm -3 and donor
no greater than ni = 1 ´ 1012 cm -3. The maximum
concentration is
temperature allowed for the silicon is ( E g = 112
. eV)
concentration p0 is
(A) 300 K
(B) 360 K
(A) 2.97 ´ 10 9 cm -3
(B) 2.68 ´ 1012 cm -3
(C) 382 K
(D) 364 K
(C) 2.95 ´ 1013 cm -3
(D) 2.4 cm -3
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N d = 0. The thermal equilibrium
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UNIT 2
Statement for Q.8-9:
Electronics Devices
15. For a particular semiconductor at T = 300 KE g = 15
.
In germanium semiconductor at T = 300 K, the
eV, mp* = 10 mn* and ni = 1 ´ 1015 cm -3. The position of
Fermi level with respect to the center of the bandgap is
impurity concentration are
N d = 5 ´ 1015 cm -3 and N a = 0
(A) +0.045 eV
(B) - 0.046 eV
(C) +0.039 eV
(D) - 0.039 eV
8. The thermal equilibrium electron concentration n0 is
(A) 5 ´ 1015 cm -3
(C) 115
. ´ 10 cm
9
(B) 115
. ´ 1011 cm -3
-3
(D) 5 ´ 10 cm
6
16. A silicon sample contains acceptor atoms at a
-3
concentration of N a = 5 ´ 1015 cm -3. Donor atoms are
9. The thermal equilibrium hole concentration p0 is
(A) 396
. ´ 1013
(B) 195
. ´ 1013 cm -3
(C) 4.36 ´ 1012 cm -3
(D) 396
. ´ 1013 cm -3
conduction band edge. The concentration of donors
atoms added are
10. A sample of silicon at T = 300 K is doped with boron
at a concentration of 2.5 ´ 1013 cm -3 and with arsenic at
a concentration of 1 ´ 1013 cm -3. The material is
(B) 4.6 ´ 1016 cm -3
(C) 39
. ´ 1012 cm -3
(D) 2.4 ´ 1012 cm -3
doped with phosphorus atoms at a concentrations of 1015
(B) p - type with p0 = 15
. ´ 10 7 cm -3
(C) n - type with n0 = 15
. ´ 10 cm
(A) 12
. ´ 1016 cm -3
17. A silicon semiconductor sample at T = 300 K is
(A) p - type with p0 = 15
. ´ 1013 cm -3
13
added forming and n - type compensated semiconductor
such that the Fermi level is 0.215 eV below the
cm -3. The position of the Fermi level with respect to the
-3
intrinsic Fermi level is
(D) n - type with n0 = 15
. ´ 10 7 cm -3
11. In a sample of gallium arsenide at T = 200 K,
n0 = 5 p0 and N a = 0. The value of n0 is
(A) 0.3 eV
(B) 0.2 eV
(C) 0.1 eV
(D) 0.4 eV
(A) 9.86 ´ 10 9 cm -3
(B) 7 cm -3
18. A silicon crystal having a cross-sectional area of
(C) 4.86 ´ 10 3 cm -3
(D) 3 cm -3
0.001 cm 2 and a length of 20 mm is connected to its ends
to a 20 V battery. At T = 300 K, we want a current of
12. Germanium at T = 300 K is uniformly doped with
an acceptor concentration of N a = 10 cm
15
-3
and a donor
100 mA in crystal. The concentration of donor atoms to
be added is
concentration of N d = 0. The position of fermi energy
(A) 2.4 ´ 1013 cm -3
(B) 4.6 ´ 1013 cm -3
with respect to intrinsic Fermi level is
(C) 7.8 ´ 1014 cm -3
(D) 8.4 ´ 1014 cm -3
(A) 0.02 eV
(B) 0.04 eV
(C) 0.06 eV
(D)0.08 eV
19. The cross sectional area of silicon bar is 100 mm 2 .
The length of bar is 1 mm. The bar is doped with
13. In germanium at T = 300 K, the donor concentration
are N d = 1014 cm -3 and N a = 0. The Fermi energy level
with respect to intrinsic Fermi level is
Page
110
(A) 0.04 eV
(B) 0.08 eV
(C) 0.42 eV
(D) 0.86 eV
arsenic atoms. The resistance of bar is
(A) 2.58 mW
(B) 11.36 kW
(C) 1.36 mW
(D) 24.8 kW
20. A thin film resistor is to be made from a GaAs film
doped n - type. The resistor is to have a value of 2 kW.
14. A GaAs device is doped with a donor concentration
of 3 ´ 1015 cm -3. For the device to operate properly, the
intrinsic carrier concentration must remain less than
5% of the total concentration. The maximum
temperature on that the device may operate is
10 -6 cm 2 . The doping efficiency is known to be 90%. The
(A) 763 K
(B) 942 K
(A) 8.7 ´ 1015 cm -3
(B) 8.7 ´ 10 21 cm -3
(C) 486 K
(D) 243 K
(C) 4.6 ´ 1015 cm -3
(D) 4.6 ´ 10 21 cm -3
The resistor length is to be 200 mm and area is to be
mobility of electrons is 8000
cm 2 V - s. The doping
needed is
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21. A silicon sample doped n - type at 1018 cm -3 have a
-6
Chap 2.1
27. For a sample of GaAs scattering time is t sc = 10 -13 s
2
and electron’s effective mass is me* = 0.067 mo . If an
and a length of 10 mm . The doping efficiency of the
electric field of 1 kV cm is applied, the drift velocity
sample is (m n = 800 cm V - s)
produced is
resistance of 10 W . The sample has an area of 10
cm
2
(A) 43.2%
(B) 78.1%
(A) 2.6 ´ 106 cm s
(B) 263 cm s
(C) 96.3%
(D) 54.3%
(C) 14.8 ´ 106 cm s
(D) 482
22.
Six
volts
is
applied
across
a
2
cm
long
semiconductor bar. The average drift velocity is 10 4
cm s. The electron mobility is
(A) 4396 cm 2 V - s
(B) 3 ´ 10 4 cm 2 V - s
(C) 6 ´ 10 4 cm 2 V - s
(D) 3333 cm 2 V - s
28. A gallium arsenide semiconductor at T = 300 K is
doped with impurity concentration N d = 1016 cm -3. The
mobility m n is 7500 cm 2 V - s. For an applied field of 10
V cm the drift current density is
23. For a particular semiconductor material following
(A) 120 A cm 2
(B) 120 A cm 2
(C) 12 ´ 10 4 A cm 2
(D) 12 ´ 10 4 A cm 2
parameters are observed:
29. In a particular semiconductor the donor impurity
m n = 1000 cm 2 V - s ,
concentration is N d = 1014 cm -3. Assume the following
m p = 600 cm V - s ,
2
N c = N v = 10 cm
19
parameters,
-3
m n = 1000 cm 2 V - s,
These parameters are independent of temperature.
æ T ö
N c = 2 ´ 1019ç
÷
è 300 ø
The measured conductivity of the intrinsic material is
s = 10 -6 (W - cm) -1 at T = 300 K. The conductivity at
æ T ö
N v = 1 ´ 1019ç
÷
è 300 ø
T = 500 K is
(A) 2 ´ 10
-4
(C) 2 ´ 10
-5
(W - cm)
(W - cm)
-1
-1
(B) 4 ´ 10
-5
-1
(W - cm)
(D) 6 ´ 10
-3
(W - cm) -1
24. An n - type silicon sample has a resistivity of 5
W - cm at T = 300 K. The mobility is m n = 1350
32
cm -3,
32
cm -3,
. eV.
E g = 11
An electric field of E = 10 V cm is applied. The
electric current density at 300 K is
cm 2 V - s. The donor impurity concentration is
(A) 2.3 A cm 2
(B) 1.6 A cm 2
(A) 2.86 ´ 10 -14 cm -3
(B) 9.25 ´ 1014 cm -3
(C) 9.6 A cm 2
(D) 3.4 A cm 2
(C) 11.46 ´ 1015 cm -3
(D) 11
. ´ 10 -15 cm -3
25.
Statement for Q.30-31:
In a silicon sample the electron concentration
18
drops linearly from 10
cm
-3
16
to 10
cm
-3
A semiconductor has following parameter
over a length
m n = 7500 cm 2 V - s,
of 2.0 mm. The current density due to the electron
diffusion current is ( Dn = 35 cm 2 s).
m p = 300 cm 2 V - s,
(A) 9.3 ´ 10 4 A cm 2
(B) 2.8 ´ 10 4 A cm 2
ni = 3.6 ´ 1012 cm -3
(C) 9.3 ´ 10 9A cm 2
(D) 2.8 ´ 10 9 A cm 2
30.
26. In a GaAs sample the electrons are moving under
an
electric
field
of
5
kV cm
and
the
carrier
concentration is uniform at 1016 cm -3. The electron
velocity is the saturated velocity of 10
7
(A) 1.6 ´ 10 A cm
(C) 1.6 ´ 108 A cm 2
conductivity
is
minimum,
the
hole
(A) 7.2 ´ 1011 cm -3
(B) 1.8 ´ 1013 cm -3
(C) 1.44 ´ 1011 cm -3
(D) 9 ´ 1013 cm -3
cm s. The drift
31. The minimum conductivity is
current density is
4
When
concentration is
2
2
(A) 0.6 ´ 10 -3 (W - cm) -1
(B) 17
. ´ 10 -3 (W - cm) -1
(D) 2.4 ´ 108 A cm 2
(C) 2.4 ´ 10 -3 (W - cm) -1
(D) 6.8 ´ 10 -3 (W - cm) -1
(B) 2.4 ´ 10 A cm
4
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UNIT 2
32.
A
particular
intrinsic
semiconductor
has
Electronics Devices
æ -x ö
÷
ç
÷
ç
15 è L p ø
a
Hole concentration p0 = 10 e
resistivity of 50 (W - cm) at T = 300 K and 5 (W - cm) at
T = 330 K. If change in mobility with temperature is
cm -3,x ³ 0
ö
æ
ç -x ÷
÷
ç
neglected, the bandgap energy of the semiconductor is
Electron concentration n0 = 5 ´ 1014 e è L n ø cm -3,x £ 0
(A) 1.9 eV
(B) 1.3 eV
Hole diffusion coefficient Dp = 10 cm 2 s
(C) 2.6 eV
(D) 0.64 eV
Electron diffusion coefficients Dn = 25 cm 2 s
a
Hole diffusion length Lp = 5 ´ 10 -4 cm,
semiconductor. If only the first mechanism were
Electron diffusion length Ln = 10 -3 cm
33.
Three
scattering
mechanism
exist
in
present, the mobility would be 500 cm V - s. If only the
2
second mechanism were present, the mobility would be
The total current density at x = 0 is
750 cm 2 V - s. If only third mechanism were present,
(A) 1.2 A cm 2
(B) 5.2 A cm 2
the mobility would be 1500 cm 2 V - s. The net mobility
(C) 3.8 A cm 2
(D) 2 A cm 2
is
(A) 2750 cm 2 V - s
(B) 1114 cm 2 V - s
37. A germanium Hall device is doped with 5 ´ 1015
(C) 818 cm 2 V - s
(D) 250 cm 2 V - s
donor atoms per cm 3 at T = 300 K. The device has the
geometry d = 5 ´ 10 -3 cm, W = 2 ´ 10 -2 cm and L = 0.1 cm.
34. In a sample of silicon at T = 300 K, the electron
The current is I x = 250 mA, the applied voltage is
concentration varies linearly with distance, as shown in
Vx = 100
fig. P2.1.34. The diffusion current density is found to be
tesla. The Hall voltage is
J n = 0.19 A cm 2 . If the electron diffusion coefficient is
(A) -0.31mV
(B) 0.31 mV
Dn = 25 cm 2 s, The electron concentration at is
(C) 3.26 mV
(D) -3.26 mV
14
Statement for Q.38-39:
n(cm-3)
5´10
A silicon Hall device at T = 300 K has the
geometry d = 10 -3 cm , W = 10 -2 cm, L = 10 -1 cm. The
following parameters are measured: I x = 0.75 mA,
n(0)
0
mV, and the magnetic flux is Bz = 5 ´ 10 -2
0.010
Vx = 15 V, V H = +5.8 mV, tesla
x(cm)
Fig. P2.1.34
38. The majority carrier concentration is
(A) 4.86 ´ 10 cm
8
-3
(C) 9.8 ´ 10 26 cm -3
-3
(A) 8 ´ 1015 cm -3, n - type
(D) 5.4 ´ 1015 cm -3
(B) 8 ´ 1015 cm -3, p - type
(B) 2.5 ´ 10 cm
13
35. The hole concentration in p - type GaAs is given by
xö
æ
p = 10 ç 1 - ÷ cm -3 for 0 £ x £ L
L
è
ø
(C) 4 ´ 1015 cm -3, n - type
(D) 4 ´ 1015 cm -3, p - type
16
39. The majority carrier mobility is
where L = 10 mm. The hole diffusion coefficient is
10 cm s. The hole diffusion current density at x = 5 mm
2
(A) 430 cm 2 V - s
(B) 215 cm 2 V - s
(C) 390 cm 2 V - s
(D) 195 cm 2 V - s
is
(A) 20 A cm 2
(B) 16 A cm 2
40. In a semiconductor n0 = 1015 cm -3 and ni = 1010 cm -3.
(C) 24 A cm 2
(D) 30 A cm 2
The excess-carrier life time is 10 -6 s. The excess hole
36. For a particular semiconductor sample consider
following parameters:
Page
112
concentration is dp = 4 ´ 1013 cm -3. The electron-hole
recombination rate is
(A) 4 ´ 1019 cm -3s -1
(B) 4 ´ 1014 cm -3s -1
(C) 4 ´ 10 24 cm -3s -1
(D) 4 ´ 1011 cm -3s -1
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Semiconductor Physics
41. A semiconductor in thermal equilibrium, has a hole
concentration
cm -3
p0 = 1016
of
and
an
SOLUTIONS
intrinsic
-3
concentration of ni = 10 cm . The minority carrier life
10
time
4 ´ 10 -7s.
is
The
thermal
equilibrium
recombination rate of electrons is
(A) 2.5 ´ 10 22 cm -3 s -1
(B) 5 ´ 1010 cm -3 s -1
(C) 2.5 ´ 1010 cm -3 s -1
(D) 5 ´ 10 22 cm -3 s -1
n-type
1. (C) ni2 = N c N v e
For Ge at 300 K,
. ´ 1019, N v = 6.0 ´ 1018 , E g = 0.66 eV
N c = 104
silicon
sample
contains
a
donor
concentration of N d = 106 cm -3. The minority carrier
hole lifetime is t p 0 = 10 m s.
-3
Þ
(A) 5 ´ 10 cm s
-1
-3
-1
3
-3
-1
(B) 10 cm s
-3
(C) 2. 25 ´ 10 cm s
9
4
-1
(D) 10 cm s
ni = 8.5 ´ 1014 cm -3
2. (C) n = N c N v e
2
i
electron is
T e
(A) 1125
.
´ 10 cm s
(C) 8.9 ´ 10
44.
-10
-3
cm s
A n -type
-1
-3
(B) 2.25 ´ 10 cm s
9
-1
sample
concentration of N d = 10
16
contains
a
-13´10 3
T
= 9.28 ´ 10 -8 , By trial T = 382 K
-1
(D) 4 ´ 10 9 cm -3 s -1
silicon
æ 1 .12 e ö
÷÷
kT ø
æ T ö - ççè
(1012 ) 2 = 2.8 ´ 1019 ´ 104
. ´ 1019ç
÷ e
è 300 ø
3
-3
æ -Eg ö
÷
- çç
÷
è kT ø
3
43. The thermal equilibrium generation rate for
9
æ 0 .66 ö
3
÷÷
- çç
æ 400 ö
è 0 .0345 ø
ni2 = 104
. ´ 1019 ´ 6.0 ´ 1018 ´ ç
÷ ´e
è 300 ø
42. The thermal equilibrium generation rate of hole is
8
æ -Eg ö
÷
- çç
÷
è kT ø
æ 400 ö
Vt = 0.0259ç
÷ = 0.0345
è 300 ø
Statement for Q.42-43:
A
Chap 2.1
-
donor
E gA
æ E - E gB ö
ç gA
÷
÷
kT
ø
-ç
n2
e kT
3. (B) iA
= E gB = e è
2
niB
e kT
= 2257.5
Þ
niA
= 47.5
niB
-3
cm . The minority carrier
hole lifetime is t p 0 = 20 m s. The lifetime of the majority
carrier is ( ni = 15
. ´ 1010 cm -3)
(A) 8.9 ´ 106 s
(B) 8.9 ´ 10 -6 s
(C) 4.5 ´ 10 -17 s
(D) 113
. ´ 10 -7 s
concentration are N a = 10
cm
-3
and N a = 0. The
equilibrium recombination rate is Rp 0 = 1011 cm -3s -1 . A
uniform generation rate produces an excess- carrier
-3
concentration of dn = dp = 10 cm . The factor, by which
14
the total recombination rate increase is
ni2 (15
. ´ 1010 ) 2
=
= 4.5 ´ 1015 cm -3
n0
5 ´ 10 4
5. (A) p0 = N v e
45. In a silicon semiconductor material the doping
16
4. (A) p0 =
6. (B) p0 = N v e
-
-
( EF - Ev )
kT
( EF - Ev )
kT
-0 .22
= 104
. ´ 1019 e 0 .0259 = 2 ´ 1015 cm -3
Þ
At 300 K, N v = 10
. ´ 1019 cm -3
æ 104
. ´ 1019 ö
÷÷ = 0. 239 eV
EF - Ev = 0.0259 lnçç
1015
ø
è
( Ec - EF )
(A) 2.3 ´ 1013
(B) 4.4 ´ 1013
n0 = N c e
(C) 2.3 ´ 10 9
(D) 4.4 ´ 10 9
At 300 K, N c = 2.8 ´ 1019 cm -3
***********
æN ö
EF - Ev = kT ln çç v ÷÷
è p0 ø
kT
Ec - EF = 112
. - 0. 239 = 0.881 eV
n0 = 4.4 ´ 10 4 cm -3
2
7. (C) p0 =
Na - Nd
N ö
æ
+ ç N a - d ÷ + ni2
2
2 ø
è
For Ge ni = 2.4 ´ 10 3
2
æ 1013 ö
1013
÷÷ + (2.4 ´ 1013) 2 = 2.95 ´ 1013 cm -3
p0 =
+ çç
2
è 2 ø
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Semiconductor Physics
=
(1.6 ´ 10
Þ n0 =
26. (A) J = evn = (1.6 ´ 10 -19)(10 7)(1016 ) = 1.6 ´ 10 4 A cm 2
0.1
= 11.36 kW
)(1100)(5 ´ 1016 )(100 ´ 10 -8 )
-19
L
,
sA
20. (A) R =
s » e m n n0 ,
R=
27. (A) vd =
L
em n n0 A
et sc E (1.6 ´ 10 -19)(10 -13)(10 5)
=
(0.067)(9.1 ´ 10 -31 )
me*
= 26.2 ´ 10 3 m s = 2.6 ´ 106 cm s
L
em n AR
28. (A) N d >> ni
n0 = 0.9 N d
=
Chap 2.1
-4
20 ´ 10
= 8.7 ´ 1015 cm -3
(0.9)(1.6 ´ 10 -19)( 8000)(10 -6 )(2 ´ 10 3)
21. (B) s » e m n n0 , R =
L
L
, n0 =
sA
em n AR
10 ´ 10 -4
= 7.81 ´ 1017 cm -3
=
(1.6 ´ 10 -19)( 800)(10 -6 )(10)
Þ
J = em n n0 E = (1.6 ´ 10
29. (D) ni2 = N c N v e
)(7500)(1016 )(10) = 120 A cm 2
æ -Eg ö
÷
- çç
÷
è kT ø
= (2 ´ 1019)(1 ´ 1019) e
Þ
n0 = N d
-19
æ 1 .1 ö
÷÷
- çç
è 0 .0259 ø
= 7.18 ´ 1019
ni = 8.47 ´ 10 9 cm -3
N d >> ni
Þ
N d = n0
Efficiency =
n0
7.8 ´ 1017
´ 100 =
´ 100 = 78.1 %
Nd
1018
J = sE = em n n0 E
22. (D) E =
V 6
= = 3 V/cm, vd = m n E,
L 2
30. (A) s = em n n0 + em p p0 and n0 =
vd 10 4
=
= 3333 cm 2 V - s
E
3
mn =
= (1.6 ´ 10 -19)(1000)(1014 )(100) = 1.6 A cm 2
Þ
10 -6 = (1.6 ´ 10 -19)(1000 + 600)ni
At T = 300 K, ni = 391
. ´ 10 cm
9
ni2 = N c N v e
ni2
+ em p p0 ,
p0
ds
( -1) em n ni2
=0 =
+ em p
dp0
p02
23. (D) s1 = eni (m n + m p )
æ Eg ö
÷
- çç
÷
è kT ø
s = em n
ni2
p0
1
-3
æN N ö
Þ E g = kT ln çç c 2 v ÷÷
è ni ø
Þ
æm
p0 = ni ç n
çm
è p
1
ö2
2
÷ = 3.6 ´ 1012 æç 7500 ö÷
÷
300
è
ø
ø
= 7. 2 ´ 1011 cm -3
ö
æ 1019
eV
÷ = 1122
Þ E g = 2(0.0259) ln çç
.
9 ÷
391
.
´
10
ø
è
31. (B) smin =
æ 500 ö
At T = 500 K , kT = 0.0259ç
÷ = 0.0432 eV,
è 300 ø
= 2 ´ 1.6 ´ 10 -19( 3.6 ´ 1012 ) (7500)( 300)
ni2 = (1019) 2 e
Þ
æ 1 .122 ö
÷÷
- çç
è 0 .0432 ø
32. (B) s =
= (1.6 ´ 10 -19)(2.29 ´ 1013)(1000 + 600)
= 5.86 ´ 10 -3(W - cm) -1
24. (B) r =
Nd =
1
1
=
s em n N d
1
1
= 9.25 ´ 1014 cm -3
=
5(1.6 ´ 10 -19)(1350)
rem n
25. (B) J n = eDn
dn
dx
æ 1018 - 1016
= (1.6 ´ 10 -19)( 35)çç
-4
è 2 ´ 10
mp + mn
= 2 en i m pm n
= 17
. ´ 10 -3(W - cm) -1
cm -3,
ni = 2.29 ´ 1013 cm -3
2 s i m pm n
1
= emni ,
r
Eg
1
ni1 e 2 kT1
r1
=
=
Eg
1
ni 2
2 kT2
e
r2
0.1 = e
-
Eg æ 1
1 ö÷
ç
2 k çè T1
T2 ÷ø
E g æ 330 - 300 ö
ç
÷ = ln 10
2 k çè 330 ´ 300 ÷ø
E g = 22( k300) ln 10 = 1.31 eV
ö
÷÷ = 2.8 ´ 10 4 A cm 2
ø
33. (D)
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1 1
1
1
=
+
+
m m1 m 2 m 3
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CHAPTER
2.2
THE PN JUNCTION
In this chapter, N d and N a denotes the net donor
5. The Fermi level on p - side is
p-region.
(A) 0.2 eV
(C) 0.4 eV
1. An abrupt silicon in thermal equilibrium at T = 300
Statement for Q.6–8:
and acceptor concentration in the individual n and
K is doped such that Ec - EF = 0.21 eV in the n - region
(B) 0.1 eV
(D) 0.3 eV
A silicon pn junction at T = 300 K with zero
and EF - Ev = 0.18 eV in the p - region. The built-in
applied bias has doping concentrations of N d = 5 ´ 1016
potential barrier Vbi is
cm -3 and N a = 5 ´ 1015 cm -3.
(A) 0.69 V
(C) 0.61 V
(B) 0.83 V
(D) 0.88 V
6. The width of depletion region extending into the
2. A silicon pn junction at T = 300 K has N d = 1014
n-region is
cm and N a = 10 cm . The built-in voltage is
(A) 4 ´ 10 -6 cm
(C) 4 ´ 10 -5 cm
(A) 0.63 V
(C) 0.026 V
7. The space charge width is
-3
17
-3
(B) 0.93 V
(D) 0.038 V
3. In a uniformly doped GaAs junction at T = 300 K, at
zero bias, only 20% of the total space charge region is to
be in the p-region. The built in potential barrier is
Vbi = 1.20 V. The majority carrier concentration in
n-region is
(A) 1 ´ 1016 cm -3
(C) 1 ´ 10
22
cm
-3
(B) 4 ´ 10 16 cm -3
(D) 4 ´ 10
22
cm
(B) 3 ´ 10 -6 cm
(D)3 ´ 10 -5 cm
(A) 32
. ´ 10 -5 cm
(C) 4.5 ´ 10 -4 cm
(B) 4.5 ´ 10 -5 cm
(D) 32
. ´ 10 -4 cm
8. In depletion region maximum electric field | Emax | is
(A) 1 ´ 10 4 V cm
(C) 3 ´ 10 4 V cm
(B) 2 ´ 10 4 V cm
(D) 4 ´ 10 4 V cm
9. An n – n isotype doping profile is shown in fig. P2.2.9.
-3
The built-in potential barrier is
(ni = 15
. ´ 1010 cm -3 )
Nd(cm-3)
Statement for Q.4–5:
10
An abrupt silicon pn junction at zero bias and
16
15
10
T = 300 K has dopant concentration of N a = 1017 cm -3
and N d = 5 ´ 1015 cm -3.
0
4. The Fermi level on n - side is
(A) 0.1 eV
(B) 0.2 eV
(C) 0.3 eV
(D) 0.4 eV
Fig. P2.2.9
(A) 0.66 V
(C) 0.03 V
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(B) 0.06 V
(D) 0.33 V
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UNIT 2
Electronics Devices
concentration are N d = 4 ´ 1016 cm -3 and N a = 4 ´ 1017
Statement for Q.10–11:
A silicon abrupt junction has dopant concentration
cm -3. The magnitude of the reverse bias voltage is
N a = 2 ´ 1016 cm -3 and N d = 2 ´ 10 15 cm -3. The applied
(A) 3.6 V
(B) 9.8 V
reverse bias voltage is VR = 8 V.
(C) 7.2 V
(D) 12.3 V
10. The maximum electric field | Emax | in depletion region is
17. An abrupt silicon pn junction has an applied
(A) 15 ´ 10 V cm
(B) 7 ´ 10 V cm
reverse bias voltage of VR = 10 V. it has dopant
(C) 35
. ´ 10 4 V cm
(D) 5 ´ 10 4 V cm
concentration of N a = 1018 cm -3 and N d = 1015 cm -3. The
4
4
pn junction area is 6 ´ 10 -4 cm 2 . An inductance of 2.2
11. The space charge region is
mH is placed in parallel with the pn junction. The
(A) 2.5 mm
(B) 25 mm
resonant frequency is
(C) 50 mm
(D) 100 mm
(A) 1.7 MHz
(B) 2.6 MHz
(C) 3.6 MHz
(D) 4.3 MHz
12.
A uniformly
doped
silicon
pn junction
has
N a = 5 ´ 1017 cm -3 and N d = 10 17 cm -3. The junction has
18. A uniformly doped silicon p+ n junction is to be
a cross-sectional area of 10 -4 cm -3 and has an applied
designed such that at a reverse bais voltage of V R = 10
reverse-bias voltage of VR = 5 V. The total junction
V the maximum electric field is limited to Emax = 106
capacitance is
V cm. The maximum doping concentration in the
(A) 10 pF
(B) 5 pF
n-region is
(C) 7 pF
(D) 3.5 pF
(A) 32
. ´ 1019 cm -3
(B) 32
. ´ 1017 cm -3
(C) 6.4 ´ 1017 cm -3
(D) 6.4 ´ 1019 cm -3
Statement for Q.13–14:
An ideal one-sided silicon n+ p junction has
19. A diode has reverse saturation current I s = 10 -10 A
uniform doping on both sides of the abrupt junction.
and non ideality factor h = 2. If diode voltage is 0.9 V,
The doping relation is N d = 50 N a . The built-in potential
then diode current is
barrier is Vbi = 0.75 V. The applied reverse bias voltage
(A) 11 mA
(B) 35 mA
is V R = 10.
(C) 83 mA
(D) 143 mA
13. The space charge width is
20. A diode has reverse saturation current I s = 10 -18 A
(A) 1.8 mm
(B) 1.8 mm
and nonideality factor h = 105
. . If diode has current of 70
(C) 1.8 cm
(D) 1.8 m
mA, then diode voltage is
14. The junction capacitance is
(A) 3.8 ´ 10 -9 F cm 2
(B) 9.8 ´ 10 -9 F cm 2
(C) 2.4 ´ 10 -9 F cm 2
(D) 5.7 ´ 10 -9 F cm 2
(A) 0.63 V
(B) 0.87 V
(C) 0.54 V
(D) 0.93 V
21. An ideal pn junction diode is operating in the
forward bais region. The change in diode voltage, that
+
15. Two p n silicon junction is reverse biased at VR = 5
will cause a factor of 9 increase in current, is
V. The impurity doping concentration in junction A are
(A) 83 mV
(B) 59 mV
(C) 43 mV
(D) 31 mV
N a = 10 cm
18
-3
and N d = 10
-15
-3
cm , and those in junction
B are N a = 1018 cm -3 and N d = 1016 cm -3. The ratio of the
22. An pn junction diode is operating in reverse bias
space charge width is
(A) 4.36
(B) 9.8
region. The applied reverse voltage, at which the ideal
(C) 19
(D) 3.13
reverse current reaches 90% of its reverse saturation
current, is
16. The maximum electric field in reverse-biased silicon
pn junction
Page
118
is
| Emax | = 3 ´ 10
5
V cm.
The
doping
(A) 59.6 mV
(B) 2.7 mV
(C) 4.8 mV
(D) 42.3 mV
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The pn Junction
p+ n junction diode the doping
23. For a silicon
cross-sectional area is 10 -3 cm 2 . The minority carrier
-3
cm .
lifetimes are t n 0 = 1 ms and t p 0 = 0.1 m s. The minority
The minority carrier hole diffusion coefficient is Dp = 12
carrier diffusion coefficients are Dn = 35 cm 2 s and
cm 2 s and the minority carrier hole life time is t p 0 = 10 -7
Dp = 10 cm 2 s. The total number of excess electron in
s. The cross sectional area is A = 10 -4 cm 2 . The reverse
the p - region, if applied forward bias is Va = 0.5 V, is
saturation current is
(A) 4 ´ 10 7 cm -3
(B) 6 ´ 1010 cm -3
(C) 4 ´ 1010 cm -3
(D) 6 ´ 10 7 cm -3
concentrations are N a = 10
18
(A) 4 ´ 10
-12
cm
-3
Chap 2.2
and N d = 10
16
(B) 4 ´ 10
A
(C) 4 ´ 10 -11 A
-15
A
(D) 4 ´ 10 -7 A
28. Two ideal pn junction have exactly the same
24. For an ideal silicon pn junction diode
electrical and physical parameters except for the band
gap of the semiconductor materials. The first has a
t no = t po = 10 -7 s ,
bandgap energy of 0.525 eV and a forward-bias current
Dn = 25 cm 2 s ,
of 10 mA with Va = 0.255 V. The second pn junction
Dp = 10 cm 2 s
diode is to be designed such that the diode current
The ratio of N a N d , so that 95% of the current in
I = 10 mA at a forward-bias voltage of Va = 0.32 V. The
the depletion region is carried by electrons, is
bandgap energy of second diode would be
(A) 0.34
(B) 0.034
(A) 0.77 eV
(B) 0.67 eV
(C) 0.83
(D) 0.083
(C) 0.57 eV
(D) 0.47 eV
Statement for Q.25–26:
29. A pn junction biased at Va = 0.72 V has DC bias
A ideal long silicon pn junction diode is shown in
current I DQ = 2 mA. The minority carrier lifetime is 1 ms
fig. P2.2.25–26. The n - region is doped with 10
is both the n and p regions. The diffusion capacitance is
organic atoms per cm 3 and the p - region is doped with
in
5 ´ 10
cm . The minority carrier
(A) 49.3 nF
(B) 38.7 nF
lifetimes are D n = 23 cm s and Dp = 8 cm s. The
(C) 77.4 nF
(D) 98.6 nF
16
16
3
boron atoms per
2
2
forward-bias voltage is Va = 0.61 V.
30. A p+ n silicon diode is forward biased at a current of
1 mA. The hole life time in the n - region is 0.1 ms.
W
Neglecting
p
Va
n
x=0
x
the
diode
(A) 38.7 + j12.1 W
(B) 235
. + j7.5 W
(C) 38.7 - j12.1 mW
(D) 235
. - j7.5 W
forward-bias current of a p+ n diode is 2.5 ´ 10 -6 F A.
(A) 6.8 ´ 1012 e -246 x cm -3, x ³ 0
The hole lifetime is
(B) 6.8 ´ 1012 e -246 x cm -3, x ³ 0
(C) 3.8 ´ 1014 e -3534 x cm -3, x ³ 0
(D) 3.8 ´ 10 e
capacitance
31. The slope of the diffusion capacitance verses
25. The excess hole concentration is
+3534 x
depletion
impedance at 1 MHz is
Fig. P.2.2.25-26
14
the
(A) 1.3 ´ 10 -7 s
(B) 1.3 ´ 10 -4 s
(C) 6.5 ´ 10 -8 s
(D) 6.5 ´ 10 -4 s
-3
cm , x ³ 0
32. A silicon pn junction with doping profile of N a = 1016
26. The hole diffusion current density at x = 3 mm is
(A) 0.6 A cm 2
(B) 0.6 ´ 10 -3 A cm 2
(C) 0.4 A cm 2
(D) 0.4 ´ 10 -3 A cm 2
cm -3 and N d = 10 15 cm -3 has a cross sectional area of
10 -2 cm 2 . The length of the p - region is 2 mm and
length of the n - region is 1 mm. The approximately
series resistance of the diode is
27. The doping concentrations of a silicon pn junction
are
N d = 10
16
cm
-3
and
N a = 8 ´ 10
15
-3
cm .
The
(A) 62 W
(B) 43 W
(C) 72 W
(D) 81 W
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UNIT 2
Electronics Devices
33. A gallium arsenide pn junction is operating in
39. A GaAs laser has a threshold density of 500 A cm 2 .
reverse-bias voltage VR = 5 V. The doping profile are
The laser has dimensions of 10 mm ´ 200 mm. The active
N a = N d = 10
16
-3
cm . The minority carrier life- time are
t p 0 = t n 0 = t 0 = 10
-8
s. The reverse-biased generation
(e r = 131
. , ni = 1.8 ´ 10 )
6
current density is
(A) 19
. ´ 10
-8
(C) 1.4 ´ 10
-8
A cm
(B) 19
. ´ 10
2
A cm
-9
(D) 1.4 ´ 10
2
-9
A cm
region is dLas = 100 A °. The electron-hole recombination
time at threshold is 1.5 ns. The current density of 5J th
is injected into the laser. The optical power emitted, if
2
emitted photons have an energy of 1.43 eV, is
2
(A) 143 mW
(B) 71.5 mW
(C) 62.3 mW
(D) 124.6 mW
A cm
34. For silicon the critical electric field at breakdown is
approximately Ecrit = 4 ´ 10 5 V cm. For the breakdown
voltage
of 25
V,
the
maximum
n - type
doping
+
concentration in an abrupt p n-junction is
(A) 2 ´ 1016 cm -3
(B) 4 ´ 10 16 cm -3
(C) 2 ´ 1018 cm -3
(D) 4 ´ 10 18 cm -3
35. A uniformly doped silicon pn junction has dopant
profile of N a = N d = 5 ´ 1016 cm -3. If the peak electric
field in the junction at breakdown is E = 4 ´ 10 5 V cm,
the breakdown voltage of this junction is
(A) 35 V
(B) 30 V
(C) 25 V
(D) 20 V
36. An abrupt silicon p+ n junction has an n - region
doping
concentration
n - region
minimum
of
N d = 5 ´ 10 15
width,
such
that
cm -3.
The
avalanche
breakdown occurs before the depletion region reaches
an ohmic contact, is (VB » 100 V)
(A) 5.1 mm
(B) 3.6 mm
(C) 7.3 mm
(D) 6.4 mm
37. A silicon pn junction diode has doping profile
N a = N d = 5 ´ 1019 cm -3. The space charge width at a
forward bias voltage of Va = 0.4 V is
(A) 102 A°
(B) 44 A°
(C) 153 A°
(D) 62 A°
38. A
GaAs
pn+
junction LED has following
parameters
Dn = 25 cm 2 s, Dp = 12 cm 2 s
N d = 5 ´ 1017 cm -3, N a = 1016 cm -3
t n 0 = 10 ns , t p 0 = 10 ns
The injection efficiency of the LED is
Page
120
(A) 0.83
(B) 0.99
(C) 0.64
(D) 0.46
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UNIT 2
= 301
. ´ 10 -5 cm,
CT =
-14
eA 117
. ´ 8.85 ´ 10 ´ 10
=
W
301
. ´ 10 -5
-4
= 35
. ´ 10 -12 F
ö
÷÷
ø
=
2
eEmax
2e
æ 1
1 ö
÷÷
çç
+
è Na Nd ø
(117
. ´ 8.85 ´ 10 -14 )( 3 ´ 10 5) 2 æ
1
1
ö
V
+
ç
16
17 ÷
2 ´ 1.6 ´ 10 -19
4
´
10
4
´
10
è
ø
= 8.008 V
1
ì 2 e ( V + VR ) é 1
1 ùü 2
W = í s bi
+
ê
úý ,
e
ë Na Na û þ
î
VR = 8.008 - 0.826 = 7.18 V
æN N ö
17. (B) Vbi = Vt ln çç a 2 d ÷÷
è ni ø
1
Þ
1
Vbi + VR =
N a = 4. 2 ´ 1015 cm -3, N d = 2.1 ´ 1017 cm -3
Nd > > Na
æ 4 ´ 1016 ´ 4 ´ 10 17 ö
÷÷ = 0.826 V
= 0.0259 ln çç
2. 25 ´ 10 20
ø
è
é2 e( Vbi + VR )
Na Nd ù 2
| Emax | = ê
e
( N a + N d ) úû
ë
æN N ö
13. (A) Vbi = Vt ln çç a 2 d ÷÷
è ni ø
æ 50 N a2
0.751 = 0.0259 ln çç
20
è 2.25 ´ 10
Electronics Devices
é2 e( Vbi + VR ) æ 1 öù 2
÷÷ú
çç
W » ê
e
è N a øû
ë
1
2
é2 ´ (117
. ´ 8.85 ´ 10 -4 )(10.752) ù
-4
=ê
ú = 1.8 ´ 10 cm
1.6 ´ 10 -19 ´ 4.2 ´ 1015
ë
û
= 1.8 mm
æ 1018 ´ 1015
= 0.0259 ln çç
20
è 2.25 ´ 10
ö
÷÷ = 0.754 V
ø
1
é
ù2
eeN a N d
C¢ = ê
ú
ë2 ( Vi + VR )( N a + N d ) û
1
é
ù
ee N a N d
14. (D) C¢ = ê
ú
ë2( Vbi + VR )( N a + N d ) û
é ee N a ù
For N d >> N a , C¢ = ê
ú
ë2( Vbi + VR ) û
é1.6 ´ 10
=ê
ë
-19
´ 117
. ´ 8.85 ´ 10
2 (10 + 0.754)
-4
é
ù2
eeN d
For N a >> N d , C¢ = ê
ú
ë2 ( Vbi + VR ) û
1
2
1
é1.6 ´ 10 -19 ´ 117
. ´ 8.85 ´ 10 -4 ´ 1015 ù 2
=ê
ú
2 (10 + 0.754)
ë
û
1
2
´ 4.2 ´ 10 ù
ú
û
15
= 5.7 ´ 10 -9 F cm 2
1
2
= 2.77 ´ 10 -9 F cm 2
C = AC¢ = 6 ´ 10 -4 ´ 2.77 ´ 10 -9 = 1.66 ´ 10 -12 F
1
1
fo =
=
= 2.6 MHz.
2p LC
2 p 2. 2 ´ 10 -3 ´ 1.66 ´ 10 -12
1
ì 2 e ( V + VR ) é 1
1 ùü 2
15. (D) W = í s bi
+
êN
úý
e
ë a Na û þ
î
1
W A é( Vbia + V R ) ( N aA + N dA ) N aB N dB ù 2
=
WB êë ( Vbib + R ) ( N aB + N dB ) N aA N dB úû
æN N ö
Vbi = Vt ln çç a 2 d ÷÷
è ni ø
æ 1018 ´ 1015
= 0.0259 ln çç
20
è 2.25 ´ 10
ö
÷÷ = 0.754 V
ø
æ 1018 ´ 1016
VbiB = 0.0259 ln çç
20
è 2.25 ´ 10
ö
÷÷ = 0.814 V
ø
W A éæ 5.754 öæ 1018 + 1015
= êç
֍
WB ëè 5.814 øçè 10 18 + 1016
öæ 1016 öù 2
.
÷÷çç 15 ÷÷ú = 313
øè 10 øû
VbiA
1
æN N ö
16. (C) Vbi = Vt ln çç a 2 d ÷÷
è ni ø
Page
122
18. (B) | Emax | =
eN a xn
e
é2e( Vbi + VR ) ù
For a p+ n junction, xn » ê
ú
eN d
ë
û
1
é2 eN d
ù2
So that | Emax | = ê
( Vbi + VR ) ú
ë es
û
Assuming Vbi << VR ,
Nd =
2
eEmax
(117
. ´ 8.85 ´ 10 -14 )(10 6 ) 2
=
2 eVR
2(1.6 ´ 10 -14 )(10)
= 3. 24 ´ 1017 cm -3
0 .9
æ VD
ö
19. (B) I D = I s ç e hVt - 1 ÷ = 10 -10 ( e 2 ´( 0 .0259) - 1 ) = 35 mA
ç
÷
è
ø
æ VD
ö
20. (B) I D = I s ç e hVt - 1 ÷
ç
÷
è
ø
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Þ
æ
I ö
VD = hVt ln çç 1 + D ÷÷
Is ø
è
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The pn Junction
æ
70 ´ 10 -6
= (105
. )(0.0259) ln çç 1 +
10 -18
è
21. (B) I d » I s e
æ Vö
÷
çç V ÷
è
tø
æI
V1 - V2 = Vt ln çç d 2
è I d1
ö
÷÷ = 0.87 V
ø
-
V1
Vt
I d1 e
= V2 = e
I d2
e Vt
,
é çæç eVa ÷ö÷ ù
27. (A) N p = ALn np 0 êe è kT ø -1ú
êë
úû
Þ
ni2 (15
. ´ 1010 ) 2
=
= 2.8 ´ 10 4 cm -3
Na
8 ´ 1015
np 0 =
( V1 - V2 )
Vt
Ln = Dn t no = 35 ´ 10 -6 = 5.9 ´ 10 -3 Cm
ö
÷÷ = 0.0259 ln 10 = 59.6 mV
ø
æ VV
ö
22. (A) I = I s ç e t - 1 ÷
ç
÷
è
ø
Chap 2.2
é çæç 0 .5 ÷ö÷
ù
N p = 10 -3 ´ 5.9 ´ 10 -3 ´ 2.8 ´ 10 4 êe è 0 .0259 ø - 1ú
êë
úû
ö
æ I
V = Vt ln çç + 1 ÷÷
I
ø
è s
= 4 ´ 10 7 cm -3
æ -Eg ö
ç
÷
ç V ÷
t ø
æ Va ö
ç ÷
æ Va ö
ç ÷
çV ÷
è tø
I
= -0.90 (–ive due to reverse current)
Is
28. (A) I µ ni2 e çè Vt ÷ø µ e è
V = 0.0259 ln (1 - 0.9) = -59.6 mV
Þ
1
23. (B) I s = Aen
Nd
ø
t
( Va1 - Va2 - E g1 + E g2 )
I1 e è
= æ V - E ö = e Vt
a2
g2 ÷
ç
I2
ç
÷
Vt
ø
eè
=
(10 -4 )(1.6 ´ 10 -19)(15
. ´ 10 10 ) 2
1016
1
J n = en
Na
2
i
Dn
N
Dn + a
Nd
Þ
Dn
,
t no
10 = e
1
J p = en
Nd
= 0.95,
Dp
10 ´ 10 -3
=e
10 ´ 10 -6
3
2
i
t po
5
= 0.95
Na
5+
10
Nd
æ
é æçç eVa ö÷÷
ù é çç - Lx
= pn 0 êe è kt ø - 1ú êe è p
êë
úû ê
ë
ö
÷
÷
ø
cm
ö
÷÷
ø
t n 0 = t p 0 = 10 -6 s,
I p 0 + I n 0 = I dQ = 2 mA
Cd =
Z=
2 Vt
Vt
=
=
10 -3
= 3.86 ´ 10 -2 S
0.0259
10 -3 ´ 10 -7
= 193
. ´ 10 -9 F
2 ´ (0.0259)
1
1
=
= 235
. - j7.5 W
Y g d + jwCd
tp 0
æ 1 ö
÷÷( I pot po),
Cd = çç
è 2 Vt ø
-3
Þ
-4
)
=
2 Vt
= 2.5 ´ 10 -6
t p 0 = 2 ´ 0.0259 ´ 2.5 ´ 10 -6 = 1.3 ´ 10 -7 s
32. (C) RP =
x = 3 mm = 3 ´ 10 -4 cm
J p = (1.6 ´ 10 -19)(18)( 3.8 ´ 1014 )( 3534) e - ( 3534 )( 3´10
I dQt p 0
I dQ
31. (A) For a p+ n diode I p 0 >> I n 0
¶ ( dpn )
26. (A) J p = - eDp
= eDp ( 3.8 ´ 1014 )( 3534) e -3534 x
¶x
= 0.6 A cm 2
æ E g2 - 0 .59 ö
ç
÷
ç 0 .0259 ÷
è
ø
30. (D) g d =
ù
ú
ú
û
æ
ö
é æç 0 .61 ö÷
ùé ç- x ÷ ù
dpn = 2.25 ´ 10 4 êe çè 0 .0259 ÷ø - 1ú êe çè 2 .83´10 -4 ÷ø ú
ë
ûë
û
= 3.8 ´ 10 e
( 0 .0259)
æ I p 0 t p 0 + I n 0 tn 0
29. (B) Cd = çç
2 Vt
è
Lp = Dp t p 0 = ( 8)(1 ´ 10 -8 ) = 2.83 ´ 10 -4 cm
-3534 x
( 0 .255- 0 .32 - 0 .525+ E g2 )
2 ´ 10 -3 ´ 10 -6
= 3.86 ´ 10 -8 F
Cd =
2( 0.0259)
ni2 (15
. ´ 1010 ) 2
=
= 2. 25 ´ 10 4 cm -3
Nd
1016
14
1
E g 2 = 0.59 + 0.0259 ln 10 3 = 0.769 EV
Dp
Na
= 0.083
Nd
25. (C) dpn = pn - pn 0
pn 0 =
12
= 394
. ´ 10 -15 A
10 -7
Jn
= 0.95,
Jn + Jp
24. (D)
æ V -Eg ö
ç a
÷
ç V ÷
ø
t
I µ eè
æ V a - E g1 ö
ç
÷
ç
÷
V
D
t po
2
i
e
rp L
A
=
L
A( em p N a )
0.2
= 26 W
(10 -2 )(1.6 ´ 10 -19)( 480)(10 16 )
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UNIT 2
Rn =
=
rn L
L
=
A
Ae(m n N d )
æ 5 ´ 1019
= 2(0.0259) ln çç
. ´ 1010
è 15
0.10
= 46.3 W
(10 )(1.6 ´ 10 -19)(1350)(1015)
1
R = Rp + Rn = 72.3 W
1
é2 ´ (117
. ´ 8.85 ´ 10 -14 )(114
. - 0.4) æ
2
öù 2
=ê
ç
-14
19 ÷ú
1.6 ´ 10
è 5 ´ 10 øû
ë
æN N ö
33. (B) Vbi = Vt ln çç a 2 d ÷÷
è ni ø
æ 10
= 2(0.0259) ln çç
6
è 1.8 ´ 10
= 6.19 ´ 10 -7 cm = 62 A°
ö
÷÷ =1.16 V
ø
38. (B) Ln = Dn t n 0 , Lp = Dp t p 0
1
ì 2 e ( V + VR )
W = í s bi
e
î
Dn np 0
é 1
1 ùü 2
ê N + N úý
ë a
a ûþ
é2 ´ (131
. ´ 8.85 ´ 10 -14 )( 6.16) æ 2 öù
-4
=ê
ç 16 ÷ú = 1.34 ´ 10 cm
-19
1
.
6
´
10
10
è
øû
ë
=
eniW
2t o
1.6 ´ 10 -19 ´ 1.8 ´ 106 ´ 1.34 ´ 10 -4
= 193
. ´ 10 -9 A cm 2
2 ´ 10 -8
34. (A) VB =
25 =
Dn np 0
Ln
35. (D) Emax =
eN d xn
e
=
np 0
Lp
Dp
Dn
+ pn 0
tn 0
tp 0
ni2 (1.8 ´ 106 ) 2
=
= 324
. ´ 10 -4 cm -3
Na
1016
pn 0 =
ni2 (1.8 ´ 106 ) 2
=
= 6.48 ´ 106 cm -3
Nd
5 ´ 1017
Dn
=
tn 0
25
= 5 ´ 10 4 ,
10 ´ 10 -9
Dp
12
= 35
. ´ 10 4
10 ´ 10 -9
tp 0
=
(5 ´ 10 4 )( 3. 24 ´ 10 -4 )
(5 ´ 10 )( 3. 24 ´ 10 -4 ) + ( 35
. ´ 10 4 )( 6.48 ´ 10 -6 )
hinj =
N B = N d = 2 ´ 1016 cm -3
+
Dp pn 0
Dn
tn 0
np 0 =
2
eEcrit
2 eN B
(117
. ´ 8.85 ´ 10 -4 )( 4 ´ 10 5) 2
2 ´ 1.6 ´ 10 -19 ´ N B
np 0
Ln
hinj =
1
2
J gen =
ö
÷÷ = 114
. V
ø
ì 2 e ( V + VR ) é 1
1 ùü 2
W = í s bi
+
êN
úý
e
ë a Na û þ
î
-3
16
Electronics Devices
4
= 0.986
Þ
xn =
eEmax
eN d
39. (B) The areal density at threshold is
n2 D =
(117
. ´ 8.85 ´ 10 -14 )( 4 ´ 10 5)
=
= 5.18 ´ 10 -5 cm
(1.6 ´ 10 -19)(5 ´ 1016 )
æN N ö
æ 5 ´ 106
Vbi = Vt ln çç a 2 d ÷÷ = 2(0.0259) ln çç
. ´ 1010
è 15
è ni ø
ö
÷÷ = 0.778 V
ø
1
J th t r (500)(15
. ´ 10 -9)
=
= 4.69 ´ 1012 cm -3
e
1.6 ´ 10 -19
The carrier density is
nth =
n2 D 4.69 ´ 1012
=
= 4.69 ´ 1018 cm -3
dLas
10 -6
ì 2 e V æ N öæ
öü 2
1
÷ý
xn = í s bi çç a ÷÷çç
÷
î e è N d øè N a + N d øþ
Once the threshold is reached, the carrier density does
é2(117
. ´ 8.85 ´ 10 -4 )( Vbi + VR ) ù
(5.18 ´ 10 ) = ê
-19
6
ú
ë (1.6 ´ 10 )(2 ´ 5 ´ 10 )
û
15
. ´ 10 -9
J th
t r ( J th ) =
= 3 ´ 10 -10 s
5
J
JA
The optical power produced is p =
hw
e
not
Vbi + VR = 20.7,
VR = 19.9 V,
xn
J > J th
the
VR = VB
1
é2 eVB ù 2 é2(117
. ´ 8.85 ´ 10 -14 )(100) ù 2
» ê
=
-19
15
ú ê
ú = 5.1 mm
ë eN d û ë (1.6 ´ 10 )(5 ´ 10 ) û
=
(5 ´ 500)(2 ´ 10 -5)(1.43 ´ 1.6 ´ 10 -19)
= 715
. MW
1.6 ´ 10 -19
****************
æN N ö
37. (D) Vbi = Vt ln çç a 2 d ÷÷
è ni ø
Page
124
electron
tr ( J ) =
36. (A) For a p+ n diode, Neglecting Vi compared to VB ,
1
When
recombination is
-5 2
Þ
change.
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CHAPTER
2.3
THE BIPOLAR JUNCTION TRANSISTOR
Statement for Q.1-2:
5. A uniformly doped npn bipolar transistor is biased in
The parameters in the base region of an npn
bipolar transistor are as follows Dn = 20 cm 2 s, nB0 = 10 4
cm -3, xB = 1 mm, ABE = 10 -4 cm 2 .
the
forward-active
region.
The
transistor
doping
concentration are N E = 5 ´ 1017 cm -3, N B = 10 16 cm -3 and
N C = 1015 cm -3. The minority carrier concentration pE0 ,
nB0 and pC0 are
1. If VBE = 0.5 V, then collector current I C is
(A) 4.5 ´ 10 2 , 2. 25 ´ 10 4 , 2. 25 ´ 10 5 cm -3
(A) 7.75 mA
(B) 1.6 mA
(B) 2. 25 ´ 10 4 , 2. 25 ´ 10 5, 4.5 ´ 10 2 cm -3
(C) 0.16 mA
(D) 77.5 mA
(C) 2. 25 ´ 10 4 , 2. 25 ´ 10 5, 4.5 ´ 10 4 cm -3
(D) 4.5 ´ 10 4 , 2.25 ´ 10 4 , 2. 25 ´ 10 5 cm -3
2. If VBE = 0.7 V, then collector current I C is
(A) 418 mA
(B) 210 mA
(C) 17.5 mA
(D) 98 mA
6. A uniformly doped silicon pnp transistor is biased in
the forward-active mode. The doping profile is N E = 10 18
cm -3, N B = 5 ´ 1016 cm -3 and N C = 1015 cm -3. For VEB = 0.6
V, the pB at x = 0 is
(See fig. P2.3.7-8)
3. In bipolar transistor biased in the forward-active
(A) 5.2 ´ 10 cm
-3
(B) 5.2 ´ 10 13 cm -3
region the base current is I B = 50 mA and the collector
(C) 5.2 ´ 1016 cm -3
(D) 5.2 ´ 10 11 cm -3
currents is I C = 2.7 mA. The a is
(A) 0.949
(B) 54
(C) 0.982
(D) 0.018
19
Statement for Q.7-8:
An npn bipolar transistor having uniform doping
of N E = 10 18 cm -3 N B = 1016 cm -3 and N C = 6 ´ 10 15 cm -3
4. A uniformly doped silicon npn bipolar transistor is to
be biased in the forward active mode with the B-C
junction reverse biased by 3 V. The transistor doping
is operating in the inverse-active mode with VBE = - 2 V
and VBC = 0.6 V. The geometry of transistor is shown in
fig P2.3.7-8.
Emitter
-n-
are N E = 1017 cm -3, N B = 1016 cm -3 and N C = 10 15 cm -3.
Base
-p-
Collector
-n-
The BE voltage, at which the minority carrier electron
concentration at x = 0 is 10% of the majority carrier hole
xE
xB
xC
concentration, is
(A) 0.94 V
(C) 0.48 V
(B) 0.64 V
x' = xE
(D) 0.24 V
x'=0
x'
x=0
x
x = xB x'' = 0
x'' = xC
x''
Fig. P2.3.7-8
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UNIT 2
Electronics Devices
7. The minority carrier concentration at x = x B is
N E = 1018 cm -3, N B = 5 ´ 1016 cm -3,
(A) 4.5 ´ 1014 cm -3
(B) 2.6 ´ 10 12 cm -3
N C = 2 ´ 1019 cm -3,
(C) 2.6 ´ 1014 cm -3
(D) 39
. ´ 10 14 cm -3
DE = 8 cm 2 s , DB = 15 cm 2 s , DC = 14 cm 2 s
xE = 0.8 mm, xB = 0.7 mm
8. The minority carrier concentration at x¢¢ = 0 is
(A) 39
. ´ 1014 cm -3
(C) 2.7 ´ 10
14
cm
(B) 2.7 ´ 10 12 cm -3
-3
(D) 4.5 ´ 10
14
cm
The emitter injection efficiency g is
-3
(A) 0.999
(B) 0.977
(C) 0.982
(D) 0.934
9. An pnp bipolar transistor has uniform doping of
N E = 6 ´ 1017 cm -3, N B = 2 ´ 10 16 cm -3 and N C = 5 ´ 1014
-3
15. A uniformly doped silicon epitaxial npn bipolar
cm . The transistor is operating is inverse-active mode.
transistor
The maximum V CB voltage, so that the low injection
N B = 3 ´ 10
condition applies, is
with N C = 5 ´ 10
fabricated
cm
-3
(A) 0.86 V
(B) 0.48 V
xB = 0.7 mm
(C) 0.32 V
(D) 0.60 V
punch-through is
Statement for Q.10-12:
The
following
currents
are
measured
in
with
a
base
doping
of
and a heavily doped collector region
cm -3. The neutral base width is
17
when
VBE = VBC = 0.
The
(A) 26.3 V
(B) 18.3 V
(C) 12.2 V
(D) 6.3 V
VBC
at
a
16. A silicon npn transistor has a doping concentration
uniformly doped npn bipolar transistor:
of
I nE = 120
. mA, I pE = 0.10 mA, I nC = 118
. mA
N B = 1017
cm -3
and
N C = 7 ´ 10 15
cm -3.
The
metallurgical base width is 0.5 mm. Let VBE = 0.6 V.
I R = 0.20 mA, I G = 1 mA, I pC0 = 1 mA
Neglecting the B–E junction depletion width the VCE at
punch-through is
10. The a is
(A) 0.667
(B) 0.733
(C) 0.787
(D) 0.8
(A) 146 V
(B) 70 V
(C) 295 V
(D) 204 V
17. A uniformly doped silicon pnp transistor is to
11. The b is
(A) 3.69
(B) 0.44
(C) 2.27
(D) 8.39
designed with N E = 1019 cm -3 and N C = 10 16 cm -3. The
metallurgical base width is to be 0.75 mm. The
minimum
base
doping,
so
that
the
minimum
punch-through voltage is Vpt = 25 V, is
12. The g is
(A) 0.816
(B) 0.923
(C) 1.083
(D) 0.440
(A) 4.46 ´ 1015 cm -3
(B) 4.46 ´ 10 16 cm -3
(C) 195
. ´ 1015 cm -3
(D) 195
. ´ 1016 cm -3
13. A silicon npn bipolar transistor has doping
18. For a silicon npn transistor assume the following
concentration of N E = 2 ´ 1018 cm -3, N B = 10 17 cm -3 and
parameters:
N C = 15
. ´ 1016 cm -3. The area is 10 -3 cm 2 and neutral
I E = 0.5 mA, b = 48
base width is 1 mm. The transistor is biased in the active
xB = 0.7 mA, xdc = 2 mm
region at VBE = 0.5 V. The collector current is
Cs = Cm = 0.08 pF, C je = 0.8 pF
(DB = 20 cm 2 s)
Page
126
is
16
(A) 9 mA
(B) 17 mA
(C) 22 mA
(D) 11 mA
Dn = 25 cm 2 s, rc = 30 W
The carrier cross the space charge region at a
speed of 10 7 cm s. The total delay time t ec is
14. A uniformly doped npn bipolar transistor has
(A) 164.2 ps
(B) 234.4 ps
following parameters:
(C) 144.2 ps
(D) 298.4 ps
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The Bipolar Junction Transistor
19. In a bipolar transistor, the base transit time is 25%
Statement for Q.24-26:
of the total delay time. The base width is 0.5 mm and
base diffusion coefficient is DB = 20 cm 2 s. The cut-off
Chap 2.3
For the transistor in circuit of fig. P2.3.24-26. The
parameters are bR = 1 , bF = 100 , and I s = 1 fA .
frequency is
(A) 637 MHz
(B) 436 MHz
(C) 12.8 GHz
(D) 46.3 GHz
5V
20. The base transit time of a bipolar transistor is 100
ps and carriers cross the 1.2 mm B–C space charge at a
speed of 10 7 cm s . The emitter-base junction charging
time is 25 ps and the collector capacitance and
resistance are 0.10 pF and 10 W, respectively. The cutoff
frequency is
(A) 43.8 GHz
(B) 32.6 GHz
(C) 3.26 GHz
(D) 1.15 GHz
Fig. P2.3.24-26
24. The current I C is
(A) 1 fA
(C) 1.384 fA
(B) 2 fA
(D) 0 A
25. The current I E is
Statement for Q.21-22:
Consider the circuit shown in fig. P2.3.21-22. If
voltage Vs = 0.63 V, the currents are I C = 275 mA and
I B = 5 mA .
(A) 1 fA
(B) -1 fA
(C) 2 fA
(D) -2 fA
26. The current I B is
(A) 2 fA
(B) -2 fA
(C) 1 fA
(D) -1 fA
27. For the transistor in fig. P2.3.27, I S = 10 -15 A,
bF = 100 , bR = 1. The current I CBO is
Vs
5V
Fig.P2.3.21-22
21. The forward common-emitter gain bF is
(A) 56
(B) 55
(C) 0.9821
(D) 0.9818
Fig.P2.3.27
(A) 101
. ´ 10 -14 A
(C) 101
. ´ 10 -15 A
22. The forward current gain a F is
(A) 0.9821
(B) 0.9818
(C) 55
(D) 56
(B) 2 ´ 10 -14 A
(D) 2 ´ 10 -15 A
Statement for Q.28-31:
Determine
23. Consider the circuit shown in fig P2.3.23. If Vs = 0.63
V, I1 = 275 mA and I 2 = 125 mA, then the value of I 3 is
the
region
of
operation
for
the
transistor shown in circuit in question.
28.
I1
6V
I2
I3
Vs
Fig. P2.3.23
(A) - 400 mA
(B) 400 mA
(C) - 600 mA
(D) 600 mA
Fig.P2.3.28
(A) Forward-Active
(C) Saturation
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(B) Reverse-Active
(D) Cutoff
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127
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GATE EC BY RK Kanodia
UNIT 2
29.
Electronics Devices
33. The current I1 is
(A) -12.75 mA
(B) 12.75 mA
(C) 12.5 mA
(D) - 12.5 mA
Statement for Q.34–35:
6V
The leakage current of a transistor are I CBO = 5 m
A and I CEO = 0.4 mA, and I B = 30 mA.
Fig. P2.3.29
(A) Forward-Active
(B) Reverse-Active
(C) Saturation
(D) Cutoff
34. The value of b is
30.
(A) 79
(B) 81
(C) 80
(D) None of the above
35. The value of I C is
(A) 2.4 mA
(B) 2.77 mA
(C) 2.34 mA
(D) 1.97 mA
6V
Statement for Q.36–37:
Fig.P2.3.30
(A) Forward-Active
(B) Reverse-Active
(C) Saturation
(D) Cutoff
For a BJT, I C = 5 mA, I B = 50 mA and I CBO = 0.5 mA.
36. The value of b is
31.
3V
(A) 103
(B) 91
(C) 83
(D) 51
37. The value of I E is
6V
(A) 5.25 mA
(B) 5.4 mA
(C) 5.65 mA
(D) 5.1 mA
Fig.P2.3.31
(A) Forward-Active
(B) Reverse-Active
(C) Saturation
(D) Cutoff
********
Statement for Q.32-33:
For the circuit shown in fig. P2.3.32-33, let the
value of b R = 0.5 and bF = 50. The saturation current is
10 -16 A.
+3 V
250 mA
I1
Fig. P2.3.32-33
32. The base-emitter voltage is
Page
128
(A) 0.53 V
(B) 0.7 V
(C) 0.84 V
(D) 0.98 V
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GATE EC BY RK Kanodia
The Bipolar Junction Transistor
SOLUTIONS
1. (A) I C = I s e
æ
çç
ö
÷÷
0 .5
2. (C) I C = 32
. ´ 10
-14
e
æ 0 .7 ö
çç
÷÷
è 0 .0259 ø
= 17.5 mA
pC (0) = 0.10 N C = 5 ´ 1013 cm -3,
ni2 (15
. ´ 1010 ) 2
=
= 4.5 ´ 10 5
NC
5 ´ 1014
ö
æ
ç VCB ÷
ç V ÷
t ø
pC (0) = pC 0 e è
10. (C) a =
ni2
(15
. ´ 1010 ) 2
=
= 2. 25 ´ 10 4 cm -3
NB
1016
æ VBE ö
÷
ç
ç V ÷
t ø
At x = 0, np (0) = np 0 e è
VBE
æ np (0) ö
÷
VBE = VT ln ç
ç n ÷
è p0 ø
=
æ
1015
= 0.0259 ln çç
4
è 2.25 ´ 10
ö
÷÷ = 0.635 V
ø
ni2 (15
. ´ 10 10 ) 2
=
= 450 cm -3
NE
5 ´ 1017
nB 0 =
ni2 (15
. ´ 10 10 ) 2
=
= 2. 25 ´ 10 4 cm -3
NB
5 ´ 1016
pC 0 =
ni2 (15
. ´ 10 10 ) 2
=
= 2. 25 ´ 10 5 cm -3
NE
5 ´ 1015
6. (B) pB 0 =
pB (0) = pB 0 e
7. (C) nB 0 =
ni2 (15
. ´ 1010 ) 2
=
= 4.5 ´ 10 3 cm -3
NB
5 ´ 1016
æ VEB
ç
ç V
è t
ö
÷
÷
ø
æ 0 .6 ö
÷÷
çç
3 è 0 .0259 ø
= 4.5 ´ 10 e
= 5. 2 ´ 1013 cm -3
ni2 (15
. ´ 10 10 ) 2
=
= 2. 25 ´ 10 4 cm -3
NB
1016
nB ( x = xB ) = nB 0 e
Þ
æ p (0) ö
VCB = Vt ln çç C ÷÷
è pC 0 ø
ö
æ
ç VBC ÷
ç V ÷
è t ø
0 .6
= 2. 25 ´ 10 4 e 0 .0259 = 2.6 ´ 10 14 cm -3
J nE
J nC
I nC
=
+ J R + J pE I nE + I R + I pE
118
.
= 0.787
12
. + 0.2 + 0.1
11. (A) b =
a
0.787
==
= 3.69
1-a
1 - 0.787
12. (B) g =
J nE
I nE
1. 2
=
=
= 0.923
J nE + J pE I nE + I pE 1. 2 + 0.1
10
1016
´ NB =
= 10 15
100
10
5. (A) pE 0 =
ö
÷÷
æ 5 ´ 1013 ö
÷ = 0.48 V
= 0.0259 ln çç
5÷
è 4.5 ´ 10 ø
IC
2.7m
=
= 0.982
I C + I B 2.7m + 50m
np (0) =
0 .6
9. (B) Low injection limit is reached when
pC 0 =
I
bF
3. (C) bF = C , a F =
IB
1 + bF
Þ
æ VBC ö
÷
ç
ç V ÷
t ø
æ
çç
I C = 3. 2 ´ 10 -14 e è 0 .0259 ø = 7.75 mA
4. (B) np 0 =
ni2 (15
. ´ 1010 ) 2
=
= 375
. ´ 10 4 cm -3
NC
6 ´ 1015
= 375
. ´ 10 4 e è 0 .0259 ø = 4.31 ´ 1014 cm -3
(1.6 ´ 10 -19)(20)(10 -4 )(10 4 )
= 32
. ´ 10 -14 A
10 -4
aF =
8. (D) pC 0 =
pC ( x¢¢ = 0) = pC 0 e è
æ VBE ö
÷
ç
ç V ÷
è b ø
eDn ABE nB 0
Is =
xB
=
Chap 2.3
13. (B) nB 0 =
ni2 (15
. ´ 10 10 ) 2
=
= 2.25 ´ 10 3 cm -3
NB
1017
æ VBE ö
÷
ç
ç V ÷
t ø
nB (0) = nB 0 e è
IC =
=
0 .5
ö
÷÷
eDB AnB (0)
xB
(1.6 ´ 10 -19)(20)(10 -3)(5.45 ´ 10 11 )
= 17.4 mA
10 -4
14. (B) g =
=
æ
çç
= 2.25 ´ 10 3 e è 0 .0259 ø = 5.45 ´ 1011 cm -3
1
N B D E xB
1+
×
×
N E D B xE
1
= 0.977
5 ´ 1016 8 0.7
1+
1018 15 0.8
æN N ö
15. (B) Vbi = Vt ln çç B 2 C ÷÷
è ni ø
æ 3 ´ 1016 ´ 5 ´ 1017
= 0.0259 ln çç
. ´ 1010 ) 2
è (15
ö
÷÷ = 0.824 V
ø
At punch-through
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The Bipolar Junction Transistor
ö
æ
æ VBE ö
ç VBC ÷ ö
æ çæç VBE ÷ö÷
ö
÷
ç
I æ çç ÷÷
27. (C) I E = I S ç e è Vt ø - e è Vt ø ÷ + S ç e è Vt ø - 1 ÷ = 0
ç
÷ bF ç
÷
è
ø
è
ø
Þ
ö
æ
ç VBE ÷
ç V ÷
t ø
eè
=
33. (A) I E = (bF + 1) I B = 12.75 mA
I1 = - I E = - 12.75 mA.
æ VBC ö
÷
Vt ÷ø
ç
ç
1
bR
+
eè
1 + bF 1 + BF
34. (A) I CEO = (b + 1) I CBO
b+1=
ö
æ
æ VBC ö
ç VBC ÷ ù
é æçç VBE ö÷÷
ù
ç V ÷
I é çç V ÷÷
I C = I S êe è Vt ø - e è t ø ú - S êe è t ø - 1ú
ê
ú bR ê
úû
ë
ë
û
I CBO =
IS é
ê1 - e
1 + bF ê
ë
æ VBC ö
÷
ç
ç V ÷
è t ø
ù I é
ú - S êe
úû bR êë
æ VBC
ç
ç V
è t
ö
÷
÷
ø
0.4m
= 80
5m
Þ
b = 79
35. (B) I C = bI B + I CEO = 79( 30m ) + 0.4m =2.77 mA
ù
- 1ú
úû
36. (A) I C = bI B + I CEO = b I B + (b + 1) I CBO
b=
VBC = - 5 V, Vt = 0.0259 V
I CBO =
Chap 2.3
Is
I
(1 - 0) - S (0 - 1) = 101
. I S = 101
. ´ 10 -15 A
101
1
I C - I CBO 5.2m - 0.5m
=
» 10396
.
I B + I CBO 50m + 0.5m
37. (A) a =
28. (D)
IE =
b
= 0.9904
b+1
I C - I CBO 5.2m - 0.5m
=
= 5.25 MA
a
0.9904
B-C Junction VBC
B-E junction VBE
Reverse Bias
Forward bias
Forward bias
Forward-Active
Saturation
Reverse Bias
Cut-off
Reverse-Active
*******
VBE = 0 , VBC < 0, Thus both junction are in reverse bias.
Hence cutoff region.
29.(A) VBE > 0 , VBC = 0, Base-Emitter junction forward
bais,
Base-collector
junction
reverse
bias,
Hence
forward-active region.
30. (B) VBE = 0 , VBC > 0, Base-Emitter junction reverse
bais , Base-collector junction forward bias, Hence
reverse-active region.
31. (C) VBE = 6 V, VBC = 3 V, Both junction are forward
biase, Hence saturation region.
32. (C) The current source will forward bias the
base-emitter junction and the collector base junction
will then be reverse biased. Therefore the transistor is
in the forward active region
IC = ISe
æ VBE ö
÷
ç
ç V ÷
è t ø
I C = bF I B = 50 ´ 250 ´ 10 -6 = 12.5 ´ 10 -3 A
æ 12.5 ´ 10 -3 ö
æI ö
VBE = Vt ln çç C ÷÷ = 0.0259 ln çç
-16
÷÷ = 0.84 V
ø
è 10
è IS ø
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UNIT 2
electronics Devices
11. In n-well CMOS fabrication substrate is
18. Monolithic integrated circuit system offer greater
(A) lightly doped n - type
reliability than discrete-component systems because
(B) lightly doped p - type
(A) there are fewer interconnections
(C) heavily doped n - type
(B) high-temperature metalizing is used
(D) heavily doped p - type
(C) electric voltage are low
(D) electric elements are closely matched
12. The chemical reaction involved in epitaxial growth
in IC chips takes place at a temperature of about
(A) 500° C
(B) 800° C
(C) 1200° C
(D) 2000° C
19. Silicon dioxide is used in integrated circuits
(A) because of its high heat conduction
(B) because it facilitates the penetration of diffusants
13. A single monolithic IC chip occupies area of about
(C) to control the location of diffusion and to protect
and insulate the silicon surface.
(A) 20 mm 2
(B) 200 mm 2
(D) to control the concentration of diffusants.
(C) 2000 mm 2
(D) 20,000 mm 2 s
20. Increasing the yield of an IC
14. Silicon dioxide layer is used in IC chips for
(A) reduces individual circuit cost
(A) providing mechanical strength to the chip
(B) increases the cost of each good circuit
(C) results in a lower number of good chips per wafer
(B) diffusing elements
(D) means that more transistor can be fabricated on
the same size wafer.
(C) providing contacts
(D) providing mask against diffusion
21. The main purpose of the metalization process is
15. The p-type substrate in a monolithic circuit should
be connected to
(A) to act as a heat sink
(B) to interconnect the various circuit elements
(A) any dc ground point
(C) to protect the chip from oxidation
(B) the most negative voltage available in the circuit
(D) to supply a bonding surface for mounting the chip
(C) the most positive voltage available in the circuit
22. In a monolithic-type IC
(D) no where, i.e. be floating
16. The collector-substrate junction in the epitaxial
(A) each transistor is diffused into a separate
isolation region
(B) all components are fabricated into a single crystal
of silicon
collector structure is, approximately
(A) a step-graded junction
(C) resistors and capacitors of any value may be
made
(B) a linearly graded junction
(C) an exponential junction
(D) all isolation problems are eliminated
(D) None of the above
23. Isolation in ICs is required
17. The sheet resistance of a semiconductor is
(A) to make it simpler to test circuits
(A) an important characteristic of a diffused region
especially when used to form diffused resistors
(B) to protect the transistor from possible ``thermal
run away’’
(B) an undesirable parasitic element
(C) to protect the components mechanical damage
(C) a characteristic whose value determines the
required area for a given value of integrated
capacitance
(D) to minimize electrical interaction between circuit
components
(D) a parameter whose value is important in a
thin-film resistance
Page
140
24. Almost all resistor are made in a monolithic IC
(A) during the base diffusion
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Integrated Circuits
Chapc2.5
(B) during the collector diffusion
30. For the circuit shown in fig. P2.5.30, the minimum
(C) during the emitter diffusion
number and the maximum number of isolation regions
(D) while growing the epitaxial layer
are respectively
25. The equation governing the diffusion of neutral
(A)
¶N
¶2 N
=D
¶t
¶x 2
(C)
¶ N
¶N
=D
2
¶t
¶x
R1
R2
atom is
(B)
¶N
¶2 N
=D
¶x
¶t 2
(D)
¶ N
¶N
=D
2
¶x
¶t
2
Vo2
Q1
Q2
2
Fig. P2.5.32
26. The true statement is
(A) thick film components are vacuum deposited
(B) thin film component are made by screen-and- fire
process
(A) 2, 6
(B) 3, 6
(C) 2, 4
(D) 3, 4
31. For the circuit shown in fig. P2.5.31, the minimum
number of isolation regions are
(C) thin film resistor have greater precision and are
more stable
(D) thin film resistor are cheaper than the thin film
resistor
27. The False statement is
(A) Capacitor of thin film capacitor made with proper
dielectric is not voltage dependent
(B) Thin film resistors and capacitor need to be
biased for isolation purpose
(C) Thin film resistors and capacitor have smaller
stray capacitances and leakage currents.
Fig. P2.5.31
(A) 2
(B) 3
(C) 4
(D) 7
(D) None of the above
*******
28. Consider the following two statements
S1 : The dielectric isolation method is superior to
junction isolation method.
S2 : The beam lead isolation method is inferior to
junction isolation method.
The true statements is (are)
(A) S1 , S2
(B) only
(C) only
(D) Neither nor S2
29. If P is passivation, Q is n-well implant, R is
metallization and S is source/drain diffusion, then the
order in which they are carried out in a standard n-well
CMOS fabrication process is
(A) S - R - Q - P
(B) R - P - S - Q
(C) Q - S - R - P
(D) P - Q - R - S
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UNIT 2
SOLUTIONS
1. (D)
2. (D)
3. (B)
4. (B)
5. (C)
6. (B)
7. (C)
8. (B)
9. (C)
10. (D)
11. (B)
12. (C)
13. (C)
14. (D)
15. (B)
16. (A)
17. (A)
18. (A)
19. (C)
20. (A)
21. (B)
22. (B)
23. (D)
24. (A)
25. (A)
26. (C)
27. (B)
28. (B)
29. (C)
30. (D) The minimum number of isolation region is 3
one containing Q1 , one containing and one containing
both and . The maximum number of isolation region is
4, or one per component.
31. (A) The minimum number of isolation region is two.
One for transistor and one for resistor.
*******
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142
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GATE EC BY RK Kanodia
CHAPTER
3.1
DIODE CIRCUITS
vo
Statement for Q.1–4:
vo
22
18
In the question a circuit and a waveform for the
t
input voltage is given. The diode in circuit has cutin
(A)
voltage Vg = 0. Choose the option for the waveform of
(B)
vo
output voltage vo .
vo
t
-3
(C)
vi
+
20
2.2 kW
3.
vo
5V
(D)
2 kW
t
_
t
-7
1.
vi
t
vi
+
16
-5
vo
vi
Fig.3.1.1
2
T
4V
20
_
15
t
t
-5
-10
(A)
(B)
Fig. P3.1.3
vo
vo
16
4
4
t
T
T
2
vo
20
(A)
20
(A)
2.
(D)
10 kW
T
T
2
-4
(D)
R
vi
+
D1
t
-5
t
4.
vi
vo
_
T
4
t
(C)
20
vi
T
2
t
16
12
2V
+
T
vo
5
t
T
2
(B)
vo
t
t
-16
12
vo
T
6V
vo
vi
8V
Fig.3.1.2
10
D2
_
t
-10
Fig.3.1.4
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UNIT 3
vo
Analog Electronics
vo
vo
vo
10
8
9.42
9.42
5.7
t
4.3
t
5.7
-6
15
vi
(C)
-10
(A)
vi
(D)
7. For the circuit shown in fig. P3.1.7, each diode has
(B)
vo
15
4.3
Vg = 0.7 V. The vo for -10 £ vs +10 V is
vo
+10 V
8
6
t
10 kW
t
-6
D1
-8
(C)
D2
vs
(D)
vo
5. For the circuit in fig. P.3.1.5, let cutin voltage Vg = 0.7
V. The plot of vo verses vi for -10 £ vi £ 10 V is
10 kW
D4
D3
10 kW
+
10 kW
20 kW
-10 V
vo
vi
Fig. P3.1.7.
10 V
10 V
vo
_
vo
8.43
7.48
Fig. P3.1.5
vo
vo
9.3
-10
10
9.3
3.33
3.33
-10
10
4.03
10
(B)
vo
vi
vo
5.68
vo
10
10
4.65
-10 -6.8
6.8 10
vs
-10
-5.68
4.65
-4.65
(C)
3.33
10
vi
4.33
-10
(C)
10
vi
(D)
clipping circuit shown in fig. P3.1.8. If diode has rf = 0
(D)
and rr = 2 MW and Vg = 0, the output waveform is
is Vg = 0.7 V. The plot of vo versus vi is
+
1 MW
+
vo
vi
1 kW
2.5 V
vo
1 kW
_
vi
15 V
_
Fig. P3.1.8
vo
Fig. P3.1.6
vo
vo
10
vo
19.6
19.6
t
-5
5
5.7
(A)
Page
146
15
(B)
vo
4.3
vi
15
4.3
vo
5
vi
(B)
t
-5
(C)
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t
-10
(A)
5.7
vs
varies between +10 V and -10 V is impressed upon the
6. For the circuit in fig. P3.1.6 the cutin voltage of diode
2 kW
10
4.65
8. A symmetrical 5 kHz square wave whose output
3.33
3.33
-10
vs
-7.48
(B)
vo
10
(A)
-10
(A)
-10
-8.43
3.33
vi
vs
t
(D)
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Diode Circuits
Chap 3.1
C
9. In the circuit of fig. P3.1.9, the three signals of fig are
vi
+
10
impressed on the input terminals. If diode are ideal
then the voltage vo is
vi
D1
v
D2
+
D3
+
v1
v2
+
vo
v3
10 kW
-
-
t
vo
R
5V
_
v3
v2
-20
Fig. P.3.1.11
v1
vo
t
35
vo
25
-
Fig. P.3.1.9
vo
5
t
vo
t
-5
(A)
(B)
vo
t
t
(A)
15
(B)
vo
(D) None of the above
t
vo
-15
(C)
t
t
(D)
12. In the circuit of fig. P3.1.12, D1 and D2 are ideal
diodes. The current i1 and i2 are
(C)
(D)
D1
10. For the circuit shown in fig. P3.1.10 the input
i1
voltage vi is as shown in fig. Assume the RC time
D2
i2
500 W
constant large and cutin voltage of diode Vg = 0. The
3V
5V
output voltage vo is
5V
C
vi
+
Fig. P3.1.12
10
vi
t
vo
R
-10
_
Fig. P.3.1.10
vo
vo
(A) 0, 4 mA
(B) 4 mA, 0
(C) 0, 8 mA
(D) 8 mA, 0
13. In the circuit of Fig. P3.1.13 diodes has cutin
20
voltage of 0.6 V. The diode in ON state are
10
t
t
(A)
D1
(B)
vo
12 W
6W
D2
vo
t
5.4 V
t
18 W
5V
-10
-20
(C)
Fig. P3.1.13
(D)
11. For the circuit shown in fig. P.3.1.11, the input
(A) only D1
(B) only D2
voltage vi is as shown in fig. Assume the RC time
(C) both D1 and D2
(D) None of the above
constant large and cutin voltage Vg = 0. The output
voltage vo is
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Diode Circuits
Chap 3.1
22. The diodes in the circuit in fig. P3.1.22 has
26. If v2 = 0, then output voltage vo is
parameters Vg = 0.6 V and rf = 0. The current iD2 is
(A) 6.43 V
(B) 9.43 V
(C) 7.69 V
(D) 8.93 V
+10 V
9.5 kW
27. If v2 = 5 V, then vo is
D2
0.5 kW
0V
iD2
vo
+5 V
D1
0.5 kW
D3
Fig. P3.1.22
(B) 10 mA
(C) 7.6 mA
(D) 0 mA
(B) 12.63 V
(C) 18.24 V
(D) 10.56 V
28. If v2 = 10 V, then vo is
+5 V
(A) 8.4 mA
(A) 8.93 V
(A) 10 V
(B) 9.16 V
(C) 8.43 V
(D) 12.13 V
Statement for Q.29–30:
The diode in the circuit of fig. P3.1.29–30 has the
Statement for Q.23–25:
non linear terminal characteristic as shown in fig. Let
The diodes in the circuit in fig. P3.1.23-25 have
linear parameter of Vg = 0.6 V and rf = 0.
the voltage be vs = cos wt V.
100 W
vi
9.5 kW
0.5 kW
D2
iD
4
+
vD
100 W
-
2V
vo
v2
~
iD(mA)
a
+10 V
b
v1
vD(V)
Fig. P3.1.29–30
D1
0.5 kW
0.5 0.7
Fig. P3.1.23–25
29. The current iD is
23. If v1 = 10 V and v2 = 0 V, then vo is
(A) 8.93 V
(B) 7.82 V
(C) 1.07 V
(D) 2.18 V
(A) 2.5(1 + cos wt) mA
(B) 5(0.5 + cos wt) mA
(C) 5(1 + cos wt) mA
(D) 5(1 + 0.5 cos wt) mA
30. The voltage vD is
24. If v1 = 10 V and v2 = 5 V, then vo is
(A) 9.13 V
(B) 0.842 V
(A) 0.25( 3 + cos wt) V
(C) 5.82 V
(D) 1.07 V
(C) 0.5( 3 + 1 cos wt) V
(B) 0.25(1 + 3 cos wt) V
(D) 0.5(2 + 3 cos wt) V
25. If v1 = v2 = 0, then output voltage vo is
31. The circuit inside the box in fig. P3.1.31. contains
(A) 0.964 V
(B) 1.07 V
only resistor and diodes. The terminal voltage vo is
(C) 10 V
(D) 0.842 V
connected to some point in the circuit inside the box.
The largest and smallest possible value of vo most
Statement for Q.26–28:
nearly to is respectively
+15 V
The diodes in the circuit of fig. P3.1.26–28 have
linear parameters of Vg = 0.6 V and rf = 0.
500 W
D2
-9 V
v2
Circuit Containing
Diode and Resistor
vo
vo
+10 V
500 W
D1
Fig. 3.1.31
9.5 kW
(A) 15 V, 6 V
(B) 24 V, 0 V
(C) 24 V, 6 V
(D) 15 V, -9 V
Fig. P3.1.26–28
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UNIT 3
32. In the voltage regulator circuit in fig. P3.1.32 the
Statement for Q.36–38:
maximum load current iL that can be drawn is
In the voltage regulator circuit in fig. P3.1.36–38
15 kW
the Zener diode current is to be limited to the range
iL
30 V
Analog Electronics
Vz = 9 V
Rz = 0
5 £ iz £ 100 mA.
12 W
RL
iL
iz
Fig. 3.1.32
(A) 1.4 mA
(B) 2.3 mA
(C) 1.8 mA
(D) 2.5 mA
power dissipation in the Zener diode is
150 W
Vz = 15 V
Rz = 0
Fig. P3.1.33
(B) 1.5 W
(C) 2 W
(D) 0.5 W
36. The range of possible load current is
(A) 5 £ iL £ 130 mA
(B) 25 £ iL £ 120 mA
(C) 10 £ iL £ 110 mA
(D) None of the above
37. The range of possible load resistance is
75 W
(A) 1 W
RL
Fig. P3.1.36–38
33. In the voltage regulator shown in fig. P3.1.33 the
50 V
Vz = 4.8 V
Rz = 0
6.3 V
(A) 60 £ RL £ 372 W
(B) 60 £ RL £ 200 W
(C) 40 £ RL £ 192 W
(D) 40 £ RL £ 360 W
38. The power rating required for the load resistor is
34. The Q-point for the Zener diode in fig. P3.1.34 is
(A) 576 mW
(B) 360 mW
(C) 480 mW
(D) 75 mW
11 kW
39. The secondary transformer voltage of the rectifier
20 V
Vz = 4 V
Rz = 0
circuit shown in fig. P3.1.39 is vs = 60 sin 2 p60 t V. Each
3.6 kW
diode has a cut in voltage of Vg = 0.6 V. The ripple
voltage is to be no more than Vrip = 2 V. The value of
filter capacitor will be
Fig. P3.1.34
(A) (0.34 mA, 4 V)
(B) (0.34 mA, 4.93 V)
(C) (0.94 mA, 4 V)
(D) (0.94 mA, 4.93 V)
+
vi
35. In the voltage regulator circuit in fig. P3.1.35 the
+ vo 10 kW
+
vs
-
power rating of Zener diode is 400 mW. The value of RL
C
-
that will establish maximum power in Zener diode is
Fig. P3.1.39
222 W
20 V
Vz = 10 V
Rz = 0
RL
(A) 48.8 mF
(B) 24.4 mF
(C) 32.2 mF
(D) 16.1 mF
40. The input to full-wave rectifier in fig. P3.1.40 is
Fig. P3.1.35
Page
150
(A) 5 kW
(B) 2 kW
(C) 10 kW
(D) 8 kW
vi = 120 sin 2 p60 t V. The diode cutin voltage is 0.7 V. If
the output voltage cannot drop below 100 V, the
required value of the capacitor is
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GATE EC BY RK Kanodia
Diode Circuits
+
1 : 1
vi
SOLUTIONS
+
+
vs
+
vs
-
2.5 kW
C
vo
-
1. (D) Diode is off for vi < 5 V. Hence vo = 5 V.
For vi > 5 V, vo = vi , Therefore (D) is correct option.
-
2. (C) Diode will be off if vi + 2 > 0.Thus vo = 0
Fig. P3.1.40
(A) 61.2 mF
(B) 41.2 mF
(C) 20.6 mF
(D) 30.6 mF
For vi + 2 < 0 V,
voltage is Vin = 0. The ripple voltage is to be no more
than vrip = 4 V. The minimum load resistance, that can
be connected to the output is
+
~
50 mF
vo
vi < - 2, vo = vi + 2 = -3 V
Thus (C) is correct option.
41. For the circuit shown in fig. P3.1.41 diode cutin
75sin 2p60t V
Chap 3.1
3. (D) For vi < 4 V the diode is ON and output vo = 4 V.
For vi > 4 V diode is off and output vo = vi .
Thus (D) is correct option.
4. (C) During positive cycle when vs < 8 V, both diode
are OFF vo = vs . For vs > 8 V , vo = 8 V, D1 is ON. During
RL
negative cycle when |vs | < 6 V, both diode are OFF,
-
vo = vs . For vs > 6 V, D2 is on vo = -6 V. Therefore (C) is
Fig. P3.1.41
correct.
(A) 6.25 kW
(B) 12.50 kW
(C) 30 kW
(D) None of the above
10 10
5. (B) For D off , vo = 20 20 = 3.33 V.
1
1
+
20 10
For vi £ 3.33 + 0.7 = 4.03 V, vo = 3.33 V
****************
For vi > 4.03 V , vo = vi - 0.7
For vi = 10 V, vo = 9.3 V
6. (C) Let v1 be the voltage at n-terminal of diode,
v1 =
15 ´ 1
=5 V
2 +1
For vi £ 5.7 V, vo = vi
v1 - 15 v1
v - vi
+
+ o
=0
2k
1k
1k
Þ
3v1 + 2 vo - 2 vi = 15
Þ
vo = 0.4 vi + 3.42
vo = v1 + 0.7
5 vo - 2 v1 = 15 + 2.1 = 17.9
7. (D) For vs > 0, when D1 is OFF, Current through D2 is
i=
10 - 0.7
= 0.465 mA, vo = 10 ki = 4.65 V
10 + 10
vo = vs for 0 < vs < 4.65 V.
For negative values of vs , the output is negative of
positive part. Thus (D) is correct option.
8. (B) The diode conducts (zero resistance) when vi < 2.5
V and vo = vi . Diode is open (2 MW resistance) when
v - 2.5
vi > 2.5 V and vo = 2.5 + i
= 5 V.
3
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UNIT 3
= 50 ´ 5 (1 + cos wt) ´ 10 -3 + 0.5
C=
= 0.75 + 0. 25 cos wt = 0. 25( 3 + cos wt) V
31. (D) The output voltage cannot exceed the positive
power supply voltage and cannot be lower than the
negative power supply voltage.
32. (A) At regulated power supply is =
vmax
2 fkVrip
RL =
30 - 9
= 1.4 mA iL
15 k
75(50)
50
V
=
75 + 150
3
50
> VZ , RTH = 150 ||75 = 50 W
3
1 æ 50
ö
iZ =
- 15 ÷ = 33 mA, P = 15 iZ = 0.5 W
ç
50 è 3
ø
3.6(20)
= 4.93 V > VZ ,
11 + 3.6
4.93 - 4
= 11 || 3.6 = 2.71 kW, iZ =
= 0.34 mA
2.71k
34. (A) vTH =
RTH
35. (B) iZ ( max ) =
iL + iZ =
400m
= 40 mA
10
20 - 10
= 45 mA
222
iL ( min ) = 45 - 40 = 5 mA, RL =
10
= 2 kW
5m
36. (B) Current through 12 W resistor is
6.3 - 4.8
i=
= 125 mA
12
iL = i - iZ = 125 - iZ
Þ
25 £ iL £ 120 mA
37. (C) 25 £ iL £ 120 mA, iL RL = 4.8 V
25 £
4.8
£ 120 mA
RL
Þ
40 £ RL £ 192 W
38. (A) PL = iL VZ = (120m)( 4.8) = 576 mV
39. (B) vs = 60 sin 2 p60 t V
vmax = 60 - 1.4 = 58.6 V
58.6
vmax
C=
=
= 24.4 mF
2 fRVrip
2( 60)10 ´ 10 3 ´ 2
40. (C) Full wave rectifier
vs = vi = 120 sin 2 p60 t V
vmax = 120 - 0.7 = 119.3 V
Vrip = 119.3 - 100 = 19.3 V
Page
154
=
41. (A) Vrip =
will remain less than 1.4 mA.
33. (D) vTH =
Analog Electronics
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119.3
= 20.6 mF
2( 60)2.5 ´ 10 3 ´ 14.4
vmax
fRL C
75
vmax
=
= 6.25 kW
fCVrip 60 ´ 50 ´ 10 -5 ´ 4
***************
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GATE EC BY RK Kanodia
CHAPTER
3.2
BASIC BJT CIRCUITS
VBE ( ON ) = 0.7
VCE ( Sat ) = 0.2
(A) 8.4 V
(B) 6.2 V
transistor if not given in problem.
(C) 4.1 V
(D) None of the above
Statement for Q.1-4:
3. I C , RC = ?
Use
V,
V
for
npn
+5 V
The common-emitter current gain of the transistor
is b =75. The voltage VBE in ON state is 0.7 V.
1. I E , RC = ?
RC
VC = 2 V
50 kW
+12 V
10 kW
IQ = 1 mA
+
VEC = 6 V
-
-5 V
Fig. P3.3.3
RC
(A) 0.987 mA, 3.04 kW
(B) 1.013 mA, 2.96 kW
-12 V
(D) 0.946 mA, 4.18 kW
Fig. P3.3.1
(A) 1.46 mA, 6.74 kW
(B) 0.987 mA, 3.04 kW
(C) 1.13 mA,, 5.98 kW
(D) None of the above
(D) 1.057 mA, 3.96 kW
4. VC = ?
2. VEC = ?
+5 V
+8 V
10 kW
20 kW
10 kW
VC
10 kW
-2 V
2 kW
3 kW
-8 V
Fig. P3.3.2
Fig. P3.3.4
(A) 1.49 V
(B) 2.9 V
(C) 1.78 V
(D) 2.3 V
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UNIT 3
Statement for Q.5-6:
Analog Electronics
9. VB = 1 V
In the circuit of fig.P3.3.5-6 VB = - 1 V
(A) 4 V
(C) 1 V
+3 V
(B) 3 V
(D) 1.9 V
10. VB = 2 V
500 kW
4.8 kW
(B) 1.5 V
(C) 2.6 V
(D) None of the above
Statement for Q.11-12:
-3 V
The transistor in circuit shown in fig. P3.3.11-12
has b = 200. Determine the value of voltage Vo for given
value of VBB .
Fig. P3.3.5-6
5. b = ?
(A) 103.4
(C) 134.5
(A) -7 V
(B) 135.5
(D) 102.4
+5 V
5 kW
6. VCE = ?
(A) 6.4 V
(C) 1.3 V
Vo
50 kW
(B) 4.7 V
(D) 4.2 V
10 kW
VBB
7. In the circuit shown in fig. P3.3.7 voltage VE = 4 V.
The value of a and b are respectively
+5 V
Fig. P3.3.11-12
11. VBB = 0
2 kW
(A) 2.46 V
(C) 3.33 V
100 kW
VE
(B) 1.83 V
(D) 4.04 V
12. VBB = 1 V
8 kW
(A) 4.11 V
(C) 2.46 V
(B) 1.83 V
(D) 3.44 V
-5 V
13. VBB = 2 V
Fig. P3.3.7
(A) 0.943, 17.54
(C) 0.914, 11.63
(B) 0.914, 17.54
(D) 0.914, 11.63
(A) 3.18 V
(C) 0.2 V
(B) 1.46 V
(D) None of the above
Statement for Q.14-16:
Statement for Q.8-10:
For the transistor in circuit shown in fig.
P3.3.8-10, b = 200. Determine the value of I E and I C for
given value of VB in question.
The transistor shown in the circuit of fig.
P3.3.14-16 has b = 150. Determine Vo for given value of
I Q in question.
+5 V
+6 V
5 kW
10 kW
Vo
VC
VB
IQ
1 kW
-5 V
Fig. P3.3.14-16
Fig. P3.3.8-10
14. I Q = 0.1 mA
8. VB = 0 V
(A) 6.43 mA, 2.4 V
(C) 0 A, 6 V
Page
156
(B) 2.18 mA, 3.4 V
(D) None of the above
(A) 1.4 V
(B) 4.5 V
(C) 3.2 V
(D) None of the above
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GATE EC BY RK Kanodia
Basic BJT Circuits
15. I Q = 0.5 mA
(A) 3.16 V
(B) 2.52 V
(C) 2.14 V
(D) 394
. V
Chap 3.2
(A) 0.991
(B) 0.939
(C) 0.968
(D) 0.914
20. For the transistor in fig. P3.3.20 , b = 50. The value
16. I Q = 2 mA
of voltage VEC is
(A) 4.9 V
(B) -4.9 V
(C) 0.5 V
(D) -0.5 V
+9 V
1 mA
17. For the circuit in fig. P3.3.17 VB = VC and b = 50. The
value of VB is
+6 V
50 kW
10 kW
4.7 kW
VC
-9 V
VB
Fig. P3.3.20
1 kW
Fig. P3.317
(A) 0.9 V
(B) 1.19 V
(C) 2.14 V
(D) 1.84 V
(A) 3.13 V
(B) 4.24 V
(C) 5.18 V
(D) 6.07 V
21. In the circuit shown in fig. P3.3.21 if b = 50, the
power dissipated in the transistor is
+9 V
18. For the circuit shown in fig. P3.3.18, VCB = 0.5 V and
b = 100. The value of I Q is
0.5 mA
+5 V
5 kW
Vo
50 kW
4.7 kW
-9 V
IQ
Fig. P3.3.21
-5 V
Fig. P3.3.18
(A) 1.68 mA
(B) 0.909 mA
(C) 0.134 mA
(D) None of the above
19. For the circuit shown in fig. P3.3.19 the emitter
voltage is VE = 2 V. The value of a is
(A) 3.87 mW
(B) 10.46 mW
(C) 7.49 mW
(D) 18.74 mW
22. For the circuit shown in fig. P3.3.22 the Q-point is
VCEQ = 12 V and I CQ = 2 A when b = 60. The value of
resistor RC and RB are
+24 V
+10 V
RC
RB
10 kW
VE
50 kW
10 kW
-10 V
Fig. P3.3.19
Fig. P3.3.22
(A) 10 kW, 241 kW
(B) 10 kW, 699 kW
(C) 6 kW, 699 kW
(D) 6 kW, 241 kW
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GATE EC BY RK Kanodia
Basic BJT Circuits
Chap 3.2
29. For the transistor in the circuit of fig. P3.3.29,
32. The current gain of the transistor shown in the
b = 100. The voltage VB is
circuit of fig.P3.3.32 is b = 100. The values of Q-point
( I CQ , VCEQ) is
+10 V
+5 V
20 kW
1 kW
5 kW
12 kW
15 kW
0.5 kW
2 kW
Fig. P3.3.29
-5 V
(A) 3.6 V
(B) 4.29 V
(C) 3.9 V
(D) 4.69 V
Fig. P3.3.32
30. The current gain of the transistor shown in the
(A) (1.8 mA, 2.1 V)
(B) (1.4 mA, 2.3 V)
(C) (1.4 mA , 1.8 V)
(D) (1.8 mA, 1.4 V)
circuit of fig. P3.3.30 is b = 125. The Q-point values
( I CQ , VCEQ) are
33. For the circuit in fig. P3.3.33, let b = 60. The value of
+24 V
VECQ is
+5 V
+10 V
58 kW
42 kW
10 kW
2 kW
20 kW
2.2 kW
10 kW
Fig. P3.3.30
-5 V
(A) (0.418 mA, 20.4 V)
(B) (0.915 mA, 14.8 V)
(C) (0 .915 mA, 16.23 V)
(D) (0.418 mA, 18.43 V)
31. For the circuit shown in fig. P3.3.31, let b = 75. The
-10 V
Fig.P3.3.33
(A) 2.68 V
(B) 4.94 V
(C) 3.73 V
(D) 5.69 V
34. In the circuit of fig. P3.3.34 Zener voltage is VZ = 5
Q-point (I CQ , VCEQ) is
+24 V
V and b = 100. The value of I CQ and VCEQ are
+12 V
25 kW
3 kW
500 W
8 kW
1 kW
Fig. P3.3.34
Fig. P3.3.31
(A) ( 4.68 mA, 16.46 V)
(B) ( 312
. mA , 1.86 V)
(C) ( 312
. mA, 8.46 V)
(D) ( 4.68 mA , 5.22 V)
(A) 12.47 mA, 4.3 V
(B) 12.47 mA, 5.7 V
(C) 10.43 A, 5.7 V
(D) 10.43 A , 4.3 V
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UNIT 3
35. The two transistor in fig. P3.2.35 are identical. If
b = 25, the current I C2 is
Analog Electronics
(A) 36.63 mA
(C) 49.32 mA
(B) 36.17 mA
(D) 49.78 mA
+5 V
IC2
39. In the bipolar current source of fig. P3.2.39 the
diode voltage and transistor BE voltage are equal. If
25 mA
base current is neglected then collector current is
Fig. P3.2.35
(A) 28 mA
(B) 23.2 mA
(C) 26 mA
(D) 24 mA
10 kW
36. In the shunt regulator of fig. P3.2.26, the VZ = 8.2 V
4.7 kW
and VBE = 0.7 V. The regulated output voltage Vo is
10 kW
120 W
Vo
+22 V
-20 V
Fig. P3.2.39
(A) 6.43 mA
(C) 1.48 mA
100 W
(B) 2.13 mA
(D) 9.19 mA
40. In the current mirror circuit of fig. P3.2.40. the
transistor parameters are VBE = 0.7 V, b = 50 and the
Early voltage is infinite.
Fig. P3.2.36
(A) 11.8 V
(B) 7.5 V
(C) 12.5 V
(D) 8.9 V
Assume transistor are
matched. The output current I o is
+5 V
37. In the series voltage regulator circuit of fig. P3.2.37
1 mA
VBE = 0.7 V, b = 50, VZ = 8.3 V. The output voltage Vo is
+25 V
Io
Vo
220 W
20 kW
50 kW
Fig. P3.2.40
50 kW
(A) 1.04 mA
(C) 962 mA
30 kW
(B) 1.68 mA
(D) 432 mA
41. All transistor in the N output mirror in fig. P3.2.41
are matched with a finite gain b and early voltage
Fig. P3.2.37
(A) 25 V
(B) 25.7 V
(C) 15 V
(D) 15.7 V
V A = ¥. The expression for each load current is
V
38. In the regulator circuit of fig. P3.2.38 VZ = 12 V,
b = 50, VBE = 0.7 V. The Zener current is
+20 V
+
Io1
Iref
Io2
Io3
R1
Vo
QS
220 W
QR
1 kW
Q1
V
Q2
-
Fig. P32.41
Fig. P3.2.38
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160
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QN
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GATE EC BY RK Kanodia
Basic BJT Circuits
(A)
(C)
I ref
(B)
æ
(1 + N ) ö
çç 1 +
÷
b
(b + 1) ÷ø
è
b I ref
(D)
æ
(1 + N ) ö
çç 1 +
÷
(b + 1) ÷ø
è
I ref
Chap 3.2
SOLUTIONS
æ
N ö
çç 1 +
÷
(
b
+ 1) ÷ø
è
1. (C) I E =
b I ref
æ
N ö
çç 1 +
÷
+ 1 ÷ø
b
è
12 - 0.7
10 k
Þ
. mA
I E = 113
æ 75 ö
I C = çç
. ) = 112
. mA
÷÷(113
è 75 + 1 ø
42. Consider the basic three transistor current source in
VCE = 12 - 1.13 ´ 10 - 1.12 RC - ( -12) = 6 V
fig. P3.2.42. Assume all transistor are matched with
RC = 5.98 kW
finite gain and early voltage V A = ¥. The expression for
I o is
V
2. (C) 8 = 10 ´ (75 + 1) I B + 0.7 + 10 I B - 2
+
IB =
Io
I C = bI B = 0.906 mA, I E = (b + 1) I B = 0.918 mA
R1
Iref
9.3
= 12.08 mA,
10 + 760
8 = 10(0.918) + VEC + 3(0.906) - 8
Þ
VEC = 4.1 V
75
æ 75 ö
3. (A) I C = ç
(1m) = 0.987 mA
÷ IE =
76
è 75 + 1 ø
V
-
RC =
Fig. P3.2.42
(A)
(C)
I ref
(B)
æ
2 ö
çç 1 +
÷
(1 + b) ÷ø
è
I ref
(D)
æ
ö
2
çç 1 +
÷÷
b
(
1
+
b
)
è
ø
I ref
æ
1 ö
çç 1 +
÷
(2 + b) ÷ø
è
4. (A) 5 = (1 + b)10 kI B + 20 kI B + 0.7 + b2 kI B
5 = (760 k + 20 k + 150 k) I B + 0.7
I ref
Þ
æ
ö
1
çç 1 +
÷÷
b
(
2
+
b
)
è
ø
VC = 5 - (b + 1) I B RC = 5 - 760 ´ 4.62 ´ 10 -3 = 1.49 V
Both of transistor are identical and b >> 1 and VBE1 = 0.7
V. The value of resistance R1 and RE to produce I ref = 1
mA and I o = 12 mA is ( Vt = 0.026)
5. (C) VB = - I B RB
Þ
IB =
-VB
1
=
= 2.0 mA
RB 500k
VE = -1 - 0.7 = - 17
. V
V - ( -3) -17
. +3
IE = E
=
= 0.271 mA
4.8 k
4.8 k
IE
0.271m
= (b + 1) =
IB
2m
+5 V
Iref
Io
Q1
I B = 4.62 mA,
I C = bI B = 0.347 mA
43. Consider the wilder current source of fig. P3.2.43.
R1
5.2
= 304
. kW
0.987m
Þ
Q2
b = 134.5
6. (B) VCE = 3 - VE = 3 - ( -17
. ) = 4.7 V
RE
-5 V
7. (C) I E =
Fig. P3.2.43
5.4
= 0.5 mA
2k
(A) 9.3 kW, 18.23 kW
(B) 9.3 kW , 9.58 kW
4 = 0.7 + I B RB + I C RC - 5, I C » I E ,
(C) 15.4 kW , 16.2 kW
(D) 15.4 kW , 32.4 kW
8.3 = 100 I B + 0.5 ´ 8
*****************
Þ
I B = 43 mA ,
IE
0.5m
=1 + b =
= 11.63
IB
43m
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UNIT 3
b = 10.63, a =
b
® a = 0.914
1+b
Vo = VCE ( sat ) - VBE = 0.2 - 0.7 = - 0.5 V
17. (B) I E =
8. (C) VB = 0 Transistor is in cut-off region.
VB - 0.7
1k
æ b ö
æ 50 ö
I C = çç
÷÷I E = ç
÷ ( VB - 0.7) mA
è 51 ø
èb + 1ø
I E = 0, VC = 6 V
9. (B) VB = 1 V , I E =
Analog Electronics
1 - 0.7
= 0.3 mA
1k
IC =
I C » I E = 0.3 mA
6 - VC
mA, VC = VB
10
50
6 - VB
( VB - 0.7) =
51
10
VC = 6 - I C RC = 6 - (0.3)(10) = 3 V
10.8 VB = 12.86 , VB = 119
. V
2 - 0.7
10. (B) VB = 2 V, I E =
= 1.3 mA,
1
18. (B) VCB = 0.5 V , VC = 0.5 V
I C » I E = 1.3 mA
VC = 6 - (1.3)(10) = - 7 V
Transistor is in saturation. The saturation voltage
IC =
5 - 0.5
æ 101 ö
=0.9 mA, I Q = ç
÷0.9 = 0.909 mA
5k
è 100 ø
VCE = 0.2 V
19. (C) I E =
11. (C) VBB = 0, Transistor is in cutoff region
VB = VE - 0.7 = 1.3 V
RL
10(5)
Vo =
VCC =
+ 5 = 3.33 V
RC + RL
10
IB =
12. (B) I B =
1 - 0.7
= 6 mA
50 k
I C = bI B = 75 ´ 6m = 0.45 mA
5 - Vo
V
= IC + o
5k
10 k
Vo Vo
(1 - 0.45) =
+
, Þ
Vo = 1.83 V
5
10
13. (C) I B =
2 - 0.7
= 26 mA
50 k
I C = b I B = 75 ´ 26 mA =1.95 mA
. = -4.75 V
VC = 5 - I C RC = 5 - 5 ´ 195
Transistor is in saturation, VCE = 0.2 V = VC = Vo
14. (B) I E = 0.1 mA
IC =
b
150
IE =
(0.1) = 0.099 mA
(b + 1)
151
VB
1.3
=
= 26 mA
RB 50 k
b+1=
a=
I E 0.8m
=
= 30.77
I B 26m
Þ
b = 29.77
b
29.77
=
= 0.968
b + 1 30.77
æ b
ö 50
20. (D) I C = çç
mA = 0.98 mA
I E ÷÷ =
b
+
1
è
ø 51
VC = I C RC - 9 = (0.98)( 4.7) - 9 = - 4.394 V
IB =
IE
1
=
mA = 19.6 mA
(b + 1) 51
VE = I B RB + VEB = 50(0.0196) + 0.7 = 1.68 V
VEC = 1.68 - ( -4.394) = 6.074 V
æ b ö
50
21. (A) I C = çç
(0.5) = 0.49 mA
÷÷I E =
b
+
1
51
ø
è
Vo = 5 - RC I C = 5 - 5(0.099) = 4.50 V
IB =
15. (B) I E = I Q = 0.5 mA
VE = I B RB + VEB = (0.0098)(50) + 0.7 = 1.19 V
æ 150 ö
IC = ç
÷ (0.5m) = 0.497 mA
è 150 + 1 ø
VC = I C RC - 9 = (0.49)( 4.7) - 9 = - 6.7 V
Vo = 5 - RC I C = 2.517 V
PQ = I C VEC + I B VEB
16. (D) Transistor is in saturation
Page
162
10 - VE
= 0.8 mA
10k
. V
VE = (1.3)(1) = 1.3 V , VC = VCE + VE = 15
0.5
= 9.8 mA
51
VEC = 119
. - ( -6.7) = 7.89 V
= (0.49)(7.89) + (0.0098)(0.7) mW = 3.87 mW
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UNIT 3
æ R2
VTH = çç
è R1 + R2
ö
æ 15 ö
÷÷VCC = ç
÷(10) = 4.29 V
è 20 + 15 ø
ø
10 = I E (1k) + VEB +
10 = I E + 0.7 +
Þ
IB =
IE
( 8.57 k) + 4.29
b+1
Analog Electronics
5.82 - 0.7 = ( 6.06 k + 76 k) I BQ
Þ
I BQ = 62.4 mA
I EQ = (b + 1) I BQ = 4.74 mA
I CQ = bI BQ = 4.68 mA
VCEQ = 24 - I CQ RC - I EQ RE
IE
( 8.57 k) + 4.29
101
= 24 - ( 4.68)( 3) - ( 4.74)(1) = 5.22 V
I E = 4.62 MA
32. (B) R1 = 12 kW, R2 = 2 kW
IE
= 0.046 mA
b+1
RTH = R1||R2 = 12 ||2 = 171
. kW
æ 2 ö
VTH = ç
. V
÷(10) - 5 = - 357
è 12 + 2 ø
VB = ( 8.57)(0.046) + 4.29 = 4.69 V
30. (B) R1 = 58 kW, R2 = 42 kW
+5 V
+24 V
5 kW
1.71 kW
-3.57 V
24.36 kW
+10.1 V
0.5 kW
10 kW
-5 V
Fig. S3.3.32
Fig. S3.3.30
-357
. = I BQ(171
. k) + VBE + (b + 1) I BQ(0.5 k) - 5
RTH = 58 ||42 = 24.36 kW
VTH
5 - 357
. - 0.7 = (171
. + 50.5)I BQ
æ 42 ö
=ç
÷(24) = 10.1 V
è 42 + 58 ø
Þ
10.1 = I BQ(24.36 k) + VBE + (b + 1) I BQ(10 k)
10.1 - 0.7 = I BQ(24.36 k + 1260 k)
I BQ = 14 mA
I EQ = (100 + 1) I BQ = 1.412 mA
I CQ = 100 I BQ = 1.4 mA
VCEQ = 5 - RC I CQ - RE I EQ + 5
I BQ = 7.32 mA
= 5 - (5)(1.4) - (0.5)(1.412) + 5 = 2.3 V
I CQ = bI BQ = 0.915 mA
I EQ = (b + 1) I BQ = 0.922 mA
33. (B) RTH = 20 ||10 = 6.67 kW
VCEQ = 24 - (0.922)(10) = 14.8 V
æ 20 ö
VTH = ç
÷10 - 5 = 1.67 V
è 10 + 20 ø
31. (D) R1 = 25 kW, R2 = 8 kW
+10 V
+24 V
2 kW
3 kW
6.67 kW
6.06 kW
+1.67 V
+5.82 V
2.2 kW
1 kW
-10 V
Fig. S3.3.33
Fig. S3.3.31
RTH = 25 ||8 = 6.06 kW, VTH
æ 8 ö
=ç
÷(24) = 5.82 V
è 25 + 8 ø
5.82 = ( 6.06 k)( I BQ) + VBE + (b + 1) I B (1k)
Page
164
10 = (1 + b) I BQ(2) + VEB + I BQ( 6.67) + 1.67
10 - 1.67 - 0.7 = I BQ( 6.67 + 122)
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UNIT 3
41. (A) I ref = I CR + I BS = I CR +
æ I ref
I o RE = Vt ln çç
è Io
I ES
1+b
I ES = I BR + I B1 + I B 2 + ....... I BN
I BR = I Bi , I CR = I Ci = I oi
(1 + N ) I CR
I ES = (1 + N ) I BR =
b
Then I ref = I CR +
æ 1 ´ 10 -3
0.026
ln çç
-6
-6
12 ´ 10
è 12 ´ 10
R1 =
V + - VBE1 - V - 5 - 7 - ( -5)
=
= 9.3 kW
I ref
1m
I ES
(1 + N ) I CR
= I CR +
b+1
b (b + 1)
+
V
Iref
Io = IC2
Q3
IC1
IE3
Q2
Q1
IB1 IB2
V
-
Fig. S3.2.42
I E 3 = (1 + b) I B 3
IE3
2 IB2
I ref = I C1 +
= I C1 +
(1 + b)
(1 + b)
I C1 = I C 2 = bI B 2
æ
ö
2 IC2
2
= I C2 çç 1 +
÷÷
b(1 + b)
b
(
1
+
b
)
è
ø
I ref
æ
ö
2
çç 1 +
÷÷
1
b
(
+
b
)
è
ø
43. (B) If b >> 1 and transistor are identical
VBE1
VBE2
I ref » I C1 = I S e
Vt
, Io = IC2 = IS e
Vt
æ I ref
VBE1 = Vt lnçç
è IS
ö
æI
÷÷ , VBE 2 = Vt lnçç o
è IS
ø
ö
÷÷
ø
æ I ref
VBE1 - VBE 2 = Vt ln çç
è Io
ö
÷÷
ø
From the circuit,
VBE1 - VBE 2 = I E 2 RE » I o RE
Page
166
ö
÷÷ = 9.58 kW
ø
***********
42. (C) I ref = I C1 + I B 3 , I B1 = I B 2 , I E 3 = 2 I B 2
IC2 = Io =
ö
÷÷
ø
RE =
æ
(1 + N ) ö
= I Oi çç 1 +
÷
b(b + 1) ÷ø
è
I ref
I oi =
æ
(1 + N ) ö
çç 1 +
÷
b(b + 1) ÷ø
è
I ref = I C 2 +
Analog Electronics
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GATE EC BY RK Kanodia
CHAPTER
3.3
BASIC FET CIRCUITS
Statement for Q.4–6:
Statement for Q.1–3:
In the circuit shown in fig. P3.3.1–3 the transistor
parameter are as follows:
parameters are as follows:
Threshold voltage
In the circuit shown in fig. P3.3.4–6 the transistor
VTN = 2 V
Conduction parameter K n = 0.5 mA / V
VTN = 2 V, k¢n = 60 mA / V 2 ,
2
W
= 60
L
+10 V
+10 V
32 kW
4 kW
14 kW
1.2 kW
18 kW
2 kW
6 kW
0.5 kW
-10 V
Fig. P3.3.1–3
Fig. P3.3.4–6
4. VGS = ?
1. VGS = ?
(A) 2.05 V
(B) 6.43 V
(A) -3.62 V
(B) 3.62 V
(C) 4.86 V
(D) 3.91 V
(C) -0.74 V
(D) 0.74 V
5. I D = ?
2. I D = ?
(A) 1.863 mA
(B) 1.485 mA
(C) 0.775 mA
(D) None of the above
(A) 13.5 mA
(B) 10 mA
(C) 19.24 mA
(D) 4.76 mA
6. VDS = ?
3. VDS =?
(A) 4.59 V
(B) 3.43 V
(A) 2.95 V
(B) 11.9 V
(C) 5.35 V
(D) 6.48 V
(C) 3 V
(D) 12.7 V
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Basic FET Circuits
Chap 3.3
16. The parameter of the transistor in fig. P3.3.16 are
19. In the circuit of fig. P.3.3.19 the PMOS transistor
VTN = 1.2 V, K n = 0.5 mA / V 2 and l = 0. The voltage VDS
has parameter VTP = -1.5 V, k¢p = 25 mA / V 2 , L = 4 mm
is
and l = 0. If I D = 0.1 mA and VSD = 2.5 V, then value of W
will be
+5 V
+9 V
50 mA
R
Fig. P3.3.16
(A) 1.69 V
(B) 1.52 V
(C) 1.84 V
(D) 0
Fig. P3.3.19
17. The parameter of the transistor in fig. P3.3.17 are
VTN = 0.6 V and K n = 0. 2 mA / V 2 . The voltage VS is
(A) 15 mm
(B) 1.6 mm
(C) 32 mm
(D) 3.2 mm
20. The PMOS transistor in fig. P3.3.20 has parameters
+9 V
VTP = -1.2 V,
24 kW
W
= 20, and k¢p = 30 mA / V 2 .
L
+5 V
Rs
0.25 mA
RD
-9 V
Fig. P3.3.17
-5 V
(A) 1.72 V
(B) -1.72 V
(C) 7.28 V
(D) -7.28 V
Fig. P3.3.20
If I D = 0.5
mA and VD = -3 V, then value of RS
18. In the circuit of fig. P3.3.18 the transistor
and RD are
parameters are VTN = 17
. V and K n = 0.4 mA / V 2 .
(A) 4 kW, 5.8 kW
(B) 4 kW, 5 kW
(C) 5.8 kW, 4 kW
(D) 5 kW, 4 kW
+5 V
21. The parameters for the transistor in circuit of fig.
RD
P3.3.21 are VTN = 2 V and K n = 0.2 mA / V 2 . The power
dissipated in the transistor is
+10 V
RS
50 kW
-5 V
Fig. P3.3.18
10 kW
If I D = 0.8 mA and VD = 1 V, then value of resistor
Fig. P3.3.21
RS and RD are respectively
(A) 2.36 kW, 5 kW
(B) 5 kW, 2.36 kW
(C) 6.43 kW, 8.4 kW
(D) 8.4 kW, 6.43 kW
(A) 5.84 mW
(B) 2.34 mW
(C) 0.26 mW
(D) 58.4 mW
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UNIT 3
Statement for Q.22–23:
Analog Electronics
25. The transistors in the circuit of fig. P3.3.25 have
Consider the circuit shown in fig. P3.2.22–33.
parameter VTN = 0.8 V, kn¢ = 40 mA / V 2 and l = 0. The
width-to-length ratio of M 2
+5 V
when Vi = 5 V, then
( )
W
L 1
is
( WL )2 = 1.
If Vo = 0.10 V
for M1 is
+5 V
M1
Vo
M1
M2
Vo
The both transistor have parameter as follows
VTN = 0.8 V,
M2
Vi
Fig. P3.3.22-23
Fig. P3.3.25
k¢n = 30 mA / V 2
22. If the width-to-length ratios of M1 and M 2 are
æW ö æW ö
ç ÷ = ç ÷ = 40
è L ø1 è L ø2
(A) 47.5
(B) 28.4
(C) 40.5
(D) 20.3
Statement for Q.26–27:
The output Vo is
All transistors in the circuit in fig. P3.3.26–27
(A) -2.5 V
(B) 2.5 V
(C) 5 V
(D) 0 V
have parameter VTN = 1 V and l = 0.
+5 V
æW ö
æW ö
23. If the ratio is ç ÷ = 40 and ç ÷ = 15, then Vo is
è L ø1
è L ø2
(A) 2.91 V
(B) 2.09 V
(C) 3.41 V
(D) 1.59 V
RD
M4
M1
ID1
24. In the circuit of fig. P3.324. the transistor
RG
ID4
M3
M2
parameters are VTN = 1 V and k¢n = 36 mA / V 2 . If I D = 0.5
mA, V1 = 5 V and V2 = 2 V then the width to-length
-5 V
Fig. P3.3.26–27
ratio required in each transistor is
+5 V
The conduction parameter are as follows:
M1
K n1 = 400 mA / V 2
V1
K n 2 = 200 mA / V 2
M2
K n 3 = 100 mA / V 2
V2
K n 4 = 80 mA / V 2
M3
Fig. P3.3.24
Page
170
26. I D1 = ?
(A) 0.23 mA
(B) 0.62 mA
(C) 0.46 mA
(D) 0.31 mA
æW ö
ç ÷
è L ø1
æW ö
ç ÷
è L ø2
æW ö
ç ÷
è L ø3
(A)
1.75
6.94
27.8
27. I D4 = ?
(B)
4.93
10.56
50.43
(A) 0.62 mA
(B) 0.31 mA
(C)
35.5
22.4
8.53
(C) 0.46 mA
(D) 0.92 mA
(D)
56.4
38.21
12.56
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Basic FET Circuits
Chap 3.3
28. For the circuit in fig. P3.3.28 the transistor
(A) 7.43 V
(B) 8.6 V
parameter are VTN = 0.8 V and kn¢ = 30 mA / V 2 . If output
(C) -1.17 V
(D) 1.17 V
voltage is Vo = 0.1 V, when input voltage is Vi = 4.2 V,
the required transistor width-to length ratio is
32. A p-channel JFET biased in the saturation region
with VSD = 5 V has a drain current of I D = 2.8 mA, and
+5 V
I D = 0.3 mA at VGS = 3 V. The value of I DSS is
10 kW
Vo
(A) 10 mA
(B) 5 mA
(C) 7 mA
(D) 2 mA
Vi
Statement for Q.33–34:
For the p-channel transistor in the circuit of fig.
Fig. P3.3.28
(A) 1.568
(B) 0.986
(C) 0.731
(D) 1.843
P3.3.33–34 the parameters are I DSS = 6 mA, VP = 4 V
and l = 0.
1 kW
29. For the transistor in fig. P3.3.29 parameters are
VTN = 1 V and K n = 12.5 mA / V 2 . The Q-point ( I D , VDS ) is
+10 V
0.4 kW
20 kW
-5 V
Fig. P3.3.33–34
10 kW
33. The value of I DQ is
Fig. P3.3.29.
(A) 8.86 mA
(B) 6.39 mA
(C) 4.32 mA
(D) 1.81 mA
(A) (1 mA, 8 V)
(B) (0.2 mA, 4 V)
34. The value of VSD is
(C) (1.17 mA, 8 V)
(D) (0.23 mA, 3.1V)
(A) -4.28 V
(B) 2.47 V
30. For an n-channel JFET, the parameters are I DSS = 6
(C) 4.28 V
(D) 2.19 V
mA and VP = -3 V. If VDS > VDS ( sat ) and VGS = -2 V, then
35. The transistor in the circuit of fig. P3.3.35 has
I D is
parameters I DSS = 8 mA and VP = -4 V. The value of VDSQ
(A) 16.67 mA
(B) 0.67 mA
(C) 5.55 mA
(D) 1.67 mA
is
31. For the circuit in fig. P3.3.32 the transistor
+20 V
140 kW
2.7 kW
parameters are Vp = - 35
. V, I DSS = 18 mA, and l = 0. The
value of VDS is
+15 V
60 kW
2 kW
0.8 kW
Fig. P3.3.35
IQ = 8 mA
(A) 2.7 V
(B) 2.85 V
(C) -1.30 V
(D) 1.30 V
-15 V
Fig. P3.3.32
******************
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UNIT 3
10 - ( 4.67 + VSG) = (0.5)(1)( VGS )
SOLUTIONS
1. (A) R1 = 32 kW,
æ R2
VG = çç
è R1 + R2
R2 = 18 kW,
Þ
VDD = 10 V
ö
æ 18 ö
÷÷VDD = ç
÷10 = 3.6 V
è 18 + 32 ø
ø
=
Þ
VGS = 2.05 V
10. (C) Assume transistor in saturation.
I D = 0.4 mA,
VG = 0,
Þ
VSG = 2 + 0.8 = 2.21 V
VSG = VS - VG = VS
11. (A) VD = I D RD - 5 = (0.4)(5) - 5 = -3 V
= 10 - 0.775( 4 + 2) = 5.35 V
VSD = VS - VD = 2. 21 - ( -3) = 5. 21 V
VDS ( sat ) = VGS - VTN = (2.05 - 0.8) = 1. 25 V
VDS > VDS ( sat ) as assumed.
12. (C) R1 = 14.5 kW, R2 = 5.5 kW,
4. (B) R1 = 14 kW, R2 = 6 kW, RS = 0.5 kW, RD = 1.2 kW
RS = 0.6 kW, RD = 0.8 kW,
æ RL
VG = çç
è R1 + R2
ö
æ 6 ö
÷÷(20) - 10 = ç
÷(20) - 10 = -4 V
è 14 + 6 ø
ø
ö
5.5
æ
ö
÷÷(10) - 5 = ç
÷(10) - 5 = -2.25 V
è 14.5 + 5.5 ø
ø
Assume transistor in saturation.
V - ( -5)
ID = S
= K n ( VGS - VTN ) 2
RS
Assume transistor in saturation
V - ( -10) VG - VGS + 10
ID = S
=
= K n ( VGS - VTN ) 2
RS
RS
VS = VG - VGS
kn¢ W
( 60)( 60 ´ 10 -6 )
=
= 1.8 mA / V 2
2 L
2
-2.25 - VGS + 5 = (0.6)(0.5) ( VGS - ( -1)) 2
Þ
- 4 - VGS + 10 = (0.5)(1.8)( VGS - 2) 2
Þ
Þ
VGS = 3.62, - 0.74 V, VGS will be positive.
VGS is positive. Thus (D) is correct option.
5. (D) I D =
VG - VGS + 10 -4 - 3.62 + 10
=
= 4.76 mA
RS
0.5 k
VGS = 124
. , - 6.58 V
13. (D) I D =
VS + 5 VG - VGS + 5 -2.25 - 124
. +5
=
=
RS
RS
0.6 k
= 2.52 mA, Therefore (D) is correct option.
6. (B) 10 = I D( RS + RD) + VDS - 10
VDS = 20 - 4.76(12
. + 0.5) = 119
. V
14. (B) 5 = I D( RS + RD) + VDS - 5
VDS ( sat ) = VGS - VTN = 3.62 - 2 = 1.62 V
VDS = 10 - I D( RS + RD) = 10 - 2.52(0.8 + 0.6) = 6.47 V
VDS = 119
. V > VDS ( sat ) , Assumption is correct.
VDS ( sat ) = VGS - VTH = 124
. - ( -1) =2.24
7. (B) R1 = 8 kW, RL = 22 kW, RS = 0.5 kW, RD = 2 kW
æ R2
VG = çç
è R1 + R2
ö
æ 22 ö
÷÷(20) - 10 = ç
÷(20) - 10 = 4.67 V
è 8 + 22 ø
ø
Assume transistor in saturation
10 - VS
ID =
= K P ( VSG + VTP ) 2
RS
VS = VG + VSG
VDS > VDS ( sat ) ,Assumption is correct.
15. (B) I S = 50 mA = I D ,I D = K n ( VGS - VTN ) 2
Þ
50 ´ 10 -6 = 0.5 ´ 10 -3( VGS - 12
. )2
V
VG = 0, VS = VG - VGS = -1516
.
VDS = VD - VS = 5 - ( -1516
.
) = 6.516 V
16. (B) I D = 50m = K n ( VGS - VTN ) 2
Þ
Page
172
0.4 = K P ( VGS + VTP ) 2
0.4 = (0.2)( VSG - 0.8) 2
3. (C) VDS = VDD - I D( RD + RS )
Kn =
10 - ( 4.67 + 377
. )
= 312
. mA
0.5
VSD = 20 - I D ( RS + RD)= 20 - 2.12(2 + 0.5) = 12.2 V
V - VGS
3.6 - 2.05
2. (C) I D = G
=
= 0.775 mA
RS
2k
æ RL
VG = çç
è R1 + R2
10 - VS 10 - ( VG + VGS )
=
RS
RS
9. (C) 10 = I D( RS + RD) + VSD - 10
K n = 0.5 mA / V 2
3.6 - VGS = (2)(0.5)( VGS - 0.8) 2
VSG = 377
. V, - 177
. V, VSG is positive voltage.
8. (A) I D =
Assume that transistor in saturation region
V
V - VGS
ID = S = G
= K n ( VGS - VTN ) 2
RS
RS
RS = 2 kW,
Analog Electronics
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50 ´ 10 -6 = 0.5 ´ 10 -3( VGS - 12
. )2
Þ
V,
VGS = 1516
.
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Basic FET Circuits
VGS = 152
. V,
æW ö æW ö
23. (A) ç ÷ > ç ÷ thus VGS1 < VGS 2
è L ø1 è L ø2
VGS = VDS
17. (B) I D = K n ( VGS - VTN ) 2
Þ
0.25 = 0.2( VGS - 0.6)
VGS = VG - VS ,
18. (A) I D =
Þ
2
VG = 0,
40( VGS1 - 0.8) 2 = 15( VGS 2 - 0.8) 2
V,
VGS = 172
.
VGS 2 = 5 - VGS1
V
VS = -172
.
5 - VD
= 0.8 mA,
RD
RD =
1.63( VGS1 - 0.8) = (5 - VGS1 - 0.8)
VGS1 = 2.09, VGS 2 = 2.91 V, Vo = VGS 2 = 2.91 V
6 -1
= 5 kW
0.8m
24. (A) Each transistor is biased in saturation because
I D = K n ( VGS - VTN ) 2
Þ 0.8 = (0.4)( VGS - 17
. )2
Þ
VDS = VGS and VDS > VGS - VTN
VGS = 311
. V
For M 3 , V2 = 2 V = VGS 3
VGS = VG - VS , VG = 0, VS = -311
. V
-311
. - ( -5)
I D = 0.8 mA =
Þ RS = 2.36 kW
RS
æ 25 ö æ W
10 -4 = ç
֍
è 2 øè 4
ö
. )2
÷ (2.5 - 15
ø
æ 30 ´ 10 -6
20. (D) K p = çç
2
è
I D = K p ( VSG + VTP ) 2
Þ
æ 36 ´ 10 -3 öæ W ö
÷÷ç ÷ (2 - 1) 2
I D = 0.5 = çç
2
øè L ø3
è
Þ
æ 36 ´ 10 -3 öæ W ö
÷÷ç ÷ ( 3 - 1) 2
I D = 0.5 = çç
2
øè L ø2
è
25. (D) M 2 is in saturation because
0.5 = 0.3( VSG - 12
. )2
VGS 2 = VDS 2 > VGS 2 - VTN
VSG = 2.49 V, VG = 0
VD - ( -5)
RD
Þ RD =
M1 is in non saturation because
VGS1 = Vi = 5 V, VDS1 = VD = 0 V
VDS1 < VGS1 - VTN , I D1 = I D2
æW ö
æW ö
2
2
ç ÷ [2( VGS1 - VTN1 ) VDS1 - VDS 2 ]= ç ÷ ( VGS 2 - VTN 2 )
L
L
è ø1
è ø2
-3 + 5
= 4 kW
0.5m
æW ö
Þ ç ÷ [2(5 - 0.8)(0.1) - (0.1) 2 ] = (1)(5 - 0.1 - 0.8) 2
è L ø1
21. (B) Assume transistor in saturation
10 - VGS
ID =
= K n ( VGS - VTN ) 2
10 k
æW ö
ç ÷ (0.83) = 16.81
è L ø1
10 - VGS = (10)(0.2)( VGS - 2) 2
Þ
. V
VGS = VDS = 377
10 - 377
.
ID =
= 0.623 mA
10 k
VGS1 = 5 - VGS 2
Þ
VDS > VGS - VTN assumption is correct.
transistor
I D1 = I D2
Þ
are
K n 1 ( VGS1 - VTN1 ) = K n 2 ( VGS 2 - VTN 2 )
K n 1 = K n 2 , VTN1 = VTN 2
5
VGS1 = VGS 2 = V
2
Vo = VGS 2 = 2.5 V
-6
(5 - VGS 2 - 1) 2 = 200 ( VGS 2 - 1) 2
VGS1 = 2.24 V
(2.24 - 1) 2 =0.62 mA
I D4 = K n 4 ( VGS 4 - VTN ) 2 = K n 3 ( VGS 3 - VTN ) 2
in
saturation.
2
æW ö
ç ÷ =20.3
è L ø1
27. (B) VGS 2 = VGS 3 = 2.76 V
22. (B) For both transistor VDS = VGS ,
both
Þ
VGS 2 = 2.76 V,
I D1 = 400 ´ 10
Power = I DVDS = 2.35 mW
Therefore
Þ
26. (B) I D1 = K n 1 ( VGS1 - VTN ) 2 = K n 2 ( VGS 2 - VTN ) 2
. V, - 0.27 V, VGS will be 3.77 V
VGS = 377
VDS > VGS - VTN
æW ö
ç ÷ = 6.94
è L ø2
æ 36
öæ W ö
æW ö
I D = 0.5 = ç
´ 10 -5 ÷ç ÷ (5 - 1) 2 Þ ç ÷ = 174
.
è 2
øè L ø1
è L ø1
VS = VSG = 2.49 V
5 - VS
5 - 2.49
ID =
Þ RS =
= 5.02 kW
RS
0.5m
ID =
Þ
For M1 , VGS1 = 10 - V1 = 10 - 5 = 5 V
W = 32 mm
ö
÷÷(20) = 0.3 mA / V 2
ø
Þ
æW ö
ç ÷ = 27.8
è L ø3
Þ
For M 2 , VGS 2 = V1 - V2 = 5 - 2 = 3 V
k¢p W
( VGS + VTP ) 2
2 L
19. (C) VSD = VSG, I D =
Chap 3.3
= 100 ´ 10 -6 (2.76 - 1) 2 = 0.31 mA
28. (C) VGS = 4.2 V,
2
VDS = 0.1 V
VDS < VGS - VTN , Thus transistor is in non saturation.
5 - 0.1
ID =
= 0.49 mA
10 k
k¢ W
ID = n
{2 ( VGS - VTN ) VDS - VDS2 }
2 L
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CHAPTER
3.4
AMPLIFIERS
1. If the transistor parameter are b = 180 and Early
3. The small signal votlage gain Av = vo vs is
voltage V A = 140 V and it is biased at I CQ = 2 mA, the
(A) -4.38
(B) 4.38
values of hybrid-p parameter g m , rp and ro are
(C) -1.88
(D) 1.88
respectively
4. The nominal quiescent collector current of a
(A) 14 A V, 2.33 kW, 90 kW
transistor is 1.2 mA. If the range of b for this transistor
(B) 14 A V, 90 kW , 2.33 kW
is 80 £ b £ 120 and if the quiescent collector current
(C) 77 mA V, 2.33 kW , 70 kW
changes by ±10 percent, the range in value for rp is
(D) 77.2 A V, 70 kW, 2.33 kW
(A) 1.73 kW < rp < 2.59 kW
(B) 1.93 kW < rp < 2.59 kW
Statement for Q.2–3.
(C) 1.73 kW < rp < 2.59 kW
Consider the circuit of fig. P3.4.2–3. The transistor
(D) 1.56 kW < rp < 2.88 kW
parameters are b = 120 and V A = ¥.
Statement for Q.5–6:
+5 V
Consider the circuit in fig. P3.4.5.6. The transistor
parameter are b = 100 and V A = ¥.
4 kW
250 kW
vs
+10 V
vo
~
RC
50 kW
2V
vs
Fig. P3.4.2–3
~
vBB
2. The hybrid-p parameter values of g m , rp and ro are
(A) 24 mA V, ¥, 5 kW
Fig. P3.4.5–6
(B) 24 mA V, 5 kW , ¥
5. If Q-point is in the center of the load line and I CQ = 0.5
(C) 48 mA V, 10 kW , 18.4 kW
mA, the values of VBB and RC are
(D) 48 mA V, 18.4 kW, 10 kW
(A) 10 kW , 0.95 V
(B) 10 kW , 1.45 V
(C) 48 kW , 0.95 V
(D) 48 kW , 1.45 V
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Amplifiers
Statement for Q.14–15:
Chap 3.4
19. For an n-channel MOSFET biased in the saturation
Consider the common Base amplifier shown in fig.
P3.4.14–15. The parameters are
g m = 2 mS
and
region, the parameters are VTN = 1 V, 12 m n Cox = 18 mA V 2
and l = 0.015 V -1 and I DQ = 2 mA. If transconductance is
ro = 250 kW. Find the Thevenin equivalent faced by load
g m = 3.4 mA V, the width-to-length ratio is
resistance RL .
(A) 80.6
(B) 43.2
(C) 190
(D) 110
Thevenin equivalent
20. In the circuit of fig. P3.4.20, the parameters are
270 W
vi
~
g m = 1mA / V, ro = 50 kW. The gain Av = vo vs is
RL
VDD
Fig. P3.4.14–15
60 kW
14. The Thevenin voltage vTH is
(A) 263vi
(B) 132vi
(C) 346vi
(D) 498vi
2 kW
vs
~
300 kW
15. The Thevenin equivalent resistance RTH is
(A) 384 kW
(B) 697 kW
(C) 408 kW
(D) 915 kW
Fig. P3.4.20
Statement for Q.16–17:
(A) -8.01
(B) 8.01
(C) 14.16
(D) -14.16
Statement for Q.21–23:
The common-base amplifier is drawn as a two-port
in fig. P3.4.16–17. The parameters are b = 100, g m = 3
mS, and ro = 800 kW.
+
For the circuit shown in fig. P3.4.21–23 transistor
parameters are VTN = 2 V, K n = 0.5 mA / V 2 and l = 0.
The transistor is in saturation.
i2
i1
v1
+
3.9 kW
10 kW
18 kW
_
+10 V
v2
10 kW
vo
_
Fig. P3.4.16–17
vi
16. The h-parameter h21 is
~
VGG
(A) 2.46
(B) 0.9
(C) 0.5
(D) 0.67
Fig. P3.4.21–23
21. If I DQ is to be 0.4 mA, the value of VGSQ is
17. The h-parameter h12 is
(A) 3.8 ´10 -4
(B) 4.83 ´10 -3
(C) 3.8 ´10 4
(D) 4.83 ´10 3
18. For an n-channel MOSFET biased in the saturation
region, the parameters are K n = 0.5 mA V 2 , VTN = 0.8 V
and l = 0.01 V -1 , and I DQ = 0.75 mA. The value of g m and
ro are
(A) 5.14 V
(B) 4.36 V
(C) 2.89 V
(D) 1.83 V
22. The values of g m and ro are
(A) 0.89 mS, ¥
(B) 0.89 mS, 0
(C) 1.48 mS, 0
(D) 1.48 mS, ¥
23. The small signal voltage gain Av is
(A) 0.68 mS, 603 kW
(B) 1.22 mS, 603 kW
(A) 14.3
(B) -14.3
(C) 1.22 mS, 133 kW
(D) 0.68 mS, 133 kW
(C) -8.9
(D) 8.9
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Amplifiers
(A) 4.44
(B) -4.44
(C) 2.22
(D) -2.22
SOLUTIONS
the
source-follower
circuit
in
fig.
P3.4.33-34. The values of parameter are g m = 2 mS and
ro = 100 kW.
+5 V
Ro
vs
~
I CQ
1. (C) g m =
Statement for Q.33–34:
Consider
Chap 3.4
IQ
500 kW
bVt
b
180
=
=
= 2.33 kW
I CQ g m 77.2m
ro =
V A 140
=
= 70 kW
I CQ 2m
gm =
Fig. P3.4.33-34
rp =
33. The voltage gain Av is
(B) -0.89
(C) 2.79
(D) -2.79
2 - 0.7
= 5.2 m A
250 k
I CQ
=
Vt
0.624
= 24 mA V
0.0259
bVt
b
120
=
=
= 5 kW, ro = ¥
I CQ g m 24m
ö
æ rp
bRC
3. (C) Av = - g m RC çç
÷÷ =
r
+
R
r
B ø
p + RB
è p
34. The output resistance Ro is
æ
ö
5k
= - (24m)( 4 k) çç
÷÷ = -1.88
5
k
+
250
k
è
ø
(A) 100 kW
(B) 0.498 kW
(C) 1.33 kW
(D) None of the above
4. (D) rp =
rp(min) =
bVT
,
I CQ
(120)(0.0259)
= 2.88 kW,
108
. m
rp( max ) =
*******************
2m
= 77.2 mA V
0.0259
I CQ = bI B = (120)(5.2m ) = 0.642 mA
4 kW
-5 V
(A) 0.89
=
rp =
2. (B) I BQ =
vo
Vt
( 80)(0.0259)
= 156
. kW
1.32m
5. (A) VECQ =
1
VCC = 5 V
2
VECQ = 10 - I CQ RC = 5
Þ
10 - (0.5m) RC = 5
RC = 10 kW,
I BQ =
I CQ
b
=
0.5
=5 mA
100
VEB ( ON ) + I BQ RB = VBB
Þ
0.7 + (5m ) (50 k) = 0.95 V
6. (D) g m =
rp =
I CQ
Vt
=
0.5
= 19.3 mA V
0.0259
bVt (100)(0.0259)
=
= 5.18 kW , ro = ¥
0.5m
I CQ
æ 100 ö
7. (B) I CQ = ç
÷(0.35) = 0.347 mA
è 1001 ø
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UNIT 3
The small-signal equivalent circuit is as shown in fig.
S3.4.7
500 W
vo
+
~
vs
C
B
10 kW
rp
Vp
_
gmVp
ro
7 kW
Analog Electronics
9. (A) DC Analysis: VTH =
RTH = 11||50 = 8.33 kW
12 - 0.7 - 10
I BQ =
= 119
. mA
8.33k + (101)1k
. mA, I EQ = 12
. mA
I CQ = bI BQ = 119
VECQ = 12 - (1. 20)1 - (119
. )2 = 8.42 V
E
AC Analysis:
Ib
Fig. S3.4.7
rp
vs
ö
æ rp ||10 k
vo
÷÷ ( ro ||7 k)
= - g m çç
vs
500
+
r
|
|
10
k
ø
p
è
I CQ 0.347m
gm =
=
= 1313
. mA V
Vt
0.0259
rp =
288 ´ 7
7.6 ´ 10
= 6.83 kW, rp ||10k =
= 4.32 kW
288 + 7
7.6 + 10
vo
10||50 kW
bIb
E
2 kW
(b+1)Ib
bVt
(100)(0.0259
=
= 2.18 kW
I CQ
119
. m
vo = bI b (2k) , vs = - (b + 1) I b (1k) + I b ( rp)
v
-b(2 k)
-(100)(2 k)
Av = o =
=
= -196
.
vs
rp + (b + 1)1k
2.18 k + (100)(1k)
10. (B) VECQ = 8.42 V,
4.32 k
æ
ö
Av = -1313
. mç
÷( 6.83k) = - 80
è 500 + 4.32 k ø
Þ Output voltage swing = 5.16 V peak to peak.
8. (C) DC Analysis: I CQ = I EQ
11. (B) Since the B–C junction is not reverse biased, the
For 1 £ vEC £ 11 V,
VCEQ = 5 = 10 - I CQ ( RC + RE )
Þ 5 = 10 - I CQ(1. 2 k + 0. 2 k)
3.57
I BQ =
= 23.8 m A
150
Ib
I CQ = 3.57 mA
-mode
C
B
vo
+
rp
vs
~
R1 || R2
Ie
+
+
vce
_
vce
rp
gmvce
_
Vp bIb
_ E
Fig. S 3.4.11
1.2 kW
bVt
(0.0259)
= (150)
= 109
. kW , ro = ¥
I CQ
357
. m
æ 1 ö
vce
1
, So rp || çç
÷÷ || ro
=
g m Vce g m
è gm ø
(100)(0.0259)
rp =
= 2.33 kW
2m
I CQ
2m
gm =
=
= 77.2 mA V
Vt 0.0259
vo -(b I b ) RC
, vs = I b rp + (b + 1) RE I b
=
vs
vs
1
V
150
= 12.95 , ro = A =
= 75 kW
gm
I CQ 2m
0.2 kW
Fig. S3.4.8
rp =
DvEC = 11 - 8.42 = 2.58 V
transistor continues to operate in the forward-active
Þ
AC Analysis:
Av =
Av =
Page
180
C
Fig. S3.4.9
VA
100
=
= 288 kW
I CQ 0.347m
ro ||7k =
~
Vp
+
1 kW
b
bV
100
rp =
= t =
= 7.6 kW
g m I CQ 1313
. m
ro =
Ic
B
_
rp ||10 k
( vs ), vo = - g m Vp( ro ||7k)
500 + rp ||10 k
Vp =
50
(12) = 10 V
10 + 50
-bRC
-(150)(12
. )k
=
= -5.75
rp + (1 + b) RE 109
. k + (151)(0.2 k)
r=
re = (2.33k)||(12.95)||(75 k) =12.87 W
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Amplifiers
VA
I CQ
12. (C) ro =
Þ
I CQ =
Vp Vp
v + Vpi
+
+ g m Vp + i
=0
rp
ro
270
VA
75
=
= 0.375 mA
ro 200 k
vi + Vpi
Vp
Vp
+
+ 2mVp +
=0
50 k 250 k
270
bV
75(0.0259)
13. (B) rp = t =
= 194
.
kW
I CQ
1m
E
bIb
Ie
Ib
C
2.7 kW
B
Þ
16. (B) The equivalent small-signal circuit is shown in
+
vo
_
fig. S3.4.16
1.5 kW
ro
Fig. S 3.4.13
i1
vi = I b ( rp + 15k
. ), I in = I e = (b + 1) I b
Vi ( rp + 15
. k) 194
.
+ 15
. k
Rin =
=
=
= 45 W
Ie
(b + 1)
76
E
~
vs
3.9 kW rp
~
rp =
gmVp
Vp
+
vi rp
Removing the RL , -Vp =
270 + rp
rp =
vi rp(1 + g m ro)
270 + rp
i2
i1
, i2 =
v2 = 0
Vp
V
V
- p - p - g m Vp ,
39
. k rp
ro
h21 =
i2
=
i1
- gm
- g m rp 39
. k
=
= 0.91
1
1
39
.
k
. k
r
+
+
g
p
m rp 39
+ + gm
39
. k rp
i1 = 0
E
3.9 kW rp
fig. S3.4.15
C
Vp
+
gmVp
18 kW
i2
~
v2
B
ro
Fig. S 3.4.17
270 W
_
rp
gmVp
Vp
+
Fig. S 3.4.15
I sc = g m Vp +
v1
v
v - v2
+ 1 + 1
= g m Vp
39
. k rp
ro
_
15. (A) The equivalent small-signal circuit is shown in
~
Vp
can be neglected
ro
ro
vi 50 k(1 + (2m)(250 k))
= 498 vi
270 + 50 k
vi
Vp
+ g m Vp
r0
i1 = -
17. (A) v1 = -Vp ,
b
100
=
= 50 kW
g m 2m
vTH =
v2 = 0
b
100
=
= 33.3 kW
gm
3m
h21 =
Fig. S 3.4.14
vTH = -ro g m Vp - Vp =
18 kW
Vp
+
Fig. S 3.4.16
270 W
_
gmVp
B
ro
rp
C
_
14. (D) The equivalent circuit is shown below
vi
Vp = -0.647 vi ,
.
mvi
I sc = 1297
vTH
498 vi
RTH =
=
= 384 kW
I SC 1297
.
mvi
rp
~
vi
Chap 3.4
Vp
Vp
= 2.004 mVp
= 2mVp +
ro
250 k
Isc
æ 1
1 1ö v
v1 çç
+ + ÷÷ - 2 = - g m v1
39
.
k
r
ro ø ro
è
p
1
1
v1
ro
800
k
=
=
1
1
1
1
1 1
v2
+
+
+ 3m
+ + + gm
39
. k 33.3k 800 k
39
. k rp ro
= 3.8 ´ 10 -4
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Amplifiers
RS = 4 kW, v gs = 0.84 vi ,
v gs = vi ,
vo = - g m v gs ( ro ||RD)
Chap 3.4
vo
= Av = - g m (7k)
vi
g m = 2 K n ( VGS - VTN )
= -(1.41m)(0.84 vi )(100 k || 5 k)
vo
Þ
= Av = - 5.6
vi
= 2 (1m)(151
. - 0.8) = 1.42 mS
28. (A) Ro = RD ||ro || 100k =4.76 kW
32. (A) The small-signal equivalent circuit is shown in
Av = -(1.42m) (7 k) = - 9.9
fig. S.3.4.34
29. (A) As shown in fig. S3.4.27, Ri = R1 || R2 = 20.6 kW
. V, I DQ = 0.5 mA
VGSQ = 15
vi
g m = 2 K n ( VGS - VTN ) = 2(1m) (15
. - 0.8) = 1.4 mA V
ro = [ lI DQ ]
vo
vgs
~
10 kW
RS
5 kW
+
RD
RL
4 kW
G
=¥
Fig. S3.4.32
The resulting small-signal equivalent circuit is shown
in fig. S5.4.30
G
D
+
~
vi
D
_
30. (C) From the DC analysis:
-1
gmvgs
S
vgs gmvgs
RTH
RD
vo = - g m v gs ( RD || RL ), vi = -v gs
v
Av = o = g m ( RD || RL ) = (2m)(5 k ||4 k) = 4.44
vi
vo
7 kW
_
33. (A) The small-signal equivalent circuit is shown in
S
fig. S3.4.33
RS
0.5 kW
G
D
+
Fig. S 3.4.30
vi
vo = - g m v gs RD, vi = v gs + g m v gs RS
vo
- g m RD
(7 k)
Þ
=
= - (1.4m)
= -5.76
vi
1 + g m RS
1 + (1.4m) (0.5 k)
~
ro
vgs gmvgs
500 kW
_
vo
S
4 kW
31. (B) Since the DC gate current is zero, VS = -VGSQ
Fig. S3.4.33
I DQ = I Q = K n ( VGSQ - VTN ) 2
Þ
0.5 = 1( VGSQ - 0.8) 2
vo = g m v gs ( RL || ro)
VGSQ = 151
. V = -VS
VDSQ = 5 - (0.5m)(7 k) - ( -151
. ) = 301
. V
The transistor is therefore biased in the saturation
region. The small-signal equivalent circuit is shown in
fig.S3.4.31.
D
G
+
vs
~
vgs
_
gmvgs
vo
7 kW
vi = v gs + vo = v gs + g m v gs ( RL || ro)
1
v gs =
1 + g m ( RL || ro)
vo
g m ( RL || ro)
= Av =
vi
1 + g m ( RL || ro)
RL ||ro = 4 k ||100 k =
Av =
(2m)( 3.85 k)
= 0.89
1 + (2m)( 3.85 k)
S
34. (B) Ro =
Fig. S3.4.31
vo = - g m v gs (7k)
100 k
= 3.86 kW
26
1
||ro
gm
æ1
= ç ö÷ ||(100) » 0.498 kW
è2ø
********************
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GATE EC BY RK Kanodia
CHAPTER
3.5
OPERATIONAL AMPLIFIERS
1. Av =
vo
=?
vi
400 kW
vi
(A) -2 sin wt m A
(B) -7 sin wt m A
(C) -5 sin wt m A
(D) 0
40 kW
4. In circuit shown in fig. P3.5.4, the input voltage vi is
vo
0.2 V. The output voltage vo is
50 kW
R
150 kW
10 kW
vi
25 kW
Fig. P3.5.1
(A) -10
(B) 10
(C) -11
(D) 11
2. Av =
vo
Fig. P3.5.4
vo
=?
vi
400 kW
40 kW
vi
vo
(A) 6 V
(B) -6 V
(C) 8 V
(D) -8 V
5. For the circuit shown in fig. P3.5.5 gain is
60 kW
Av = vo vi = -10. The value of R is
R
Fig. P3.5.2
(A) -10
(B) 10
(C) 13.46
(D) -13.46
100 kW
vi
3.
The
input
to
the
circuit
in
fig.
100 kW
P3.5.3
100 kW
vo
is
vi = 2 sin wt mV. The current io is
10 kW
vi
Fig. P3.5.5
1 kW
io
vo
(A) 600 kW
(B) 450 kW
(C) 4.5 MW
(D) 6 MW
4 kW
Fig. P3.5.3
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Operational Amplifiers
6. For the op-amp circuit shown in fig. P3.5.6 the
Chap 3.5
10. In the circuit of fig. P3.5.10 the output voltage vo is
voltage gain Av = vo vi is
20 kW
R
R
20 kW
+0.5 V
R
40 kW
-1 V
R
R
vi
vo
60 kW
+2 V
R
vo
Fig. P3.5.10
Fig. P3.5.6
(A) -8
(B) 8
(C) -10
(D) 10
(B) -2.67 V
(C) -6.67 V
(D) 6.67 V
11. In the circuit of fig. P3.5.11 the voltage vi1 is
7. For the op-amp shown in fig. P3.5.7 open loop
differential gain is Aod = 10 3. The output voltage vo for
vi = 2 V is
(A) 2.67 V
(1 + 2 sin wt) mV and vi2 = -10 mV. The output voltage
vo is
20 kW
20 kW
100 kW
vi1
vi
2 kW
1 kW
100 kW
vo
vo
vi2
1 kW
Fig. P3.5.11
Fig. P3.5.7
(A) -1.996
(B) -1.998
(A) -0.4(1 + sin wt) mV
(B) 0.4(1 + sin wt) mV
(C) -2.004
(D) -2.006
(C) 0.4(1 + 2 sin wt) mV
(D) -0.4(1 + 2 sin wt) mV
8. The op-amp of fig. P3.5.8 has a very poor open-loop
12. For the circuit in fig. P3.5.12 the output voltage is
voltage gain of 45 but is otherwise ideal. The closed-loop
vo = 2.5 V in response to input voltage vi = 5 V. The finite
gain of amplifier is
open-loop differential gain of the op-amp is
100 kW
vi
2 kW
500 kW
vo
vo
1 kW
vi
Fig. P3.5.8
(A) 20
(B) 4.5
(C) 4
(D) 5
Fig. P3.5.12
(A) 5 ´ 10 4
(B) 250.5
(C) 2 ´ 10
(D) 501
4
9. For the circuit shown in fig. P3.5.9 the input voltage
13. vo = ?
vi is 1.5 V. The current io is
10 kW
vi
100 kW
6 kW
100 kW
8 kW io
vo
vo
20 kW
+18 V
5 kW
40 kW
+15 V
Fig. P3.5.13
Fig. P3.5.9
(A) -1.5 mA
(B) 1.5 mA
(A) 34 V
(B) -17 V
(C) -0.75 mA
(D) 0.75 mA
(C) 32 V
(D) -32 V
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UNIT 3
28. For the circuit shown in fig. P3.5.28 the input
Analog Electronics
31. io = ?
resistance is
6 kW
io
2 kW
vo
6A
2 kW
4 kW
is
Fig. S3.5.31
2 kW
10 kW
(A) -18 A
(B) 18 A
(C) -36 A
(D) 36 A
Fig. P3.5.28
(A) 38 kW
(B) 17 kW
(C) 25 kW
(D) 47 kW
Statement for Q.32–33:
Consider the circuit shown below
29. In the circuit of fig. P3.5.29 the op-amp slew rate is
SR = 0.5 V ms. If the amplitude of input signal is 0.02 V,
then the maximum frequency that may be used is
vi
3 kW
D1
6 kW
D2
2 kW
vo
240 kW
vi
10 kW
vo
Fig. P3.5.32–33
32. If vi = 2 V, then output vo is
Fig. P3.5.29
(A) 0.55 ´ 106 rad/s
(B) 0.55 rad/s
(C) 1.1 ´ 106 rad/s
(D) 1.1 rad/s
(B) -4 V
(C) 3 V
(D) -3 V
33. If vi = -2 V, then output vo is
30. In the circuit of fig. P3.5.30 the input offset voltage
and input offset current are Vio = 4 mV and I io = 150 nA.
The total output offset voltage is
(A) 4 V
(A) -6 V
(B) 6 V
(C) -3 V
(D) 3 V
34. vo( t) = ?
500 kW
vi
8 mF
5 kW
vo
5u(t) mA
250 W
vo
50 W
1 kW
5 kW
Fig. P3.5.34
Fig. P3.5.30
(A) e
(A) 479 mV
(C) 168 mV
Page
188
(B) 234 mV
(D) 116 mV
(C) e
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-
t
10
-
t
1 .6
-
u( t) V
(B) -e
u( t) V
(D) -e
t
10
-
t
1 .6
u( t) V
u( t) V
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Operational Amplifiers
Chap 3.5
35. The circuit shown in fig. P3.5.35 is at steady state
(A) vs vss
(B) -vs vss
before the switch opens at t = 0. The vC ( t) for t > 0 is
t=0
v
(C) - s
vss
(D)
20 kW
39. If the input to the ideal comparator shown in fig.
vs
vss
P3.5.39 is a sinusoidal signal of 8 V (peak to peak)
20 kW
20 kW
without any DC component, then the output of the
+
4 mF
5V
comparator has a duty cycle of
vC
-
Input
Output
Fig. P3.5.35
(A) 10 - 5 e -12 .5t V
(C) 5 + 5 e
-
t
12 .5
Vref = 2 V
(B) 5 + 5 e -12 .5t V
(D) 10 - 5 e
V
-
t
12 .5
Fig. P3.5.39
V
36. The LED in the circuit of fig. P3.5.36 will be on if vi
is
10 kW
+10 V
470W
10 kW
vi
(A)
1
2
(B)
1
3
(C)
1
6
(D)
1
12
40. In the op-amp circuit given in fig. P3.5.40 the load
current iL is
Fig. P3.5.36
(A) > 10 V
(B) < 10 V
(C) > 5 V
(D) < 5 V
R1
R1
vs
37. In the circuit of fig. P3.5.37 the CMRR of the
op-amp is 60 dB. The magnitude of the vo is
R2
2V
IL
R2
RL
100 kW
R
R
1 kW
Fig. P3.5.40
R
R
vo
1 kW
(A) -
vs
R2
(B)
vs
R2
(C) -
vs
RL
(D)
vs
RL
100 kW
Fig. P3.5.37
(A) 1 mV
(B) 100 mV
(C) 200 mV
(D) 2 mV
41. In the circuit of fig. P3.5.41 output voltage is |vo| = 1
V for a certain set of w, R, an C. The |vo| will be 2 V if
R1
38. The analog multiplier X of fig. P.3.5.38 has the
R1
characteristics vp = v1 v2 . The output of this circuit is
vss
vi = sin wt V
vo
10 kW
X
vs
R
R
C
vo
Fig. P3.5.41
Fig. P3.5.38
(A) w is doubled
(B) w is halved
(C) R is doubled
(D) None of the above
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UNIT 3
Analog Electronics
42. In the circuit of fig. P3.5.42. the 3 dB cutoff
+15 V
frequency is
50 nF
47 kW
6 kW
3 kW
vo
vi
vo
Fig. P3.5.42
(A) 10 kHz
(B) 1.59 kHz
(C) 354 Hz
(D) 689 Hz
Vz = 5 V
100 W
Fig. P3.5.45
43. The phase shift oscillator of fig. P3.5.43 operate at
46. In the circuit in fig. P3.5.46 both transistor Q1 and
f = 80 kHz. The value of resistance RF is
Q2 are identical. The output voltage at T = 300 K is
RF
100 pF
100 pF
100 pF
R
R
R1
R2
v1
R
v2
vo
333 kW
20 kW
Fig. P3.5.43
(A) 148 kW
(B) 236 kW
(C) 438 kW
(D) 814 kW
vo
20 kW
333 kW
44. The value of C required for sinusoidal oscillation of
Fig. P3.5.46
frequency 1 kHz in the circuit of fig. P3.5.44 is
2.1 kW
1 kW
C
1 kW
æv R ö
(A) 2 log10 çç 2 1 ÷÷
è v1 R2 ø
æv R ö
(B) log10 çç 2 1 ÷÷
è v1 R2 ø
æv R ö
(C) 2.303 log10 çç 2 1 ÷÷
è v1 R2 ø
æv R ö
(D) 4.605 log10 çç 2 1 ÷÷
è v1 R2 ø
47. In the op-amp series regulator circuit of fig. P8.3.47
Vz = 6.2 V, VBE = 0.7 V and b = 60. The output voltage vo is
C
1 kW
vo
+36 V
1 kW
Fig. P3.5.44
(A)
(C)
1
mF
2p
1
2p 6
30 kW
(B) 2p mF
mF
(D) 2 p 6 mF
10 kW
45. In the circuit shown in fig. P3.5.45 the op-amp is
ideal. If bF = 60, then the total current supplied by the
15 V source is
(A) 123.1 mA
(B) 98.3 mA
(C) 49.4 mA
(D) 168 mA
Fig. P3.5.47
(A) 35.8 V
(B) 24.8 V
(C) 29.8 V
(D) None of the above
*******
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Operational Amplifiers
26. (C) v2 + = v2 - = 0 V, current through 6 V source
i=
6
= 2 mA, vo = -2m( 3k + 2 k) = -10 V
3k
33. (D) If vi < 0, then vo > 0, D2 blocks and D1 conduct
Av = -
vo(1) vo
v (2) vo(1)
= , v- = i
+
1+ 3 4
2+1 2 +1
2v v
v
v
v+ = v- , o = o + i , o = -8
4
3
3
vi
Chap 3.5
3k
= -15
. , vo = ( -2)( -15
. )=3 V
2k
34. (A) Voltage follower vo = v- = v+
27. (D) v+ =
v+ (0 + ) = 5m(250 ||1000) = 1 V, v+ ( ¥) = 0
t = 8m(1000 + 250) = 10 s
35. (A) vc (0 - ) = 5 V = vc (0 + ) = 5 V
28. (B) Since op-amp is ideal
For t > 0 the equivalent circuit is shown in fig. S3.5.35
20 kW
i1
is
+
vC
–
is
2 kW
i2
Fig. S3.5.35
10 kW
t = 20 k ´ 4m = 0.08 s
vc = 10 + (5 - 10) e
Fig. S3.5.28
v- = v+ , 2 kis = 4 ki1
Þ
is = 2 i1
vs = 2 kis + 10 ki2
36. (C) v- =
i2 = is + i1 , vs = 2 kis + 10 k( is + i1 ), i1 =
i ö
æ
vs = 2 kis + 10 kç is + s ÷
2ø
è
Þ
is
2
-
t
0 .08
= 10 - 5 e -12 .5t V for t > 0
(10)(10 k)
=5 V
10 k + 10 k
When v+ > 5 V, output will be positive and LED will be
on. Hence (C) is correct.
vs
= 17k = Rin
is
R
R
= 1 V, v- = (2)
= 1 V, vd = 0
2R
2R
v + vR
VCM
VCM = +
= 1, vo = F
2
1 CMRR
100 1
CMRR = 60 dB = 10 3 , vo =
= 100 mV
1 10 3
37. (B) v+ = (2)
½R ½ 240 k
29. (C) Closed loop gain A =½ F½ =
= 24
½ R1½ 10 k
The maximum output voltage vom = 24 ´ 0.02 = 0.48 V
w £
4 mF
10 V
4 kW
SR
0.5 / m
=
= 11
. ´ 106 rad/s
vom
0.48
38. (C) v+ = 0 = v- ,
æ
R ö
30. (A) The offset due to Vio is vo = çç 1 + 1 ÷÷Vio
R1 ø
è
Let output of analog multiplier be vp .
vp
vs
=Þ vs = -vp , vp = vss vo
R
R
v
vs = -vss vo , vo = - s
vss
500 ö
æ
= ç1 +
÷ 4m = 404 mV
5 ø
è
Due to I io, vo = RF I io = (500 k)(150n) = 75 mV
Total offset voltage vo = 404 + 75 = 479 mV
39. (B) When vi > 2 V, output is positive. When vi < 2 V,
-vo
v
, io = - 6 + o
6k
3k
-6( 6 k)
io = - 6 +
= -18 A.
3k
output is negative.
31. (A) 6 =
V
4V
2V
32. (B) If vi > 0, then vo < 0, D1 blocks and D2 conducts
Av = -
6k
= -2
3k
p
6
Þ
2p
5p
6
t
vo = ( -2)(2) = -4 V
Fig. S3.5.39
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UNIT 3
5p p
TON
1
Duty cycle =
= 6 6 =
T
2p
3
vs - v- v- - vo
=
R1
R1
40.(A)
v+
v
v - vo
+ + + +
=0
R2 RL
R2
Þ
Analog Electronics
44. (A) This is Wien-bridge oscillator. The ratio
R2 2.1k
=
= 2.1 is greater than 2. So there will be
R1
1k
oscillation
2 v1 = vs + vo
R2
R1
æ
R ö
vo = çç 2 + 2 ÷÷ v+
R
è
L ø
Þ
æ
R ö
2 v- = vs + çç 2 + 2 ÷÷ v+ , v- = v+
R
è
L ø
R
Þ 0 = vs + 2 v+
RL
v+ = -
RL
vs ,
R2
iL =
v+
,
RL
iL = -
R
R
vs
R2
Fig. S3.5.44
Frequency =
41. (D) This is a all pass circuit
vo
1 - jwRC
,
= H ( jw) =
vi
1 + jwRC
| H( jw)| =
1 + ( wR 2 C) 2
1 + ( wRC) 2
C
C
=1
C=
1
2pRC
Þ
1 ´ 10 3 =
1
mF
2p
Thus when w and R is changed, the transfer function is
45. (C) v+ = 5 V = v- = vE ,
unchanged.
The input current to the op-amp is zero.
i+15V = iZ + iC = iZ + a F iE
42. (B) Let R1 = 3 kW , R2 = 6 kW , C = 50 nF
vi
v - vo
+ i
=0
1
R2
æ
ö
R1 ||ç
÷
è sC ø
Þ
vi
v
v
+ i = o
ö R2 R2
æ
R1
÷÷
çç
1
+
sR
C
ø
1
è
éR
ù
vi ê 2 (1 + sR1 C) + 1ú = vo
R
ë 1
û
vi
[R2 + R1 + sR1 R2 C ] = vo
R1
vo
R + R1
= 2
vi
R1
Þ
f3dB
=
é
sR1 R2 C ù
ê1 + R + R ú
1
2 û
ë
vo æ
R ö
= ç 1 + 2 ÷÷(1 + s( R1 || R2 ) C)
vi çè
R1 ø
1
=
2 p( R1 || R2 ) C
1
1
=
= 159
. kHz
2 p( 3k ||6 k)50n 2 p(2 k)50n
Þ R=
1
( 80 k)(2 p 6 )(100 p)
RF
= 29
R
Þ
=
15 - 5 60 æ 5 ö
+
ç
÷ = 49.4 mA
47 k
61 è 100 ø
46. (B) vo =
333
( vo1 - vo2 )
20
æi
vo1 = -vBE1 - Vt ln çç c1
è is
43. (B) The oscillation frequency is
1
1
f =
Þ
80 k =
2 p 6 RC
2 p 6 R(100 p)
= 8.12 kW
RF = ( 8.12 k)(29) = 236 kW
æi
vo1 - vo2 = -Vt ln çç c1
è ic 2
v1
,
R1
ic1 =
ic 2 =
ö
æi
÷÷, vo2 = -vBE 2 - Vt ln çç c 2
ø
è is
æi
ö
÷÷ = Vt ln çç c 2
è ic1
ø
ö
÷÷
ø
ö
÷÷
ø
v2
R2
æv R ö
vo1 - vo2 = Vt ln çç 2 1 ÷÷, Vt = 0.0259 V
è R2 v1 ø
vo =
æv R ö
333
333
( vo1 - vo2 ) =
(0.0259) ln çç 2 1 ÷÷
20
20
è v1 R2 ø
æv R ö
æv R ö
= 0.4329 lnçç 2 1 ÷÷ = 0.4329(2.3026) log10 çç 2 1 ÷÷
è v1 R2 ø
è v1 R2 ø
æv R ö
= log10 çç 2 1 ÷÷
è v1 R2 ø
47. (B) v+ = v- , vZ =
10 vo
v
= o
10 + 30 4
vo = 4 vz = 6.2 ´ 4 = 24.8 V
************
Page
194
1
2 p(1k) C
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GATE EC BY RK Kanodia
CHAPTER
4.1
NUMBER SYSTEMS & BOOLEAN ALGEBRA
1. The 100110 2 is numerically equivalent to
1. 2616
2. 3610
3. 468
5. A computer has the following negative numbers
stored in binary form as shown. The wrongly stored
4. 212 4
number is
The correct answer are
(A) 1, 2, and 3
(B) 2, 3, and 4
(C) 1, 2, and 4
(D) 1, 3, and 4
(B) 5
(C) 7
(D) 9
(B) -89 as 1010 0111
(C) -48 as 1110 1000
(D) -32 as 1110 0000
6. Consider the signed binary number A = 01010110
2. If (211) x = (152)8 , then the value of base x is
(A) 6
(A) -37 as 1101 1011
and B = 1110 1100 where B is the 1’s complement and
MSB is the sign bit. In list-I operation is given, and in
list-II resultant binary number is given.
3. 11001, 1001 and 111001 correspond to the 2’s
List–I
complement representation of the following set of
List-II
numbers
(A) 25, 9 and 57 respectively
P. A + B
(B) -6, -6 and -6 respectively
Q. B - A
(C) -7, -7 and -7 respectively
R. A - B
(D) -25, -9 and -57 respectively
1. 0 1 0 0
2. 0 1 1 0
3. 0 1 0 0
4. 1 0 0 1
5. 1 0 1 1
6. 1 0 0 1
7. 1 0 1 1
8. 0 1 1 0
S. - A - B
4. A signed integer has been stored in a byte using 2’s
0011
1001
0010
0101
1100
0110
1101
1010
complement format. We wish to store the same integer
The correct match is
in 16-bit word. We should copy the original byte to the
less significant byte of the word and fill the more
P
Q
R
S
significant byte with
(A) 0
(A)
3
4
2
5
(B) 1
(B)
3
6
8
7
(C) equal to the MSB of the original byte
(C)
1
4
8
7
(D) complement of the MSB of the original byte.
(D)
1
6
2
5
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UNIT 4
28.
The
simplified
form
of
a
logic
Digital Electronics
A
function
A
B
Y = A( B + C( AB + AC)) is
Z
C
(A) A B
(B) AB
(C) AB
(D) AB
D
(B) A + B
(C) AB + AB
(D) A B + AB
B
(C) 0
(D) 1
Z
D
B
Z
C
D
(C)
(D)
35. In fig. P.4.1.35 the input condition, needed to
30. If X Y + XY = Z then XZ + XZ is equal to
(B) Y
A
C
(A) A + B
(B)
A
Y = ( AB ) × ( AB) is
Z
D
(A)
29. The reduced form of the Boolean expression of
(A) Y
B
C
produce X = 1, is
A
B
31. If XY = 0 then X Å Y is equal to
(A) X + Y
(B) X + Y
(C) XY
(D) X Y
Fig. P4.1.34
32. From a four-input OR gate the number of input
condition, that will produce HIGH output are
(A) 1
(B) 3
(C) 15
(D) 0
(A) A = 1, B = 1, C = 0
(B) A = 1, B = 1, C = 1
(C) A = 0, B = 1, C = 1
(D) A = 1, B = 0, C = 0
36. Consider the statements below:
33. A logic circuit control the passage of a signal
according to the following requirements :
1. Output X will equal A when control input B and
1. If the output waveform from an OR gate is the same
as the waveform at one of its inputs, the other input is
being held permanently LOW.
2. If the output waveform from an OR gate is always
HIGH, one of its input is being held permanently
HIGH.
C are the same.
The statement, which is always true, is
2. X will remain HIGH when B and C are
different.
The logic circuit would be
A
X
C
X
C
(B)
A
(C) Only 2
(D) None of the above
is
(A) 3
(B) 4
(C) 5
(D) 6
A
B
X
C
B
X
number of two-input NAND required to implement
(D)
34. The output of logic circuit is HIGH whenever A and
B are both HIGH as long as C and D are either both
LOW or both HIGH. The logic circuit is
38. If the X and Y logic inputs are available and their
complements X and Y are not available, the minimum
C
(C)
Page
200
(B) Only 1
NAND gates, minimum number of requirement of gate
B
(A)
(A) Both 1 and 2
37. To implement y = ABCD using only two-input
A
B
X
C
X Å Y is
(A) 4
(B) 5
(C) 6
(D) 7
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Number Systems & Boolean Algebra
Statement for Q.39–40:
Chap 4.1
Assuming complements of x and y are not
A Boolean function Z = ABC is to be implement
available, a minimum cost solution for realizing f using
using NAND and NOR gate. Each gate has unit cost.
2-input NOR gates and 2-input OR gates (each having
Only A, B and C are available.
unit cost) would have a total cost of
39. If both gate are available then minimum cost is
(A) 2 units
(B) 3 units
(C) 4 units
(D) 6 units
(A) 1 units
(B) 2 units
(C) 3 units
(D) 4 units
44. The gates G1 and G2 in Fig. P.4.2.44 have
propagation delays of 10 ns and 20 ns respectively.
40. If NAND gate are available then minimum cost is
(A) 2 units
(B) 3 units
(C) 4 units
(D) 6 units
G1
1
Vi
0
G2
Vo
Vi
to
Fig. P4.1.44
If the input Vi makes an abrupt change from logic
41. In fig. P4.1.41 the LED emits light when
0 to 1 at t = t0 then the output waveform Vo is
VCC = 5 V
1 kW
1 kW
[t1 = t0 + 10 ns, t2 = t1 + 10 ns, t3 = t2 + 10 ns]
(B)
(A)
1 kW
t0
1 kW
t2
t1
t3
t1
t2
t3
t0
t1
t2
t3
(D)
(C)
t0
Fig. P4.1.41
(A) both switch are closed
t0
t2
t1
t3
45. In the network of fig. P4.1.45 f can be written as
(B) both switch are open
X0
(C) only one switch is closed
1
2
X1
(D) LED does not emit light irrespective of the switch
positions
3
X2
42. If the input to the digital circuit shown in fig.
n-1
X3
Xn-1
n
F
Xn
Fig. P4.1.45
P.4.1.42 consisting of a cascade of 20 XOR gates is X,
(A) X 0 X1 X 3 X 5 + X 2 X 4 X 5 .... X n -1 + .... X n -1 X n
then the output Y is equal to
(B) X 0 X1 X 3 X 5 + X 2 X 3 X 4 .... X n + .... X n -1 X n
(C) X 0 X1 X 3 X 5 .... X n + X 2 X 3 X 5 K X n + .... + X n -1 X n
1
(D) X 0 X1 X 3 X 5... X n -1 + X 2 X 3 X 5 K X n +..+ X n -1 X n - 2 + X n
Y
X
Fig. P4.1.42
(A) X
(B) X
(C) 0
(D) 1
*******
43. A Boolean function of two variables x and y is
defined as follows :
f (0, 0) = f (0, 1) = f (1, 1) = 1; f (1, 0) = 0
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UNIT 4
Digital Electronics
A - B = A + B,
SOLUTIONS
A
010 10110
B
+ 00010011
0110 1001
1. (D) 100110 2 = 2 5 + 2 2 + 21 = 3810
- A - B = A + B,
2616 = 2 ´ 16 + 6 = 3810
468 = 4 ´ 8 + 6 = 3810
A
1010 1001
B
+ 00010011
10111100
212 4 = 2 ´ 4 2 + 41 = 3810
7. (B) Here A , B are 2’s complement
So 3610 is not equivalent.
2. (C) 2 x 2 + x + 1 = 64 + 5 ´ 8 + 2
Þ
A + B,
x =7
A
0100 0110
B
+ 1101 0011
1 0001 1001
3. (C) All are 2’s complement of 7
Þ
11001
Discard the carry 1
00110
+
1
Þ
1001
Þ
B - A,
= 710
+ 0010 1101
B
1101 0011
A
+ 1011 1010
1 1000 1101
1
= 710
000111
Discard the carry 1
- A - B = A + B,
4. (C) See a example
42 in a byte
00101010
42in a word
0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0
-42 in a byte
11010110
-42 in a word
1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 0
A
1011 1010
B
+ 0010 1101
1110 0111
8. (B) 1110 = 10112
0.3
2Fi-1
Bi
Fi
0.6
0
0.6
1.2
1
0.2
1
0.4
0
0.4
11010000
0.8
0
0.8
1.6
1
0.6
Therefore (C) is correct.
00110000 2
1100 1111
+
6. (D) Here A , B are 1’s complement
A + B,
B
0111 0011
000110
+
-4810 =
010 0 0110
1
0111
5. (C) 4810 =
A
0110
+
111001
A - B = A + B,
= 710
00111
A
01010110
Repeat from the second line 0.310 = 0.01001
2
B + 1110 1100
10100 0010 ,
+
1
0100 0011
B - A = B + A,
9. (C)
B
1110 1100
A
+ 1010 1001
110010101
1
+
Received
b3
b2
p3
b1
p2
p1
1
1
0
1
1
0
0
C1* = b4 Å b2 Å b1 Å p1 = 0
C2* = b4 Å b3 Å b1 Å p2 = 1
C3* = b4 Å b3 Å b2 Å p3 = 1
10010110
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GATE EC BY RK Kanodia
Number Systems & Boolean Algebra
Chap 4.1
C3* C2* C1* = 110 which indicate position 6 in error
= ( AA + AB)( A + B + C) = A( A + B + C) = A
Transmitted code 1001100.
Therefore No gate is required to implement this
function.
10. (D) X = MNQ + M NQ + M NQ
23. (A)
= MQ + M NQ = Q( M + M N ) = Q( M + N )
11. (A) The logic circuit can be modified as shown in fig.
S. 4.1.11
A
B
Z
C+D
E
Fig. S4.1.11
Now Z = AB + ( C + D) E
A
B
C
( A + BC)
( A + B)( A + C)
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
0
1
1
1
1
1
0
0
1
1
1
0
1
1
1
Fig. S 4.1.23
12. (D) You can see that input to last XNOR gate is
same. So output will be HIGH.
24. (B) X = ABC + ABC + ABC = BC + ABC
13. (D) Z = A + ( AB + BC) + C
25. (B) ( A + B)( B + C ) = ( AB)( BC) = ABC
= A + ( A + B + B + C) + C = A + B + C
( A + B)( B + C ) = ( A + B) + ( B + C) = A + B + C
ABC = A + B + C
( A + B)( B + C) = ( A + B) + ( B + C)
AB + BC + AC = A + B + B + C + A + C = A + B + C
= AB + B + C = A + B + C
14. (C) ( X + Y )( X + Y ) = XY + X Y
From truth table Z = A + B + C
( X + Y )( X + Y )( X + Y ) = ( X + Y )( XY + X Y )
Thus (B) is correct.
= XY + XY = XY
26. (D) AC + BC = AC( B + B) + ( A + A) BC
= ABC + ABC + ABC + ABC
15. (B) Using duality
( A + B)( A + C)( B + C) = ( A + B)( A + C)
27. (D) F = A + AB + A BC + A B C( D + DE)
Thus (B) is correct option.
= A + AB + A B( C + C( D + E))
= A + A( B + B( C + D + E)) = A + B + C + D + E
16. (B) Z = ( AB)( CD)( EF ) = AB + CD + EF
17. (A) X = ( A B + AB)( A + B) = ( AB + A B)( AB) = AB
18. (B) Y = ( A Å B) × C = ( AB + AC) × C
= ( AB + AB) + C = A B + AB + C
19. (C) Z = A( A + A) BC = ABC
20. (A) Z = AB( B + C) = ABC
21. (A) Z = ( A + B ) × BC = ( AB) × BC = ABC
28. (B) A( B + C ( AB + AC)) = AB + AC ( AB × AC)
= AB + AC[( A + B)( A + C)]
= AB + AC ( A + AC + AB + B C) = AB
29. (C) ( AB ) × ( AB) = AB + AB = AB + AB
30. (B) X Z + XZ = X ( XY + XY ) + X ( X Y + XY )
= X ( XY + X Y ) + XY = XY + XY = Y
31. (A) X Å Y = X Y + XY = ( XY + XY ) = ( XY ) = X + Y
22. (A) A( A + B)( A + B + C)
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UNIT 4
Digital Electronics
(A) ( w + y)( x + y + z)( w + x + z)
(A) AC + AB
(B) AC + AB + BC
(B) ( w + x)( w + z )( x + y)( y + z )
(C) AB + BC + ABC
(D) Above all
(C) ( x + z)( w + y)
12. In the logic circuit of fig. P4.2.12 the redundant gate
(D) ( x + z )( w + y)
is
7. A function with don’t care condition is as follows
x
y
z
w
x
z
w
y
z
f ( a, b, c, d) = Sm(0, 2, 3, 5, 7, 8, 9, 10, 11) + Sdc ( 4, 15)
The minimized expression for this function is
(A) ab + b d + cd + abc
(B) ab + b d + cd + abd
(C) ab + b d + b c + abd
(D) Above all
8. A function with don’t cares is as follows :
g( X , Y , Z ) = Sm(5, 6) + Sdc (1, 2, 4)
1
2
4
X
3
Fig. P4.2.12
(A) 1
(B) 2
(C) 3
(D) 4
For above function consider following expression
1. XYZ + XYZ
2. XY + XZ
3. XZ + XZ + YZ
4. YZ + YZ
13. If function W, X, Y, and Z are as follow
W = R + PQ + RS
The solution for g are
(A) 1, 2, and 3
(B) 1, 2, and 4
(C) 1, and 4
(D) 1, and 3
X = PQRS + P Q RS + PQ R S
Y = RS + PR + PQ + P Q
Z = R + S + PQ + P Q R + PQ S
9. A logical function of four variable is given as
Then
f ( A, B, C, D) = ( A + B C)( B + CD)
The function as a sum of product is
(A) A + BC + ACD + BCD
(B) A + BC + ACD + BCD
(A) W = Z , X = Z
(B) W = Z , X = Y
(C) W = Y
(D) W = Y = Z
14. In a certain application four inputs A, B, C, D are
(C) AB + BC + ACD + BCD
fed to logic circuit, producing an output which operates
(D) AB + AB + ACD + BCD
a relay. The relay turns on when f(A, B, C, D) =1 for the
10. A combinational circuit has input A, B, and C and
its K-map is as shown in fig. P4.2.10. The output of the
circuit is given by
00
00
01
0101, 0110, 1101 and 1110. States 1000 and 1001 do not
occur, and for the remaining states the relay is off. The
minimized Boolean expression f is
CD
A
following states of the inputs (ABCD) : 0000, 0010,
01
11
1
10
1
1
(A) ACD + BCD + BCD
(B) BCD + BCD + ACD
(C) ABD + BCD + BCD
(D) ABD + BCD + BCD
15. There are four Boolean variables x1 , x2 , x3 and x4 .
1
The following function are defined on sets of them
f ( x3 , x2 , x1 ) = Sm( 3, 4, 5)
Fig. P4.2.1
Page
206
(A) ( AB + AB ) C
(B) ( AB + AB ) C
(C) ABC
(D) A Å B Å C
g( x4 , x3 , x2 ) = Sm(1, 6, 7)
h( x4 , x3 , x2 , x1 ) = fg
Then h( x4 , x3 , x2 , x1 ) is
11. The Boolean Expression Y = ( A + B)( A + C) is equal
(A) Sm(3, 12, 13)
(B) Sm(3, 6)
to
(C) Sm(3, 12)
(D) 0
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Combinational Logic Circuits
Statement for Q.16–17:
Chap 4.2
21. For a binary half subtractor having two input A and
A switching function of four variable, f ( w, x y, z) is
to equal the product of two other function f1 and f2 , of
B, the correct set of logical expressions for the outputs
D = ( A - B) and X (borrow) are
the same variable f = f1 f2 . The function f and f1 are as
(A) D = AB + AB , X = AB
follows :
(B) D = AB + AB , X = AB
f = Sm( 4, 7, 15)
(C) D = AB + AB , X = AB
f1 = Sm(0, 1, 2, 3, 4, 7, 8, 9, 10, 11, 15)
(D) D = AB + AB , X = AB
22. f1 f2 = ?
16. The number of full specified function, that will
satisfy the given condition, is
(A) 32
(B) 16
x0
I0
(C) 4
(D) 1
x1
I1
17. The simplest function for f2 is
x2
I2
(A) x
(B) x
(C) y
(D) y
D0
3-to-8
Decoder D1
D2
D3
D4
D5
D6
D7
and m9. If the literals in these minterms are
complemented, the corresponding minterm numbers are
(A) m3 and m0
(B) m9 and m6
(C) m2 and m0
(D) m6 and m9
(A) x0 x1 x2
(B) x0 Å x1 Å x2
(C) 1
(D) 0
23. The logic circuit shown in fig. P4.2.23 implements
A
I0
B
I1
C
I2
19. The minimum function that can detect a “divisible
by 3’’ 8421 BCD code digit (representation D8 D4 D2 D1 ) is
given by
(A) D8 D1 + D4 D2 + D8 D2 D1
(B) D8 D1 + D4 D2 D1 + D4 D2 D1 + D8 D4 D2 D1
D0
3-to-8
D
Decoder 1
D2
D3
D4
D5
D6
D7
EN
Fig. P4.2.23
(D) D4 D2 D1 + D4 D2 D1 + D8 D4 D2 D1
(A) D( A
I1
x2
I2
u C + AC )
(B) D( B Å C + AC )
(C) D( B Å C + AB )
20. f ( x2 , x1 , x0 ) = ?
x1
Z
D
(C) D4 D1 + D4 D2 + D8 D1 D2 D1
I0
f2
Fig. P4.2.22
18. A four-variable switching function has minterms m6
x0
f1
(D) D( B
u C + AB )
Statement for Q.24–25:
D0
3-to-8
D
Decoder 1
D2
D3
D4
D5
D6
D7
The building block shown in fig. P4.2.24–25 is a
f
active high output decoder.
Fig. P4.2.21
(A) P(1, 2, 4, 5, 7)
(B) S(1, 2, 4, 5, 7)
(C) S(0, 3, 6)
(D) None of Above
A
I0
B
I1
C
I2
D0
3-to-8
Decoder D1
D2
D3
D4
D5
D6
D7
X
Y
Fig. P4.2.24-25
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UNIT 4
24. The output X is
Digital Electronics
28. Z 2 = ?
(A) AB + BC + CA
(B) A + B + C
(A) ab + bc + ca
(B) a + b + c
(C) ABC
(D) None of the above
(C) abc
(D) a u b u c
25. The output Y is
29. This circuit act as a
(A) A + B
(B) B + C
(A) Full adder
(B) Half adder
(C) C + A
(D) None of the above
(C) Full subtractor
(D) Half subtractor
26. A logic circuit consist of two 2 ´ 4 decoder as shown
30. The network shown in fig. P4.2.30 implements
in fig. P4.2.26.
x
1 MUX
1
0
f
D0
A0
D1
D2
y
A
A1
D0
A1
D3
f
D1
B
1 MUX
0
0
S0
D2
z
A0
D3
S0
C
Fig. P4.2.26
Fig. P4.2.30
The output of decoder are as follow
D0 = 1 when A0 = 0,
A1 = 0
D1 = 1 when A0 = 1,
A1 = 0
D2 = 1 when A0 = 0,
A1 = 1
D3 = 1 when A0 = 1,
A1 = 1
(A) NOR gate
(B) NAND gate
(C) XOR gate
(D) XNOR gate
31. The MUX shown in fig. P4.2.31 is 4 ´ 1 multiplexer.
The output Z is
C
The value of f ( x, y, z) is
I3
I2
(A) 0
(B) z
(C) z
(D) 1
MUX
I0
Statement for Q.27-29:
+5 V
S1 S0
A
B) A Å B Å C
(A) ABC
c
1
0
MUX
Z1
B
Fig. P4.2.31
A MUX network is shown in fig. P4.2.27-29.
c
Z
I1
(C) A u B
uC
(D) A + B + C
32. The output of the 4 ´ 1 multiplexer shown in fig.
S0
P4.2.32 is
a
1
a
0
MUX
Z0
S0
b
Y
c
b
1
S0
MUX
+5 V
Z2
X
0
Page
208
(A) a + b + c
(B) ab + ac + bc
(C) a u b u c
(D) a Å b Å c
MUX
I1
I0
Z
S1 S0
Y
Fig. P4.2.32
Fig. P4.2.27-29
27. Z1 = ?
I3
I2
(A) X + Y
(B) X + Y
(C) XY + X
(D) XY
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Combinational Logic Circuits
Chap 4.2
33. The MUX shown in fig. P4.2.33 is a 4 ´ 1
36. The 4–to–1 multiplexer shown in fig. P4.2.36
multiplexer. The output Z is
implements the Boolean expression
I3
C
I3
I2
MUX
I1
C
I0
Z
uC
(D) B u C
0
1
0
1
1
The input to I1 and I 3 will be
(A) yz , y + z
(B) y + z, y u z
(C) y + z, y Å z
(D) x + y ,
yÅ z
37. The 8-to-1 multiplexer shown in fig. P4.2.37 realize
0
1
1
x
I0
I1 MUX
I2
I3
EN
S1 S0
w
x
f ( w, x, y, z) = Sm( 4, 5, 7, 8, 10, 12, 15)
34. f = ?
w
S1 S0
w
(B) A
(C) B Å C
f
Fig. P4.2.36
Fig. P4.2.33
I0
I1 MUX
I2
I3
EN
S1 S0
I0
0
B
(A) A Å C
0
1
1
0
1
MUX
I1
S1 S0
A
I2
z
I0
I1 MUX
I2
I3
EN
S1 S0
y
the following Boolean expression
1
z
z
f
0
z
0
1
z
z
I0
I1
I2
I3
I4
I5
I6
I7
MUX
EN S2 S1 S0
0
x
w
x
y
Fig. P4.2.37
Fig. P4.2.34
(A) wxyz + wx yz + xy + yz
(A) wxz + w x z + wyz + xy z
(B) wxyz + wxyz + x y + yz
(B) wxz + wyz + wyz + w x y
(C) wx yz + w x yz + yz + zx
(C) w x z + wy z + w yz + wxz
(D) wxyz + wxyz + gz + zx
(D) MUX is not enable
35. For the logic circuit shown in fig. P4.2.35 the output
Statement for Q.38-40:
Y is
A PLA realization is shown in fig. P4.2.38–40
x0
1
0
I0 I1 I2 I3 I4 I5 I6 I7
C
B
A
EN
S2
S1
S0
x1
x2
MUX
Y
X
Fig. P4.2.35
(A) A Å B
(B) A Å B
(C) A Å B Å C
(D) A Å B Å C
X
X
X
X
X
X
X
f1
f2
f3
Fig. P.4.2.38-40
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UNIT 4
38. f1 ( x2 , x1 , x0 ) = ?
Digital Electronics
The terms P1 , P2 , P3 , P4 and P5 are
(A) x2 x0 + x1 x0
(B) x2 x0 + x1 x2
(C) x2 Å x0
(D) x2
ux
0
39. f2 ( x2 , x1 , x0 ) = ?
(A) Sm(1, 2, 5, 6)
(B) Sm(1, 2, 6, 7)
(C) Sm(2, 3, 4)
(D) None of the above
(A) ab, ac, bc, bc, ab
(B) ab , bc, ac, ab , bc
(C) ac, ab, bc, ab , bc
(D) Above all
43. The circuit shown in fig. P.4.2.43 has 4 boxes each
described by input P, Q, R and output Y , Z with
Y = P Å Q Å R and Z = RQ + PR + QP .
Q
40. f3( x2 , x1 , x0 ) = ?
(A) P M(0, 4, 6, 7)
(B) P M(2, 4, 5,7)
(C) P M(1, 2, 3, 5)
(D) P M(2, 3, 4, 7)
P
P
Q
P
Q
P
Q
P
Q
Z
R
Z
R
Z
R
Z
R
41. If the input X 3 X 2 X1 X 0 to the ROM in fig. P4.2.41
are 8–4–2–1 BCD numbers, then output Y3Y2 Y1 Y0 are
X3
X2
X1
X0
Output
Fig. P4.2.43
The circuit act as a 4 bit
BCD to Decimal Decoder
(A) adder giving P + Q
D0 D1 D2 D3 D4 D5 D6 D7 D8 D9
X X X X Y3
(C) subtractor giving Q - P
X X X X X X Y2
(D) adder giving P + Q + R
X X
X
(B) subtractor giving P - Q
X
X X Y1
X
X
44. The circuit shown in fig. P4.2.44 converts
X Y0
MSB
Fig. P4.2.41
(A) 2–4–2–1 BCD number
(B) gray code number
(C) excess 3 code converter
(D) none of the above
+
42. It is desired to generate the following three Boolean
+
function
f1 = ab c + abc + bc
MSB
Fig. P4.2.44
f2 = ab c + ab + abc,
f3 = ab c + abc + ac
(A) BCD to binary code
by using an OR gate array as shown in fig. P4.2.42
where P1 and P5 are the product terms in one or more
of the variable a, a , b, b , c and c.
P1
X
P2
X
(B) Binary to excess
(C) Excess–3 to Gray Code
(D) Gray to Binary code
X
X
P3
*******
X
P4
P5
X
F1
F2
F3
Fig. P4.2.42
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210
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Combinational Logic Circuits
16. (A) f = Sm(4, 7, 15),
RS
00
01
11
00
PQ
Chap 4.2
f1 = Sm(0, 1, 2, 3, 4, 7, 8, 9, 10, 11, 15)
10
f2 = Sm(4, 7, 15) + Sdc(5, 6, 12, 13, 14)
1
01
1
1
1
11
1
1
1
There are 5 don't care condition. So 2 5 = 32 different
1
functions f2 .
17. (A) f2 = Sm(4, 7, 15) + Sdc(5, 6, 12, 13, 14), f2 = x
1
10
yz
00
01
11
10
00
0
0
0
0
01
1
´
1
´
11
´
´
1
´
10
0
0
0
0
Fig. S 4.2.13c
Z = R + S + PQ + P Q R + PQ S= R + S + PQP Q R PQ S
wx
= R + S + ( P + Q )( P + Q + R)( P + Q + S)
= R + S + ( PQ + PR + Q P + Q R)( P + Q + S)
= R + S + PQ + PQS + PR + PRQ + PRS + Q PS + Q PR + Q RS
Fig. S 4.2.17
RS
00
01
11
10
1
1
1
1
1
1
11
1
1
1
10
1
1
1
00
PQ
01
1
18. (B) m6 = ABCD ,
m9 = ABCD
After complementing literal
m6¢ = ABCD = m9 ,
m¢9 = ABCD = m6
19. (B) 0, 3, 6 and 9 are divisible by 3
D2 D1
00
00
Fig. S 4.2.13d
D8 D4 01
11
= R + S + PQ
We can see that W = Z , X = Z
01
11
1
10
1
1
´
10
´
´
´
1
´
´
14. (D) ABD + BCD + BCD
Fig. S 4.2.19
CD
00
00
AB
01
11
1
10
1
01
1
1
11
1
1
10
´
f = D8 D1 + D4 D2 D1 + D4 D2 D1 + D8 D4 D2 D1
20. (B) f = Sm(0, 3, 6) = Sm(1, 2, 4, 5, 7)
21. (C) D = AB + AB,
´
X = AB
A
B
D
X
0
0
0
0
Fig. S 4.2.14
0
1
1
1
15. (A) f = x3 x2 x1 + x3 x2 x1 + x3 x2 x1 = x3 x2 x1 + x3 x2
1
0
1
0
g = x4 x3 x2 + x4 x3 x2 + x4 x3 x2 = x4 x3 x2 + x4 x3
1
1
0
0
fg = x4 x3 x2 x1 + x4 x3 x2
Fig. S 4.2.20
= x4 x3 x2 x1 + x4 x3 x2 x1 + x4 x3 x2 x1
h = Sm(3, 12, 13)
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UNIT 4
Digital Electronics
22. (D) f1 = Sm(0, 2, 4, 6),
29. (A) The equation of Z1 is the equation of sum of A
f2 = Sm(1, 3, 5, 7), f1 f2 = 0
and B with carry and equation of 2 is the resultant
carry. Thus, it is a full adder.
23. (D) Z = D( ABC + ABC + ABC + ABC + ABC)
= D( AB ( C + C) + BC( A + A) + ABC )
30. (B) f1 = CD + CB = CB ,
= D( AB + BC + ABC ) = D( B ( A + AC ) + BC)
f = f 1 + f1 A = CB + CBA = CB + A
= D( BA + BC + BC) = D( B
u C + AB )
24. (A) X = Sm(3, 5, 6, 7),
S = F1
= C + B + A = ABC
X = AB + BC + CA
31. (D) Z = ABC + AB + AB + AB
= A ( BC + B) + A( B + B) = A ( B + C) + A = A + B + C
25. (D) Y = Sm(1, 3, 5, 7),
Y =C
32. (A) Z = XY + XY + XY ,
Z=X +Y
BC
00
A
01
11
00
33. (A) Z = ABC + ABC + ABC + ABC
10
= AC + AC = A Å C
1
01
1
1
1
34. (A) The output from the upper first level
multiplexer is fa
multiplexer is fb
Fig. S4.2.24
fa = wx + wx,
26. (D) D0 = A1 A0 , D1 = A1 A0 ,
and from the lower first level
fb = wx + wx = x
f = fa yz + fb yz + yz = ( wx + wx) yz + xyz + yz
D2 = A1 A0 ,
= wxyz + wx yz + xy + yz
BC
00
A
01
11
00
1
1
01
1
1
10
35. (D) Output is 1 when even parity
BA
00
00
C
Fig. S4.2.25
01
11
1
10
1
1
01
1
D3 = A1 A0
Fig. S4.2.35
For first decoder A0 = x , A1 = y, D2 = yx , D3 = xy
For second decoder A1 = D2 D3 = yxxy = 0,
A0 = z
Therefore Y = A Å B Å C
f = D0 + D1 = A1 A0 + A1 A0 = A1 = 1
36. (B) I1 = y + z ,
27. (D) The output of first MUX is
Z o = ab + ab = ( a Å b)
28. (A) Z 2 = bS0 + cS0
= b( ab + ab ) + c( ab + ab ) = ab + ab c + abc
z
yz
00
01
11
10
00
0
0
0
0
I0 = 0
01
1
1
1
0
I1 = y + z
11
1
0
1
0
10
1
0
0
1
This is input to select S0 of both second-level MUX
Z1 = CS0 + CS0 = C Å S0 = a Å b Å c
I3 = y
S1 S0
bb
wx
= a( b + b c) + abc = ab + ac + abc
= ab + ac + bc
Fig. S 4.2.36
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214
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I2 = z
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Combinational Logic Circuits
Chap 4.2
37. (C) Let z = 0, Then
42. (A) f1 = ab c + abc + bc = ac + ab
f = w x y + w xy + wx y + wxy = w x + w y
f2 = ab c + ab + abc = ac + b c
If we put z = 0 in given option then
f3 = ab c + abc + ac = ab + bc
(A) = w x + xy
(B) = wy + w x y
(C) = w x + wy
Since MUX is enable so option (C) is correct.
Thus P1 = ab, P2 = ac,
P3 = bc, P4 = bc, P5 = ab
43. (B) Let P = 1001 and Q = 1010 then
38. (C)f = x0 x2 + x0 x1 x2 + x0 x2 = x0 x2 (1 + x1 ) + x0 x2
Yn = Pn Å Qn Å Rn ,
= x0 x2 + x0 x2
Z n = Rn Qn + Pn Rn + Qn Pn
output is 1111 which is 2’s complement of –1. So it gives
39. (B) f2 = x0 x1 + x1 x2 + x0 x1 x2
P - Q . Let another example
= x0 x1 x2 + x0 x1 x2 + x1 x2 x0 + x1 x2 x0 + x0 x1 x2
then output is 00111. It gives P–Q.
P = 1101 and Q = 0110
So (B) is correct.
= x2 x1 x0 + x2 x1 x0 + x2 x1 x0 + x2 x1 x0
f2 ( x2 , x1 , x0 ) = Sm(1, 2, 6, 7)
Pn
Qn
Rn
Zn
Yn
40. (C) f3 = x0 x1 + x1 x2
n =1
1
0
0
0
1
= x2 x1 x0 + x2 x1 x0 + x2 x1 x0 + x2 x1 x0
n =2
0
1
0
1
1
f3( x2 , x1 , x0 ) = Sm(0, 4, 6, 7)
n=3
0
0
1
1
1
f3( x2 , x1 , x0 ) = PM(1, 2, 3, 5)
n=4
1
1
1
1
1
1
41. (A)
Fig. S4.2.43a
Let X 3 X 2 X1 X 0 be 1001 then Y3Y2 Y1 Y0 will be 1111.
Let X 3 X 2 X1 X 0 be 1000 then Y3Y2 Y1 Y0 will be 1110
Let X 3 X 2 X1 X 0 be 0110 then Y3Y2 Y1 Y0 will be 1100
bc
00
01
00
a
1
01
Pn
Qn
Rn
Zn
Yn
n =1
1
0
0
0
1
11
10
n =2
0
1
0
1
1
1
1
n=3
1
1
1
1
1
n=4
1
0
1
0
0
1
0
Fig. S4.2.43b
Fig. S4.2.42a
44. (D) Let input be
1010
10
output will be
1101
1
Let input be
0110
1
output will be 0100
bc
00
01
11
00
a
01
1
1
This convert gray to Binary code.
Fig. S4.2.42b
bc
00
a
00
01
11
1
1
1
01
*******
10
1
Fig. S4.2.42b
So this converts 2–4–2–1 BCD numbers.
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Sequential Logic Circuits
1
(B)
Q1
D1
D2
13. The counter shown in fig. P4.3.13 counts from
Y
Q2
Chap 4.3
J
C
X
Q1
(C)
Q1
D1
D2
Q1
(D)
Y
Q2
1
D2
A CLR K
Fig.P4.3.13
Q2
Q1
D1
B CLR K
A
B
C
X
1
J
A
Q2
C CLR K
1
J
B
Y
Q2
X
(A) 0 0 0 to 1 1 1
(B) 1 1 1 to 0 0 0
(C) 1 0 0 to 0 0 0
(D) 0 0 0 to 1 0 0
14. The mod-number of the asynchronous counter
Q1
Q2
shown in fig. P4.2.13 is
J
Q0
J
Q1
J
Q2
Q3
J
J
Q4
11. The circuit shown in fig. P4.3.11 is
K CLR
T
T
Q
K CLR
K CLR
K CLR
K CLR
Q
All J.K. input are HIGH
CLK
CLK
A
B
Q
Q
Fig.P4.3.14
Fig.P4.3.11
(A) a MOD–2 counter
(A) 24
(B) 48
(C) 25
(D) 36
(B) a MOD–3 counter
15. The frequency of the pulse at z in the network
(C) generate sequence 00, 10, 01, 00.....
shown in fig. P4.3.15. is
(D) generate sequence 00, 10, 00, 10, 00 ......
160 kHZ
12. The counter shown in fig. P4.3.12 is a
10-Bit
Ring Counter
w 4-Bit Parallel
Counter
x
Mod-25
Ripple Counter
y
4-Bit Johnson
Counter
z
Fig.P4.3.15
J
Q
Q
B
C
Q
J
Q
K
Q
J
1
Q
K
1
(B) 160 Hz
(C) 40 Hz
(D) 5 Hz
16. The three-stage Johnson counter as shown in fig.
A
K
(A) 10 Hz
P4.2.16 is clocked at a constant frequency of fc from the
CLK
starting state of Q2Q1Q0 = 101. The frequency of output
Q2Q1Q0 will be
Fig.P4.3.12
(A) MOD–8 up counter
J2
Q2
J1
Q1
J0
Q0
K2
Q2
K1
Q1
K0
Q0
(B) MOD–8 down counter
(C) MOD–6 up counter
CLK
(D) MOD–6 down counter
Fig.P4.3.16
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UNIT 4
(A)
fc
8
fc
6
21. In the circuit shown in fig. P4.3.21 is PIPO 4-bit
f
(D) c
2
input lines are connected to a 4 bit bus. Its output acts
(B)
f
(C) c
3
register, which loads at the rising edge of the clock. The
as the input to a 16 ´ 4 ROM whose output is floating
17. The counter shown in the fig. P4.3.17 has initially
Q2Q1Q0 = 000. The status of Q2Q1Q0 after the first pulse
is
J2
Q1
J1
Q2
Digital Electronics
J0
when the enable input E is 0. A partial table of the
contents of the ROM is as follows
Address
0
2
4
6
8
10
12
Data
0011
1111
0100
1010
1011
1000
0010
The clock to the register is shown below, and the
Q0
data on the bus at time t1 is 0110.
K2
Q2
Q1
K1
K0
MSB
Q0
CLK
Fig.P4.3.17
(A) 0 0 1
(B) 0 1 0
(C) 1 0 0
(D) 1 0 1
CLK
A
18. A 4 bit ripple counter and a 4 bit synchronous
1
counter are made by flips flops having a propagation
E
delay of 10 ns each. If the worst case delay in the ripple
ROM
counter and the synchronous counter be R and S
respectively, then
(A) R = 10 ns, S = 40 ns
(B) R = 40 ns, S = 10 ns
(C) R = 10 ns, S = 30 ns
(D) R = 30 ns, S = 10 ns
19. A 4 bit modulo–6 ripple counter uses JK flip-flop. If
the propagation delay of each FF is 50 ns, the
maximum clock frequency that can be used is equal to
(A) 5 MHz
(B) 10 MHz
(C) 4 MHz
(D) 20 Mhz
t2
Fig. P4.3.21
right-shift, register shown in fig. P4.3.20 is 0 1 1 0.
After three clock pulses are applied, the contents of the
CLK
t
t1
20. The initial contents of the 4-bit serial-in-parallel-out
shift register will be
CLK
The data on the bus at time t2 is
(A) 1 1 1 1
(B) 1 0 1 1
(C) 1 0 0 0
(D) 0 0 1 0
22. A 4-bit right shift register is initialized to value
1000 for (Q3 , Q2 , Q1 , Q0 ). The D input is derived from
0
1
1
0
Q0 , Q2 and Q3 through two XOR gates as shown in fig.
P4.2.22. The pattern 1000 will appear at
Fig.P4.3.20
Page
219
(A) 0 0 0 0
(B) 0 1 0 1
(C) 1 1 1 1
(D) 1 0 1 0
D
Q0
Q1
Fig.P4.3.22
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Q3
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Sequential Logic Circuits
Chap 4.3
(A) 3rd pulse
(B) 7th pulse
28. To count from 0 to 1024 the number of required
(C) 6th pulse
(D) 4th pulse
flip-flop is
Statement for Q.23–24:
The 8-bit left shift register and D-flip-flop shown
in fig. P4.3.22–23 is synchronized with same clock. The
(A) 10
(B) 11
(C) 12
(D) 13
29. Four memory chips of 16 ´ 4 size have their address
buses connected together. This system will be of size
b7
b6 b5 b4 b3 b2 b1 b0
D
(A) 64 ´ 4
(B) 32 ´ 8
(C) 16 ´ 16
(D) 256 ´ 1
Q
30. The address bus width of a memory of size 1024 ´ 8
CLK
bits is
Q
Fig.P4.3.23-24
(A) 10 bits
(B) 13 bits
(C) 8 bits
(D) 18 bits
31. For the circuit of Fig. P4.3.31 consider the
D flip-flop is initially cleared.
statement:
23. The circuit act as
Assertion (A) : The circuit is sequential
(A) Binary to 2’s complement converter
Reason (R) : There is a loop in circuit
(B) Binary to Gray code converter
d
(C) Binary to 1’s complement converter
(D) Binary to Excess–3 code converter
24. If initially register contains byte B7, then after 4
c
(B) 72
(C) 7E
(D) 74
b
e
z0
clock pulse contents of register will be
(A) 73
z1
a
b
Fig.P4.3.131
Choose correct option
Statement for Q.25–26:
A Mealy system produces a 1 output if the input
has been 0 for at least two consecutive clocks followed
immediately by two or more consecutive 1’s.
(A) Both A and R true and R is the correct
explanation of A
(B) Both A and R true but R is not a correct
explanation on of A
(C) A is true but R is false
25. The minimum state for this system is
(A) 4
(B) 5
(C) 8
(D) 9
(D) A is false
*****************
26. The flip-flop required to implement this system are
(A) 2
(B) 3
(C) 4
(D) 5
27. The output of a Mealy system is 1 if there has been
a pattern of 11000, otherwise 0. The minimum state for
this system is
(A) 4
(B) 5
(C) 6
(D) 7
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UNIT 4
Digital Electronics
6. (D) Q + = LM + LMQ
SOLUTIONS
= L( M + MQ )
1. (C) Given FF is a negative edge triggered T flip-flop.
= L M + LQ
So at the negative edge of clock Vi FF will invert the
L
M
Q+
0
0
0
2. (A) At first rising edge of clock, D is HIGH. So Q will
0
1
0
be high till 2nd rising edge of clock. At 2nd rising edge,
1
0
1
D is low so Q will be LOW till 3rd rising edge of clock.
1
1
Q1
output if there is 1 at input.
At 3rd rising edge, D is HIGH, so Q will be HIGH till
Fig. S4.3.6
4th rising edge. At 4th rising edge D is HIGH so Q will
be HIGH till 5th rising. edge. At 5th rising edge, D is
LOW, so Q will be LOW till 6th rising edge.
7. (D)
Initially
3. (C)
J
Q
Q
1
0
1
Qn + 1
Qn +1
x
Q
S
R
Q+
Clock 1st
1
1
0
1
1
0
0
0
0
1
0
2nd
0
1
1
0
0
1
0
1
1
0
1
3rd
1
1
0
1
1
0
1
0
1
0
1
4th
0
1
1
0
0
1
1
1
0
1
0
5th
1
1
0
1
1
0
Fig. S4.3.3
Fig. S4.3.7
4. (D) Q + = x Å Q
Therefore sequence is 010101.
Q1+ = x1 Å Q0 = x1 0 + x1 0 = x1
8. (A)
Q2+ = x2 Å x1 ,
Q3+ = x3 Å x2 Å x1
Q4+ = x4 Å x3 Å x2 Å x1
5. (C)
A
B
X
Y
1
1
0
1
1
0
0
1
X and Y are fixed at 0 and 1.
So this generate the even parity and check odd parity.
9. (D) Z = XQ + YQ
A
B
S
R
Q
Q+
0
0
1
0
0
1
0
0
1
0
1
1
0
1
0
1
0
0
0
1
0
1
1
0
1
0
0
0
0
0
1
0
0
0
1
1
1
1
1
1
0
´
Comparing from the truth table of J–K FF
1
1
1
1
1
´
Y = J, X = K
X
Y
Z
0
0
Q
0
1
0
1
0
1
1
1
Q1
Fig. S4.3.9
Fig. S4.3.5
Q + = AB + AQ = AB + BQ
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221
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Sequential Logic Circuits
Chap 4.3
14. (A) It is a 5 bit ripple counter. At 11000 the output
10. (C)
of NAND gate is LOW. This will clear all FF. So it is a
t0
t2
t1
Mod–24 counter. Note that when 11000 occur, the
t3
CLR input is activated and all FF are immediately
Q1
cleared. So it is a MOD 24 counter not MOD 25.
0
1
15. (D) 10-bit ring counter is a MOD–10, so it divides
D2=Q1
the 160 kHz input by 10. therefore, w = 16 kHz. The
Q2
four-bit parallel counter is a MOD–16. Thus, the
t0
frequency at x = 1 kHz. The MOD–25 ripple counter
t1
produces a frequency at y = 40 Hz. (1 kHz/25 = 40 Hz).
Fig.S4.3.10
11. (B)
The four-bit Johnson Counter is a MOD-8. This, the
Present State
FF Input
Next State
Q A QB
TA TB
Q +A QB+
0
0
0
1
0
1
0
1
1
1
1
0
1
0
1
0
0
0
1
1
1
1
0
0
frequency at z = 5 Hz.
16. (D)
Q0 Q0
Q2 Q2
Q1 Q1
J2 K2
J1 K 1
J0 K0
Q2+ Q1+ Q0+
1 0 1
Fig. S4.3.11
From table it is clear that it is a MOD–3 counter.
12. (B) It is a down counter because 0 state of previous
0 1
1 0
0 1
0 1 0
1 0
0 1
1 0
1 0 1
0 1
1 0
0 1
0 1 0
1 0
0 1
1 0
1 0 1
FFs change the state of next FF. You may trace the
following sequence, let initial state be 0 0 0
Fig. S4.3.16
We see that 1 0 1 repeat after every two cycles, hence
frequency will be fc / 2 .
FF C
FF B
FF A
JK C
JK B
JK A
C+ B+ A+
111
111
111
111
J 2K 2 = 1 0
Þ
Q2 = 1,
000
000
110
110
J1 K 1 = 0 0
Þ
Q1 = 1,
000
110
111
101
J 0K 0 = 0 0
Þ
Q0 = 0
000
001
110
100
18. (B) In ripple counter delay 4Td = 40 ns.
111
111
111
011
The synchronous counter are clocked simultaneously,
001
000
110
010
then its worst delay will be equal to 10 ns.
001
110
111
001
19. (A) 4 bit uses 4 FF
000
001
110
000
Total delay Ntd = 4 ´ 50 ns = 200 ´ 10 -9
1
f =
= 5 Mhz
200 ´ 10 -9
Fig. S4.3.12
13. (C) It is a down counter because the inverted FF
17. (C) At first cycle
output drive the clock inputs. The NAND gate will clear
20. (D) At pulse 1 input,
FFs A and B when the count tries to recycle to 111. This
So contents are 1 0 1 1,
will produce as result of 100. Thus the counting
At pules 2 input 1 Å 1 = 0‘
sequence will be 100, 011, 010, 001, 000, 100 etc.
So contents are 0 1 0 1,
1 Å 0 =1
At pules 3 input 0 Å 1 = 1, contents are 1 0 1 0
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CHAPTER
4.4
DIGITAL LOGIC FAMILIES
Statement for Q.1–2:
4. In the circuit shown in fig. P.4.4.4. the output Z is
+5 V
Consider the DL circuit of fig. P4.4.1–2.
+5 V
+5 V
+5 V
Z
A
B
+
C
Fig. P4.4.4
+
V1
+
Vo
-
V2
-
-
Fig. P4.4.1-2
1. For positive logic the circuit is a
(A) AB + C
(B) ABC
(C) ABC
(D) ABC
Statement for Q.5–7:
(A) AND
(B) OR
(C) NAND
(D) NOR
Consider the AND circuit shown in fig. P4.4.5–7.
The binary input levels are V(0) = 0 V and V(1) = 25 V.
Assume ideal diodes. If V1 = V (0) and V2 = V (1), then Vo
2. For negative logic the circuit is a
is to be at 5 V. However, if V1 = V2 = V (1), then Vo is to
(A) AND
(B) OR
(C) NAND
(D) NOR
rise above 5 V.
Vss
3. The diode logic circuit of fig. P4.4.3 is a
20 kW
1 kW
D2
D1
V1
V1
Vo
V2
V2
Vo
1 kW
D1
D2
D0
+5 V
Fig. P4.4.5-7
Fig. P4.4.3
5. If Vss = 20 V
and V1 = V2 = V (1), the diode current
I D1 , I D2 , and I D0 are
Page
224
(A) AND
(B) OR
(A) 1 mA, 1 mA, 4 mA
(B) 1 mA, 1 mA, 5 mA
(C) NAND
(D) NOR
(C) 5 mA, 5mA, 1 mA
(D) 0,
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Digital Logic Families
Chap 4.4
6. If Vss = 40 V and both input are at HIGH level then,
+VCC
diode current I D1 , I D2 and I D0 are respectively
(A) 0.4 mA, 0.4 mA, 0
RC
(B) 0, 0, 1 mA
(C) 0.4 mA, 0.4 mA, 1 mA
Vo
RB1
(D) 0, 0, 0
RB2
V1
V2
Q2
Q1
7. The maximum value of Vss which may be used is
(A) 30 V
(B) 25 V
(C) 125 V
(D) 20 V
Fig. P4.4.10-11
10. For positive logic the gate is
8. The ideal inverter in fig. P4.4.8 has a reference
(A) AND
(B) OR
voltage of 2.5 V. The forward voltage of the diode is 0.75
(C) NAND
(D) NOR
V. The maximum number of diode logic circuit, that
may be cascaded ahead of the inverter without
producing logic error, is
+5 V
+5 V
11. For negative logic the gate is
(A) AND
(B) OR
(C) NAND
(D) NOR
+5 V
Statement for Q.12–13:
Consider the RTL circuit of fig. P4.4.12–13.
+5 V
A
Z
+VCC
B
C
RC
D
n Stages of Diode Logic
RB1
RB2
V1
Fig. P4.4.8
(B) 4
(C) 5
(D) 9
Vo2
RB3
V2
Q2
Q1
(A) 3
Vo1
Q3
Fig. P4.4.12-13
9. Consider the TTL circuit in fig. P4.4.9. The value of
12. If Vo1 is taken as the output, then circuit is a
V H and VL are respectively
(A) AND
(B) OR
(C) NAND
(D) NOR
+5 V
13. If Vo2 is taken as output, then circuit is a
2 kW
4 kW
Vo
(A) AND
(B) OR
(C) NAND
(D) NOR
Vi
Statement for Q.14–15:
Consider the TTL circuit of fig. P4.4.14. If either or
Fig. P4.4.9
(A) 5 V, 0 V
(B) 4.8 V, 0 V
(C) 4.8 V, 0.2 V
(D) 5 V, 0.2 V
both V1
and V2
are logic LOW, Q1
saturation.
+VCC
Statement Q.10–11:
R1
R2
Consider the resistor transistor logic gate of fig.
Vo2
R3
V1
V2
P4.4.10-11.
is driven to
Q2
Q1
Vo1
Fig. P4.4.14-15
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Digital Logic Families
22. The circuit shown in fig. P4.4.22 is
25. The circuit shown in fig. P4.4.25. implements the
function
+VDD
M2
B
M1
+VDD
C
Y
A
Chap 4.4
A
B
D
Y
Fig. P4.4.22
(A) NAND
(B) NOR
(C) AND
(D) OR
Fig. P4.4.25
+VDD
Y
M2
(A) ( A + B) C + D
(B) ( AB + C) D
(C) ( A + B) C + D
(D) ( AB + C) D
26. Consider the CMOS circuit shown in fig. P4.4.26.
M3
M1
D
C
23. The circuit shown in fig. P4.4.23 acts as a
A
B
A
The output Y is
+VDD
B
C
A
Fig. P4.4.23
B
(A) NAND
Y
(B) NOR
C
(C) AND
(D) OR
B
A
24. The circuit shown in fig. P4.4.24 implements the
function
Fig. P4.4.26
+VDD
A
B
C
A
B
C
+VDD
(A) ( A + C) B
(B) ( A + B) C
(C) AB + C
(D) AB + C
27. The CMOS circuit shown in fig. P4.4.27 implement
Y
A
A
B
B
C
C
+5 V
Logic Input
A to E
PMOS
Network
Y
A
B
C
D
Fig. P4.4.24
E
(A) ABC + ABC
(B ABC + ( A + B + C)
(C) ABC + ( A + B + C)
(D) None of the above
Fig. P4.4.27
(A) AB + CD + E
(B) ( A + B)( C + D) E
(C) AB + CD + E
(D) ( A + B)( C + D) E
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UNIT 4
V1
V2
Digital Electronics
For n attached gate I o = nI B ( sat ) .
Vo
To assure no logic error Vo = VCC - I o RC > V H = 35
. V
V - 35
.
5 - 35
.
n £ CC
=
= 15.6 Þ n £ 15
RC I B ( sat ) 640(0.15m)
Actual
Logic
Actual
Logic
Actual
Logic
VH
1
VH
1
VCE ( sat )
0
VL
0
VL
0
VCE
1
VH
1
VL
0
VCE ( sat )
0
19. (A) Let V1 = V2 = 0 V, then M 3 will be ON, M1 and
VL
0
VH
1
VCE ( sat )
0
M 2 OFF and M 4 ON, hence Vo = -VDD . Let V1 = 0 V and
V2 = -VDD then M 3 will be ON, M1 OFF M 4 OFF, M 2
13. (B) The Q3 stage is simply an inverter (a NOT gate).
ON, hence Vo = -VDD. Let V1 = -VDD and V2 = 0 V, then
Thus output Vo2
M 3 OFF, M 4 ON, M 2 OFF hence Vo = -VDD. Finally if
is the logic complement of Vo1 .
Therefore this is a OR gate.
V1 = V2 = -VDD, M 3 and M 4 will be OFF and M1 , M 2
14. (A) When Q1
satisfies the function of a negative NAND gate.
will be ON, hence Vo = 0 V. Thus the given CMOS gate
is saturated, Vo1 is logic LOW
otherwise Vo1 is logic HIGH. The following truth table
shows AND logic
20. (C) If V A = -VDD then M1
is ON and VY = 0 V. If
VB = VC = -VDD and V A = 0 V then M 3 and M 2 are ON
V1
V2
Vo1
but M1 is OFF hence VY = 0 V. If V A = 0 V and either or
1
1
1
both VB , VC are 0 V then M1 is OFF and either or both
0
1
0
1
0
0
0
0
0
M2
and M 3
will be OFF, which implies no current
flowing through M 4
hence VY = -VDD . Thus given
circuit satisfies the logic equation A + BC .
21. (A) Let V1 = V2 = 0 V = V (0) then M 4 and M 3 will be
OFF hence Vo = VDD = V (1). Let
15. (C) The Q2 stage is simply an inverter. Thus output
ON and M 2 , M1
Vo2 is the logic complement of Vo1 .
V1 = 0 V, V2 = VDD then M 4 and M 2 will be ON but M 3
and
16. (C) If V1 = V2 £ VL , Vo1 » VCC . If V1 ( V2 ) > V H , while
V2 ( V1 ) £ VL , Q1 (Q2 )
is ON and Q2 (Q1 )
is OFF and
Vo1 » VCC . If V1 = V2 > V H , both Q1 and Q2 are ON and
Vo1 » 2 VCE ( sat ) . The truth table shows NAND logic
M1
will
be
OFF
hence
Vo = 0 = V (0).
Let
V1 = VDD , V2 = 0 V , then M 4 and M 3 will be OFF and
M1 ON hence Vo = 0 V = V (0). Finally if V1 = V2 = VDD ,
M1
and M 2
will be ON but M 4 will be OFF hence
Vo = 0 V = V (0). Thus the given CMOS satisfy the
function of a positive NOR gate.
V1
V2
Vo
Actual
Logic
Actual
Logic
Actual
Logic
VL
0
VL
0
VCC
1
VH
1
VL
0
VCC
1
VL
0
VH
1
VCC
1
VH
1
VH
1
2VCE ( sat )
0
22. (A) If either one or both the inputs are V(0) = 0 V
the corresponding FET will be OFF, the voltage across
the load FET will be 0 V, hence the output is VDD . If
boths inputs are V (1) = VDD , both M1 and M 2 are ON
and the output is V(0) = 0 V. It satisfy NAND gate.
23. (B) If both the inputs are at V(0) = 0 V,
17. (A) The Q3 stage is simple an inverter. Hence AND
transistor M1
and M 2 are OFF, hence the output is
logic.
V (1) = VDD . If either one or both of the inputs are at
18. (C) For each successive gate, that has a transistor in
output will be V (0) = 0 V. Hence it is a NOR gate.
V (1) = VDD , the corresponding FET will be ON and the
saturation, the current required is
I B ( sat ) =
Page
230
the
I C ( sat ) VCC - VCE ( sat ) 5 - 0. 2
=
=
= 0.15 mA
b
bRC
50( 640)
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Digital Logic Families
Chap 4.4
24. (B) If all inputs A, B and C are HIGH, then input to
30. (A) When an output is LOW, it may be as high as
invertor is LOW and output Y is HIGH. If all inputs are
VOL ( max ) = 0.4 V. The maximum voltage that an input will
LOW, then input to inverter is also LOW and output Y
respond to as a LOW is V IL ( max ) = 0.8 V. A positive noise
is HIGH. In all other case the input to inverter is HIGH
spike can drive the actual voltage above the 0.8 V level
and output Y is LOW.
if its amplitude is greater than
V NL = V IL ( max ) - VOL ( max ) = 0.8 - 0.4 = 0.4 V
Y = ABC + ABC = ABC + ( A + B + C)
Hence
31. (B) A positive noise spike can drive the voltage
25. (C) The operation of circuit is given below
ABCD
PA
PB
´´´ 1
´
´
´ ´0 0
´
´
0010
PC
PD
above 1.0 V level if the amplitude is greater than
Y
NA NB NC ND
´ OFF
V NL = V IL ( max ) - VOL ( max ) = 1 - 0.1 = 0.9 V,
A negative noise spike can drive the voltage below 3.5 V
´
´
´
´
´
OFF OFF
HIGH
ON ON OFF ON
OFF OFF ON OFF
HIGH
0110
ON OFF OFF ON
OFF ON ON OFF
LOW
32. (B) V IH ( min ) = VOH ( min ) - V NH = - 0.8 - 0.5 = - 1.3 V
1010
OFF ON OFF ON
ON OFF ON OFF
LOW
V IL ( max ) = VOL ( max ) + V NL = 0.5 + ( -2) = -15
. V
1110
OFF OFF OFF ON
ON ON ON OFF
LOW
ON
ON
ON
LOW
if the amplitude is greater than
V NH = VOH ( min ) - V IH ( min ) = 4.9 - 35
. = 1.4 V
33. (C) V NH = VOH ( min ) - V IH ( min ) , V NL = V IL ( max ) - VOL ( max )
V NH = 2.7 (for LS) -2.0 (for ALS) = 0.7 V
Y = ( A + B)C + D
26. (B) The operation of this circuit is given below :
V NL = 0.8 (for ALS) -0.5 (for LS) = 0.3 V
34. (B) V NH = 2.5 (for ALS) - 2.0 (for LS) = 0.5 V
A B C
PA
PB
PC
N A NB NC
´
´ 0
´
´
ON
´
0
0 1
´
1 1
1 ´
1
´
Y
V NL = 0.8 (for LS) - 0.4 (for ALS) = 0.4 V
OFF
HIGH
ON ON OFF
OFF OFF ON
HIGH
´
´
OFF OFF
OFF ´
OFF
ON
ON
LOW
´
ON
LOW
ON
35. (D) V NH ( min ) = 0.5 V ,
V NL ( min ) = 0.3 V
36. (B) fanout (LOW) =
I OL ( max )
8m
=
= 80
I IL ( max ) 0.1m
fanout (HIGH) =
Y = ( A + B) C
I OH ( max ) 400m
=
= 20
I IH ( max )
20m
The fanout is chosen the smaller of the two.
27. (B) If input E is LOW, output will not be LOW. It
37. (B) In HIGH state the loading on the output of gate
must be HIGH. Option (B) satisfy this condition.
1 is equivalent to six 74LS input load.
28. (A) In this circuit parallel combination are OR gate
and series combination are AND gate.
Hence load = 6 ´ I IH = 6 ´ 20m = 120 mA
38. (C) The NAND gate represent only a single input
Hence Y = ( A + B)( C + D)( E + F )
load in the LOW state. Hence only five loads in the
29. (A) When an output is HIGH, it may be as low as
VOH ( min ) = 2.4 V. The minimum voltage that an input will
LOW state.
load = 5I IL = 5 ´ 0.4 = 2 mA
respond to as a HIGH is V IH ( min ) = 2.0 V. A negative noise
spike that can drive the actual voltage below 2.0 V if its
amplitude is greater than
*******
V NH = VOH ( min ) - V IH ( min ) = 2.4 - 2.0 = 0.4 V
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GATE EC BY RK Kanodia
CHAPTER
4.6
MICROPROCESSOR
1. After an arithmetic operation, the flag register of
8085 mP has the following contents
D7
D6
D5
D4
D3
D2
D1
D0
1
0
´
1
´
0
´
1
HLT
DSPLY : XRA
OUT
HLT
A
PORT1
The output at PORT1 is
(A) 00
(B) FEH
(C) 01H
(D) 11H
The contents of accumulator after operation may be
(A) 75
(B) 6C
5. Consider the following 8085 assembly program
(C) DB
(D) B6
MVI
A, DATA1
MOV
B, A
SUI
51H
JC
DLT
MOV
A, B
SUI
82H
JC
DSPLY
DLT :
XRA
A
OUT PORT1
HLT
DSPLY : MOV
A, B
OUT PORT2
HLT
2. In an 8085 microprocessor, the instruction CMP B
has been executed while the contents of accumulator is
less than that of register B. As a result carry flag and
zero flag will be respectively
(A) set, reset
(B) reset, set
(C) reset, reset
(D) set, set
3. Consider the following 8085 instruction
MVI
MVI
ADD
ORA
This program will display
A, A9H
B, 57H
B
A
(A) the bytes from 51H to 82H at PORT2
(B) 00H AT PORT1
The flag status (S, Z, CY) after the instruction
ORA A is executed, is
(C) all byte at PORT1
(D) the bytes from 52H to 81H at PORT 2
(A) (0, 1, 1)
(B) (0, 1, 0)
(C) (1, 0, 0)
(D) (1, 0, 1)
6. It is desired to mask is the high order bits ( D7 - D4 ) of
the data bytes in register C. Consider the following set
4. Consider the following set of 8085 instructions
MVI
ADI
JC
OUT
A, 8EH
73H
DSPLY
PORT1
of instruction
(a)
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MOV
ANI
MOV
HLT
A, C
F0H
C, A
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UNIT 4
(b)
(c)
(d)
Digital Electronics
MOV
MVI
ANA
MOV
HLT
A, C
B, F0H
B
C, A
(C) 8529H are complemented and stored at location
529H
MOV
MVI
ANA
MOV
HLT
A, C
B, 0FH
B
C, A
10. Consider the sequence of 8085 instruction
MOV
ANI
MOV
HLT
A, C
0FH
C, A
(D) 5829H are complemented and stored at location
85892H
MVI
ADI
MOV
HLT
A, 5EH
A2H
C, A
The initial contents of resistor and flag are as
follows
The instruction set, which execute the desired
A
´´
operation are
(A) a and b
(B) c and d
(C) only a
(D) only d
S
0
CY
0
After execution of the instructions the contents of
A
A
B, 4AH
4FH
B
The contents of register A and B are respectively
C
S
Z
CY
(A) 10H
10H
0
0
1
(B) 10H
10H
1
0
0
(C) 00H
00H
1
1
0
(D) 00H
00H
0
1
1
(A) 05, 4A
(B) 4F, 00
11. It is desired to multiply the number 0AH by 0BH
(C) B1, 4A
(D) None of the above
and store the result in the accumulator. The numbers
8. Consider the following 8085 assembly program :
MVI
MOV
MOV
MVI
OUT
HLT
B, 89H
A, B
C, A
D, 37H
PORT1
are available in register B and C respectively. A part of
the 8085 program for this purpose is given below :
A, 00H
MVI
LOOP : ----------------------------------------------------------------------HLT
END
The output at PORT1 is
(A) 89
(B) 37
(C) 00
(D) None of the above
The sequence of instruction to complete the
program would be
9. Consider the sequence of 8085 instruction given
(A)
JNZ
ADD
DCR
LOOP
B
C
(B)
ADD
JNZ
DCR
B
LOOP
C
(C)
DCR
JNZ
ADD
C
LOOP
B
(D)
ADD
DCR
JNZ
B
C
LOOP
below
LXI
MOV
CMA
MOV
H, 9258H
A, M
M, A
By this sequence of instruction the contents of
memory location
(A) 9258H are moved to the accumulator
(B) 9258H are compared with the contents of the
accumulator
Page
240
Z
0
register and flags are
7. Consider the following 8085 instruction
XRA
MVI
SUI
ANA
HLT
C
´´
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GATE EC BY RK Kanodia
Microprocessor
Chap 4.6
12. Consider the following assembly language program:
(A) A7H
(B) 98H
MVI
MOV
START : JMP
MVI
XRA
OUT
HLT
NEXT : XRA
JP
OUT
HLT
(C) 47H
(D) None of the above
B, 87H
A, B
NEXT
B, 00H
B
PORT1
15. The memory requirement for this program is
B
START1
PORT2
16. The instruction, that does not clear the accumulator
(A) 20 Byte
(B) 21 Byte
(C) 23 Byte
(D) 18 Byte
of 8085, is
The execution of the above program in an 8085
will result in
(A) XRA A
(B) ANI 00H
(C) MVI A, 00H
(D) None of the above
(A) an output of 87H at PORT1
17. The contents of some memory location of an 8085 mP
(B) an output of 87H at PORT2
based system are shown
(C) infinite looping of the program execution with
accumulator data remaining at 00H
Address Hex.
Contents (Hex.)
(D) infinite looping of the program execution with
accumulator data alternating between 00H and 87H.
3000
02
3001
30
13. Consider the following 8085 program
3002
00
MVI
ORA
JM
OUT
CMA
DSPLY : ADI
OUT
HLT
3003
30
A, DATA1
A,
DSPLY
PORT1
01H
PORT1
Fig. P4.6.17
The program is as follows
If DATA1 = A7H, the output at PORT1 is
(A) 47H
(B) 58H
(C) 00
(D) None of the above
LHLD
MOV
INX
MOV
LDAX
MOV
INX
LDAX
MOV
Statement for Q.14–15:
Consider the following program of 8085 assembly
language:
LXI
LDA
MOV
LDA
CMP
JZ
JC
MOV
JMP
MOV
FNSH : HLT
3000H
E, M
H
D, M
D
L, A
D
D
H, A
The contents if HL pair after the execution of the
H 4A02H
4A00H
B, A
4A01H
B
FNSH
GRT
M, A
FNSH
M, B
program will be
(A) 0030 H
(B) 3000 H
(C) 3002 H
(D) 0230H
18. Consider the following loop
14. If the contents of memory location 4A00H, 4A01H
and 4A02H, are respectively A7H, 98H and 47H, then
XRA
LXI
LOOP : DCX
JNZ
A
B, 0007H
B
LOOP
This loop will be executed
after the execution of program contents of memory
(A) 1 times
(B) 8 times
location 4A02H will be respectively
(C) 7 times
(D) infinite times
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GATE EC BY RK Kanodia
UNIT 4
19. Consider the following loop
LXI
LOOP : DCX
MOV
ORA
JNZ
24. Consider the following program
MVI
RRC
RRC
H, 000AH
B
A, B
C
LOOP
(A) 1 time
(B) 10 times
(C) 11 times
(D) infinite times
execution of program will be
(A) 08H
(B) 8CH
(C) 12H
(D) None of the above
25. Consider the following program
20. The contents of accumulator after the execution of
following instruction will be
A, A7H
A
(A) CFH
(B) 4FH
(C) 4EH
(D) CEH
RJCT :
21. The contents of accumulator after the execution of
following instructions will be
MVI
ORA
RAL
A, BYTE1
If BYTE1 = 32H, the contents of A after the
This loop will be executed
MVI
ORA
RLC
Digital Electronics
MVI
MVI
MVI
CMP
JC
CMP
JNC
OUT
HLT
SUB A
OUT
HLT
PORT1
If the following sequence of byte is loaded in
accumulator,
A, B7H
A
DATA (H)
(A) 6EH
(B) 6FH
(C) EEH
(D) EFH
58
64
73
C8
FA
(A) 00, 00, 73, B4, 00, FA
(B) 58, 64, 00, 00, C8, FA
(C) 58, 00, 00, 00, C8, FA
of the following program will be
(D) 00, 64, 73, B4, 00, FA
A, C5H
A
26. Consider the following instruction to be executed by
a 8085 mp. The input port has an address of 01H and
has a data 05H to input:
(A) 45H
(B) C5H
(C) C4H
(D) None of the above
IN
ANI
23. Consider the following set of instruction
MVI
RLC
MOV
RLC
RLC
ADD
B4
then sequence of output will be
22. The contents of the accumulator after the execution
MVI
ORA
RAL
01H
80H
After execution of the two instruction the contents
A, BYTE1
of flag register are
B, A
(A)
1
0
´
1
´
1
´
0
B
(B)
0
1
´
0
´
1
´
0
(C)
0
1
´
1
´
1
´
0
(D)
0
1
´
1
´
0
´
0
If BYTE1 = 07H, then content of A, after the
execution of program will be
Page
242
A, DATA
B, 64H
C, C8H
B
RJCT
C
RJCT
PORT1
(A) 46H
(B) 70H
(C) 38H
(D) 68H
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Microprocessor
ORA
JM
A
DSPLY
OUT
DSPLY : CMA
ADI
OUT
HLT
PORT1
01H
PORT1
;Set flag
;If negative jump to
;DSPLY
;A ® PORT1
;Complement A
;A+1 ® A
;A ® PORT1
Chap 4.6
18. (A) The instruction XRA will set the Z flag. LXI and
DCX does not alter the flag. Hence this loop will be
executed 1 times.
19. (B) LXI
LOOP : DCX
This program displays the absolute value of DATA1. If
MOV
ORA
JNZ
DATA1 is negative, it determine the 2’s complements
and display at PORT1.
B, 000AH
B
A, B
C
LOOP
;00 ® C, 0AH ® B
; CB – 1 ® B,
;flag not affected
;B ® A
;A OR C ® A, set flag
Hence this loop will be executed 0AH or ten times.
14. (A)
LXI
H, 4A02H
LDA
4A00H
MOV
LDA
B, A
4A01H
CMA
JZ
B
FNSH
JC
GRT
MOV
JMP
GRT :
MOV
FNSH : HLT
M, A
FNSH
M, B
;Store destination address
;in HL pair
;Load A with contents of
;memory location A00H
;A ® B
;Load A with contents of
;memory location 4A01H
;Compare A and B
;Jump to FNSH if two
;number are equal
;If CY = 1, (A <B) jump
;to GRT
;Otherwise A ® (4A02H)
MVI
ORA
RLC
A, B7H
A
;B7H ® A
;Set Flags, CY = 1
;Rotate accumulator left
The contents of bit D7 are placed in bit D0 .
Accumulator
Before RLC
10100111
After RLC
01001111
21. (A) RAL instruction rotate the accumulator left
through carry.
D7 ® CY ,
CY ® D0 ,
ORA reset the carry.
Accumulator
This program find the larger of the two number stored
in location 4A00H and 4A01H and store it in memory
location 4A002.
A7H > 98H
20. (B)
Thus A7H will be stored at 4A02H.
15. (C) Operand R, M or implied : 1–Byte instruction
Operand 8–bit
: 2–Byte instruction
Operand 16–bit
: 3–Byte instruction
CY
Before RAL
10110111
0
After RAL
01101110
1
22. (A) RRC instruction rotate the accumulator right
and D0 is placed in D7 .
MVI
ORA
RAL
3–Byte instruction are: LXI, LDA, JZ, JC, JMP
A, C5H
A
RRC
;C5H ® A
;Reset Carry flag
;Rotate A left through
;carry, A = 8AH
;Rotate A right, A = 45H
P–Byte instruction are : MOV, CMP, HLT
Hence memory = 3 ´ 6 + 1 ´ 5 = 23
Byte.
23. (A) This program multiply BYTE1 by 10. Hence
content of A will be 46H.
16. (D) All instruction clear the accumulator
17. (C)
XRA
ANI
MVI
A
00H
A
LHLD
MOV
INX
MOV
LDAX
MOV
INX
LDAX
MOV
3000H
E, M
H
D, M
D
L, A
D
D
H, A
;A Å A
;A AND 00
;00 ® A
;(3000A) ® HL = 3002H
;(3002H) ® E = 00
;HL +1 ® HL = 3003H
;M ® D=(3003H) = 30H
;(DE) ® A=(3000H) = 02H
;A ® L = 02H
;DE +1 ® DE = 3001H
;(DE) ® A = (3001) = 30H
;A ® H = 30H
Hence HL pair contain 3002H.
07H = 0710 ,7 ´ 10 = 70, 7010 = 46H
24. (B) Contents of Accumulator A = 0011 0010
After First RRC
= 0001 1001
After second RRC
= 1000 1100
25. (D) This program will display the number between
64H to C8H including 64H. C8H will not be displayed.
Thus (D) is correct option.
26. (C) 05H AND 80H =00
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UNIT 4
After the ANI instruction S, Z and P are modified to
reflect the result of operation. CY is reset and AC is set
. Thus,
S = 0, Z = 1, AC = 1, P =1, CY = 0
27. (B)
ACI
;A + 56H + CY ® A
56H
37H + 56H + 1 =8EH
28. (C) Instruction load the register pairs HL with
01FFH. SHLD instruction store the contents of L in the
memory location 2050H and content of H in the memory
location 2051H. Contents of HL are not altered.
29. (B) At a time 8085 can drive only a digit. In a second
each digit is refreshed 500 times. Thus time given to
1
each digit =
= 0.4 ms.
(5 ´ 500)
30. (C) The stack pointer register SP point to the upper
memory location of stack. When data is pushed on
stack, it stores above this memory location.
31. (B) Line 5 push the content of HL register pair on
stack. The contents of L will go to 03FFH and contents
of H will go to 03FEH. Hence memory location 03FEH
contain 22H.
32. (C) Contents of register pair B lie on the top of stack
when POP H is executed, HL pair will be loaded with
the contents of register pair B.
33. (C) The instruction PUSH B store the contents of
BC at stack. The POP PSW instruction copy the
contents of BC in to PSW. The contents of register C
will be copied into flag register.
D0 = 1 = carry flag,
D6 = 0 = zero flag.
Hence zero flag will be reset and carry will be set.
34. (A)
MVI A DATA1
ORA
A
JP
DSPLY
XRA
DSPLY OUT
HLT
A
PORT1
;DATA1 ® A
; Set flag
;If A is positive, then
;jump to DSPLY
; Clear A
; A ® PORT2
If DATA1 is positive, it will be displayed at port1
otherwise 00.
********************
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GATE EC BY RK Kanodia
UNIT 5
ì e2 t, t < 0
30. y( t) = u( t) * h( t) , where h( t) = í -3t
îe , t >0
1
5 1
(A) e -2 t u( - t - 1) + - e -3t u( -t)
2
6 3
Signal & System
(A)
5
3
1
13 -4 t
sin t +
cos t + e - t e , t ³ 0
34
34
6
61
(B)
5
3
13 -4 t 1 - t
sin t +
cos t e + e , t ³ 0
34
34
51
6
(B)
1 2t
5 1
e u( -t - 1) + - e-3t u( -t)
2
6 3
(C)
3
5
13 -4 t 1 - t
sin t +
cos t e + e , t ³ 0
34
34
51
6
(C)
1 2t 1
e + [5 - 3e 2 t - 2 e -3t ]u( t)
2
6
(D)
3
5
1
13 -4 t
sin t +
cos t + e -4 t e , t ³ 0
34
34
6
51
(D)
1 2t 1
e + [5 - 3e 2 t - 2 e -3t ]u( -t)
2
6
37.
d 2 y ( t)
dy ( t)
+6
+ 8 y( t) = 2 x( t),
dt 2
dt
Statement for Q.31-34:
y (0 - ) = -1,
The impulse response of LTI system is given.
Determine the step response.
(A)
2 - t 5 -2 t 5 -4 t
e - e + e , t ³0
3
2
6
31. h( t) = e - |t |
(B)
2 5 -2 t 5 -4 t
+ e + e , t ³ 0
3 2
6
(A) 2 + e t - e - t
(B) e t u( -t + 1) + 2 - e - t
(C) e t u( -t + 1) + [2 - e - t ]u( t)
(D) e t + [2 - e - t - e t ]u( t)
(C) 4 + 5( 3e -2 t + e -4 t ) , t ³ 0
(D) 4 - 5( 3e -2 t + e -4 t ), t ³ 0
32. h( t) = d( 2 ) ( t)
(A) 1
(B) u( t)
(C) d( 3) ( t)
(D) d( t)
38.
d 2 y( t)
3dx( t)
,
+ y( t) =
dt 2
dt
y (0 - ) = -1,
33. h( t) = u( t) - u( t - 4)
(A) tu( t) + (1 - t) u( t - 4)
(B) tu( t) + (1 - t) u( t - 4)
(C) 1 + t
(D) (1 + t) u( t)
(A) sin t + 4 cos t - 3te -3t + t, t ³ 0
(B) 4 sin t - cos t - 3te - t , t ³ 0
(D) 4 sin t + cos t - 3te - t , t ³ 0
(A) u( t)
(B) t
(C) 1
(D) tu( t)
39. The raised cosine pulse x( t) is defined as
ì 1
ï (cos wt + 1) ,
x ( t) = í 2
ïî
0,
Statement for Q.35-38:
The system described by the differential equations
has been specified with initial condition. Determine the
output of the system and choose correct option.
dy( t)
+ 10 y( t) = 2 x( t), y(0 - ) = 1, x( t) = u( t)
dx
(A) 15 (1 + 4 e -10 t ) u( t)
(B) 15 (1 + 4 e -10 t )
(C) - 15 (1 + 4 e -10 t ) u( t)
(D) - 15 (1 + 4 e -10 t )
-
p
p
£t£
w
w
otherwise
The total energy of x ( t) is
3p
3p
(B)
(A)
4w
8w
(C)
3p
w
(D)
3p
2w
40. The sinusoidal signal x( t) = 4 cos (200 t + p 6) is
d 2 y( t)
dy( t)
dx( t)
,
36.
+5
+ 4 y( t) =
2
dt
dt
dt
dy( t)
= 1, x( t) = sin t u( t)
y (0 - ) = 0,
dt 0 -
Page
252
dy( t)
= 1, x( t) = 2 te- t u( t)
dt 0 -
(C) sin t - 4 cos t + 3te -3t + t, t ³ 0
34. h( t) = y( t)
35.
dy( t)
= 1, x( t) = e - t u( t)
dt 0 -
passed through a square law device defined by the
input output relation y ( t) = x 2 ( t). The DC component in
the signal is
(A) 3.46
(B) 4
(C) 2.83
(D) 8
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Continuous-Time Systems
41. The impulse response of a system is h( t) = d( t - 0.5).
If two such systems are cascaded, the impulse response
Chap 5.1
46. The y( t) = x( t) * h( t) is
y(t)
y(t)
of the overall system will be
(A) 0.58( t - 0.25)
(B) d( t - 0.25)
(C) d( t - 1)
(D) 0.5 d( t - 1)
a
t
1+a
a
t
y(t)
h(t)
a
1
t
5
3
t
1+a
2
1+a
t
a
1-a
(C)
Fig P5.1.42
The output of the system is zero every where
except for the
value of a is
(A) 1
(B) 2
(C) 1 < t < 5
(D) 1 < t < 8
(C) 3
(D) 0
S1 : h1 ( t) = e - (1 - 2 j ) t u( t)
48. Consider the signal x( t) = d( t + 2) - d( t - 2).The value
of E¥ for the signal y( t) =
S2 : h2 ( t) = e cos 2 t u( t)
The stable system is
(A) S1
(B) S2
(C) Both S1 and S2
(D) None
ò x( t) dt
is
(A) 4
(B) 2
(C) 1
(D) ¥
49. The
response of a system S to a complex input
x( t) = e j 5t is specified as y( t) = te j 5t . The system
44. The non-invertible system is
(B) y( t) =
t
ò
(A) is definitely LTI
x( t) dt
(B) is definitely not LTI
-¥
dx( t)
dt
t
-¥
-t
(A) y( t) = x( t - 4)
t
47. If dy( t) dt contains only three discontinuities, the
(B) 0 < t < 8
43. Consider the impulse response of two LTI system
1
(D)
(A) 0 < t < 5
(C) y( t) =
1
(B)
y(t)
impulse response h( t) of the system.
1-a
(A)
42. Fig. P5.1.40 show the input x( t) to a LTI system and
x(t)
1
(D) None of the above
(C) may be LTI
(D) information is insufficient
45. A continuous-time linear system with input x( t) and
output y( t) yields the following input-output pairs:
x( t) = e j 2 t Û y( t) = e j 5t
(B) is definitely not LTI
If x1 ( t) = cos (2 t - 1), the corresponding y1 ( t) is
(B) e- j cos (5 t - 1)
(D) e j cos (5 t - 1)
Suppose that
0 £ t £1
elsewhere
(C) may be LTI
(D) information is insufficient.
51. The auto-correlation of the signal x( t) = e - t u( t) is
Statement for Q.46–47:
ì 1,
x( t) = í
î 0,
response of a system S to a complex input
x( t) = e j8 t is specified as y( t) = cos 8 t. The system
(A) is definitely LTI
x( t) = e - j 2 t Û y( t) = e - j 5t
(A) cos (5 t - 1)
(C) cos 5( t - 1)
50. The
and
ætö
h( t) = xç ÷, where 0 < a £ 1.
èaø
(A)
1 t
1
e u( -t) + e - t u( t)
2
2
(B)
et 1 - t
+ ( e - e t ) u( t)
2 2
(C)
1 -t
1
e u( -t) + e - t u( t)
2
2
(D)
1 t
1
e u( -t) - e - t u( t)
2
2
*********************
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UNIT 5
Signal & System
11. (C)
SOLUTIONS
x(10t)
x(10t-5)
1
1. (A)
2p
= 60 p
T
2. (C) T1 =
Þ
T=
p
30
1
-0.5 -0.4
0.4 0.5
t
0.1
0.9
1
t
Fig S5.1.11
12. (D) Multiplication by 5 will bring contraction on
2p
2p
æ 2p 2p ö
s, T2 =
s, LCM ç
,
÷ = 2p
5
7
7 ø
è 5
time scale. It may be checked by x(5 ´ 0.8) = x( 4).
3. (D) Not periodic because of t.
4. (D) Not periodic because least common multiple is
13. (A) Division by 5 will bring expansion on time scale.
æ 20 ö
It may be checked by y( t) = xç
÷ = x( 4).
è 5 ø
infinite.
5. (C) y( t) is not periodic although sin t and 6 cos 2 pt are
independently periodic. The fundamental frequency
can’t be determined.
¥
ò | x ( t)|dt < ¥
=
-¥
¥
¥
-¥
0
-4 t
-4 t
ò e u( t) dt = ò e dt =
¥
7. (A) | x( t)| = 1, E¥ =
ò | x( t)| dt
2
1
4
1
T® ¥ 2T
T
5
-5
4
for
4 < t <5
otherwise
=8 +
=¥
-¥
ò | x( t)| dt
2
5
4
5
0
0
4
15. (D) E = 2 ò x 2 ( t) dt = 2 ò (1)1 dt + 2 ò (5 - t) 2 dt
2 26
=
3 3
16. (B) Let x1 ( t) = v( t) then y1 ( t) = u{v( t)}
So this is a power signal not a energy.
P¥ = lim
-4
for - 5 < t < - 4
E = ò (1) 2 dt + ò ( -1) 2 dt = 2
6. (C) This is energy signal because
E¥ =
ì 1,
ï
14. (C) y( t) = í -1,
ï 0,
î
Let x2 ( t) = kv( t) then y2 ( t) = u{kv( t)} ¹ ky1 ( t)
=1
(Not homogeneous not linear)
-T
8. (D) v( t) is sum of 3 unit step signal starting from, 1, 2,
and 3, all signal ends at 4.
y1 ( t) = u{v( t)},
y2 ( t) = u{v( t - to)} = y1 ( t - to)
(Time invariant)
The response at any time depends only on the
9. (A) The function 1 does not describe the given pulse.
excitation at time t = to and not on any future value.
(Causal)
It can be shown as follows :
u(a-t)
u(t-b)
17. (C) y1 ( t) = v( t - 5) - v( 3 - t)
u(a-t) - u(t-b)
y2 ( t) = kv( t - 5) - kv( 3 - t) = ky1 ( t)
t
a
t
b
t
(Homogeneous)
Let x1 ( t) = v( t) then y1 ( t) = v( t - 5) - v( 3 - t)
Let x2 ( t) = 2 w( t) then y2 ( t) = w( t - 5) - w( 3 - t)
Fig S5.1.3.9
Let x3( t) = x( t) + w( t)
10. (B)
Then
r(t-4)
r(t-6)
r(t-4) - r(t-6)
2
2
2
y3( t) = v( t - 5) + w( t - 5) - v( 3 - t) - w( 3 - t)
= y1 ( t) + y2 ( t)
(Additive)
Since it is both homogeneous and additive, it is also
linear.
4
6
8
t
4
6
8
Fig S5.1.10
Page
254
t
4
6
8
t
y1 ( t) = v( t - 5) - v( 3 - t)
y2 ( t) = v( t - to - 5) - v( 3 - t + to) = y1 ( t - to)
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(Time invariant)
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Continuous-Time Systems
At
time,
t = 0, y(0) = x( -5) - x( 3).
Therefore
the
Chap 5.1
21. (C) All option are linear. So it is not required
response at time, t = 0 depends on the excitation at a
to check linearity.
later time t = 3.
d
y1 ( t) - 8 y1 ( t) = v( t)
dt
d
Let x2 ( t) = v( t - to) then t
y2 ( t) - 8 y2 ( t) = v( t - to)
dt
(Not causal)
If x( t) is bounded then x( t - 5) and x( 3 - t) are bounded
and so is y( t).
(Stable)
ætö
ætö
18. (D) y1 ( t) = vç ÷ , y2 ( t) = kvç ÷ = ky1 ( t)
è2 ø
è2 ø
(Homogeneous)
Let x1 ( t) = v( t) then t
The first equation can be written as
d
( t - to)
y( t - to) - 8 y( t - to) = v( t - to)
dt
This equation is not satisfied if y2 ( t) = y1 ( t - to) therefore
x3 = v( t) + w( t) then
ætö
ætö
y3( t) = vç ÷ + wç ÷ = y1 ( y) + y2 ( t)
è2 ø
è2 ø
(Additive)
Since it is both homogeneous and additive, it is also
y2 ( t) ¹ y1 ( t - to)
(Time Variant)
The system can be written as
y( t) =
linear
x( l)
y( l)
dl + 8 ò
dl
l
-¥
-¥ l
t
t
ò
So the response at any time, t = to depends on the
ætö
æt
ö
æ t - to ö
y1 ( t) = vç ÷ , y2 ç - to ÷ ¹ y( t - to) = vç
÷
2
2
è ø
è
ø
è 2 ø
excitation at t £ to , and not on any future values.
(Time variant)
(Causal)
At time t = -2, y( -2) = x( -1), therefore, the response at
The Homogeneous solution to the differential equation
time t = -2, depends on the excitation at a later time,
is of the form y( t) = kt8 . If there is no excitation but the
t = -1.
zero excitation, response is not zero. The response will
(Not causal)
It x( t) is bounded then y( t) is bounded.
(Stable)
increases without bound as time increases.
(Unstable)
19. (C) y1 ( t) = cos 2 pt v( t)
y2 ( t) k cos 2 pt v( t) = ky1 ( t)
(Homogeneous)
22. (C) y1 ( t) =
x3( t) = v( t) + w( t)
t+ 3
ò v(l)dl
-¥
y3( t) = cos 2 pt [ v( t) + w( t)] = y1 ( t) + y2 ( t)
(Additive)
t+ 3
t+ 3
-¥
-¥
ò kv(l)dl = k ò v(l)dl = ky1 ( t)
y2 ( t) =
Since it is both homogeneous and additive. It is also
linear.
x3( t) = v( t) + w( t)
y1 ( t) = cos 2 pt v( t)
y3( t) =
y2 ( t) = cos 2 pt ( t - to) ¹ y( t - to)
= cos [2p( t - to)]v( t - to)
(Time Variant)
(Homogeneous)
t+ 3
t+ 3
t+ 3
-¥
-¥
-¥
ò [ v( l) + w( l)]dl =
ò v( l)dl +
ò w( l)dl
= y1 (t) + y2 (t)
(Additive)
The response at any time t = to depends only on the
excitation at that time and not on the excitation at any
Since it is Homogeneous and additive, it is also linear.
later time.
(Causal)
y1 ( t) =
If x( t) is bounded then y( t) is bounded.
(Stable)
ò v(l)dl
-¥
y2 ( t) =
20. (C) y1 ( t) = |v( t)|, y2 ( t) = |kv( t)| = k y1 ( t)
t+ 3
t+ 3
t - to + 3
-¥
-¥
ò v(l - to )dl =
ò v(l)dl = y (t - t )
o
1
If k is negative k y1 ( t) ¹ ky1 ( t)
(Time invariant)
(Not Homogeneous Not linear).
y1 ( t) = |v( t)|, y2 ( t) = | y( t - to)| = y1 ( t - to)
The response at any time, t = to , depends partially on
the excitation at time to to < t < ( to + 3) which are in
(Time Invariant)
The response at any time t = to depends only on the
excitation at that time and not on the excitation at any
later time.
If x( t) is bounded then y( t) is bounded.
(Causal)
future.
If x( t) is a constant k, then y ( t) =
(Not causal)
t+ 3
t+ 3
-¥
-¥
ò kdl = k ò dl and as
t ® ¥ , y ( t) increases without bound.
(unstable)
(Stable)
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UNIT 5
Signal & System
æ pn ö
æ pn ö
æ pn p ö
14. x[ n] = cos ç
+ ÷
÷ - sin ç
÷ + 3 cos ç
3ø
è 2 ø
è 8 ø
è 4
11. x[ n + 2 ] y[ n - 2 ]
x[n]
3
(A) periodic with period 16
2
(A)
(B) periodic with period 4
1
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
n
(C) periodic with period 2
(D) Not periodic
x[n]
3
15. x[ n] = 2 e
2
(B)
1
-6
-5
-4
-3
-2
-1
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
1
2
3
4
5
6
n
(A) periodic with 12p
(B) periodic with 12
(C) periodic with 11p
(D) Not periodic
16. The sinusoidal signal has fundamental period
n
N = 10 samples. The smallest angular frequency, for
which x[ n] is periodic, is
1
(A)
rad/cycle
10
-1
(C)
æn
ö
j çç - p÷÷
è6
ø
-2
-3
(C) 5 rad/cycle
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
n
-1
(D)
(B) 10 rad/cycle
(D)
p
rad/cycle
5
17. Let x[ n], - 5 £ n £ 3 and h[ n], 2 £ n £ 6 be two finite
duration signals. The range of their convolution is
-2
-3
(A) -7 £ n £ 9
(B) -3 £ n £ 9
(C) 2 £ n £ 3
(D) -5 £ n £ 6
Statement for Q.12–15:
A discrete-time signal is given. Determine the
Statement for Q.18–26:
period of signal and choose correct option.
x[ n] and h[ n] are given in the question. Compute
the convolution y[ n] = x[ n] * h[ n]
pn
æ pn 1 ö
12. x[ n] = cos
+ sin ç
+ ÷
9
2ø
è 7
(A) periodic with period N = 126
(B) periodic with period N = 32
Page
260
and choose correct
option.
18. x[ n] = {1, 2, 4}, h[ n] = {1, 1, 1, 1, 1}
(A) {1, 3, 7, 7, 7, 6, 4}
(B) {1, 3, 3, 7, 7, 6, 4}
(C) periodic with period N = 252
(C) {1, 2, 4}
(D) Not periodic
(D) {1, 3, 7}
æ nö
æ pn ö
13. x[ n] = cos ç ÷ cos ç
÷
è8ø
è 8 ø
19. x[ n] = {1, 2, 3, 4, 5}, h [ n] = {1}
(A) {1, 3, 6, 10, 15}
(B) {1, 2, 3, 4, 5}
(A) Periodic with period 16p
(C) {1, 4, 9, 16, 20}
(D) {1, 4, 6, 8, 10}
(B) periodic with period 16( p + 1)
20. x[ n] = {1, 2, -1}, h [ n] = x [ n]
(C) periodic with period 8
(A) {1, 4, 1}
(B) {1, 4, 2, -4, 1}
(D) Not periodic
(C) {1, 2, -1}
(D) {2, 4, -2}
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Discrete-Time Systems
21. x[ n] = {1, -2, 3}, h [ n] = {0, 0, 1, 1, 1, 1}
­
Chap 5.2
25. x[ n] = {1, 4, -3, 6, 4}, h[ n] = {2, -4, 3}
­
­
­
(A) {2, 4, -19, 36, -25, 2, 12}
(A) {1, -2, 4, 1, 1, 1}
­
­
(B) {4, -19, 36, -25}
(B) {0, 0, 3}
­
­
(C) {1, 4, -3, 6, 4}
(C) {0, 0, 3, 1, 1, 1, 1}
­
­
(D) {1, 4, -3, 6, 4}
(D) {0, 0, 1, -1, 2, 2, 1, 3}
­
­
22. x[ n] = {0, 0, 1, 1, 1, 1}, h[ n] = {1, -2, 3}
­
­
(A) {0, 0, 1, -1, 2, 2, 1, 3}
ì 1,
ï
26. x[ n] = í 2,
ï0
î
n = -2, 0, 1
n = -1
elsewhere
h (n) = d[ n ] - d[ n - 1] + d[ n - 4]
­
(A) d[ n ] - 2 d[ n - 1 ] + 4 d[ n - 4 ] + d[ n - 5 ]
(B) {0, 0, 1, -1, 2, 2, 1, 3}
(B) d[ n + 2 ] + d[ n + 1 ] - d[ n ] + 2 d[ n - 3 ] + d[ n - 4 ] + d[ n - 5 ]
­
(C) d[ n + 2 ] - d[ n + 1 ] + d[ n ] + 2 d[ n - 3 ] - d[ n - 4 ] + 2 d[ n - 5 ]
(C) {1, -2, 3, 1, 1, 2, 1, 1}
(D) d[ n ] + 2 d[ n - 1 ] + 4 d[ n - 5 ] + d[ n - 5 ]
­
Statement for Q.27–30:
(D) {1, -2, 3, 1, 1, 1, 1}
­
In question y[ n] is the convolution of two signal.
23. x[ n] = {1, 1, 0, 1, 1}, h[ n] = {1, -2, - 3, 4}
­
­
(A) {1, -1, - 2, 4, 1, 1}
­
(B) {1, -1, - 2, 4, 1, 1}
­
(C) {1, -1, - 5, 2, 3, -5, 1, 4}
­
(D) {1, -1, - 5, 2, 3, -5, 1, 4}
­
24. x[ n] = {1, 2, 0, 2, 1}, h[ n] = x[ n]
­
(A) {1, 4, 4, 4, 10, 4, 4, 4, 1}
­
(B) {1, 4, 4, 4, 10, 4, 4, 4, 1}
­
(C) {1, 4, 4, 10, 4, 4, 4, 1}
­
(D) {1, 4, 4, 10, 4, 4, 4, 1}
­
Choose correct option for y[ n].
27. y[ n] = ( -1) n * 2 n u[2 n + 2 ]
(A)
4
6
(B)
4
u[ -n + 2 ]
6
(C)
8
( -1) n u[ -n + 2 ]
3
(D)
8
( -1) n
3
28. y[ n] =
1
u[ n] * u[ n + 2 ]
4n
æ1 1 ö
(A) ç - n ÷u[ n]
è3 4 ø
æ 1 12 ö
(B) ç - n ÷u[ n + 2 ]
è3 4 ø
æ 4 1 æ 1 ön ö
÷u[ n + 2 ]
(C) ç ç 3 12 çè 4 ÷ø ÷
è
ø
1 ö
æ 16
(D) ç
- n ÷u[ n + 2 ]
è 3 4 ø
29. y[ n] = 3n u[ -n + 3] * u[ n - 2 ]
ì 3n
,
ïï
(A) í 2
ï 83 ,
ïî 2
ì 3n
,
ïï
(C) í 2
ï 81 ,
ïî 2
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n £5
n³6
n £5
n³6
ì 3n ,
ï
(B) í 83
ïî 2 ,
ì 3n
,
ïï
(D) í 6
ï 81 ,
ïî 2
n £5
n³6
n £5
n³6
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UNIT 5
30. y[ n] = u[ n + 3] * u[ n - 3]
37. x[ n] as shown in fig. P5.2.37
(A) ( n + 1) u[ n]
(B) nu[ n]
(C) ( n - 1) u[ n]
(D) u[ n]
31.
The
convolution
of
Signal & System
x[ n] = cos ( 2p n) u[ n]
5
x[n]
and
+
10
y[n]
Fig. P5.2.37
h[ n] = u[ n - 1] is f [ n]u[ n - 1]. The function f [ n] is
n = 4 m + 1, 4 m + 2
ì 1,
(A) í
n = 4 m, 4 m + 3
î 0,
(A) P, Q, R, S
(B) Q, R, S
(C) P, R, S
(D) P, Q, S
n = 4 m + 1, 4 m + 2
ì 0,
(B) í
î 1,
n = 4 m, 4 m + 3
ì 1,
(C) í
î 0,
n = 4 m + 1, 4 m + 3
n = 4 m, 4 m + 2
ì 0,
(D) í
î 1,
n = 4 m + 1, 4 m + 3
38. x[ n] as shown in fig. P5.2.38
x[n]
+
y[n]
+
Fig. P5.2.38
n = 4 m, 4 m + 2
Statement for Q.32–38:
Let P be linearity, Q be time invariance, R be
causality and S be stability. In question discrete time
(A) P, Q, R, S
(B) P, Q, R
(C) P, Q
(D) Q, R, S
Statement for Q.39–41:
and S2
Two discrete time systems S1
input x[ n] and output y[ n] relationship has been given. In
the option properties of system has been given. Choose
the option which match the properties for system.
fig. P5.2.39–41.
32. y[ n] = rect ( x[ n])
x[n]
(A) P, Q, R
(B) Q, R, S
(C) R, S, P
(D) S , P, Q
S2
S1
y[n]
Fig. P5.2.39–41.
39. Consider the following statements
33. y[ n] = nx[ n]
(A) P, Q, R, S
(B) Q, R, S
(a) If S1 and S2 are linear, the S is linear
(C) P, R
(D) Q, S
(b) If S1 and S2 are nonlinear, then S is nonlinear
34. y[ n] =
(c) If S1 and S2 are causal, then S is causal
n +1
å
u[ m ]
(d) If S1 and S2 are time invariant, then S is time
invariant
m = -¥
(A) P, Q, R, S
(B) R, S
(C) P, Q
(D) Q, R
35. y[ n] = x[ n]
(A) Q, R, S
(B) R, S, P
(C) S, P, Q
(D) P, Q, R
36. x[ n] as shown in fig. P5.2.36
x[n]
2
y[n]
True statements are :
(A) a, b, c
(B) b, c, d
(C) a, c, d
(D) All
40. Consider the following statements
(a) If S1 and S2 are linear and time invariant, then
interchanging their order does not change the system.
(b) If S1 and S2 are linear and time varying, then
interchanging their order does not change the system.
True statement are
Fig. P5.2.36
Page
262
are
connected in cascade to form a new system as shown in
(A) P, Q, R, S
(B) Q, R, S
(C) P, Q
(D) R, S
(A) Both a and b
(B) Only a
(C) Only b
(D) None
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Discrete-Time Systems
41. Consider the statement
Chap 5.2
(D) Above all
(a) If S1 and S2 are noncausal, the S is non causal
(b) If S1 and/or S2 are unstable, the S is unstable.
45. The system shown in fig. P5.2.45 is
(A) Both a and b
y[n]
+
1
4
x[n]
True statement are :
(B) Only a
(C) Only b
D
y[n-2]
D
+
+
(D) None
1
4
+
-1
2
42. The following input output pairs have been
Fig. P5.2.45
observed during the operation of a time invariant
(A) Stable and causal
system :
x1 [ n] = {1, 0, 2}
S
¬¾
®
y1 [ n] = {0, 1, 2}
(B) Stable but not causal
­
(C) Causal but unstable
­
x2 [ n] = {0, 0, 3}
¬¾®
S
y2 [ n] = {0, 1, 0, 2}
­
­
x3[ n] = {0, 0, 0, 1}
¬¾®
S
46. The impulse response of a LTI system is given as
y3[ n] = {1, 2, 1}
­
(D) unstable and not causal
n
æ 1ö
h[ n] = ç - ÷ u[ n].
è 2ø
­
The conclusion regarding the linearity of the
The step response is
system is
(A) System is linear
1 æç
æ 1ö
2 -ç- ÷
ç
3è
è 2ø
(C)
1 æç
æ 1ö
2 + ç- ÷
3 çè
è 2ø
(B) System is not linear
(C) One more observation is required.
n +1
(A)
(D) Conclusion cannot be drawn from observation.
n +1
ö
÷u[ n]
÷
ø
(B)
1 æç
æ 1ö
2 -ç- ÷
ç
3è
è 2ø
ö
÷u[ n]
÷
ø
(D)
1 æç
æ 1ö
2 + ç- ÷
3 çè
è 2ø
n
n
ö
÷u[ n]
÷
ø
ö
÷u[ n]
÷
ø
43. The following input output pair have been observed
47. The difference equation representation for a system
during the operation of a linear system:
is
x1 [ n] = { -1, 2, 1}
¬¾®
S
y1 [ n] = {1, 2, - 1, 0, 1}
­
x2 [ n] = {1, - 1, - 1}
­
S
¬¾
®
y2 [ n] = { - 1, 1, 0, 2}
­
­
x3[ n] = {0, 1, 1}
y[ n] -
S
¬¾
®
­
(A)
3æ 1ö
ç - ÷ u[ n]
2è 2ø
(C)
3æ 1ö
ç ÷ u[ n]
2 è2 ø
(A) System is time-invariant
(A) y[ n] = x[ n] + 11
. y[ n - 1]
(B) y[ n] = x[ n] -
1
( y[ n - 1] + y[ n - 2 ])
2
(D)
2æ1ö
ç ÷ u[ n]
3è2 ø
n
y[ n] - 2 y[ n - 1] + y[ n - 2 ] = x[ n] - x[ n - 1]
If y[ n] = 0 for n < 0 and x[ n] = d[ n], then y[2 ] will
(C) One more observation is required
44. The stable system is
2æ 1ö
ç - ÷ u[ n]
3è 2 ø
48. The difference equation representation for a system is
(B) System is time variant
(D) Conclusion cannot be drawn from observation
n
(B)
n
The conclusion regarding the time invariance of
the system is
y [ -1] = 3
The natural response of system is
n
y3[ n] = {1, 2, 1}
­
1
y[ n - 1] = 2 x[ n],
2
be
(A) 2
(B) -2
(C) -1
(D) 0
49. Consider a discrete-time system S whose response
to a complex exponential input e jpn
S : e jpn
(C) y[ n] = x[ n] - (15
. y[ n - 1] + 0.4 y[ n - 2 ])
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2
Þ
e jp3n
2
is specified as
2
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UNIT 5
24. (B) y[ n] = {1, 4, 4, 10, 4, 4, 4, 1}
Signal & System
28. (C) For n + 2 < 0
­
for
n + 2 ³0
or
n < - 2, y [ n] = 0
n ³ - 2, y[ n] =
or
n+2
1
å4k
4 1 1
,
3 12 4 n
=
k=0
1
2
0
2
1
1
1
2
0
2
1
2
2
4
0
4
2
Þ
æ 4 1 æ 1 ön ö
÷u[ n + 2 ]
y[ n] = ç ç 3 12 çè 4 ÷ø ÷
è
ø
n -2
0
0
0
0
0
k=¥
0
2
2
4
0
4
2
1
1
2
0
2
1
for n - 2 ³ 4 or n ³ 6, y[ n] =
Fig. S5.2.24
25. (A) y[ n] = {2, 4, -19, 36, -25, 2, 12}
­
-3
6
4
2
2
8
-6
12
8
-4
-4
-16
12
-24
-16
3
3
12
ì 3n
,
ïï
y[ n] = í 6
ï 81 ,
ïî 2
-9
18
=
k
81
,
2
n³6
n < 0,
y[ n] = 0
n -3
n > 0, y[ n] =
for n - 3 ³ - 3 or
å 1 = n + 1,
k = -3
y[ n] = ( n + 1) u[ n]
31. (A) For n - 1 < 0
12
For
Fig. S5.2.25
26. (B) x[ n] = {1, 2, 1, 1}, h[ n] = {1, -1, 0, 0, 1}
­
å3
n £5
30. (A) For n - 3 < - 3 or
4
3
k = -¥
Þ
1
3n
6
å 3k =
29. (D) For n - 2 £ 3 or n £ 5 , y[ n] =
Þ
n -1 ³ 0
ì 1,
y[ n] = í
î 0,
or
or
n <1 ,
y[ n] = 0
æp ö
n ³ 1, y[ n] = å cos ç k ÷
k=0
è2 ø
n -1
n = 4 m + 1, 4 m + 2
n = 4 m, 4 m + 3
­
32. (B) y1 [ n] = rect ( v[ n]) ,
y2 [ n] = rect ( kv[ n])
1
2
1
1
y2 [ n] ¹ k y1 [ n]
1
1
2
1
-1
y1 [ n] = rect ( v[ n]), y2 [ n] = rect ( v[ n - no ])
-1
-1
-2
-1
-1
0
0
0
0
0
0
0
0
0
0
(Not Homogeneous not linear)
y1 [ n - no ] = rect ( v[ n - no ]) = y2 [ n]
(Time Invariant)
At any discrete time n = no , the response depends only
on the excitation at that discrete time.
(Causal)
No matter what values the excitation may have the
response can only have the values zero or one.
1
1
2
1
(Stable)
1
Fig. S5.2.26
33. (C) y1 [ n] = nv[ n] ,
y[ n] = {1, 1, -1, 0, 0, 2, 1, 1}
ky1 [ n] = y2 [ n]
­
y[ n ] = d[ n + 2 ] + d[ n + 1 ] - d[ n ] + 2 d[ n - 3 ] + d[ n - 4 ] + d[ n - 5 ]
¥
å ( -1)
27. (D) y[ n] =
k
k=n -2
æ 1ö
2n ç - ÷
è 2ø
=
1
1+
2
Page
266
n -2
=
8
( -1) n
3
2n -k = 2n
¥
å
k=n -2
y2 [ n] = nkv[ n]
æ 1ö
ç- ÷
è 2ø
k
(Homogeneous)
Let x1 [ n] = v[ n]
then
y1 [ n] = nv[ n]
Let x2 [ n] = w[ n]
then
y2 [ n] = nw[ n]
Let x3[ n] = v[ n] + w[ n] then
y3[ n] = n( v[ n] + w[ n]) = nv[ n] + nw[ n]
= y1 [ n] + y2 [ n]
(Additive)
Since the system is homogeneous and additive, it is also
linear.
y1 [ n - no ] = ( n - no) v[ n - no ] ¹ yn [ n] = nv[ n - no ]
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Discrete-Time Systems
(Time variant)
Chap 5.2
If the excitation is bounded, the response is bounded.
At any discrete time, n = no the response depends only
on the excitation at that same time.
(Causal)
(Stable).
If the excitation is a constant, the response is
37. (B) y1 [ n] = 10 v[ n] - 5,
unbounded as n approaches infinity.
y2 [ n] ¹ ky1 [ n]
n +1
34. (C) y1 [ n] =
å v[ m ] ,
y2 [ n ] =
m = -¥
(Unstable)
n +1
å
å kv[ m ]
n = -¥
y3[ n] =
y1 [ n - no ] = 10 v[ n - no ] - 5, = y2 [ n]
m = -¥
(Homogeneous)
v[ m ], y2 [ n] =
(Not Homogeneous so not linear)
y1 [ n] = 10 v[ n] - 5, y2 [ n] = 10 v[ n - no ] - 5
n +1
y2 [ n] = ky1 [ n]
y1 [ n] =
y2 [ n] = 10 kv[ n] - 5
n +1
å w[ m ]
(Time Invariant)
At any discrete time, n = no the response depends only
on the excitation at that discrete time and not on any
future excitation.
n = -¥
(Causal)
If the excitation is bounded, the response is bounded.
n +1
å ( v[ n] + w[ m ])
(Stable).
m = -¥
=
n +1
n +1
m = -¥
m = -¥
å v[ m ] + å w[ n] = y [ n] + y [ n]
1
(Additive)
2
Since the system is homogeneous and additive it is also
linear
y1 [ n] =
n +1
å v[ n] , y2 [ n] =
m = -¥
y1 [ n - no ] =
y[ n] = x[ n] + x[ n - 1] + y[ n - 2 ], Then by induction
¥
y[ n] = x[ n - 1] + x[ n - 2 ] + K x[ n - k] + K = å x[ n - k]
k=0
n +1
å v[ m - no ]
n +1
m = -¥
q = -¥
å v[ q - no ] = y2 [ n]
y1 [ n] =
n
å v[ m ] ,
n
m =n
m = -¥
n
å kv[ m ] = ky [ n]
y2 [ n ] =
m = -¥
-¥
å x[ m ] = å x[ m ]
Let m = n - k then y[ n] =
m = -¥
n -no +1
å v[ m ] =
38. (B) y[ n] = x[ n] + y[ n - 1], y[ n - 1] = x[ n - 1] + y[ n - 2 ]
1
m = -¥
(Time Invariant)
At any discrete time, n = no , the response depends on
the excitation at the next discrete time in future.
(Anti causal)
(Homogeneous)
y3[ n] =
n
¥
n
m = -¥
m = -¥
m = -¥
å (v[ m ] + w[ m ]) = å v[ m ] + å w[ m ]
= y1 [ n] + y2 [ n]
(Additive)
If the excitation is a constant, the response increases
System is Linear.
without bound.
y1 [ n] =
(Unstable)
¥
å v[ m ]
m = -¥
35. (A) y1 [ n] = v[ n] , y2 = kv[ n] = k v[ n]
y1 [ n] = v[ n] , y2 [ n] = v[ n - no ]
(Time Invariant)
At any discrete time n = no , the response depends only
on the excitation at that time
o
m = -¥
y1 [ n] can be written as
ky1 [ n] = k v[ n] ¹ y2 [ n] (Not Homogeneous Not linear)
y1 [ n - no ] = v[ n - no ] = y2 [ n]
n
å v[ n - n ]
, y2 =
(Causal)
If the excitation is bounded, the response is bounded.
(Stable).
36. (B) y[ n] = 2 x [ n]
y1 [ n - no ] =
n -no
n
m = -¥
q = -¥
å v[ m ] = å v[ q - n ] = y [ n]
o
2
(Time Invariant)
At any discrete time n = no the response depends only
on the excitation at that discrete time and previous
discrete time.
(Causal)
If the excitation is constant, the response increase
without bound.
(Unstable)
2
Let x1 [ n] = v[ n]
Let x2 [ n] = kv[ n] then
ky[ n] ¹ y2 [ n]
Let x1 [ n] = v[ n]
39. (C) Only statement (b) is false. For example
then y1 [ n] = 2 v 2 [ n]
S1 : y[ n] = x[ n] + b, and S2 : y[ n] = x[ n] - b , where b ¹ 0
y2 [ n] = 2 k2 v 2 [ n]
S{x[ n]} = S2 {S1 {x[ n]}} = S2 {x[ n] + b} = x[ n]
(Not homogeneous Not linear)
then
y1 [ n] = 2 v 2 [ n]
Hence S is linear.
Let x2 [ n] = v[ n - no ] then y2 [ n] = 2 v 2 [ n - no ]
40. (B) For example
y1 [ n - no ] = 2 v[ n - no ] = y2 [ n]
S1 : y[ n] = nx[ n]
(Time invariant)
At any discrete time, n = no , the response depends only
on the excitation at that time.
(Causal)
and S2 : y[ n] = nx[ n + 1]
If x[ n] = d[ n] then S2 {S1 {d[ n]}} = S2 [0 ] = 0,
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CHAPTER
5.3
THE LAPLACE TRANSFORM
Statement for Q.1-12:
6. x( t) = u( t) - u( t - 2)
Determine the Laplace transform of given signal.
1. x( t) = u( t - 2)
-2 s
(A)
-e
s
(B)
e
-2 s
e -2 s - 1
s
(B)
1 - e -2 s
s
(C)
2
s
(D)
-2
s
s
7. x( t) =
-2 s
e
(C)
1+ s
(A)
(D) 0
(A)
1
s( s + 1) 2
(B)
s
( s + 1) 2
(C)
e- s
s+1
(D)
e- s
( s + 1) 2
2. x( t) = u( t + 2)
(A)
1
s
(B) -
(C)
e -2 s
s
(D)
1
s
-e - 2 s
s
3. x( t) = e -2 t u( t + 1)
(A)
(C)
1
s+2
(B)
e- ( s + 2 )
s+2
(D)
e- s
s+2
-e - s
s+2
4. x( t) = e 2 t u( -t + 2)
(A)
e
2 (s - 2 )
-1
s -2
-2 ( s - 2 )
(C)
1-e
s -2
8. x( t) = tu( t) * cos 2pt u( t)
(A)
1
s( s 2 + 4 p2 )
(B)
2p
s 2 ( s 2 + 4 p2 )
(C)
1
s ( s + 4 p2 )
(D)
s3
s + 4 p2
-3
s4
e
s+2
-2 s
(D)
2
2
2
9. x( t) = t 3u( t)
(A)
3
s4
(B)
(C)
6
s4
(D) -
-2 s
(B)
d
{ te - t u( t)}
dt
e
s -2
6
s4
10. x( t) = u( t - 1) * e -2 t u( t - 1)
5. x( t) = sin 5 t
5
(A) 2
s +5
s
(B) 2
s +5
(A)
e -2 ( s + 1 )
2s + 1
(B)
e -2 ( s + 1 )
s+1
5
(C) 2
s + 25
s
(D) 2
s + 25
(C)
e- ( s + 2 )
s+2
(D)
e -2 ( s + 1 )
s+2
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UNIT 5
t
11. x( t) = ò e -3t cos 2 t dt
17. X ( s) =
0
s2 - 3
( s + 2)( s 2 + 2 s + 1)
(A)
-( s + 3)
s(( s + 3) 2 + 4)
(B)
( s + 3)
s(( s + 3) 2 + 4)
(A) ( e -2 t - 2 te - t ) u( t)
(B) ( e -2 t + 2 te - t ) u( t)
(C) ( e - t - 2 te -2 t ) u( t)
(D) ( e - t + 2 te -2 t ) u( t)
(C)
s( s + 3)
( s + 3) 2 + 4
(D)
-s( s + 3)
( s + 3) 2 + 4
18. X ( s) =
12. x( t) = t
( s + 2 s + 2)
( s 2 + 4 s + 2) 2
(C) ( 3e - t cos 3t - e - t sin 3t) u( t)
-( s + 2 s + 2)
( s 2 + 4 s + 2) 2
2
(D)
(D) ( 3e - t sin 3t + 3e - t cos 3t) u( t)
19. X ( s) =
Statement for Q.13–24:
Determine the time signal x( t) corresponding to
given X ( s) and choose correct option.
(B) (2 e - t - e -2 t ) u( t)
(C) (2 e -2 t - e - t ) u( t)
(D) (2 e - t + e -2 t ) u( t)
(A) 2 d( t) + ( e -3t - e-2 t ) u( t)
(A) (2 e -2 t + 2 e - t sin 2 t - 2 e - t cos 2 t) u( t)
(B) (2 e -2 t + 2 e - t cos 2 t - 2 e - t sin 2 t) u( t)
(D) (2 e -2 t + 2 e - t sin 2 t - e - t cos 2 t) u( t)
(A) (2 e -2 t + e - t ) u( t)
2 s 2 + 10 s + 11
s2 + 5 s + 6
20. X ( s) =
(C) 2 d( t) + ( e
+ e ) u( t)
(D) 2 d( t) - ( e
-2 t
+ e -3t ) u( t)
15. X ( s) =
-t
-3t
2s - 1
s2 + 2 s + 1
-t
(A) ( 3e - 2 te ) u( t)
(B) ( 3e- t + 2 te- t ) u( t)
(B) ( e -2 t + 2 e -3t cos t - 6 e -3t sin t) u( t)
(C) ( e -2 t + 2 e -3t cos t - 2 e -3t sin t) u( t)
(D) (9 e -2 t - 6 e -3t cos t + 3e -3t sin t) u( t)
21. X ( s) =
16. X ( s) =
5s + 4
s 3 + 3s 2 + 2 s
2 s 2 + 11s + 16 + e -2 s
( s2 + 5 s + 6)
(A) 2 d( t) + ( 3e -2 t - 2 e-3t ) u( t - 2)
(B) 2 d( t) + (2 e -2 t - e -3t + e -2 ( t - 2 ) + e -3( t - 2 ) ) u( t)
(C) 2 d( t) + (2 e -2 t - e -3t ) u( t) + ( e -2 t - e -3t ) u( t - 2)
(D) 2 d( t) + (2 e -2 t - e -3t ) u( t) + ( e -2 ( t - 2 ) - e -3( t - 2 ) ) u( t - 2)
22. X ( s) = s
(C) (2 e- t - 3te- t ) u( t)
(D) (2 e- t + 3te- t ) u( t)
3s 2 + 10 s + 10
( s + 2)( s 2 + 6 s + 10)
(A) ( e -2 t + 2 e -3t cos t + 2 e -3t sin t) u( t)
(B) 2 d( t) + ( e -2 t - e-3t ) u( t)
-2 t
4 s 2 + 8 s + 10
( s + 2)( s 2 + 2 s + 5)
(C) (2 e -2 t + 2 e - t cos 2 t - e - t sin 2 t) u( t)
s+3
13. X ( s) = 2
s + 3s + 2
14. X ( s) =
2
1
æ
ö
(B) ç 3e- t sin 3t - e- t cos 3t ÷u( t)
3
è
ø
( s 2 + 4 s + 2)
(B) 2
( s + 2 s + 2) 2
2
(C)
3s + 2
s + 2 s + 10
1
æ -t
ö
(A) ç 3e cos 3t - e- t sin 3t ÷u( t)
3
è
ø
d -t
{ e cos t u( t)}
dt
-( s 2 + 4 s + 2)
(A) 2
( s + 2 s + 2) 2
d2
ds 2
æ 1 ö
1
çç 2
÷÷ +
s
+
9
s
+
3
è
ø
ö
æ
2t
t2
(A) çç e -3t +
sin 3t + cos 3t ÷÷u( t)
3
9
ø
è
(B) ( e -3t + 2 t sin 3t + t 2 cos 3t) u( t)
(A) (2 + e - t + 3e -2 t ) u( t)
2t
æ
ö
(C) ç e -3t +
sin 3t + t 2 cos 3t ÷u( t)
3
è
ø
(B) (2 + e - t - 3e -2 t ) u( t)
(D) ( e -3t + t 2 sin 3t + 2 t cos 3t) u( t)
(C) ( 3 + e - t - 3e -2 t ) u( t)
(D) ( 3 + e - t + 3e -2 t ) u( t)
Page
270
Signal & System
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The Laplace Transform
23. X ( s) =
(A) e
(C)
-0 .5t
Chap 5.3
Statement for Q.30–33:
1
(2 s + 1) 2 + 4
Given the transform pair
1
(B) e - t sin t u( t)
2
sin t u( t)
1 -0 .5t
e
sin t u( t)
4
x( t) u( t)
(D) e - t sin t u( t)
L
¬¾
®
2s
.
s2 + 2
Determine the Laplace transform Y ( s) of the given
time signal in question and choose correct option.
24. X ( s) = e
-2 s
d æ
1
ç
ds çè ( s + 1) 2
ö
÷÷
ø
30. y( t) = x( t - 2)
(A) - te - t u(1 - t)
(B) -te - t u( t - 1)
(C) -( t - 2) 2 e - ( t - 2 ) u( t - 2)
(D) te - t u( t - 1)
Statement for Q.25–29:
(A)
2 se -2 s
s2 + 2
(B)
2 se 2 s
s2 + 2
(C)
2( s - 2)
( s - 2) 2 + 1
(D)
2( s + 2)
( s + 2) 2 + 1
Given the transform pair below. Determine the
31. y( t) = x( t) *
time signal y( t) and choose correct option.
cos 2t u( t)
L
¬¾
®
X ( s).
dx( t)
dt
(A)
4 s3
( s 2 + 2) 2
(B)
4
( s 2 + 2) 2
-4 s 3
( s + 2) 2
(D)
4
( s 2 + 2) 2
25. Y ( s) = ( s + 1) X ( s)
(A) [cos 2 t - 2 sin 2 t ]u( t)
sin 2 t ö
æ
(B) ç cos 2 t +
÷u( t)
2 ø
è
(C)
(C) [cos 2 t + 2 sin 2 t ]u( t)
sin 2 t ö
æ
(D) ç cos 2 t ÷u( t)
2 ø
è
32. y( t) = e - t x( t)
(A)
2( s + 1)
( s + 1) 2 + 2
(B)
2( s + 1)
s2 + 2 s + 2
(C)
2( s + 1)
s2 + 2 s + 4
(D)
2( s + 1)
s2 + 2 s
26. Y ( s) = X ( 3s)
æ2 ö
(A) cos ç t ÷u( t)
è3 ø
1
æ2 ö
(B) cos ç t ÷u( t)
3
è3 ø
(C) cos 6t u( t)
(D)
1
cos 6 t u( t)
3
33. y( t) = 2 tx( t)
(A)
8 - 4 s2
( s 2 + 2) 2
(B)
4 s2 - 8
( s 2 + 2) 2
(C)
4 s2
s2 + 1
(D)
s2
s +1
27. Y ( s) = X ( s + 2)
(A) cos 2( t - 2) u( t)
(B) e2 t cos 2 t u( t)
(C) cos 2( t + 2) u( t)
(D) e-2 t cos 2 t u( t)
2
2
Statement for Q.34–43:
X ( s)
28. Y ( s) = 2
s
(A) 4 cos 2 t u( t)
2
(C) t cos 2 t u( t)
29. Y ( s) =
Determine the bilateral laplace transform and
(B)
1 - cos 2 t
u( t)
4
cos 2 t
(D)
u( t)
t2
d -3s
[ e X ( s)]
ds
(A) t cos 2( t - 3) u( t - 3)
(B) t cos 2( t - 3) u( t)
(C) -t cos 2( t - 3) u( t - 3)
(D) -t cos 2( t - 3) u( t)
choose correct option.
34. x( t) = e - t u( t + 2)
(A)
e2 ( s + 1 )
,
s+1
Re ( s) > -1
(B)
1
,
1+ s
Re ( s) < - 1
(C)
e2 ( s + 1 )
,
s+1
Re ( s) < - 1
(D)
1
,
1+ s
Re ( s) > -1
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UNIT 5
35. x( t) = u( -t + 3)
1 - e -3s
,
(A)
s
(B)
1-e
(C)
s
(D)
-3s
Re ( s) > 3
(B)
e 3s
,
s-3
Re ( s) < 3
(C)
e 3( s -1 )
,
s-3
Re ( s) > 3
(D)
e 3( s -1 )
,
s-3
Re ( s) < 3
Re ( s) < 0
,
-e -3s
,
s
Re ( s) > 0
(A) e , Re( s) > 0
s
(B) e , Re ( s) < 0
s
(D) None of above
s
(C) e , all s
37. x( t) = sin t u( t)
1
,
(A)
(1 + s 2 )
Re ( s) > 0
-1
(C)
,
(1 + s 2 )
Re ( s) < 0
-1
,
(1 + s 2 )
Re ( s) > 0
-
t
2
41. x( t) = cos 3t u( -t) * e - t u( t)
-s
,
(A)
( s + 1)( s 2 + 9)
-s
,
( s + 1)( s 2 + 9)
-1 < Re ( s) < 0
(C)
s
,
( s + 1)( s 2 + 9)
-1 < Re ( s) < 0
(D)
s
,
( s + 1)( s 2 + 9)
Re ( s) > 0
42. x( t) = e t sin (2 t + 4) u( t + 2)
-t
38. x( t) = e u( t) + e u( t) + e u( -t)
t
(A)
6 s2 + 2 s - 2
,
(2 s + 1)( s 2 - 1)
Re ( s) < - 0.5
(B)
6 s2 + 2 s - 2
,
(2 s + 1)( s 2 - 1)
-1 > Re ( s) > 1
(C)
1
1
1
,
+
+
s + 0.5 s + 1 s - 1
-1 < Re ( s) < 1
(D)
1
1
1
,
+
s + 0.5 s + 1 s - 1
-0.5 < Re ( s) < 1
(A)
e 2 ( s -1 )
,
( s - 1) 2 + 4
Re ( s) > 1
(B)
e 2 ( s -1 )
,
( s - 1) 2 + 4
Re ( s) < 1
(C)
e( s - 2 )
,
( s - 1) 2 + 4
Re ( s) > 1
(D)
e( s - 2 )
,
( s - 1) 2 + 4
43. x( t) = e t
(A)
1-s
,
s+1
Re ( s) > -1
(C)
s -1
,
s+1
Re ( s) < - 1
s -1
,
s+1
Re ( s) > -1
(1 - s)
1
1
, 0.5 < Re ( s) < 1
+
+
2
( s - 1) + 4 s + 1 s - 0.5
(D)
(B)
(1 - s)
1
1
, -1 < Re ( s) < 1
+
+
( s - 1) 2 + 4 s + 1 s - 0.5
Statement for Q.44–49:
(C)
( s - 1)
1
1
, 0.5 < Re ( s) < 1
+
+
( s - 1) 2 + 4 s + 1 s - 0.5
bilateral Laplace transform.
40. x( t) = e ( 3t + 6 ) u( t + 3)
Page
272
Re ( s) < - 1
(A)
( s - 1)
1
1
, -1 < Re ( s) < 1
+
+
( s - 1) 2 + 4 s + 1 s - 0.5
Re ( s) < 1
d -2 t
[ e u( -t)]
dt
1-s
,
s+1
(B)
t
39. x( t) = e t cos 2 t u( -t) + e - t u( t) + e 2 u( t)
(D)
Re ( s) > 0
(B)
Re ( s) < 0
1
,
(1 + s 2 )
(D)
e 3s
,
s-3
Re ( s) < 0
36. y( t) = d( t + 1)
(B)
(A)
Re ( s) > 0
-e -3s
,
s
Signal & System
Determine the corresponding time signal for given
44. X ( s) =
e 5s
with ROC: Re ( s) < -2
s+2
(A) e -2 ( t + 5) u( t + 5)
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UNIT 5
(A)
1 -t
e sin t u( t)
2
(D)
58.
62. A stable system has input
(B) 2e- t cos t u( t)
(C) 2 e - t cos t u( t) +
is
(A) d( t) - ( e -2 t cos t + e -2 t sin t) u( t)
1 -t
e cos t u( t - 1) + 2 e - t sin t u( t - 1)
2
(B) d( t) - ( e -2 t cos t + e -2 t sin t) u( t - 2)
(C) d( t) - ( e 2 t cos t + e 2 t sin t) u( t)
d 3 y( t)
d 2 y( t)
dy( t)
+
4
+3
= x( t)
3
2
dt
dt
dt
(D) d( t) - ( e 2 t cos t + e 2 t sin t) u( t + 2)
All initial condition are zero, x( t) = 10 e -2 t
63. The relation ship between the input x( t) and output
y( t) of a causal system is described by the differential
5
é5
ù
(A) ê + 5 e - t - 5 e -2 t + e -3t ú u( t)
3
ë3
û
equation
5
é5
ù
(B) ê - 5 e - t + 5 e -2 t + e -3t ú u( t)
3
ë3
û
5
5
u( t) - 5 u( t - 1) + 5 u( t - 2) + u( t - 3)
3
3
(D)
5
5
u( t) + 5 u( t - 1) - 5 u( t - 2) + u( t - 3)
3
3
dy( t)
+ 10 y( t) = 10 x( t)
dt
The impulse response of the system is
59. The transform function H ( s) of a causal system is
(A) -10 e -10 t u( -t + 10)
(B) 10 e -10 t u( t)
(C) 10 e -10 t u( -t + 10)
(D) -10 e -10 t u( t)
64. The relationship between the input x( t) and output
y( t) of a causal system is defined as
2 s2 + 2 s - 2
H ( s) =
s2 - 1
d 2 y( t) dy( t)
dx( t)
.
- 2 y( t) = -4 x( t) + 5
2
dt
dt
dt
The impulse response is
The impulse response of system is
(A) 2d( t) - ( e- t + et ) u( -t)
-t
(A) 3e - t u( t) + 2 e 2 t u( -t)
(B) 2d( t) - ( e + e ) u( t)
t
-t
(B) ( 3e - t + 2 e 2 t ) u( t)
(C) 2d( t) + e u( t) - e u( -t)
t
(C) 3e - t u( t) - 2 e 2 t u( -t)
(D) 2d( t) + ( e- t + et ) u( t)
(D) ( 3e - t - 2 e 2 t ) u( -t)
60. The transfer function H ( s) of a stable system is
2s - 1
H ( s) = 2
s + 2s + 1
The impulse response is
(A) 2 u( -t + 1) - 3tu( -t + 1)
(B) ( 3te- t - 2 e- t ) u( t)
(C) 2 u( t + 1) - 3tu( t + 1)
(D) (2 e- t - 3te- t ) u( t)
61. The transfer function H ( s) of a stable system is
H ( s) =
s2 + 5 s - 9
( s + 1)( s 2 - 2 s + 10)
The impulse response is
(A) -e - t u( t) + ( e t sin 3t + 2 e t cos 3t) u( t)
(B) -e - t u( t) - ( e t sin 3t + 2 e t cos 3t) u( -t)
(C) -e - t u( t) - ( e t sin 3t + 2 e t cos 3t) u( t)
(D) -e - t u( t) + ( e t sin 3t + 2 e t cos 3t) u( -t)
Page
274
x( t) and output
y( t) = e -2 t cos t u( t). The impulse response of the system
1 -t
e sin t u( t)
2
(C)
Signal & System
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The Laplace Transform
SOLUTIONS
¥
¥
0
2
1. (B) X ( s) = ò x( t) e - st dt = ò e - st dt =
Chap 5.3
r ( t) = e -2 t u( t)
L
¬¾
®
v( t) = e -2 t u( t - 1)
e -2 s
s
L
¬¾
®
x( t) = q( t) * v( t)
¥
¥
¥
0
0
0
2. (A) X ( s) = ò x( t) e -3t dt = ò u( t + 2) -3t dt = ò e -3t dt =
1
s
L
¬¾
®
3. (A) X ( s) = ò e -2 t e - st dt =
0
1
s+2
¥
¥
0
0
- 1 1 - e -2 ( 2 - s )
=
2-s
s -2
p( t) dt
1
P ( s)
p( t) dt +
ò
s -¥
s
L
¬¾
®
Þ
X ( s) =
5
( e j 5t - e - j 5t ) - st
e dt = 2
2
25
j
s
+
0
q( t) =
d
p( t)
dt
x( t) = tq( t)
6. (B) X ( s) = ò e - st dt =
1 - e -2 s
s
7. (B) p( t) = te - t u( t)
L
¬¾
®
2
0
d
p( t)
dt
L
¬¾
®
Þ
P ( s) =
1
( s + 1) 2
s
( s + 1) 2
X ( s) =
8. (A) p( t) = tu( t)
L
¬¾
®
P ( s) =
1
s2
q( t) = cos 2 pt u( t)
L
¬¾
®
Q( s) =
s
s + 4 p2
x( t) = p( t) * q( t)
X ( s) =
L
¬¾
®
q( t) = - tp( t)
t n u( t)
X ( s) = P ( s)Q( s)
L
¬¾
®
P ( s) =
Q( s) =
X ( s) =
1
s2
d
-2
P ( s) = 3
ds
s
d
6
Q( s) = 4
ds
s
n!
sn + 1
L
¬¾
®
10. (D) p( t) = u( t)
q( t) = u( t - 1)
L
¬¾
®
L
¬¾
®
Q( s) =
X ( s) = -
P ( s) =
s+1
( s + 1) 2 + 1
s( s + 1)
( s + 1) 2 + 1
d
Q( s)
ds
-( s 2 + 4 s + 2)
( s 2 + 2 s + 2) 2
13. (B) X ( s) =
A=
L
¬¾
®
s+3
A
B
=
+
( s + 3s + 2) s + 1 s + 2
2
s+3
s+3
= 2, B =
= -1
s + 2 s = -1
s + 1 s = -2
x( t) = [2 e - t - e -2 t ]u( t)
14. (A) X ( s) = 2 -
1
1
1
=2 +
( s + 2) ( s + 3)
( s + 2) ( s + 3)
x( t) = 2 d( t) + ( e-3t - e-3t ) u( t)
2s - 1
A
B
=
+
s 2 + 2 s + 1 ( s + 1) ( s + 1) 2
B = (2 s - 2) s = -1 = -3, A = 2
x( t) = x( t) = [2 e - t - 3te - t ]u( t)
16. (B) X ( s) =
5s + 4
A
B
C
=
+
+
2
s + 3s + 2 s s s + 1 s + 2
3
A = sX ( s) s = 0 = 2, B = ( s + 1) X ( s) s = -1 = 1,
C = ( s + 2) X ( s) s = -2 = -3
x( t) = [2 + e - t - 3e -2 t ]u( t)
L
¬¾
®
L
¬¾
®
L
¬¾
®
15. (C) X ( s) =
2
9. (C) p( t) = tu( t)
X ( s) =
L
¬¾
®
2
1
s( s + 4 p2 )
x( t) = - tq( t)
( s + 3)
s[( s + 3) 2 + 4 ]
12. (A) p( t) = e - t cos t u( t)
¥
s+3
( s + 3) 2 + 4
0
ò
5. (C) X ( s) = ò
Þ
X ( s) = Q( s) V ( s)
2 (2 - s )
0
x( t) =
e- ( s + 2 )
s2
e
X ( s) =
s+2
Þ
t
4. (C) X ( s) = ò x( t) e - st dt = ò e 2 t u( -t + 2) e - st dt
e
V ( s) =
-2 ( s + 1 )
-¥
= ò e t ( 2 - s ) dt =
1
s+2
L
11. (B) p( t) = e -3t cos 2 t u( t) ¬¾
® P ( s) =
¥
2
R( s) =
P ( s) =
Q( s) =
e- s
s2
1
s
17. (C) X ( s) =
=
s2 - 3
( s + 2)( s 2 + 2 s + 1)
A
B
C
+
+
( s + 2) ( s + 1) ( s + 1) 2
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UNIT 5
A = ( s + 2) X ( s) s = -2 = 1, C = ( s + 1) 2 X ( s)
A + B =1
x( t) = [ e
-2 t
Þ
B =0
æsö
L
23. (C) P ç ÷ ¬¾
® ap( at)
èaø
1
1 -t
L
¬¾
®
e sin 2 t u( t)
( s + 1) 2 + 4
2
= -2
s = -1
-t
- te ]u( t)
3s + 2
3( s + 1)
1
=
2
2
( s + 1) 2 + 32
s + 2 s + 10 ( s + 1) + 3
18. (A) X ( s) =
2
1
é
ù
x( t) = ê3e - t cos 3t - e - t sin 3t ú u( t)
3
ë
û
Q( s) =
Þ
A = ( s + 2) X ( s) s = -2 = 2
Þ
Þ
C = -2
Þ
x( t) = [2 e -2 t + 2 e - t cos 2 t - e - t sin 2 t ]u( t)
A = ( s + 2) X ( s) s = -2 = 1, A + B = 3
x( t) = [ e
-2 t
+ 2e
-3t
21. (D) X ( s) =
Þ
Þ
-3t
sin t ]u( t)
( s + 2)(2 s 2 + 11s + 16)
=2
( s2 + 5 s + 6)
s = -2
B=
( s + 3)(2 s + 11s + 16)
= -1
( s2 + 5 s + 6)
s = -3
R( s) = sQ( s)
L
¬¾
®
e-2 t x( t)
X ( s)
s
L
¬¾
®
p( t) =
ò x( t) dt
-¥
t
ò cos 2 t u( t) dt =
P ( s)
s
1
sin 3t u( t)
3
sin 2 t
1 - cos 2 t
dt =
u( t),
2
4
0
ò
L
¬¾
®
31. (A) p( t) =
q( t) = - p( t)
32. (A) e - t x ( t)
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L
¬¾
®
d
x( t)
dt
y( t) = x( t) * p( t)
ù
é2 t
x( t) = v( t) + r ( t) = ê sin 3t u( t) + t 2 cos 3t u( t) + e -3t ú u( t)
ë3
û
p( t) = x( t - 3)
= -t cos 2( t - 3) u( t - 3).
t2
sin 3t u( t)
3
2t
sin 3t u( t) + t 2 cos 3t u( t)
3
1
L
V ( s) =
¬¾
® v( t) = e -3t u( t)
s+3
L
¬¾
®
= cos 2( t - 3) u( t - 3)
d
L
Q( s) =
P ( s) ¬¾
®
ds
30. (A) x( t - 2)
d
q( t) - q(0 - )
dt
sin 2 t
2
t
29. (C) P ( s) = e -3s X ( s)
q( t) = ( -1) 2 t 2 p( t) =
r ( t) =
t
L
¬¾
®
-¥
=
Page
276
L
¬¾
®
L
¬¾
®
x( t) = 2 d( t) + [2 e -2 t - e-3t ]u( t) + [ e -2 ( t - 2 ) - e -3( t - 2 ) ]u( t - 2)
L
¬¾
®
ax( at)
1
æ2 ö
cos ç t ÷ u( t)
3
è3 ø
L
¬¾
®
28. (B) P ( s) =
2
d2
P ( s)
ds 2
L
¬¾
®
2 s 2 + 11s + 16 + e -2 s
( s 2 + 5 s + 6)
A=
Q( s) =
dx( t)
+ x( t)
dt
L
¬¾
®
C = -6
cos t - 6 e
1
s2 + 9
x( t) = q( t - 2)
y( t) = ( -2 sin 2 t + cos 2 t ) u( t)
27. (D) X ( s + 2)
B =2
A
B
e -2 s
e -2 s
=2 +
+
+
( s + 2) ( s + 3) ( s + 2) ( s + 3)
22. (C) P ( s) =
L
¬¾
®
x( t) = - ( t - 2) e ( t - 2 ) u( t - 2)
X ( 3s)
A
B( s + 3)
C
=
+
+
( s + 2) ( s + 3) 2 + 1 ( s + 3) 2 + 1
q( t) = -tp( t) = -t 2 e - t u( t)
L
¬¾
®
æsö
26. (B) X ç ÷
èaø
3s 2 + 10 s + 10
20. (B) X ( s) =
( s + 2)( s 2 + 6 s + 10)
10 A + 6 B + 2 C = 10
d
P ( s)
ds
p( t) = te - t u( t)
L
¬¾
®
25. (A) sX ( s) + X ( s)
B =2
5 A + 2 B + 2 C = 10
1
( s + 1) 2
X ( s) = e -2 sQ( s)
A
B( s + 1)
C
=
+
+
2
2
( s + 2) ( s + 1) + 2
( s + 1) 2 + 2 2
A + B=4
1 -0 .5t
e
sin t u( t)
4
L
¬¾
®
x( t)
24. (C) P ( s) =
4 s 2 + 8 s + 10
( s + 2)( s 2 + 2 s + 5)
19. (C) X ( s) =
Signal & System
e -2 s X ( s), Y ( s) =
L
¬¾
®
L
¬¾
®
L
¬¾
®
2 se -2 s
s2 + 2
P ( s) = sX ( s)
Y ( s) = P ( s) X ( s) = s( X ( s)) 2
X ( s + 1) =
2( s + 1)
( s + 1) 2 + 2
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The Laplace Transform
33. (B) 2 tx( t)
L
¬¾
®
¥
ò x( t) e
34. (A) X ( s) =
-2
- st
d
4 s2 - 8
X ( s) = 2
ds
( s + 2) 2
Chap 5.3
42. (A) x( t) = e -2 e t + 2 sin (2 t + 4) u( t + 2)
p( t + 2)
X ( s) =
dt
L
¬¾
®
e 2 s P ( s),
e 2 s e -2
,
( s - 1) 2 + 4
Re ( s) > 1
-¥
¥
¥
-2
-2
= ò e - t e - st dt = ò e - t ( s + 1 ) dt =
e2 ( s + 1 )
, Re( s) > - 1
s+1
¥
-3s
-e
35. (B) X ( s) = ò u( -t + 3) e - st dt = ò e - st dt =
s
-¥
-¥
36. (C) Y ( s) =
3
¥
ò d( t + 1) e
- st
dt = e s ,
43. (A) p( t) = e -2 t u ( -t)
q( t) =
Re ( s) < 0
( e - e ) - st
e dt
2j
0
37. (B) X ( s) = ò
¥
X ( s) = e 5s P ( s)
¥
1
1
1
,
e t ( j - s ) dt e - t ( j + s ) dt =
=
2 j ò0
2 j ò0
1 + s2
¥
-
¥
t
2
38. (D) X ( s) = ò e e - st + ò e - t e - st dt +
0
=
0
Þ
Re ( s) > 0
òee
t - st
dt
-¥
òe
-¥
¥
X ( s) = -
0.5 < Re ( s) < 1
q( t) = p( t + 3)
X ( s) =
e 3( s -1 )
,
s-3
¬¾®
L
L
¬¾
®
Re ( s) > 3
41. (B) p( t) * q( t)
d2
P ( s)
ds 2
1
R( s)
s
u( -( t + 5))
p( t) = e 3t u( t)
L
¬¾
®
x( t) = t 2 e 3t u( t)
Þ
L
¬¾
®
Re ( s) > -1, Re ( s) < 0
-1 < Re ( s) < 0
P ( s)Q( s)
p( t) = u( t)
L
¬¾
®
r ( t) = -tq( t) = -tu( t - 3)
v( t) =
t
ò r( t) dt
-¥
t
3
e 3s
s-3
L
¬¾
®
q( t) = p( t - 3) = u( t - 3)
v( t) = -ò tdt = -
X ( s) =
1
s
L
¬¾
®
L
¬¾
®
1
v( s)
s
1 2
( t - 9)
2
L
¬¾
®
x( t) = -
t
1
2
ò ( t - 9)
2 -¥
9
é -1
ù
x( t) = ê ( t 3 - 27) + ( t - 3) ú u( t - 3)
6
2
ë
û
48. (B) X ( s) =
-s æ 1 ö
X ( s) = 2
ç
÷
s + 9 è s + 1ø
Þ
x( t) = p( t + 5)
47. (C) Right-sided, P ( s) =
Þ
1
P ( s) =
s-3
Q( s) = e 3s P ( s) =
-
V ( s) =
40. (C) x( t) = e -3e -3( t + 3) u( t + 3)
p( t) = e u( t)
-2 ( t + 5)
Q( s) = e-3s P ( s)
d
R( s) =
Q( s)
ds
s -1
1
1
+
+
( s - 1) 2 + 4 ( s + 1) s - 0.5
3t
L
¬¾
®
t
2
( e - e ) - st
e dt + ò e - t e- st dt + ò e e - st dt
2j
0
0
Re ( s) < 1, Re ( s) > -1, Re ( s) > 0.5
Therefore
p( t) = -e -2 t u( -t)
x( t) = -u( -t) + u ( -t + 1) + d( t + 2)
¥
- jt
jt
t
1-s
1+ s
46. (D) Left-sided
-0.5 < Re ( s) < 1
39. (A) X ( s) =
x( t) = - e
X ( s) =
Re ( s) > -0.5, Re ( s) > -1, Re ( s) < 1
0
X ( s) = Q( s - 1) =
45. (A) Right-sided
1
L
P ( s) =
¬¾
®
( s - 3)
0
1
1
1
+
s + 0.5 s + 1 s - 1
Þ
L
¬¾
®
-1
, Re ( s) < - 2
s+2
Q( s) = sP ( s)
44. (C) Left-sided
1
L
P ( s) =
¬¾
®
s+2
- jt
jt
x( t) = e t q( t)
L
¬¾
®
P ( s) =
Re ( s) < - 1 thus left-sided .
All s
-¥
¥
d
p( t)
dt
L
¬¾
®
-s - 4
-3
2
=
+
s 2 + 3s + 2 ( s + 1) s + 2
Left-sided, x( t) = 3e - t u( -t) - 2 e -2 t u( -t)
49. (A) X ( s) =
Left-sided,
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5
1
( s + 1) ( s + 1) 2
x( t) = -5 u( -t) + te - t u( -t)
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UNIT 5
50. (D) x(0 + ) = lim sX ( s) =
x ®¥
h( t) = (2 e - t - 3te- t ) u( t).
s
=0
s + 5s - 2
2
61. (A) H ( s) =
s2 + 2 s
51. (A) x(0 + ) = lim sX ( s) = 2
=1
s ®¥
s + 2s - 3
52. (D) x(0 + ) = lim sX ( s) =
s ®¥
53. (A) x( ¥) = lim sX ( s) =
s ®0
e
-2 s
h( t) = - e - t u( t) + (2 e t cos 3t + et sin 3t) u( -t)
(6s + s )
=0
s2 + 2 s - 2
3
s ®0
2
62. (A) X ( s) =
2 s3 + 3s
=0
s2 + 5 s + 1
H ( s) =
=1 -
1
,
s+1
Y ( s) =
( s + 2)
( s + 2) 2 + 1
Y ( s) ( s + 1)( s + 2)
=
X ( s) ( s + 2) 2 + 1
( s + 2)
1
( s + 2) 2 + 1 ( s + 2) 2 + 1
h( t) = d( t) - ( e -2 t cos t + e -2 t sin t) u( t)
e -3s (2 s2 + 1) 1
=
s2 + 5 s + 4 4
63. (B) sY ( s) + 10 Y ( s) = 10 X ( s)
H ( s) =
-
56. (C) sY ( s) - y(0 ) + 10 Y ( s) = 10( s)
Þ
1
y(0 ) = 1, X ( s) =
s
10
1
1
Y ( s) =
+
=
s( s + 1) ( s + 1) s
-
Þ
-1
2( s - 1)
3
+
+
( s + 1) ( s - 1) 2 + 32 ( s - 1) 2 + 32
System is stable
s+2
54. (C) x( ¥) = lim sX ( s) = 2
=2
s ®0
s + 3s + 1
55. (B) x( ¥) = lim sX ( s) =
Signal & System
Y ( s)
10
=
X ( s) s + 10
h( t) = 10 e-10 t u( t)
64. (B) Y ( s)( s 2 - s - 2) = X ( s)(5 s - 4)
H ( s) =
y( t) = u( t)
Y ( s)
5s - 4
3
2
=
=
+
X ( s) s 2 - s - 2 s + 1 s - 2
h( t) = 3e - t u( t) + 2 e 2 t u( t).
57. (C) s Y ( s) - 2 s + 2 sY ( s) - 2 + 5 Y ( s) = 1
2
( s 2 + 2 s + 5) Y ( s) = 3 + 2 s
2s + 3
2( s + 1)
1
Y ( s) = 2
=
+
s + 2 s + 5 ( s + 1) 2 + 2 2 ( s + 1) 2 + 2 2
Þ
y( t) = 2 e - t cos t u( t) +
1 -t
e sin t u( t)
2
58. (B) s 3Y ( s) + 4 s 2 Y ( s) + 3sY ( s) =
Y ( s) =
***********
10
( s + 2)
10
A
B
C
D
=
+
+
+
s( s + 1)( s + 2)( s + 3) s ( s + 1) ( s + 2) s + 3
5
A = sY ( s) s = 0 = , B = ( s + 1) Y ( s) s = -1 = -5,
3
C = ( s + 2) Y ( s) s = -2 = 5, D = ( s + 3) Y ( s) s = 0 =
Þ
5
é5
ù
y( t) = ê - 5 e - t + 5 e -2 t + e -3t ú u( t)
3
3
ë
û
59. (D) For a causal system
H ( s) = 2 +
Þ
h( t) = 0
for t < 0
1
1
+
s + 1 s -1
h( t) = 2d( t) + ( e- t + e t ) u( t)
60. (D) H ( s) =
Page
278
5
3
2
3
, System is stable
s + 1 ( s + 1) 2
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GATE EC BY RK Kanodia
CHAPTER
5.4
THE Z-TRANSFORM
Statement forQ.1-12:
Determine the z-transform and choose correct
option.
1. x[ n] = d[ n - k] , k > 0
k
(A) z ,
k
(C) z ,
z >0
z ¹0
(C)
(B) z
-k
,
z >0
(D) z
-k
,
z ¹0
(A) z - k , z ¹ 0
(B) z k , z ¹ 0
(C) z - k , all z
(D) z k , all z
(C)
z
, |z|< 1
1 - z -1
(B)
(D)
1
, |z|< 1
1 - z -1
z
, |z|> 1
1 - z -1
(A)
z - 0.25
, z > 0.25
z 4 ( z - 0.25)
(B)
z - 0.25
, z >0
z 4 ( z - 0.25)
(C)
z 5 - 0.25 5
, z < 0.25
z 3( z - 0.25)
(D)
z 5 - 0.25 5
, all z
z 4 ( z - 0.25)
5
(B)
-5 z
2
3
,
<|z|<
(2 z - 3)( 3z - 2)
3
2
(C)
5z
2
2
,
<|z |<
(2 z - 3) ( 3z - 2)
3
3
(D)
5z
3
2
, - <z <(2 z - 3)( 3z - 2)
2
3
(A)
4z
1
, |z|>
4z - 1
4
(B)
4z
1
, |z|<
4z - 1
4
(C)
1
1
, |z|>
1 - 4z
4
(D)
1
1
, |z|<
1 - 4z
4
n
æ1ö
æ1ö
8. x[ n] = ç ÷ u[ n] + ç ÷ u[ -n - 1]
è2 ø
è4ø
(A)
(B)
4
æ1ö
5. x[ n] = ç ÷ u[ -n]
è4ø
3
, |z|< 3
3-z
-5 z
3
2
, - <z <(2 z - 3)( 3z - 2)
2
3
n
5
(D)
(A)
n
5
z
, |z|< 3
3-z
|n|
æ1ö
4. x[ n] = ç ÷ ( u[ n] - u[ n - 5 ])
è4ø
5
3
, |z|> 3
3-z
(B)
æ2 ö
7. x[ n] = ç ÷
è 3ø
2. x[ n] = d[ n + k] , k > 0
3. x[ n] = u[ n]
1
(A)
, |z|> 1
1 - z -1
6. x[ n] = 3n u[ -n - 1]
z
(A)
, |z|> 3
3-z
(C)
1
1
,
1 -1
1 -1
1- z
1- z
2
4
1
1
<|z|<
4
2
1
1
+
,
1 -1
1 -1
1- z
1- z
2
4
1
1
<|z|<
4
2
1
1
1
, |z|>
1 -1
1 -1
2
1- z
1- z
2
4
(D) None of the above
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UNIT 5
33. X ( z) =
Signal & System
37. Consider three different signal
1
1
, |z|<
-2
1 - 4z
4
n
é
æ1ö ù
x1 [ n] = ê2 n - ç ÷ ú u[ n]
è 2 ø úû
êë
¥
(A) - å 2 2 ( k + 1 ) d[ -n - 2( k + 1)]
k=0
¥
(B) - å 2 2 ( k + 1 ) d[ -n + 2( k + 1)]
x2 [ n] = -2 n u[ -n - 1] +
1
u[ -n - 1]
2n
x3[ n] = - 2 n u[ -n - 1] -
1
u[ n]
2n
k=0
¥
(C) - å 2 2 ( k + 1 ) d[ n + 2( k + 1)]
k=0
¥
(D) - å 2 2 ( k + 1 ) d[ n - 2( k + 1)]
Fig. P.5.4.37 shows the three different region.
k=0
Choose the correct option for the ROC of signal
34. X ( z) = ln (1 + z -1 ) , |z|> 0
( -1)
(A)
k
k -1
Im
( -1)
(B)
k
d[ n - 1]
( -1) k
(C)
d[ n - 1]
k
k -1
R1
z - plane
d[ n + 1]
R2
R3 2
( -1) k
(D)
d[ n + 1]
k
Re
1
2
35. If z-transform is given by
X ( z) = cos ( z -3), |z|> 0,
Fig. P5.4.37
R1
R2
R3
(A)
x1 [ n]
x2 [ n ]
x3[ n]
(B)
x2 [ n ]
x3[ n]
x1 [ n]
(C)
x1 [ n]
x3[ n]
x2 [ n ]
(D)
x3[ n]
x2 [ n ]
x1 [ n]
The value of x[12 ] is
(A) -
1
24
(B)
(C) -
1
6
(D)
1
24
1
6
36. X ( z) of a system is specified by a pole zero pattern
in fig. P.5.4.36.
38. Given
7 -1
z
6
X ( z) =
1 -1 öæ
1 -1 ö
æ
ç 1 - z ÷ç 1 + z ÷
2
3
è
øè
ø
Im
1+
z - plane
1
3
Re
2
For three different ROC consider there different
Fig. P.5.4.36
solution of signal x[ n] :
Consider three different solution of x[ n]
(a) |z|>
n
é
æ1ö ù
x1 [ n] = ê2 n - ç ÷ ú u[ n]
è 3 ø úû
êë
n
é 1
1
æ -1 ö ù
, x[ n] = ê n -1 - ç
÷ ú u[ n]
2
è 3 ø úû
êë2
(b) |z|<
é - 1 æ - 1 ön ù
1
, x[ n] = ê n -1 + ç
÷ ú u[ -n + 1]
3
è 3 ø úû
êë2
1
u[ n]
3n
1
x3[ n] = - 2 n u[ n - 1] + n u[ -n - 1]
3
Correct solution is
x2 [ n] = - 2 n u[ n - 1] -
Page
282
n
(c)
1
1
1
æ -1 ö
<|z |< , x[ n] = - n -1 u[ -n - 1] - ç
÷ u[ n]
3
2
2
è 3 ø
Correct solutions are
(A) x1 [ n]
(B) x2 [ n]
(A) (a) and (b)
(B) (a) and (c)
(C) x3[ n]
(D) All three
(C) (b) and (c)
(D) (a), (b), (c)
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GATE EC BY RK Kanodia
The z-Transform
39. X ( z) has poles at z = 1 2 and z = -1. If
x [1] = 1
Chap 5.4
44. The transfer function of a causal system is given as
x [ -1] = 1, and the ROC includes the point z = 3 4. The
H ( z) =
time signal x[ n] is
(A)
1
u[ n] - ( -1) n u[ -n - 1]
2 n -1
The impulse response is
(A) ( 3n + ( -1) n 2 n + 1 ) u[ n]
1
(B) n u[ n] - ( -1) n u[ -n - 1]
2
(C)
(D)
1
2 n -1
(B) ( 3n + 1 + 2 ( -2) n ) u[ n]
(C) ( 3n -1 + ( -1) n 2 n + 1 ) u[ n]
u[ n] + u[ -n + 1]
(D) ( 3n -1 - ( -2) n + 1 ) u[ n]
1
u[ n] + u[ -n + 1]
2n
45. A causal system has input
40. x[ n] is right-sided, X ( z) has a signal pole, and
x[0 ] = 2, x[2 ] = 1 2. x[ n] is
u[ -n]
(A) n -1
2
(C)
u[ n]
(B) n -1
2
u[ -n]
2n +1
as
3 -1
z
2
(1 - 2 z -1 ) (1 +
1
1
d[ n - 1] - d[ n - 2 ] and output
4
8
y[ n] = d[ n] -
3
d[ n - 1] .
4
(A)
n
n
1 é æ -1 ö
æ1ö ù
5
2
÷
ç ÷ ú u[ n]
ê ç
3 êë è 2 ø
è 4 ø úû
(B)
1 é æ1ö
æ -1 ö
÷
ê5ç ÷ + 2ç
3 êë è 2 ø
è 4 ø
(C)
1
3
(D)
1 é æ1ö
æ1ö
ê5ç ÷ + 2ç ÷
3 êë è 2 ø
è4ø
41. The z-transform function of a stable system is given
2-
x[ n] = d[ n] +
The impulse response of this system is
u[ -n]
2n +1
(D)
H ( z) =
1 -1
z )
2
n
n
æ1ö
(A) 2 n u[ -n + 1] - ç ÷ u[ n]
è2 ø
ù
ú u[ n]
úû
n
ù
ú u[ n]
úû
46. A causal system has input x[ n] = ( -3) n u[ n] and
n
æ -1 ö
(B) 2 n u[ -n - 1] + ç
÷ u[ n]
è 2 ø
output
n
é
æ1ö ù
y[ n] = ê4(2) n - ç ÷ ú u[ n].
è 2 ø úû
êë
n
æ -1 ö
(C) -2 n u[ -n - 1] + ç
÷ u[ n]
è 2 ø
The impulse response of this system is
n
æ1ö
(D) 2 n u[ n] - ç ÷ u[ n]
è2 ø
Let
n
n
é æ 1 ön
æ -1 ö ù
5
2
ç
÷
ç
÷
ê
ú u[ n]
è 4 ø úû
êë è 2 ø
n
The impuse response h[ n] is
42.
5z2
z2 - z - 6
x[ n] = d[ n - 2 ] + d[ n + 2 ].
The
unilateral
z-transform is
(A) z -2
(B) z 2
(C) -2 z -2
(D) 2 z 2
n
é æ 1 ön
æ1ö ù
(A) ê7ç ÷ - 10ç ÷ ú u[ n]
è 2 ø úû
êë è 2 ø
n
é
æ1ö ù
(B) ê7(2 n ) - 10ç ÷ ú u[ n]
è 2 ø úû
êë
é æ 1 ö2
ù
(C) ê10ç ÷ - 7(2) n ú u[ n]
êë è 2 ø
úû
n
é
æ1ö ù
(D) ê10 (2 n ) - 7ç ÷ ú u[ n]
è 2 ø úû
êë
47. A system has impulse response
h[ n] =
43. The unilateral z-transform of signal x[ n] = u[ n + 4 ]
is
(A) 1 + z + z 2 + 3z + z 4
(B)
1
1-z
(C) 1 + z -1 + z -2 + z -3 + z -4
(D)
1
1 - z -1
1
u[ n]
2n
The output y[ n] to the input x[ n] is given by
y[ n] = 2 d[ n - 4 ]. The input x[ n] is
(A) 2 d[ -n - 4 ] - d[ -n - 5 ]
(B) 2 d[ n + 4 ] - d[ n + 5 ]
(C) 2 d[ -n + 4 ] - d[ -n + 5 ]
(D) 2 d[ n - 4 ] - d[ n - 5 ]
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UNIT 5
48. A system is described by the difference equation
y[ n] -
1
2
n -2
53. The z-transform of a signal x[ n] is given by
1
y[ n - 1] = 2 x[ n - 1]
2
X ( z) =
The impulse response of the system is
-1
1
(B) n - 2 u[ n + 1]
(A) n - 2 u[ n - 1]
2
2
(C)
Signal & System
u[ n - 2 ]
(D)
If X ( z) converges on the unit circle, x[ n] is
1
(A) -
-1
u[ n - 2 ]
2n -2
(B)
49. A system is described by the difference equation
(C)
y[ n] = x[ n] - x[ n - 2 ] + x[ n - 4 ] - x[ n - 6 ]
3n -1 8
1
3n -1 8
1
3n -1 8
The impulse response of system is
(D) -
(A) d[ n] - 2 d[ n + 2 ] + 4 d[ n + 4 ] - 6 d[ n + 6 ]
(B) d[ n] + 2 d[ n - 2 ] - 4 d[ n - 4 ] + 6 d[ n - 6 ]
u[ n] -
3n + 3
u[ -n - 1]
8
u[ n] -
3n + 3
u[ -n]
8
1
n -1
3
3n + 3
u[ -n - 1]
8
u[ n] -
8
u[ n] -
3n + 3
u[ -n]
8
54. The transfer function of a system is given as
(C) d[ n] - d[ n - 2 ] + d[ n - 4 ] - d[ n - 6 ]
(D) d[ n] - d[ n + 2 ] + d[ n + 4 ] - d[ n + 6 ]
H ( z) =
50. The impulse response of a system is given by
h[ n] =
3
u[ n - 1].
4n
4 z -1
1 -1 ö
æ
ç1 - z ÷
4
è
ø
2
, |z|>
1
4
The h[ n] is
The difference equation representation for this
system is
(A) 4 y[ n] - y[ n - 1] = 3 x[ n - 1]
(A) Stable
(B) Causal
(C) Stable and Causal
(D) None of the above
55. The transfer function of a system is given as
(B) 4 y[ n] - y[ n + 1] = 3 x[ n + 1]
1ö
æ
2ç z + ÷
2ø
è
.
H ( z) =
1 öæ
1ö
æ
ç z - ÷ç z - ÷
2 øè
3ø
è
(C) 4 y[ n] + y[ n - 1] = - 3 x[ n - 1]
(D) 4 y[ n] + y[ n + 1] = 3 x[ n + 1]
Consider the two statements
51. The impulse response of a system is given by
Statement(i) : System is causal and stable.
h[ n] = d[ n] - d[ n - 5 ]
The difference equation representation for this
Statement(ii) : Inverse system is causal and stable.
system is
The correct option is
(A) y[ n] = x[ n] - x[ n - 5 ]
(B) y[ n] = x[ n] - x[ n + 5 ]
(A) (i) is true
(C) y[ n] = x[ n] + 5 x[ n - 5 ]
(D) y[ n] = x[ n] - 5 x[ n + 5 ]
(B) (ii) is true
52. The transfer function of a system is given by
H ( z) =
z( 3z - 2)
1
z2 - z 4
(C) Both (i) and (ii) are true
(D) Both are false
56. The impulse response of a system is given by
n
(A) Causal and Stable
(B) Causal, Stable and minimum phase
(C) Minimum phase
(D) None of the above
n
æ -1 ö
æ -1 ö
h[ n] = 10ç
÷ u[ n] - 9ç
÷ u[ n]
è 2 ø
è 4 ø
The system is
Page
284
3
10 -1
1z + z -2
3
For this system two statement are
Statement (i): System is causal and stable
Statement (ii): Inverse system is causal and stable.
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The z-Transform
Chap 5.4
62. The impulse response of the system shown in fig.
The correct option is
(A) (i) is true
(B) (ii) is true
(C) Both are true
(D) Both are false
P5.4.62 is
X(z)
Y(z)
z-1
+
z-1
57. The system
z-1
y[ n] = cy[ n - 1] - 0.12 y[ n - 2 ] + x[ n - 1] + x[ n - 2 ]
Fig. P5.4.62
is stable if
(A) c < 112
.
(B) c > 112
.
(C) |c|< 112
.
(D) |c|> 112
.
æn
ö
çç - 2 ÷÷
ø
(A) 2 è 2
(B)
58. Consider the following three systems
y2 [ n] = x[ n] - 0.1 x[ n - 1]
(C) 2 è 2
(D)
y3[ n] = 0.5 y[ n - 1] + 0.4 x[ n] - 0.3 x[ n - 1]
(B) y2 [ n] and y3[ n]
(C) y3[ n] and y1 [ n]
(D) all
1
d[ n]
2
2n
1
[1 + ( -1) n ] u [ n] - d [ n]
2
2
H ( z) =
z
z2 + z + 1
is shown in fig. P5.4.63. This system diagram is a
X(z)
59. The z-transform of a causal system is given as
X ( z) =
(1 + ( -1) n ) u[ n] -
63. The system diagram for the transfer function
The equivalent system are
(A) y1 [ n] and y2 [ n]
1
d[ n]
2
2n
1
(1 + ( -1) n ) u[ n] + d[ n]
2
2
æn
ö
çç - 2 ÷÷
ø
y1 [ n] = 0.2 y[ n - 1] + x[ n] - 0.3 x[ n - 1] + 0.02 x[ n - 2 ]
(1 + ( -1) n ) u[ n] +
. z -1
2 - 15
-1
1 - 15
. z + 0.5 z -2
Y(z)
+
+
+
The x[0 ] is
z-1
z-1
(A) -15
.
(B) 2
(C) 1.5
(D) 0
Fig. P5.4.63
(A) Correct solution
60. The z-transform of a anti causal system is
X ( z) =
12 - 21z
3 - 7 z + 12 z 2
The value of x[0 ] is
7
(A) 4
(B) Not correct solution
(C) Correct and unique solution
(D) Correct but not unique solution
(B) 0
(C) 4
*****************
(D) Does not exist
61. Given the z-transforms
X ( z) =
z( 8 z - 7)
4z2 - 7z + 3
The limit of x[ ¥ ] is
(A) 1
(B) 2
(C) ¥
(D) 0
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The z-Transform
z 2 - 3z
1 - 3z -1
=
3
3
z 2 + z -1 1 + z -1 - z -2
2
2
2
1
1
, ROC : <|z|< 2
=
1 + 2 z -1 1 - 1 z -1
2
2
1
Þ x[ n] = -2(2) n u[ -n - 1] - n u[ n]
2
Chap 5.4
ìæ 1 ö 2
ï
, n even and n ³ 0
= í çè 4 ÷ø
ï
n odd
î 0 ,
n
24. (A) X ( z) =
ì2-n ,
=í
î 0,
n even and n ³ 0
n odd
¥
¥
k=0
k=0
33. (C) X ( z) = -4 z 2 å (2 z) 2 k = - å 2 2 ( k + 1 ) z 2 ( k + 1 )
25. (A) x[ n] is right sided
1
49
47
z - z -1
4
32
32
=
+
X ( z) =
1 - 16 z -1 1 + 4 z -1 1 - 4 z -1
47 n ù
é 49
Þ x[ n] = ê
( -4) n +
4 ú u[ n]
32
ë 32
û
Þ
¥
x[ n] = - å 2 2 ( k + 1 ) d[ n + 2 ( k + 1)]
k=0
34. (A) ln (1 + a) =
( -1) k -1
( a) k
k
k =1
¥
å
( -1) k -1
( z -1 ) k
k
k =1
¥
X ( z) = å
26. (C) x[ n] is right sided
( -1) k -1
d[ n - 1]
k
k =1
¥
x [ n] = å
1
-1 ö 2
æ
X ( z) = ç 2 +
+
÷z
-1
1+ z
1 - z -1 ø
è
Þ
Þ
35. (B) cos a = å
x[ n] = 2 d[ n + 2 ] + (( -1) n - 1) u[ n + 2 ]
( -1) k 2 k
a
k = 0 (2 k) !
¥
27. (A) d[ n] + 2 d[ n - 6 ] + 4 d[ n - 8 ]
( -1) k -3k 2 k
(z )
k = 0 (2 k) !
¥
X ( z) = å
10
1
å k d[ n - k]
28. (B) x[ n] is right sided, x[ n] =
Þ
k=5
x [ n] =
( -1) k
¥
å (2 k) ! d [ n - 6 k]
k=0
n = 12
29. (D) x[ n] is right sided signal
X ( z) = 1 + 3z
Þ
-1
+ 3z
-2
+z
x[ n] = d[ n] + 3d[ n - 1] + 3d[ n - 2 ] + d[ n - 3]
12 - 6 k = 0, k = 2,
36. (D) All gives the same z transform with different
30. (A)
x[ n] = d[ n + 6 ] + d[ n + 2 ] + 3d[ n] + 2 d [ n - 3] + d[ n - 4 ]
31. (B) X ( z) = 1 + z +
z2 z3
+
+ .........
2 ! 3!
1
1
1
.........
+1 + +
+
2
z 2!z
3! z3
d[ n + 2 ] d[ n + 3]
x[ n] = d[ n] + d[ n + 1] +
+
......
2!
3!
d[ n - 2 ] d[ n - 3]
+d[ n] + d[ n - 1] +
+
.........
2!
3!
1
x [ n] = d [ n] +
n!
32. (A) X ( z) = 1 +
Þ
Þ
( -1) 2
1
x[12 ] =
=
4!
24
-3
¥
k
-2
-2
æz
z
+ çç
4
è 4
æ1ö
x [ n] = å ç ÷ d[ n - 2 k]
k=0 è 4 ø
2
ö
æ1
ö
÷÷ = å ç z -2 ÷
k=0 è 4
ø
ø
¥
k
ROC. So all are the solution.
37. (C) x1 [ n] is right-sided signal
z1 > 2 , z1 >
1
gives z1 > 2
2
x2 [ n] is left-sided signal
1
1
z 2 < 2, z 2 < gives z 2 <
2
2
x3[ n] is double sided signal
1
1
z 3 > and z 3 < 2 gives
< z3 < 2
2
2
38. (B) X ( z) =
2
-1
,
+
1 -1
1
1- z
1 + z -1
2
z
n
2
æ -1 ö
u[ n] - ç
÷ u[ n]
2n
è 3 ø
|z|>
1
(Right-sided) Þ
2
x[ n] =
|z|<
1
(Left-sided) Þ
3
é -2 æ -1 ön ù
x[ n] = ê n + ç
÷ ú u[ -n - 1]
è 3 ø úû
êë 2
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UNIT 5
n
1
1
2
æ -1 ö
<|z|< (Two-sided) x[ n] = - n u[ -n - 1] - ç
÷ u[ n]
3
2
2
è 3 ø
So (b) is wrong.
3
39. (A) Since the ROC includes the z = , ROC is
4
1
<|z|< 1,
2
A
B
X ( z) =
+
1 -1 1 + z -1
1- z
2
A
Þ x[ n] = n u[ n] × B ( -1) n u[ -n - 1]
2
A
1=
Þ A =2 ,
2
Signal & System
-2
5
Y ( z)
3
3
,
H ( z) =
=
+
X ( z ) 1 - 1 z -1 1 + 1 z -1
4
2
Þ
h[ n] =
n
n
1 é æ -1 ö
æ1ö ù
5
2
ç
÷
ç
÷
ê
ú u[ n]
3 êë è 2 ø
è 4 ø úû
Y ( z) =
H ( z) =
4
1
3
=
-1
1
1
1 - 2z
æ
ö
1 - z -1
(1 - 2 z -1 )ç 1 - z -1 ÷
2
2
è
ø
10
Y ( z)
-7
=
+
X ( z ) 1 - 2 z -1 1 - 1 z -1
2
x[ -1] = 1 = ( -1) B( -1) Þ B = 1
1
Þ x[ n] = n -1 u[ n] - ( -1) n u[ -n - 1]
2
Þ
40. (B) x[ n] = Cpn u[ n] , x[0 ] = 2 = C
47. (D) H ( z) =
x[2 ] =
1
= 2 p2
2
Þ
p=
X ( z) =
n
41. (B) H ( z) =
Þ
1
1
+
-1
1
1 - 2z
1 + z -1
2
n
æ -1 ö
h[ n] = (2) u[ -n - 1] + ç
÷ u [ n]
è 2 ø
¥
-n
Y ( z)
= 2 z -4 - z -5
H ( z)
x[ n] = 2[ d - 4 ] - d[ n - 5 ]
H ( z) =
Þ
n
å x[ n]z
= d[ n - 2 ]z - n = z -2
Y ( z)
2 z -1
=
z -1
X ( z)
12
æ1ö
h[ n] = 2ç ÷
è2 ø
49. (C) H ( z) =
n =0
+
43. (D) X ( z) =
¥
å x[ n]z
n =0
-n
=
¥
åz
n =0
Þ
-n
1
=
1 - z -1
3
2
44. (B) H ( z) =
+
-1
1 - 3z
1 + 2 z -1
h[ n] is causal so ROC is |z|> 3,
Þ
h[ n] = [ 3n + 1 + 2 ( -2) n ]u[ n]
45. (A) X ( z) = 1 +
1
, Y ( z ) = 2 z -4
1 -1
1- z
2
é
z -1 ù
-1
48. (A) Y ( z) ê1 ú = 2 z X ( z)
2
ë
û
h[ n] is stable, so ROC includes |z|= 1
1
ROC : <|z|< 2 ,
2
42. (A) X + ( z) =
n
é
æ1ö ù
h[ n] = ê10(2) n - 7ç ÷ ú u( n)
è 2 ø úû
êë
1
,
2
æ1ö
x[ n] = 2ç ÷ u( n)
è2 ø
Page
288
1
1 + 3z -1
46. (D) X ( z) =
z -1 z -2
3z -1
, Y ( z) = 1 4
8
4
n -1
u[ n - 1]
Y ( z)
= (1 - z -2 + z -4 - z -6 )
X ( z)
h[ n] = d[ n] - d[ n - 2 ] + d[ n - 4 ] - d[ n - 6 ]
50. (A) h[ n] =
3æ1ö
ç ÷
4è4ø
n -1
u[ n - 1]
3 -1
z
Y ( z)
4
H ( z) =
=
X ( z ) 1 - 1 z -1
4
1
3
Y ( z) - z -1 Y ( z) = z -1 X ( z)
4
4
1
3
Þ
y[ n] - y[ n - 1] = x[ n - 1]
4
4
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The z-Transform
51. (A) H ( z) =
Þ
Y ( z)
= 1 - z -5
X ( z)
So y1 and y2 are equivalent.
59. (B) Causal signal x[0 ] = lim X ( z) = 2
y[ n] = x[ n] - x[ n - 5 ]
52. (D) Zero at : z = 0,
Chap 5.4
z ®¥
1± 2
2
, poles at z =
3
2
60. (C) Anti causal signal, x[0 ] = lim X ( z) = 4
z ®¥
(i) Not all poles are inside |z|= 1, the system is not
causal and stable.
3
61. (A) The function has poles at z = 1, . Thus final
4
(ii) Not are poles and zero are inside |z|= 1, the system
value theorem applies.
is not minimum phase.
7
)
4
=1
lim x( n) = lim( z - 1) X ( z) = ( z - 1)
n ®¥
z ®1
3ö
æ
( z - 1) ç z - ÷
4ø
è
z(2 z -
3
27
8
8
53. (A) X ( z) =
+
1 -1 1 - 3z -1
1- z
3
-
62. (C) [2 Y ( z) + X ( z)] z -2 = Y ( z)
Since X ( z) converges on |z|= 1. So ROC must include
this circle.
1
ROC : <|z|< 3,
3
H ( z) =
Þ
3n + 3
x[ n] = - n -1 u[ n] u[ -n - 1]
3 8
8
1
=-
z -2
1 - 2 z -2
1
1
1
4
4
h[ n] = - +
+
2 1 - 2 z -1 1 + 2 z -1
1
1
d[ n] + {( 2 ) n + ( - 2 ) n } u[ n]
2
4
n
æ1ö
54. (C) h[ n] = 16 nç ÷ u[ n]. So system is both stable and
è4ø
63. (D) Y ( z) = X ( z) z -1 - { Y ( z) z -1 + Y ( z) z -2 }
causal. ROC includes z = 1.
Y ( z)
z -1
z
=
= 2
-1
-2
X ( z) 1 + z + z
z + z +1
1 1
,
2 3
1
Pole of inverse system at : z = 2
So this is a solution but not unique. Many other correct
55. (C) Pole of system at : z = -
diagrams can be drawn.
For this system and inverse system all poles are inside
***********
|z|= 1. So both system are both causal and stable.
56. (A) H ( z) =
=
10
9
1 -1
1
1+ z
1 + z -1
2
4
1 - 2 z -1
1 -1 öæ
1 -1 ö
æ
ç 1 + z ÷ç 1 + z ÷
2
4
è
øè
ø
Pole of this system are inside |z|= 1. So the system is
stable and causal.
For the inverse system not all pole are inside |z|= 1. So
inverse system is not stable and causal.
57. (C) |a2|= 0.12 < 1, a1 =|-c|< 1 + 0.12, |c|< 112
.
58. (A) Y1 ( z) = 1 - 0.1z -1 , Y2 ( z) = 1 - 0.1z -1
Y3( z) =
0.4 - 0.3z -1
1 - 0.5 z -1
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CHAPTER
5.6
THE DISCRETE-TIME FOURIER TRANSFORM
-n
Statement for Q.1–9:
Determine the discrete-time Fourier Transform for
the given signal and choose correct option.
|n|£ 2
ì 1,
1. x[ n] = í
î 0,
sin 5W
(A)
sin W
(C)
otherwise
(B)
sin 2.5W
sin W
sin 4W
sin W
(C)
(
3
4
e - jW
)
4
(B)
1 - 43 e - jW
3
4
e - jW
1+
3
4
)
(
3
4
e jW
)
(B)
2 e jW
2 - e - jW
(C)
e jW
2 - e jW
(D)
2 e jW
2 - e jW
(A) 2 e - j 2 W
(B) 2 e j 2 W
(C) 1
(D) None of the above
(A) pd(W) -
1
1 + e - jW
(B)
1
1 - e - jW
(C) pd(W) +
1
1 - e - jW
(D)
1
1 + e - jW
4
1 - 43 e jW
4
(D) None of the above
e jW
8. x[ n] = {-2, - 1,
3. x[ n] = u[ n - 2 ] - u[ n - 6 ]
(A) e 3 jW + e 3 jW + e 4 jW + e 5 jW
(C) e -2 jW + e -3 jW + e -4 jW + e -5 jW
(B)
(D)
e -2 jW(1 - e 3 jW)
1 - e jW
e
-2 jW
-3 jW
(1 - e
1 - e - jW
)
0, 1, 2}
­
(A) 2 j(2 sin 2W + sin W)
(B) 2(2 cos 2W - cos W)
(C) -2 j(2 sin 2W + sin W)
(D) -2(2 cos 2W - cos W)
æp ö
9. x[ n] = sin ç n ÷
è2 ø
4. x[ n] = a|n| , |a|< 1
(A) p( d[W - p 2 ] - d[W + p 2 ])
(A)
1-a
1 - 2 a sin W + a 2
(B)
(C)
1 - a2
1 - 2 ja sin W + a 2
(D) None of the above
2
Page
300
e jW
2 - e - jW
7. x[ n] = u[ n]
n
(
(A)
6. x[ n] = 2 d[ 4 - 2 n]
(D) None of the above
æ 3ö
2. x[ n] = ç ÷ u[ n - 4 ]
è4ø
(A)
æ1ö
5. x[ n] = ç ÷ u[ -n - 1]
è2 ø
1-a
1 - 2 a cos W + a 2
2
(B)
j
( d[W + p 2 ] - d[W - p 2 ])
2
(C) 2 p( d[W - p 2 ] - d[W + p 2 ])
(D) jp( d[W + p 2 ] - d[W - p 2 ])
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The Discrete-Time Fourier Transform
16. X ( e jW) = j 4 sin 4W - 1
Statement for Q.10–21:
Determine
the
signal
having
the
Fourier
(C) d[ n + 4 ] - d[ n - 4 ] - d[ n]
1
, |a|< 1
(1 - ae - jW) 2
(A) ( n - 1) a n u[ n]
(D) None of the above
(B) ( n + 1) a n u[ n]
n
(D) None of the above
(C) na u[ n]
(A) (d[ n + 2 ] + 2 d[ n] + d[ n - 2 ])
(C) -4(d[ n + 2 ] + d[ n] + d[ n - 2 ])
(d[ n + 2 ] + d[ n] + d[ n - 2 ])
(C)
ì 2 j, 0 <W £ p
12. X ( e jW) = í
î - 2 j, - p < W £ 0
4
æ pn ö
(A) sin 2 ç
÷
pn
è 2 ø
(C)
2-n
5
2
+ e - jW + 6
n +1
æ
ö
ç 1 + æç -2 ö÷ ÷u[ n]
ç
è 3 ø ÷ø
è
n +1
ö
÷u[ n]
÷
ø
n +1
æ
ö
ç ( -1) n + æç 2 ö÷ ÷u[ n]
ç
è 3 ø ÷ø
è
(D) None of the above
4
æ pn ö
(B)
sin 2 ç
÷
pn
è 2 ø
8
æ pn ö
sin 2 ç
÷
pn
è 2 ø
5
2-n
-e
- j2 W
æ
æ -2 ö
(B) 2 - n ç 1 - ç
÷
ç
è 3 ø
è
(B) 2(d[ n + 2 ] + 2 d[ n] + d[ n - 2 ])
1
2
17. X ( e jW) =
(A)
11. X ( e jW) = 8 cos 2 w
(D)
(A) 4 pd[ n + 4 ] - 4 pd[ n - 4 ] - 2 pd[ n]
(B) 2 d[ n + 4 ] - 2 d[ n - 4 ] - d[ n]
transform given in question.
10. X ( e jW) =
Chap 5.6
(D) -
18. X ( e jW) =
2+
- 18 e - j 2 W
e - jW
+ 14 e - jW + 1
1
4
(A) 2 - n + 1 [1 + ( -2) - n ]u[ n]
8
æ pn ö
sin 2 ç
÷
pn
è 2 ø
(B) 2 - n [1 + ( -2) - n ]u[ n]
(C) 2 - n + 1 [( -1) n + 2 - n ]u[ n]
ì
ï 1,
jW
13. X ( e ) = í
ï 0,
î
(A)
(B)
3p
p
£|W|<
4
4
p 3p
0 £|W|< ,
£|W|£ p
4
4
(D) 2 - n [( -1) n + 2 - n ]u[ n]
19. X ( e jW) =
2æ
æ 3pn ö
æ pn ö ö
ç sin ç
÷ - sin ç
÷÷
nè
è 4 ø
è 4 øø
(A) 2 n -1 [1 + ( -1) n ]u[ n]
1 æ
æ 3pn ö
æ pn ö ö
ç sin ç
÷ - sin ç
÷÷
pn è
è 4 øø
è 4 ø
(C) 21 - n [1 - ( -1) n ]u[ n]
(B) 21 - n [1 + ( -1) n ]u[ n]
(D) 2 n -1 [1 - ( -1) n ]u[ n]
2æ
æ 3pn ö
æ pn ö ö
(C) ç cos ç
÷ + cos ç
÷÷
nè
è 4 ø
è 4 øø
(D)
20. X ( e jW) =
1 æ
æ 3pn ö
æ pn ö ö
ç cos ç
÷ + cos ç
÷÷
pn è
è 4 ø
è 4 øø
14. X ( e jW) = e
-
æ2
(A) ç
ç9
è
jW
2
for
-p £ W £ p
(A) pd[ n - 1 2 ]
(C)
2 e - jW
1 - 14 e - j 2 W
( -1) n + 1
p(n - 12 )
(B) 2pd[ n - 1 2 ]
(D) None of the above
1 - 13 e - jW
1 - 14 e - jW - 18 e -2 jW
n
7 æ 1ö
æ1ö
ç ÷ + ç- ÷
2
9 è 4ø
è ø
n
ö
÷u[ n]
÷
ø
æ 2 æ 1 ön 7 æ 1 ön ö
(B) ç ç - ÷ + ç ÷ ÷u[ n]
ç9 è 2 ø
9 è 4 ø ÷ø
è
æ 2 æ 1 ön 7 æ 1 ön ö
(C) ç ç - ÷ - ç ÷ ÷u[ n]
ç9 è 2 ø
9 è 4 ø ÷ø
è
15. X ( e jW) = cos 2W + j sin 2W
(A) 2 pd[ n + 2 ]
(B) d[ n + 2 ]
(C) 0
(D) None of the above
æ2
(D) ç
ç9
è
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n
7 æ 1ö
æ1ö
ç ÷ - ç- ÷
2
9 è 4ø
è ø
n
ö
÷u[ n]
÷
ø
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301
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UNIT 5
n
n
æ 1ö
æ1ö
x[ n] = ç - ÷ u[ n] + ç ÷ u[ n] = 2 - n [( -1) n + 2 - n ]u[ n]
2
è
ø
è4ø
Signal & System
¥
å x[ n] , This condition is satisfied only
27. (D) X ( e j 0 ) =
n = -¥
if the samples of the signal add up to zero. This is true
2 e - jW
2
2
19. (C) X ( e ) =
=
1 - j2 W
1 - jW
1 - jW
1- e
1- e
1+ e
4
2
2
for signal (b) and (h).
jW
n
28. (A) X ( e j 0 ) =
n
æ1ö
æ 1ö
x[ n] = 2ç ÷ u[ n] - 2ç - ÷ u[ n]
è2 ø
è 2ø
1
= n -1 [1 - ( -1) n ]u[ n] = 21 - n [1 - ( -1) n ]u[ n]
2
n = -¥
29. (A) y[ n] = x[ n + 2 ]
Y ( e ) is real and even.
Y ( e jW) = e j 2 W X ( e jW)
Thus arg{X ( e jW)} = -2W
30. (C)
=
p
ò X (e
jW
) dW = 2 px[0 ] = 4 p
-p
¥
å ( -1)
31. (A) X ( e jp) =
2æ1ö
7æ 1ö
x[ n] = ç ÷ u[ n] + ç - ÷ u[ n]
9 è2 ø
9è 4ø
e j2 W
X ( e jW) = e - j 2 WY ( e jW),
Þ
Since Y ( e jW) is real. This imply arg{ Y ( e jW)} = 0
n
21. (C) X ( e jW) =
is an even signal. Therefore
jW
1 - jW
1- e
jW
3
20. (A) X ( e ) =
1
1
1 - e - jW - e -2 jW
4
8
2
7
9
9
=
+
1 - jW
1 - jW
1- e
1+ e
2
4
n
¥
å x[ n] = 6
n
x[ n] = 2
n = -¥
DTFT
32. (C) Ev{x[ n]} ¬ ¾¾
® Re{X ( e jW)}
( b - a) e jW
- ( a + b) e jW + ab
Ev{x[ n] =
- jW
( b - a) e
1
-1
=
+
1 - ( a + b) e - jW + abe- j 2 W 1 - be - jW 1 - ae - jW
( x[ n] + x[ -n])
2
1ü
1
1
ì 1
, 1, 0, 0, 1, 2, 1, 0, 0, 1, , 0, - ý
= í - , 0,
2
2
2
2
î
þ
­
x[ n] = bn u[ n] + a n u[ -n - 1] .
22. (D) The signal must be read and odd. Only signal ( h)
33. (D)
23. (A) The signal must be real and even. Only signal (c)
and (e) are real and even.
24. (A) Y ( e jW) = e jaW X ( e jW),
y[ n] = x[ n + a ]
real.). Therefore x [ n + a ] is even and x [ n] has to be
symmetric about a.This is true for signal (a), (c), (e), (f)
and (g).
ò X (e
2
n = -¥
DTFT
¬ ¾¾
®
j
) dW = 2 px[0 ] ,
35. (A) Y ( e jW) = e - j 4 W X ( e jW)
|n - 4|
æ 3ö
y[ n] = x[ n - 4 ] = ( n - 4) ç ÷
è4ø
imaginary. Thus y[ n] = 0.
x[0 ] = 0 is for signal (c), (f), (g) and (h).
37. (D) X 2 ( e jW) = X ( e j 2 W)
is always periodic with period 2p.
X ( e j 2 W)
Therefore all signals satisfy the condition.
DTFT
¬ ¾¾
®
ì x[ n] , n even
x 2 [ n] = í
otherwise
î 0,
|n|
ì
2æ 3 ö
jn
ï
ç ÷ , n even
y[ n] = í
è4ø
ï0 ,
otherwise
î
Page
306
d X ( e jW)
dW
36. (C) Since x[ n] is real and odd, X ( e jW) is purely
-p
26. (D) X ( e jW)
34. (C) nx[ n]
¥
= 2 p å | x[ n]| = 28 p
¥
2
½dX ( e jW)½
½
|n|2 x[ n] = 316 p
ò- p½ dW ½½ = 2 pnå
= -¥
If Y ( e ) is real, then y[ n] is real and even (if x[ n] is
25. (D)
2
p
jW
jW
jW
-p
is real and odd.
p
p
ò | X ( e )|
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The Discrete-Time Fourier Transform
38. (B) Y ( e jW) = X ( e jW ) * X ( e j ( W- p 2 ) )
y[ n] = 2 px[ n] x1 [ n], x1 [ n] = e
y[ n] = 2 pn2 e jpn
Þ
39. (C) Y ( e jW) =
æ 3ö
ç ÷
è4ø
2
jpn 2
This is possible only if b = -a.
x[ n] ,
44. (A) For x[ n] = d[ n], X ( e jW) = 1,
2|n|
h[ n] =
d
X ( e jW)
dW
=
|n|
æ 3ö
y[ n] = - jnx[ n] = - jn2 ç ÷
è4ø
Þ
41. (C) For a real signal x[ n]
)
3=
2
dW =
å | x[ n]|
2
n = -¥
x[0 ] = ± 1,
å | x[ n]|
2
= ( x [0 ]) + 2
But
x[0 ] = 0,
Hence
n
n
n =0
=
-p
n
æ1ö
h2 [ n] = ç ÷ u[ n]
è 3ø
¬ ¾¾®
DTFT
1
2 - jW
1- e
3
d
j
dW
æ
ç
1
ç
çç 1 - 2 e - jW
è
3
2 - jW
ö
e
÷
÷= 3
÷÷ 1 - 2 e - jW
ø
3
2 - jW
e
Y ( e jW )
3
47. (B) H ( e ) =
=
X ( e jW) 1 - 2 e - jW
3
2
2
æ
ö
Þ ç 1 - e - jW ÷ Y ( e jW) = e- jW X ( e jW)
3
3
è
ø
jW
x[0 ] = 1
1
æ1ö
ç ÷
1 - jW è 4 ø
1- e
4
DTFT
¬ ¾¾
®
n
1 - jW
æ
ö
n
e
÷
d ç
1
æ1ö
DTFT
4
ç
÷=
nç ÷ u[ n] ¬ ¾¾® j
2
dW ç 1 - 1 e - jW ÷ æ
è2 ø
ç
÷ ç 1 - 1 e - jW ö÷
è
4
ø è
4
ø
1
e jWn dW
2
æ1ö
42. (C) ç ÷ u[ n]
è4ø
¥
- jW
2 - jW
e
2 e - jW
H( e ) = 3
=
- jW
2
1 - e - jW 3 - 2 e
3
x[ n] = d[ n] + d[ n + 1] - d[ n + 2 ]
å næçè 2 ö÷ø
p
òe
jW
n =- ¥
-¥
-1
¥
DTFT
¬ ¾¾
®
n
æ2 ö
nç ÷ u[ n]
è 3ø
Using Parseval’s relation
jW
1
2p
Y ( e jW)
,
X ( e jW)
n
x[ n] = 2od{ x[ n] } = d[ n + 1] - d[ n + 2 ] For n < 0
ò | X ( e )|
1
,
1 - jW
1- e
3
æ2 ö
ç ÷ u[ n]
è 3ø
Since x[ n] = 0 for n > 0,
¥
-p
46. (D) H ( e jW) =
Therefore od{x[ n]} = F -1 { jIm{X ( e jW)}}
1
= ( d[ n + 1] - d[ n - 1] - d[ n + 2 ] + d[ n - 2 ])
2
x[ n] - x[ -n]
Od{ x[ n]} =
2
1
2p
) e jWn dW =
1
sin p( n - 1)
e jW ( n -1 ) dW =
2 p -òp
p( n - 1)
H 2 ( e jW) =
jIm{X ( e jW)}= j sin W - j sin 2W,
1 jW
=
e - e - j W - e 2 jW + e -2 jW
2
(
jW
p
jIm{X ( e jW)}
DTFT
¬ ¾¾
®
p
ò Y (e
-12 + 5 e - jW
1
-2
=
+
1 - jW
1
12 - 7 e - jW + e - j 2 W
1- e
1 - e - jW
3
4
y[ n] = x[ n] + x[ -n] = 0
od{x[ n]}
1
2p
dX ( e jW)
=0
dW
45. (B) H ( e jW) = H1 ( e jW) + H 2 ( e jW)
40. (B) Y ( e jW) = X ( e jW ) + X ( e - jW)
Þ
Chap 5.6
¥
å x[ n] = X ( e
j0
)=
n = -¥
y[ n] -
Þ
3 y[ n] - 2 y[ n - 1] = 2 x[ n - 1] .
4
9
|
2
2
y[ n - 1] = x[ n - 1]
3
3
Þ
*********
|
43. (A) For all pass system H ( e jW) = 1 for all W
H ( e jW) =
- jW
b+ e
, b + e - jW = 1 - a e - jW
1 - ae- jW
|
| |
|
1 + b2 + 2 b cos W = 1 + a 2 - 2 a cos W
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CHAPTER
5.7
THE CONTINUOUS-TIME FOURIER SERIES
x(t)
Statement for Q.1-5:
A
Determine the Fourier series coefficient for given
periodic signal x( t).
-2p
3
1. x( t) as shown in fig. P5.7.1
Fig. P5.7.3
10
-5
0
5
10
(C) - j
t
Fig. P5.7.1
(A) 1
æp ö
(B) cosç k ÷
è2 ø
æp ö
(C) sinç k ÷
è2 ø
(D) 2
A æç - j ççè
e
2 pk ç
è
æ 4 pk ö
÷÷
3 ø
(B) j
ö
- 1÷
÷
ø
(D)
A æç - j ççè
e
2 pk ç
è
æ 4 pk ö
÷÷
3 ø
- A æç - j ççè
e
2 pk ç
è
æ 4 pk ö
÷÷
3 ø
ö
- 1÷
÷
ø
ö
- 1÷
÷
ø
4. x( t) as shown in fig. P5.7.4
x(t)
A
1
-1
-A
2. x( t) as shown in fig. P5.7.2
Fig. P5.7.4
x(t)
A
-T
0
T
4
T
2
T
t
A
(A)
(1 - ( - 1 ) k )
kp
A
(C)
(1 - ( - 1 ) k )
jkp
A
(1 + ( - 1 ) k )
kp
A
(D)
(1 + ( - 1 ) k )
jkp
(B)
5. x( t) = sin 2 t
Fig. P5.7.2
(A)
A
æp ö
sin ç k ÷
jpk
è2 ø
(B)
A
æp ö
cos ç k ÷
jpk
è2 ø
(A) -
1
1
1
d[ k - 1] + d[ k] - d[ k + 1]
4
2
4
(C)
A
æp ö
sin ç k ÷
pk
è2 ø
(D)
A
æp ö
cos ç k ÷
pk
è2 ø
(B) -
1
1
1
d [ k - 2 ] + d[ k] - d[ k + 2 ]
4
2
4
(C) -
1
1
d[ k - 1] + d[ k] - d[ k + 1]
2
2
(D) -
1
1
d[ k - 2 ] + d[ k] - d[ k + 2 ]
2
2
3. x( t) as shown in fig. P5.7.3
Page
308
t
4p
2p
æ 4 pk ö
ö
A æç - j ççè 3 ÷÷ø
(A)
e
- 1÷
÷
2 pk ç
è
ø
x(t)
-10
-4p
3
0
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The Continuous-Time Fourier Series
Statement for Q.6-11:
(A)
sin 9pt
sin pt
(B)
(C)
sin 18 pt
2 sin pt
(D) None of the above
In the question, the FS coefficient of time-domain
signal have been given. Determine the corresponding
time domain signal and choose correct option.
7. X [ k] = jd[ k - 1] - jd[ k + 1] + d[ k + 3] + d[ k - 3], wo = 2 p
(A) 2(cos 3pt - sin pt)
(B) -2(cos 3pt - sin pt)
(C) 2(cos 6 pt - sin 2 pt)
(D) -2(cos 6 pt - sin 2 pt)
Chap 5.7
sin 9pt
p sin pt
11. X [ k] As depicted in fig. P5.7.11, wo = p
X [k]
3
2
|k|
æ -1 ö
8. X [ k] = ç
÷ , wo = 1
3 ø
4è
(A)
5 + 3 sin t
5
(B)
4 + 3 sin t
5
(C)
4 + 3 cos t
4
(D)
5 + 3 cos t
1
k
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Fig. P5.7.11
(A) 3 cos 3pt + 2 cos 2 pt + cos pt
(B) 3 sin 3pt + 2 sin 2 pt + sin pt
9. X [ k] as shown in fig. P5.7.9 , wo = p
|{X [k]}|
(C) 6 sin 3pt + 4 sin 2 pt + 2 sin pt
2
(D) 6 cos 3pt + 4 cos 2 pt + 2 cos pt
1
k
-6
-5
-4
-3
-2
-1
1
0
2
3
4
5
6
Statement for Q.12-16:
Ð{X [k]}
p
4
Consider a continuous time periodic signal x( t)
k
-p
4
with
coefficients
(C) e
-1
1
0
Determine
the
Fourier
series
2
3
4
5
- to
X [ k] + e to X [ -k]
æ 2p
ö
(B) 2 sin ç
kto ÷ X [ k]
T
è
ø
(D) e
- to
X [ -k] + e to X [ k]
(A)
X [ k] + X [ - k]
2
(B)
X [ k] - X [ - k]
2
6
(C)
X [ k] + X *[ - k]
2
(D)
X [ k] + X *[ - k]
2
Ð{X [k]}
8p
6p
4p
2p
14. y( t) =Re{x( t)}
k
-2p
-4p
-6p
-8p
series
13. y( t) = Ev{x( t)}
k
-2
Fourier
æ 2p
ö
(A) 2 cos ç
kto ÷ X [ k]
T
è
ø
1
-3
and
12. y( t) = x( t - to ) + x ( t - to )
|{X [k]}|
-4
T
choose correct option.
10. X [ k] As shown in fig. P5.7.10 , wo = 2 p
-5
X [ k].
period
coefficient of the signal y( t) given in question and
Fig. P5.7.9
pö
pö
æ
æ
(A) 6 cos ç 2 pt + ÷ - 3 cos ç 3pt - ÷
4
4
è
ø
è
ø
pö
pö
æ
æ
(B) 4 cos ç 4 pt - ÷ - 2 cos ç 3pt + ÷
4ø
4ø
è
è
pö
pö
æ
æ
(C) 2 cos ç 2 pt + ÷ - 2 cos ç 3pt - ÷
4ø
4ø
è
è
pö
pö
æ
æ
(D) 4 cos ç 4 pt + ÷ + 2 cos ç 3pt - ÷
4
4
è
ø
è
ø
-6
fundamental
(A)
X [ k] + X [ - k]
2
(B)
X [ k] - X [ - k]
2
(C)
X [ k] + X *[ - k]
2
(D)
X [ k] + X *[ - k]
2
Fig. P5.7.10
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UNIT 5
d 2 x( t)
dt 2
15. y( t) =
2
æ 2 pk ö
(A) ç
÷ X [ k]
è T ø
2
æ 2 pk ö
(C) jç
÷ X [ k]
è T ø
Signal & System
Statement for Q.20-21:
Let x1 ( t) and x2 ( t) be continuous time periodic
2
æ 2 pk ö
(B) - ç
÷ X [ k]
è T ø
signal with fundamental frequency w1 and w2 , Fourier
series coefficients X1 [ k] and X 2 [ k] respectively. Given
2
æ 2 pk ö
(D) - jç
÷ X [ k]
è T ø
that
x2 ( t) = x1 ( t - 1) + x1 (1 - t)
20. The relation between w1 and w2 is
16. y( t) = x( 4 t - 1)
8p
(A)
X [ k]
T
(C) e
æ 8 pö
- jk çç ÷÷
èTø
(A) w2 =
4p
(B)
X [ k]
T
X [ k]
(D) e
æ 8 pö
jk çç ÷÷
èTø
w1
2
(B) w2 = w12
(C) w2 = w1
X [ k]
(D) w2 = w1
21. The Fourier coefficient X 2 [ k] will be
17. Consider a continuous-time signal
(A) ( X1 [ k] - jX1 [ -k]) e - jw 1 k
x( t) = 4 cos 100 pt sin 1000 pt
1
with fundamental period T =
. The nonzero FS
50
coefficient for this function are
(B) ( X1 [ -k] - jX1 [ k]) e - jw 1 k
(A) X[ -4 ], X[ 4 ], X[ -7 ], X[7 ]
(C) ( X1 [ k] + jX1 [ -k]) e - jw 1 k
(D) None of the above
Statement for Q.22-23:
(B) X[ -1], X[1], X[ -10 ], X[10 ]
Consider three continuous-time periodic signals
(C) X[ -3], X[ 3], X[ -4 ], X[ 4 ]
whose Fourier series representation are as follows.
(D) X[ -9 ], X[9 ], X[ -11], X[11]
2p
k
100
æ 1 ö - jk t
x1 ( t) = å ç ÷ e 50
k=0 è 3 ø
18. A real valued continuous-time signal x( t) has a
fundamental period T = 8. The nonzero Fourier series
x2 ( t) =
coefficients for x( t) are
100
å cos kp e
- jk
2p
t
50
k = -100
x3( t) =
X [1] = X [ -1] = 4, X [ 3] = X *[ -3] = 4 j
100
å
k = -100
2p
æ kp ö - jk 50 t
j sinç
÷e
è 2 ø
The signal x( t) would be
æp ö
æ 3p ö
(A) 4 cos ç t ÷ + 4 j sin ç
t÷
è4 ø
è 4 ø
22. The even signals are
(A) x2 ( t) only
(B) x2 ( t) and x3( t)
æp ö
æ 3p ö
(B) 4 cos ç t ÷ - 4 j cos ç t ÷
è4 ø
è 4 ø
(C) x1 ( t) and x3( t)
(D) x1 ( t) only
pö
æp ö
æ 3p
(C) 8 cos ç t ÷ + 8 cos ç t + ÷
2ø
è4 ø
è 4
23. The real valued signals are
(A) x1 ( t) and x2 ( t)
(B) x2 ( t) and x3( t)
(D) None of the above
(C) x3( t) and x1 ( t)
(D) x1 ( t) and x3( t)
19. The continuous-time periodic signal is given as
æ 2p ö
æ 5p ö
x( t) = 4 + cos ç
t ÷ + 6 sin ç
t÷
è 3 ø
è 3 ø
The nonzero Fourier coefficients are
(A) X [0 ], X [ -1], X [1], X [ -5 ], X [5 ]
24. Suppose the periodic signal x( t) has fundamental
period T and Fourier coefficients X [ k]. Let Y [ k] be the
Fourier coefficient of y( t) where
Fourier coefficient X [ k] will be
(A)
TY [ k]
,k¹0
j2 pk
(B)
TY [ k]
j2pk
(C)
TY [ k]
,k¹0
jk
(D)
TY [ k]
jk
(B) X [0 ], X [ -2 ], X [2 ], X [ -5 ], X [5 ]
(C) X [0 ], X [ -4 ], X [ 4 ], X [ -10 ], X [10 ]
(D) None of the above
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310
y( t) = dx( t) dt . The
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The Continuous-Time Fourier Series
25. Suppose we have given the following information
The signal will be
æp ö
æp ö
(A) 4 cos ç t ÷ - 2 sin ç t ÷
è4 ø
è4 ø
about a signal x( t) :
1. x( t) is real and odd.
æp ö
æp ö
(B) 2 cos ç t ÷ + 4 sin ç t ÷
è4 ø
è4 ø
2. x( t) is periodic with T = 2
3. Fourier coefficients X [ k] = 0 for |k|> 1
æp ö
æp ö
(C) 2 cos ç t ÷ + 2 sin ç t ÷
è4 ø
è4 ø
2
4.
Chap 5.7
2
1
x( t) dt = 1
ò
20
(D) None of the above
The signal, that satisfy these condition, is
(A)
2 sin pt and unique
(B)
2 sin pt but not unique
Statement for Q.29-31:
Consider the following three continuous-time
(C) 2 sin pt and unique
signals with a fundamental period of T = 1
(D) 2 sin pt but not unique
x( t) = cos 2pt , y( t) = sin 2pt , z( t) = x( t) y( t)
26. Consider a continuous-time LTI system whose
29. The Fourier series coefficient X [ k] of x( t) are
frequency response is
(A)
1
2
( d[ k + 1] + d[ k - 1])
(B)
1
2
( d[ k + 1] - d[ k - 1])
(C)
1
2
( d[ k - 1] - d[ k + 1])
H ( jw) =
¥
ò h( t) e
- jwt
dt =
-¥
sin 4 w
w
(D) None of the above
The input to this system is a periodic signal
ì 2, 0 £ t £ 4
x( t) = í
î -2, 4 £ t £ 8
30. The Fourier series coefficient of y( t), Y [ k] will be
with period T = 8. The output y( t) will be
æ pt ö
(A) 1 + sin ç ÷
è4ø
æ pt ö
(B) 1 + cos ç ÷
è4ø
æ pt ö
æ pt ö
(C) 1 + sin ç ÷ + cos ç ÷
è4ø
è4ø
(D) 0
2
2
(A)
j
2
( d[ k + 1] + d[ k + 1])
(B)
j
2
( d[ k + 1] - d[ k - 1])
(C)
j
2
( d[ k - 1] - d[ k + 1])
(D)
1
2j
( d[ k + 1] + d[ k + 1])
31. The Fourier series coefficient of z( t) , Z [ k] will be
27. Consider a continuous-time ideal low pass filter
having the frequency response
ì 1, |w|£ 80
H ( jw) = í
î 0, |w|> 80
(A)
1
4j
( d[ k - 2 ] - d[ k + 2 ])
(B)
1
2j
( d[ k - 2 ] - d[ k + 2 ])
(C)
1
2j
d[ k + 2 ] - d[ k - 2 ])
(D) None of the above
When the input to this filter is a signal x( t) with
fundamental frequency wo = 12 and Fourier series
32. Consider a periodic signal x( t) whose Fourier series
coefficients are
ì 2,
ï
X [ k] = í æ 1 ö|k|
ï jç 2 ÷ ,
î è ø
S
coefficients X [ k], it is found that x( t) ¬¾
® y( t) = x( t).
The largest value of|k,
| for which X [ k] is nonzero, is
(A) 6
(B) 80
(C) 7
(D) 12
28.
A
continuous-time
periodic
otherwise
Consider the statements
signal
has
a
fundamental period T = 8. The nonzero Fourier series
coefficients are as,
X [1] = X [ -1] = j , X [5 ] = X [ -5 ] = 2,
*
k =0
1. x( t) is real.
2. x( t) is even
3.
dx( t)
is even
dt
The true statements are
(A) 1 and 2
(B) only 2
(C) only 1
(D) 1 and 3
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UNIT 5
Statement for Q.33-36:
(A)
A 2A
1
1
+
(sin t - sin 3t + sin 5 t....)
2
p
3
5
(B)
A 2A
1
1
+
(cos t - cos 2 t + cos 3t....)
2
p
2
3
(C)
A 2A
1
1
+
(cos t - cos 3t + cos 5 t....)
2
p
3
5
(D)
A 2A
1
1
+
(sin t + cos t + sin 3t + cos 3t ....)
2
p
3
3
A waveform for one peroid is depicted in figure in
question. Determine the trigonometric Fourier series
and choose correct option.
33.
x(t)
1
-p
p
Signal & System
t
36.
-1
x(t)
Fig. P5.7.33
2
2
1
1
1
(A) (cos t + cos 2 t + cos 3t + cos 4 t +....)
p
2
3
4
2
1
1
1
(sin t - sin 2 t + sin 3t - sin 4 t +....)
p
2
3
4
(C)
2
1
1
1
(sin t + cos t - sin 2 t - cos 2 t + sin 3t +....)
p
2
2
3
Fig. P5.7.36
(A)
A
1 12
1
1
+
(cos pt + cos 3pt +
cos 5 pt +....)
2 p2
9
25
(B) 3 +
(C)
x(t)
p
*****
Fig. P5.7.34
(A)
A 4A æ
1
1
ö
+
ç sin t + sin 2 t + sin 3t +.... ÷
2
p è
2
3
ø
(B)
A 4A æ
1
1
ö
+
ç cos t + cos 3t + cos 5 t +.... ÷
2
p è
3
5
ø
(C)
4A æ
1
1
ö
ç sin t + sin 3t + sin 5 t + .... ÷
p è
3
5
ø
(D)
4A æ
1
1
ö
ç cos t + cos 2 t + cos 3t +.... ÷
p è
2
3
ø
35.
x(t)
A
-p
2
p
2
p
t
Fig. P5.7.35
Page
312
12
1
1
(sin pt - sin 3pt +
sin 5 pt -....)
2
p
9
25
t
-A
-p
12
1
1
(cos pt + cos 3pt +
cos 5 pt +....)
p2
9
25
1 12
1
1
+
(sin pt - sin 3pt +
sin 5 pt -....)
2 p2
9
25
(D) 3 +
-p
t
-1
2
1
1
1
(D) (sin t + cos t + sin 3t + cos 3t + sin 5 t + ....)
p
3
3
5
34.
1
-1
(B)
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The Continuous-Time Fourier Series
SOLUTIONS
1
T
T 2
- jkw t
ò Ad( t) e o dt =
-T 2
- jkw ot
-T 2
1
dt =
T
ò Ae
- jkw ot
- jkt
ò x( t) e dt =
0
dt
1
2p
Ae- jkt dt =
0
jA é êe
2 pk ê
ë
ù
- 1ú
úû
æ 4 pk ö
÷÷
j çç
è 3 ø
å X [ k]e
k = -¥
å X [ k]e
=
sin 9 pt
sin pt
jpkt
1
T
X1 [ k] =
ò x( t - t ) e
o
-k
e-
T
j kw o to
T
ò x( t) e
- jkw ot
dt
T
Y [ k] = X1 [ k] + X 2 [ k] = e - jkw oto X [ k] + e jkw oto X [ k]
= 2 cos ( wo kto ) X [ k]
13. (A) Ev{ x( t)} =
2p
=2 ,
p
x( t) + x( - t)
,
2
The FS coefficients of x( t) are
1
1
X1 [ k] = ò x( -t) e - jkw ot dt = ò x( t) e jkw ot dt = X [ -k]
TT
T T
Therefore, the FS
X [ k] + X [ -k]
Y [ k] =
2
j 2 pt
æ -1 ö jkt
= åç
÷ e +
k = -¥ è 3 ø
-1 - jt
e
4
1
= 3
+
=
1
1
1 + e - jt 1 + e j t 5 + 3cos t
3
3
dt =
The FS coefficients of x( t - to ) + x( t + to ) are
2
-1
- jkw ot
X 2 [ k] = e jkw oto X [ k]
ö
- jkpt
ò0 Ae dt ÷÷
ø
1
= je j 2 p t - je -
series coefficients X1 [ k] of x( t - to ) are
+ e j6 pt + e -
k
æ -1 ö jkt
÷ e
ç
å
k=0 è 3 ø
¥
coefficients
of
Ev{ x( t)}
are
j6 pt
14. (C) Re{ x( t)} =
jkt
- j 2 pk ( t -1 )
k = -4
Similarly, the FS coefficients of x( t + to ) are
ö
-1 2 jt
÷÷ =
( e - 2 + e -2 jt )
4
ø
j 2 pkt
4
åe
e jpkt =
= e - jkw oto X [ k]
= - 2 sin 2 pt + 2 cos 6 pt
8. (D) x( t) =
p
12. (A) x( t - to ) is also periodic with T. The Fourier
k = -¥
¥
j
e j ( 3) pt + 2 e 4 e j (4 )pt
= 6 cos 3pt + 4 cos 2 pt + 2 cos pt
-1
1
1
d[ k - 1] + d[ k] - d[ k + 1]
4
2
4
¥
4
= 3e j ( -3) pt + 2 e j ( -2 ) pt + e j ( -1 ) pt + e j (1 ) pt + 2 e j ( 2 ) pt + 3e j ( 3) pt
4p
0<t<
3
4p
< t < 2p
3
The fundamental period of sin 2 ( t) is p and wo =
å X [ k]e
jp
k = -¥
A æ 1 - e jkp e - jkp -1 ö
A
÷÷ =
(1 - ( -1) k )
= çç
+
2 è jkp
- jkp ø jkp
7. (C) x( t) =
- j 2 pk
¥
x( t) =
1
0
1
1æ
X [ k] = ò x( t) e - jkt dt = çç ò - Ae - jkpt dt +
2 -1
2 è -1
X [ k] =
-
11. (D) X [ k] =|k|, - 3 £ k £ 3
4p 3
ò
4
k = -4
ì - A, - 1 < t < 0
2p
4. (C) T = 2, wo =
= p, x( t) = í
0< t<1
2
î A,
æ e jt - e - jt
5. (A) sin 2 t = çç
è 2j
p
-T 4
ì
ï A,
2p
3. (B) T = 2p , wo =
= 1, x( t) = í
2p
ï0,
î
2p
j
e j ( -4 ) pt + e 4 e j ( -3) pt + e
åe
x( t) =
T 4
4
1
2p
p
4
= 4 cos ( 4 pt + p 4) + 2 cos( 3pt - p 4)
é e - jkw o t ù 4
A
æ pk ö
ê - jkw ú - T = pk sinç 2 ÷
è ø
o û
ë
X [ k] =
- j
= 2( e - j ( 4 pt + p 4 ) + e j ( 4 pt + p 4 ) ) + ( e - j ( 3pt - p 4 ) + e j ( 3pt - p 4 ) )
A
,
T
T
A
=
T
jpkt
10. (A) X [ k] = e - j 2 pk , -4 £ k £ 4
T 2
ò x( t) e
å X [ k]e
k = -¥
A = 10 , T = 5, X [ k] = 2
1
2. (C) X [ k] =
T
¥
9. (D) x( t) =
= 2e
1. (D) X [ k] =
Chap 5.7
x ( t) + x *( t)
,
2
The FS coefficient of x *( t) is
1
X1 [ k] = ò x *( t) e - jkw ot dt = X1*[ -k]
TT
X1*[ k] =
1
T
ò x( t) e
jkw ot
dt = X [ -k]
T
X1 [ k] = X *[ -k]
Y [ k] =
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X [ k] + X *[ - k]
2
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UNIT 5
æ 10 p ö
æ 4p ö
13. X [ k] = cos ç
k ÷ + 2 j sin ç
k÷
è 19 ø
è 19 ø
18. y[ n] = x[ n] - x[ n - N 2 ] , (assume that N is even)
(A) (1 - ( -1) k + 1 ) X [2 k]
(B) (1 - ( -1) k ) X [ k]
(D) (1 - ( -1) k\ ) X [2 k]
(A)
19
( d[ n + 5 ] + d[ n - 5 ]) + 19(d[ n + 2 ] - d[ n - 2 ]),|n|£ 9
2
(C) (1 - ( -1) k + 1 ) X [ k]
(B)
1
( d[ n + 5 ] + d[ n - 5 ]) + (d[ n + 2 ] - d[ n - 2 ]),|n|£ 9
2
19. y[ n] = x *[ -n]
9
(C) ( d[ n + 5 ] + d[ n - 5 ]) + 9(d[ n + 2 ] - d[ n - 2 ]),|n|£ 9
2
(D)
1
( d[ n + 5 ] + d[ n - 5 ]) + (d[ n + 2 ] - d[ n - 2 ]),|n|£ 9
2
æ pk ö
14. X [ k] = cos ç
÷
è 21 ø
(A) - X *[ k ]
(B) - X *[ -k ]
(C) X *[ k ]
(D) X *[ -k ]
20. y[ n] = ( -1) n x[ n], (assume that N is even)
N ù
é
(A) X êk 2 úû
ë
N ù
é
(B) X êk +
2 úû
ë
N
é
ù
(D) X êk +
- 1ú
2
ë
û
(A)
21
(d[ n + 4 ] + d[ n - 4 ]),|n|£ 10
2
N
é
ù
(C) X êk + 1ú
2
ë
û
(B)
1
(d[ n + 4 ] + d[ n - 4 ]),|n|£ 10
2
Statement for Q.21-23:
(C)
21
(d[ n + 4 ] - d[ n - 4 ]),|n|£ 10
2
(D)
1
(d[ n + 4 ] - d[ n - 4 ]),|n|£ 10
2
Consider a discrete-time periodic signal
ì 1, 0 £ n £ 7
x[ n] = í
î 0, 8 £ n £ 9
with period N = 10. Also y[ n] = x[ n] - x[ n - 1 ]
Statement for Q.15-20:
Consider a periodic signal x[ n] with period N and
21. The fundamental period of y[ n] is
FS coefficients X [ k]. Determine the FS coefficients Y [ k]
(A) 9
(B) 10
of the signal y[ n] given in question.
(C) 11
(D) None of the above
15. y[ n] = x[ n - no ]
(A) e
æ 2 pö
÷÷ n ok
j çç
è Nø
(B) e
X [ k]
æ 2 pö
÷÷ n ok
- j çç
è Nø
22. The FS coefficients of y[ n] are
(A)
æ 8 pö
j çç ÷÷ k ö
1 æç
1-e è 5ø ÷
÷
10 ç
è
ø
(B)
æ 8 pö
- j çç ÷÷ k ö
1 æç
1-e è 5ø ÷
÷
10 ç
è
ø
(C)
1
10
æ 4 pö
÷k ö
æ
jç
ç 1 - e çè 5 ÷ø ÷
ç
÷
è
ø
(D)
1
10
X [ k]
(D) -kX [ k]
(C) kX [ k]
16. y[ n] = x[ n] - x[ n - 2 ]
æ 4p ö
(A) sin ç
k ÷ X [ k]
è N ø
æ 4p ö
(B) cos ç
k ÷ X [ k]
è N ø
æ 4 pö
÷÷ k ö
æ
- j çç
(C) ç 1 - e è N ø ÷ X [ k]
ç
÷
è
ø
æ 4 pö
÷÷ k ö
æ
j çç
(D) ç 1 - e è N ø ÷ X [ k]
ç
÷
è
ø
23. The FS coefficients of x[ n] are
æ pk ö
(A) -
17. y[ n] = x[ n] + x[ n + N 2 ] , (assume that N is even)
æN
ö
(A) 2 X [2 k - 1], for 0 £ k £ ç
- 1÷
è 2
ø
(B) 2 X [2 k - 1], for 0 £ k £
Page
318
Signal & System
N
2
(C) 2 X [2 k],
æN
ö
for 0 £ k £ ç
- 1÷
è 2
ø
(D) 2 X [2 k],
for 0 £ k £
N
2
j - j ççè 10 ÷÷ø
æ pk ö
e
cosec ç
÷ Y [ k], k ¹ 0
2
è 10 ø
æ pk ö
(B)
j - j ççè 10 ÷÷ø
æ pk ö
e
cosec ç
÷ Y [ k], k ¹ 0
2
è 10 ø
æ pk ö
(C) -
1 - j ççè 10 ÷÷ø
æ pk ö
e
sec ç
÷ Y [ k]
2
è 10 ø
æ pk ö
(D)
1 - j ççè 10 ÷÷ø
æ pk ö
e
sec ç
÷ Y [ k]
2
è 10 ø
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æ 4 pö
÷k ö
æ
- jç
ç 1 - e çè 5 ÷ø ÷
ç
÷
è
ø
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GATE EC BY RK Kanodia
The Discrete-Time Fourier Series
Statement for Q.24-27:
29. A real and odd periodic signal x[ n] has fundamental
Consider a discrete-time signal with Fourier
period N = 7 and FS coefficients X [ k]. Given that
X [15 ] = j,
representation.
x[ n]
¬ ¾ ¾¾®
X [ k]
Determine the corresponding signal y[ n] and choose
correct option.
æp ö
(B) 2 cos ç n ÷ x[ n]
è5 ø
æp ö
(C) 2 sin ç n ÷ x[ n]
è2 ø
æp ö
(D) 2 cos ç n ÷ x[ n]
è2 ø
(A) 0, j, 2 j, 3 j
(B) 1, 1, 2, 3
(C) 1, -1, -2, -3
(D) 0, - j , -2 j, -3 j
(B)
4.
1
( x[ n + 2 ] + x[ n - 2 ])
2
9
1
10
å X [ k]
The signal x[ n] is
æ p ö
(A) 5 cos ç
n÷
è 10 ø
æ p ö
(B) 5 sin ç
n÷
è 10 ø
æp ö
(C) 10 cos ç n ÷
è5 ø
æp ö
(D) 10 sin ç n ÷
è5 ø
31. Each of two sequence x[ n] and y[ n] has a period
N = 4. The FS coefficient are
(B) j2 p( x[ n]) 2
The FS coefficient Z [ k] for the signal
z[ n] = x[ n] y[ n] will be
x [ n] - x[ -n]
2
(B)
x [ n] + x[ -n]
2p
(D)
(A) 6
(B) 6|k|
(C) 6|k|
(D) e
æ 11p ö
sin ç
n÷
20 ø
è
x[ n] =
æ p ö
sin ç
n÷
è 20 ø
1. x[ n] is periodic with N = 6
5
å x[ n] = 2
n =0
n
with a fundamental period N = 20. The Fourier
x[ n] = 1
n -2
4. x[ n] has the minimum power per period among the
set of signals satisfying the preceding three condition.
The sequence would be..
ì 1 1 1 1 1 ü
(A) í ... , , , , , ...ý (B)
î 2 6 2 6 2 þ
­
ì 1 1 1 1 1 ü
(C) í ... , , , , , ...ý
î 3 6 3 6 3 þ
­
p
j k
2
32. Consider a discrete-time periodic signal
28. Consider a sequence x[ n] with following facts :
7
1
1
X [1] = X [2 ] = 1 and
2
2
Y [0 ], Y [1], Y [2 ], Y [ 3] = 1
(D) 2 p( x[ n]) 2
x [ n] - x[ -n]
2p
å ( -1)
= 50
X [0 ] = X [ 3] =
27. Y [ k] = Re{ X [ k]}
x [ n] + x[ -n]
(A)
2
3.
2
n =0
2
(C) ( x[ n]) 2
2.
of
30. Consider a signal x[ n] with following facts
26. Y [ k] = X [ k] * X [ k]
(C)
values
3. X[11] = 5
1
( x[ n + 10 ] + x[ n + 10 ]) (D) None of the above
2
( x[ n])
2p
The
2. The period of x[ n] is N = 10
æp ö
(A) 2 sin ç n ÷ x[ n]
è5 ø
(A)
X [17 ] = 3 j.
1. x[ n] is a real and even signal
24. Y [ k ] = X [ k - 5 ] + X [ k + 5 ]
æ pk ö
25. Y [ k] = cos ç ÷ X [ k]
è 5 ø
1
(A) ( x[ n + 5 ] + x[ n + 5 ])
2
X [16 ] = 2 j,
X [0 ], X [ -1], X [ -2 ], and X [ -3] will be
p
DTFS ;
10
In question the FS coefficient Y [ k] is given.
(C)
Chap 5.8
series coefficients of this function are
1
(A)
( u[ k + 5 ] - u[ k - 6 ]), |k|£ 10
20
(B)
1 1 1 ü
ì
í ...0, 1, , , , ...ý
2
3 4 þ
î
­
1
( u[ k + 5 ] - u[ k - 5 ]), |k|£ 10
20
(C) ( u[ k + 5 ] - u[ k + 6 ]), |k|£ 10
(D) ( u[ k + 5 ] - u[ k - 6 ]), |k|£ 10
(D) {...0, 1, 2, 3, 4, ...}
­
************
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319
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The Discrete-Time Fourier Series
11. (D) N = 7, W o =
x[ n] =
3
å X [ k]e
2p
,
7
æ 2 pö
÷÷ kn
j çç
è 7 ø
17. (C) Note that y[ n] = x[ n] + x [ n + N 2 ] has a period
= 2e
æ 2 pö
÷÷ n
j ( -1 ) çç
è 7 ø
- 1 + 2e
of N 2 and N has been assumed to be even,
æ 2 pö
÷÷ n
j (1 ) çç
è 7 ø
Y [ k] =
n = -3
æ 2p ö
= 4 cos ç
n÷ - 1
è 7 ø
6
æ pö
- j çç ÷÷ k
è6 ø
e
æ pö
j çç ÷÷ kn
è6 ø
e
=
9p
j ( n -1 )
6
p
j ( n -1 ) ö
æ
÷
ç 1-e 6
÷
ç
ø
è
13. (A) N = 19, W o =
ö
÷ sin æç 3p ( n - 1) ö÷
÷
ø =
è 4
ø
æ p
ö
sin ç
( n - 1) ÷
è 12
ø
æ 4 pö
÷÷ kn
- j çç
è Nø
n =0
19. (C) y[ n] = x *[ -n]
1
Y [ k] =
N
2p
19
N -1
å x [ -n]e
*
æ 2 pö
÷÷ kn
- j çç
è Nø
= X *[ k]
n =0
20. (A) With N even
æ 10 p ö
æ 10 p ö
X [ k] = cos ç
k ÷ + 2 j sin ç
k÷
19
ø
è 19 ø
è
=
( x[ n] + x[ n + N 2 ]) e
k even
ì 0,
=í
2
X
[
k
],
k
odd
î
k = -6
æ
ç 1-e
ç
è
å
æ 2 pö N
÷÷ k ö
æ
- j çç
Y [ k] = ç 1 - e è N ø 2 ÷ X [ k] = (1 - e - jpk ) X [ k]
ç
÷
è
ø
æ pö
j çç ÷÷ k ( n -1 )
è6 ø
6
åe
=
k = -6
p
j ( -4 ) ( n -1 )
6
N 2 -1
18. (B) y[ n] = x[ n] - x [ n - N 2 ]
æ pö
åe
2
N
= 2 X [2 k] for 0 £ k £ ( N 2 - 1)
- j çç ÷÷ k
p
12. (C) N = 12, W o = , X [ k] = e è 6 ø
6
x[ n] =
Chap 5.8
y[ n] = ( -1) n x[ n] = e jpn x[ n] = e
2p
2p
2p
2p
- j ( 5)
kö
k
- j(2 )
kö
æ - j ( -2 ) 19
1 æ - j ( -5) 19 k
19 ÷
19 ÷
çe
çe
+
e
+
+
e
÷ ç
÷
2 çè
ø è
ø
Y [ k] =
By inspection
19
x[ n] =
( d[ n + 5 ] + d[ n - 5 ]) + 19 ( d[ n + 2 ] - d [ n - 2 ]),
2
Where |n|£ 9
=
1
N
1
N
N -1
åe
æ 2 pö N
÷÷
j çç
è Nø 2
x[ n]e
æ 2 pö N
÷÷
j çç
è Nø 2
x[ n]
æ 2 pö
÷÷ kn
- j çç
è Nø
n =0
N -1
å x[ n]e
æ 2 pö æ
Nö
÷÷ n çç k - ÷÷
- j çç
2 ø
è Nø è
= X [k - N 2]
n =0
21. (B) y[ n] is shown is fig. S5.8.21. It has fundamental
y[n]
14. (A) N = 21, W o =
2p
21
1
9
2p
- j ( -4 ) k
21
2p
- j(4 ) k
21
ö
æ 8 p ö 1 æç
÷
X [ k] = cos ç
k÷ = ç e
+e
÷
21
2
è
ø
è
ø
1
Since X [ k] =
å x[ n]e- jkWon , By inspection
N n=N
-1
15. (B) Y [ k] =
æ 2 pö
=
1 - j ççè N ÷÷ø kn o
e
N
1
N
N -1
å x[ n - n ]e
o
n =0
å x[ n]e
æ 2 pö
÷÷ kn
- j çç
è Nø
æ 2 pö
÷÷ kn o
- j çç
è Nø
22. (B) Y [ k] =
=
1 æç
1-e
10 ç
è
16. (C) Y [ k] = X [ k] - e
æ
X [ k] = ç 1 - e
ç
è
5
7
8
n
10 11
9
å y[ n]e
æ 2 pö
÷÷ kn
- j çç
è 10 ø
n =0
ö 1
÷=
÷ 10
ø
æ
ç1 - e
ç
è
æ 8 pö
- j çç ÷÷ k
è 5ø
ö
÷
÷
ø
23. (A) y [ n] = x [ n] - x [ n - 1]
X [ k]
æ 4 pö
÷÷ k
- j çç
è Nø
1
10
æ 2 pö
÷÷ k8
- j çç
è 10 ø
n =0
æ 2 pö
÷÷ 2 k
- j çç
è Nø
4
period of 10.
æ 2 pö
÷÷ kn
- j çç
è Nø
=e
3
Fig. S5.8.21
ì 21
ï , n = ±4
x[ n] = í 2
ïî 0, otherwise n Î { -10, - 9, ......9. 10}
N -1
2
1
ö
÷ X [ k]
÷
ø
Y [ k] = X [ k] - e
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æ 2 pö
÷÷ k
- j çç
è 10 ø
X [ k]
Þ
X [ k] =
Y [ k]
1-e
æ pö
- j çç ÷÷ k
è 5ø
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UNIT 5
Þ
e
X [ k] =
e
=
-j
e
2
æ p ö
- j çç k ÷÷
è 10 ø
æ pö
j çç ÷÷ k
è 10 ø
Y [ k]
-e
æ pö
- j çç ÷÷ k
è 10 ø
æ p
cosec ç
è 10
24. (D) W o =
Þ
æ pö
j çç ÷÷ k
è 10 ø
=
e
æ pö
j çç ÷÷ k
è 10 ø
Signal & System
29. (D) Since the FS coefficient repeat every N. Thus
Y [ k]
æ pk ö
2 j sin ç
÷
è 10 ø
X [1] = X [15 ], X [2 ] = X [16 ], X [ 3] = X [17 ]
The signal real and odd, the FS coefficient X [ k] will be
purely imaginary and odd. Therefore X[0 ] = 0
ö
k ÷ Y [ k]
ø
X [ -1] = - X [1], X [ -2 ] = - X [2 ], X [ -3] = - X [ 3]
Therefore (D) is correct option.
p
, Y [ k] = X [ k - 5 ] + X [ k + 5 ]
10
30. (C) Since N = 10, X [11] = X [1] = 5
p
j ( -5) n ö
æ j ( 5) p n
æp ö
y[ n] = çç e 10 + e 10 ÷÷ x[ n] = 2 cos ç n ÷ x[ n]
è2 ø
è
ø
Since x[ n] is real and even X [ k] is also real and even.
Therefore X [1] = X [ -1] = 5.
Using Parseval’s relation
æ
çe
æp ö
25. (B) Y [ k] = cos ç k ÷ X [ k] = ç
è5 ø
ç
è
p
p
j k
5
+e
2
p
-j k
5
ö
÷
÷ X [ k]
÷
ø
2
2
2
X [ -1] + X [1] + X [0 ] +
2
X [0 ] +
8
å X [ k]
2
27. (A) Y [ k] =Re{ X [ k]} Þ y[ n] =Ev{ x[ n]} =
x[ n] = å X [ k]e
2
x[ n] + x[ -n]
2
æ
= 5ç e
ç
è
æ 2 pö
÷÷ kn
j çç
è Nø
=
2p
,
6
åe
k = -1
1
Þ X [0 ] = ,
3
Þ
1
6
åe
n =0
3
Z[ k ] = X [ 0 ] Y [ k ] + X [ 1 ] Y [ k - 1 ] + X [ 2 ] Y [ k - 2 ] + X [ 3 ] Y [ k - 3 ]
32. (A) N = 20 We know that
1
1
x[ n] = , X [ 3] =
6
6
P = å X [ k] ,
|n|£ 5
ì 1,
í
î 0, 5 <|n|£ 10
The value of P is minimized by choosing
X [1] = X [2 ] = X [ 4 ] = X [5 ] = 0
Therefore
x[ n] = X [0 ] + X [ 3]e
p
10
¬ ¾ ¾¾®
ö
÷÷ 3n
ø
=
1
1 1
1
+ ( -1) n = + ( -1) n
3
6 3
6
Using duality
æ 11p ö
sin ç
n÷
p
DTFS ;
è 20 ø ¬ ¾
10
¾¾
®
æ p ö
sin ç
n÷
è 20 ø
ì 1 1 1 1 1 ü
x[ n] = í ... , , , , , ...ý
î 2 6 2 6 2 þ
æ 11p ö
sin ç
k÷
è 20 ø
æ p ö
sin ç
k÷
è 20 ø
|k|£ 5
1 ì 1,
í
20 î 0, 5 <|k|£ 10
*********
­
Page
322
DTFS ;
2
k=0
æ 2p
çç
è 6
å X [ l ]Y [ k - l ]
Thus Z [ k] = 6, for all k.
By Parseval’s relation, the average power in x[ n] is
5
DTFS
¬ ¾¾
®
Since Y [ k] is 1 for all values of k.
å ( -1) n x[ n] = 1
æ 2 pö
÷÷ ( 3) k
j çç
è 6 ø
ö
÷ = 10 cos æç p n ö÷
÷
è5 ø
ø
= Y [ k] + 2 Y [ k - 1] + 2 Y [ k - 2 ] + Y [ k - 3]
1
x[ n] =
3
n =2
5
æ 2 pö
÷÷ n
j çç
è 10 ø
l= 0
Þ
7
From fact 3,
æ 2 pö
÷÷ kn
j çç
è 10 ø
Z [ k] = å X [ l ]Y [ k - l ]
Þ
å x[ n] = 2
n =0
+e
å X [ k]e
k =< N>
n =0
Þ
æ 2p ö
- j çç ÷÷ n
è 10 ø
8
31. (A) z[ n] = x[ n] y[ n]
5
1
6
= 50
Therefore X [ k] = 0 for k = 0, 2, 3, ..... 8.
26. (C) Y [ k] = X [ k] * X [ k] Þ y[ n] = x[ n] x[ n] = ( x[ n])
æ 2 pö
÷÷ ( 0 ) k
j çç
è 6 ø
2
2
k = -1
=0
N
5
8
å X [ k]
8
å X [ k]
k=2
1
y[ n] = ( x[ n - 2 ] + x[ n + 2 ])
2
From fact 2,
= 50 =
k=2
p
28. (A) N = 6, W o =
2
N
j ( -2 )
kö
1 æ j(2 ) k
10 ÷
= çç e 10 + e
÷ X [ k]
2è
ø
Þ
å X [ k]
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GATE EC BY RK Kanodia
CHAPTER
6.1
TRANSFER FUNCTION
1. The equivalent transfer function of three parallel
R(s)
+
C(s)
G(s)
blocks
G1 ( s) =
1
1
s+3
, G2 ( s) =
and G3( s) =
is
s+1
s+4
s+5
H(s)
Fig. P.6.1.3
( s 3 + 10 s 2 + 34 s + 37)
(A)
( s + 1)( s + 4)( s + 5)
(A)
( s + 3)
( s + 1)( s + 4)( s + 5)
s( s + 2)( s + 3)
s + 7 s 2 + 12 s + 3
(B)
-( s 3 + 10 s 2 + 34 s + 37)
( s + 1)( s + 4)( s + 5)
s( s + 2)( s + 3)
s + 5 s2 + 4 s - 3
(C)
-( s + 3)
( s + 1)( s + 4)( s + 5)
( s + 1)( s + 4)
s + 7 s 2 + 12 s + 3
(D)
( s + 1)( s + 4)
s + 5 s2 + 4 s - 3
(B)
(C)
(D)
2. The block having transfer function
3
3
3
3
4. A feedback control system is shown in fig. P.6.1.4.
1
1
s+1
, G2 ( s) =
, G3( s) =
G1 ( s) =
s+2
s+5
s+3
The transfer function for this system is
R
are cascaded. The equivalent transfer function is
(A)
(B)
( s 3 + 10 s 2 + 37 s 2 + 31)
( s + 2)( s + 3)( s + 5)
1
G1
G3
C
Fig. P.6.1.4
-( s 3 + 10 s 2 + 37 s 2 + 31)
(C)
( s + 2)( s + 3)( s + 5)
(A)
G1 G2
1 + H1 G1 G2 G3
-( s + 1)
(D)
( s + 2)( s + 3)( s + 5)
(B)
G2 G3
G1 (1 + H1 G2 G3)
(C)
G2 G3
1 + H1 G1 G2 G3
(D)
G2 G3
G1 (1 + H1 G2 G3)
3. For a negative feedback system shown in fig. P.6.1.3
s+1
s+3
and H ( s) =
s( s + 2)
s+4
G2
H2
s+1
( s + 2) ( s + 3) ( s + 5)
G( s) =
+
The equivalent transfer function is
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UNIT 6
5. Consider the system shown in fig. P.6.1.5.
R(s)
+
G1
+
G2
+
(D)
Control Systems
G1 G2
1 + G1 H1 + G2 H 2 + G1 G2 H1 H 2
C(s)
+
8. The closed loop gain of the system shown in fig.
P6.1.8 is
Fig. P.6.1.5
R
+
C
6
+
The input output relationship of this system is
R(s)
C(s)
G1G2
R(s)
1 + G1 + G1G2
(A)
R(s)
Fig.P6.1.8
(B)
C(s)
G1 + G2
R(s)
1 + G2 + G1G2
(C)
1
3
C(s)
C(s)
(D)
(A) -2
(B) 6
(C) -6
(D) 2
9. The block diagrams shown in fig. P.6.1.9 are
6. A feedback control system shown in fig. P.6.1.6 is
equivalent if G is equal to
R(s)
C(s)
s+2
s+1
subjected to noise N ( s).
N(s)
R(s)
R(s)
+
G1
+
+
1
s+1
C(s)
G2
Fig. P.6.1.6
C ( s)
The noise transfer function N
is
N ( s)
G2
1 + G1 G2 H
(B)
G2
(C)
1 + G2 H
C(s)
+
Fig. P.6.1.9
H2
(A)
+
G
(A) s + 1
(B) 2
(C) s + 2
(D) 1
10. Consider the systems shown in fig. P.6.1.10. If the
G2
1 + G1 H
forward path gain is reduced by 10% in each system,
then the variation in C1 and C2 will be respectively
(D) None of the above
R1
R2
7. A system is shown in fig. P6.1.7. The transfer
16
C1
3
+
C2
10
function for this system is
H1
R(s)
+
G1
Fig. P.6.1.10
+
G2
C(s)
(A) 10% and 1%
(B) 2% and 10%
(C) 10% and 0%
(D) 5% and 1%
H2
Fig. P.6.1.7
(A)
(B)
(C)
Page
326
G1 G2
1 + G1 G1 H 2 + G2 H1
11. The transfer function
C
R
of the system shown in the
fig. P.6.1.11 is
R
G1 G2
1 + G1 G2 + H1 H 2
1
H1
+
H2
G1
G2
G1 G2
1 - G1 H1 - G2 H 2 + G1 G2 H1 H 2
Fig. P.6.1.11
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Transfer Function
(A)
(C)
G1 H 2
H1 (1 + G1 G2 H 2 )
(B)
G2 G1
1 + H1 H 2 G1 G2
(D)
Chap 6.1
15. A closed-loop system is shown in fig. P.6.1.15. The
G1 G2 H 2
H1 (1 + G1 G2 H 2 )
noise transfer function Cn ( s) N ( s) is approximately
G1 G2
H1 (1 + G1 G2 H 2 )
R(s) 1
G1 C(s)
-H2
H1
12. In the signal flow graph shown in fig. P6.1.12 the
1
sum of loop gain of non-touching loops is
t32
x2
x1
Fig. P.6.1.15
t56
t45
t34
t23
t12
N(s)
t44
t43
x5
x4
x3
x6
(A)
1
For |G1 ( s) H1 ( s) H 2 ( s)| << 1
G1 ( s) H1 ( s)
(B)
1
For |G1 ( s) H1 ( s) H 2 ( s)| >> 1
-H1 ( s)
(C)
1
For |G1 ( s) H1 ( s) H 2 ( s)| >> 1
H1 ( s) H 2 ( s)
(D)
1
For |G1 ( s) H1 ( s) H 2 ( s)| << 1
G1 ( s) H1 ( s) H 2 ( s)
t24
t25
Fig. P.6.1.12
(A) t32 t23 + t44
(B) t23t32 + t34 t43
(C) t24 t43t32 + t44
(D) t23t32 + t34 t43 + t44
13. For the SFG shown in fig. P.6.1.14 the graph
determinant D is
-c
16. The overall transfer function
of the system shown
in fig. P.6.1.16 will be
b
a
C
R
d
1
-H1
-H2
1
R
i
e
G
1
C
h
j
Fig. P.6.1.16
f
(A) G
-g
(B)
G
1 + H2
(D)
G
1 + H1 + H 2
Fig. P.6.1.13
(A) 1 - bc - fg - bcfg + cigj
(C)
G
(1 + H1 )(1 + H 2 )
(B) 1 - bc - fg - cigj + bcfg
(C) 1 + bc + fg + cig j - bcfg
17. Consider the signal flow graphs shown in fig.
(D) 1 + bc + fg + bcfg - cigj
P6.1.17. The transfer 2 is of the graph
1
1
14. The sum of the gains of the feedback paths in the
1
signal flow graph shown in fig. P.6.1.13 is
1
a
b
c
f
e
d
1
1
1
1
1
2
1
Fig. P.6.1.14
(C) af + be + cd + abef + abcdef
(D) af + be + cd + cbef + bcde + abcdef
1
1
2
1
2
(A) af + be + cd + abef + bcde
(B) af + be + cd
1
2
1
Fig. P.6.1.17
(A) a
(B) b
(C) b and c
(D) a, b and c
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GATE EC BY RK Kanodia
UNIT 6
Control Systems
18. Consider the List I and List II
List I
-H3
List II
(Signal Flow Graph)
R(s) 1
(A)
(Transfer Function)
1
1
2
-H2
3
G1
4
1 C(s)
4
1
4
1 C(s)
4
1
G2
-H1H2
P.
Q.
a
xi
b
xi
xo
-H2
2
3
G1
C(s)
G2
-H2
a xo
a
3.
(1 - ab)
S.
b
R.
1
-H2
1
xi 1
1
2. ab
b
xi 1
R(s) 1
(B)
xo
a
-H3
1. a + b
1 xo
a
4.
b
R(s) 1
(C)
1
1
3
2
G2
G1
-H3
-H1H2
a
1-b
H3
R(s) 1
(D)
1
1
3
2
G2
C(s)
G1
The correct match is
H2
P
Q
R
S
(A)
2
1
3
4
(B)
2
1
4
3
(C)
1
2
4
3
21. The block diagram of a system is shown in fig.
(D)
1
2
3
4
P.6.1.21. The closed loop transfer function of this system
-H1
19. For the signal flow graph shown in fig. P6.1.19 an
is
H1
equivalent graph is
ta
R(s)
tc
e1
G1
e1
Fig. P.6.1.21
tc+ td
tatb
e1
e4
(A)
e4
e3
(B)
ta + tb
e1
tatb
tctd
e2
e4
e1
(C)
1
+
G1 G2 G3
1 + G1 G2 G3 H1
(B)
G1 G2 G3
1 + G1 G2 G3 H1 H 2
(C)
G1 G2 G3
1 + G1 G2 H1 + G2 G3 H 2
(D)
G1 G2 G3
1 + G1 G2 H1 + G1 G3 H 2 + G2 G3 H1
tc+ td
e4
e2
block diagram shown in figure
P.6.1.20
R(s)
(A)
(D)
20. Consider the
3
2
+
C(s)
H2
tctd
e3
G3
e4
e3
Fig. P.6.1.19
ta + tb
G2
+
td
e2
tb
+
G2
+
4
G2
22. For the system shown in fig. P6.1.22 transfer
C(s)
function C( s) R( s) is
G3
H2
H1
R(s)
+
G1
+
+
G2
H3
H1
Fig. P.6.1.20
H2
For this system the signal flow graph is
Fig. P.6.1.22
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328
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Transfer Function
(A)
26. The transfer function of the system shown in fig.
G3
1 - H1 G2 - H 2 G3 - G1 G2 H 2
(B)
G3 + G1 G2
1 + H1 G2 + H 2 G3 + G1 G2 H 2
(C)
G3
1 + H1 G2 + H 2 G3 + G1 G2 H 2
Chap 6.1
P.6.1.26 is
R(s)
+
+
+
G1
C(s)
G2
H1
H2
Fig. P.6.1.26
G3
(D)
1 - H1 G2 - H 2 G3 - G1 G2 H 2
(A)
G1 G2
1 - G1 G2 H1 - G1 G2 H 2
(B)
G1 G2
1 - G2 H 2 - G1 G2 H1
23. In the signal flow graph shown in fig. P6.1.23 the
(C)
G1 G2
1 - G2 H 2 + G1 G2 H1 H 2
(D)
G1 G2
1 - G1 G2 H1 H 2
transfer function is
3
2
C
27. For the block diagram shown in fig. P.6.1.27
transfer function C( s) R( s) is
-3
Fig. P.6.1.23
(B) -3
(C) 3
(D) -3.75
R(s)
+
G1
+
-1
-1
-1
2
3
4
1
1
G3
G4
24. In the signal flow graph shown in fig. P6.1.24 the
R
G2
C
+
‘(A) 3.75
gain C R is
C(s)
G8
G5
+
5
G7
G6
+
R
Fig. P.6.1.27
5
(A)
G1 G2
1 + G1 G2 + G1 G7G3 + G1 G2 G8 G6 + G1 G2 G3G7G5
Fig. P.6.1.24
44
(A)
23
(B)
29
19
(B)
G1 G2
1 + G1 G4 + G1 G2 G8 + G1 G2 G5G7 + G1 G2 G3G6 G7
44
19
(D)
29
11
(C)
G1 + G2
1 + G1 G4 + G1 G2 G8 + G1 G2 G5G7 + G1 G2 G3G6 G7
(D)
G1 + G2
1 + G1 G2 + G3G6 G7 + G1 G3G4 G5 + G1 G2 G3G6 G7G8
(C)
25. The gain C( s) R( s) of the signal flow graph shown in
fig. P.6.1.25 is
28. For the block diagram shown in fig. P.6.1.28 the
G4
G3
R(s) 1
1 C(s)
G1
numerator of transfer function is
G2
G1
-H1
Fig. P.6.1.25
(A)
G1 G2 + G2 G3
1 + G1 G2 H1 + G2 G3 H1 + G4
(B)
G1 G2 + G2 G3
1 + G1 G3 H1 + G2 G3 H1 - G4
(C)
G1 G3 + G2 G3
1 + G1 G3 H1 + G2 G3 H1 - G4
(D)
G1 G3 + G2 G3
1 + G1 G3 H1 + G2 G3 H1 + G4
R(s)
+
G2
+
+
G5
+
+
G6
C(s)
G3
G4
+
+
Fig. P.6.1.28
(A) G6 [ G4 + G3 + G5( G3 + G2 )]
(B) G6 [ G2 + G3 + G5( G3 + G4 )]
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UNIT 6
Control Systems
33. The closed loop transfer function of the system
(C) G6 [ G1 + G2 + G3( G4 + G5)]
(D) None of the above
G1
29. For the block diagram shown in fig. P.6.1.29 the
R
+
transfer function C( s) R( s) is
50( s - 2)
50( s - 2)
(A) 3
(B) 3
2
s + s + 150 s - 100
s + s 2 + 150 s
(C)
50 s
s + s + 150 s - 100
3
(D)
2
+
+
G2
C
G3
H1
50
s + s + 150
Fig. P.6.1.32
2
shown in fig. P6.1.33 is
30. For the SFG shown in fig. P.6.1.30 the transfer
function
C
R
G2
is
R
G2
G1
-H1
G3
-H2
R
C
1
+
-H3
+
+
G1
+
+
G3
H3
G4
H2
Fig. P.6.1.30
H1
Fig. P.6.1.33
(A)
G1 + G2 + G3
1 + G1 H1 + G2 H 2 + G3 H 3
(A)
G1 G2 G3 + G2 G3G4 + G1 G4
1 + G1 G3G4 H1 H 2 H 3 + G2 H 4 H1 H 2 + G4 H1
(B)
G1 + G2 + G3
1 + G1 H1 + G2 H 2 + G3 H 3 + G1 G3 H1 H 3
(B)
G2 G4 + G1 G2 G3
1 + G1 G3 H1 H 2 H 3 + G4 H1 + G3G4 H1 H 2
(C)
G1 G2 G3
1 + G1 H1 + G2 H 2 + G3 H 3
(C)
G1 G3G4 + G2 G4
1 + G3G4 H1 H 2 + G4 H1 + G1 G3 H 3 H 2
(D)
G1 G2 G3
1 + G1 H1 + G2 H 2 + G3 H 3 + G1 G3 H1 H 3
(D)
G1 G3G4 + G2 G3G4 + G2 G4
1 + G1 G3G4 H1 H 2 H 3 + G3G4 H1 H 2 + G4 H1
31. Consider the SFG shown in fig. P6.1.31. The D for
this graph is
Statement for Q.34-37:
A block diagram of feedback control system is
G4
shown in fig. P6.1.34-37
-H3
R
1
G2
G1
G3
1
C
-H1
R1(s)
+
+
G
R2(s)
+
+
+
G
C1(s)
-H2
Fig. P.6.1.31
(A) 1 + G1 H1 + G2 G3 H 3 + G1 G3 H 2
(B) 1 + G1 H1 - G2 G3 H 3 - G1 G3 H 3 + G2 G4 H 2 H 3
(C) 1 + G1 H1 + G2 G3 H 3 + G1 G3 H 3 - G2 G4 H 2 H 3
32. The transfer function of the system shown in fig.
(C)
Page
330
G2 G3 + G1 G3
1 + G3 H1 + G2 G3
C2(s)
Fig. P.6.1.34-37
(D) 1 + G1 H1 + G2 G3 H 3 + G1 G3 H 3 + G2 G4 H 2 H 3
P.6.1.32 is
G2 G3 + G1 G3
(A)
1 - G3 H1 + G2 G3
C
34. The transfer function
C1
R1
is
R2 = 0
(B)
G2 G3 + G1 G3
1 + G3 H1 - G2 G3
(A)
G
1 - 2G2
(B)
G(1 - G)
1 - 2G2
(D)
G2 G3 + G1 G3
1 - G3 H1 - G2 G3
(C)
G(1 - 2 G)
1 - G2
(D)
G
1 - G2
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Transfer Function
35. The transfer function
(A)
(C)
C1
R2
R1 = 0
(B)
G2
1 - 2G2
(D)
C2
R1
G (1 + G)
(A)
1 - 2G2
5
G
1 - G2
G
1 - G2
+
+
2s
C2(s)
Fig. P.6.1.40-41
is
R2 = 0
G
1 - G2
(D)
G(1 + G)
(A)
1 - 2G2
1
s
+
2
G2
(B)
1 - 2G2
C
37. The transfer function 2
R2
(C)
R1(s)
G2
1 - G2
2
(C)
40. The transfer function for this system is
is
G
1 - 2G2
36. The transfer function
Chap 6.1
(A)
2 s(2 s + 1)
2 s 2 + 3s + 5
(B)
2 s(2 s + 1)
2 s 2 + 13s + 5
(C)
2 s(2 s + 1)
4 s 2 + 13s + 5
(D)
2 s(2 s + 1)
4 s 2 + 3s + 5
41. The pole of this system are
is
R1 = 0
(A) -0.75 ± j1.39
(B) -0.41, - 6.09
(C) -0.5,
(D) -0.25 ± j0.88
- 1.67
G
(B)
1 - 2G2
G
1+ G
(D)
G
1 - G2
********
Statement for Q.38–39:
A signal flow graph is shown in fig. P.6.1.38–39.
G4
Y1
1
Y2
G1
Y3
-H1
Y4 G3
G2
Y5
1
Y5
-H2
-H3
Fig. P.6.1.38-39
38. The transfer function
Y2
is
Y1
1 + G2 H 2
D
(A)
1
D
(B)
(C)
G1 G2 G3
D
(D) None of the above
39. The transfer function
Y5
is
Y2
(A)
G1 G2 G3 + G4 G3
D
(B) G1 G2 G3 + G4 G3
(C)
G1 G2 G3 + G4 G3
G1 G2 G3
(D)
G1 G2 G3 + G4 G3
1 + G2 H 2
Statement for Q.40–41:
A block diagram is shown in fig. P6.1.40–41.
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GATE EC BY RK Kanodia
UNIT 6
SOLUTIONS
C2 =
C¢2 =
9
9
= , C2 is reduced by 1%.
9 + 1 10
11. (A) Apply the feedback formula and then multiply
1
by
,
H1
1. (A) Ge ( s) = G1 ( s) + G2 ( s) + G3( s)
=
1
1
s+3
+
+
( s + 1) ( s + 4) ( s + 5)
=
s 2 + 9s + 20 + s 2 + 6s + 5 + s 3 + 5s 2 + 4s + 3s 2 + 15s + 12
(s + 1)(s + 4)(s + 5)
=
s 3 + 10 s 2 + 34 s + 37
( s + 1)( s + 4)( s + 5)
æ 1 ö
( H 2 G1 ) çç
÷÷
C
H 2 G1
è H1 ø =
=
R
H1 (1 + G1 G2 H 2 )
1 + H 2 G1 G2
12. (A) There cannot be common subscript because
2. (B) Ge ( s) = G1 ( s) G2 ( s) G3( s) =
3. (C)
10
10
,
=
10 + 1 11
Control Systems
( s + 1)
( s + 2)( s + 5)( s + 3)
subscript refers to node number. If subscript is common,
that means that node is in both loop.
C( s)
G( s)
=
R( s) 1 + H ( s) G( s)
13. (D) L1 = -bc, L2 = - fg, L3 = jgic, L1 L3 = bcfg
D = 1 - ( -bc - fg + cigj) + bcfg = 1 + bc + fg - cig j + bcfg
s+1
( s + 1)( s + 4)
s( s + 2)
=
= 3
( s + 3) ( s + 1)
s + 7 s 2 + 12 s + 3
1+
( s + 4) s( s + 2)
14. (A) In this graph there are three feedback loop. abef
is not a feedback path because between path x2 is a
summing node.
4. (B) Multiply G2 and G3 and apply feedback formula
and then again multiply with
T( s) =
1
.
G1
P1 = - H 2 G1 , L1 = - G1 H 2 H1 , D1 = 1,
G2 G3
G1 (1 + G2 G3 H1 )
if |G1 H 2 H1 | >> 1, Tn ( s) =
5. (D) T( s) = G2 (1 + G1 ) + 1 = 1 + G1 + G1 G2
TN ( s) =
T( s) =
G2
1 + G1 G2 H
7. (D) Apply the feedback formula to both loop and then
=
öæ
G2
÷÷çç
+
1
G2 H 2
øè
ö
÷÷
ø
- H 2 G1
-1
=
G1 H 2 H1 H1
G
G
=
1 + H1 + H 2 + H1 H 2 (1 + H1 )(1 + H 2 )
17. (B) Ga = 1, Gb = 1 + 1 = 2, Gc =
1 1 1 1
+ + + =1
4 4 4 4
18. (B)
P. P1 = ab, D = 1, L = 0 , T = ab
G1 G2
1 + G1 H1 + G2 H 2 + G1 G2 H1 H 2
8. (C) For positive feedback
Q. P1 = a, P2 = b , D = 1, L = Dk = 0, T = a + b
a
R. P1 = a, L1 = b, D = 1 - b, D1 = 1, T =
a-b
a
S. P1 = a, L1 = ab, D = 1 - ab, D1 = 1, T =
1 - ab
C
6
=
=-6
R 1 - 6 ´31
9. (D) For system (b) closed loop transfer function
G
G + s+1 G + s+1 s+2
,
,
Hence G = 1
+1=
=
s+1
s+1
s+1
s+1
Page
332
- H 2 G1
1 + G1 H 2 H1
There are no loop in any graph. So option (B) is correct.
multiply
æ
G1
T( s) = çç
1
+
G1 H1
è
Tn ( s) =
16. (C) P1 = G , L1 = - H1 , L2 = - H 2 , L1 L2 = H1 H 2 , D1 = 1
6. (A) Open-loop gain = G2
Feed back gain = HG1
15. (B) By putting R ( s) = 0
19. (A) Between e1 and e2 , there are two parallel path.
Combining them gives ta + tb . Between e2 and e4 there
is a path given by total gain tc td . So remove node e3 and
10. (A) In open loop system change will be 10% in C1
place gain tc td of the branch e2 e4 . Hence option (A) is
also but in closed loop system change will be less
correct.
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Transfer Function
20. (A) Option (A) is correct. Best method is to check the
Chap 6.1
28. (A) SFG:
signal flow graph. In block diagram there is feedback
G3
from 4 to 1 of gain -H1 H 2 . The signal flow graph of
G4
G3
option (A) has feedback from 4 to 1 of gain -H1 H 2 .
R
1
21. (C) Consider the block diagram as SFG. There are
G5
G2
-G1
22. (B) Consider the block diagram as a SFG. Two
forward path G1 G2 and G3 and three loops
P1 = G2 G5G6 , P2 = G3G5G6 , P3 = G3G6 , P4 = G4 G6
If any path is deleted, there would not be any loop.
Hence D1 = D2 = D3 = D4 = 1
C G4 G6 + G3G6 + G3G5G6 + G2 G5G6
=
D
R
-G1 G2 H 2 , - G2 H1 , - G3 H 2 .
There are no nontouching loop. So (B) is correct.
23. (C) P1 = 5 ´ 3 ´ 2 = 30, D = 1 - ( 3 ´ - 3) = 10
29. (A)
C 30
D1 = 1 ,
=
=3
R 10
R
1
-2
s
1
s2
L3 = -4, L4 = -5,
C
s
-1
L1 L3 = 8, D = 1 - ( -2 - 3 - 4 - 5) + 8 = 23,
Fig. S6.1.29
D1 = 1, D2 = 1 - ( -3) = 4,
C 24 + 5 ´ 4 44
=
=
R
24
23
P1 =
1
50
50
×
×s=
s2 ( s + 1)
s( s + 1)
P2 =
1 50
-100
×
× ( -2) = 2
s2 s + 1
s ( s + 1)
L1 =
50 -2
-100
×
=
s+1 s
s( s + 1)
L2 =
1 50
-50
×
× s × ( -1) =
s2 s + 1
s( s + 1)
L3 =
1 50
100
×
× ( -2) × ( -1) = 2
s2 s + 1
s ( s + 1)
P2 = G3G2
L1 = -G3G2 H1 , L2 = -G1 G2 H1 ,
-2
1
50
(s + 1)
24. (A) P1 = 2 ´ 3 ´ 4 = 24 , P2 = 1 ´ 5 ´ 1 = 5
25. (B) P1 = G1 G2 ,
C
Fig. S6.1.28
forward path G1 G2 G3 . So (D) is correct option.
L2 = -3,
1
-1
two feedback loop -G1 G2 H1 and -G2 G3 H 2 and one
L1 = -2 ,
G6
L3 = G4 , D1 = D2 = 1
There are no nontouching loop.
P1 D1 + P2 D2
G1 G2 + G2 G3
T( s) =
=
1 - ( L1 + L2 + L3) 1 + G1 G2 H1 + G2 G3 H1 - G4
26. (C) P1 = G1 G2 , L1 = -G1 G2 H1 H 2 , L2 = G2 H 2
C( s)
G1 G2
=
R( s) 1 + G1 G2 H1 H 2 - G2 H 2
27. (B) There is one forward path G1 G2 .
D =1 +
100
50
100
+
s( s + 1) s( s + 1) s 2 ( s + 1)
D1 = D2 = 1
C P1 + P2
50( s - 2)
=
= 3
R
D
s + s2 + 150 s - 100
Four loops -G1 G4 , - G1 G2 G8 , - G1 G2 G5G7
30. (D) P1 = G1 G2 G3
and -G1 G2 G3G6 G7 .
L1 = - G1 H1 , L2 = - G2 H 2 , L3 = - G3 H 3
R(s) 1
G2
G1
1 C(s)
D = 1 - ( - G1 H1 - G2 H 2 - G3 H 3 ) + G1 G3 H1 H 3
1
-H1
L1 L3 = G1 G3 H1 H 3
H2
Fig. S6.1.27
There is no nontouching loop. So (B) is correct.
D = 1 + G1 H1 + G2 H 2 + G3 H 3 + G1 G3 H1 H 3
D1 = 1
C
G1 G2 G3
=
R 1 + G1 H! + G2 H 2 + G3 H 3 + G1 G3 H1 H 3
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CHAPTER
6.2
STABILITY
1. Consider the system shown in fig. P6.2.1. The range
of K for the stable system is
R(s)
+
E(s)
C(s)
K(s2 - 2s + 2)
The range of K to ensure stability is
6
3
(A) K >
(B) K < - 1 or K >
8
4
(C) K < - 1
(D) -1 < K <
3
4
1
s2 + 2s + 1
5. Consider a ufb system with forward-path transfer
Fig. P6.2.1
(A) -1 < K < -
1
2
function
(B) -
(C) -1 < K < 1
1
< K <1
2
G( s) =
(D) Unstable
The system is stable for the range of K
2. The forward transfer function of a ufb system is
G( s) =
K ( s + 3)
s 4 ( s + 2)
K ( s 2 + 1)
( s + 1)( s + 2)
(A) K > 0
(B) K < 0
(C) K > 1
(D) Always unstable
6. The open-loop transfer function of a ufb
The system is stable for
control
system is
(A) K < - 1
(B) K > -1
(C) K < - 2
(D) K > -2
G( s) =
3. The open-loop transfer function with ufb are given
below for different systems. The unstable system is
For K > 6, the stability characteristic of the
open-loop and closed-loop configurations of the system
(A)
2
s+2
(B)
2
s 2 ( s + 2)
are respectively
(C)
2
s( s + 2)
(D)
2( s + 1)
s( s + 2)
(B) stable and stable
(A) stable and unstable
(C) unstable and stable
4. Consider a ufb system with forward-path transfer
function
K ( s + 2)
( s + 1)( s - 7)
(D) unstable and unstable
7. The forward-path transfer function of a ufb system is
G( s) =
K ( s + 3)( s + 5)
( s - 2)( s - 4)
G( s) =
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K ( s 2 - 4)
s2 + 3
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UNIT 6
For the system to be stable the range of K is
3
(A) K > -1
(B) K <
4
(C) -1 < K <
3
4
Control & System
13. The open-loop transfer function of a ufb system is
R(s)
+
C(s)
1
s(s + 1)(s + 5)
K
(D) marginal stable
Fig. P6.2.12
8. A ufb system have the forward-path transfer function
G( s) =
K ( s + 6)
s( s + 1)( s + 3)
G( s) =
K ( s + 10)( s + 20)
s 2 ( s + 2)
The closed loop system will be stable if the value of
The system is stable for
(A) K < 6
(B) -6 < K < 0
K is
(C) 0 < K < 6
(D) K > 6
(A) 2
(B) 3
(C) 4
(D) 5
9. The feedback control system shown in the fig. P6.2.8.
R(s)
+
K(1 + Ts)
s2(1 + s)
Statement for Q.14–15:
C(s)
A feedback system is shown in fig. P6.14-15.
2
s
Fig. P6.2.9
R(s)
+
is stable for all positive value of K , if
(A) T = 0
(B) T < 0
(C) T > 1
(D) 0 < T < 1
+
C(s)
+
s2
1
s+1
Fig. P6.2.14–15
10. Consider a ufb system with forward-path transfer
function
14. The closed loop transfer function for this system is
K
G( s) =
( s + 15)( s + 27)( s + 38)
The system will oscillate for the value of K equal to
(A) 23690
(B) 2369
(C) 144690
(D) 14469
(A)
s 5 + s 4 + 2 s 3 + ( K + 2) s 2 + ( K + 2) s + K
s3 + s2 + 2 s + K
(B)
2 s 4 + ( K + 2) s 3 + Ks 2
s3 + s2 + 2 s + K
(C)
s3 + s2 + 2 s + K
s 5 + s 4 + 2 s 3 + ( K + 2) s 2 + ( K + 2) s + K
(D)
s3 + s2 + 2 s + K
2 s + ( K + 2) s 3 + Ks 2
11. The forward-path transfer function of a ufb system is
G( s) =
K ( s - 2)( s + 4)( s + 5)
( s 2 + 3)
1
3
<K <
54
40
4
15. The poles location for this system is shown in fig.
For system to be stable, the range of K is
1
3
(A) K >
(B) K <
54
40
(C)
K
s2
P6.2.15. The value of K is
jw
(D) Unstable
s
12. The closed loop system shown in fig. P6.2.12 become
marginally stable if the constant K is chosen to be
Page
336
(A) 30
(B) -30
(C) 10
(D) -10
Fig. P6.2.15
(A) 4
(B) -4
(C) 2
(D) -2
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UNIT 6
(A) stable
Control & System
32. The closed loop transfer function of a system is
(B) unstable
(C) marginally stable
T( s) =
(D) More information is required.
( s + 8)( s + 6)
s 5 - s 4 + 4 s 3 - 4 s 2 + 3s - 2
The number of poles in RHP and in LHP are
27. The forward path transfer of ufb system is
1
G( s) = 2 2
4 s ( s + 1)
(A) 4, 1
(B) 1, 4
(C) 3, 2
(D) 2, 3
33. The closed loop transfer function of a system is
The system is
(A) stable
(B) unstable
T( s) =
(C) marginally stable
s 3 + 3s 2 + 7 s + 24
s - 2 s 4 + 3s 3 - 6 s 2 + 2 s - 4
5
The number of poles in LHP, in RHP, and on jw
(D) More information is required
axis are
28. The forward-path transfer function of a ufb system is
G( s) =
G( s)
2 s4 + 5 s3 + s2 + 2 s
(B) 0, 1, 4
(C) 1, 0, 4
(D) 1, 2, 2
34. For the system shown in fig. P6.2.34. the number
The system is
(A) stable
(A) 2, 1, 2
of poles on RHP, LHP, and imaginary axis are
(B) unstable
R(s)
(C) marginally stable
+
C(s)
507
s4 + 3s3 + 10s2 + 30s + 169
(D) more information is required.
1
s
29. The open loop transfer function of a system is as
Fig. P6.2.34
K ( s + 0.1)
s( s - 0.2)( s 2 + s + 0.6)
G( s) H ( s) =
(A) 2, 3, 0
(B) 3, 2, 0
(C) 2, 1, 2
(D) 1, 2, 2
The range of K for stable system will be
(A) K > 0.355
(B) 0.149 < K < 0.355
35. A Routh table is shown in fig. P6.2.36. The location
(C) 0.236 < K < 0.44
(D) K > 0.44
of pole on RHP, LHP and imaginary axis are
30. The open-loop transfer function of a ufb control
system is given by
G( s) =
K
s( sT1 + 1)( sT2 + 1)
s7
1
2
s5
1
2
s5
3
4
s4
1
-1
For the system to be stable the range of K is
æ 1
1
(A) 0 < K < çç
+
è T1 T2
ö
÷÷
ø
æ 1
1
(B) K > çç
+
è T1 T2
(C) 0 < K < T1 T2
ö
÷÷
ø
(D) K > T1 T2
(A) 1, 2, 4
(B) 1, 6, 0
(C) 1, 0, 6
(D) None of the above
36. For the open loop system of fig. P6.2.35 location of
31. The closed loop transfer function of a system is
poles on RHP, LHP, and an jw-axis are
s + 4 s + 8 s + 16
s + 3s 4 + 5 s 2 + s + 3
3
T( s) =
Fig. P6.2.35
2
R(s)
5
Page
338
(A) 3, 2
(B) 2, 3
(C) 1, 4
(D) 4, 1
C(s)
Fig. P6.2.35
The number of poles in right half-plane and in left
half-plane are
-8
s6 + s5 - 6s4 + s2 + s - 6
(A) 3, 3, 0
(B) 1, 3, 2
(C) 1, 1, 4
(D) 3, 1, 2
************
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UNIT 6
11. (C) T( s) =
=
Control & System
K 2
æK 2ö
+ . Then apply feedback formula with ç 2 + ÷ and
2
s
s
sø
ès
1
, and then multiply with s 2 .
( s + 1)
G( s)
1 + G( s)
K ( s - 2)( s + 4)( s + 5)
Ks + (7 K + 1) s 2 + 2 Ks + ( 3 - 40 K )
3
æK 2ö
s2 ç 2 + ÷
2 s 4 + ( K + 2) s 3 + Ks 2
sø
ès
T( s) =
=
1 æK 2ö
s3 + s2 + 2 s + K
1+
ç 2 + ÷
s + 1è s
sø
Routh table is as shown in fig. S.6.211
s3
K
2K
s2
7K + 1
3 - 40 K
2
s1
54 K - K
7K + 1
s0
3 - 40 K
15. (C) Denominator = s 3 + s 2 + 2 s + K
Routh table is as shown in fig. S.6.2.15
Fig. S.6.2.11
K > 0,
7K + 1 > 0
Þ
54 K 2 - K
>0
7K + 1
Þ
3 - 40 K > 0
Þ
12. (A) T( s) =
1ü
K >- ï
7
ï
1 ï
K >
ý
54 ï
3 ï
K <
40 ïþ
Þ
1
3
<K <
54
40
1
5
s2
1
K
s1
2-K
s0
K
Fig. S.6.2.15
Row of zeros when K = 2,
s 2 + 2 = 0,
1
s + 6s + 5s + K
3
s3
Þ
s = -1, j 2 , - j 2
2
16. (D) Applying the feedback formula on the inner loop
Routh table is as shown in fig. S.6.212
s3
1
5
s2
6
K
s1
30 - K
s0
K
and multiplying by K yield
K
,
Ge ( s) =
s( s 2 + 5 s + 7)
T( s) =
K
s + 5s + 7s + K
3
2
17. (B) Routh table is as shown in fig. S.6.2.17
Fig. S.6.2.12
K ( s + 10)( s + 20)
13. (D) T( s) = 3
s + ( K + 2) s 2 + 30 Ks + 200 K
s3
1
7
s2
5
K
Routh table is as shown in fig. S.6.2.13
s1
35- K
5
s0
K
s3
1
30K
s2
K +2
200K
s1
30 K 2 - 140 K
s0
200K
Fig. S.6.2.17
K >0 ,
K < 35
Auxiliary equation 5 s 2 + 35 = 0,
200 K > 0 ®
K > 0, 30 K 2 - 140 K > 0
14
Þ K > , 5 satisfy this condition.
3
Page
340
Þ
18. (C) At K = 35 system will oscillate.
Fig. S.6.2.13
14. (B) First combine the parallel loop
35 - K
>0
5
K
2
and
giving
2
s
s
Þ
19. (B) For inner loop
K
K
,
Gi ( s) =
=
( s - a)( s + 3a)( s + 4 a) P ( s)
For outer loop, Go( s) = Ti ( s) =
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s=± j 7
Ti ( s) =
K
,
P ( s) + K
K
P ( s) + K
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Stability
To( s) =
K
,
P ( s) + 2 K
Therefore if inner loop is stable for X < K < Y , then
outer loop will be stable for X < 2 K < Y
X
Y
.
Þ
<K <
2
2
20. (D) T( s) =
2K - 1
>0
2
18 K - 109
>0
2K - 1
Chap 6.2
Þ
Þ
1 ü
2 ï
ý
109 ï
K >
18 þ
K >
K ( s + 2)
s 4 + 3s 3 - 3s 2 + ( K + 3) s + (2 K - 4)
Routh table is as shown in fig. S.6.2.23
2K - 4
s4
1
20
s3
9
K
K
1
-3
s3
3
K +3
s2
180 - K
9
s2
- ( K +312 )
2K - 4
s1
K ( K - 99)
K -180
s1
K ( K + 33)
K + 12
s0
K
s0
2K - 4
For stability 0 < K < 99
-( K + 12)
> 0 Þ K < - 12, 2 K - 4 > 0
3
24. (C) T( s) =
Þ K > 2 and K > -33, These condition can not be met
simultaneously. System is unstable
for any value of K
.
2
2
s4
1
-3
s3
3
K +3
s2
- K + 12
3
2K - 4
4
1
1
3
K
1
s1
K ( K + 33)
K + 12
K -1
K
1
s0
2K - 4
s
s
K ( s + 2)
s + 3s - 3s + ( K + 3) s + (2 K - 4)
4
Routh table is as shown in fig. S.6.2.24
21. (D) Routh table is as shown in fig. S.6.2.21
1
1
K -1 - K 2
K -1
0
2K - 4
Fig. S.6.2.24
For K < - 33, 1 sign change
1
For -33 < K < - 12, 1 sign change
Fig. S.6.2.21
K > 0, K - 1 > 0 Þ
K
Fig. S.6.2.23
Fig. S.6.2.20
s2
109
18
s4 + 9 s3 + 20 s2 + Ks + K = 0
s4
s
K >
23. (B) Characteristic equation
Routh table is as shown in fig. S.6.2.20
s
Þ
K >1 ,
For -12 < K < 0, 1 sign change
K -1 - K2
> 0,
K -1
For 0 < K < 2,
3 sign change
But for K > 1 third term is always -ive. Thus the three
For K > 2,
condition cannot be fulfilled simultaneously.
Therefore K > 2 yield two RHP pole.
22. (D) Routh table is as shown in fig. S.6.2.22
25. (B) Routh table is as shown in fig. S.6.2.25
2 sign change
s4
1
8
9
s3
4
20
25
s2
3
15
18 K -109
2 K -1
s1
6
25
s0
15
s4
1
4+K
s3
2
s2
2 K -1
2
s1
s0
25
Fig. S.6.2.22
15
ROZ
Fig. S.6.2.25
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UNIT 6
P ( s) = 3s 2 + 15,
d P ( s)
= 6 s, No sign change from s 2 to s 0
ds
on jw-axis 2 roots, RHP 0,
Þ
s
s
3
2
s1
s
0
Routh table is as shown in fig. S.6.2.29
2
0.4
s3
0.8
K - 0.12
0.1K
35
10
50
s2
0.55 - 125
. K
30
264
s1
-1 .25K 2 + 0 .63K - 0 .066
0 .55-1 .25K
s0
0.1K
-386
264
s4
1
ROZ
264
K > 0, 0.55 - 125
. K > 0 Þ K < 0.44
Two sign change. RHP-2 poles. System is not stable.
27. (C) Closed loop transfer function
G( s)
1
T( s) =
=
1 + G( s) 4 s 4 + 4 s 2 + 1
-125
. K 2 + 0.63K - 0.066 > 0
( K - 0.149)( K - 0.355) < 0, 0.149 < K < 0.355
30. (A) Characteristic equation
Routh table is as shown in fig. S.6.2.27
s( sT1 + 1)( sT2 + 1) + K = 0
s4
4
4
1
s3
16
8
ROZ
s2
2
1
s1
46
s0
1
T1 T2 s3 + ( T1 + T2 ) s2 + s + K = 0
Routh table is as shown in fig. S.6.2.30
s3
T1 T2
1
s2
T1 + T2
K
Fig. S.6.2.27
s1
( T1 + T2 ) - T1 T2K
T1 + T2
dp( s)
= 16 s 3 + 3s
ds
s0
K
ROZ
Fig. S.6.2.30
There is no sign change. So all pole are on jw–axis. So
æ 1
1 ö
÷÷
0 < K < çç
+
T
T
è 1
2 ø
system is marginally stable.
K > 0, ( T1 + T2 ) - T1 T2 K > 0
28. (B) Closed loop transfer function
G( s)
1
T( s) =
= 4
3
1 + G( s) 2 s + 5 s + s 2 + 2 s + 1
31. (B) Routh table is as shown in fig. S.6.2.31
Routh table is as shown in fig. S.6.2.28
Þ
s5
1
5
1
s4
3
4
3
s4
2
1
s3
5
2
s3
3.67
0
s2
1
5
1
s2
4
3
s1
-23
s1
-2.75
s0
1
s0
3
Fig. S.6.2.28
Page
342
1
Fig. S.6.2.29
Fig. S.6.2.26
P ( s) = 4 s 4 + 4 s + 1,
s( s - 0.2)( s 2 + s + 0.6) + K ( s + 0.1) = 0
s 4 + 0.8 s 3 + 0.4 s 2 + ( K - 0.12) s + 0.1K = 0
Routh table is as shown in fig. S.6.2.26
s
2 RHP poles so unstable.
29. (B) The characteristic equation is 1 + G( s) H ( s) = 0
LHP 2.
26. (B) Closed-loop transfer function is
G( s)
240
T( s) =
= 4
3
s + 10 s + 35 s 2 + 50 s + 264
1 + G( s)
4
Control & System
1
In RHP -2 poles. In LHP -3 poles.
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Stability
32. (C) Routh table is as shown in fig. S.6.2.32
e
+
-
+
+
s5
s
- -
4
3
4
-1
-4
-2
3
e
1
-2
s
+
-
s2
1-4e
e
+
+
s1
2e2 +1-4e
1-4e
-
-
s0
-2
From s 4 row down to s 0 there is one sign change. So
LHP–1 + 1 = 2 pole. RHP–1 pole, jw-axis -2 pole.
35. (A) Notice that in s 5 row there would be zero. In this
row coefficient of
Corresponding to this pole there is a pole on LHP. Rest
4 out of 6 poles are on imaginary axis. Rest 1 pole is on
LHP.
36. (A) Routh table is as shown in fig. S.6.2.36
1
3
2
-2
-6
-4
-2
-3
ROZ
2
-3
-4
1
-
s
s
s
3
s
s
e
1
3
-4
0
Fig. S.6.2.33
dP ( s)
P ( s) = - 2 s - 6 s ,
= - 8 s 3 - 12 s , -2 , -3
ds
4
2
4
0
No sign change exist from the s row down to the s row.
+ + +
+
+
-
-
-
-
+
-
-
+
-
-
Thus, the even polynomial does not have RHP poles. Therefore
4 pole
RHP
1 pole
LHP
0 pole
s6
1
-6
s5
1
0
s4
-6
0
s3
-24
0
s2
e
s1
- 144
e
s0
-6
-6
ROZ
Fig. S.6.2.36
P ( s) = - 6 s 4 - 6,
because of symmetry all four poles must be on jw-axis.
jw-axis
P ( s) = s6 + 2 s 4 - s 2 - 2
Corresponding to this pole there is a pole on LHP.
33. (B) Routh table is as shown in fig. S.6.2.33
4
, where
there is one sign change. So there is one pole on RHP.
2 LHP poles.
s5
dP ( s )
ds
have been entered. From s6 to row down to the s 0 row,
Fig. S.6.2.32
3 RHP,
dP ( s)
= 12 s 3 + 60
ds
P ( s) = 3s 4 + 30 s 2 + 507,
1
-
+
Chap 6.2
dP ( s)
= -24 s 3 ,
ds
There is two sign change from the s 4 row down to the s 0
(1 sign change)
row. So two roots are on RHS. Because of symmetry rest
two roots must be in LHP. From s6 to s 4 there is 1 sign
change so 1 on RHP and 1 on LHP.
34. (D) Closed loop transfer function
G( s)
T( s) =
1 + G( s) + H ( s)
=
Total
LHP 3 root,
RHP 3 root.
507 s
s 5 + 3s 4 + 10 s 3 + 30 s 2 + 169 s + 507
Routh table is as shown in fig. S.6.2.34
***********
5
1
10
69
s4
3
30
57
s3
12
60
ROZ
s2
15
507
s1
-345.6
s0
507
s
Fig. S.6.2.33
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Time Response
8. A system is shown in fig. P6.3.8. The rise time and
Chap 6.3
d. T( s) =
s+2
s2 + 9
e. T( s) =
( s + 5)
( s + 10) 2
settling time for this system is
R(s)
1
s
10
(s + 10)
C(s)
Consider the following response
Fig. P6.3.8
(A) 0.22s, 0.4s
(B) 0.4s, 0.22s
(C) 0.12s, 0.4s
(D) 0.4s, 0.12s
1. Overdamped
2. Under damped
3. Undamped
4. Critically damped.
The correct match is
9. For a second order system settling time is Ts = 7 s and
1
2
3
4
peak time is Tp = 3 s. The location of poles are
(A)
a
c
d
e
(A) -0.97 ± j0.69
(B) -0.69 ± j0.97
(B)
b
a
d
e
(C) -1.047 ± j0.571
(D) -0.571 ± j1.047
(C)
c
a
e
d
(D)
c
b
e
d
10. For a second order system overshoot = 10% and
peak time Tp = 5 s. The location of poles are
(A) -0.46 ± j0.63
(B) -0.63 ± j0.46
(C) -0.74 ± j0.92
(D) -0.92 ± j0.74
15. The forward-path transfer of a ufb control system is
G( s) =
11. For a second-order system overshoot = 12 % and
settling time = 0.6 s. The location of poles are
(A) -9.88 ± j6.67
(B) -6.67 ± j9.88
(C) -4.38 ± j6.46
(D) -6.46 ± j4.38
The step, ramp, and parabolic error constants are
(A) 0, 1000, 0
(B) 1000, 0, 0
(C) 0, 0, 0
(D) 0, 0, 1000
16. The open-loop transfer function of a ufb control
Statement for Q.12–13:
A system has a damping ratio of 1.25, a natural
system is
G( s) =
frequency of 200 rad/s and DC gain of 1.
12. The response of the system to a unit step input is
(A) 1 +
(C) 1 +
1000
(1 + 0.1s)(1 + 10 s)
5 -50 t 2 -150 t
e
- e
3
3
(B) 1 -
1 -100 t 4 -400 t
e
- e
3
3
(D) 1 +
4 -100 t 1 -400 t
e
+ e
3
3
2 -50 t 5 -150 t
e
- e
3
3
K (1 + 2 s)(1 + 4 s)
s 2 ( s 2 + 2 s + 8)
The position, velocity and acceleration error
constants are respectively
K
, 0
8
(A) 0, 0, 4K
(B) ¥ ,
(C) 0, 4 K , ¥
(D) ¥, ¥,
K
8
13. The system is
(A) overdamped
(B) under damped
17. The open-loop transfer function of a unit feedback
(C) critically damped
(D) None of the above
system is
G( s) =
14. Consider the following system
a. T( s) =
5
( s + 3)( s + 6)
10( s + 7)
b. T( s) =
( s + 10)( s + 20)
20
c. T( s) = 2
s + 6 s + 144
50
(1 + 0.1s)(1 + 2 s)
The position, velocity and acceleration error
constants are respectively
(A) 0, 0, 250
(B) 50, 0, 0
(C) 0, 250, ¥
(D) ¥, 50, 0
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UNIT 5
Control Systems
Statement for Q.18–19:
D(s)
The forward-path transfer function of a unity
R(s)
feedback system is
G( s) =
1
(s + 5)
+
100
s(s + 2)
+
+
C(s)
K
s ( s + a)
n
Fig. P6.3.22
The system has 10% overshoot and velocity error
constant K v = 100.
(A) -
49
11
(B)
49
11
18. The value of K is
(C) -
63
11
(D)
63
11
(A) 237 ´ 10
3
(B) 144
(C) 14.4 ´ 10
(D) 237
3
23. The forward path transfer function of a ufb system
19. The value of a is
(A) 237
. ´ 10
is
3
(C) 14.4 ´ 10
(B) 237
3
G( s) =
(D) 144
20. For the system shown in fig. P6.3.20 the steady
state error component due to unit step disturbance is
0.000012 and steady state error component due to unit
ramp input is 0.003. The values of K 1
and K 2
are
K
s( s + 4)( s + 8)( s + 10)
If a unit ramp is applied, the minimum possible
steady-state error is
(A) 0.16
(B) 6.25
(C) 0.14
(D) 7.25
respectively
24. The forward-path transfer function of a ufb system
D(s)
R(s)
K1(s + 2)
(s + 3)
+
is
K2
s(s + 4)
+
+
C(s)
G( s) =
1000( s 2 + 4 s + 20)( s 2 + 20 s + 15)
s 3( s + 2)( s + 10)
The system has r ( t) = t 3 applied to its input. The
steady state error is
Fig. P6.3.20
(A) 16.4, 1684
(B) 1250, 2.4
(C) 125 ´ 10 , 0.016
(D) 463, 3981
3
21. The transfer function for a single loop nonunity
(A) 4 ´ 10 -4
(B) 0
(C) ¥
(D) 2 ´ 10 -5
25. The transfer function of a ufb system is
feedback control system is
G( s) =
G( s) =
1
1
, H ( s) =
s2 + s + 2
( s + 1)
10 5( s + 3)( s + 10)( s + 20)
s( s + 25)( s + a)( s + 30)
The value of a to yield velocity error constant
The steady state error due to unit step input is
6
6
(A)
(B)
7
5
2
(C)
3
(B) 0
(C) 8
(D) 16
26. A system has position error constant K p = 3. The
error due to a unit step input and a unit step
Page
346
(A) 4
(D) 0
22. For the system of fig. P6.3.22 the total steady state
disturbance is
K v = 10 4 is
steady state error for input of 8tu( t) is
(A) 2.67
(B) 2
(C) ¥
(D) 0
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Time Response
27. The forward path transfer function of a unity
If
Chap 6.3
the
r ( t) = 1 + t +
feedback system is
G( s) =
For input of 60u( t) steady state error is
(B) 300
(C) ¥
(D) 10
E(s)
G(s)
subjected
(B) 0.1
(C) 10
(D) ¥
to
an
input
32. The system shown in fig. P6.3.32 has steady-state
error 0.1 to unit step input. The value of K is
R(s)
function is
+
is
(A) 0
28. For ufb system shown in fig. P6.3.28 the transfer
R(s)
system
t 2 , t > 0 the steady state error of the
system will be
1000
( s + 20)( s 2 + 4 s + 10)
(A) 0
1
2
+
C(s)
K
(s + 1)(0.1s + 1)
C(s)
Fig. P6.3.32
Fig. P6.3.28
G( s) =
20( s + 3)( s + 4)( s + 8)
s 2 ( s + 2)( s + 15)
(B) 0
(C) ¥
(D) 64
(B) 0.9
(C) 1.0
(D) 9.0
Statement for Q.33–34:
If input is 30 t 2 , then steady state error is
(A) 0.9375
(A) 0.1
Block diagram of a position control system is
shown in fig.P6.3.33–34.
29. The forward-path transfer function of a ufb control
R(s)
+
Ka
1
s(0.5s + 1)
+
C(s)
system is
G( s) =
450( s + 8)( s + 12)( s + 15)
s( s + 38)( s 2 + 2 s + 28)
sKt
Fig. P6.3.33–34
The steady state errors for the test input 37tu( t) is
(A) 0
(B) 0.061
(C) ¥
(D) 609
unit ramp input is
30. In the system shown in fig. P6.3.30, r ( t) = 1 + 2 t,
t > 0. The steady state error e( t) is equal to
r(t)
+
10(s + 1)
s2(s + 2)
e(t)
1
5
(B) 5
(C) 0
(D) ¥
G( s) =
10(1 + 4 s)
s 2 (1 + s)
(B) 0.2
(C) ¥
(D) 0
0.7 without affecting the steady state error, then the
c(t)
value of K a and K t are
(A) 86, 12.8
(B) 49, 9.3
(C) 24.5, 3.9
(D) 43, 6.4
35. A system has the following transfer function
G( s) =
31. A ufb control system has a forward path transfer
function
(A) 5
34. If the damping ratio of the system is increased to
Fig. P6.3.30
(A)
33. If K t = 0 and K a = 5, then the steady state error to
100( s + 15)( s + 50)
s 4 ( s + 12)( s 2 + 3s + 10)
The type and order of the system are respectively
(A) 7 and 5
(B) 4 and 5
(C) 4 and 7
(D) 7 and 4
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Time Response
SOLUTIONS
Chap 6.3
10
1
1
= s( s + 10) s s + 10
8. (A) C( s) =
Þ
1. (D) Characteristic equation is s + 9 s + 18.
c( t) = 1 - e -10 t
2
a = 10,
w = 18, 2 xwn = 9
2
n
Therefore x = 106
. ,
wn = 4.24 rad/s
Settling time Ts =
1
0.6
6 ( s + 0.8) 2 + (0.6) 2
2. (A) T( s) =
wn 1 - x2 = 0.6 ,
Hence wn = 1,
9. (D) xwn =
xwn = 0.8
Ds = { s - ( -3 + j 4)}{ s - ( -3 - j 4)} = ( s + 3) + 4 .
2
= s + 6 s + 25, w = 25
Þ
2 xwn = 6 ,
x=
wn = 2 ,
xp
1-x
2
T( s) =
=3
xp
1 - x2
Þ
=
x = 0.59
-
xp
1 - x2
Þ
x = 0.56, wn =
4
= 1192
.
xTs
Note :
32
.
, For 0 < x < 0.69
Ts =
xwn
5
= 0.05,
100
Ts =
4.5
, For
xwn
12. (B) T( s) =
1
= 1.45
0.69
=
Peak time,
Tp =
Þ
Therefore Poles = - xwn ± jwn 1 - x2 = - 6.67 ± j9.88
x = 0.5
2 xwn = 2,
x = 0.69,
wn =
xp
1 - x2
1 - x2 = 0.779,
11. (B) 0.12 = e
1
K
= 2
1 + G( s) s + 2 s + K
2 xwn = 2,
p
Tp
-
Poles = - xwn ± jwn 1 - x2 = - 0.46 ± j0.63
w2n = 4
5. (D) M p = e
wn =
6
= 0.6
2´5
16
4
=
( 4 s 2 + 8 s + 16) ( s 2 + 2 s + 4)
-
2
wn = 5 rad/s
4. (A) T( s) =
Þ
4
= 0.4s
a
4
p
= 0.571, wn 1 - x2 =
= 1047
.
Ts
Tp
10. (A) 0.1 = e
3. (A) Characteristic equation is
2
n
2.2 2.2
=
= 0.22s
a
10
Poles = - 0.571 ± j1.047
x = 0.8
2
Rise time Tr =
p
p
p
=
=
= 3 sec
wd wn - x2
1.45 (1 - 0.69 2 )
40000
w2n
= 2
2
s + 2 xwn s + wn
s + 500 s + 40000
40000
( s + 100)( s + 400)
R( s) =
But the peak time Tp given is 1 sec. Hence these two
x > 0.69
40000
1
4
1
= +
s( s + 100)( s + 400) s 3( s + 100) 3( s + 400)
r ( t) = 1 -
4 -100 t 1 -400 t
e
+ e
3
3
specification cannot be met.
13. (A) System has two different poles on negative real
K1
,
6. (C) T( s) = 2
s + ( K 2 + s) + K 1
w2n = K 1 ,
14. (A) 1. Overdamped response (a, b)
2 xwn = 1 + K 2
wd = 0.10,
wn = 12.5
axis. So response is over damped.
x = 0.6,
Þ
wd = wn 1 - 0.6 2 = 10
2. Underdamped response (c)
K 1 = 156.25,
Poles : Two complex in left half plane
2 wn 3 = K 2 + 1
2 ´ 12.5 ´ 0.6 = K 2 + 1
7. (A) M p = e
-
Þ
K 2 = 14
xp
1 - x2
Poles : Two real and different on negative real axis.
3.Undamped response (d)
Poles : Two imaginary.
4.Critically damped (e)
, At x = 0,
M p = 1 = 100%
Poles : Two real and same on negative real axis.
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UNIT 5
15. (B) K p = lim G ( s) = 1000
s ®0
K v = lim sG( s) = 0
s ®0
ess = lim sE( s) = lim
s ®0
R( s) =
K a = lim s 2 G( s) = 0
s ®0
16. (D) H ( s) = 1,
K p = lim G( s) × H ( s) = ¥
s ®0
s®
s ®0
17. (B) H ( s) = 1,
K
8
K p = lim G( s) × H ( s) = 50
s ®0
K v = lim sG( s) × H ( s) = 0
ess = lim
s ®0
sR( s)
,
1 + G ( s)
1
6
=
K 1 K 2 ( s + 2)
K
1K 2
s+
( s + 3)( s + 4)
6
= 0 .003
125 ´ 10 3 K 2
Þ
K 2 = 0.016
21. (C) E( s) = R( s) - C( s) H ( s)
= R( s) -
s ®0
s ®0
1
K 1 K 2 ( s + 2)
, G( s) =
s2
s( s + 3)( s + 4)
K v = lim sG( s) × H ( s) = ¥
K a = lim s 2 G ( s) × H ( s) =
Control Systems
K a = lim s 2 G( s) × H ( s) = 0.
R( s) G( s) H ( s)
R( s)
=
1 + G( s) H ( s)
1 + G( s) H ( s)
s ®0
18. (C) System type = 1, so n = 1
K v = lim sG( s) =
s ®0
ess = lim sE( s) = lim
s ®0
K
= 100
a
s ®0
For 10% overshoot,
0.1 = = e
T( s) =
-
22. (A) ess = lim
xp
1 - x2
Þ
s ®0
x = 0.6
G( s)
K
= 2
1 + G ( s) s + as + K
2 xwn = a ,
wn2 = K
K
´ 0.6 K = 100
2
Þ
Þ
2 ´ 0.6 K = a
R( s) = D( s) =
K = 14400
K
= 100, K = 14400,
a
14400
= 100
a
Þ
a = 144
Error in output due to disturbance
E( s) = TD( s) D ( s),
1
,
s
essD = lim sE( s) = lim s ×
3
= 0.000012
2 K1
s ®0
Þ
1
3
× TD( s) = lim TD( s) =
s ®0
s
2 K1
1
s+5
and
G2 ( s) =
100
s+2
1
s
23. (A) Using Routh-Hurwitz Criterion, system is stable
2000
= 6. 25
4 ´ 8 ´ 10
1
1
=
= 0.16
K v 6.25
K v = lim sG( s) =
s ®0
24. (A) R( s) =
6
,
s4
E( s) =
R( s)
1 + G ( s)
ess = lim sE( s)
s ®0
6s
s4
= lim
2
s ®0
1000 ( s + 4 s + 20) ( s 2 + 20 s + 15)
1+
s 3( s + 2) ( s + 10)
= lim
s ®0
K 1 = 125 ´ 10 3
Error due to ramp input
Page
350
sR( s) - sD( s) G2 ( s)
1 + G1 ( s) G2 ( s)
minimum possible error
K 2 ( s + 3)
s( s + 3)( s + 4) + K 1 K 2 ( s + 2)
s ®0
2
3
100
-49
2
=
=
1 100
11
1+ ´
5
2
maximum
K2
s( s + 4)
TD( s) =
K 1 K 2 ( s + 2)
1+
s( s + 4)( s + 3)
If D( s) =
=
for 0 < K < 2000
20. (C) If R ( s) = 0
=
1
1
1+ 2
( s + s + 2) ( s + 1)
1-
ess
19. (D)
G1 ( s) =
where
s
s
=
6
1000 ( s 2 + 4 s + 20) ( s 2 + 20 s + 15)
s +
( s + 2) ( s + 10)
3
6
0+
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Time Response
25. (A) K v = lim sG( s)
10 4 ´ 3 ´ 10 ´ 20
25 ´ a ´ 30
10 4 =
Þ
a=4
1
= ¥
K v = 0, ess =
Kv
26. (C) System is zero type
34. (C) The equivalent open-loop transfer function
Ka
Ka
s(0.5 s + 1)
Ge =
=
sK t
s(0.5 s + 1 + K t )
1+
s(0.5 s + 1)
27. (D) K p = lim G( s) = 5
s ®0
60
= 10
1 + Kp
T( s) =
28. (A) K a = lim s 2 G( s) = 64
s ®0
ess =
30 ´ 2
= 0.9375
64
=
G( s)
Ka
=
1 + G( s) 0.5 s 2 + s(1 + K t ) + K a
2K a
s 2 + 2 s(1 + K t ) + 2 K a
w2n = 2 K a Þ wn =
29. (B) K v = lim sG( s) = 609.02
s ®0
ess =
1
1
s2
= lim
=
= 0.2
s ®0
5
5
1+
s(0.5 s + 1 )
s×
ess
For input 60u( t), ess =
5
1
, H ( s) = 1, R( s) = 2
s(0.5 s + 1)
s
G( s) =
s ®0
Chap 6.3
2K a
2 x wn = 2(1 + K t )
Kt
x =1 +
= 0.7
2K a
37
= 0.0607
Kv
30. (C) The system is type 2. Thus to step and ramp
ess = lim
s ®0
input error will be zero.
E( s) = R( s) - C( s) = R( s) -
G( s) R( s)
1 + G ( s)
R( s)
1 + G( s)
=
... (i)
sR( s)
1
, R ( s) = 2
1 + Ge ( s)
s
1
1 + Kt
=
Ka
ö
æ
Ka
÷÷
s çç 1 +
s(0.5 s + 1 + K t ) ø
è
1 + Kt
=
= 0. 2
Ka
ess = lim
s ®0
1 2
s+2
+
=
s s2
s2
s+2
E( s) =
10( s + 1)
s2 +
( s + 2)
R( s) =
ess
...(ii)
Solving (i) and (ii)
K a = 24.5 ,
ess ( t) = lim sE( s) = 0
s ®0
.
K t = 39
35. (C) The s has power of 4 and denominator has order
31. (C) System is type 2. Therefore error due to 1 + t
of 7. So Type 4 and Order 7.
t2
1
would be
.
would be zero and due to
2
Ka
36. (D) For 8u( t),
K a = lim s 2 G( s) = 10,
ess ( t) =
s ®0
1
= 0.1
10
ess =
8
= 2.
1 + Kp
For 8tu( t), ess = ¥, since the system is type 0.
Note that you may calculate error from the formula
ess ( t) = lim sE( s) =
s ®0
37. (A) For equivalent unit feedback system the forward
sR( s)
1 + G( s)
transfer function is
Ge =
32. (D) K p = lim G( s) = K
s ®0
1
1
ess ( t) =
=
= 0.1
1 + Kp 1 + K
Þ
K = 9.
s ®0
When K t = 0 and K a = 5
10 ( s + 10 )
s (s + 2 )
10 ( s + 10 )( s + 3)
s (s + 2 )
10( s + 10)
11s + 132 s + 300
2
The system is of Type 0. Hence step input will produce a
33. (B) ess = lim e( t) = lim sE( s) = lim
t ®¥
=
G( s)
=
1 + G( s) H ( s) - G( s) 1 +
s ®0
sR( s)
1 + G( s) H ( s)
constant error constant.
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GATE EC BY RK Kanodia
CHAPTER
6.5
FREQUENCY-DOMAIN ANALYSIS
Statement for Q.1–2:
4. The gain-phase plots of open-loop transfer function of
An under damped second order system having a
correct sequence of the increasing order of stability of
transfer function of the form
T( s) =
four different system are shown in fig. P6.5.4. The
these four system will be
Kw2n
2
s + 2 xwn s + w2n
dB
B
A
C
40 dB
has a frequency response plot shown in fig.
D
30 dB
P6.5.1–2.
20 dB
10 dB
|T( jw )|
-270o
2.5
-225
o
o
-135
o
-90
o
-45
-20 dB
-30 dB
1.0
w
wn
Fig. P6.5.4
Fig. P6.5.1-2
1. The system gain K is
(A) 1
(C)
(B) 2
(D)
2
(A) D, C, B, A
(B) A, B, C, D
(C) B, C, A, D
(D) A, D, B, C
1
5. The open-loop frequency response of a
2
feedback system is shown in following table
2. The damping factor x is approximately
w
|G ( jw)|
ÐG ( jw)
(A) 0.6
(B) 0.2
2
8.5
-119°
(C) 1.8
(D) 2.4
3
6.4
-128°
4
4.8
-142°
5
2.56
-156°
6
1.4
-164°
8
1.00
-172°
10
0.63
-180°
3. For the transfer function
G( s) H ( s) =
1
s( s + 1)( s + 0.5)
the phase cross-over frequency is
(A) 0.5 rad/sec
(B) 0.707 rad/sec
(C) 1.732 rad/sec
(D) 2 rad/sec
Fig. P6.5.5
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Frequency-Domain Analysis
The gain margin and phase margin of the system
Chap 6.5
10. The gain margin of the ufb system
are
(A) 2 dB, 8°
(B) 2 dB, -172 °
(C) 4 dB, 8°
(D) 4 dB, -172 °
G( s) =
Statement for Q.6–7:
Consider the gain-phase plot shown in fig.
2
is
( s + 1)( s + 2)
(A) 1.76 dB
(B) 3.5 dB
(C) -3.5 dB
(D) -1.76 dB
11. The open-loop transfer function of a system is
K
s(1 + 2 s)(1 + 3s)
G( s) H ( s) =
P6.5.6–7.
dB
w=2
2 dB
0
The phase crossover frequency is
(B) 2.46 rad/sec
(C) 0.41 rad/sec
(D) 3.23 rad/sec
ÐG( jw)
w = 10
-2 dB
(A) 6 rad/sec
12. The open-loop transfer function of a ufb system is
w = 100
o
-270
o
G( s) =
o
-180
-140
o
-90
Fig. P6.5.6-7
1+ s
s(1 + 0.5 s)
The corner frequencies are
6. The gain margin and phase margin are
(A) 0 and 2
(B) 0 and 1
(A) -2 dB, 40°
(B) 2 dB, 40°
(C) 0 and -1
(D) 1 and 2
(C) 2 dB, 140°
(D) -2 dB, 140°
13. In the Bode-plot of a unity feedback control system,
7. The gain crossover and phase crossover frequency are
the value of magnitude of G( jw) at the phase crossover
respectively
frequency is
(A) 10 rad/sec, 100 rad/sec
(A) 2
1
2
(B) 100 rad/sec, 10 rad/sec
(C)
(C) 10 rad/sec, 2 rad/sec
. The gain margin is
1
(B)
2
1
3
(D) 3
(D) 100 rad/sec, 2 rad/sec
8. The phase margin of a system with the open loop
transfer function
14. In the Bode-plot of a ufb control system, the value of
phase of G( jw) at the gain crossover frequency is -120 °.
The phase margin of the system is
G( s) H ( s) =
(1 - s)
is
(1 + s)( 3 + s)
(A) 68.3°
(B) 90°
(C) 0
(D) ¥
(A) -120 °
(B) 60°
(C) -60 °
(D) 120°
15. The transfer function of a system is given by
G( s) =
9. Consider a ufb system having an open-loop transfer
function
K
1
; K <
s( sT + 1)
T
The Bode plot of this function is
G( s) =
K
s(0.2 s + 1)(0.05 s + 1)
dB
-20 dB/dec
For K = 1, the gain margin is 28 dB. When gain
margin is 20 dB, K will be equal to
(A) 2
(B) 4
(C) 5
(D) 2.5
dB
-40 dB/dec
-40 dB/dec
0 dB
0.1
T
1
T
(A)
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0.1
T
1
T
w
(B)
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UNIT 6
dB
dB
-20 dB/dec
(A)
100
s + 10
(B)
(C)
1
s + 10
(D) None of the above
-20 dB/dec
0 dB
w
1
T
0.1
T
0 dB
1
T
0.1
T
w
Control & System
10
s + 10
-40 dB/dec
19. Consider the asymptotic Bode plot of a minimum
(C)
(D)
phase linear system given in fig. P6.5.19. The transfer
function is
dB
16. The asymptotic approximation of the log-magnitude
versus frequency plot of a certain system is shown in
32
fig. P6.5.16. Its transfer function is
-20 dB/dec
dB
54 dB
6
-40 dB/dec
-20 dB/dec
w1
0.1
-60 dB/dec
w2
w
10
-40 dB/dec
-40 dB/dec
-60 dB/dec
0.1
5
2
w
25
(C)
Fig. P6.5.16
50( s + 5)
(A) 2
s ( s + 2)( s + 25)
(C)
10 s 2 ( s + 5)
( s + 2)( s + 25)
Fig. P6.5.19
8 s( s + 2)
(A)
( s + 5)( s + 10)
(B)
20( s + 5)
s 2 ( s + 2)( s + 25)
(D)
20( s + 5)
s( s + 2)( s + 25)
4( s + 2)
s( s + 5)( s + 10)
(B)
4( s + 5)
( s + 2)( s + 10)
(D)
8 s( s + 5)
( s + 2)( s + 10)
20. The Bode plot shown in fig. P6.5.20 represent
dB
100 dB
17. For the Bode plot shown in fig. P6.5.17 the transfer
-60 dB/dec
function is
40 dB/dec
dB
4
-4
100
( s + 4)( s + 10)
100 s 2
(1 + 0.1s) 3
(B)
1000 s 2
(1 + 0.1s) 3
100 s 2
(1 + 0.1s) 5
(D)
1000 s 2
(1 + 0.1s) 5
c
-2
0
de
B/
dB
/d
0d
(C)
w
Fig. P6.5.20
(A)
(B)
100( s + 4)
s( s + 10) 2
(C)
(D)
100
s ( s + 4)( s + 10)
Statement for Q.21–22:
Fig. P6.5.17
100 s
(A)
( s + 4)( s + 10) 2
w = 10
w
10
ec
0 dB
2
18. Bode plot of a stable system is shown in the fig.
P6.3.18. The open-loop transfer function of the ufb
The Bode plot of the transfer function K (1 + sT) is
given in the fig. P6.5.21–22.
dB
Phase
system is
dB
-20 dB/dec
20 dB
-20 dB/dec
1
T
w
w
-45
10
T
°/d
e
c
Fig. P6.5.21-22
Fig. P6.5.18
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364
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T
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GATE EC BY RK Kanodia
UNIT 6
Statement for Q.29–30:
Control & System
Im
Im
2
2
Consider the Bode plot of a ufb system shown in
fig. P6.5.29–30.
w= 0
w=¥
Re
w= 0
w=¥
Re
dB
32 dB
0 dB
(A)
-20 dB/dec
18 dB
-40 dB/dec
w1
0.1
(B)
Im
w
Im
2
2
Fig. P6.5.29-30
Re
w= 0
w=¥
w= 0
w=¥
Re
29. The steady state error corresponding to a ramp
input is
(A) 0.25
(B) 0.2
(C) 0
(D) ¥
(C)
(D)
35. Consider a ufb system whose open-loop transfer
30. The damping ratio is
function is
(A) 0.063
(B) 0.179
(C) 0.483
(D) 0.639
G( s) =
K
s( s 2 + 2 s + 2)
The Nyquist plot for this system is
31. The Nyquist plot of a open-loop transfer function
Im
G( jw) H ( jw) of a system encloses the ( -1, j0) point. The
Im
gain margin of the system is
(A) less than zero
(B) greater than zero
(C) zero
(D) infinity
w=¥
Re
Re
w=¥
w= 0
32. Consider a ufb system
G( s) =
K
s(1 + sT1 )(1 + sT2 )(1 + sT3)
w= 0
(A)
(B)
The angle of asymptote, which the Nyquist plot
approaches as w ® 0, is
Im
(A) -90 °
(B) 90°
(C) 180°
(D) -45 °
Im
w=¥
33. If the gain margin of a certain feedback system is
Re
Re
w=¥
w= 0
given as 20 dB, the Nyquist plot will cross the negative
real axis at the point
w= 0
(A) s = -0.05
(B) s = -0.2
(C) s = -0.1
(D) s = -0.01
(C)
34. The transfer function of an open-loop system is
G( s) H ( s) =
36. The open loop transfer function of a system is
s+2
( s + 1)( s - 1)
G( s) H ( s) =
K (1 + s) 2
s3
The Nyquist plot for this system is
The Nyquist plot will be of the form
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366
(D)
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Frequency-Domain Analysis
Chap 6.5
Im
Im
Im
w= 0
w= 0
Re
w=¥
10.64
Re
w=¥
w=¥
Re
w= 0
(A)
(B)
Im
Im
Fig. P6.5.38
The no. of poles of closed loop system in RHP are
w=¥
w=¥
Re
Re
(C)
(B) 1
(C) 2
D) 4
Statement for Q.39–40:
w= 0
w= 0
(A) 0
The open-loop transfer function of a feedback
(D)
control system is
37. For the certain unity feedback system
G( s) H ( s) =
K
G( s) =
s( s + 1)(2 s + 1)( 3s + 1)
-1
2 s(1 - 20 s)
39. The Nyquist plot for this system is
The Nyquist plot is
Im
Im
Im
Im
w= 0
w= 0
w=¥
w=¥
Re
Re
w=¥
Re
Re
w=¥
w= 0
w= 0
(A)
(A)
(B)
Im
(B)
Im
Im
Im
w=¥
w=¥
Re
Re
w=¥
Re
Re
w=¥
w= 0
w= 0
w= 0
w= 0
(C)
(C)
(D)
(D)
38. The Nyquist plot of a system is shown in fig.
40. Regarding the system consider the statements
P6.5.38. The open-loop transfer function is
1. Open-loop system is stable
G( s) H ( s) =
4s + 1
s ( s + 1)(2 s + 1)
2
2. Closed-loop system is unstable
3. One closed-loop poles is lying on the RHP
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UNIT 6
Control & System
Im
The correct statements are
(A) 1 and 2
(B) 1 and 3
(C) only 2
-2
3
Im
w=¥
-3
2
Re
w=¥
Re
(D) All
41. The Nyquist plot shown in the fig. P6.5.41 is for
(A) type–0 system
(B) type–1 system
(C) type–2 system
(D) type–3 system
w= 0
w= 0
(C)
Statement for Q.42–43:
The open-loop transfer function of a feedback
system is
(D)
45. The phase crossover and gain crossover frequencies
are
K (1 + s)
(1 - s)
(A) 1.414 rad/sec, 0.57 rad/sec
42. The Nyquist plot of this system is
(C) 0.707 rad/sec, 0.57 rad/sec
G( s) H ( s) =
(B) 1.414 rad/sec, 1.38 rad/sec
Im
(D) 0.707 rad/sec, 1.38 rad/sec
Im
46. The gain margin and phase margin are
w=¥
w= 0
Re
w=¥
w= 0
(A)
Re
(A) -3.52 dB, -168.5 °
(B) -3.52 dB, 11.6 °
(C) 3.52 dB, -168.5 °
(D) 3.52 dB, 11.6 °
(B)
****************
Im
Im
w=¥
w= 0
Re
w=¥
w= 0
(C)
Re
(D)
43. The system is stable for K
(A) K > 1
(B) K < 1
(C) any value of K
(D) unstable
Statement for Q.44–46:
A unity feedback system has open-loop transfer
function
G( s) =
1
s(2 s + 1)( s + 1)
44. The Nyquist plot for the system is
Im
Im
w=¥
w= 0
2
3
w=¥
(A)
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368
Re
3
2
w= 0
Re
(B)
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Frequency-Domain Analysis
At w = 10 rad/sec gain is 0 dB. Gain cross over frequency
SOLUTIONS
1. (A) T( s) =
w =10 rad/sec.
Kw2n
s + 2 xwn s + w2n
8. (D) |GH( jw)|¹ 1, for any value of w. Thus phase
2
margin is ¥.
Kw2n
T( jw) =
-w2 + 2 jxwn w + w2n
|T( jw)|
2
=
9. (D) For 28 dB gain Nyquist plot intersect the real
axis at a,
1
20 log = 28
a
K w
( w - w ) + 4 x2 w2n w2
2
2
n
4
n
2 2
From the fig. P6.5.1–2, |T( j0)| = 1
|T( j0)|
2
=
K 2 wn4
= K2 =1
wn4
Þ
denominator of function |T( jw)| is minimum i.e. when
2
Þ
w = wn
K w
K2
|T( jwn )| = 2 = 2
4x w
4x
2
x=
2
4
n
4
n
Þ
= 15
. = 35
. dB
ÐGH ( jw) = -180 °
GH ( jw) =
K
jw(1 + 2 jw)(1 + 3 jw)
-90 °- tan -1 2 wp - tan -1 3wp = -180 °
- tan -1 2 w - tan -1 w - 90 ° = -180 °
tan -1 2 wp + tan -1 3wp = 90 °
2 wp + 3wp
= tan 90 °
1 - (2 wp)( 3wp)
tan -1 2 w + tan -1 w = 90 °
2w + w
= tan 90 ° = ¥
1 - (2 w)( w)
2
-1
11. (C) For phase crossover frequency
At phase cross over point f = -180 °
w=
1
2
-1
f = -90 °- tan -1 2 w - tan -1 w
1 - (2 w) w = 0 Þ
This is achieved if the system gain is increased by factor
0.1
= 2.5. Thus K = 2.5.
0.04
é KT1 T2 ù
é(2)(0.5) ù
Gain Margin = ê
ú = ê 1 + 0.5 ú
T
+
T
ë
û
2û
ë 1
1
jw( jw + 1)(0.5 + jw)
1
a = 0.04
10. (B) Here K = 2, T1 = 1, T2 =
K
|T( jwn )| = = 2.5
2x
K
= 0.2
5
3. (B) G( jw) H ( jw) =
Þ
For 20 dB gain Nyquist plot should intersect at b,
1
20 log
= 20 Þ b = 0.1.
b
K =1
2. (B) The peak value of T( jw) occurs when the
w2n - w2 = 0
Chap 6.5
1 - 6 w2p = 0
= 0.707 rad/sec
Þ
12. (D) G( s) =
4. (B) For a stable system gain at 180° phase must be
wp = 0.41 rad/s
s+1
s+1
=
s(1 + 0.5 s)
æs
ö
sç + 1 ÷
è2
ø
negative in dB. More magnitude more stability.
The Bode plot of this function has break at w = 1 and
5. (C) At 180° gain is 0.63. Hence gain margin is
1
= 20 log
= 4 dB
0.63
w = 2. These are the corner frequencies.
13. (A) G.M. =
At unity gain phase is -172 °,
Phase margin = 180 °-172 ° = 8 °
14. (B) P.M. = 180 ° + ÐGH ( jw1 ) = 180 °-120 ° = 60 °
6. (A) At ÐG( jw) = 180 ° gain is -2 dB. Hence gain
margin is 2 dB. At 0 dB gain phase is -140 °. Hence
phase margin is 180 °-140 ° = 40 °.
7.
(A) At w = 100 rad/sec phase is 180°. Phase cross-
over frequency wp = 100 rad/sec.
1
1
= =2
GH ( jwp) 1 2
15. (D) Due to pole at origin initial plot has a slope of
1
-20 dB/decade. At s = jw = . Slope increases to -40
T
1
dB/decade. At w = ,
T
|G( jw)| » KT < 1 ,Gain
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Frequency-Domain Analysis
27. (C) Initially slope is -20 dB/decade. Hence there is a
Chap 6.5
ÐGH ( jw) = - tan -1
pole at origin and system type is 1. For type–1 system
position error coefficient is ¥.
20 log K = 6
Þ
GH ( jw) =
K = 2,
w (2 - w2 ) 2 + 4 w2
At w = 0,
28. (B) The system is type –0,
At w = ¥
1
1
.
20 log K p = 40, K p = 100, estep ( ¥) =
=
1 + K p 101
At w = 1,
-20 dB/dec
0.5
0.1
1.4
K
2 18
Ð - 206.6 °,
correct option.
4
w
36. (B) GH ( jw) =
Fig.S6.5.29
K v = 4, eramp ( ¥) =
GH ( jw) = 0 Ð - 270 °,
K
GH ( jw) =
Ð - 153.43°,
5
Due to s there will be a infinite semicircle. Hence (C) is
-40 dB/dec
18 dB
0 dB
GH ( jw) = ¥ Ð - 90 °,
At w = 2, GH ( jw) =
29. (A) The Bode plot is as shown in fig. S6.5.29
32 dB
2w
- 90 °
2 - w2
K
K (1 + w2 )
w3
|GH( jw)| =
1
1
= = 0.25
Kv 4
K (1 + jw) 2
( jw) 3
ÐGH ( jw) = -270 °+ 2 tan -1 w
For w = 0, GH ( jw) = ¥Ð - 270 °
0.5
w
30. (B) From fig. S6.5.29 x = 2 =
= 0.179
2 w3 2(1.4)
For w = 1 , ÐGH ( jw) = -180 °
For w = ¥, GH ( jw) = 0 Ð - 90 °
31. (A) If Nyquist plot encloses the point ( -1, j0), the
As w increases from 0 to ¥, phase goes -270 ° to -90 °.
Due to s 3 term there will be 3 infinite semicircle.
system is unstable and gain margin is negative.
37. (A) |GH ( jw)| =
K
32.(A) GH ( jw) =
jw(1 + jwT1 )(1 + jwT2 ) (1 + jwT2 )
,
ÐGH ( jw) = -90 °- tan -1 w - tan -1 2 w - tan -1 3w ,
K
K
lim GH ( jw) = lim
= lim
Ð - 90 °
w ®0
w ®0
jw w ®0 w
For w = 0, GH ( jw) = ¥Ð - 90 °,
Hence, the asymptote of the Nyquist plot tends to an
angle of -90 ° as w ® 0.
K
1 + w2 1 + 4 w2 1 + 9 w2
For w = ¥, GH ( jw) = 0 Ð - 360 °,
Hence (A) is correct option.
38. (C) The open-loop poles in RHP are P = 0. Nyquist
33. (C) 20 log
1
= 20
GH ( jw)
path enclosed 2 times the point ( -1 + j0). Taking
1
= 10
GH ( jw)
Þ
N = P - Z,
GH ( jw) = 0.1
Since system is stable, it will cross at s = -0.1.
clockwise encirclements as negative N = -2.
-2 = 0 - Z ,
Z = 2 which implies that two
poles of closed-loop system are on RHP.
-1
,
2 s(1 - 20 s)
s+2
34. (B) GH ( s) = 2
( s - 1)
39. (B) G( s) H ( s) =
jw + 2
GH ( jw) =
( -1 - w2 )
GH ( jw) =
At w = 0 ,
GH ( jw) = 2 Ð - 180 °
ÐGH ( jw) = 180 °-90 ° - tan -1
At w = ¥,
GH ( jw) = 0 Ð - 270 °
Hence (B) is correct option.
35. (C) GH ( jw) =
K
jw( -w + 2 jw + 2)
2
1
2 w 1 + 400 w2
-20 w
,
1
At w = 0 GH ( jw) = ¥Ð90 °
At w = ¥ GH ( jw) = 0 Ð180 °
At w = 0.1 GH ( jw) = 2.24 Ð153.43°
At w = 0.01 GH ( jw) = 49 Ð9115
. °
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UNIT 6
40. (C) One open-loop pole is lying on the RHP. Thus
Control & System
The frequency at which magnitude unity is
open-loop system is unstable and P = 1. There is one
Im
clockwise encirclement. Hence N = -1.
Z = P - N = 1 - ( -1) = 2,
-2
3
Hence there are 2 closed-loop poles on the RHP and
Re
-1 wp
system is unstable.
w1
41. (B) There is one infinite semicircle. Which represent
Phase Margin
single pole at origin. So system type is1.
K 1 + w2
42. (D) |GH ( jw)| =
1 + w2
ÐGH ( jw) = tan -1 w - tan -1 -
=K
Fig.S6.5.44
w (1 + w ) (1 + 4 w ) = 1
2
1
-w
1
At w = 2 GH ( jw) = KÐ127 °,
43. (A) RHP poles of open-loop system P = 1, Z = P - N .
For closed loop system to be stable,
w ®0
At unit gain w1 = 0.57 rad/sec,
Phase margin = -168.42 °+180 ° = 11.6 °
Note that system is stable. So gain margin and phase
margin are positive value. Hence only possible option is
H ( s) = 1
(D).
***************
1
jw(2 jw + 1)( jw + 1)
w ®0
1
= ¥Ð - 90 °
jw
lim GH ( jw) = lim w
w ®¥
w ®¥
1
= 0 Ð - 270 °
2( jw) 3
The intersection with the real axis can be calculated as
Im{ GH ( jw)} = 0, The condition gives w (2 w2 - 1) = 0
i.e. w = 0,
1
2
,
æ 1 ö -2
GH ç j
÷=
2ø 3
è
With the above information the plot in option (C) is
correct.
45. (C) The Nyquist plot crosses the negative real axis
1
rad/sec. Hence phase crossover frequency is
at w =
2
wp =
Page
372
1
2
3
= 352
. dB.
2
ÐGH ( jw) = -90 °- tan -1 w - tan -1 2 w ,
1
s(2 s + 1)( s + 1)
lim GH ( jw) = lim
2
3
ÐGH ( jw1 ) = -90 °- tan -1 0.57 - tan -1 2(0.57) = -168.42 °
( -1 + j0). It is possible when K > 1.
GH ( jw) =
|GH( jwp)| =
Phase at this frequency is
N =1
There must be one anticlockwise rotation of point
GH ( s) =
,
|GH( jwp)|
Gain Margin = 20 log
At w = ¥ GH ( jw) = KÐ180 °,
1
,
s(2 s + 1)( s + 1)
1
46. (D) G.M. = 20 log
At w = 1 GH ( jw) = KÐ90 °,
44. (C) G( s) =
2
1
w2 = 0.326, w1 = 0.57 rad/sec
At w = 0 GH ( jw) = KÐ0 °,
Z = 0, 0 = 1 - N Þ
2
1
= 0.707 rad/sec.
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CHAPTER
6.6
DESIGN OF CONTROL SYSTEMS
1. The term ‘reset control’ refers to
(C) is predictive in nature
(A) Integral control
(B) Derivative control
(D) increases the order of the system
(C) Proportional control
(D) None of the above
6. Consider the List I and List II
2. If stability error for step input and speed of response
List I
List II
be the criteria for design, the suitable controller will be
P. Derivative control
1. Improved overshoot response
(A) P controller
(B) PI controller
Q. Integral control
2. Less steady state errors
(C) PD controller
(D) PID controller
R. Rate feed back control
3. Less stable
S. Proportional control
4. More damping
3. The transfer function
1 + 0.5 s
represent a
1+ s
The correct match is
(A) Lag network
(B) Lead network
(C) Lag–lead network
(D) Proportional controller
P
Q
R
S
(A)
1
2
3
4
(B)
4
3
1
2
(C)
2
3
1
4
(D)
1
2
4
3
4. A lag compensation network
7. Consider the List–I (Transfer function) and List–II
(a) increases the gain of the original network without
affecting stability.
(Controller)
List I
List II
(b) reduces the steady state error.
P.
1. P–controller
(c) reduces the speed of response
Q.
(d) permits the increase of gain of phase margin is
acceptable.
2. PI–controller
R.
K1s + K2
K 3s
3. PD–controller
S.
K1
K 2s
4. PID–controller
In the above statements, which are correct
(A) a and b
(B) b and c
(C) b, c, and d
(D) all
The correct match is
P
Q
R
S
(A)
3
4
2
1
5. Derivative control
(B)
4
3
1
2
(A) has the same effect as output rate control
(C)
3
2
1
4
(D)
4
1
2
3
(B) reduces damping
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UNIT 6
Control Systems
8. The transfer function of a compensating network is of
(B) Two–position controller
form (1 + aTs) (1 + Ts). If this is a phase–Lag network,
(C) Floating controller
the value of a should be
(D) Proportional–position controller
(A) greater than 1
(B) between 0 and 1
14. In case of phase–lag compensation used is system,
(C) exactly equal to 1
gain crossover frequency, band width and undamped
(D) exactly equal to 0
frequency are respectively
(A) decreased, decreased, decreased
9.
The
poll–zero
configuration
of
a
phase–lead
compensator is given by
jw
(A)
(B) increased, increased, increased
(C) increased, increased, decreased
jw
(B)
(D) increased, decreased, decreased
s
s
15. A process with open–loop model
(C)
jw
(D)
jw
s
G( s) =
s
Ke - s TD
ts + 1
is controlled by a PID controller. For this purpose
10. While designing controller, the advantage of pole–
zero cancellation is
(A) the derivative mode improves transient
performance
(A) The system order is increased
(B) the derivative mode improves steady state
performance
(B) The system order is reduced
(C) the integral mode improves transient performance
(C) The cost of controller becomes low
(D) the integral mode improves steady state
performance.
(D) System’s error reduced to optimum levels
The correct statements are
11. A proportional controller leads to
(A) (a) and (c)
(B) (b) and (c)
(A) infinite error for step input for type 1 system
(C) (a) and (d)
(D) (b) and (d)
(B) finite error for step input for type 1 system
(C) zero steady state error for step input for type 1
system
(D) zero steady state error for step input for type 0
system
12. The transfer function of a phase compensator is
given by (1 + aTs) (1 + Ts) where a > 1 and T > 0. The
16. A lead compensating network
(a) improves response time
(b) stabilizes the system with low phase margin
(c) enables moderate increase in gain without
affecting stability.
(d) increases resonant frequency
In the above statements, correct are
maximum phase shift provided by a such compensator
is
æ a + 1ö
(A) tan -1 çç
÷÷
è a -1ø
(C) tan
-1
æ a -1ö
çç
÷÷
è a + 1ø
(D) cos
-1
(B) (a) and (c)
(C) (a), (c) and (d)
(D) All
17. A Lag network for compensation normally consists
æ a -1ö
çç
÷÷
è a + 1ø
of
(A) R, L and C elements
13. For an electrically heated temperature controlled
(B) R and L elements
liquid heater, the best controller is
(C) R and C elements
(A) Single–position controller
Page
374
æ a -1ö
(B) sin -1 çç
÷÷
è a + 1ø
(A) (a) and (b)
(D) R only
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GATE EC BY RK Kanodia
Design of Control Systems
18. The pole–zero plot given in fig.P6.6.18 is that of a
Chap 6.6
SOLUTIONS
jw
s
1. (A)
2. (D)
3. (A)
4. (D)
5. (B)
6. (D)
7. (A)
8. (B)
9. (A)
10. (B)
11. (C)
12. (B)
13. (C)
14. (D)
15. (C)
16. (D)
17. (C)
18. (D)
19. (D)
20. (D)
Fig. P6.6.18
(A) PID controller
(B) PD controller
(C) Integrator
21. (D)
(D) Lag–lead compensating network
19. The correct sequence of steps needed to improve
system stability is
(A) reduce gain, use negative feedback, insert
derivative action
(B) reduce gain, insert derivative action, use negative
feedback
(C) insert derivative action, use negative feedback,
reduce gain
(D) use negative feedback, reduce gain, insert
derivative action.
20. In a derivative error compensation
(A) damping decreases and setting time decreases
(B) damping increases and setting time increases
(C) damping decreases and setting time increases
(D) damping increases and setting time decreases
21. An ON–OFF controller is a
(A) P controller
(B) PID controller
(C) integral controller
(D) non linear controller
**********
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CHAPTER
6.7
THE STATE-VARIABLE ANALYSIS
1. Consider the SFG shown in fig. P6.7.1
u
1
s
1
1
s
x3
1
s
x2
x1
1
y
-1
-2
-3
Fig. P6.7.1
For this
é x& 1 ù é3
(A) ê x& 2 ú = ê0
ê ú ê
êë x& 3 úû êë3
system dynamic equation is
1 2 ù é x1 ù é0 ù
1 1ú ê x2 ú + ê0 ú u
úê ú ê ú
2 1úû êë x3 úû êë1 úû
1 0ù
é-2
ê
(A) 0 -2 0 ú
ê
ú
0 -3úû
êë 0
é2 -1 0 ù
(B) ê0 2 0 ú
ê
ú
êë0 0 3úû
1 -2 ù
é 2
ê
(C) 0 -2
0ú
ê
ú
0 úû
êë-3 0
é0 -1 2 ù
(D) ê0 2 0 ú
ê
ú
êë3 0 0 úû
3.
5
1
s
1 0 ù é x1 ù é0 ù
é x& 1 ù é 0
ê
ú
ê
(B) x& 2 = 0
0
1ú ê x2 ú + ê0 ú u
ê ú ê
úê ú ê ú
êë x& 3 úû êë-3 -2 -1ûú êë x3 úû êë1 ûú
u
-2
1
s
1
1
s
x2
1
x1
5
y
-2
1
x3
5
-3
é x& 1 ù é0 -1 0 ù é x1 ù é0 ù
(C) ê x& 2 ú = ê0 0 -1ú ê x2 ú + ê0 ú u
ê ú ê
úê ú ê ú
1úû êë x3 ûú êë1 ûú
êë x& 3 úû êë3 2
Fig. P6.7.3
(D) None of the above
Statement for Q.2–4:
Represent the given system in state-space
equation x& = A ×x + B× u. Choose the correction option for
é 1 0 -2 ù
(A) ê 0 -2
0ú
ê
ú
0 úû
êë-3 0
é-1 0 2 ù
(B) ê 0 2 0 ú
ê
ú
êë 3 0 0 úû
0
1ù
é-2
(C) ê 0 -2 0 ú
ê
ú
0 -3úû
êë 0
é2 0 -1ù
(D) ê0 2 0 ú
ê
ú
3úû
êë0 0
matrix A.
4.
2.
1
s
x2
1
s
1
2
x1
5
y
1
u
-2
1
s
u
-2
x3
1
s
x3 1
1
s
x2
Fig. P6.7.4
Fig. P6.7.2
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1
s
-4
-3
5
-3
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x1 1
y
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The State-Variable Analysis
Chap 6.7
é 0 1 -4 ù
(A) ê 1 0
0ú
ê
ú
0 úû
êë-3 0
é 0 -1 4 ù
(B) ê-1 0 0 ú
ê
ú
êë 3 0 0 úû
8. The F( t ) is
é cos 2 t sin 2 t ù
(A) ê
ú
ë- sin 2 t cos 2 t û
écos 2 t - sin 2 t ù
(B) ê
cos 2 t úû
ësin 2 t
é- 4 1 0 ù
(C) ê 0 0
1ú
ê
ú
êë 0 0 -3úû
é4 - 1 0 ù
(D) ê0 0 -1ú
ê
ú
3úû
êë0 0
é sin 2 t cos 2 t ù
(C) ê
ú
ë- cos 2 t sin 2 t û
ésin 2 t - cos 2 t ù
(D) ê
sin 2 t úû
ëcos 2 t
9. The q( t ) is
5. The state equation of a LTI system is represented by
1ù
é 0
é0 1ù
x& = ê
ú x + ê1 0 ú u
2
1
ë
û
ë
û
The Eigen values are
(A) -1, + 1
(B) -0.5 ± j1.323
(C) -1, - 1
(D) None of the above
é-3 0 ù
é0 ù
x& = ê
x + ê úu
ú
ë 0 -3û
ë1 û
The state-transition matrix F( t) is
é-e -3t
(C) ê
ë 0
0 ù
ú
e -3t û
0 ù
ú
-e -3t û
é-e -3t
(B) ê
ë 0
0 ù
ú
e -3t û
ée -3t
(D) ê
ë0
0 ù
ú
-e -3t û
ésin 2 t ù
(B) ê
ú
ëcos 2 t û
é0.5(1 - cos 2 t) ù
(C) ê
ú
ë 0.5 sin 2 t û
écos 2 t ù
(D) ê
ú
ësin 2 t û
10. From the following matrices, the state-transition
6. The state equation of a LTI system is
ée -3t
(A) ê
ë0
é0.5(1 - sin 2 t) ù
(A) ê
ú
ë 0.5 cos 2 t û
matrices can be
é -e - t
0 ù
(A) ê
ú
0
1
e- t û
ë
é1 - e - t
(B) ê
ë 0
0ù
ú
e- t û
é 1
(C) ê
-t
ë1 - e
é1 - e - t
(D) ê
ë 0
e- t ù
ú
e- t û
0ù
e - t úû
Statement for Q.11–13:
A system is described by the dynamic equations
x& ( t) = A ×x ( t) + B ×u( t), y( t) = C × x( t) where
1 0ù
é 0
é0 ù
ê
ú
A= 0
0
1 , B = ê0 ú, C = [1 0 0 ]
ê
ú
ê ú
êë-1 -2 -3ûú
êë1 ûú
7. The state equation of a LTI system is
é 0 2ù
é0 ù
x& = ê
x + ê úu
ú
ë-2 0 û
ë1 û
11. The Eigen values of A are
(A) 0.325, -1.662 ± j0.562
The state transition matrix is
é cos 2 t sin 2 t ù
(A) ê
ú
ë- sin 2 t cos 2 t û
écos 2 t - sin 2 t ù
(B) ê
cos 2 t úû
ësin 2 t
(B) 2.325, 0.338 ± j0.562
é sin 2 t cos 2 t ù
(C) ê
ú
ë- cos 2 t sin 2 t û
ésin 2 t - cos 2 t ù
(D) ê
sin 2 t úû
ëcos 2 t
(D) -0.325, 1.662 ± j0.562
(C) -2.325, -0.338 ± j0.562
12. The transfer-function relation between X ( s) and
Statement for Q.8–9:
The state-space representation of a system is given
by x& ( t) = A ×x ( t) + B ×u( t), where
U ( s) is
(A)
é 1ù
1
ê -s ú
s 3 + 3s 2 + 2 s - 1 ê 2 ú
êë s úû
(B)
(C)
é 1ù
1
ê -s ú
s 3 + 3s 2 + 2 s + 1 ê 2 ú
êë s úû
(D) None of the above
é0 ù
é 0 2ù
, B=ê ú
A =ê
ú
ë1 û
ë-2 0 û
If x(0) is the initial state vector, and the
component of the input vector u( t) are all unit step
é 1ù
1
ê sú
s 3 + 3s 2 + 2 s + 1 ê 2 ú
êës úû
function, then the state transition equation is given by
x& ( t) = F( t )x (0) + q( t ), where F( t ) is a state transition
13. The output transfer function Y ( s) U ( s) is
matrix and q( t ) is a vector matrix.
(A) s( s 3 + 3s 2 + 2 s + 1) -1
(B) s( s 3 + 3s 2 + 2 s - 1) -1
(C) ( s 3 + 3s 2 + 2 s + 1) -1
(D) None of the above
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UNIT 6
14. A system is described by the dynamic equation
x& ( t) = A ×x ( t) + B ×u( t), y( t) = C × x( t) where
é-1 0 ù
é1 ù
, B = ê ú and C = [1 1]
A =ê
ú
ë 0 -2 û
ë0 û
(C)
( s + 1)
( s + 2) 2
(B)
s+1
s+2
( s + 2)
( s + 1)
(D) None of the above
18.
15. The state-space representation of a system is given
by
(C) s( s 2 + 3s + 2) -1
(D) ( s + 1) -1
1 0ù
é 0
é10 ù
x& = ê 0
0
1ú x + ê 0 ú u, y = [1 0 0 ]x
ê
ú
ê ú
êë-1 -2 -3úû
êë 0 úû
(A)
10(2 s 2 + 3s + 1)
s 3 + 3s 2 + 2 s + 1
(B)
10(2 s 2 + 3s + 1)
s 3 + 2 s 2 + 3s + 1
(C)
10(2 s 2 + 3s + 2)
s 3 + 3s 2 + 2 s + 1
(D)
10(2 s 2 + 3s + 2)
s 3 + 2 s 2 + 3s + 1
1
é0
ê0 0
(C) x& = ê
ê0 0
ê20 10
ë
0 ù
é0 ù
ú
ê0 ú
1 0
úx + ê úr
0
1 ú
ê0 ú
ú
ê1 ú
7 100 û
ë û
0
1
0
0 ù
é 0
é0 ù
ê0 ú
ê 0
ú
0
1
0
úx + ê úr
(D) x& = ê
0
0
1 ú
ê 0
ê0 ú
ê-20 -10 -7 -100 ú
ê ú
ë
û
ë1 û
y = [100 0 0 0 ]x
19. A state-space representation of a system is given by
Statement for Q.17–18:
Determine the state-space representation for the
transfer function given in question. Choose the state
variable as follows
2
3
dc
d c
d c
= c& , x3 = 2 = &&c , x4 = 2 = &&&
c
dt
dt
dt
C( s)
24
= 3
R( s)
s + 9 s 2 + 26 s + 24
1
0 ù é x1 ù é 0 ù
é x& 1 ù é 0
ê
ú
ê
(A) x& 2 = 0
0
1 ú ê x2 ú + ê 0 ú r
ê ú ê
úê ú ê ú
êë x& 3 úû êë-24 -26 -9 ûú êë x3 úû êë24 úû
1 0 ù é x1 ù é 0 ù
é x& 1 ù é 0
(B) ê x& 2 ú = ê 0 0 1 ú ê x2 ú + ê 0 ú r
ê ú ê
úê ú ê ú
êë x& 3 úû êë24 26 9 úû êë x3 úû êë24 úû
Page
378
0
1
y = [100 0 0 0 ]x
The transfer function Y ( s) U ( s) is
17.
0ù
é 0 ù
ê 0 ú
ú
0
úx + ê
ú r,
0 0
1ú
ê 0 ú
ê100 ú
7 10 20 úû
ë
û
1
0
y = [1 0 0 0 ]x
16. The state-space representation for a system is
x1 = c = y, x2 =
é 0
ê 0
(A) x& = ê
ê 0
ê100
ë
1
0
0 ù
é 0
é 0 ù
ê 0
ú
ê 0 ú
0
1
0
úx + ê
úr
(B) x& = ê
0
0
1 ú
ê 0
ê 0 ú
ê-100 -7 -10 -20 ú
ê
ú
ë
û
ë100 û
The transfer function of this system is
(B) ( s + 2) -1
C( s)
100
= 4
R( s)
s + 20 s 3 + 10 s2 + 7 s + 100
y = [1 0 0 0 ]x
é-1 0 ù
é1ù
x& ( t) = ê
x( t) + ê ú u( t), y( t) = [1 1]x( t)
ú
ë 0 -2 û
ë1û
(A) ( s 2 + 3s + 2) -1
é x& 1 ù é0 1 0 ù é x1 ù é 0 ù
(C) ê x& 2 ú = ê0 0
1 ú ê x2 ú + ê 0 ú r
ê ú ê
úê ú ê ú
êë x& 3 úû êë9 26 24 úû êë x3 úû êë24 úû
1
0 ù é x1 ù é 0 ù
é x& 1 ù é 0
(D) ê x& 2 ú = ê 0
0
1 ú ê x2 ú + ê 0 ú r
ê ú ê
úê ú ê ú
êë x& 3 úû êë-9 -26 -24 úû êë x3 úû êë24 úû
The output transfer function Y ( s) U ( s) is
(A)
Control & System
é 0 1ù
é0 ù
x& = ê
ú x, y = [1 - 1]x, and x(0) = ê1 ú
2
0
ë
û
ë û
The time response of this system will be
3
(A) sin 2t
(B)
sin 2 t
2
(C) -
1
2
(D)
sin 2 t
3 sin 2 t
20. For the transfer function
Y ( s)
s+3
=
U ( s) ( s + 1)( s + 2)
The state model is given by x& = A ×x + B× u,
y = C × x. The A , B, C are
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UNIT 6
-4 -2 ù
é-5
é1 ù
ê
ú
29. x& = -3 -10
0 x + ê1 ú r, y = [ -1 2 1]x
ê
ú
ê ú
1 -5 úû
êë -1
êë0 ûú
estep ( ¥)
Statement for Q.34–36:
Consider the system shown in fig. P6.7.34-36.
1
s
eramp ( ¥)
(A) 1.0976
1
u
0
(B) 1.0976
¥
(C) 0
1.0976
(D) ¥
1.0976
0.714
(B) ¥
0.714
(C) 0
4.86
(D) ¥
4.86
x2
1
s
1
-1
1
s
y
-10
x3
34. The controllability matrix for this system is
10 ù
1 -2 ù
é 10 -10
é0
(A) ê-10
(B) ê1 -1
0 -20 ú
1ú
ê
ú
ê
ú
40 úû
4 úû
êë 10 -20
êë1 -2
10 ù
é 10 -10
(C) ê-10
0
20 ú
ê
ú
êë 10 -20 -40 úû
1
-2
10
Fig. P6.7.34-36
Consider the system shown in fig. P6.7.31-33
u
x1
-2
Statement for Q.31–33:
1
s
1
s
1
10
eramp ( ¥)
(A) 0
x2
-1
1
1 0ù
é 0
é0 ù
ê
ú
30. x& = -5 -9 7 x + ê0 ú r, y = [1 0 0 ]x
ê
ú
ê ú
êë -1 0 0 úû
êë1 ûú
estep ( ¥)
Control & System
x1
1
y
1 -1ù
é0
(D) ê1 6 -1ú
ê
ú
êë1 -4 -4 úû
35. The observability matrix is
10 ù
1 -2 ù
é 10 -10
é0
(A) ê-10
(B) ê1 -1
0 -20 ú
1ú
ê
ú
ê
ú
40 úû
4 úû
êë 10 -10
êë1 -2
10 ù
é 10 -10
(C) ê-10
0
20 ú
ê
ú
êë 10 -10 -40 úû
1 2ù
é0
(D) ê1 -1 1ú
ê
ú
êë1 -2 4 úû
-5
36. The system is
-6
Fig. P6.7.31-33
31. The controllability matrix is
é 1 -2 ù
é1 0 ù
(A) ê
(B) ê
ú
4 úû
ë0 1û
ë-2
é1 0 ù
(C) ê
ú
ë0 -1û
é 1 2ù
(D) ê
ú
ë-2 -4 û
(A) Controllable and observable
(B) Controllable only
(C) Observable only
(D) None of the above
Statement for Q.37–38:
A state flow graph is shown in fig. P6.7.37-38
32. The observability matrix is
é1 0 ù
é 1 -2 ù
(A) ê
(B) ê
ú
0
1
4 úû
ë
û
ë-2
é1 0 ù
(C) ê
ú
ë0 -1û
(A) Controllable and observable
(C) Observable only
u
1
s x2
1
s
x1
5
y
-5
Fig. P6.7.37-38
37. The state and output equation for this system is
éx ù
é x& ù é0 -1ù é x ù é0 ù
(A) ê 1 ú = ê 21 ú ê 1 ú + ê ú u, y = [5 4 ]ê 1 ú
5
&
x
x
1
ë 2 û êë
ë x2 û
4 úû ë 2 û ë û
(D) None of the above
Page
380
1
-21
4
é 1 2ù
(D) ê
ú
ë-2 -4 û
33. The system is
(B) Controllable only
4
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The State-Variable Analysis
Chap 6.7
év& ù é-1 -1ù év1 ù é1 ù
év ù
(B) ê &1 ú = ê
+ ê ú vi , iR = [ 4 1]ê 1 ú
ê
ú
ú
ë i3 û ë-3 -1û ë i3 û ë0 û
ë i3 û
1ù é x ù é0 ù
é x& ù é 0
éx ù
21 ú ê 1 ú + ê ú u, y = [5 4 ]ê 1 ú
(B) ê 1 ú = ê
5
&
x
1
ë x2 û êë
ë x2 û
4 úû ë 2 û ë û
é v& ù é1 -3ù é v1 ù é 1 ù
év ù
(C) ê 1 ú = ê
+ ê ú vi , iR = [1 4 ]ê 1 ú
ê
ú
ú
&
ëv2 û ë1 6 û ëv2 û ë-1û
ëv2 û
éx ù
é x& ù é0 -1ù é x ù é1ù
(C) ê 1 ú = ê 21 ú ê 1 ú + ê ú u, y = [ 4 5 ]ê 1 ú
5
x
1
ë x2 û
ë x& 2 û êë
4 úû ë 2 û ë û
é v& ù é 1 3ù é v1 ù é 1 ù
é v1 ù
(D) ê 1 ú = ê
ú êv ú + ê-1ú vi , iR = [1 4 ]êv ú
&
v
1
6
û ë 2û ë û
ë 2û ë
ë 2û
1ù é x ù é1ù
é x& ù é 0
éx ù
21 ú ê 1 ú + ê ú u, y = [ 4 5 ]ê 1 ú
(D) ê 1 ú = ê
5
&
x
x
1
ë 2 û êë
ë x2 û
4 úû ë 2 û ë û
Statement for Q.41–43:
38. The system is
Consider the network shown in fig. P6.7.41-43.
This system may be represented in state space
representation x& = A × x + B × u
(A) Observable and controllable
(B) Controllable only
(C) Observable only
iC
(D) None of the above
iR1
39. Consider the network shown in fig. P6.7.39. The
state-space representation for this network is
iL
iL
iR2
1
2F
1W
is
1W
4H
vs
2W
Fig. P6.7.41-43
iC
iR
41. The state variable may be
1F
Fig. P6.7.39
év& ù é-0.25 1ù évC ù é 1 ù
évC ù
(A) ê & C ú = ê
ú ê i ú + ê0.25 ú vs , iR = [0.5 0 ]ê i ú
i
0
5
0
.
û
ûë Lû ë
ë Lû ë
ë Lû
(A) iR1 , iR 2
(B) iL , iC
(C) vC , iL
(D) None of the above
42. If state variable are chosen as in previous question,
then the matrix A is
é 1 -1ù
(A) ê
ú
ë-1 3û
é 1 -3ù
(B) ê
ú
ë-1 1û
é-1 3ù
(C) ê
ú
ë 1 -1û
é 3 -1ù
(D) ê
ú
ë-1 1û
év& ù é-1 0.25 ù évC ù é0.25 ù
év ù
(D) ê & C ú = ê
+ê
vs , iR = [0.5 0 ]ê C ú
ê
ú
ú
ú
ë iL û ë 0 -0.5 û ë iL û ë 0 û
ë iL û
43. The matrix B is
é 3ù
(A) ê ú
ë-1û
é-1ù
(B) ê ú
ë 3û
40. For the network shown in fig. P6.7.40. The output is
é-3ù
(C) ê ú
ë 1û
é 1ù
(D) ê ú
ë-3û
év& ù é -0.5 1ù évC ù é0.25 ù
évC ù
(B) ê & C ú = ê
ú ê i ú + ê 1 ú vs , iR = [0.5 0 ]ê i ú
i
0
25
0
.
û
ûë Lû ë
ë Lû ë
ë Lû
év& ù é1 0.25 ù évC ù é0.25 ù
év ù
(C) ê & C ú = ê
+ê
vs , iR = [0.5 0 ]ê C ú
ê
ú
ú
ú
ë iL û ë0 0.5 û ë iL û ë 0 û
ë iL û
i1
1W
1H
v1
i3
v2
iR
i2
vi
1F
4vL
Statement for Q.44–47:
4v1
1W
Consider the network shown in fig. P6.7.44-47
i1
Fig. P6.7.40
iR ( t). The state space representation is
vi
év& ù é 1 -1ù év1 ù é1 ù
év ù
(A) ê &1 ú = ê
+ ê ú vi , iR = [ 4 1]ê 1 ú
ê
ú
ú
ë i3 û ë-3 1û ë i1 û ë0 û
ë i3 û
1W
i3
1W
i5
i2
i4
1H
1H
1W
1F
+
vo
-
Fig. P6.7.44-47
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UNIT 6
Control & System
44. The state variable may be
(A) i2 , i4
(B) i2 , i4 , vo
(C) i1 , i3
(D) i1 , i3 , i5
45. In state space representation matrix A is
1
1ù
é 2
ê- 3 - 3
3ú
ê 1
2
2ú
(A) êú
3
3ú
ê 3
ê- 1 - 2 - 1 ú
êë 3
3
3 úû
1
2ù
é 1
ê 3 - 3 - 3ú
ê 2
2
1ú
(B) ê
- ú
3
3ú
ê 3
ê- 1 - 2 - 1 ú
êë 3
3
3 úû
1
1ù
é 2
ê 3 - 3 - 3ú
ê 1
2
2ú
(C) ê
ú
3
3ú
ê 3
1ú
ê- 1 - 2
êë 3
3
3 úû
1
2ù
é 1
ê- 3 - 3
3ú
ê 2
2
1ú
(D) êú
3
3ú
ê 3
ê- 1 - 2 - 1 ú
êë 3
3
3 úû
46. The matrix B is
é 2ù
ê 3ú
ê 1ú
(A) ê- ú
ê 3ú
ê- 1 ú
êë 3 úû
é2 ù
ê3ú
ê1 ú
(B) ê ú
ê3ú
ê1 ú
êë 3 úû
é 1ù
ê- 3 ú
ê 1ú
(C) ê- ú
ê 3ú
ê 1ú
êë 3 úû
é2 ù
ê3ú
ê1 ú
(D) ê ú
ê3ú
ê2 ú
êë 3 úû
47. If output is vo , then matrix C is
(A) [-1 0 0]
(B) [1 0 0]
(C) [0 0 -1]
(D) [0 0 1]
SOLUTIONS
1. (B) From the SFG
x& 3 = -3 x1 - 2 x2
x
x2 = 3
Þ
s
x
x1 = 2
Þ
s
- x3 + u
x& 2 = x3
x& 1 = x2
2. (A) x& 1 = - 2 x1 + x3 , x& 2 = - 2 x2 + u , x& 3 = -3 x3 + u
1 0 ù é x1 ù é0 ù
é x& 1 ù é-2
ê x& ú = ê 0 -2 0 ú ê x ú + ê0 ú u
ê 2ú ê
ú ê 2ú ê ú
&
x
0
0
3
êë 3 úû êë
ûú êë x3 úû êë1 ûú
3. (C) x& 1 = - 2 x1 + x3 , x& 2 = - 2 x2 + u , x& 3 = -3 x3 + u
y = 5 x1 + 5 x2 + 5 x3
0
1ù é x1 ù é0 ù
é x& 1 ù é-2
ê x& ú = ê 0 -2 0 ú ê x ú + ê1 ú u
ê 2ú ê
ú ê 2ú ê ú
0 -3úû êë x3 úû êë1 úû
êë x& 3 úû êë 0
4. (C) x& 1 = -4 x1 + x2 , x& 2 = x3 + 2 u , x& 3 = - 3 x3 + u
é x& 1 ù é-4 1 1ù é x1 ù é0 ù
ê x& ú = ê 0 0
1ú ê x2 ú + ê2 ú u
ê 2ú ê
úê ú ê ú
êë x& 3 úû êë 0 0 -3úû êë x3 úû êë1 úû
5. (B) Ds = |sI - A| = s 2 + s + 2 = 0 Þ s = - 0.5 ± j1.323
0
és + 3
6. (A) ( sI - A) = ê
s+
ë 0
( sI - A)
-1
é 1
ê
= ês + 3
ê 0
ë
ù
0 ú
1 ú
ú
s + 3û
ée -3t
F( t ) = L-1 {( sI - A )} == ê
ë0
************************
ù
1
, |sI - A|=
3úû
( s + 3) 2
0 ù
ú
e -3t û
é s -2 ù
2
7. (A) ( sI - A) = ê
ú, |sI - A|= s + 4
ë2 s û
( sI - A)
-1
é s
1 é s 2 ù ê s2 + 4
= 2
=
s + 4 êë-2 s úû ê -2
ê
ë s2 + 4
2 ù
s2 + 4 ú
s ú
ú
2
s + 4û
é cos 2 t sin 2 t ù
F( t ) = L-1 {( sI - A )} = ê
ú
ë- sin 2 t cos 2 t û
é s -2 ù
2
8. (A) ( sI - A) = ê
ú, Ds =|sI - A|= s + 4
ë2 s û
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382
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The State-Variable Analysis
( sI - A)
-1
é s
1 é s 2 ù ê s2 + 4
= 2
=
s + 4 êë-2 s úû ê -2
ê
ë s2 + 4
é1ù
é1 ù
15. (D) T( s) = ê ú( sI - A) -1 ê ú
ë1û
ë0 û
2 ù
s + 4ú
s ú
ú
2
s + 4û
2
( sI - A)
é cos 2 t sin 2 t ù
F( t ) = L-1 {( sI - A )} = ê
ú
ë- sin 2 t cos 2 t û
ì 1 é s 2 ù é0 ù é1ù 1 ü
é2 ù ü
1
-1 ì
= L-1 í 2
ý =L í 2
ê
ú
ê
ú
ê
ú
ê úý
î s + 4 ë-2 s û ë1 û ë1û s þ
î s( s + 4) ë s û þ
2
é
ù
ê s( s 2 + 4) ú
ê
ú
1
ê 2
ú
ë ( s + 4) û
10. (C) (A) is not a state-transition matrix, since F(0) ¹ I
(C) is a state-transition matrix since F(0) = I and
[ F( t)]-1 = F( -t)
0 ù
é s -1
ê
11. (C) ( sI - A) = 0 s
-1 ú
ê
ú
êë1 2 s + 3úû
|sI - A| = s3 + 3s2 + 2 s + 1,
Þ
s = - 2.325, - 0.338 ± j0.562
X ( s)
= ( sI - A) -1 B
U ( s)
12. (B)
és 3 + 3s + 2
s+3
1 ù é0 ù
1
ê
ú
s( s + 3) s ê0 ú
= 3
-1
úê ú
s + 3s 2 + 2 s + 1 ê
2
-s
-2 s - 1 s úû êë1 úû
êë
é1 ù
1
êsú
= 3
s + 3s 2 + 2 s + 1 ê 2 ú
êës úû
13. (C)
Y ( s) CX ( s)
=
U ( s)
U ( s)
1
ù
é
ê s 3 + 3s 2 + 2 s + 1 ú
ú
ê
s
1
= [1 0 0 ]ê 3
ú= 3
2
2
ê s + 3s 2+ 2 s + 1 ú s + 3s + 2 s + 1
s
ú
ê
êë s 3 + 3s 2 + 2 s + 1 úû
14. (D)
Y ( s)
= C ( sI - A) -1 B
U ( s)
B = [1 1]
1 ù é0 ù
s+2
1 és + 1
=
ê
s + 1úû êë1 úû ( s + 1) 2
Ds ë 0
é 1
ê
= ês + 1
ê 0
ë
ù
0 ú
1 ú
ú
s + 2û
ù
0 ú é1 ù
1
=
1 ú êë0 úû s + 1
ú
s + 2û
16. (C) x& = A ×x + B× u, y = C × x + Du
ü
ïï é0.5(1 - cos 2 t) ù
ý= ê
ú
ï ë 0.5 sin 2 t û
ïþ
(B) is not a state-transition matrix since F(0) ¹ I
-1
é 1
é1ù ê s + 1
T( s) = ê ú ê
ë1û ê 0
ë
9. (C) q( t) = L-1 {( sI - A) -1 BR( s)}
ì
ïï
= L-1 í
ï
ïî
Chap 6.7
1 0ù
é 0
é10 ù
A =ê 0
0
1ú, B = ê 0 ú, C = [1 0 0 ], D = 0
ê
ú
ê ú
êë-1 -2 -3úû
êë 0 úû
Y ( s)
T( s) =
= C ( sI - A ) -1 B + D
U ( s)
( sI - A)
-1
és 3 + 3s + 2
s+3
1ù
1
ê
ú
= 3
1
s
(
s
+
3
)
s
ú
s + 3s 2 + 2 s + 1 ê
-s
-2 s - 1 s 2 úû
êë
Substituting the all values,
T( s) =
10(2 s 2 + 3s + 2)
s 3 + 3s 2 + 2 s + 1
17. (A)
C( s)
b0
24
=
=
R( s) s 3 + a2 s 2 + a1 s + a0 ( s 3 + 9 s 2 + 26 s + 24)
( s 3 + a2 s 2 + a1 s + a0 ) C( s) = b0 R( s)
Taking the inverse Laplace transform assuming zero
initial conditions
&&&
c + a2 &&c + a1 c& + a0 c = b0 r
x1 = c = y, x2 = c& , x3 = &&c
x& 1 = c& = x2 , x& 2 = &&c = x3
x& 3 = &&&
c = b0 r - a2 &&c - a1 c& - a0 c
= - a0 x1 - a1 x2 - a2 x3 + b0 r ,
1
0 ù é x1 ù é 0 ù
é x& 1 ù é 0
ê x& ú = ê 0
0
1 ú ê x2 ú + ê 0 ú r
ê 2ú ê
úê ú ê ú
êë x& 3 úû êë-a0 -a1 -a1 úû êë x3 úû êëb0 úû
a0 = 24 , a1 = 26, a2 = 9, b0 = 24
1
0 ù é x1 ù é 0 ù
é x& 1 ù é 0
ê x& ú = ê 0
0
1 ú ê x2 ú + ê 0 ú r
ê 2ú ê
úê ú ê ú
êë x& 3 úû êë-24 -26 -9 úû êë x3 úû êë24 úû
é x1 ù
y = [1 0 0 ] ê x2 ú
ê ú
êë x3 úû
18. (B) Fourth order hence four state variable
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UNIT 6
é 0
ê 0
x& = ê
ê 0
ê -a
ë 0
1
0
0
1
0
0
-a1
-a 2
0 ù
é0 ù
ú
ê0 ú
0
ú x = ê ú r, y = [1 0 0 0 ]x
1 ú
ê0 ú
ú
ê ú
-a 3 û
ëb0 û
a0 = 100, a1 = 7, a2 = 10,
a3 = 20 ,
=
b0 = 100
Y ( s) = [0 1] X ( s) =
é
sin 2 t ù
cos 2 t
ú
F( t ) = L-1 {( sI - A)} = ê
2 ú
ê
êë- 2 sin 2 t cos 2 t úû
Þ
20. (C) Find the transfer function of option
Y ( s)
1
For (A) ,
,
=
U ( s) s - 2
y( t) =
0 ù é1ù
és + 2
Y ( s)
1
= [0 1]
s + 1úû êë1úû
U ( s)
( s + 1)( s + 2) êë 2
|sI - A| = s
- 8 s - 11s + 8
Þ
-1
æ é2 ù é1ù 3 ö
ç ê ú+ê ú 2
÷
ç 1
÷
è ë û ë1û s + 9 ø
é2 s + 4 s + 21s + 45 ù
1
2
( s + 5)( s + 9) êë s 3 - 7 s 2 + 12 s - 7 úû
3
1 1 -2 t
- e
2 2
A
-1
0.05 -0.05 ù
é-0.4
ê
=
-1 -0.25 -0.25 ú
ê
ú
15
.
-0.5 ûú
êë -2
1ù
é
-2 - ú
27. (A) estep ( ¥) = 1 + CA -1B, A -1 = ê
3
êë 1
0 úû
s = 9.11, 0.53, - 1.64
23. (D) X ( s) = ( sI - A) -1 (x(0) + B × u)
és - 1 -2 ù
=ê
s + 1úû
ë 3
1
s( s + 2)
= 1 - 0.2 = 0.8
s = -5.79, - 121
.
-2
-3 ù
és
22. (B) ( sI - A) = ê 0 s - 6 -5 ú
ê
ú
êë-1 -4 s - 2 úû
2
é0 ù é1 ù ö
ê0 ú + ê0 ú 1 ÷
ê ú ê ú s÷
êë0 úû êë0 ûú ÷ø
1
é
ù
ê
ú
s( s + 2)
ê
ú
1
ú
=ê
2
ê s ( s + 2) ú
1
ê
ú
êë s 2 ( s + 1)( s + 2) úû
é-0.4 0.05 -0.05 ù é0 ù
estep ( ¥) = 1 + [ -1 1 0 ]ê -1 0.25 -0.25 ú ê0 ú
ê
úê ú
15
.
-0.5 úû êë1 úû
êë -2
é-2 -1ù
21. (C) A = ê
ú,
ë -3 -5 û
3
æ
ç
ç
ç
è
X ( s) =
1 0ù
é- 5
ê
A= 0
-2 1ú,
ê
ú
êë20 -10 1úû
So (C) is correct option.
Þ
-1
26. (D) For a unit step input estep ( ¥) = 1 + CA -1B
és + 2 ù
s+3
1
= [0 1]
=
( s + 1)( s + 2) êës + 3úû ( s + 1)( s + 3)
|sI - A| = s2 + 7 s + 7
1
1
- e - t + e -2 t
2
2
Y ( s) = [1 0 0 ],
Y ( s)
1
For (B) ,
=
U ( s) s - 2
For (C),
y( t) =
1
s( s + 1)( s + 2)
0 ù
és + 2 -1
ê
= 0
-1 ú
s
ê
ú
0 s + 1úû
êë 0
sin 2 t
2
é1 ù é1ù 1 ö
ê0 ú + ê1ú s ÷÷
ë û ë û ø
25. (B) X ( s) = ( sI - A) -1 (x(0) + B × u)
é
sin 2 t ù
cos 2 t +
ê
ú
x( t) = F( t )x(0) =
2
ê
ú
êë- 2 sin 2 t + cos 2 t úû
3
0 ùæ
és + 1
1
ç
( s + 1)( s + 2) êë -1 s + 2 úû çè
( s + 1)
é
ù
ê s( s + 2) ú
=ê
ú
1
ê
ú
ë s( s + 1)( s + 2) û
é 0 1ù
1 é s 1ù
19. (B) A = ê
, ( sI - A) -1 = 2
ú
s + 2 êë-2 súû
ë-2 0 û
y = x1 - x2 =
Control & System
2
1ù
é
1 2
-2 - ú é0 ù
estep ( ¥) = 1 + [1 1] ê
3 ê ú =1 - =
3 3
êë 1
0 úû ë1 û
28. (C) eramp ( ¥) = lim [(1 + CA -1B) t + C( A -1 ) 2 B]
t ®¥
2
1 + CA -1B = ,
3
é2
ù
eramp ( ¥) = lim ê t + C( A -1 ) 2 Bú = ¥
t ®¥ ë 3
û
2
4 s 3 - 10 s 2 + 45 s - 105
Y ( s) = [1 2 ] X ( s) =
( s 2 + 5)( s 2 + 9)
24. (B) X ( s) = ( sI - A) -1 (x(0) + B × u)
Page
384
29. (B) estep ( ¥) = 1 + CA -1B
-4 -2 ù
é-5
é1 ù
A = ê -3 -10
0 ú, B = ê1 ú, C = [ -1 2 1]
ê
ú
ê ú
1 -5 úû
êë -1
êë0 úû
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The State-Variable Analysis
A
-1
é-1 1 0 ù
A = ê 0 -1 0 ú, C = [10 - 10 10 ],
ê
ú
êë 0 0 -2 ûú
0.134
0.122 ù
é-0.305
ê
= 0.091 -0.140 -0.037 ú
ê
ú
êë -0.079 -0.055 -0.232 úû
é-1 1 0 ù
CA = [10 - 10 10 ] ê 0 -1 0 ú = [10
ê
ú
êë 0 0 -2 úû
.
estep ( ¥) = 1 + 0.0976 = 10976
-1
-1 2
eramp ( ¥) = lim [(1 + CA B) t + C( A ) B] = ¥
t ®¥
0
-1 ù
é 0
30. (B) A -1 = ê 1
0
0 ú
ê
ú
êë1.286 0.143 -0.714 úû
CA 2 = [10
0
-1 ù é0 ù
é 0
ê
estep ( ¥) = 1 + [1 0 0 ] 1
0
0 ú ê0 ú = 0
ê
úê ú
êë1.286 0.143 -0.714 úû êë1 úû
.
-0.143 0.714 ù
é -1286
ê
(A ) =
0
0
-1 ú
ê
ú
ëê-0.776 -0.102 -0.776 úû
nonsingular and system is controllable
1ù
é 0
é 1ù
A =ê
, B=ê ú
ú
ë-6 -5 û
ë-2 û
10 ù
é 10 -10
det O M = detê-10
0 -10 ú = -3000
ê
ú
40 úû
êë 10 -20
1ù é 1ù ù é 1 -2 ù
é 0
ê-6 -5 ú ê-2 ú ú = ê-2
4 úû
ë
ûë ûû ë
The rank of O M is 3. Hence system is observable.
32. (A) y = x1 , y = [1 0 ]x ,
0 ], CA = [1
37. (B) x& 2 = - 5 x1 -
é C ù é1 0 ù
0 ] , OM = ê
ú =ê
ú
ëCA û ë0 1 û
21
x2 + u , x& 1 = x2 , y = 5 x1 + 4 x2
4
1ù é x ù é0 ù
é x& 1 ù é 0
é x1 ù
1
ê x& ú = ê-5 - 21 ú ê x ú + ê1 ú u, y = [5 4 ]ê x ú
ë 2 û êë
ë 2û
4 úû ë 2 û ë û
33. (C) det CM = 0. Hence system is not controllable. det
O M = 1. Hence system is observable.
é C ù é 5 4ù
38. (B) O M = ê
ú =ê
ú
ëCA û ë-20 1 û
34. (B) x& 1 = - x1 + x2 , x& 2 = - x2 + u , x& 3 = - 2 x3 + u
det O M = 0. Thus system is not observable
é-1 1 0 ù
é0 ù
é-1 1 0 ù
é0 ù
&x = ê 0 -1 0 ú x + ê1 ú u, A = ê 0 -1 0 ú , B = ê1 ú
ê
ú
ê ú
ê
ú
ê ú
êë 0 0 -2 úû
êë1 úû
êë 0 0 -2 úû
êë1 úû
1ù
é0
21 ú
CM = [B AB] = ê
1 êë
4 úû
é-1 1 0 ù é0 ù é 1ù
AB = ê 0 -1 0 ú ê1 ú = ê -1ú
ê
úê ú ê ú
êë 0 0 -2 úû êë1 úû êë-2 úû
det CM = -1. Thus system is controllable.
39. (B)
é-1 1 0 ù é 1ù é-2 ù
A 2B = ê 0 -1 0 ú ê -1ú = ê 1ú
ê
úê ú ê ú
êë 0 0 -2 úû êë-2 úû êë 4 úû
1 -2 ù
é0
ê
1ú
Cm = [B AB A B] = 1 -1
ê
ú
4 úû
êë1 -2
2
35. (A) y = 10 x1 - 10 x2 + 10 x3 , y = [10 - 10
40 ]
Since the determinant is not zero, the 3 ´ 3 matrix is
31. (B) x& 1 = x2 + u , x& 2 = - 5 x2 - 6 x1 - 2 u
C = [1
é-1 1 0 ù
0 - 20 ] ê 0 -1 0 ú = [10 - 10
ê
ú
êë 0 0 -2 úû
1 -2 ù
é0
36. (A) det Cm = det ê1 -1
1ú = -1,
ê
ú
4 úû
êë1 -2
C( A -1 ) 2 B = 0.714, eramp ( ¥) = 0.714
éé 1ù
CM = [B AB] = êê ú
ëë-2 û
0 - 20 ]
10 ù
é C ù é 10 -10
ê
ú
ê
O M = CA = -10
0 -20 ú
ê
ú ê
ú
2
40 ûú
êëCA úû êë 10 -10
-1 2
1ù
é 0
é 1ù
x& = ê
x + ê ú u,
ú
ë-6 -5 û
ë-2 û
Chap 6.7
10 ]x
dvc
di
v
= ic , L = L = 0.25 vL
dt
dt
4
vC and iL are state variable.
v
iL = iC + iR , iC = iL - iR = iL - C , vL = vs - vC
2
dvL
v
Hence equations are
= iL - C = - 0.5 vC + iL
dt
2
diL
= 0.25( vs - vC ) = - 0.25 vC + 0.25 vs
dt
év& C ù é -0.5 1ù évC ù é0.25 ù
ê i& ú = ê-0.25 0 ú ê i ú + ê 1 ú vs ,
û
ûë Lû ë
ë Lû ë
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UNIT 6
iR =
év ù
iR = [0.5 0 ] ê C ú
ë iL û
vC
= 0.5 vC ,
2
45. (A)
Control & System
di2
di
dvo
= v2 , 4 = v4 ,
= i5
dt
dt
dt
i1
40. (B)
dv1
di
= i2 , 3 = vL
dt
dt
1W
vi
Hence v1 and i3 are state variable.
1W
v2 i3
v4 i5
i2
i4
1H
1H
1W
1F
+
vo
-
i2 = i1 - i3 = ( vi - v1 ) - i3 , i2 = - v1 - i3 + vi
vL = v1 - v2 = v1 - iR , = v1 - ( i3 + 4 v1 ) = -3v1 - i3
dv1
di
= - v1 - i3 + vi , 3 = - 3v1 - i3,
y = iR = 4 v1 + i3
dt
dt
év&1 ù é-1 -1ù év1 ù é1 ù
év1 ù
ê i& ú = ê-3 -1ú ê i ú + ê0 ú vi , iR = [ 4 1]ê i ú
û ë 3û ë û
ë 3û ë
ë 3û
41. (C) Energy storage elements are capacitor and
inductor. vC and iL are available in differential form and
linearly independent. Hence vC and iL are suitable for
state-variable.
1 dvC
dvC
= iC Þ
= 2 iC
2 dt
dt
1 diL
diL
= vL Þ
= 2 vL
2 dt
dt
42. (B)
iC
iR1
is
1W
+
iL
+ vC 1
2F
vL
iR2
+
vR2
-
1W
4vL
-
Fig. S6.7.42
vL = vC + vR 2 = vC + iR2 , iC + 4 vL = iR 2
vL = vC + iC + 4 vL , -3vL = vC + iC
v
iC = is - iR1 - iL , iC = is - L - iL
1
...(i)
...(ii)
Solving equation (i) and (ii)
-3 ( is - iL - iC ) = vC + iC , 2 iC = vC - 3iL + 3is
Fig. S6.7.45
Now obtain v2 , v4 and i5 in terms of the state variable
-vi + i1 + i3 + i5 + vo = 0
But i3 = i1 - i2 and i5 = i3 - i4
-vi + i1 + ( i1 - i2 ) + ( i3 - i4 ) + vo = 0
2
1
1
1
i1 = i2 + i4 - vo + vi
3
3
3
3
2
1
1
2
v2 = vi - i1 = - i2 - i4 + vo + vi
3
3
3
3
1
1
1
1
i3 = i1 - i2 = - i2 + i4 - vo + vi
3
3
3
3
1
2
1
1
i5 = i3 - i4 = - i2 - i4 - vo + vi
3
3
3
3
1
2
2
1
v4 = i5 + vo = - i2 - i4 + vo + vi
3
3
3
3
1
1ù
é2 ù
é 2
é i&2 ù ê 3
3
3 ú é i2 ù ê 3 ú
ê& ú ê 1
2
2 ú ê ú ê1 ú
ú i4 + ê ú vi
ê i4 ú = ê- 3 - 3
3úê ú ê3ú
êv& o ú ê 1
2
1 êëvo úû ê 1 ú
ë û ê- ú
êë 3
êë 3 úû
3
3 úû
1
1ù
é 2
ê- 3 - 3
3ú
ê 1
2
2ú
A = êú,
3
3ú
ê 3
ê- 1 - 2 - 1 ú
êë 3
3
3 úû
47. (D) vo is state variable
-3vL = vC + is - vL - iL , 2vL = - vC + iL - is
dvC
di
= vC - 3iL + 3is , L = - vC + iL - is
dt
dt
év& C ù é 1 -3ù évC ù é 3 ù
ê i& ú = ê-1 1ú ê i ú + ê-1ú is
ûë Lû ë û
ë Lû ë
y = vo ,
é i2 ù
y = [0 0 1] = ê i4 ú
ê ú
êëvo úû
********
é 3ù
43. (A) B = ê ú
ë-1û
44. (B) There are three energy storage elements, hence
3 variable. i2 , i4 and vo are available in differentiated
form hence these are state variable.
Page
386
é2 ù
ê3ú
ê1 ú
46. (B) B = ê ú
ê3ú
ê1 ú
êë 3 úû
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UNIT 7
10. If machine is not properly adjusted, the product
Statement for Question 15-16:
resistance change to the case where ax = 1050 W. Now
the rejected fraction is
Communication System
The life time of a system expressed in weeks is a
Rayleigh random variable X for which
(A) 5046%
(C) 2.18%
(B) 10.57%
(D) 6.43%
ì x -x
e 400
ïï
200
f X ( x) = í
ï
ïî 0
2
11. Cannon shell impact position, as measured along
the line of fire from the target point is described by a
0£x
x <0
gaussian random variable X. It is found that 15.15% of
15. The probability that the system will not last a full
shell falls 11.2 m or farther from the target in a
week is
direction toward the cannon, while 5.05% fall farther
(A) 0.01%
(B) 0.25%
from the 95.6 m beyond the target. The value of ax and
(C) 0.40%
(D) 0.60%
sx
for X is
(Given that F(103
. ) = 0.8485 and
16. The probability that the system lifetime will exceed
F(1.64) = 0.9495)
(A) T + 40 m and 50 m
(B) T + 40 m and 30 m
in year is
(C) T + 10 m and 50 m
(D) T + 30 m and 40 m
(A) 0.01%
(B) 0.05%
(C) 0.12%
(D) 0.22%
12. A gaussian random voltage X for which a X = 0 and
s X = 4.2 V appears across a 100 W resistor with a power
rating of 0.25 W. The probability, that the voltage will
17. The cauchy random variable has the following
probability density function
f X ( x) =
cause an instantaneous power that exceeds the
resistor's rating, is
æ 5 ö
(A) 2Qç
÷
è 4.2 ø
æ 5 ö
(B) Qç
÷
è 4.2 ø
æ 5 ö
(C) 1 + Qç
÷
è 4.2 ø
æ 5 ö
(D) 1 - Qç
÷
è 4.2 ø
For real numbers 0 < b and -¥ < a < ¥. The
distribution function of X is
1
æ x -aö
(A) tan -1 ç
÷
p
è b ø
Statement for Question 13 -14 :
Assume that the time of arrival of bird at
Bharatpur
sanctuary
on
a
migratory
route,
first day), is approximated as a gaussian random
variable X with a X = 200 and sx = 20 days. Given that :
F(10
. ) = 0.8413.,
(B)
1
æ x -aö
cot-1 ç
÷
p
è b ø
(C)
1 1
æ x -aö
+ tan -1 ç
÷
2 p
è b ø
(D)
1 1
æ x -aö
+ cot-1 ç
÷
2 p
è b ø
as
measured in days from the first year (January 1 is the
F(0.5) = 0.6915,
b/ p
b2 + ( x - a) 2
F(15
. ) = 0.8531,
Statement for Question 18 - 19
The number of cars arriving at ICICI bank
F(155
. ) = 0.9394 and F(2.0) = 0.9773.
drive-in window during 10-min period is Poisson
13. What is the probability that birds arrive after 160
days but on or before the 210
th
(A) 0.6687
(B) 0.8413
(C) 0.8531
(D) 0.9773
day ?
18. The probability that more than 3 cars will arrive
during any 10 min period is
14. What is the probability that bird will arrive after
231st day ?
(A) 0.0432
(B) 0.1123
(C) 0.0606
(D) 0.0732
Page
390
random variable X with b = 2.
(A) 0.249
(B) 0.143
(C) 0.346
(D) 0.543
19. The probability that no car will arrive is
(A) 0.516
(C) 0.246
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(B) 0.459
(D) 0.135
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Random Variables
Chap 7.1
20. The power reflected from an aircraft of complicated
26. The mean of random variable X is
shape that is received by a radar can be described by an
(A) 1/4
(B) 1/6
exponential random variable W . The density of W is
(C) 1/3
(D) 1/5
ì 1 - w W0
e
ï
fW ( w) = í W0
ïî 0
ü
w > 0ï
ý
w < 0 ïþ
27. The variance of random variable X is
where W0 is the average amount of received power.
(A) 1/10
(B) 3/80
(C) 5/16
(D) 3/16
The probability that the received power is larger than
the power received on the average is
(A) e -2
(C) 1 - e
28. A Random variable X is uniformly distributed on
X
(B) e -1
-1
(D) 1 - e
the interval (-5, 15). Another random variable Y = e
-2
-
5
is formed. The value of E[ Y ] is
Statement for Question 21-23:
(A) 2
(B) 0.667
(C) 1.387
(D) 2.967
Delhi averages three murder per week and their
29. A random variable X has X = -3, x 2 = 11 and s2X = 2
occurrences follow a poission distribution.
2
For a new random variable Y = 2 x - 3, the Y , Y and s2Y
21. The probability that there will be five or more
are
murder in a given week is
(A) 0, 81, 8
(B) - 6, 8, 89
(C) - 9, 89, 8
(D) None of the above
(A) 0.1847
(B) 0.2461
(C) 0.3927
(D) 0.4167
22. On the average, how many weeks a year can Delhi
Statement for Question 31-32 :
A joint sample space for two random variable X
expect to have no murders ?
(A) 1.4
(B) 1.9
(C) 2.6
(D) 3.4
and Y has four elements (1, 1), (2, 2), ( 3, 3) and (4, 4).
Probabilities of these elements are 0.1, 0.35, 0.05 and
0.5 respectively.
23. How many weeds per year (average) can the Delhi
expect the number of murders per week to equal or
exceed the average number per week ?
(A) 15
(B) 20
(C) 25
(D) 30
30. The probability of the event{ X £ 2.5, Y £ 6} is
(A) 0.45
(B) 0.50
(C) 0.55
(D) 0.60
31. The probability of the event { X £ 3} is
24. A discrete random variable X has possible values
(A) 0.45
(B) 0.50
xi = i , i = 1, 2, 3, 4 which occur with probabilities 0.4,
(C) 0.55
(D) 0.60
2
0.25, 0.15, 0.1,. The mean value X = E[ X ] of X is
(A) 6.85
(C) 3.96
(B) 4.35
(D) 1.42
Statement for Question 32-34 :
Random variable X and Y have the joint distribution
25. The random variable X is defined by the density
x
1
f X ( x) = u( x) e 2
2
The expected value of g( X ) = X 3 is
(A) 48
(B) 192
(C) 36
(D) 72
ì 5 æ x + e- ( x + 1 ) y2
2 ö
- e - y ÷u( y), 0 £ x £ 4
ï çç
÷
x+1
ï4è
ø
ï
FX , Y ( x, y) = í 0
x < 0 or y < 0
ï
1 -5 y 2 5 - y 2
- e ,
4 £ x and any y ³ 0
ï1 + e
4
4
ï
î
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UNIT 7
32. The marginal distribution function FX ( x) is
37. The marginal distribution function FX ( x) is
x, > 0
ì 0,
ïï 5 x
(A) í
, -4< x£4
ï 4( x + 1)
ïî 1,
x £ -4
x <0
ì 0,
ïï 5 x
(B) í
, 0 £ x<4
ï 4( x + 1)
ïî 1,
x³4
x >0
ì 1,
ïï 5 x
(C) í
, - 4 < x £0
ï 4( x + 1)
ïî 0,
x £ -4
x <0
ì 1,
ïï 5 x
(D) í
, 0 £ x<4
ï 4( x + 1)
ïî 0,
x³4
33. The marginal distribution function FY ( y) is
ì 5 - y2
ï- e ,
(A) í 4
2
ï 1 + 4 e -5 y ,
î
4
(A)
5
1
[ 4( x) - 4( x - 4)]
4 ( x + 1) 2
(B)
5
1
[ 4( x) - 4( x - 4)]
2 ( x + 1) 2
(C)
5
1
[ 4( x) - 4( x - 4)]
4 ( x + 1)
(D) None of the above
38. The marginal distribution function FY ( y) is
y >0
2
(A) [1 - e - y ]u( y)
y £0
(C)
y >0
ì 0,
ï
(B) í
1 -5 y 2 5 - y 2
- e , y £0
ïî 1 + 4 e
4
5
4
(A) 0.001
(B) 0.002
(C) 0.003
(D) 0.004
ì 5 é x + e- ( x + 1 ) y2
2ù
- e - y ú,
ï ê
ï 8
x+1
û
(C) í ë
2
1 -5 y 2
ï
- 5 e - y ],
ïî 1 + 4 [ e
Statement for Question 35-39 :
Two random variable X and Y have a joint density
2
10
FX , Y ( x, y) =
[ u( x) - u( x - 4)]u( y) y 3e - ( x + 1 ) y
4
35. The marginal density f X ( x) is
u( x) - u( x - 4)
u( x) - u( x - 4)
(A) 5
(B) 5
( x + 1) 2
( x + 1)
5 u( x) - u( x - 4)
(D)
4
( x + 1)
2
2
2
2
(B)
5
2
y 2 [ e - y - e -5 y ]u( y)
(C)
5
4
y[ e - y - e -5 y ]u( y)
(D)
5
2
y[ e - y - e -5 y ]u( y)
Page
392
2
2
2
2
0 £ x £ 4 and y > 0
x > 4 and y > 0
0 £ x £ 4 and y > 0
x > 4 and y > 0
0 £ x £ 4 and y > 0
x > 4 and y > 0
ì 5 é x + e- ( x + 1 ) y2
2ù
- e - y ú,
ï ê
ï4
x+1
û
(D) í ë
2
1 -5 y 2
ï
- 5 e - y ],
ïî 1 + 2 [ e
0 £ x £ 4 and y > 0
x > 4 and y > 0
40. The function
FX , Y ( x, y) =
36. The marginal density fY ( y) is
y 2 [ e - y - e -5 y ]u( y)
2
[1 - e - y ]u( y)
(D) None of the above
ì 5 é x + e- ( x + 1 ) y2
2ù
- e - y ú,
ï ê
ï 8
x+1
û
(B) í ë
2
1 -5 y 2
ï
- 5 e - y ],
ïî 1 + 2 [ e
34. The probability P { 3 < X £ 5, 1 < y £ 2} is
5
4
2
[1 - e - y ]u( y)
ì 5 é x + e- ( x + 1 ) y2
2ù
- e - y ú,
ï ê
ï4
x+1
û
(A) í ë
2
1 -5 y 2
ï
- 5 e - y ],
ïî 1 + 4 [ e
y <0
ì 0,
ï
(D) í
1 -5 y 2 5 - y 2
- e , y ³0
ïî 1 + 4 e
4
(A)
5
2
(B)
39. The joint distribution function is
ì -5 - y 2
e ,
y <0
ï
(C) í 4
2
2
ï 1 + 1 e -5 y - 5 e - y , y ³ 0
î
4
4
5 u( x) - u( x - 4)
(C)
4
( x + 1) 2
Communication System
a ép
æ x öù é p
æ y öù
+ tan -1 ç ÷ú ê + tan -1 ç ÷ú
2 êë2
2
2
è øû ë
è 3 øû
is a valid joint distribution function for random
variables X and Y if the constant a is
1
2
(A) 2
(B) 2
p
p
(C)
4
p2
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(D)
8
p2
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Random Variables
41. Random variable
and Y
X
have the joint
(A)
a2
1 - e- a
(B)
(C)
1
1 - e- a
(D) None of the above
distribution function
ì 0,
ï
ï 27
ï 26
ï
ï 27
FX , Y ( x, y) = í
ï 26
ï 27
ï
ï 26
ï 1,
î
The
x < 0 or y < 0
æ
y ö
÷,
xçç 1 27 ÷ø
è
æ
y2 ö
yçç 1 ÷,
27 ÷ø
è
Chap 7.1
2
a
1 - e- a
0 £ x < 1 and 1 £ y
Statement for Question 46-47 :
1 £ x and 0 £ y < 1
æ
x2 y2
xyçç 1 27
è
ö
÷÷,
ø
Random variable X and Y have the joint density
x y
- 1
f X , Y ( x, y) = u( x) u( y) e 4 3
2
0 £ x < 1 and 0 £ y < 1
46. The probability of the event {2 < X £ 4, - 1 < Y £ 5}
1 £ x and 1 £ y
probability
of
the
is
event
{0 < X £ 0.5, 0 < Y £ 0.25} is
(A) 0.13
(B) 0.24
(C) 0.69
(D) 1
(A) 0.1936
(B) 6.2964
(C) 0
(D) None of the above
47. The probability of the event {0 < X < ¥, < y £ -2} is
Statement for question 42-43 :
(A) 0.2349
(B) 0.3168
(C) 0.4946
(D) None of the above
The joint probability density function of random
48. Let X and Y be two statistically independent
variable X and Y is given by
pXY ( x, y) = xye
-
random variables uniformly distributed in the ranges (
( x 2 + y2 )
2
-1, 1) and (-2, 1) respectively. Let Z = X + Y . Then the
u( x) u( y)
probability that ( Z £ -2) is
42. The pX ( x) is
2
(A) 2 xe - x u( x)
(B) xe
-
x2
2
x
2
(C) xe - x u( x)
(A) zero
(C) 1/3
u( x)
49. The probability density function of two statistically
2
(D) 2 xe 2 u( x)
independent random variable X and Y are
f X ( x) = 5 u( x) e -5x
43. The pY / X ( y/x) is
(A)
1
2
2
2
ye - y u( y)
(C) ye
-
(B) ye - y u( y)
y2
2
(D)
u( y)
(B) 1/6
(D) 1/12
1
2
ye
-
fY ( y) = 24( y) e -2 y
The density of the sum W = X + Y is
y2
2
u( y)
(A)
10
6
[ e -2 w - e 5w ]u( w)
10
8
[ e -2 w - e 5w ]u( w)
44. The probability density function of a random
((B)
variable X is given as f X ( x) . A random variable Y is
(C)
10
13
[ e -2 w - e -5w ]u( w)
defined as y = ax + b where a < 0. The PDF of random
(D)
10
2
[ e -2 w - e -5w ]u( w)
variable Y is
æ y - bö
(A) bf X ç
÷
è a ø
(C)
æ y - bö
(B) af X ç
÷
è a ø
1 æ y - bö
f Xç
÷
a è a ø
(D)
50. The density function of two random variable X and
Y is
ì 1
0 < x < 6 and 0 < y < 4
ï
f X , Y ( x, y) = í 24
ïî 0
else where
1 æ y - bö
f Xç
÷
b è a ø
45. The function
ì be
f X , Y ( x, y) = í
î0
The expected value of the function g( x, y) = ( XY )
- ( x + y)
0 < x < a and 0 < y < ¥
else where
is a valid joint density function if b is
is
(A) 64
(B) 96
(C) 32
(D) 48
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UNIT 7
51. The density function of two random variable X and
56. The value of s2X , s2Y , R XY and r are respectively
Y is
f X , Y ( x, y) =
e
æ x 2 + y2 ö
÷
-ç
ç 2 s2 ÷
ø
è
2 ps2
with s2 a constant. The mean value of the function
g( X , Y ) = X 2 + Y 2 is
(A) s2
(B) s
(C) 2s2
(D) 2s
have
Y = E[ Y ] = Y .
mean
They
X = E[ X ] = 2
values
have
11 27 1 æ
1 ö
33
, .ç 2 +
÷, and -3
4 2 2 è
2
3ø
(B)
11 11 1 æ
1 ö
33
, ç2 +
÷, and -3
4 2 2è
2
3ø
(C)
9 11 1 æ
1 ö
1 2
, ç2 ÷, and 3 33
4 2 2è
3ø
(D)
9 11 1 æ
1 ö
1 2
, , ç2 ÷, and 4 2 2è
3 33
3ø
W = ( X + 3Y ) 2 + 2 X + 3 is
The statistically independent random variable X
Y
(A)
57. The mean value of the random variable
Statement for Question 52-54 :
and
Communication System
second
and
moments
(A) 98 + 3
(B) 98 - 3
(C) 49 - 3
(D) 49 + 3
X 2 = E[ X 2 ] = 8 and Y 2 = E[ Y 2 ] = 25. Consider a random
***********
variable W = 3 X - Y .
52. The mean value E[W ] is
(A) 2
(B) 4
(C) 8
(D) 25
53. The second moment of W is
(A) 145
(B) 49
(C) 97
(D) 0
54. The variance of the random variable is
(A) 4
(B) 45
(C) 49
(D) 54
55. Two random variable X and Y have the density
function
ì xy
,
ï
f X , Y ( x, y) = í 9
ïî 0
0 < x < 2 and 0 < y < 3
elsewhere
The X and Y are
(A) Correlated but statistically independent
(B) uncorrelated but statistically independent
(C) Correlated but statistically dependent
(D) Uncorrelated but statistically dependent
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Random Variables
= FX (900) + [1 - FX (1100)]
SOLUTION
æ 900 - ax
= F çç
sx
è
¥
1
¥
1
x
2
e
-x
dx =
¥
1
x e - x dx = 0.368
2 ò1
0
2. (C) P { -1 < X £ 2} = ò -1
=1 -
= 2 - 2(0.9938) = 0.012 or 1.2 %
æ 900 - 1050 ö
10. (B) P(resistor rejected) = F ç
÷+1
40
è
ø
2
1 x
1
xe dx + ò xe - x dx
2
2
0
æ 1100 - 1050 ö
- Fç
. ) + 1 - F (125
. )
÷ = F ( -375
40
è
ø
= 1 - F ( 375
. ) + 1 - F (125
. )
3. (A) Test 1: f X ( x) ³ 0 is true
= 2 - 0.9999 - 0.8944 = 0.1057 or 20.57 %
l 3x
1 é l 3b 1 ù
Test 2: area must be 1 i.e. ò
dx = ê
- ú =1
4
4ë 3
3û
0
b
4. (C)
11. (D) P { x > T + 95.6} = 0.0505
æ T + 95.6 - ax
( T + 95.6 - ax )
= F çç
sx
sx
è
T + 95.6 - ax
This occurs when
= 1.64
sx
1
ln 13
3
¥
ò r( v) = 1 Þ
-¥
ö
÷÷
ø
= F ( -2.5) + 1 - F (2.5) = 1 - F (2.5) + 1 - F (2.5) = 2 - 2 F(2.5)
1
3
= 0.429
e 2 e2
Thus b =
æ 1100 - ax
ö
÷÷ + 1 - F çç
sx
è
ø
æ 900 - 1000 ö
æ 11000 - 1000 ö
= Fç
÷ + 1 - Fç
÷
40
40
è
ø
è
ø
1. (A) P { X > 1} = ò pX ( x) dx
=ò
Chap 7.1
=1 - F
1
1
4k = 1 Þ k =
2
2
¥
4
2
2
ò x r( x) dx = ò x
-¥
0
x
dx = 8
8
This occur when -
P ( A) = FX ( 3) - FX (1) = e
1
2
-e
6. (D) P ( B) = FX (2.5) = 1 - e
-
-
3
2
2 .5
2
12. (A) 0.25 exceeds when
P ( C) = FX (2.5) - FX (1) = e
-e
x
100
> 0.21 or x > 5 v
P(0.25 W exceeded) = P { x > 5}
= 0.7135
= P { x > 5} + P { x < -5} = 1 - P ( x £ 5) + P { x < -5}
æ5 -0 ö
æ -5 - 0 ö
æ 5 ö
æ -5 ö
= 1 - Pç
÷ + Pç
÷ = 1 - Fç
÷ + Fç
÷
è 4.2 ø
è 4.2 ø
è 4.2 ø
è 4.2 ø
2 .5
2
æ 5 ö
æ 5 ö æ
æ 5 öö
æ 5 ö
= 1 - Fç
÷ + 1 - Fç
÷ = 2ç 1 - F ç
÷ ÷ = 2Qç
÷
4
.
2
4
.
2
4
.
2
è
ø
è
ø è
è
øø
è 4.2 ø
= 0.3200
13. (A) P {160 < X £ 120} = FX (210) - FX (160)
æ 210 - 200 ö
æ 160 - 200 ö
= Fç
÷ - Fç
÷
20
20
è
ø
è
ø
8. (A) P ( X > 2) = P {2 < x} + P { x < -2}
= 1 - P { x £ 2} + P { x < -2} = 1 - F (2) + F ( -2)
We know that for gaussian function F ( - x) = 1 - F ( x)
= F (0.5) - F ( -2) = F (0.5) + F (2) - 1
Thus P ( X > 2)= 1 - F (2) + 1 - F (2)
= 0.6915 + 0.9773 - 1 = 0.6687
= 2 - 2 F(2) = 2 - 2(0.9772) = 0.0456
9. (C) Rejected resistor corresponds to { x < 900 W} and
{ x > 1100 W}.
Fraction
rejected
corresponds
...(ii)
2
= 0.3834
7. (D) C = A Ç B = {1 < X < 2.5}
1
2
. - ax
T - 112
= 103
.
sx
Solving (i) and (ii) we get ax = T + 30 and sx = 40
x
x
- ù
é
1 5. (B) For x = 0, FX ( x) = ò e 2 dx = ê1 - e 2 ú u( x)
0 2
ë
û
x
-
...(i)
P { x £ T - 112
. } = 0.1515
( T - 112
. - ax )
( T - 112
. - ax )
=F
=1 - F
= 8485
sx
sx
kv v
Thus r( v) =
=
4
8
Mean Square Value =
ö
÷÷ = 0.9495
ø
to
14. (C) P { X > 231}= 1 - P { X £ 231} = 1 - FX (231)
æ 231 - 200 ö
= 1 - Fç
. ) = 1 - 0.9394 = 0.0606
÷ = 1 - F(155
20
è
ø
probability of rejection.
P {resistor rejected} = P { X < 900} + P { X < 1100}
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UNIT 7
15. (B) We use the Rayleigh distribution with a = 0 and
b = 400
For probability P { X £ 1}= FX (1) = 1 - e
-
1
400
17 -3 ö
æ
= 52ç 1 e ÷ = 29.994 weeks
2
è
ø
4
24. (B) E[ X ] = X = å xi P ( xi )
= 0.0025 or 0.25 %
i =1
= 10
. (0.4) + 4(0.25) + 9(0.15) + 16(0.1) = 4.35
16. (C) P { X ³ 52} = 1 - FX (52)
52 ù
52
é
= 1 - ê1 - e 400 ú = 1 - e 400 = 0.00116 or 0.12 %
êë
úû
2
2
x
òf
17. (C) FX =
Communication System
X
( u) du =
-¥
x
òb
2
-¥
x
1 3 -2 1 é 6 ù
xe = ê
ú = 48
2 êë(1 2) 4 úû
0 2
¥
25. (A) E[ g( X )] = E[ X 3 ] = ò
( b p) du
+ ( u - a) 2
26. (A) Mean of X =
x
1
-¥
0
2
ò xf X ( x) dx = ò x 3(1 - x) dx =
1
4
Let v = u - a and dv = du to get
b
FX ( x) =
p
=
x -a
ò
-¥
27. (B) Variance of X is s2x = E[ X 2 ] - m 2x
x -a
dv
b é1
æ v öù
=
tan -1 ç ÷ú
b2 + v 2 p êë b
è b øû -¥
E[ X 2 ] =
1 1
æ x -aö
+ tan -1 ç
÷
2 p
è b ø
òx
1
2
-¥
f X ( x) dx = ò x 2 3(1 - x) 2 dx =
0
s2x =
1 æ1ö
3
-ç ÷ =
10 è 4 ø
80
Hence B is correct option
P { x > 0} = 1 - P { x £ 3}
28. (B) Here Y = g( X ) = e
= 1 - P ( x = 0) - P ( x = 1) - P ( x = 2) - P ( x = 3)
So E[ Y ] = E[ g( Y )] =
ì 2 0 21 2 2 2 3 ü
æ 19 ö
=1 - e í
+
+
+ ý = 1 - e -2 ç
÷ = 0.1429
0
!
2
!
2
!
3
!
è 3 ø
î
þ
-2
-
=
0
2
= 0.135
0!
1
20
X
3
¥
ò g( X ) g
-¥
15
X
( x) dx = ò e
-5
x
5
1
dx
15 - ( -5)
15
x
- ù
é
1 1
-3
5
ê-5 e ú = [ e - e ] = 0.667
5
ë
û -5
E[ Y 2 ] = E[(2 X - 3) 2 ] = 4 X 2 - 12 X - 9
æ
ö
= 1 - ç 1 - e W0 ÷ = e -1
ç
÷
è
ø
W0
= 4(11) - 12( -3) - 9 = 89
2
s2Y = Y 2 - Y = 89 - 9 2 = 8
21. (A) P {5 or more}= 1 - P (0) - P (1) - P (2) - P ( 4)
é 30 31 32 33 34 ù
131 -3
= 1 - e -3 ê +
+
+
+
=1 e = 0.1847
ú
8
ë 0 ! 11 2 ! 3 ! 4 ! û
30. (A) F
XY
(x, y) = 0.1u (x - 1)u ( y - 1) + 0.35u (x - 2)u ( y - 2)
+0.05u (x - 3)u ( y - 3) + 0.5u (x - y)u ( y - 4)
P { X £ 2.5, Y £ 6.0} = f XY (2.5, 6.0) = 0.1 + 0.35 = 0.45
-3
22. (C) P(0) = e = 0.0498
average number of week, per year with no murder
52 e -3 = 2.5889 week.
31. (B) P { X £ 30
. }= FX ( 30
. ) = FXY ( 30
. , ¥)
= 0.1 + 0.35 + 0.05 = 0.5
32. (B) FX ( x) = FX , Y ( x, ¥)
23. (D) P {3 or more}= 1 - P (0) - P (1) - P (2)
2 ö
5x
5 æ x + e- ( x + 1 ) y
lim ç
- e - y ÷u( y) =
÷
y ®¥ 4 ç
x
+
4
(
x
+ 1)
1
è
ø
2
é
32 ù
17 -3
= 1 - e ê1 + 3 +
ú = 1 - 2 e = 0.5768
2
ë
û
-3
Average number of weeks per year that number of
murder exceeds the average
-
29. (C) E[ Y ] = E[2 X - 3] = 2 X - 3 = 2( -3) - 3 = -9
20. (B) P {W > W0 } = 1 - P {W £ W0 } = 1 - FW (W0 )
2
2 ö
1
5
æ
limç 1 + e -5 y - e - y ÷ = 1
y ®¥
4
4
è
ø
33. (B) FY ( y)= FX , Y ( ¥, y)
Page
396
1
10
2
¥
æ 3k ö
18. (B) Here f X ( x) = e -2 å çç ÷÷d( x - k)
k=0 è k! ø
19. (D) P ( x = 0) = e -2
¥
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UNIT 7
¥
w
= ò 5 e - ( w- y) u( w - y) u( y)2 e -2 ydy = 10 ò e -5w+ 3 ydy, w > 0
-¥
0
10
=
u( w)[ e -2 w - e -5w ]
3
=
6
2
y= 0 x = 0
òòx
2
y 2 f X , Y ( x, y) dx dy
2
51. (C) E[ g( X , Y ) ] =
=
¥
ò
-¥
x2 e
x2
-
2 s2
2 ps2
¥
dx ò
-¥
e
-
R XY = XY = C XY + X Y = -
¥ ¥
ò ò(x
2
+ y2 )
e
- y2
2 ps2
dy +
¥
ò
-¥
y2 e 2 s
¥
2
2 ps2
r=
æ x 2 + y2 ö
÷
-ç
2 ÷
ç
è 2s ø
2 ps2
-¥ -¥
y2
2 s2
dy ò
-¥
dx dy
e2s
2 ps2
dx
variance s2 .
Thus E[ g( x, y)]= E[( x 2 + y 2 )]2s2
52. (A) E[W ] = E[ 3 X - Y ] = 3 X - Y = 6 - 4 = 2
53. (B) E[W 2 ] = E[( 3 X - Y ) 2 ] = E[9 X 2 - 6 XY + Y 2 ]
= 9 X 2 - 6 XY + Y 2 = 9 X 2 - 6 X Y + Y 2
= 9( 8) - 6(2)( 4) + 25 = 49
2
54. (B) s2W = E[(W - W ) 2 ] = E[W 2 - 2W W + W ]
2
= W 2 - W = 49 - 4 = 45
-¥ -¥
E[ X ] = ò ò
0 0
3 2
E[ Y ] = ò ò
0 0
3 2
f X , Y ( x, y) dxdy = ò ò
0 0
x2 y2
8
dx dy =
9
3
2
4
x y
dx dy =
9
3
x2 y
dx dy = 2
9
Since R XY =
8
æ4ö 8
= E[ X ] E[ Y ] = 2ç ÷ = , we have X and Y
3
è 3ø 3
uncorrelated form
3
From marginal densities f X ( x) = ò
0
2
fY ( y) = ò
0
Page
398
xy
x
dy = , 0 < x < 2
9
2
xy
2y
dy =
, 0< y<3
9
9
we have f X ( x) fY ( y) =
1
2 3
+
1
1æ
1 ö
(2) = ç 2 ÷
2
2è
3ø
1 ö
æ1ö 5
æ 1 öæ
æ 19 ö
= 3 + 2ç ÷ + + 6ç ÷ç 2 ÷ + 9ç
÷ = 98 - 3
3ø
è2 ø 2
è 2 øè
è 2 ø
second moment of gaussian density with zero mean and
3 2
Y
2
= 3 + 2 X + X 2 + 6 XY + 9 Y 2
2
density. The first factor equal s2 because they are
ò ò xy
and
C XY
-1/2 3
-1 2
=
=
s X sY ( 914 )( 11/2 )
3 33
factors equal 1 because they are area of a gaussian
¥ ¥
X
57. (B) W = ( X + 3Y ) 2 + 2 X + 3
-x 2
Both double integral are of the same form. the second
55. (B) R XY =
and
5 æ1ö
9
-ç ÷ =
2 è2 ø
4
2
11
11
2
- (2) =
s2Y = Y 2 - Y =
2
2
x y
dx dy = 64
24
ò ò
= f X ( x) fY ( y)
2
-¥ -¥
4
f X , Y ( x, y)
statistically independent.
56. (C) s2X = X 2 - X =
¥ ¥
50. (A) E [( XY ) 2 ] =
Thus
Communication System
xy
, 0 < x < 2 and 0 < y < 3
9
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CHAPTER
7.2
RANDOM PROCESS
Statement for Question 1 - 4 :
3. The value of E[ X 2 ( t)] is
A random process X ( t) has periodic sample
(A)
t0 A 2
T
(B)
(C)
2 t0 A 2
3T
(D) 0
functions as shown in figure where A, T and 4 t0 £ T are
constant but Î is random variable uniformly distributed
on the interval (0, T).
4. The value of s2X is
X (t)
t
0
Fig. P7.2.1-4
1. The first order density function is
2t
ìï T - 2 t0
d( x) + 0
(A) í T
AT
ïî 0
(B)
0£x<A
t0 A
4T
(B)
t0 A 2 t0
T
T
(C)
t0 A
T
é2 t0 ù
êë 3 + T úû
(D)
t0 A 2
T
é2 t0 ù
êë 3 + T úû
5. An ergodic random power x( t) has an auto-correlation
function
R XX ( t) = 18 +
2
1 + 4 cos(12 t)
6 + t2
-
The X is
(A) ± 18
(C) ± 17
(B) ± 13
(D) ± 18 ± 17
6. For random process X = 6 and
else where
R XX ( t, t + t) = 36 + 25 e -|t|.
Consider following statements :
1. X ( t) is first order stationary.
2. X ( t) has total average power of 36 W.
2. The value of E[ X ( t)] is
(C)
t0 A é2 t0 ù
T êë 3 T úû
else where
T - 2 t0
2 T0
d( x) +
T
AT
t A
(A) 0
2T
(A)
0£x<A
T - 2 t0 2 t0
+
[ u( x) - u( x - A)]
T
AT
2t
ìï T + 2 t0
d( x) + 0
(C) í T
AT
ïî 0
(D)
t0 A 2
3T
3. X ( t) is a wide sense stationary.
t A
(B) 0
T
(D) 0
4. X ( t) has a periodic component.
The true statement is/are
(A) 1, 2, and 4
(B) 2, 3, and 4
(C) 2 and 3
(D) only 3
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UNIT 7
Communication System
2
7. A random process is defined by X ( t) + A where A is
The variance of random variable Y = ò X ( t) dt will be
continuous random variable uniformly distributed on
(A) 1
(B) 2.31
(0,1). The auto correlation function and mean of the
(C) 4.54
0
(D) 0
process is
(A) 1/2 & 1/3
(B) 1/3 & 1/2
13. A random process is defined by X ( t) = A cos( pt)
(C) 1 & 1/2
(D) 1/2 & 1
where A is a gaussian random variable with zero mean
and variance s 2p . The density function of X(0)
Statement for Question 8 - 9 :
A
random
process
(A)
is
defined
by
Y ( t) = X ( t) cos( w0 t + q) where X ( t) is a wide sense
1
2 ps A
-
e
x2
2 s 2A
(B)
(C) 0
2 ps A e
-
x2
2 s 2A
(D) 1
stationary random process that amplitude modulates a
carrier of constant angular frequency w0 with a random
Statement for Question 14-15 :
phase q independent of X ( t) and uniformly distributed
on ( -p / p).
The two-level semi-random binary process is
defined by
X ( t) = A or -A
8. The E[ Y ( t)] is
where ( n - 1) T < t < nt and the levels A and -A
(B) -E[ X ( t)]
(D) 0
(A) E[ X ( t)]
(C) 1
occur with equal probability. T is a positive constant
9. The autocorrelation function of Y ( t) is
1
(A) R XX ( t) cos( w0 t)
(B) R XX ( t) cos( w0 t)
2
(D) None of the above
(C) 2 R XX ( t) cos( w0 t)
Statement for Question 10 - 11 :
Consider a low-pass random process with a
white-noise power spectral density S X ( w) = N/2 as
shown in fig.P7.2.10-11.
and n = 0, ± 1, ± 2
14. The mean value E[ X ( t)] is
(A) 1/2
(B) 1/4
(C) 1
(D) 0
15. The auto correlation R XX ( t1 = 0.5 T, t2 = 0.7 T) will be
(A) 1
(B) 0
(C) A
2
(D) A 2/2
16. A random process consists of three samples function
X ( t, s1 ) = 2, X ( t, s2 ) = 2 cos t1 and X ( t, s3) = 3 sin t- each
occurring with equal probability. The process is
(A) First order stationary
(B) Second order stationary
Fig.P7.2.10-11
(C) Wide-sense stationary
10. The auto correlation function R X ( t) is
(D) Not stationary in any sense
(A) 2 NB sinc (2pbt)
(B) pNB sinc (2pbt)
(C) NB sinc (2pbt)
(D) None of the above
Statement for Question 17 - 19 :
(A) 2 NB
(B) pNB
ergodic random process is shown in fig.P.7.2.17-19
(C) NB
D) D
The auto correlation function of a stationary
11. The power PX is
NB
2p
50
12. If X ( t) is a stationary process having a mean value
E[ X ( t)] = 3
R XX ( t) = 9 + 2 e
Page
400
and
autocorrelation
function
-|t|
.
20
-10
10
Fig. P7.2.17-19
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Random Process
17. The mean value E[ X ( t)] is
Chap 7.2
24. A complex random process Z ( t) = X ( t) + jY ( t) is
(A) 50
(B)
50
defined by jointly stationary real process X ( t) and Y ( t).
(C) 20
(D)
20
The E[|Z ( t)|2 ] will be
(A) 2 R XY (0) + R XX (0) + RYY (0) (B) R XX (0) + RYY (0)
(C) R XX (0) - RYY (0)
18. The E[ X 2 ( t)] is
(A) 10
(B)
10
(C) 50
(D)
50
(D) RYY (0) - R XX (0)
25. Consider random process X ( t) = A0 cos( w0 t + q)
where A0 and w0 are constant and q is a random
variable uniformly distributed on the interval (0, p). The
19. The variance s2X is
power in X ( t) is
(A) 20
(B) 50
(A) A 2
(C) 70
(D) 30
(C)
20. Two zero mean jointly wide sense stationary
random process
X ( t) and Y ( t) have no periodic
components. It is know that s 2X = 5 and s2Y = 10. The
function, that can apply to the process is
2
(A) R XX ( t) = 6 u( t) e -3t
(C) R XY ( t) = 9(1 + 2 e 2 ) -1
ésin( 3t) ù
(B) RYY ( t) = 5 ê
ë 3t úû
(D) None of the above
A2
1
4
(D) 1
real random process is
(A) d( w + w0 ) - d( w - w0 )
(C) d( w) +
w2
w + 16
2
w2
w2 + 25
w2
(D) 2
w + 16
(B)
27. The valid power density spectrum is
21. A stationary zero mean random process X ( t) is
w2
1 + w2 + jw2
(C) e - ( w -1 )
2
(B)
w2
- d( w)
w +1
(D)
w2
w + 3w2 + 3
component. The valid auto correlation function is
(A) 16 + 18 cos( 3t)
(B) 24 da 2 (2 p)
e ( -6 t )
(C)
(1 + 3t 2 )
(D) 24d( t - t)
A2
1
2
26. The non valid power spectral density function of a
(A)
ergodic has average power of 24 W and has no periodic
(B)
4
6
28. A power spectrum is given as
P
é
r XX ( w) = ê1 + ( w / W ) 2
ê
0
ë
22. Air craft of Jet Airways at Ahmedabad airport
w < KW
w > KW
where P , W , and K are real positive constants. The
arrive according to a poisson process at a rate of 12 per
hour. All aircraft are handled by one air traffic
sums bandwidth of power spectrum is
controller. If the controller takes a 2 - minute coffee
(A) W
tan+ k
-1
k
(B) W
(C) W
tan -1 k
+1
k
(D) ¥
break, what is the probability that he will miss one or
more arriving aircraft ?
(A) 0.33
(B) 0.44
(C) 0.55
(D) 0.66
29. Consider the power spectrum given by
ìP
r XX ( w) = í
î0
23. Delhi airport has two check-out lanes that develop
waiting lines if more than two passengers arrives in
any one minute interval. Assume that a poission
process describes the number of passengers that arrive
(A) 0.16
(C) 0.32
(B) 0.29
(D) 0.49
w <W
w >W
where P and W are real positive constants. The
rms bandwidth of the power spectrum is
for check-out. The probability of a waiting line if the
average rate of passengers is 2 per minute, is
k
-1
tan -1 k
(A)
W
(C)
W
2
3
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(B)
W2
3
(D)
W
2
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UNIT 7
Communication System
30. For a random process R XX ( t) = P cos 4 ( w0 t) where P
37. A random process X ( t) has an autocorrelation
and w0 are constants. The power in process is
function R XX ( t) = A 2 + Be -|t| Where A and B are
(A) P
(B) 2P
constants. A system have an input response
(C) 3P
(D) 4P
ì e - Wt 0 < t
h( t) = í
î 0 t <0
31. A random process has the power density spectrum
r XX ( w) =
6w2
[1 + w 2 ] 3
where W is a real positive constant, which X ( t) is
. The average power in process is
(A) 1/4
(B) 3/8
(C) 5/8
(D) 1/2
its input. The mean value of the response is
A
A
(B)
(A)
W
2W
32. A deterministic signal A cos( w0 t), where A and w0
(C)
are real constants is added to a noise process N ( t) for
which r NN ( w) =
W2
W2+ w2
and W > 0 is a constant. The ratio
2A
W
38. In previous question if impulse response of system is
of average signal power to average noise power is
(C)
ì e - Wt sin( w0 t) 0 < t
h( t) = í
0
t <0
î
A
(B)
W
A2
(D)
W
(A) 1
2A
W
where W and w0 are real positive constants, the
mean value of response is
33. The autocorrelation function of a random process
(A)
Aw0
w + W2
(C)
2A æ
1
çç 2
w0 è w0 + W 2
X ( t) is
2
R XX ( t, t + t) = 12 e Yt cos 2 (24 t)
The R XX ( t) is
(A) 6 e -4 t
(C) 48 e
2
(B) 12 e -4 t
-4 t 2
power density spectrum f XX ( w) is
(B)
(C) 3 + jw2
35.
The
4 e -3|t|
1 + w2
of jointly
wide
sense
- Wt
where A > 0 and W > 0 are constants. The r XX ( w) is
A
W - w2
(B)
A
W + w2
(C)
A
W + jw
(D)
A
W - jw
2
E[ X ( t)] = 2, the mean value of the system's response
Y ( t) is
(A)
1
128
(B)
1
64
(C)
3
128
(D)
1
32
A random process X ( t) is applied to a network with
stationary process X ( t) and Y ( t) is R XY ( t) = Au( t) e
(A)
impulse response h( t) = u( t) te - at
Y ( t) is known to have the same form R XY ( t) = u( t) te - at
40. The auto correlation of Y ( t) is
4 + at - a|t|
1 + at - a|t|
(A)
(B)
e
e
3
4a
3a 2
2
36. A random process X ( t) is applied to a linear time
system.
A response
Y ( t) = X ( t) - X ( t - t)
occurs when t is a real constant. The system's transfer
function is
(A) 1 - e jwt
(C) 2 je - jwt/ 2 cos
Page
402
wt
2
(B) 2 je - jwt/ 2 sin
(D) 1 + e - jwt
where a > 0 is a
constant. The cross correlation of X ( t) with the output
(C)
invariant
ö
÷÷
ø
ö
1
÷
2 ÷
+W ø
1
+ W2
Statement for Question 40-41 :
(D) 18d( w)
cross correlation
ö
÷÷
ø
A
2 w0
input of a system for which h( t) = 3u( t) t 2 e -8 t . If
(D) None of the above
6
6 + 7 w3
æ
çç 2
è w0
A æ
(D)
ç
2 w0 çè w02
(B)
2
0
39. A stationary random process X ( t) is applied to the
2
34. If X ( t) and Y ( t) are real random process, the valid
(A)
(D) 0
wt
2
4 + at - a|t|
e
8a2
(D)
1 + at - a|t|
e
4 a3
41. The average power in Y ( t) is
1
1
(A)
(B) 3
a
4a3
(C)
1
3a 2
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Random Process
Chap 7.2
Statement for Question 42 - 43 :
SOLUTION
A random noise X ( t) having a power spectrum
r XX ( w) =
3
49+ w 2
is applied to a differentiator that has a
transfer function H ( w) = jw. The output is applied to a
network for which h( t) = u( t) t 2 e -7t
1.
(A)
Î
Let
have
value
Now
e.
P { X £ x Î= e} = FX ( X|Î= e) and for any Î must be zero
for x < 0 because x( t) is never negative. The event
{ X £ 0} is satisfied whenever x( t) is zero. This happens
42. The average power in X ( t) is
(A) 5/21
(B) 5/24
(C) 5/42
(D) 3/14
during
the
fraction
of
time ( T - 2 t0 ) / T.
FX ( x|Î= e) = [( T - 2 T0 ) / T} u( x).
For
Hence
0£x<A
the
additional time interval or fraction of time where X £ x
becomes 2 to 2 t0 x / AT.
43. The power spectrum of Y ( t) is
2t x
æ T - 2 t0 ö
Thus FX ( x Î= e) = ç
÷u( x) + 0 , 0 £ x < A
T
AT
è
ø
(A)
4 w2
( 49 + w2 ) 3
(B)
12 w2
( 49 + w2 ) 4
(C)
42 w3
( 49 + w2 ) 2
(D) None of the above
= 1,A £ x
= 0,x < 0
By differentiation
44. White noise with power density N0/2 is applied to a
lowpass network for which H(0) = 2. It has a noise
2t
æ T - 2 t0 ö
f X ( x Î= e) = ç
÷d( x) + 0 , 0 £ x < A
AT
è T ø
= 0 else where
bandwidth of 2 MHz. If the average output noise power
f X , e ( x, e) = f X ( x Î= e) fÎ( e)
is 0.1 W in a 1 - W resistor, the value of N0 is
(A) 12.5 nW/Hz
(B) 12.5 mW/Hz
(C) 25 nW/Hz
(D) 25 mW/Hz
2 T0
æ T - 2 t0 ö
,0 £ x < A and 0 < e < T
=ç
÷d( x) +
2
AT 2
è T
ø
45. An ideal filter with a mid-band power gain of 8 and
f X ( x) =
power spectrum r XX ( w) =
e
- w 2 /8
the network's output is ( F (2) = 0.9773)
(B) 90.3
(C) 20.2
(D) 100.4
( x, e) de
X ,Î
2t
æ T - 2 t0 ö
=ç
÷d( x) + 0 .0 £ x < A
AT
è T ø
. The noise power at
(A) 60.8
òf
-¥
bandwidth of 4 rad/s has noise X ( t) at its input with
50
8p
¥
= 0 elsewhere.
¥
ò xf
2. (B) E[ X ( t)] =
( x) dx
X
-¥
46. White noise with power density N0/2 = 6 mW/Hz is
applied to an ideal filter of gain 1 and bandwidth W
=
¥
2t x
t A
æ T - 2t ö
ò-¥ xçè T 0 ÷ød( x) dx + ò0 AT0 dx = 0T
A
rad/s. If the output's average noise power is 15 watts,
the bandwidth W is
3. (C) E[ X 2 ( t)] =
(A) 2.5 ´ 10 -6
(B) 2.5 p ´ 10 -6
(C) 5 ´ 10 -6
(D) p5 ´ 10 -6
òx
-¥
A
2
2 t0 x 2 2 t0 A 2
=
AT
3T
0
f X ( x) dx = ò
4. (D) s2X = E[ X 2 ( t)] - { E[ X ( t)]} 2
2
47. A system have the transfer function H ( w) = 1 + ( w1/W ) 4
where W is a real positive constant. The noise
bandwidth of the system is
=
2 t0 A 2 t02 A 2 t0 A 2
=
3T
T2
T
é2 t0 ù
êë 3 - T úû
5. (A) We know that ( i) if X ( t) has a periodic component
(A)
1
3
pW 2
(B)
(C)
1
6
pW 2
(D) None of the above
1
4
¥
pW 2
then R XX ( t) will have a periodic component with the
-
same period. (ii) if E[ X ( t)] = X m 0 and X ( t) is ergodic
with no periodic components then lim R XX ( t) = X
2
|t|®¥
2
************
Thus we get X = 18 or X= ± 18
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UNIT 7
6. (C) X = Constant and R XX ( t) is not a function of t, so
X ( t) is a wide sense stationary. So 1 is false & 3 is true.
Communication System
13. (A) For t = 0,X(0)= A,
1
So f X ( x) =
2 ps A
-
e
x2
2 s 2A
PXX = R XX (0) = 36 + 25 = 61.
Thus 2 is false if X ( t) has a periodic component, then
14. (D) E[ X ( t)] = AP ( A) + ( - A) P ( - A) =
R XX( t) will have a periodic component with the same
period. Thus 4 is false.
15. (C) Here R XX ( t1 , t2 )= A 2
1
7. (B) R XX ( t, t + t) = E[ X ( t) X ( t + t)] = E[ A 2 ] = ò a 2 da =
0
1
X= E[ X ( t)] = E[ A ] = ò a da =
0
1
3
= E X [ X ( t)] ò cos( w0 t + q)
-X
If
both
and
t1
are
t2
in
the
same
interval
( n - 1) T < t, t2 < nT, n = 0, ±, ±2...
and R XX ( t1 , t2 ) = 0 otherwise
1
2
Hence R XX (0.5 T, 0.7 T) = A 2
16. (D) Let x1 = 2, x2 = 2 cos t and x3 = 3 sin( t)
1
1
1
Then f X ( x) = d( x - x1 ) + d( x - x2 ) + d( x - x3)
2
3
3
8. (D) E[ Y ( t)} = E[ X ( t) cos( w0 t + q)]
X
A A
- =0
2 2
1
dq = 0
2p
where E X [×] represent expectation with respect to X
¥
and E[ X ( t)] =
ò xf
X
( x) dx
-¥
only
=
9. (B) RYY ( t, t + t)
é1
1
ù
1
ò x êë 3 d( x - x ) + 3 d( x - x ) + 3 d( x - x ) úû
1
2
2
-¥
= E[ X ( t) cos( w0 t + q) X ( t + t) cos( w0 t + q + w0 t)]
1
= R XX ( t) [cos( w0 t) + cos(2 w0 t + 2 q + w0 t)]
2
1
= R XX ( t) cos( w0 t)
2
10. (C) S X ( w) =
¥
=
1
[2 + 2 cos t + 3 sin t ]
3
The mean value is time dependent so X ( t) is not
stationary in any sense.
17. (D) We know that for ergodic with no periodic
N
æ w ö
rectç
÷
2
è 4 pb ø
component
2
lim R XX ( t) = X ,
We know that R X ( t)¬¾® S X ( w)
W
æ w ö
sin(Wt)¬¾® rect ç
÷
p
è 2W ø
|t|®¥
2
Thus X = 20
or
X = 20
18. (C) R XX (0) = E[ X 2 ( t)] = R XX (0) = 50 = X 2
Here W = 2pB
Hence R X ( t) =
2
2 pB N
sin (2pBt) = NB sinc (2pBt)
p 2
19. (D) s2X = X 2 - X = 50 - 20 = 30
20. Here X = 0, Y = 0, R XX (0) = 5, sY2 = RYY (0) = 10
11. (C) PX = X = R X (0) = NB since sinc (0) = 1
2
2
2
For (A) : Function does not have even symmetry
For (B) : Function does not satisfy RYY (0) = 10
2
12. (C) E[ Y ] = E[ ò X ( t) dt = ò E[ X ( t)]dt = 3ò dt = 6
0
0
2
2
0
0
For
0
2 2
E[ Y 2 ] = E[ ò X ( t) dt ò X ( u) du]] = ò ò E[ X ( t) X ( u) du dt ]
(C)
:
Function
not
satisfy
|R XY ( t)|£ R XX (0) RYY (0) = 50
0 0
2 2
2 2
21. (D) For (A) : It has a periodic component.
0 0
0 0
For (B) ; It is not even in t, total power is also incorrect.
= ò ò R XX ( t - u) dt du = ò ò [9 + 2 e -|t - u| ]dt du
2 2
= 36 + 2 ò ò e -|t - u|dt du = 4(10 + e -2 )
0 0
s2Y = E[ Y 2 ] - ( E[ Y ]) 2 = 4(1 + e -2 ) = 4.541
For (C) It depends on t not even in t and average power
is ¥.
22. (A) P (miss/or more aircraft)= 1 - P(miss 0)
= 1 - P (0 arrive) = 1 -
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404
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UNIT 7
¥
¥
-¥
0
E[ Y ( t)] = A ò h( x)dx = A ò e - Wt dt =
So W = 2.5 p ´ 106
A
W
¥
38. (A) X = A
¥
¥
-¥
0
E[ Y ( t)] = Y = X ò h( t) dt = A ò e
¥
Aw
sin( w0 t) dt = 2 0 2
w0 + W
- Wt
47. (B) Noise bandwidth Wn =
¥
ò H( w)
2
H (0)
¥
¥
òR
XY
2 -8 t
*********
( t + x)h( x) dx
-¥
¥
ò u( x)u( x + t)( tx + x ) e
= e - at
2
-2 ax
dx
-¥
There are two cases of interest t ³ 0 and t < 0 Since
RYY ( t) is an even function we solve only the ease t ³ 0
¥
RYY ( t) = e - at ò ( tx + x2 )e -2 ax dx =
0
1 + at - ax
e
4 a3
41. (B) Power in y( t) = RYY (0) =
1
4a3
¥
¥
dw
1
3
3
ò r XX ( w) dw = 2 p -¥ò 49 + w2 = 14
2p -¥
42. (D) PXX =
F
43. (B)h2 = 49 t) t 2 e -7t ¬¾
®
2
= H 2 ( w)
(7 + jw) 3
2
sYY ( w) = s XX ( w) = H1 ( w) H 2 ( w) =
12 w2
( 49 + w2 ) 4
2
44. (A) PYY =
So N0 =
N0 H (0) Wn
2p
2 p(0.1)
2
H (0) Wn
=
= 0.1
2 p(0.1)
(2) 2 2 p ´ 2 ´ 106
. ´ 10 -8 W/Hz = 12.5 nW/Hz
= 125
45. (A) PYY =
4
¥
2
1
r XX ( w) H ( w) dw
ò
2 p -¥
w2
1
50 - 8
=
e
ò
2p 4 8p
=
Page
406
200
=
p
4
ò
4
e
-
w2
2 (u)
2 p( 4)
200
200
[ F (2) - F ( -2)] =
[2 F (2) - 1)] = 60.8
p
p
46. (B) PYY =
=
(8 ) dw
¥
2
1
ò r XX ( w) H( w) dw
2 p -¥
1
6 ´ 10 -6 W 6 ´ 10 -6 W
-6
d
w
6
´
10
=
=
= 15
2 p -òW
p
p
W
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dw
2
dw
pW
=
4
(
)
1
+
w
/W
2
2
0
3
39. (C)Y = X ò h( t) dt = 2 ò 3t e dt =
128
-¥
0
40. (D) RYY ( t)=
2
0
Wn = ò H ( w) dwsince H(0) = 1 = ò
0
¥
Communication System
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GATE EC BY RK Kanodia
CHAPTER
7.3
NOISE
1. The power spectral density of a bandpass white noise
5. A mixer stage has a noise figure of 20 dB. This mixer
n( t) is N / 2 as shown in fig.P.7.3.1. the value of n is
stage is preceded by an amplifier which has a noise
2
figure of 9 dB and an available power gain of 15 dB.
The overall noise figure referred to the input is
Fig. P7.3.1
(A) 11.07
B) 18.23
(C) 56.48
(D) 97.38
6. A system has three stage cascaded amplifier each
(A) NB
(B 2 NB
(C) 2pNB
(D)
stage having a power gain of 10 dB and noise figure of 6
NB
p
dB. the overall noise figure is
2. In a receiver the input signal is 100 mV, while the
(A) 1.38
(B) 6.8
(C) 4.33
(D) 10.43
internal noise at the input is 10 mV. With amplification
7. A signal process m( t) is mixed with a channel noise
the output signal is 2 V, while the output noise is 0.4 V.
n( t). The power spectral density are as follows
The noise figure of receiver is
Sm ( w) =
(A) 2
(B) 0.5
(C) 0.2
(D) None of the above
3. A receiver is operated at a temperature of 300 K. The
transistor used in the receiver have an average output
resistance of 1 kW. The Johnson noise voltage for a
receiver with a bandwidth of 200 kHz is
(A) 1.8 mV
(B) 8.4 mV
(C) 4.3 mV
(D) 12.6 mV
(C) 4 m V
(D) 16 ´ 10 -18 V
(C)
w2 + 10
w2 + 9
(B)
1
w + 10
2
(D) None of the above
A sonar echo system on a sub marine transmits a
noise voltage generated in a bandwidth of 10 kHz is
(B) 0.4 m V
The optimum Wiener-Hopf filter is
w2 + 9
(A) 2
w + 10
Statement for Question 8-9
4. A resistor R = 1 kW is maintained at 17 o C. The rms
(A) 16 ´ 10 -14 V
6
, Sn ( w) = 6
9 + w2
random noise n( t) to determine the distance to another
targeted submarine. Distance R is given by vt R / 2
where v is the speed of the sound wave in water and t R
is the time it takes the reflected version of n( t) to
return. Assume that n( t) is a sample function of an
ergodic random process N ( t) and T is very large.
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GATE EC BY RK Kanodia
UNIT 7
8. The V will be
Communication System
14. The standard spot noise figure of amplifier is
(A) 2 R NN ( t R - t T )
(B) R NN ( t R - 2 t T )
(A) 4 dB
(B) 5 dB
(C) R NN ( t R - t T )
(D)
R NN ( t R - t T )
(C) 7 dB
(D) 9 dB
1
2
9. What value of the delay t T will cause v to be
15. If a matched attenuator with a loss of 3.2 dB is
maximum ?
placed between the source and the amplifier's input,
what is the operating spot noise figure of the attenuator
(A) t R
(B) 2t R
(C) 3t R
(D) None of the above
amplifier
cascade
if
the
attenuator's
temperature is 290 K ?
10. Two resistor with resistance R1 and R2 are
(A) 9 dB
(B) 10.4 dB
connected in parallel and have Physical temperatures
(C) 11.3 dB
(D) 13.3 dB
T1 and T2 respectively. The effective noise temperature
Ts of an equivalent resistor is
physical
16. In previous question what is the standard spot noise
figure of the cascade ?
(A)
T1 R1 + T2 R2
R1 + R2
(B)
T1 R1 + T2 R1
R1 + R2
(C)
T1 T2 ( R1 + R2 ) 2
( T1 + T2 ) R1 R2
(D)
( T1 + T2 ) R1 R2
T1 + T2 ( R1 + R2 ) 2
(A) 10.3 dB
(C) 14.9 dB
(B) 12.2 dB
(D) 17.6 dB
17. Omega Electronics sells a microwave receiver (A)
having an operating spot noise figure of 10 dB when
Statement for Question 11-12 :
driven by a source with effective noise temperature 130
An amplifier has a standard spot noise figure
K Digilink (B) sells a receiver with a standard spot
F0 = 6.31 (8.0 dB). The amplifier, that is used to amplify
noise figure of 6 dB. Microtronics (C) sells a receiver
the output of an antenna have antenna temperature of
with standard spot noise figure of 8 dB when driven by
Ta = 180 K
a source with effective noise temperature 190 K. The
best receiver to purchase is
11. The effective input noise temperature of this
(A) A
(B) B
amplifier is
(C) C
(D) all are equal
(A) 2520 K
(B) 2120 K
(C) 2710 K
(D) 1540 K
Statement for Question 18-20 :
An amplifier has three stages for which Te1 = 150 K
12. The operating spot noise figure is
(first stage), Te 2 = 350 K, and Te 3 = 600 K (output stage).
(A) 3.2 dB
(B) 6.4 dB
Available power gain of the first stage is 10 and overall
(C) 9.8 dB
(D) 11.9 dB
input effective noise temperature is 190 K
13. An amplifier has three stages for which Te1 = 200 K
18. The available power gain of the second stage is
(first stage), Te 2 = 450 K, and Te 3 = 1000K (last stage). If
(A) 12
(B) 14
the available power gain of the second stage is 5, what
(C) 16
(D) 18
gain must the first stage have to guarantee an effective
input noise temperature of 250 K ?
19. The cascade's standard spot noise figure is
(A) 10
(C) 16
(A) 1.3 dB
(B) 2.2 dB
(C) 4.3 dB
(D) 5.3 dB
(B) 13
(D) 19
Statement for Question 14-16
20. What is the cascade's operating spot noise figure
An amplifier has an operating spot noise figure of
Page
408
when used with a source of noise temperature Ts = 50 K
10 dB when driven by a source of effective noise
(A) 1.34 dB
(B) 3.96 dB
temperature 225 K.
(C) 6.81 dB
(D) None of the above.
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GATE EC BY RK Kanodia
Noise
Chap 7.3
21. Three network are cascaded. Available power gains
elements
are G1 = 8, G2 = 6 and G3 = 20. Respective input effective
matched attenuator with an overall loss of 2.4 dB and a
spot noise temperature are Te1 = 40 K, Te 2 = 100 K and
physical temperatures of 275 K. The overall system
Te 3 = 180 K.
noise temperature of the receiver T sys = 820 K.
(A) 58.33 K
(B) 69.41 K
(C) 83.90 K
(D) 98.39 K
that can be modeled as an impedance
26. The average effective input noise temperature of the
receiver is
22. Three identical amplifier, each having a spot
(A) 420.5 K
(B) 320.5 K
effective input noise temperature of 125 K and available
(C) 220.5 K
(D) 10.5 K
power G are cascaded. The overall spot effective input
noise temperature of the cascade is 155 K. The G is
27.
(A) 3
(B) 5
attenuator-receiver cascade is
(C) 7
(D) 9
(A) 13.67 d
(B) 11.4 dB
(C) 1.4 dB
(D) 1.367 dB
23. Three amplifier that may be connected in any order
in a cascade are defined as follows:
The
average
operating
noise
figure
of
the
28. If receiver has an available power gain of 110 dB
and a noise bandwidth of 10 MHz, the available output
Amplifier
Effective Input Noise
Temperature
Available Power
Gain
A
110 K
4
B
120 K
6
C
150 K
12
noise power of receiver is
(A) 6.5 mW
(B) 8.9 mW
(C) 10.3 mV
(D) 11.4 mV
29. If antenna attenuator cascade is considered as a
noise source, its average effective noise temperature is
The sequence of connection that will give the
lowest overall effective input noise temperature for the
cascade is
(A) 63 K
(B) 149 K
(C) 263 K
(D) 249 K
Statement for question 30-32 :
(A) ABC
(B) CBA
(C) ACB
(D) BAC
An amplifier when used with a source of average
noise temperature 60 K, has an average operating noise
24. What is the maximum average effective input noise
temperature that an amplifier can have if its average
standard noise figure is to not exceed 1.7 ?
(A) 203 K
(B) 215 K
(C) 235 K
(D) 255 K
figure of 5.
30. The T e is
(A) 70 K
(B) 110 K
(C) 149 K
(D) 240 K
25. An amplifier has an average standard noise figure
31. If the amplifier is sold to engineering public, the
of 2.0 dB and an average operating noise figure of 6.5
noise figure that would be quoted in a catalog is
dB when used with a source of average effective source
(A) 0.46
(B) 0.94
temperature Ts . The Ts is
(C) 1.83
(D) 2.93
(A) 156.32 K
(B) 100.81 K
(C) 48.93 K
(D) None of the above
32. What average operating noise figure results when
the amplifier is used with an antenna of temperature
Statement for Question
30 K ?
An antenna with average noise temperature 60 K
connects to a receiver through various microwave
(A) 9.54 dB
(B) 10.96 dB
(C) 11.23 dB
(D) 12.96 dB
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GATE EC BY RK Kanodia
UNIT 7
Communication System
33. An engineer of RS communication purchase an
(A) 1 / W
(B) 2.5 / W
amplifier with average operating noise figure of 1.8
(C) 4.5 / W
(D) 6 / W
when used with a 60 W broadband source having
average source temperature of 80 K. When used with a
different 60 W source the average operating noise figure
39. White noise, for which R XX ( t) = 10 -2 8( t) is applied to
a network with impulse response h( t) = 4( t) 3 - te-4 t The
is 1.25. The average noise temperature of the source is
network's output noise power in a 1 W resistor is
(A) 125 K
(B) 156 K
(A) 0.15 mW
(B) 0.35 mW
(C) 256 K
(D) 292 K
(C) 0.55 mW
(D) 0.95 mW
34. The Te for unit 1 and 2 unit are, respectively
(A) 126.4 K and 256.9 K
40. White noise with power density N0/2 = 6(10 -6 ) W/Hz
is applied to an ideal fitter (gain= 1) with bandwidth W
(B) 256.9 K and 126.4 K
(rad/sec). For output's average noise power to be 15 W,
(C) 527.8 K and 864.2 K
the W must be
(D) 864.2 K and 527.8 K
(A) 2.5 p(10 -6 )
(B) -2.5 p(106 )
(C) 4.5 p(10 -2 )
(D) 4.5 p(106 )
35. The excess noise power of unit 1 and unit 2 are
respectively
41. An ideal filter with a mid-band power gain of 8 and
(A) 15.4 nW and 27.1 nW
bandwidth of 4 rad/s has noise X ( t) at its input with
(B) 23.8 nW and 21.1 nW
power spectrum ( F (2) = 0.9773)
(C) 23.8 nW and 27.1 nW
w2
æ 50 ö - 8
r XX ( w) = ç
÷e
è 8p ø
(D) 15.4 nW and 21.1 nW
36. Consider following statement
S1 : If the source noise temperature T s is very small,
The noise power at the network's output is
164
343
(A)
(B)
p
p
unit-2 is the best to purchase
S2 : If the source noise temperature T s is very small
(C)
211
p
(D)
191
p
unit - 1 is the best to purchase.
42. A system has the power transfer function
correct statement is
(A) S1
(B) S2
(C) both S1 and S2
(D) None
37. A source has an effective noise temperature of
and feeds an amplifier that has an
Ts ( w) = 100800
+ w2
available power gain of Ga ( w) =
(
8
10 + jw
1
2
H( w) =
) . The T
2
s
for this
source is
(A) 10 K
(B) 20 K
(C) 30 K
(D) 40 K
4
where W is a real positive constant. The noise
bandwidth of the system is
pW
(A)
2 2
(C)
38. A system have an impulse response
æ wö
1+ç ÷
èW ø
pW
2
(B)
pW
2
(D) None of the above
43. White noise with power density N0/2 is applied to a
low pass network for which H(0) = 2. It has a noise
- Wt
ìe
e<t
h=í
t
<0
0
î
bandwidth of 2 MHz. If the average output noise power
where W is a real positive constant. White noise
is 8.1 W in a 1 W resistor, the N0 is
with power density 5w/Hz is applied to this system. The
(A) 6.25 ´ 108 W/Hz
(B) 6.25 ´ 10 -8 W/Hz
mean-squared value of response is
(C) 125
. ´ 108 W/Hz
(D) 125
. ´ 10 -8 W/Hz
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GATE EC BY RK Kanodia
Noise
Chap 7.3
Statement for Question 44-46 :
SOLUTION
An amplifier has a narrow bandwidth of 1 kHz and
standard spot noise figure of 3.8 at its frequency of
operation. The amplifier's available output noise power
fc + B
ò
1. (B) n2 = 2
fc + B
N
df = 2 NB
2
is 0.1 mW when its input is connected to a radio
receiving antenna having an antenna temperature of 80
K.
2. (A) NF =
S/N
100/10
i
i
=
=2
So/N o
2/0.4
44. The amplifier's input effective noise temperature Te
3. (A) vn2 = 4kTBR
is
= 4 ´ 1.38 ´ 10 -23 ´ 300 ´ 200 ´ 10 3 ´ 10 3 = 3.3 ´ 10 -12
(A) 812 K
(B) 600 K
(C) 421 K
(D) 321 K
vnrms = 1.8m V
4. (B) vn2 = 4kTBR,
45. Its operating spot noise figure Fop is
(A) 5.16
(B) 7.98
(C) 11.15
(D) 16.23
46. Its available power gain Ga is
(A) 2 ´ 1012
(B) 4 ´ 1012
(C) 8 ´ 1012
(D) 11 ´ 1012
T = (273 + 17) K = 290 K,
R = 1000W, B = 10 Hz, k = 1.38 ´ 10 -23 J/K
4
vn2 = 4 ´ 1.38 ´ 10 -23 ´ 290 ´ 10 3 ´ 10 4 = 16 ´ 10 -14 V 2
vnrms = 0.4m V
5. (A) F1 = 9 dB = 7.94,
F2 = 20 dB = 100
A1 = 15 dB = 31.62,
F -1
100 - 1
= 7.94 +
= 1107
.
F = F1 + 2
A
31.62
6. (C) Gain of each stage A1 = A2 = A3 = 10 dB
Noise figure of each stage
F1 = F2 = F3 = 6 dB or F1 = F2 = F3 = 4 db
F - 1 F3 - 1
4 -1 4 -1
+
=4+
+
= 4.33
F = F1 + 2
A1
A1 A2
10
100
7. (B) H op ( w) =
8. (C) V =
1
2T
Sm ( w)
=
Sm ( w) + Sm ( w)
6
9+ w 2
6
9+ w 2
+6
=
1
10 + w2
T
ò n( t - t
T
)n( t - t R ) dt
-T
Since T is very large
1
T ®¥ 2 T
V = lim
T
ò n( t - t
T
)n( t - t R ) dt = A[ n( t - t T ) n( t - t R )]
-T
Since N ( t) is ergodic,
V » R NN ( t R - t T )
9. (A) Because R NN ( t) £ R NN (0) for any auto correlation
function, V will be maximum if t R = t T
10. (B) Use the current form of equivalent circuit
2 kT1 dw 2 kT2 dw
2kTs dw
where in2 =
,
in2 = i12 + i22 =
+
pR1
pR2
pR
æT
T
Thus Ts = çç 1 + 2
è R1 R2
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ö
T R + T2 R1
÷÷ R = 1 2
R1 + R2
ø
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411
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GATE EC BY RK Kanodia
UNIT 7
11. (D) Te = T0 ( F0 - 1) = 290( 6.31 - 1) = 1539.9 K
13. (B) Te = 250 = Te1 +
6 G 2 - 25 G - 25 = 0 or G = 5
23. (A) Sequence Te
120 150
ABC110 +
+
= 146.25
4
4( 6)
Te 2 Te 3
+
G1
G2
450 1000
250 = 200 +
+
G1
5 G2
14. (D) F0 = 1 +
(125 - 155) G 2 + 125 G + 125 = 0
Te
1540
=1 +
= 9.56 or 9.8 dB
Ta
180
12. (C) Fop = 1 +
or
Communication System
ACB110 +
G1 = 13
150
120
+
= 150.00
4
4(12)
BAC 120 +
Ts
225
( Fop - 1) = 1 +
(10 - 1)
T0
290
CBA150 +
110 150
+
= 144.583 ¬ Best
6
6( 4)
120
110
+
= 161528
.
12
(12)( 6)
= 7.98 or 9.0 dB
24. (A) Te = T0 ( F - 1) £ 290(17
. - 1) = 203 K
15. (D) Here L = 2.089 or 3.2 dB, TL = 290 K
Te = Te1 +
Te 2
T ( F - 1)
= TL ( L - 1) + 0 0
G1
1/ L
(or 2.0 dB) and F OP » 4.467 (or
25. (D) Here F0 » 1585
.
6.5 dB)
= 290[(2.089 - 1) + (2.089)(7.98 - 1)] = 4544.4 K
Ts =
4544.4
= 212
. or 13.3 dB
Fop = 1 +
225
16. (B) F0 = 1 +
T0 ( F0 - 1) 290(1585
.
- 1)
=
= 48.93 K
4.467 - 1
F op - 1
(or 2.4 dB), TL = 275 K
26. (B) Here Ta = 60 K, L = 1738
.
4544.4
= 16.67 or 12.2 dB
290
and T sys = 820 K.
17. (B) For A: Fop = 10(or 10 dB) when Ts = 130 K
TR =
[ T sys - Ta - TL ( L - 1)] 820 - 60 - 275(1738
.
- 1)
=
L
1738
.
= 320.5 K
TeA = 130(10 - 1) = 1170 K
For B: Fo = 398
. (or 6 dB) when Ts = 290 K
27. (B) F op = 1 +
. - 1) = 364.2 K
TeB = 290( 398
For C: Fo = 6.3(or 8 dB) when Ts = 190 K
TeC = 190( 6.3 - 1) = 1007 K, (B) is better as TeB is less.
18. (A) Te = Te1 +
Te1
Te 3
+
G1 G1 G2
Te 3
G2 =
G1 ( Te - Te1 -
600
=
Te 2
) 10(190 - 150 G1
19. (B) F0 = 1 +
350
10
)
N clo =
= 12
1
1 ù
Te 2
Te 3
é
= Te1 ê1 +
+ 2ú
+
G
G
G1
G1 G2
ë
û
Page
412
kT sys GR ( w)Wn 1.38(10 -23)( 820)(1011 )(10 7)
=
2 pL
1738
.
-5
or 6.51 mW
.
= 651110
29. (C) dN ao = k[ Ta + TL ( L - 1)]
dw
dw
= kTs
2p
2p
Thus Ts = Ta + TL ( L - 1)
= 60 + 275(1738
.
- 1) = 263 K
T
Te 3
100 280
21. (A) Te = Te1 + e 2 +
= 40 +
+
= 58.33 K
G1
G1 G2
8
8( 6)
or ( Te1 - Te ) G + Te1 G + Te1 = 0
= 13.67 or 11.4 dB
and WPV = 2 p(10 7) Hz
T
190
20. (C) Fop = 1 + e = 1 +
= 4.8 or 6.81 dB
Ts
50
2
Te
T sys - Ta
820 - 60
=1 +
=1 +
60
Ts
Ts
28. (A) Here GR ( w0 ) = 1011 (or 110 dB)
Te
190
=1 +
= 1.655 or 2.19 dB
T0
290
22. (B) Te = Te1 +
We know that
30. (D) T e = T s ( F op - 1) = 60(5 - 1) = 240 K
31. (C) F o = 1 +
Te
240
=1 +
= 1.8276
290
290
32. (A) F op = 1 +
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Te
240
=1 +
= 9 or 9.54 dB
30
Ts
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GATE EC BY RK Kanodia
CHAPTER
7.4
AMPLITUDE MODULATION
7. An AM broadcast station operates at its maximum
allowed total output of 50 kW with 80% modulation.
The power in the intelligence part is
Statement for Question 1 - 3
An AM signal is represented by
x( t) = (20 + 4 sin 500 pt) cos(2 p ´ 10 5 t) V
1. The modulation index is
(A) 20
(C) 0.2
(B) 4
(D) 10
(B) 204 W
(D) 416 W
3. The total sideband power is
(A) 4 W
(C) 16 W
AM
signal
has
(C) 6.42 kW
(D) None of the above
(A) 0.68
(B) 0.73
(C) 0.89
(D) None fo the above
9. A modulating signal is amplified by a 80% efficiency
amplifier before being combined with a 20 kW carrier to
generate an AM signal. The required DC input power to
the amplifier, for the system to operate at 100%
modulation, would be
(B) 8 W
(D) 2 W
Statement for Question 4 - 5 :
An
(B) 31.12 kW
8. The aerial current of an AM transmitter is 18 A when
unmodulated but increases to 20 A when
modulated.The modulation index is
2. The total signal power is
(A) 208 W
(C) 408 W
(A) 12.12 kW
the
form
(A) 5 kW
(B) 8.46 kW
fc = 10 5 Hz.
(C) 12.5 kW
(D) 6.25 kW
4. The modulation index is
10. A 2 MHz carrier is amplitude modulated by a 500
Hz modulating signal to a depth of 70%. If the
unmodulated carrier power is 2 kW, the power of the
modulated signal is
x( t) = [20 + 2 cos 3000 pt + 10 cos 6000 pt ]cos 2 pfc t
(A)
201
400
(B) - 201
400
(C)
199
400
(D) - 199
400
where
5. The ratio of the sidebands power to the total power is
(A)
43
226
(C)
26
226
(B)
26
226
(D)
43
224
(A) 2.23 kW
(B) 2.36 kW
(C) 1.18 kW
(D) 1.26 kW
11. A carrier is simultaneously modulated by two sine
6. A 2 kW carrier is to be modulated to a 90% level. The
total transmitted power would be
waves with modulation indices of 0.4 and 0.3. The
(A) 3.62 kW
(C) 1.4 kW
(A) 1.0
(C) 0.5
Page
414
(B) 2.81 kW
(D) None of the above
resultant modulation index will be
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(B) 0.7
(D) 0.35
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GATE EC BY RK Kanodia
Amplitude Modulation
Chap 7.4
12. In a DSB-SC system with 100% modulation, the
power saving is
Which of the following frequencies will NOT be present
in the modulated signal?
(A) 50%
(B) 66%
(A) 990 KHz
(B) 1010 KHz
(C) 75%
(D) 100%
(C) 1020 KHz
(D) 1030 KHz
13. A 10 kW carrier is sinusoidally modulated by two
carriers corresponding to a modulation index of 30%
and 40% respectively. The total radiated power(is
(A) 11.25 kW
(B) 12.5 kW
(C) 15 kW
(D) 17 kW
14. In amplitude modulation, the modulation envelope
has a peak value which is double the unmodulated carrier
value. What is the value of the modulation index ?
(A) 25%
(C) 75%
(A) 695 kHz
(B) 700 kHz
(C) 705 kHz
(D) 710 kHz
21. A message signal m( t) = sinc t + sinc 2 ( t) modulates
the carrier signal ( t) = A cos 2pfc t. The bandwidth of the
modulated signal is
(A) 2 fc
(B) 50%
(D) 100%
(C) 2
15. If the modulation index of an AM wave is changed
from 0 to 1, the transmitted power
(A) increases by 50%
(C) increases by 100%
20. For an AM signal, the bandwidth is 10 kHz and the
highest frequency component present is 705 kHz. The
carrier frequency used for this AM signal is
(B) increases by 75%
(D) remains unaffected
(B)
1
2
(D)
1
4
fc
22. The signal
m( t) = cos 2000 pt + 2 cos 4000 t is
multiplied by the carrier c( t) = 100 cos 2pfc t where fc = 1
MHz to produce the DSB signal. The expression for the
upper side band (USB) signal is
(A) 100 cos(2 p( fc + 1000) t) + 200 cos(2 p( fc + 200) t)
16. A diode detector has a load of 1 kW shunted by a
10000 pF capacitor. The diode has a forward resistance
(B) 100 cos(2 p( fc - 1000) t) + 200 cos(2 p( fc - 2000) t)
of 1 W. The maximum permissible depth of modulation,
(C) 50 cos(2 p( fc + 1000) t) + 100 cos(2 p( fc + 2000) t)
so as to avoid diagonal clipping, with modulating signal
(D) 50 cos(2 p( fc - 1000) t) + 100 cos(2 p( fc - 100) t)
frequency fo 10 kHz will be
(A) 0.847
(C) 0.734
(B) 0.628
(D) None of the above
Statement for Question 23-26 :
The Fourier transform
M ( f ) of a signal m( t) is
17. An AM signal is detected using an envelop detector.
shown in figure. It is to be transmitted from a source to
The carrier frequency and modulating signal frequency
destination. It is known that the signal is normalized,
are 1 MHz and 2 kHz respectively. An appropriate value
meaning that -1 £ m( t) £ 1
M( f)
for the time constant of the envelope detector is.
(A) 500 m sec
(B) 20 m sec
(C) 0.2 m sec
(D) 1 m sec
-10000
18. An AM voltage signal s( t), with a carrier frequency
of 1.15 GHz has a complex envelope g( t) = AC [1 + m( t)],
where Ac = 500 V, and the modulation is a 1 kHz
sinusoidal test tone described by m( t) = 0.8 sin(2 p ´ 10 3 t)
appears across a 50 W resistive load. What is the actual
power dissipated in the load ?
(A) 165 kW
(B) 82.5 kW
(C) 3.3 kW
(D) 6.6 kW
10000
f
Fig.P7.4.23-26
23. If USSB is employed, the bandwidth of the
modulated signal is
(A) 5 kHz
(B) 20 kHz, 10 kHz
(C) 20 kHz
(D) None of the above
24. If DSB is employed, the bandwidth of the modulated
signal is
19. A 1 MHz sinusoidal carrier is amplitude modulated
by a symmetrical square wave of period 100 m sec.
(A) 5 kHz
(B) 10 kHz
(C) 20 kHz
(D) None of the above
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GATE EC BY RK Kanodia
UNIT 7
25. If an AM modulation scheme with a = 0.8 is used,
the bandwidth of the modulated signal is.
(A) 5 kHz
(B) 10 kHz
(C) 20 kHz
(D) None of the above
m(t)
Communication System
S
x(t) Square- Law
Device
y(t)
Filter
AM Signal
cos wct
Fig.P7.4.30-31
26. If an FM signal with kf = 60 kHz is used, then the
bandwidth of the modulated signal is
30. The filter should be a
(A) 5 kHz
(B) 10 kHz
(A) LPP with bandwidth W
(C) 20 kHz
(D) None of the above
(B) LPF with bandwidth 2W
27. A DSB modulated signal x( t) = Am( t) cos 2pfc t is
mixed
(multiplied)
with
a
local
carrier
xL ( t) = cos(2pfc t + q) and the output is passed through a
LPF with a bandwidth equal to the bandwidth of the
message m( t). If the power of the signal at the output of
the low pass filter is pout and the power of the
modulated signal by pu , the
p out
pu
is
(A) 0.5 cos q
(B) cos q
(C) 0.5 cos q
(D)
2
cos 2 q
28. A DSB-SC signal is to be generated with a carrier
frequency fc = 1 MHz using a non-linear device with the
input-output characteristic vo = a0 vi + a1 vi3 where a0
and a1 are constants. The output of the non-linear
device can be filtered by an appropriate band-pass filter.
Let vi = Ac¢ cos(2pfc¢t) + m( t) where m( t) is the message
signal. Then the value of fc¢(in MHz) is
(A) 1.0
(B) 0.333
(C) 0.5
(D) 3.0
31. The modulation index is
2b
2a
(A)
(B)
Am
Am
a
b
a
Am
b
b
Am
a
(D)
32. A message signal is periodic with period T, as shown
in figure. This message signals is applied to an AM
modulator with modulation index a = 0.4. The
modulation efficiency would be
m(t)
K1
T
t
-K1
Fig.P7.4.32
29. A non-linear device with a transfer characteristic
given by i = (10 + 2 vi + 0.2 vi2 ) mA is supplied with a
carrier of 1 V amplitude and a sinusoidal signal of 0.5 V
amplitude in series. If at the output the frequency
component of AM signal is considered, the depth of
modulation is
(A) 18 %
(D) a BPF with center frequency f0 and BW = W such
that f0 - Wm > f0 - W2 > Wm
(C)
2
1
2
(C) a BPF with center frequency f0 and BW = W such
that f0 - Wm > f0 - W2 > 2Wm
(A) 51 %
(B) 11.8 %
(C) 5.1 %
(D) None of the above
Statement for Question 33-36
The figure 6.54-57 shows the positive portion of
the envelope of the output of an AM modulator. The
message signal is a waveform having zero DC value.
m(t)
(B) 10 %
45
(C) 20 %
(D) 33.33 %
30
15
Statement for Question 30-31
t
Consider the system shown in figP7.4.30-31. The
modulating signal m( t) has zero mean and its maximum
(absolute) value is Am = max m( t) . It has bandwidth Wm .
The nonlinear device has a input-output characteristic
y( t) = ax( t) + bx 2 ( t).
Page
416
Fig.P7.4.33-36
33. The modulation index is
(A) 0.5
(B) 0.6
(C) 0.4
(D) 0.8
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GATE EC BY RK Kanodia
Amplitude Modulation
Chap 7.4
34. The modulation efficiency is
41. The lower sideband of the SSB AM signal is
(A) 8.3 %
(B) 14.28 %
(A) -100 cos(2 p( fc - 1000) t) + 200 sin(2 p( fc - 1000) t)
(C) 7.69 %
(D) None of the above
(B) -100 cos(2 p( fc - 1000) t) - 200 sin(2 p( fc - 1000) t)
35. The carrier power is
(C) 100 cos(2 p( fc - 1000) t) - 200 sin(2 p( fc - 1000) t)
(A) 60 W
(B) 450 W
(C) 30 W
(D) 900 W
(D) 100 cos(2 p( fc - 1000) t) + 200 sin(2 p( fc - 1000) t)
Statement for Question 42-43
36. The power in sidebands is
(A) 85 W
Consider the system shown in figure 6.69-70. The
(B) 42.5 W
(C) 56 W
average value of m( t) is zero and maximum value of
(D) 37.5 W
m( t) is M. The square-law device is defined by
37. In a broadcast transmitter, the RF output is
represented as
e( t) = 50[1 + 0.89 cos 5000 t + 0.30 sin 9000 t ]cos( 6 ´ 106 t)V
y( t) = 4 x( t) + 10 x( t).
S
m(t)
x(t) Square- Law
Device
y(t)
Filter
AM Signal
What are the sidebands of the signals in radians ?
cos wct
(A) 5 ´ 10 3 and 9 ´ 10 3
(B) 5.991 ´ 106 , 5.995 ´ 106 , 6.005 ´ 106 and 6.009 ´ 106
(C) 4 ´ 10 3, 1.4 ´ 10 4
(D) 1 ´ 10 , 11
. ´ 10 , 3 ´ 10 , and 15
. ´ 10
6
7
6
42. The value of M, required to produce modulation
index of 0.8, is
7
38. An AM modulator has output
x( t) = 40 cos 400 pt + 4 cos 360 pt + 4 cos 440 pt
The modulation efficiency is
(A) 0.01
(B) 0.02
(C) 0.03
(D) 0.04
Fig. P7.4.42-43
(A) 0.32
(B) 0.26
(C) 0.52
(D) 0.16
43. Let W be the bandwidth of message signal m( t). AM
signal would be recovered if
39. An AM modulator has output
(A) fc > W
(B) fc > 2W
(C) fc ³ 3W
(D) fc > 4W
x( t) = A cos 400 pt + B cos 380 pt + B cos 420 pt
The carrier power is 100 W and the efficiency is
40%. The value of A and B are
(A) 14.14, 8.16
(B) 50, 10
(C) 22.36, 13.46
(D) None of the above
44. A super heterodyne receiver is designed to receive
transmitted signals between 5 and 10 MHz. High-side
tuning is to be used. The tuning range of the local
oscillator for IF frequency 500 kHz would be
(A) 4.5 MHz - 9.5 MHz
Statement for Question 40-41
(B) 5.5 MHz - 10.5 MHz
A single side band signal is generated by
modulating signal of 900-kHz carrier by the signal
(C) 4.5 MHz - 10.5 MHz
m( t) = cos 200 pt + 2 sin 2000 pt. The amplitude of the
(D) None of the above
carrier is Ac = 100.
(B) - sin(2 p1000 t) + 2 cos(2000 pt)
45. A super heterodyne receiver uses an IF frequency of
455 kHz. The receiver is tuned to a transmitter having
a carrier frequency of 2400 kHz. High-side tuning is to
be used. The image frequency will be
(C) sin(2 p1000 t) + 2 cos(1000 t)
(A) 2855 kHz
(B) 3310 kHz
(C) 1845 kHz
(D) 1490 kHz
$ ( t) is
40. The signal m
(A) - sin(2 p1000 t) - 2 cos(2000 pt)
(D) sin(2 p1000 t) - 2 cos(2 p1000 t)
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GATE EC BY RK Kanodia
UNIT 7
46. In the circuit shown in fig.P7.4.46, the transformers
are center tapped and the diodes are connected as
shown in a bridge. Between the terminals 1 and 2 an
a.c. voltage source of frequency 400 Hz is connected.
Another a.c. voltage of 1.0 MHz is connected between 3
and 4. The output between 5 and 6 contains components
at
Communication System
49. In fig.P7.4.49
m( t) =
2 sin 2 pt
sin 199pt
, s( t) = cos 200 pt and n( t) =
t
t
Multiplier
3
LPF
1 Hz
y(t)
n(t)
s(t)
Fig.P7.4.49
The output y( t) will be
sin 2pt
(A)
t
4
6
2
Multiplier
|H(jw)|=1
s(t)
5
1
Adder
S
m(t)
Fig.P7.4.46
(A) 400 Hz, 1.0 MHz, 1000.4 kHz, 999.6 kHz
(B)
sin 2 pt sin pt
+
cos 3pt
t
t
(C)
sin 2 pt sin 0.5 pt
+
cos 15
. pt
t
t
(D)
sin 2 pt sin pt
+
cos 0.75 pt
t
t
(B) 400 Hz, 1000.4 kHz, 999.6 kHz
(C) 1 MHz, 1000.4 kHz, 999.6 kHz
(D) 1000.4 kHz, 999.6 kHz
47. A superheterodyne receiver is to operate in the
frequency range 550 kHz-1650 kHz, with the
denote
intermediate frequency of 450 kHz. Let R = CCmax
min
the required capacitance ratio of the local oscillator and
I denote the image frequency (in kHz) of the incoming
signal. If the receiver is tuned to 700 kHz, then
50. 12 signals each band-limited to 5 kHz are to be
transmitted over a single channel by frequency division
multiplexing. If AM -SSNB modulation guard band of 1
kHz is used, then the bandwidth of the multiplexed
signal will be
(A) 51 kHz
(B) 61 kHz
(D) 81 kHz
(A) R = 4.41, I = 1600
(B) R = 2.10, I = 1150
(C) 71 kHz
(C) R = 3, I = 1600
(D) R = 9.0, I = 1150
51. Let x( t) be a signal band-limited to 1 kHz.
Amplitude modulation is performed to produce signal
A
proposed
demodulation
g( t) = x( t) sin 2000pt.
technique is illustrated in figure 6.83. The ideal low
pass filter has cutoff frequency 1 kHz and pass band
gain 2. The y( t) would be
48. Consider a system shown in Figure . Let X ( f ) and
Y ( f ) denote the Fourier transforms of x( t) and y( t)
respectively. The ideal HPF has the cutoff frequency 10
kHz. The positive frequencies where Y ( f ) has spectral
peaks are
(A) 2 y( t)
x(t)
HPF
10 kHz
Balanced
Modulator
Balanced
Modulator
~
~
10 kHz
13 kHz
X(f )
-3
-1
1
Fig.P7.4.48
(A) 1 kHz and 24 kHz
(B) 2 kHz and 24 kHz
(C) 1 kHz and 14 kHz
(D) 2 kHz and 14 kHz
Page
418
3
f (kHz)
y(t)
(C)
1
2
y( t)
(B) y( t)
(D) 0
52. Suppose we wish to transmit the signal
x( t) = sin 200 pt + 2 sin 400 pt using a modulation that
create the signal g( t) = x( t) sin 400pt. If the product
g( t) sin 400pt is passed through an ideal LPF with
cutoff frequency 400p and pass band gain of 2, the
signal obtained at the output of the LPF is
1
2
(A) sin 200pt
(B)
(C) 2 sin 200pt
(D) 0
sin 200 pt
53. In a AM signal the received signal power is 10 -10 W
with a maximum modulating signal of 5 kHz. The noise
spectral density at the receiver input is 10 -18 W/Hz. If
the noise power is restricted to the message signal
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GATE EC BY RK Kanodia
Amplitude Modulation
bandwidth only, the signals-to-noise ratio at the input
to the receiver is
(A) 43 dB
(B) 66 dB
(C) 56 dB
(D) 33 dB
Chap 7.4
SOLUTION
1. (C) u( t) = (20 + 4 sin 500 pt) cos(2 p ´ 10 5 t) V
= 20(1 + 0.2 sin 500 pt) cos(2 p ´ 10 5 t) V,
a = 0.2
Statement for Question 54-55
Consider the following Amplitude Modulated (AM)
2. (B) Pc =
signal, where fm < B
x AM ( t) = 10(1 + 0.5 sin 2 pfm t) cos 2 pfc t.
(B) 12.5
(C) 6.25
(D) 3.125
25
2 N0 B
(D)
4. (B) x( t)
= [20 + 2 cos(2 p1500 t) + 10 cos(2 p3000 t)]cos(2 pfc t)
1
1
æ
ö
= 20ç 1 +
cos(2 p1500 t) + cos(2 p3000 t) cos(2 pfc t) ÷
10
2
è
ø
55. The AM signal gets added to a noise with Power
Spectra Density Sn ( f ) given in the figure below. The
ration of average sideband power to mean noise power
would be
25
25
(B)
(A)
8 N0 B
4 N0 B
(C)
25
N0 B
Statement for Question 56-57
A certain communication channel is characterized
by 80 dB attenuation and noise power-spectral density
of 10 -10 W/Hz. The transmitter power is 40 kW and the
message signal has a bandwidth of 10 kHz.
56. In the case of conventional AM modulation, the
predetecion SNR is
This is the form of a conventional AM signal with
message signal
1
1
m( t) =
cos(2 p1500 t) + cos(2 p3000 t)
10
2
1
1
= cos 2 (2 p1500 t) +
cos(2 p1500 t) 10
2
1
1
The minimum of g( z) = z 2 +
is achieved for
z10
2
1
201
1
and it is min( g( z)) = . Since z = is in
z =20
400
20
the range of cos (2 p1500 t), we conclude that the
201
minimum value of m( t) is . Hence, the modulation
400
201
index is a = 400
5. (B) x( t) = 20 cos(2 pfc t) + cos(2 p( fc - 1500) t)
+ cos(2 p( fc - 1500) t)
(A) 108
(B) 2 ´ 108
= 5 cos(2 p( fc 3000) t) + 5 cos(2 p( fc + 3000) t)
(C) 10 2
(D) 2 ´ 10 2
The power in the sidebands is
1 1 25 25
+
= 26
Psidebands = + +
2 2
2
2
57. In case of SSB, the predetecion SNR is
(A) 2 ´ 10 2
(B) 4 ´ 10 2
(C) 2 ´ 10 3
(D) 4 ´ 10 3
*************
ö
÷÷ = 204 W
ø
3. (A) Psb = Pt - Pc = 204 - 200 = 4 W
54. The average side-band power for the AM signal
given above is
(A) 25
æ
(0.2) 2
Pt = Pc çç 1 +
2
è
20 2
= 200 W,
2
The total power is Ptotal = Pcarrier + Psidebands = 200 + 26 = 226
The ratio of the sidebands power to the total power is
Psidebands 26
=
Ptotal
226
æ
æ
0.9 2 ö
a2 ö
6. (B) Pt = Pc çç 1 +
÷÷ = 2000çç 1 +
÷ = 2810 W
2 ÷ø
2 ø
è
è
æ
a2 ö
7. (A) Pt = Pc çç 1 +
÷
2 ÷ø
è
Pc = 37.88 kW,
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or
æ
0.8 2 ö
÷
50 ´ 10 3 = Pc çç 1 +
2 ÷ø
è
Pi = ( Pt - Pc ) = (50 - 37.88) = 12.12 kW
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UNIT 7
1
1
æ
æ
a 2 ö2
a 2 ö2
8. (A) I t = I c çç 1 +
÷÷ or 20 = 18çç 1 +
÷ or a = 0.68
2 ÷ø
2 ø
è
è
18. (A) Pt =
Pt =
500 2
2
Pi = 30 - 20 = 10 kW
The DC input power =
10
= 12.5 kW.
0.8
m( t) =
1
é 4 ¥ ( -1) n -1
ù
x( t) = ê å
cos[2 pfm (2 n - 1)]ú sin(2 p1000 ´ 10 3 t)
ë p n =1 2 n - 1
û
So frequency component (106 ± fm (2 n - 1) will be present
2
P. If carrier is
3
2
suppressed then P or 66% power will be saved.
3
where n = 1, 2, 3, ....
For fm = 10 kHz and n = 1 & 2 frequency present is 990,
970, 1030 kHz.
Thus 1020 kHz will be absent.
20. (B) fc + fm = 705 kHz,
ö
÷÷
ø
BW = 2 fm = 10 kHz or fm = 5 kHz
fc = 705 - 5 = 700 kHz
= 1125
. kW
21. (C) x( t) = m( t) c( t) = A(sinc ( t) + sinc 2 ( t) cos(2pfc t)
14. (D) x( t) = Ac (1 + a cos 2 pfm t) cos 2 pfc t
Thus a =1, therefor modulation
Here Ac (1 + a) = 2 Ac ,
4 ¥ ( -1) n -1
cos[2 pfm (2 n - 1) ]
å
p n =1 2 n - 1
The modulated output
11. (C) a 2 = a1a + a 22 = 0.32 + 0.4 2 = 0.5 2 or a = 0.5
ö
æ
0.32 0.4 2
÷÷ = 10çç 1 +
+
2
2
ø
è
fc = 1000 kHz, x( t) = c( t) m( t)
Expressing square wave as modulating signal m( t)
æ
æ
0.7 2 ö
a 2 ö2
÷÷ = 2çç 1 +
÷ = 2.23 kW
Pt = Pc çç 1 +
2 ÷ø
2 ø
è
è
æ
a2 a2
13. (A) Pt = Pc çç 1 + 1 + 2
2
2
è
é
0.8 2 ù
1
+
= 165 kW
ê
2 úû
ë
19. (C) c( t) = sin 2p fc t,
a = 70% = 0.7
12. (B) In previous solution Pc =
2
a 2 m( t) ù
Ac2 é
ê1 +
ú
2 ê
2
úû
ë
Here modulation index a =1. Thus
1ö
æ
9. (C) Pt = 20000ç 1 + ÷, Pt = 30 kW,
2ø
è
10. (A) Pc = 2 kW,
Communication System
Taking the Fourier transform of both sides, we obtain
index is 1 or 100% modulation.
X( f ) =
15. (A) If modulation index a is 0, then
=
A
[ P( f ) + L( f )] * ( d( f - fc ) + d( f + fc ))
2
A
[ P( f - fc ) + L( f - fc ) + P( f + fc ) + L( f - fc )]
2
æ
0 2 ö Ac2
÷=
çç 1 +
2 ÷ø 2
è
1
Since P( f - fc ) ¹ 0 for f - fc < , whereas L( f - fc ) ¹ 0
2
If modulation index is 1 then
for f - fc < 1. Hence, the bandwidth of the bandpass
Pt1 =
Pt2 =
Ac2
2
Ac2
2
æ
12 ö 3 2
÷ = Ac ,
çç 1 +
2 ÷ø 4
è
filter is 2.
Pt 2 3
=
Pt1 2
22. (C) x( t) = m( t) c( t)
Thus Pt2 = 15
. Pt1 and Pt2 is increases by 50%
= 100[cos(2 p1000 t) + 2 cos(2 p2000 t)]cos(2 pfc t)
= 100 cos(2 p1000 t) cos(2 pfc t) + 200 cos(2 p2000 t) cos(2 pfc t)
100
=
[cos(2 p( fc + 1000) t) + cos(2 p( fc - 1000) t)]
2
16. (A) fm = 10 kHz, R = 1000 W, C = 10000 pF
Hence 2pfm RC = 2 p ´ 10 4 ´ 10 3 ´ 10 -8 = 0.628
a max = (1 + (0.628) 2 )
17. (B)
-
1
2
= 0.847
1
1
,
£ RC £
fc
BWm
+
200
[cos(2 p( fc + 2000) t) + cos(2 p( fc - 2000) t)]
2
Thus, the upper sideband (USB) signal is
Here fc = 1 MHz
xu ( t) = 50 cos[2 p( fc + 1000) t ] + 100(2 p( fc + 2000) t)
Signal Bandwidth BWm = 2 fm = 2 ´ 2 ´ 10 3 = 4 kHz
Thus
1
1
or 10 -6 £ RC £ 250 ms
£ RC £
106
4 ´ 10 3
Thus appropriate value is 20 m sec
Page
420
23. (B) When USSB is employed the bandwidth of the
modulated signal is the same with the bandwidth of the
message signal. Hence WUSSB = W = 10 4 Hz
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UNIT 7
37.
(B)
Sidebands
are
( 6 ´ 106 ± 5000)
and
Communication System
43. (C) The filter characteristic is shown in fig.S7.4.43
( 6 ´ 10 ± 9000)
6
H(f )
Thus 6.005 ´ 10 , 5.995 ´ 10 , 5.991 ´ 10 or 5.991 ´ 10 ,
6
6
6
6
m(t)
6.005 ´ 106 and 6.009 ´ 106
38. (B) x( t) can be written as
f
W
x( t) = ( 40 + 8 cos 40 pt) cos 400 pt
8
modulation index a =
= 0.2
40
1
Pc = ( 40) 2 = 800 W
2
fc + W < 2 f or fc > W
1
1
Psb = B 2 + B 2 = 66.67
2
2
the
Hilbert
44. (B) Since High-side tuning is used
fLO = fm + f IF = 500 kHz,
fLOL = 5 + 0.5 = 5.5 MHz,
45. (B) fimage = fL + 2 f IF = 2400 = 3310 kHz
46. (D) The given circuit is a ring modulator. The output
or B = 8.161
is DSB-SC signal. So it will contain m( t) cos( nwc t) where
n = 1, 2, 3...... Therefore there will be only (1 MHz ± 400
transform
of
sin (2 p1000 t) is cos (2 p1000 t)
Hz) frequency component.
47. (A) fmax = 1650 + 450 = 2100 kHz
fmin = 550 + 450 = 1000 kHz.
41. (D) The expression for the LSSB AM signal is.
frequency is minimum, capacitance will be maximum
x l ( t) = Ac m( t) cos(2 pfc t) + Ac m( t) sin(2 pfc t)
R=
$ ( t) = sin(2 p1000 t) - 2 cos(2 p1000 t)
and m
output of first modulator spectral peak will be at (10 + 1)
xl ( t) = 100[cos(2 p1000 t) + 2 sin(2 p1000 t) cos(2 pfc t)]
kHz and (10 - 1) kHz. After passing the HPF frequency
+100[sin(2 p1000 t) - 2 cos(2 p1000 t) sin(2 pfc t)]
= 100[cos(2 p1000 t) cos(2 pfc t) + sin(2 p1000 t) sin(2 pfc t)]
+200[cos(2 pfc t) sin(2 p1000 t) - sin(2 pfc t) cos(2 p1000 t)]
= 100 cos(2 p( fc - 1000) t) - 200 sin(2 p( fc - 1000) t)
= 4m(t) + 4 cos wc t + 10m2 (t) + 20m(t) cos wc t + 5 + 5 cos 2wc t
m( t) = Mmn ( t)
xc ( t) = 4[1 + 5 Mmn ( t)]cos wc t
5M = 0.8
or
M = 0.16
component of 11 kHz will remain. The output of 2nd
modulator will be (13 ± 11) kHz. So Y ( f ) has spectral
peak at 2 kHz and 24 kHz.
49. (C) m( t) s( t) = y1 ( t)
42. (D) y( t) = 4( m( t) + cos wc t) + 10( m( t) + cos wc t) 2
xc ( t) = 4[1 + 5 m( t)]cos wc t
or R = 4.41
48. (B) Since X ( f ) has spectral peak at 1 kHz so at the
we obtain
The AM signal is,
2p LC
fi = fc + 2 f IF = 700 + 2( 455) = 1600 kHz
Ac = 100, m( t) = cos(2 p1000 t) + 2 sin(2 p1000 t)
= 5 + 4m(t) + 10m2 (t) + 4[1 + 5m(t)] cos wc t + 5 cos 2wc t
2
Cmax fmax
= 2 = (2.1) 2
Cmin
fmin
or
f =
1
$ ( t)= sin (2 p1000 t) - cos (2 p1000 t)
Thus m
Substituting
Page
422
Therefore fc > 3W
fLOU = 10 + 0.5 = 10.5 MHz
40. (D) The Hilbert transform of cos (2 p1000 t) is
whereas
2fc
fc+W
fc - W > 2W or fc > 3W ,
A2
39. (A) Carrier power Pc =
= 100 W, A = 14.14
2
40
Psb
Psb
or
Eeff =
= 0.4
=
100 + Psb
Pc + Psb 100
sin (2 p1000 t),
fc-W fc
Fig.S7.4.43
The components at 180 Hz and 220 Hz are side band
1
1
Psb = ( 4) 2 + ( 4) 2 = 16 W,
2
2
Psb
16
Eeff =
=
Pc + Psb 800 + 16
Psb = 66.67 W,
2W
=
2 sin(2 pt) cos(200 pt) sin(202 pt) - sin(198 pt)
=
t
t
y1 ( t) + n( t) = y2 ( t) =
sin 202 pt - sin 198 pt sin 198 pt
+
t
t
y2 ( t) s( t) = y( t)
=
[sin 202 pt - sin 198 pt + sin 199 pt ]cos 200 pt
t
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Amplitude Modulation
=
Chap 7.4
55. (D) Noise power = Area rendered by the spectrum
1
[sin( 402 pt) + sin(2 pt) - {sin( 398 pt) - sin(2 pt)}
2
+ sin( 399pt) - sin( pt)]
After filtering
sin(2 pt) + sin(2 pt) - sin( pt)
2t
sin(2 pt) + 2 sin(0.5 t) cos(15
. pt)
=
2t
sin 2 pt sin 0.5 pt
=
+
cos 15
. pt
2t
t
y( t) =
= N0 B
Ratio of average sideband power to mean noise
6.25
25
Power =
=
N0 B 4 N0 B
56. (C) Since the channel attenuation is 80 db, then
P
10 log T = 80
PR
or PR = 10 -8 PT = 10 -8 ´ 40 ´ 10 3 = 4 ´ 10 -4 Watts
If the noise limiting filter has bandwidth B, then the
50. (D) The total signal bandwidth = 5 ´ 12 = 60 kHz
pre-detection noise power is
There would be 11 guard band between 12 signal. So
fc +
Pn = 2
guard band width = 11 kHz
Total band width = 60 + 11 = 71 kHz
ò
fc -
B
2
B
2
N0
df = N 0 B = 2 ´ 10 -10 B Watts
2
In the case of DSB or conventional AM modulation,
51. (D) x1 ( t) = g( t) cos(2000pt)
B = 2W = 2 ´ 10 4
1
= x( t) sin(2000 pt) cos(2000 pt) = x( t) sin( 4000 pt)
2
1
X1 ( jw) =
X ( j( w - 4000 p)) - X ( j( w + 4000 p))
4j
Hz, whereas in SSB modulation
B = W = 10 . Thus, the pre-detection signal to noise
4
ratio in DSB and conventional AM is
SNR DSB,AM =
This implies that X1 ( jw) is zero for w £ 2000 p because
w < 2 pfm = 2 p1000. When x1 ( t) is passed through a LPF
with cutoff frequency 2000p, the output will be zero.
PR
4 ´ 10 -4
=
= 10 2
Pn 2 ´ 10 -10 ´ 2 ´ 10 4
57. (A) In SSB modulation B = W = 10 4
SNR SSB =
52. (A) y( t) = g( t) sin( 400 pt) = x( t) sin 2 ( 400 pt)
(1 - cos)( 800 pt)
= (sin(200 pt) + 2 sin( 400 pt)
2
1
= [sin(200 pt) - sin(200 pt) cos( 800 pt) + 2 sin( 400 pt)
2
4 ´ 10 -4
= 2 ´ 10 2
2 ´ 10 -10 ´ 10 4
***********
-sin( 400 pt) cos( 800 pt)
1
1
= sin(200 pt) - [sin(1000 pt) - sin( 6000 pt)]
2
4
1
+ sin( 400 pt) - [sin(1200 pt) - sin( 400 pt)]
4
If this signal is passed through LPF with frequency
400p and gain 2, the output will be sin(200pt)
53. (A) Message signal BW fm = 5 kHz
Noise power density is 10 -18 W/Hz
Total noise power is 10 -18 ´ 5 ´ 10 3 = 5 ´ 10 -15 W
Input signal-to-noise ratio
SNR=
10 -10
= 2 ´ 10 4 or 43 dB
5 ´ 10 -15
54. (C) Average side band power is
Ac2 a 2 10 2 (0.5) 2
=
= 6.25 W
4
4
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CHAPTER
7.6
DIGITAL TRANSMISSION
Statement for Question 1-5
Statement for Question 6-7
Fig. P7.6.1-5 shows fourier spectra of signal x( t)
and y(t). Determine the Nyquist sampling rate for the
A signal x( t) is multiplied by rectangular pulse
train c( t) shown in fig.P7.6.6-7..
c(t)
given function in question.
X( jw)
0.25 ms
Y( jw)
-2´10-3
-10-3
w
w
Fig.P7.6.1-5
10-3
2´10-3
t
Fig.P7.6.6-7
6. x( t) would be recovered form the product. x( t) c( t) by
using an ideal LPF if X ( jw) = 0 for
1. x( t)
(A) 100 kHz
(B) 200 kHz
(A) w > 2000 p
(B) w > 1000 p
(C) 300 kHz
(D) 50 kHz
(C) w < 1000 p
(D) w < 2000 p
7. If X ( jw) satisfies the constraints required, then the
2. y( t)
pass band gain A of the ideal lowpass filter needed to
(A) 50 kHz
(B) 75 kHz
recover x( t) from e( t) x( t) is
(C) 150 kHz
(D) 300 kHz
(A) 1
(B) 2
(C) 4
(D) 8
3. x 2 ( t)
(A) 100 kHz
(B) 150 kHz
(C) 250 kHz
(D) 400 kHz
4. y 3( t)
(A) 100 kHz
(B) 300 kHz
(C) 900 kHz
(D) 120 kHz
8. Consider a set of 10 signals xi ( t), i = 1, 2, 3, ...10.. Each
signal is band limited to 1 kHz. All 10 signals are to be
time-division multiplexed after each is multiplied by a
carrier e( t) shown in Figure. If the period T of e( t) is
chosen the have the maximum allowable value, the
largest value of D would be
c(t)
D
5. x( t) y( t)
(A) 250 kHz
(B) 500 kHz
(C) 50 kHz
(D) 100 kHz
Page
434
-2T
-T
0
Fig.P7.6.8
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t
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GATE EC BY RK Kanodia
Digital Transmission
(A) 5 ´ 10 -3 sec
-5
(C) 5 ´ 10 sec
(B) 5 ´ 10 -4 sec
Chap 7.6
15. A CD record audio signals digitally using PCM. The
audio signal bandwidth is 15 kHz. The Nyquist samples
-6
(D) 5 ´ 10 sec
are quantized into 32678 levels and then binary coded.
9. A compact disc recording system samples a signals
The minimum number of binary digits required to
with a 16-bit analog-to-digital convertor at 44.1 kbits/s.
encode the audio signal
The CD can record an hours worth of music. The
(A) 450 k bits/sec
(B) 900 k bits/sec
approximate capacity of CD is
(C) 980 340 k bits/sec
(D) 490 170, k bits/sec
(A) 705.6 M Bytes
(B) 317.5 M Bytes
(C) 2.54 M Bytes
(D) 5.43 M Bytes
16. The American Standard Code for Information
Interchange has 128 characters, which are binary
10. An analog signal is sampled at 36 kHz and
coded. If a certain computer generates 1,000,000
quantized into 256 levels. The time duration of a bit of
character per second, the minimum bandwidth required
the binary coded signal is
to transmit this signal will be
(A) 5.78 ms
(B) 3.47 ms
(A) 1.4 M bits/sec
(B) 14 M bits/sec
(C) 6.43 ms
(D) 7.86 ms
(C) 7 M bits/sec
(D) 0.7 M bits/sec
11. An analog signal is quantized and transmitted using
17. A binary channel with capacity 36 k bits/sec is
a PCM system. The tolerable error in sample amplitude
available for PCM voic transmission. If signal is band
is 0.5% of the peak-to-peak full scale value. The
limited to 3.2 kHz, then the appropriate values of
minimum binary digits required to encode a sample is
quantizing level L and the sampling frequency will be
(A) 5
(B) 6
(A) 32, 3.6 kHz
(B) 64, 7.2 kHz
(C) 7
(D) 8
(C) 64, 3.6 kHz
(D) 32, 7.2 kHz
18. Fig.P7.4.18 shows a PCM signals in which
Statement for Question 12-13.
Ten telemetry signals, each of bandwidth 2kHz,
are to be transmitted simultaneously by binary PCM.
The maximum tolerable error in sample amplitudes is
0.2% of the peak signal amplitude. The signals must be
amplitude level of + 1 volt and - 1 volt are used to
represent binary symbol 1 and 0 respectively. The code
word used consists of three bits. The sampled version of
analog signal from which this PCM signal is derived is
sampled at least 20% above the Nyquist rate. Framing
and synchronizing requires an additional 1% extra bits.
Fig.P7.4.18
12. The minimum possible data rate must be
(A) 272.64 k bits/sec
(B) 436.32 k bits/sec
(A) 4 5 1 2 1 3
(B) 8 4 3 1 2
(C) 936.32 k bits/sec
(D) None of the above
(C) 6 4 3 1 7
(D) 1 2 3 4 5
13. The minimum transmission bandwidth is
19. A PCM system uses a uniform quantizer followed by
(A) 218.16 kHz
(B) 468.32 kHz
a 8-bit encoder. The bit rate of the system is equal to 108
(C) 136.32 kHz
(D) None of the above
bits/s. The maximum message bandwidth for which the
system operates satisfactorily is
14. A Television signal is sampled at a rate of 20%
above the Nyquist rate. The signal has a bandwidth of 6
MHz. The samples are quantized into 1024 levels. The
(A) 25 MHz
(B) 6.25 MHz
(C) 12.5 MHz
(D) 50 MHz
minimum bandwidth required to transmit this signal
20. Twenty-four voice signals are sampled uniformly at
would be
a rate of 8 kHz and then time-division multiplexed. The
(A) 72 M bits/sec
(B) 144 M bits/sec
(C) 72 k bits/sec
(D) 144 k bits/sec
sampling process uses flat-top samples with 1 ms
duration. The multiplexing operating includes provision
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UNIT 7
Communication System
for synchronization by adding and extra pulse of 1 ms
quantized into 256 level using a m-low quantizer with
duration. The spacing between successive pulses of the
m = 225.
multiplexed signal is
(A) 4 ms
(B) 6 ms
(C) 7.2 ms
(D) 8.4 ms
26. The signal-to-quantization-noise ratio is
21. A linear delta modulator is designed to operate on
speech signals limited to 3.4 kHz. The sampling rate is
10 time the Nyquist rate of the speech signal. The step
size d is 100 m V. The modulator is tested with a this
test signal required to avoid slope overload is
(A) 2.04 V
(B) 1.08 V
(C) 4.08 V
(D) 2.16 V
a
linear
DM
(B) 38.06 dB
(C) 42.05 dB
(D) 48.76 dB
27. It was found that a sampling rate 20% above the
rate wou7ld be adequate. So the maximum SNR, that
can be realized without increasing the transmission
bandwidth, would be
(A) 60.4 dB
(B) 70.3 dB
(C) 50.1 dB
(D) None of the above
28. For a PCM signal the compression parameter
Statement fo Question 22-23 :
Consider
(A) 34.91 dB
m = 100 and the minimum signal to quantization-noise
system
designed
to
ratio is 50 dB. The number of bits per sample would be.
accommodate analog message signals limited to bandwidth
(A) 8
(B) 10
of 3.5 kHz. A sinusoidal test signals of amplitude Amax = 1
(C) 12
(D) 14
V and frequency fm = 800 Hz is applied to system. The
29. A sinusoid massage signal m( t) is transmitted by
sampling rate of the system is 64 kHz.
22. The minimum value of the step size to
binary
avoid
overload is
PCM
without
compression.
the
signal
to-quantization-noise ratio is required to be at least 48
dB, the minimum number of bits per sample will be
(A) 240 mV
(B) 120 mV
(A) 8
(B) 10
(C) 670 mV
(D) 78.5 mV
(C) 12
(D) 14
23. The granular-noise power would be
(A) 1.68 ´ 10 -3 W
(C) 2.48 ´ 10
If
-3
W
30. A speech signal has a total duration of 20 sec. It is
(B) 2.86 ´ 10 -4 W
(D) 112
. ´ 10
-4
sampled at the rate of 8 kHz and then PCM encoded.
The signal-to-quantization noise ratio is required to be
W
40 dB. The minimum storage capacity needed to
24. The SNR will be
(A) 298
(C) 4.46 ´ 10
3
accommodate this signal is
(B) 1.75´10 -3
(A) 1.12 KBytes
(B) 140 KBytes
(D) 201
(C) 168 KBytes
(D) None of the above
25. The output signal-to-quantization-noise ratio of a 10-bit
31. The input to a linear delta modulator having fa
PCM was found to be 30 dB. The desired SNR is 42 dB. It
step-size D = 0.628 is a sine wave with frequency fm and
can be increased by increasing the number of quantization
peak amplitude Em . If the sampling frequency fs = 40
level.In this way the fractional increase in the transmission
kHz, the combination of the sinc-wave frequency and
bandwidth would be (assume log 2 10 = 0.3)
the peak amplitude, where slope overload will take
(A) 20%
(B) 30%
piace is
(C) 40%
(D) 50%
Em
(A) 0.3 V
fm
8 kHz
(B) 1.5 V
4 kHz
A signal has a bandwidth of 1 MHz. It is sampled
(C) 1.5 V
2 kHz
at a rate 50% higher than the Nyquist rate and
(D) 3.0 V
1 kHz
Statement for Question 26-27.
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436
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Digital Transmission
Chap 7.6
32. A sinusoidal signal with peak-to-peak amplitude of
38. Four signals g1 ( t), g 2 ( t), g s ( t) and g 4 ( t) are to be
1.536 V is quantized into 128 levels using a mid-rise
multiplexed and transmitted. g1 ( t) and g 4 ( t) have a
uniform quantizer. The quantization-noise power is
bandwidth of 4 kHz, and the remaining two signals
-6
(A) 0.768 V
(B) 48 ´ 10 V
(C) 12 ´ 10 -6 V 2
(D) 3.072 V
have bandwidth of 8 kHz,. Each sample requires 8 bit
2
for encoding. What is the minimum transmission bit
rate of the system.
33. A signal is sampled at 8 kHz and is quantized using
(A) 512 kbps
(B) 16 kbps
for a
(C) 192 kbps
(D) 384 kbps
8 bit uniform quantizer. Assuming SNR q
sinusoidal signal, the correct statement for PCM signal
39. Three analog signals, having bandwidths 1200 Hz,
with a bit rate of R is
600 Hz and 600 Hz, are sampled at their respective
(A) R = 32 kbps, SNR q = 25.8 dB
Nyquist rates, encoded with 12 bit words, and time
(B) R = 64 kbps, SNR q = 49.8 dB
division multiplexed. The bit rate for the multiplexed
(C) R = 64 kbps, SNR q = 55.8 dB
signal is
(D) R = 32 kbps, SNR q = 49.8 dB
(A) 115.2 kbps
(B) 28.8 kbps
(C) 57.6 kbps
(D) 38.4 kbps
34. A 1.0 kHz signal is flat-top sampled at the rate of
180 samples sec and the samples are applied to an ideal
rectangular LPF with cat-off frequency of 1100 Hz, then
the output of the filter contains
40. The minimum sampling frequency (in samples/sec)
required to reconstruct the following signal form its
samples without distortion would be
(A) only 800 Hz component
3
2
(B) 800 and 900 Hz components
æ sin 2 p1000 t ö
æ sin 2 p1000 t ö
x( t) = 5ç
÷ + 7ç
÷
pt
pt
è
ø
è
ø
(C) 800 Hz and 1000 Hz components
(A) 2 ´ 10 3
B) 4 ´ 10 3
(D) 800 Hz, 900 and 1000 Hz components
(C) 6 ´ 10 3
(D) 8 ´ 10 3
35. The Nyquist sampling interval, for the signal
sinc (700 t) + sinc (500 t) is
1
(A)
sec
350
p
(B)
sec
350
1
(C)
sec
700
p
(D)
sec
175
41.
The
minimum
step-size
required
for
a
Delta-Modulator operating at32 K samples/sec to track
the signal (here u( t) is the nuit function)
x( t) = 125 t( u( t) - u( t - 1)) + (250 - 125 t)( u( t - 1) - u( t - 2)s
so that slope overload is avoided, would be
36. A signal x( t) = 100 cos(24 p ´ 10 ) t is ideally sampled
3
(A) 2 -10
(B) 2 -8
(C) 2 -6
(D) 2 -4
with a sampling period of 50 m sec and then passed
through an ideal lowpass filter with cutoff frequency of
42. Four signals each band limited to 5 kHz are
15 KHz. Which of the following frequencies is/are
sampled at twice the Nyquist rate. The resulting PAM
present at the filter output
samples are transmitted over a single channel after
(A) 12 KHz only
(B) 8 KHz only
time division multiplexing. The theoretical minimum
(C) 12 KHz and 9 KHz
(D) 12 KHz and 8 KHz
transmissions bandwidth of the channel should be
equal to.
37. In a PCM system, if the code word length is
(A) 5 kHz
(B) 20 kHz
increased form 6 to 8 bits, the signal to quantization
(C) 40 kHz
(D) 80 kHz
noise ratio improves by the factor.
(A) 8/6
(B) 12
(C) 16
(D) 8
43. Four independent messages have bandwidths of 100
Hz, 100 Hz, 200 Hz and 400 Hz respectively. Each is
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GATE EC BY RK Kanodia
UNIT 7
Communication System
sampled at the Nyquist rate, time division multiplexed
50. A speech signal occupying the bandwidth of 300 Hz
and transmitted. The transmitted sample rate, in Hz,
to 3 kHz is converted into PCM format for use in digital
is given by
communication. If the sampling frequency is8 kHz and
(A) 200
(B) 400
each sample is quantized into 256 levels, then the
(C) 800
(D) 1600
output bit the rate will be
44.
The
Nyquist
sampling
rate
for
the
signal
g( t) = 10 cos(50 pt) cos 2 (150 pt). Where ' t ' is in seconds, is
(A) 3 kb/s
(B) 8 kb/s
(C) 64 kb/s
(D) 256 kb/s
(A) 150 samples per second
51. If the number of bits in a PCM system is increased
(B) 200 samples per second
from n to n + 1, the signal-to-quantization noise ratio
(C) 300 samples per second
will increase by a factor.
(D) 350 samples per second
(A)
45. A TDM link has 20 signal channels and each
(C) 2
channel is sampled 8000 times/sec. Each sample is
represented by seven binary bits and contains an
additional bit for synchronization. The total bit rate for
the TDM link is
(A) 1180 K bits/sec
(B) 1280 K bits/sec
(C) 1180 M bits/sec
(D) 1280 M bits/sec
46. In a CD player, the sampling rate is 44.1 kHz and
the samples are quantized using a 16-bit/sample
quantizer. The resulting number of bits for a piece of
music with a duration of 50 minutes is
(A) 1.39 ´ 10 9
(B) 4.23 ´ 10 9
(C) 8.46 ´ 10 9
(D) 12.23 ´ 10 9
( n + 1)
n
(B)
(D) 4
52. In PCM system, if the quantization levels are
increased
form
2
to
8,
the
256
quantization
levels.
(A) remain same
(B) be doubled
(C) be tripled
(D) become four times
53. Assuming that the signal is quantized to satisfy the
condition of previous question and assuming the
approximate bandwidth of the signal is W. The
minimum required bandwidth for transmission of a
will be.
(A) 5 W
(B) 10 W
(C) 20 W
(D) None of the above
The
bit
transmission rate for the time-division multiplexed
signal will be
(A) 8 kbps
(C) 256 kbps
************
(B) 64 kbps
(D) 512 kbps
48. Analog data having highest harmonic at 30 kHz
generated by a sensor has been digitized using 6 level
PCM. What will be the rate of digital signal generated?
(A) 120 kbps
(C) 240 kbps
(B) 200 kbps
(D) 180 kbps
49. In a PCM system, the number of quantization levels
is 16 and the maximum lsignal frequency is 4 kHz.; the
bit transmission rate is
(A) 32 bits/s
(C) 32 kbits/s
Page
438
(B) 16 bits/s
(D) 64 dbits/s
bandwidth
requirement will.
sampled at Nyquist rate are converted into binary PCM
using
relative
binary PCM signal based on this quantization scheme
47. Four voice signals. each limited to 4 kHz and
signal
( n + 1) 2
n2
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UNIT 7
16. (C) 128= 2 7,. We need 7 bits/character. For 1,000,000
character we need 7 Mbits/second. Thus minimum
bandwidth = 7 Mbits/sec.
26. (B)
Communication System
So
3 L2
=
= 6394 = 38.06 dB
N o [ln(m + 1)]2
27. (C) Nyquist Rate = 2 MHz
17. (D) fs > 2 fm = 6400 Hz, nfs £ 63000
36000
36000
n£
= 5.63, n = 5, L = 2 n = 32, fs =
= 7.2 kHz.
6400
5
50% higher rate = 3 MHz,
Thus transmission bandwidth is 3 MHz ´ 8 = 24 Mbits/s.
New sampling rate is at 20% above the Nyquist rate.
Sampling rate= 12
. ´ 2 = 2.4 MHz.
18. (D) The transmitted code word are
bits per second=
0
0
1
1
0
0
0
1
1
1
0
0
1
0
bits
24 M sec
= 10 bits
2.4 MHz
1
So 3(1024) 2
=
= 102300 = 50.1 dB
N o (ln 256) 2
Level = 210 = 1024,
Fig.S.7.6.18
In 1st word 001(1)
28. (B)
In 2nd word 010(2)
In 3rd word 011(3)
So
3 L2
=
³ 50 dB, m = 100
N o [ln(m + 1)]2
3 L2
= 100 000 or
[ln 101) 2
In 4th word 100(4)
In 5th word 101 (5)
L = 842.6
Because L is power of 2, we select L = 1024 = 210 .
19. (B) Message bandwidth= W , Nyquist rate = 2W
Thus 10 bits are required.
Bandwidth = 2W ´ 8 = 16W bit/s
29. (A) So = m 2 ( t), N o =
108
W=
= 6.25 MHz
16
16W = 10 , or
8
1
20. (A) Sampling interval T, =
= 125 ms. There are 24
8k
channels and 1 sync pulse, so the time allotted to each
channel is Tc =
T
25
= 5 ms. The pulse duration is 1 ms. So
the time between pulse is 4 ms.
21. (B) Amax =
22. (D) Amax =
23. (D) N o =
24. (C)
L = 256 = 28
So 3 L2 m 2 ( t)
,
=
3 L2
No
mp2
since signal is sinusoidal
m 2 ( t) 1
= ,
mp2
2
3 L2
= 48 dB = 63096, L = 205.09
2
Since L is power of 2, so we select L = 256
Hence 256 = 28 ,
dfs 0.1 ´ 68 k
=
= 108
. V
wm 2 p ´ 10 3
3mp2
So 8 bits per sample is required .
30. (B) ( SNR) q = 176
. + 6.02( n) = 40 dB, n = 6.35
We take the n = 7.
df s
A w
1 ´ 2 p ´ 800
or d = max m =
= 78.5 mV
wm
fs
50 ´ 10 3
d2 B (0.0785) 2 ´ 3500
=
= 1122
.
´ 10 -4 W
3 fs
3 ´ 64000
So
0.5
=
= 4.46 ´ 10 3
N o 112
. ´ 10 -4
Capacity = 20 ´ 8 k ´ 7 = 112
. Mbits = 140 Kbytes
31. (B) For slope overload to take place Em ³
df s
2 pfm
This is satisfied with Em = 15
. V and fm = 4 kHz.
32. (C) Step size d=
2 mp
L
=
1536
.
= 0.012 V
128
quantization noise power
S
25. (A) o µ L2 ,
No
æ S ö
L = 2 n çç o ÷÷ = 10 log( C2 2 n )
è N o ødB
= log C + 20 n log 2 = a + 6 n dB.
=
This equation shows
that increasing n by one bits increase the by 6 dB.
Hence an increase in the SNR by 12 dB can be
accomplished by increasing 9is form 10 to 12, the
transmission bandwidth would be increased by 20%
Page
440
d2 (0.012) 2
= 12 ´ 10 -6 V 2
=
12
12
33. (B) Bit Rate = 8 k ´ 8 = 64 kbps
(SNR) q = 176
. + 6.02 n dB = 176
. + 6.02 ´ 8 = 49.8 dB
34. (B) fs = 1800 samples/sec,
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Digital Transmission
Since the sampling rate is 1800 samples/sec the highest
frequency that can be recovered is 900 Hz.
The
maximum
frequency
component
will
be
150 + 25 = 175 Hz.
x( t) is band limited with fm = 350 Hz,
Thus Nyquist
1
rate is 2 fm = 700 Hz, Sampling interval =
sec
700
1
1
=
= 20 kHz,
T 50 ´ 10 -6
æ 1 + cos 300 pt ö
44. (D) g( t) = 10 cos 50 ptç
÷
2
è
ø
= 5 cos 50 pt + 5 cos 50 pt cos 300 pt
35. (C) x( t) = sinc 700t + sinc 500t
1
= [sin 700 pt + sin 500 pt ]
pt
36. (D) fs =
Chap 7.6
Thus fs = 2 ´ 175 = 350
. sample per second.
45. (B) Total sample = 8000 ´ 20 = 160 k sample/sec
Bit for each sample = 7 + 1 = 8
Bit Rate = 160 k ´ 8 = 1280 ´ 10 3 bits/sec
fc = 12 kHz
The frequency passed through LPF are fc , fs - fm or 12
kHz, 8 kHz
46. (B) The sampling rate is fs = 44100 meaning that we
take 44100 samples per second. Each sample is
quantized using 16 bits so the total number of bits per
37. (C) P =
( SNR)1 2
, Here n = code word length,
=
(SNR)2 2 2 n 1
2n 2
16
n1 = 61 n2 = 8,
Thus rate =
second is 44100´16. For a music piece of duration 50
min = 3000- sec the resulting number of bits per
channel
2
= 16
212
(left
= 2.1168 ´ 10
9
and
right)
is 44100 ´ 16 ´ 3000
and the overall number of bits is
2.1168 ´ 10 ´ 2 = 4.2336 ´ 10 9
9
38. (D) Signal g1 ( t), g 2 ( t), g s ( t) and g 4 ( t) will have 8 k, 8
k, 16 k and 16 k sample/sec at Nyquist rate. Thus
47. (C) Nyquist Rate = 2 ´ 4k = 8 kHz
48000 sample/sec bit rate 48000 ´ 8 = 384 kbps
Total sample = 4 ´ 8 = 32 k sample/sec
39. (C) Analog signals, having bandwidth 1200 Hz, 600
Hz and 600 Hz have 2400, 1200 samples/sec at Nyquist
256 = 28 , so that 8 bits are required
Bit Rate = 32 k ´ 8 = 256 kbps
rate. Hence 48000 sample/sec
48. (D) Nyquist Rate = 2 ´ 30 k = 60 kHz
bit rate = 48000 sample/sec ´12 = 57.6 kbps
2 n ³ 6 Thus n = 3,
3
æ sin 2 p1000 t ö
æ sin 2 p1000 t ö
40. (C) x( t) = 5ç
÷ + 7ç
÷
pt
pt
è
ø
è
ø
2
49. (C) Nyquist rate= 2 ´ 4 = 8 kHz
2 n = 16 or n = 4,
Maximum frequency component = 3 ´ 1000 = 3 kHz
Sampling rate = 2 fm = 6 kHz
50. (C) fs = 8 kHz,
41. (B) Here fs = 32 k sample/sec
1 1
fm = =
Em = 125,
T 2
51. (D)
For slope-overload to be averted Em ³
D£
Em fm
fs
or D £
Bit Rate = 4 ´ 8 = 32 kbits/sec
2 n = 256 Þ n = 8
Bit Rate = 8 ´ 8k = 64 kb/x
1
2
So
a 2 2 n , If PCM is increased form n to n + 1,
No
the ratio will increase by a factor 4. Which is
fs
fm
independent of n.
1
2
125( )
(128)( )
or D £ 2 -8
or D £
3
32 ´ 10
32 ´ 1024
42. (D) fm = 5 kHz,
Bit Rate = 60 ´ 3 = 18 kHz
Nyquist Rate = 2 ´ 5 = 10 kHz
Since signal are sampled at twice the Nyquist rate so
52. (C) If L = 2, then 2 = 2 n or n = 1 ND If L = 8, then
8 = 2 n or n = 3. So relative bandwidth will be tripled.
53. (B) The minimum bandwidth requirement for
transmission of a binary PCM signal is BW= vW. Since
sampling rate = 2 ´ 10 = 20 kHz.
v = 10, we have BW = 10 W
Total transmission bandwidth = 4 ´ 20 = 80 kHz
43. (D) Signal will be sampled 200, 200, 400 and 800
sample/sec thus 1600 sample per second,
***********
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CHAPTER
7.8
SPREAD SPECTRUM
Statement for Question 1-3 :
5. The Antijam margin is
A pseudo-noise (PN) sequance is generated using a
(A) 47.5 dB
(B) 93.8 dB
feedback shift register of length m = 4. The chip rate is
(C) 86.9 dB
(D) 12.6 dB
7
10 chips per second
6.
1. The PN sequance length is
A slow
FH/MFSK
system
has
the
following
parameters.
(A) 10
(B) 12
Number of bits per MFSK symbol = 4
(C) 15
(D) 18
Number of MFSK symbol per hop = 5
The processing gain of the system is
2. The chip duration is
(A) 1ms
(B) 0.1 ms
(C) 0.1 ms
(D) 1 ms
(B) 15 ms
(C) 6.67 ns
(D) 0.67 ns
(B) 37.8 dB
(C) 6 dB
(D) 26 dB
7.
A fast
FH/MFSK
system
has
the
following
parameters.
3. The period of PN sequance is
(A) 15
. ms
(A) 13.4 dB
Number of bits per MFSK symbol = 4
Number of pops per MFSK symbol = 4
The processing gain of the system is
Statement for Question 4-5:
(A) 0 dB
(B) 7 dB
(C) 9 dB
(D) 12 dB
A direct sequence spread binary phase-shiftkeying system uses a feedback shift register of Length
Statement for Question 8-9:
19 for the generation of PN sequence . The system is
required to have an average probability of symbol error
due to externally generated interfering signals that
does not exceed 10 -5
A rate 1/2 convolution code with dfrec = 10 is used
to encode a data requeence occurring at a rate of 1 kbps.
The modulation is binary PSK. The DS spread
spectrum sequence has a chip rate of 10 MHz
4. The processing gain of system is
8. The coding gain is
(A) 37 dB
(B) 43 dB
(A) 7 dB
(B) 12 dB
(C) 57 dB
(D) 93 dB
(C) 14 dB
(D) 24 dB
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Spread Spectrum
9. The processing gain is
Chap 7.8
Statement for question 16-18 :
(A) 14 dB
(B) 37 dB
(C) 58 dB
(D) 104 dB
A CDMA system consist of 15 equal power user
that transmit information at a rate of 10 kbps, each
using a DS spread spectrum signal operating at chip
10. A total of 30 equal-power users are to share a
rate of 1 MHz. The modulation scheme is BPSK.
common communication channel by CDM. Each user
transmit information at a rate of 10 kbps via DS spread
16. The Processing gain is
spectrum and binary PSK. The minimum chip rate to
(A) 0.01
(B) 100
(C) 0.1
(D) 10
obtain a bit error probability of 10
-5
(A) 1.3 ´ 106 chips/sec
(B) 2.9 ´ 10 5 chips/sec
(C) 19
. ´ 106 chips/sec
(D) 1.3 ´ 10 5 chips/sec
11. A CDMA system is designed based on DS spread
17. The value of e b/J 0 is
(A) 8.54 dB
(B) 7.14 dB
(C) 17.08 dB
(D) 14.28 dB
spectrum with a processing gain of 1000 and BPSK
modulation scheme. If user has equal power and the
18. How much should the processing gain be increased
desired level of performance of an error probability of
to allow for doubling the number of users with affecting
10 -6 , the number of user will be
the autopad SNR
(A) 89
(B) 117
(A) 1.46 MHz
(B) 2.07 MHz
(C) 147
(D) 216
(C) 4.93 MHz
(D) 2.92 MHz
12. In previous question if processing gain is changed to
19.
500, then number of users will be
processing gain of 500. If the desired error probability is
(A) 27 users
(B) 38 users
10 -5 and ( e b / J 0 ) required to obtain an error probability
(C) 42 users
(D) 45 users
of 10 -5 for binary PSK is 9.5 dB, then the Jamming
A DS/BPSK
spread
spectrum
signal
has
a
margin against a containers tone jammer is
Statement for Question 13-15 :
A DS spread spectrum system transmit at a rate of
(A) 23.6 dB
(B) 17.5 dB
(C) 117.4 dB
(D) 109.0 dB
1 kbps in the presets of a tone jammer. The jammer
power is 20 dB greater then the desired signal, and the
required Îb / J 0 to achieve satisfactory performance is
10 dB.
Statement for Question 20-21 :
An m = 10 ML shift register is used to generate the
pre hdarandlm sequence in a DS spread spectrum
13. The spreading bandwidth required to meet the
specifications is
system. The chip duration is Tc = l ms and the bit
duration is Tb = NTc , where N is the length (period of
the m sequence).
(A) 10 7 Hz
(B) 10 3 Hz
(C) 10 5 Hz
(D) 106 Hz
20. The processing gain of the system is
14. If the jammer is a pulse jammer, then pulse duty
cycle that results in worst case jamming is
(A) 0.14
(B) 0.05
(C) 0.07
(D) 0.10
(A) 10 dB
(B) 20 dB
(C) 30 dB
(D) 40 dB
21. If the required e b/J 0 is 10 and the jammer is a tone
jammer with an average power J av, then jamming
15. The correspond probability of error is
margin is.
(A) 4.9 ´ 10 -3
(B) 6.3 ´ 10 -3
(A) 10 dB
(B) 20 dB
(C) 9.4 ´ 10 -4
(D) 8.3 ´ 10 -3
(C) 30 dB
(D) 40 dB
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UNIT 7
Statement for Question 22-23 :
An FH binary orthogonal FSK system employs an
m = 15 stage liner feedback shift register that generates
an ML sequence. Each state of the shift register selects
one of L non over lapping frequency bands in the
hopping pattern. The bit rate is 100 bits/s. The
demodulator employ non coherent detection.
22. If the hop rate is one per bit, the hopping bandwidth
for this channel is
Communication System
(A) 0.4 GHz
(B) 0.6 GHz
(C) 0.7 GHz
(D) 0.9 GHz
28. The probability of error for the worst-case partial
band jammer is
(A) 0.2996
(B) 0.1496
(C) 0.0368
(D) 0.0298
29. The minimum hop rate for a FH spread spectrum
system that will prevent a jammer from operating five
onives away from the receiver is
(A) 6.5534 MHz
(B) 9.4369 MHz
(C) 2.6943 MHz
(D) None of the above
(A) 3.2 bHz
(B) 3.2 MHz
(C) 18.6 MHz
(D) 18.6 kHz
23. Suppose the hop rate is increased to 2 hops/bit and
the receiver uses square law combining the signal over
two hops. The hopping bandwidth for this channel is
(A) 3.2767 MHz
(B) 13.1068 MHz
(C) 26.2136 MHz
(D) 1.6384 MHz
***********
Statement for Qquestion 24-25 :
In a fast FH spread
spectrum system, the
information is transmitted via FSK with non coherent
detection. Suppose there are N = 3 hops/bit with hard
decision decoding of the signal in each hop. The channel
is AWGN with power spectral density
1
2
N0 and an SNR
20-13 dB (total SNR over the three hops)
24. The probability of error for this system is
(A) 0.013
(B) 0.0013
(C) 0.049
(D) 0.0049
25. In case of one hop per bit the probability of error is
(A) 196
. ´ 10 -5
(B) 196
. ´ 10 -7
(C) 2.27 ´ 10 -5
(D) 2.27 ´ 10 -7
Statement for Question 26-29 :
A slow FH binary FSK system with non coherent
detection
operates
at
e b / J 0 = 10,
with
hopping
bandwidth of 2 GHz, and a bit rate of 10 kbps.
26. The processing gain of this system is
(A) 23 dB
(B) 43 dB
(C) 43 dB
(D) 53 dB
27. If the jammer operates as a partial band jammar,
the bandwidth occupancy for worst case jamming is
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Spread Spectrum
Chap 7.8
W/R
W/R
Î
=
= b
J av/Pav N u - 1 J 0
SOLUTION
æÎ
W/R = çç b
è J0
1. (C) The PN sequence length is
N= 2 m - 1 = 2 4 - 1 = 15
æÎ
W = Rçç b
è J0
2. (B) The chip duration is
1
TC = 7 s = 0.1 ms
10
ö
÷÷( N u - 1)
ø
ö
÷÷( N u - 1)
ø
where R = 10 4 bps, N u = 30 and Îb / J 0 = 10
Therefore, W = 2.9 ´ 106 Hz
The minimum chip rate is 1 / Tc = W = 2.9 ´ 106 chips/sec
3. (A) The period of the PN sequence is
T= NTC = 15 ´ 0.1 = 15
. ms
11. (D) To achieve an error probability of 10 -6 , we
4. (C) m= 19
æÎ
required çç b
è J0
n= 2 m - 1 = 219 - 1 = 219
ö
÷÷ = 10.5 dB
ødB
Then, the number of users of the CDMA system is
The processing gain is 10 log10 N= 10 log10 219
W/R
1000
+1=
+ 1 = 89 users
Îb /J 0
11.3
= 190 ´ 0.3 or 57 dB
Nu =
æ Eb ö
5. (A) Antijam margin = (Processing gain) - 10 log10 çç
÷÷
è N0 ø
12. (D) If the processing gain is reduced to W/R = 500,
The probability of error is
1
Pe = erfc
2
æ
ç
ç
è
Eb
N0
N u=
ö
÷
÷
ø
500
+ 1 = 45 users
11.3
13. (D) We have a system where ( J av/Pav) dB = 20 dB,
With Pe = 10 -5, we have Eb / N 0 = 9.
R = 1000 bps and (Îb /J 0 ) dB = 10 dB
Hence, Antijam margin = 57 - 10 log10 9 = 57 - 9.5
æÎ
æJ ö
æW ö
Hence, we obtain ç ÷ = çç av ÷÷ + çç b
è R ødB è Pav ødB è J 0
or =47.5 dB
6. (D) The precessing gain (PG) is
FH Bandwidth W c
PG =
=
= 5 ´ 4 = 20
Symbol Rate
Rs
ö
÷÷ = 30 dB
ødB
W
= 1000
R
W = 1000 R = 106 Hz
Hence, expressed in decibels, PG= 10 log10 20 = 26 db
7. (D) The processing gain is
PG = 4 ´ 4 = 16
Hence, in decibels,
14. (C) The duty cycle of a pulse jammer of worst-case
0.71
0.7
jamming is a =
=
= 0.07
Îb /J 0 10
15. (D) The corresponding probability of error for this
PG = 10 log10 16 = 12 dB
8. (A) The coding gain is Rcd min =
then
worst-case jamming is
1
´ 10 = 5 or 7 dB
2
9. (B) The processing gain
P2 =
0.083 0.083
=
= 8.3 ´ 10 -3
e b/J 0
10
16. (B) Precessing gain is
W
10 7
=
= 5 ´ 10 3 or 37 dB
R 2 ´ 10 3
W 106
=
= 100
R 10 4
17. (A) We have N u = 15 users transmitting at a rate of
10. (C) We assume that the interference is characterized
10,000 bps each, in a bandwidth of W = 1MHz.
as a zero-mean AWGN process with power spectral
The e b/J 0 is.
-5
density J 0 . To achieve an error probability of 10 , the
required Îb /J 0 = 10 we have
e b W/R 106 / 10 4 100
=
=
=
14
14
J 0 Nu - 1
= 7.14 or 8.54 dB
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UNIT 7
18. (B) With N u = 30 and e b/J 0 = 7.14, the processing
Communication System
ec
gain should be increased to
1 P= e 2 N 0
2
W/R= (7.14)(29) = 207
where e c / N 0 = 20 / 3. The probability of a bit error is
W= 207 ´ 104 = 2.07 MHz
Pb = 1 - (1 - p) 2 = 1 - (1 - 2 p + p2 ) = 2 p - p2
Hence the bandwidth must be increased to 2.07 MHz
19. (B) The processing gain is given as
W
= 500 or 27 dB
R
=e
-
ec
2 N0
ec
-
1 - 2 N0
e
= 0.0013
2
25. (C) In the case of one hop per bit, the SNR per bit is
ec
The ( e b/J 0 ) required to obtain an error probability of
20, Hence,
Pb =
1 - 2 N 0 1 -10
e
= e = 2.27 ´ 10 -5
2
2
10 -5 for binary PSK is 9.5 dB. Hence, the jamming
26. (D) We are given a hopping bandwidth of 2 GHz and
margin is
a bit rate of 10 kbs.
æe ö
æ J av ö
æW ö
÷÷ = ç ÷ - çç b ÷÷ = 27.95 or 17.5 dB
çç
R
P
è
ø
è J 0 ødB
è av ødB
dB
Hence,
20. (C) The period of the maximum length shift register
sequence is
a* W = 2/( e b/J 0 ) = 0.2
Since Tb = NTc then the processing gain is
T
N b = 1023 or 30 dB
Tc
Hence a* W = 0.4 GHz
28. (C) The probability of error with worst-case
partial-band jamming is
21. (B) A Jamming margin is
æe ö
ö
÷÷ - çç b ÷÷ = 30 - 10 = 20 dB
ødB è J 0 ødB
where J av = J 0W » J 0/Tc = J 0 ´ 10
22. (A) The length of the shift-register sequence is
L = 2 m - 1215 - 1 = 32767 bits
For binary FSK modulation, the minimum frequency
separation is 2/T, where 1/T is the symbol (bit) rate.
The hop rate is 100 hops/sec. Since the shift register
e -1
e -1
= 3.68 ´ 10 -2
=
( e b/J 0 ) 10
Dd= 2 ´ 8050 = 16100
Dd
Dd= x ´ t or t =
t
Dd 16100
Þ t=
=
= 5.367 ´ 10 5
x
3 ´ 108
1
1
f= =
= 18.63 kHz
t 5.367 ´ 10 -5
L = 32767 states and each state utilizes a
bandwidth of 2/T = 200 Hz, then the total bandwidth
for the FH signal is 6.5534 MHz.
23. If the hopping rate is 2 hops/bit and the bit rate is
100 bits/sec, then, the hop rate is 200 hops/sec. The
minimum
frequency
separation
for
orthogonality
2/T = 400 Hz. Since there are N = 32767 states of the
shift register and for each state we select one of two
frequencies
separated
by
400
Hz,
the
hopping
bandwidth is 13.1068 MHz.
24. (B) The total SNR for three hops is 20 ~ 13 dB.
Therefore the SNR per hop is 20/3. The probability of a
chip error with non-coherent detection is
Page
454
P2 =
29. (D) d= 5 miles = 8050 meters
6
has
27. (A) The bandwidth of the worst partial-band
jammer is a* W , where
N = 210 - 1 = 1023
æW
æ J av ö
÷÷ = çç
çç
P
è av ødB è Rb
W 2 ´ 10 9
=
= 2 ´ 10 5or 53 dB
R
10 4
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UNIT 8
10. A field is given as
G=
Electromagnetics
16. A field is given as
13
( yu x + 3u y + xu z )
x2 + y2
G =
25
( xu x + y u y )
x2 + y2
The unit vector in the direction of G at P(3, 4, -2)
The field at point (-2, 3, 4) is
(A) 13( -2 u x + 3u y + 4 u z )
(B) -2 u x + 3u y + 4 u z
(C) 13( 3u x + 4 u y - 2 u z )
(D) 3u x + 4 u y - 2 u z
11. A field is given as F = yu x + zu y + xu z The angle
between G and u x at point (2, 2, 0) is
(A) 45°
(B) 30°
(C) 60°
(D) 90°
is
(A) 0.6 u x + 0. 8 u y
(B) 0. 8 u x + 0.6 u y
(C) 0.6 u y + 0. 8 u z
(D) 0.6 u z + 0.6 u x
17. A field is given as F = xyu x + yzu y + zxu x The value of
4 2
the double integral I = ò ò F × u ydzdx in the plane y = 7 is
0 0
12. A vector field is given as
G = 12 xyu x + 6( x 2 + 2) u y + 18 z 2 u z
(A) 128
(B) 56
(C) 190
(D) 0
18. Two vector extending from the origin are given as
The equation of the surface M on which |G | = 60 is
(A) 4 x 2 y 2 + 4 x 4 + 9 z 4 + 2 x 2 = 96
(B) 2 x 2 y 2 + x 4 + 9 z 4 + 2 x 2 = 96
(C) 2 x 2 y 2 + 4 x 4 + 9 z 4 + 2 x 2 = 96
(D) 4 x 2 y 2 + x 4 + 9 z 4 + 2 x 2 = 96
R1 = 4 u x + 3u y - 2 u z and R 2 = 3u x - 4 u y - 6 u z . The
area of the triangle defined by R1 and R 2 is
(A) 12.47
(B) 20.15
(C) 10.87
(D) 15.46
19. The four vertices of a regular tetrahedron are
located at O (0, 0, 0), A(0, 1, 0), B(0.5
13. A vector field is given by
(
E = 4 zy u z + 2 y sin 2 x u y + y sin 2 x u z
2
(C) Plane x =
3p
2
, 0.5,
2
3
). The unit vector perpendicular (outward) to
(A) 0.41u x + 0.71u y + 0.29 u z
(B) 0.47 u x + 0.82 u y + 0.33u z
(B) Plane x = 0
(C) -0.47 u x - 0.82 u y - 0.33u z
(D) all
(D) -0.41u x - 0.71u y - 0.29 u z
20. The two vector are R AM = 20 u x + 18 u y - 18 u z and
14. The vector field E is given by
E = 6 zy 2 cos 2 x u x + 4 xy sin 2 x u y + y 2 sin 2 x u z
The region in which E = 0 is
R AN = - 10 u x + 8 u y + 15 u z . The unit vector in the plane
of the triangle that bisects the interior angle at A is
(A) 0.168 u x + 0.915 u y + 0.367 u z
(A) y = 0
(B) x = 0
(B) 0.729 u x + 0.134 u y - 0.672 u z
(C) z = 0
np
(D) x =
2
(C) 0.729 u x + 0.134 u y + 0.672 u z
(D) 0.168 u x + 0.915 u y - 0.367 u z
15. Two vector fields are F = -10 u x + 20 x( y - 1) u y and
G = 2 x 2 yu x - 4 u y + 2 u z . At point A(2, 3, -4) a unit
vector in the direction of F - G is
21. Two points in cylindrical coordinates are A( r = 5,
f = 70 ° , z = -3) and B(r = 2, f = 30 ° , z = 1). A unit vector
at A towards B is
(A) 0.18 u x + 0.98 u y - 0.05 u z
(A) 0.03u x - 0.82 u y + 0.57 u z
(B) -0.18 u x - 0.98 u y + 0.05 u z
(B) 0.03u x + 0.82 u y + 0.57 u z
(C) -0.37 u x + 0.92 u y + 0.02 u z
(C) -0.82 u x + 0.003u y + 0.57 u z
(D) 0.37 u x - 0.92 u y - 0.02 u z
Page
458
3, 0.5, 0) and C
the face ABC is
2
The surface on which E y = 0 is
(A) Plane y = 0
0 .5
3
(D) 0.003u x - 0.82 u y + 0.57 u z
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Vector Analysis
22. A field in cartesian form is given as
D = xu x +
29. The vector
yu y
x +y
2
B=
2
10
u r + r cos q u q + u f
r
in cartesian coordinates at (-3, 4, 0) is
In cylindrical form it will be
u
u
uf
(B) D = r +
(A) D = r
r
r
cos f
(C) D = ru r
Chap 8.1
(D) D = ru r + cos f u f
(A) u x - 2 u y
(B) - 2u x + u y
(C) 1.36 u x + 2.72 u y
(D) -2.72 u x + 1.36 u x
30. The two point have been given A (20, 30 ° , 45 ° ) and
23. A vector extends from A(r = 4, f = 40 ° , z = -2) to B(
B ( 30, 115 ° , 160 ° ). The |R AB | is
r = 5, f = -110 ° , z = 1). The vector R AB is
(A) 22.2
(B) 44.4
(A) 4.77 u x + 7.30 u y + 4 u z
(C) 11.1
(D) 33.3
(B) -4.77 u x - 7.30 u y + 4 u z
(C) -7.30 u x - 4.77 u y + 4 u z
31. The surface r = 2 and 4, q = 30 ° and 60°, f = 20 ° and
(D) 7.30 u x + 4.77 u y + 4 u z
80° identify a closed surface. The enclosed volume is
24. The surface r = 3, r = 5, f = 100 °, f = 130 °, z = 3 and
(A) 11.45
(B) 7.15
(C) 6.14
(D) 8.26
z = 4.5 define a closed surface. The enclosed volume is
(A) 480
(B) 5.46
32. The surface r = 2 and 4, q = 30 ° and 50° and f = 20 °
(C) 360
(D) 6.28
and 60° identify a closed surface. The total area of the
enclosing surface is
25. The surface r = 2, r = 4, f = 45 °, f = 135 °, z = 3 and
z = 4 define a closed surface. The total area of the
enclosing surface is
(A) 6.31
(B) 18.91
(C) 25.22
(D) 12.61
(A) 34.29
(B) 20.7
33. At point P(r = 4, q = 0.2 p , f = 0.8 p), u r in cartesian
(C) 32.27
(D) 16.4
component is
26. The surface r = 3, r = 5, f = 100 °, f = 130 °, z = 3 and
(A) 0.48 u x + 0.35 u y + 0.81u z
z = 4.5 define a closed volume. The length of the longest
(B) 0.48 u x - 0.35 u y - 0.81u z
straight line that lies entirely within the volume is
(C) -0.48 u x + 0.35 u y + 0.81u z
(A) 3.21
(B) 3.13
(D) 0.48 u x - 0.35 u y - 0.81u z
(C) 4.26
(D) 4.21
34. The expression for u y in spherical coordinates at P(
27. A vector field H is
r = 4, q = 0.2 p, f = 0.8 p) is
æ fö
H = rz 2 sin f u r + e - z sin ç ÷ u f + r 3u z
è2 ø
(A) 0.48 u r + 0.35 u q - 0.81u f
(B) 0.35 u r + 0.48 u q - 0.81u f
p
æ
ö
At point ç 2, , 0 ÷ the value of H × u x is
3
è
ø
(A) 0.25
(B) 0.433
(C) -0.433
(D) -0.25
(C) -0.48 u r + 0.35 u q - 0.81u f
(D) -0.35 u r + 0.48 u q - 0.81u f
35. Given a vector field
28. A vector is A = yu x + ( x + z) u y. At point P(-2, 6, 3)
D = r sin f u r -
A in cylindrical coordinate is
(A) -0.949 u r - 6.008 u f
(B) 0.949 u r - 6.008 u f
(C) -6.008 u r - 0.949 u f
(D) 6.008 u r + 0.949 u f
1
sin q cos f u q + r 2 u f
r
The component of D tangential to the spherical
surface r = 10 at P(10, 150°, 330°) is
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UNIT 8
Electromagnetics
(A) 0 .043u q + 100 u f
40. The gradient of the functionG = r 3 sin 2 q sin 2 f sin q
(B) -0 .043u q - 100 u f
at point P ( 12 ,
(C) 110 u q + 0.043u f
(A) 1.41u r + 3u z
(B) u x + u y + u z
(D) 0 .043u q - 100 u f
(C) 3.46 u r + 9.3u q
(D) All
36. The circulation of F = x 2 u x - xzu y - y 2 u z around the
41.
path shown in fig. P8.1.36 is
The
1
2
1
2
,
) is
directional
derivative
of
function
F = xy + yz + zx at point P(3, -3, - 3) in the direction
toward point Q(4, -1, -1) is
z
1
(A) -3
(B) 1
(C) -2
(D) 0
42. The temperature in a auditorium is given by
T = 2 x 2 + y 2 - 2 z 2 . A mosquito located at (2, 2, 1) in the
y
auditorium desires to fly in such a direction that it will
1
get warm as soon as possible. The direction, in that it
1
x
must fly is
(A) 8 u x + 8 u y - 4 u z
Fig. P8.1.36
(A) -
1
3
(B)
1
6
(B) 2 u x + 2 u y - u z
(C) -
1
6
(D)
1
3
(D) -(2 u x + 2 u y - u z )
(C) 4 u x + 4 u y - 4 u z
37. The circulation of A = r cos f u r + z sin f u z around
the edge L of the wedge shown in Fig. P8.1.37 is
43. The angle between the normal to the surface
x2 y + z = 3
x ln z - y 2 = -4
and
at
the
point
of
intersection (-1, 2, 1) is
y
(A) 73.4°
(B) 36.3°
(C) 16.6°
(D) 53.7°
L
44. The divergence of vector A = yzu x + 4 xyu y + yu z at
60o
point P(1, -2, 3) is
x
0
2
Fig. P8.1.37
(A) 2
(B) -2
(D) 4
(A) 1
(B) -1
(C) 0
(C) 0
(D) 3
45.
The
divergence
of
the
38. The gradient of field f = y 2 x + xyz is
(A) y( y + z) u x + x(2 y + z) u y + xyu z
(B) y(2 x + z) u x + x( x + z) u y + xyu z
(C) y 2 u x + 2 yxu y + xyu z
(A) 2.6
(B) 1.5
(C) 4.5
(D) -4.5
46. The divergence of the vector
(D) y(2 y + z) u x + x(2 y + z) u y + xyu z
A = rz 2 cos f u r + z sin 2 f u z is
39. The gradient of the field f = r z cos 2 f at point
2
(1, 45 ° , 2) is
(A) 4u f
(B) 4 2u f
(C) -4u f
(D) -4 2u f
Page
460
vector
A = 2 r cos q cos f u r + r u f at point P(1, 30°, 60°) is
1 2
(A) 2rz 2 cos f u r + sin 2 f u z
(B) 2rz 2 cos f u r + sin 2 f u z
(C) 2 z 2 cos f u r + sin 2 f u z
(D) z sin 2
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Vector Analysis
Chap 8.1
47. The flux of D = r 2 cos 2 f u r + 3 sin f u f over the
54. If V = x 2 y 2 z 2 , the laplacian of the field V is
closed surface of the cylinder 0 £ z < 3,
(A) 2( xy 2 + yz 2 + zx 2 )
r = 3 is
(A) 324
(B) 81p
(B) 2( x 2 y 2 + y 2 z 2 + z 2 x 2 )
(C) 81
(D) 64p
(C) ( x 2 y 2 + y 2 z 2 + z 2 x 2 )
48. The curl of vector A = e xyu x + sin xy u y + cos 2 xz u z
is
(D) 0
55. The value of Ñ 2 V at point P(3, 60°, -2) is if
(A) y e xyu x + x cos xy u y - 2 x sin 2 xz u z
V = r 2 z(cos f + sin f)
(B) z sin 2 xy u y + ( y cos xy - xe xy) u z
(C) z sin 2 xy u y + ( x cos xy - xe xy) u z
(D) xy e xyu x + xy cos xy u y - 2 xz sin 2 xz u z
(A) -8.2
(B) 12.3
(C) -12.3
(D) 0
56. If the scalar field V = r 2 (1 + cos q sin f) then Ñ 2 V is
49. The curl of vector field
A = rz sin f u r + 3rz 2 cos f u r at point (5, 90°, 1) is
(A) 0
(B) 12u q
(C) 6u r
(D) 5u f
(A) 1 + 2(1 - r 2 ) cos q sin f
(B) 6 + 4 cos q sin f - cot q cosec q sin f
(C) 2 + 2(1 - r 2 ) cos q sin f
(D) 0
50. The curl of vector field
1
A = r cos q u r - sin q u q + 2 r 2 sin q u f is
r
1
(A) cos q u r - cos q u q
r
57. Ñ ln r is equal to
æ 1
ö
(B) 2 r 2 cos q u r - 4 r sin q u q + ç 2 sin q - r sin q ÷u f
r
è
ø
(A) Ñ ´ ( fu z )
(B) Ñ ´ ( zu f)
(C) Ñ ´ (ru f)
(D) Ñ ´ (ru z )
58. If r = xu x + yu y + zu x then ( r × Ñ ) r 2 is equal to
(A) 2 r 2
(B) 3r 2
(D) 0
(C) 4 r 2
(D) 0
51. If A = ( 3 y 2 - 2 z) u x - 2 x 2 z u y + ( x + 2 y) u z , the value
59. If r = xu x + yu y + zu x is the position vector of point
of Ñ ´ Ñ ´ A at P(-2, 3, -1) is
P( x, y, z) and r =|r| then Ñ × r n r is equal to
(C) 4 r cos q u r - 6 r sin q u q + sin q u f
(A) -( 6 u x + 4 u y)
(B) 8 ( u x + u y)
(A) nr n
(B) ( n + 3) r n
(C) -8( u x + u y)
(D) 0
(C) ( n + 2) r n
(D) 0
52. The grad × Ñ ´ A of a vector field
60. If F = x 2 yu x - yu y, the circulation of vector field F
A = x 2 yu x + y 2 zu y - 2 xzu z is
around closed path shown in fig. P8.1.60 is,
(A) 2 xy + 2 yz - 2 x
y
(B) x 2 y + y 2 z - 2 xz
1
(C) 2 x 2 y + 2 y 2 z - 2 xz
S
(D) 0
53. If V = xy - x 2 y + y 2 z 2 , the value of the div grad V
0
is
L
2
1
x
Fig. P8.1.60
(A) 0
(B) z + x 2 + 2 y 2 z
(C) 2 y( z 2 - yz - x)
(D) 2( z - y - y)
2
2
(A)
7
3
(B) -
7
6
(C)
7
6
(D) -
7
3
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UNIT 8
Electromagnetics
61. If A = r sin f u r + r 2 u r , and L is the contour of fig.
SOLUTIONS
ò A. dL is
P8.1.61, then circulation
L
1. (D) d = ( x1 - x2 ) 2 + ( y1 - y2 ) 2 + ( z1 - z 2 ) 2
y
2
= ( 4 - 2) 2 + ( -6 - 3) 2 + ( 3 - ( -1)) 2 = 4 + 81 + 16 = 101
L
1
2. (A) R AB = R B - R A
= ( 3u x + 0 u y + 2 u z ) - (5 u x - u y + 0 u z )
-2
-1
1
2
= -2 u x + u y + 2 u z
x
|R AB | =
Fig. P8.1.61
(A) 7 p + 2
(B) 7 p - 2
(C) 7p
(D) 0
uR = -
22 + 1 + 22 = 3
2
1
2
ux + u y + uz
3
3
3
3. (C) The component of F parallel to G is
62. The surface integral of vector
F = 2r 2 z 2 u r + r cos 2 f u z
=
over the region defined by 2 £ r £ 5, -1 < z < 1,
F ×G
(10, - 6, 5) × (0.1, 0.2, 0.3)
G =
(0.1, 0.2, 0.3)
2
G
0.12 + 0.2 2 + 0.32
= 9.3(0.1, 0.2, 0.3) = (0.93, 1.86, 2.79)
0 < f < 2 p is
(A) 44
(B) 176
(C) 88
(D) 352
63. If D = xyu x + yzu y + zxu z , then the value of òò A × dS
is, where S is the surface of the cube defined by
0 £ x £ 1, 0 £ y £ 1, 0 £ z £ 1
(A) 0.5
(B) 3
(C) 0
(D) 1.5
4. (C) The vector component of F perpendicular to G is
F ×G
( 3, 2, 1) × ( 4, 4, - 2)
=F G = ( 3, 2, 1) ( 4, 4, - 2)
G2
42 + 42 + 22
= (3, 2, 1)- (2, 2, - 1) = (1, 0, 2) = u x + 2 u z
5. (C) R = 3u x + 4 M - N
= 3u x + 4(2 u x + 3u y - 4 u z ) -( -4 u x + 4 u y + 3u z )
= 15 u x + 8 u y - 19 u z
64. If D = 2rzu r + 3z sin f u r - 4r cos f u z and S is the
|R| =
15 2 + 8 2 + 19 2 = 25.5 = 25.5
surface of the wedge 0 < r < 2, q < f < 45 °, 0 < z < 5, then
6. (B) R = - M + 2N
the surface integral of D is
(A) 24.89
(B) 131.57
= - ( 8 u x + 4 u y - 8 u z ) + 2( 8 u x + 6 u y - 2 u z )
(C) 63.26
(D) 0
= 8u x + 8u y + 4u z
8u x + 8u y + 4u z
uR =
82 + 82 + 42
65. If the vector field
F = ( axy + bz 3) u x + ( 3 x 2 - gz) u y + ( 3 xz 2 - y) u z
is irrotational, the value of a , b and g is
=
2
2
1
æ2 2 1ö
ux + u y + uz = ç , , ÷
3
3
3
è 3 3 3ø
(A) a = b = g = 1
(B) a = b = 1, g = 0
-M + 2N = -( 8, 4, - 8) + 2( 8, 6, - 2) = (8, 8, 4)
(C) a = 0,
(D) a = b = g = 0
uR =
b = g =1
**************
uR =
æ2 2 1ö
=ç , , ÷
è 3 3 3ø
8 +8 +4
( 8, 8, 4)
2
2
2
2
2
1
ux + u y + uz
3
3
3
æ 1 + 7 -6 - 2 4 + 0 ö
7. (C) Mid point is ç
,
,
÷ = (4, -4, 2)
2
2 ø
è 2`
uR =
Page
462
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( 4, - 4, 2)
4 + 4 +2
2
2
2
2 1ö
æ2
=ç , - , ÷
3
3 3ø
è
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Vector Analysis
=
2
2
1
ux - u y + uz
3
3
3
8. (A) G = 24(1)(2) u x + 12(1 + 2) u y + 18( -1) 2 u z
= 48 u x + 36 u y + 18 u z
2 1ö
æ2
9. (A) A = ( 6, - 2, - 4), B = k ç , - , ÷
3 3ø
è3
|B - A |= 10
2
2
2
2 ö
2 ö
1 ö
æ
æ
æ
ç 6 - k ÷ + ç -2 + k ÷ + ç -4 - k ÷ = 100
3
3
3 ø
è
ø
è
ø
è
k2 - 8 k - 44 = 0
Þ
k = 1175
. ,
Chap 8.1
= -34 u x + 84 u y - 2 u z
-34 u x + 84 u y - u z
uR =
34 2 + 84 2 + 2 2
= - 0.37 u x + 0.92 u y - 0.02 u z
16. (A) At P(3, 4, -2)
25
G= 2
( 3u x + 4 u y) = 3u x + 4 u y
3 + 42
3u x + 4 u y
uG =
= 0.6 u x + 0.8 u y
32 + 4 2
17. (B) F × u y = Fy = yz
2 1ö
æ2
B = 1175
. ç ,- , ÷
3 3ø
è3
4 2
I=
ò ò yzdzdx =
0 0
= (7.83, -7.83, -3.92)
æ2
ò0 çç ò0 yzdz
è
4
4
ö
÷dx = ò 2 ydx = 2( 4) y = 8 y
÷
0
ø
At y = 7, I = 8(7) = 56
13
10. (D) G =
( 3u x + 4 u y - 2 u z )
2
( -2) + ( 3) 2
= 3u x + 4 u y - 2 u z
éu x u y u z ù
1
1ê
18. (B) Area = R1 ´ R 2 =
4
3 -2 ú
ú
2
2ê
êë 3 -4 -6 úû
11. (A) Let q be the angle between F and u x ,
= u x ( -18 - 8) - u y( -24 + 6) + u z ( -16 - 9)
Magnitude of F is |F|= y + z + x
= -26 u x + 18 u y - 25 u z
2
F × u x = (F)(1) cos q = y
y
cos q =
=
2
y + z 2 + x2
2
2
R1 ´ R 2 = 26 2 + 18 2 + 25 2 = 40.31
2
2 +2
2
2
=
1
area =
2
q = 45 °
40.31
= 20.15
2
19. (B) R BA = (0, 1, 0) - (0.5 3 , 0.5, 0) = ( -0.5 3 , 0.5, 0)
æ 0.5
R BC = çç
, 0.5,
è 3
12. (D) |G|= 60
(12 xy) 2 + ( 6( x 2 + 2)) 2 + (18 z 2 ) 2 = 60
Þ
144 x 2 y 2 + 36( x 4 + 4 x 2 + 4) + 324 z 4 = 3600
Þ
4 x y + ( x + 4 x + 4) + 9 z = 100
Þ
4 x 2 y 2 + x 4 + 9 z 4 + 4 x 2 = 96
2
2
4
2
4
R BA ´ R BC
13. (D) For E y = 0, 2 y sin 2 x = 0
Þ
sin 2 x = 0, Þ
3p
x = 0,
2
2 x = 0, p , 3p , Þ
y =0
Hence (D) is correct.
14. (A)
E = y( 6 zy cos 2 x u x + 4 x sin 2 x u y + y sin 2 x u z )
Hence in plane y = 0,
E = 0.
15. (C) R = F - G
= ( -10 u x + 20 x( y - 1) u y) - (2 x 2 yu x - 4 u y + 2 u z )
At P(2, 3, - 4) ,
R = F - G = ( -10 u x + 80 u y) - (24 u x - 4 u y + 2 u z )
= u x 0.5
2
3
ö
÷ - (0.5 3 , 0.5, 0)
÷
ø
é
uy
ê ux
= ê0.5 3 0.5
ê
ê 1
0
êë 3
æ 1
= çç , 0,
3
è
2
3
ö
÷
÷
ø
ù
uz ú
0 ú
ú
2ú
3 úû
2
0.5
- u y (0.5 2 ) + u z
3
3
= 0.41u x - 0.71u y + 0.29 u z
The required unit vector is
0.41u x + 0.71u y + 0.29 u z
=
0.412 + 0.712 + 0.29 2
= 0.47 u x + 0.81u y + 0.33u z
20. (A) The non-unit vector in the required direction is
1
= ( u AN + u AM )
2
( -10, 8, 15)
u AN =
= ( -0.507, 0.406, 0.761)
100 + 64 + 225
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UNIT 8
(20, 18, - 10)
u AM =
400 + 324 + 100
Electromagnetics
26. (A) A(r = 3, f = 100 ° , z = 3) = A( -0.52, 2.95, 3)
= (0.697, 0.627, -0.348)
B(r = 5, f = 130 ° , z = 4.5) = B( -321
. , 3.83, 4.5)
1
( u AM + u AN )
2
1
= [(0.697, 0.627, - 0.348) + ( -0.507, 0.406, 0.761) ]
2
length = |B - A|
B - A = ( -321
. , 3.83, 4.5) - ( -0.52, 2.95, 3)
= ( -2.69, 0.88, 15
. )
= (0.095, 0.516, 0.207)
(0.095, 0.516, 0.207)
= (0.168, 0.915, 0.367)
u bis =
0.095 2 + 0.516 2 + 0.2612
Hence (A) is correct.
|B - A| = |( -2.69,
0.88, 15
. )| = 2.69 2 + 0.88 2 + 15
. 2 = 3.21
æ p ö
27. (C) At P ç 2, , 0 ÷, H = 0.5 u f + 8 u z
è 3 ø
u x = cos f u r - sin f u f =
21. (D) In cartesian coordinates
A (5 cos 70 ° , 5 sin 70 ° , - 3) = A(171
. , 4.70, - 3)
1
( u r - 3u f)
2
æ
3 ö
÷ = -0.433
H × u x = (0.5)çç ÷
2
è
ø
B (2 cos ( -30 ° ), 2 sin ( -30 ° ), - 3) = B(173
. , - 1, 1)
R AB = R B - R A = (173
. , - 1, 1) - (171
. , 4.70, - 3)
é Ar ù é cos f sin f 0 ù é Ax ù
28. (A) ê Af ú = ê- sin f cos f 0 ú ê A y ú
ê ú ê
úê ú
0
1 úû êë Az úû
êë Az úû êë 0
= (0.02, - 5.70, 4)
(0.02, - 5.70, 4)
= (0.003, -0.82, 0.57)
u AB =
0.02 2 + 5.70 2 + 4 2
22. (A) x = r cos f ,
At P (-2, 6, 3)
y = r sin f
æ 6 ö
A = 6 u x + u y , f = tan -1 ç
÷ = 108.43°
è -2 ø
r cos f u x + r sin f u y 1
D= 2
= (cos f u x + sin f u y)
r cos 2 f + r 2 sin 2 f r
cos f = - 0.316, sin f = 0.948
1
Dr = D × u r = [cos f ( u x × u r) + sin f ( u y × u r)]
r
=
1
1
[cos 2 f + sin 2 f] =
r
r
Df = D × u f =
=
0.948 0 ù é6 ù
é Ar ù é-0.316
ê A ú = ê -0.948 -0.316 0 ú ê1 ú
ê fú ê
úê ú
0
1 úû êë0 úû
êë Az úû êë 0
Ar = 6 ( -0.316) + 0.948 = -0.949,
1
[cos f ( u x × u f) + sin f ( u y × uf)]
r
Af = 6( -0.948) - 0.316 = -6.008, Az = 0
Hence (A) is correct option.
1
[cos f ( - sin f) + sin f (cos f)] = 0
r
Therefore D =
29.(B) At P (-3, 4, 0)
1
ur
r
r = x 2 + y 2 + z 2 = 32 + 4 2 + 0 2 = 5
23. (B) A( 4 cos 40 ° , 4 sin 40 ° , - 2) = A( 306
. , 2.57, - 2)
B (5 cos ( -110 ° ), 5 sin ( -110 ° ), 2) = B ( -171
. , - 4.7, 2)
R AB = R B - R A = ( -171
. , - 4.7, 2) - ( 306
. , 2.57, - 2)
4 .5 130 ° 5
ò ò ò rdrdfdz
éBx ù ésin q cos f cos q cos f - sin q ù éBr ù
êB ú = êsin q sin f cos q cos f cos f ú êB ú
ê yú ê
ú ê qú
- sin q
1 úû êëBf úû
êë Bz úû êë cos q
= 6.28
3 100 ° 3
25. (C) Area is
=2
135° 4
ò
ò rdrdf +
45° 2
4
4 135°
ò
ò 2 dfdz +
3 45°
4 135°
4 4
3 45°
3 2
ò ò 4 dfdz + 2 ò ò drdz
ér ù æ p ö
æ pö
= 2 ê ú ç ÷ + (2)(1)ç ÷ = 32.27
è2 ø
ë 2 û2è 2 ø
2
Page
464
f = tan -1
x2 + y2 p
=
z
2
y
4
= tan -1
= 126.87 °
x
-3
B = 2u r + u f
= ( -4.77, - 7.3, 4)
24. (D) Vol =
q = tan -1
é-0.6 0 -0.8 ù é2 ù
= ê 0.8
0 -0.6 ú ê0 ú
ê
úê ú
-1
0 úû êë1 úû
êë 0
Bx = 2( -0.6) + 0.8 = -2
B y = 2(0.2) - 0.6 = 1
Bz = 0
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UNIT 8
2
C3 = ò r cos f dr
Along 3,
0
ò A × dL
=
f=60 °
r2
= -1
2
Electromagnetics
44. (D) Ñ × A =
¶ Ax ¶ A y ¶ Az
+
+
=0 + 4x + 0
¶x
¶y
¶z
At P (1, -2, 3), Ñ × A = 4
= C1 + C2 + C3 = 1
L
45. (A)
38. (A) Ñf = u x
¶f
¶f
¶f
+ uy
+ uz
¶x
¶y
¶z
Ñ× A =
= y( y + z) u x + x(2 y + z) u y + xyu z
39. (C) Ñf = u r
=
¶f
1 ¶f
¶f
+ uf
+ uz
¶x
r ¶y
¶z
1
( 6 r 2 cos q cos f) + 0 + 0
r2
At P (1, 30°, 60°), Ñ × A = 6(1)(cos 30 ° )(cos 60 ° ) = 2.6
= 2 r 2 z cos 2 f u r - 2rz sin 2 f u f + r 2 cos 2 f u z
46. (C) Ñ × A =
At P (1, 45°, 2), Ñf = - 4u f
40. (B) r sin q cos f = x , r sin q sin f = y , r cos q = z
G = r 3 sin 2 q sin 2 f sin q
=
1 ¶(rrz 2 cos f) ¶( z sin 2 f)
= 2 z 2 cos f + sin 2 f
+
r
¶r
¶z
ò D × dS
¶( 4 xyz)
¶( 4 xyz)
¶( 4 xyz)
ÑG = u x
+ uy
+ uz
¶x
¶x
¶z
S
Ñ ×D =
= 4 yzu x + 4 xzu y + 4 xyu z
æ1 1 1ö
At P ç , , ÷ , ÑG = u x + u y + u z
è2 2 2 ø
ÑF = -6u x , R PQ = ( 4, - 1, - 1) - ( 3, - 3, - 3) = (1, 2, 2)
¶T
¶T
¶T
+ uy
+ uz
¶x
¶y
¶z
= 2 xu x + 2 yu z - 4 z u z
v
1 ¶(rr 2 cos 2 f) 1 ¶( z sin f)
3
+
= 3r cos 2 f + sin f
r
r
r
¶r
¶r
2
f+
v
z
sin
r
ö
f ÷÷rdfdzdr
ø
3
3
2p
3
3
2p
0
0
0
0
0
0
= ò dz ò zr 2 dr ò cos 2 f df + ò zdz ò dr ò sin f df = 81p
At P(3, -3, - 3),
= -2
ò Ñ × D dv
æ
41. (C) ÑF = ( y + z) u x + ( x + z) u y + ( x + y) u z
( -6 u x ) × ( u x + 2 u y + 2 u z )
=
ò Ñ × D dv = ò ççè 3r cos
v
42. (C) ÑT = u x
By divergence theorem
S
= 4( r sin q cos f)( r sin q sin f)( r cos q) = 4 xyz
3
1 ¶(rAr) 1 ¶( Af) ¶( Az )
+
+
r ¶r
r ¶q
¶z
47. (B) The flux is ò D × dS ,
= r 3(2 sin q cos q)(2 sin f cos f) sin q
ÑF × u R =
1 ¶(r 2 Ar )
1 ¶(sin q Aq)
1 ¶(sin q Af)
+
+
r2
r sin q
r sin q
¶f
¶r
¶q
éu x
ê¶
48. (B) Ñ ´ A = ê
ê ¶x
êë Ax
uy
¶
¶y
Ay
u z ù éu x
uy
uz ù
¶ú ê¶
¶
¶ ú
ú =ê
ú
¶z ú ê ¶x
¶y
¶z ú
Az úû êë e xy sin xy cos 2 xz úû
u x (0 - 0) - u y(2 cos xz ( - sin xz) z) + u z ( y cos xy - xexy)
= z sin 2 xy u y + ( y cos xy - xe xy) u z
At P(2, 2, 1), ÑT = 4 u x + 4 u y - 4 u z
43. (A) Let f = x 2 y + z - 3, g = x ln z - y 2 + 4
Ñf = 2 xyu x + x 2 u y + u z
Ñg = ln z u x - 2 yu y +
x
uz
z
At point P (-1, 2, 1)
uf =
-4 u x + u y + u z
18
cos q = ± u f × u g = ±
-1
q = cos 0.28 = 73.4 °
Page
466
, ug =
-5
18 ´ 17
-4 u y - u z
17
= 0.286
éu r
1ê ¶
49. (D) Ñ ´ A = ê
r ê ¶r
ëê Ar
uf uz ù
¶
¶ú
ú
¶f ¶z ú
Af Az úû
é ur
uf
uz ù
ê
¶
¶
¶ú
1
= ê
ú
r ê ¶r
¶f
¶z ú
êërz sin f 2r 2 z 2 cos f 0 úû
=
1
1
u r (-6r 2 z cos f) - u f (-r sin f) u z (6rz 2 cos f - rz cos f)
r
r
At point P(5, 90°, 1), Ñ ´ A = 5 u f
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Vector Analysis
éu r
ê¶
1
50. (C) Ñ ´ A = 2
ê
r sin q ê ¶r
êë Ar
56.(B)
r sin qu f ù
ú
¶
ú
¶f
ú
r sin q Af úû
ru q
¶
¶q
rAq
Ñ2 V =
1
r 2 sin q
u r ( 2r 3 sin q - 0) -
¶2 V
1 ¶ æ 2 ¶V ö
1
¶ æ
¶V ö
1
çr
÷+ 2
ç sin q
÷+ 2
2
2
r ¶r è
¶r ø r sin q ¶q è
¶q ø r sin q ¶f2
= 6 + 4 cos q sin f - cot q cosec q sin f
57. (A) r = x 2 + y 2
1
1
u q ( 6r 2 sin 2 q - 0) + u f ( 0 + r sin q)
r sin q
r
Ñ ln r = u x
= 4 r cos q u r - 6 r sin q u q + sin q u f
ux
uy
uz ù
é
ê
ú
¶
¶
¶
51. (A) Ñ ´ A = ê
ú
¶x
¶y
¶z ú
ê
2
2
êë( 3 y - 2 z) ( -2 x z) ( x + 2 y) úû
ux
uy
¶
¶
¶x
¶y
+ 2 x2 ) 3
=
uz
ù
ú
¶
ú
¶z
ú
-( 4 xz + 6 y) úû
Ñ ´ Ñ ´ A = -6 u x - 4 u y = -( 6 u x + 4 u y)
uz ù
¶ ú
ú
¶z ú
-2 xz úû
n
-1
æ nö
Ñ × r n r = 2 x 2 ç ÷( x 2 + y 2 + z 2 ) 2
è2 ø
n
2
n
= n( x 2 + y 2 + z 2 )( x 2 + y 2 + z 2 ) 2
2
2
2
L
2
¶
55. (A) Ñ V =
¶r
2
2
æ ¶V
çç r
è ¶r
2
2
S
Ñ ´ F = -x uz
dS = dxdy ( -u z )
ò (Ñ ´ F) × dS = - òò ( - x ) dxdy
2
= 2( y z + x z + x y ) = 2( x y + y z + z x )
2
ò (Ñ ´ F) × dS
2
¶2V ¶2V ¶2V
54. (B) Ñ 2 V =
+
+
¶x 2
¶y 2
¶z 2
2
+ 3r n
60. (C) By Stokes theorem ò F × dL =
= -2 y + 2 z - 2 y = 2( z - y - y)
2
-1
= nr n + 3r n = ( n + 3) r n
¶( z - 2 xy) ¶(2 yz - x ) ¶( x - 2 y z)
+
+
¶x
¶y
¶z
2
2
n
+ rn + rn + rn
= ( z - 2 xy) u x + (2 yz - x ) u y + ( x - 2 y z) u z
2
¶( xr n ) ¶( yr n ) ¶( zr n )
+
+
¶x
¶y
¶z
-1
-1
æ nö
æ nö
+ 2 y 2 ç ÷( x 2 + y 2 + z 2 ) 2 + 2 z 2 ç ÷( x 2 + y 2 + z 2 ) 2
è2 ø
è2 ø
2
2
x
y
x
y
ux + 2
u y = ux + u y
2
2
r
r
x +y
x +y
n
¶V
¶V
¶V
53. (D) ÑV = u x
+ uy
+ uz
¶x
¶y
¶z
2
tan -1
where r n = ( x 2 + y 2 + z 2 ) 2
Ñ (Ñ ´ A) = 0
2
0
ù
ú
ú
ú
ú
yú
x úû
uz
¶
¶z
2
59. (B) Ñ × r n r =
= - y 2 u x + 2 zu y - x 2 u z
Ñ × (ÑV ) =
uy
¶
¶y
= x(2 x) + y(2 y) + z(2 z) = 2( x 2 + y 2 + z 2 ) = 2 r 2
At P(-2, 3, -1),
uy
¶
¶y
y2 z
é
êu x
ê¶
y
uz = ê
x
ê ¶x
ê
0
êë
æ ¶
¶
¶ ö 2
58. (A) ( r × Ñ ) r 2 = çç x
+y
+z
÷÷( x + y 2 + z 2 )
x
y
z
¶
¶
¶
è
ø
= u x ( -6) - u y( -4 z) + u z (0) = -6 u x + 4 zu y
é ux
ê ¶
52. (D) Ñ ´ A = ê
ê ¶x
2
ëê x y
¶ ln r
¶ ln r
¶ ln r
x
y
+ uy
+ uz
= 2 ux + 2 u y
r
r
¶x
¶y
¶z
Ñ ´ f u z = Ñ ´ tan -1
= u x (2 + 2 x 2 ) - u y(1 - ( -2)) + u z ( -4 xz - 6 y)
é
ê
Ñ ´Ñ ´ A =ê
ê
êë(2
2 sin q
cos q sin f
cos q sin f sin q
sin 2 q
= 6(1 + cos q sin f) -
é ur
ru q
r sin q u f ù
ú
ê ¶
¶
¶
1
= 2
ê
ú
r sin q ê ¶r
¶q
¶f
ú
3
2
ëêr cos q - sin q 2 r sin q úû
=
Chap 8.1
2
2
2
S
2
ö 1 ¶2V ¶2V
+
÷÷ + 2
2
¶z 2
ø r ¶f
ò ò x dydx + ò ò x dydx
2
0 0
2
1
1
= 4 z(cos f + sin f) - z(cos f + sin f) + 0 = 3z(cos f + sin f)
æ1
3ö
÷= -8.2
At P(3, 60°, -2), Ñ 2 V = 3( -2)çç +
2 ÷ø
è2
2 -x + 2
1 x
=
0
1
=
2
ò x dx + ò x (2 - x) dx
3
0
2
1
2
é x4 ù
1 ù
é2
= ê ú + ê x3 - x4 ú
4
3
4 û1
ë
ë û0
=
1 é16
1 7
ù é2 1 ù 1 14
+
- 4ú - ê - ú = +
-4+ =
4 êë 3
3
4 6
û ë3 4 û 4
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UNIT 8
æ
ò A × dL = ççè ò
61. (C)
C
ab
+
ò
+
bc
ò
ò
+
cd
da
ò D × dS = ò (Ñ × D)dv
ö
÷÷A × dL
ø
S
2
2
= 4 ò rdr
L
1
0
a
b
d
-2
-1
1
2
=0
p
c
ò A × dL
ò r df
=
3
b
A × dL = r 3df
0
= r 3 ò df = (1) 3( -p) = - p
p
62. (B) Using divergence theorem
Ñ ×F =
ò
=
Ñ × Fdv
v
1 ¶
(2r 2 z 2 ) = 4 z 2 = 4 z 2
r ¶r
1
ò Ñ × F dv = òòò 4 z rdrdfdz = 4 ò z
2
2
0
¶( xy) ¶( yz) ¶( zx)
+
+
x =y+z+ x
¶x
¶y
¶z
S
æ1
= 3çç ò xdx
è0
1
1
1
0
0
0
( y + z + x) dxdydz
1
ö
æ1ö
dy
ò0 ò0 dz ÷÷ = 3çè 2 ÷ø = 15.
ø
1
64. (B) Ñ × D =
Page
468
2p
v
ò A × dS = ò ò ò
= 4z +
5
dz ò rdr ò df = 176
ò A × dS = ò (Ñ × A)dv
S
Ñ ×A =
2
-1
v
63. (D)
0
0
0
uy
¶
¶y
3 x 2 - gz
*******
=0
ò A × dL = 0 + 8 p + 0 - p = 7 p
ò F × dS
0
f = p, A × dL = 0,
Along da, dr = 0,
d
0
0
c
ò A × dL
45°
i.e. a = 1 = b = g.
= (2) 3 p = 8 p
d
a
5
If F is irrotational, Ñ ´ F = 0
A × dL = r 3df
Along cd, df = 0,
ò A × dL
2
uz
¶
¶z
3 xz 2 -
= ( -1 + g) u x + ( 3bz 2 - 3z 2 ) u y + ( 6 x - ax) u z
a
Along bc, dr = 0,
45°
é ux
ê
¶
65. (A) Ñ ´ F = ê
¶
x
ê
a
x
bz 2
+
êë
f = 0,
ò A × dL
5
ò zdz ò df + 3ò dr ò zdz ò cos f df
x
b
A × dL = 0,
æ
ö
3z
cos f ÷÷ rdrdfdz
çç 4 z +
r
è
ø
æ 4 öæ 25 öæ p ö
æ 25 öæ 1 ö
= 4ç ÷ç
.
÷ = 13157
÷ç ÷ + 3(2)ç
֍
è 2 øè 2 øè 4 ø
è 2 øè 2 ø
Fig. S8.1.61
Along ab, df = 0,
v
ò D × dS = ò ò ò
y
c
Electromagnetics
1 ¶(rDr) 1 ¶( Df) ¶( Dz )
+
+
r ¶r
r ¶q
¶z
3
z cos f
r
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CHAPTER
8.2
ELECTROSTATICS
1. Let Q1 = 4 mC
be located at P1 (3, 11, 8) while
Q2 = -5 mC is at P2 (6, 15, 8). The force F2 on Q2 will be
6. A 2 mC point charge is located at A (4, 3, 5) in free
space. The electric field at P(8, 12, 2) is
(A) -( 4.32 u x + 5.76 u y) N
(B) 4.32 u x + 5.76 u y N
(A) 131.1u r + 159.7 u f - 49.4 u z
(C) -( 4.32 u x + 5.76 u y) mN
(D) 4.32 u x + 5.76 u y mN
(B) 159.7 u r + 27.4 u f - 49.4 u z
2. Four 5 nC positive charge are located in the z = 0
plane at the corners of a square 8 mm on a side. A fifth
(C) 1311
. u r + 27.4 u f - 49.4 u z
(D) 159.7 u r + 137.1u f - 49.4 u z
5 nC positive charge is located at a point 8 mm distant
from each of the other charge. The magnitude of the
7. A point charge of -10 nC is located at P1 (0, 0, 0.5)
total force on this fifth charge is
and a charge of 2 mC at the origin. The E at P(0, 2, 1) is
(A) 2 ´ 10 -4 N
(C) 0.014 N
(B) 4 ´ 10 -4 N
(D) 0.01 N
3. Four 40 nC are located at A(1, 0, 0), B(-1, 0, 0), C(0,
(A) 68.83u r + 14.85 u f
(B) 68.83u r + 64.01u f
(C) 68.83u r - 14.85 u f
(D) 68.83u r - 64.01u f
8. Charges of 20 nC and -20nC are located at (3, 0, 0)
1, 0) and D(0, -1, 0) in free space. The total force on the
and (-3, 0, 0) and (-3, 0, 0), respectively. The magnitude
charge at A is
of E at y axis is
(A) 24.6u x mN
(C) -13.6u x mN
(B) -24.6u x mN
(D) 13.76u x mN
4. Let a point charge 41 nC be located at P1 (4, -2, 7) and
a charge 45 nC be at P2 (–3, 4, –2). The electric field E at
P3(1, 2, 3) will be
1080
(9 + y 2 ) 3 2
(B)
1080
(9 + y 2 ) 3
(C)
108
(9 + y 2 ) 3 2
(D)
108
(9 + y 2 ) 3
9. A charge Q0 located at the origin in free space
(A) 0.13u x + 0.33u y + 0.12 u z
produces a field for which E2 = 1 kV m at point P(–2, 2,
(B) -0.13u x - 0.33u y - 0.12 u z
–1). The charge Q0 is
(C) 115
. u x + 2.93u y + 109
. uz
(D) -115
. u x - 2.93u y - 109
. uz
5. Let a point charge 25 nC be located at P1 (4, -2, 7) and
a charge 60 nC be at P2 (-3, 4, -2). The point, at which on
the y axis, is Ex = 0, is
(A) -7.46
(C) -6.89
(A)
(B) -22.11
(D) (B) and (C)
(A) 2 mC
(B) -3 mC
(C) 3 mC
(D) -2 mC
10. The volume charge density r n = r oe -|x|-| y|-|z| exist over
all free space. The total charge present is
(A) 2r o
(B) 4r o
(C) 8r o
(D) 3r o
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UNIT 8
Electromagnetics
11. A uniform volume charge density of 0.2 mC m 2 is
18. Two identical uniform charges with r l = 80 nC m are
present throughout the spherical shell extending from
located in free space at x = 0, y = ± 3 m. The force per
r = 3 cm to r = 5 cm. If r = 0 elsewhere, the total charge
unit length acting on the line at positive y arising from
present throughout the shell will be
the charge at negative y is
(A) 41.05 pC
(B) 257.92 pC
(A) 9.375u y mN
(B) 37.5u y mN
(C) 82.1 pC
(D) 129.0 pC
(C) 19.17u y mN
(D) 75u y mN
12.
If r v =
1
z 2 + 10
5 e -0 .1 r( p-|f|) mC m 3 in
the region
19. A uniform surface charge density of 10 nC m 2 is
0 £ r £ 10, -p < f < p and all z, and r v = 0 elsewhere,
present in the region x = 0, - 2 < y < 2 and all z if e = e o
the total charge present is
, the electric field at P(3, 0, 0) has
(A) 1.29 mC
(B) 2.58 mC
(C) 0.645 mC
(D) 0
(A) x component only
(B) y component only
13. The region in which 4 < r < 5, 0 < q < 25 °, and
(C) x and y component
0.9 p < f < 11
. p contains the volume charge density of
(D) x, y and z component
f
2
r v = 10 ( r - 4) ( r - 5) sin q sin . Outside the region,
r v = 0. The charge within the region is
20. The surface charge density is r s = 5 nC m 2 , in the
(A) 0.57 C
(B) 0.68 C
region r < 0.2 , z = 0, and is zero elsewhere. The electric
(C) 0.46 C
(D) 0.23 C
field E at A(r = 0, z = 0.5) is
14. A uniform line charge of 5 nC m is located along the
(A) 5.4 V m
(B) 10.1 V m
(C) 10.5 V m
(D) 20.2 V m
line defined by y = -2, z = 5. The electric field E at P(1,
21. Three infinite charge sheet are positioned as
2, 3) is
(A) -9 u y + 4.5 u z
(B) 9 u y - 4.5 u z
follows: 10 nC m 2 at x = -3, - 40 nC m 2 at y = 4 and 50
(C) -18 u y + 9 u z
(D) 18 u y - 9 u z
nC m 2 at z = 2. The E at (4, 3, -2) is
15. A uniform line charge of 6.25 nC m is located along
the line defined by y = -2, z = 5. The E at that point in
the z = 0 plane where the direction of E is given by
(
1
3
u y - 23 u z ), is
(A) 4.5 u y + 9 u z
(B) 4.5 u y - 9 u z
(C) 9 u y - 18 u z
(D) 18 u y - 36 u z
(C) -48u y V m
(D) 24u y V m
(D) -0.56 u x - 2.23u y + 2.8 u z kV m
Let E = 5 x 3u x - 15 x 2 yu y . The equation of the
(A) y =
m
respectively. The E at P(6, 0, 6) will be
(B) 48u y V m
(C) 0.56 u x + 2.23u y + 2.8 u z kV m
stream line that passes through P(4, 2, 1) is
and y = -3
(A) -24u y V m
(B) 0.56 u x - 2.23u y + 2.8 u z kV m
22.
16. Uniform line charge of 20 nC m and -20 nC m are
located in the x = 0 plane at y = 3
(A) 0.56 u x + 2.23u y - 2.8 u z kV m
128
x3
(B) x =
128
y3
64
x2
(D) x =
64
y2
(C) y =
23. A point charge 10 nC is located at origin. Four
17. Uniform line charges of 100 nC m lie along the
uniform line charge are located in the x = 0 plane as
entire extent of the three coordinate axes. The E at
follows : 40 nC m
at y = 1 and -5 m, -60 nC m
P(-3, 2, -1) is
y = -2 and - 4 m. The D at P(0, -3, 4) is
(A) -192
. u x + 2 u y - 108
. u z kV m
(A) -19.1u y + 25.5 u z pC m 2
(B) -0.96 u x + u y - 0.54 u z kV m
(B) 19.1u y - 25.5 u z pC m 2
(C) 0.96 u x - u y + 0.54 u z kV m
(C) -16.4 u y + 219
. u z pC m 2
(D) 192
. u x - 2 u y + 108
. u z kV m
(D) 16.4 u y - 219
. u z pC m 2
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GATE EC BY RK Kanodia
Electrostatics
24. A point charge 20 nC
Chap 8.2
is located at origin. Four
31. A spherical surface of radius of 3 mm is centered at
uniform line charge are located as follows 40 nC m at
P(4, 1, 5) in free space. If D = xu x C m 2 the net electric
y = ± 1 and 50 nC m at y = ± 2. The electric flux that
flux leaving the spherical surface is
leaves the surface of a sphere, 4 m in radius, centered
at origin is
(A) 1.33 nC
(B) 1.89 mC
(C) 1.33 mC
(D) 1.89 mC
(A) 113.1 mC
(B) 339.3 nC
(C) 113.1 nC
(D) 452.4 nC
32. The electric flux density is
25. The cylindrical surface r = 8 C contains the surface
charge density, r s = 5 e
-20|z|
D=
2
nC m . The flux that leaves
the surface r = 8 cm, 1 cm < z < 5 cm
30 ° < f < 90 ° is
(A) 270.7 nC
(B) 9.45 nC
(C) 270.7 pC
(D) 9.45 pC
1
[10 xyzu x + 5 x 2 zu y + (2 z 3 - 5 x 2 y) u z ]
z2
The volume charge density r v at (-2, 3, 5) is
(A) 6.43
(B) 8.96
(C) 10.4
(D) 7.86
26. Let D = 4 xy u x + 2( x 2 + z 2 ) u y + 4 yzu z C m 2 . The
total charge enclosed in the rectangular parallelepiped
33. If D = 2ru r C m 2 , the total electric flux leaving the
0 < x < 2, 0 < y < 3, 0 < z < 5 m is
surface of the cube, 0 < x , y, z < 0.4 is
(A) 360 C
(B) 180 C
(A) 0.32
(B) 0.34
(C) 100 C
(D) 560 C
(C) 0.38
(D) 0.36
27. Volume charge density is located in free space as
rv = 2e
-1000 r
nC m
for 0 < r < 1 mm
3
and
rv = 0
34. If E = 4 u x - 3u y + 5 u z in the neighborhood of point
P(6, 2, -3). The incremental work done in moving 5 C
elsewhere. The value of Dr on the surface r = 1 mm is
charge a distance of 2 m in the direction u x + u y + u z is
(A) 1.28 pC m 2
(B) 0.28 pC m 2
(A) -60 J
(B) 34.64 J
(C) 0.78 pC m 2
(D) 0.32 pC m 2
(C) -34.64 J
(D) 60 JJ
and 4 carry uniform
35. If E = 100 u r V m , the incremental amount of work
charge densities of 20 nC m and -4 nC m . The Dr at
done in moving a 60 mC charge a distance of 2 mm from
2 < r < 4 is
P(1, 2, 3) toward Q(2, 1, 4) is
28. Spherical surfaces at r = 2
2
(A) -
16
nC m 2
r2
2
(B)
80
(C) 2 nC m 2
r
16
nC m 2
r2
80
(D) - 2 nC m 2
r
+ 6 z 3u z C m 2 . The total charge enclosed in the volume
0 < x, y , z < a is
5 4
a
3
(B) a 5 + 6 a 4
(C) 6 a 5 + a 4
(D)
(B) 3.1 mJ
(C) -31
. mJ
(D) 0
36. A 10 C charge is moved from the origin to P(3, 1, -1)
29. Given the electric flux density, D = 2 xyu x + x 2 u y
(A) 6 a 5 +
(A) -5.4 mJ
5 5
a + 6a4
3
in the field E = 2 xu x - 3 y 2 u y + 4 u z V m along the
straight line path x = -3 y, y = x + 2 z. The amount of
energy required is
(A) -40 J
(B) 20 J
(C) -20 J
(D) 40 J
37. A uniform surface charge density of 30 nC m 2 is
present on the spherical surface r = 6 mm in free space.
30. Let D = 5 x y z u y . The flux enclosed by volume
The V AB between A ( r = 2 cm, q = 35 ° , f = 55 ° )
x = 3 and 3.1, y = 1 and 1.1, and z = 2 and 2.1 is
( r = 3 cm, q = 40 ° , f = 90 ° ) is
(A) 49.6
(B) 24.8
(A) 2.03 V
(B) 10.17 V
(C) 35.4
(D) 36.4
(C) 4.07 mV
(D) -10.17 V
4
4
4
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UNIT 8
Electromagnetics
38. A point charge is located at the origin in free space.
(A) -4.94 nC
(B) -4.86 nC
The work done in carrying a charge 10 C from point A
(C) -5.56 nC
(D) -3.68 nC
( r = 4, q = p 6 , f = p 4) to B( r = 4, q = p 3 , p 6) is
(A) 0.45 mJ
(B) 0.32 mJ
(C) -0.45 mJ
(D) 0
46. A dipole having
Qd
= 100 V × m 2
4 pe o
39. Let a uniform surface charge density of 5 nC m 2 be
present at the z = 0 plane, a uniform line charge density
of 8 nC m be located at x = 0, z = 4 and a point charge
of 2 mC be present at P(2, 0, 0). If V = 0 at A(0, 0, 5),
the V at B(1, 2, 3) is
(A) 10.46 kV
(B) 1.98 kV
(C) 0.96 kV
(D) 3.78 kV
is located at the origin in free space and aligned so
that its moment is in the u z direction. The E at point
( r = 1, 45 ° , f = 0) is
(A) 158.11 V m
(B) 194.21 V m
(C) 146.21 V m
(D) 167.37 V m
47. A dipole located at the origin in free space has a
40. A non uniform linear charge density, r L = 6 ( z + 1)
moment p = 2 ´ 10 -9 u z C.m. The points at which | E|q = 1
nC m lies along the z axis. The potential at P(r = 1, 0, 0)
mV m on line y = z, x = 0 are
in free space is ( V¥ = 0)
(A) y = z = ± 23.35
(B) y = z = ± 16.5, x = 0
(C) y = z = 16.5
(D) y = 0, z = 23.35, x = 0
2
(A) 0 V
(B) 216 V
(C) 144 V
(D) 108 V
48. A dipole having a moment p = 3u x - 5 u y + 10 u z
41. The annular surface, 1 cm < r < 3 cm carries the
nC.m is located at P(1, 2, -4) in free space. The V at Q
nonuniform surface charge density r s = 5r nC m . The V
(2, 3, 4) is
2
at P(0, 0, 2 cm) is
(A) 81 mV
(B) 90 mV
(C) 63 mV
(D) 76 mV
42. If V = 2 xy 2 z 3 + 3 ln( x 2 + 2 y 2 + 3z 2 ) in free space the
magnitude of electric field E at P (3, 2, -1) is
(A) 72.6 V/m
(B) 79.6 V/m
(C) 75 V/m
(D) 70.4 V/m
V. The volume charge density at r = 0.6 m is
(B) -1.79 nC m 3
(C) 1.22 nC m 3
(D) -1.22 nC m 3
(B) 1.26 V
(C) 2.62 V
(D) 2.52 V
49. A potential field in free space is expressed as
V = 40 xyz . The total energy stored within the cube
1 < x, y, z < 2 is
43. It is known that the potential is given by V = 70 r 0 .6
(A) 1.79 nC m 3
(A) 1.31 V
(A) 1548 pJ
(B) 0
(C) 774 pJ
(D) 387 pJ
50. Four 1.2 nC point charge are located in free space at
the corners of a square 4 cm on a side. The total
potential energy stored is
44. The potential field V = 80 r 2 cos q V. The volume
(A) 1.75 mJ
(B) 2 mJ
(C) 3.5 mJ
(D) 0
charge density at point P(2.5, q = 30 °, f = 60 °) in free
51. Given the current density
space is
(A) -2.45 nC m 3
(C) -1.42 nC m
3
(B) 1.42 nC m 3
(D) 2.45 nC m
J = 10 5[sin (2 x) e -2 yu x + cos (2 x) e -2 yu y ] kA m 2
3
The total current crossing the plane y = 1 in the
45. Within the cylinder r = 2, 0 < z < 1 the potential is
u y direction in the region 0 < x < 1, 0 < z < 2 is
given by V = 100 + 50r + 150r sin f V. The charge lies
(A) 0
(B) 12.3 mA
within the cylinder is
(C) 24.6 mA
(D) 6.15 mA
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UNIT 8
Electromagnetics
64. In a certain region where the relative permitivity is
69. A potential field exists in a region where e = f ( x). If
2.4, D = 2 u x - 4 u y + 5 u z nC m 2 . The polarization is
r v = 0, the Ñ 2 V is
1 dF ¶V
(A) f ( x) dx ¶x
(A) 2.8 u x - 5.6 u y + 7 u z nC m 2
(B) 3.4 u x - 6.9 u y + 8.6 u z nC m 2
(C) 1.2 u x - 2.3u y + 2.9 u z nC m 2
(C)
(D) 3.89 u x - 6.43u y + 8.96 u z nC m 2
65. Medium 1 has the electrical permitivity e1 = 15
. eo
and occupies the region to the left of x = 0 plane.
Medium 2 has the electrical permitivity e 2 = 2.5 e o and
occupies the region to the right of x = 0 plane. If E1 in
medium 1 is E1 = (2 u x - 3u y + 1u z ) V m
then E2
in
(B) f ( x)
1 df ¶V
f ( x) dx ¶x
(D) -f ( x)
where r v = 0. It is know that both Ex and V are zero at
the origin. The V ( x, y) is
(A) 3( x 2 - y 2 )
(B) 3( y 2 - x 2 )
(C) 4 x 2 - 3 y 2
(D) 4 y 2 - 3 x 2
(A) (2.0 u x - 1.8 u x + 0.6 u z ) V m
(B) (1.67 u x - 3u y + u z ) V m
(C) (1.2 u x - 3u y + u z ) V m
(D) (1.2 u x - 1.8 u y + 0.6 u z ) V m
*********
66. Two perfect dielectrics have relative permitivities
e r1 = 2 and e r 2 = 8. The planner interface between them
is the surface x - y + 2 z = 5. The origin lies in region 1.
If E1 = 100 u x + 200 u y - 50 u z V m then E2 is
(A) 400 u x + 800 u y - 200 u z V m
(B) 400 u x + 200 u y - 50 u z V m
(C) 25 u x + 200 u y - 50 u z V m
(D) 125 u x + 175 u z V m
67. The two spherical surfaces r = 4 cm and r = 9 cm are
for
4 < r < 6 and e r 2 = 5 for 6 < r < 9. If E1 = 1000
u r then E2
r2
is
(A)
5000
ur V m
r2
(B)
400
ur V m
r2
(C)
2500
ur V m
r2
(D)
2000
ur V m
r2
68. The surface x = 0 separate two perfect dielectric. For
x > 0,
let
e r1 = 3,
while
er 2 = 5
where
x < 0.
If
E1 = 80 u x - 60 u y - 40 u z V m then E2 is
(A) (133.3u x - 100 u z - 66.7 u z ) V m
(B) (133.3u x - 60 u z - 40 u z ) V m
(C) ( 48 u x - 36 u y - 24 u z ) V m
(D) ( 48 u x - 60 u y - 40 u z ) V m
Page
474
df ¶V
dx ¶x
70. If V ( x, y) = 4 e 2 x + f ( x) - 3 y 2 in a region of free space
medium 2 is
separated by two perfect dielectric shells, e r1 = 2
df ¶V
dx ¶x
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Electrostatics
SOLUTIONS
1. (C) F2 =
=
=
f = tan -1
= 65.9 cos 56.3° + 148.3 sin 56.3 = 159.7
(4, 4), (-4, 4), (4, -4), (-4, 4), the fifth charge will be on
the z-axis at location z = 4 2. By symmetry, the force on
the fifth charge will be z
directed, and will be four
times the z component
=
2
12
= 56.3°, and z = 2
8
Er = E p × u p = 65.9( u x × u r) + 148.3 ( u y × u r)
2. (D) Arranging the charge in the xy plane at location
´
é 4 u x + 9 u y - 3u z ù
ê
ú
(106) 3 / 2
ë
û
Then at point P, r = 8 2 + 12 2 = 14.4,
= ( 4.32 u x + 5.76 u y) mN
4
2 ´ 10 -6
4 pe o
= 65.9 u x + 148.3u y - 49.4 u z
Q1Q2 R12
4 pe o |R12|3
( 4 ´ 10 -6 )( -5 ´ 10 -6 ) ( 3u x + 4 u y)
´
4 pe o
53
F=
Chap 8.2
q2
4 pe o d 2
Ef = E r × u f = 65.9( u x × u f) + 148.3 ( u y × u f)
= -65.9 sin 56.3° + 148.3 cos 56.3° = 27.4 ,
Ez = -49.4
7. (C) Ep =
2 ´ 10 -8
4 pe o
é R1
2R 2 ù
ê- |R |3 + |R |3 ú
ë
1
2 û
R1 = (0, 2, 1) -(0, 0, 0.5) =(0, 2, 0.5)
(5 ´ 10 -9) 2
´
= 10 -2 N
4 p ´ ( 8.85 ´ 10 -12 )( 8 ´ 10 -3) 2
2
4
R 2 = (0, 2, 1) - (0, 0, 0) =(0, 2, 1)
é - (2 u y + 0.5 u z ) 2(2 u y + u z ) ù
Ep = 9 ´ 10 9 ´ 10 -8 ê
+
ú
( 4.25) 3 2
( 5)3 2 û
ë
E p = 54.9 u y + 44.1u z
3. (D) The force will be
( 40 ´ 10 -9) 2 é R CA
R DA
R BA ù
F=
+
+
3
3
3ú
ê
4 pe o
ë|R CA| |R DA| |R BA| û
At P, r = 5 , q = cos -1
1
5
= 63.4 ° and f = 90 °
So Er = Ep × u r = 54.9[ u y × u r ] + 44.1[ u z × u r ]
where R CA = u x - u y , R DA = u x + u y , R BA = 2 u x
= 54.9 sin q sin f + 44.1 cos q = 68.83
|R CA|=|R DA|= 2,
Eq = E r × u q = 54.9[ u y × u q ] + 44.1[ u z × u q ]
F=
|R BA|= 2
éu x - u y u x + u y 2u x ù
( 40 ´ 10 -9) 2
+
+
-9 ê
4 p ´ ( 8.85 ´ 10 ) ë 2 2
8 úû
2 2
= 54.9 cos q sin f + 44.1 ( - sin q) = -14.85
Ef = E r × u f = 54.9( u y × u f) + 44.1( u z × u f) =54.9 cos f = 0
= 1376
. u x mN
8. (A) Let a point on y axis be P(0, y, 0)
10 -9 é 41R13
45R 23 ù
4. (C) E =
+
2
ê
|R 23|2 úû
4 pe o ë |R13|
Ep =
R13 = -3u x + 4 u y - 4 u z ,
R 23 = 4 u x - 2 u y + 5 u z
é 41( -3u x + 4u y - 4u z 45( 4u x + -2u y + 5u z ù
E = 9 ´ 10 9 ´ 10 -9 ´ ê
+
ú
( 41) 3 / 2
( 45) 3 / 2
ë
û
= 1152
.
u x + 2.93u y + 1089
.
uz
5. (D) The point is P3(0, y, 0)
R13 = -4 u x + ( y + 2) u y - 7 u z ,
R12 = 3u x + ( y - 4) u y + 2 u z
Ex =
ù
10 -9 é
25 ´ ( -4)
60 ´ 3
+
4 pe o êë[ 65 + ( y + 2) 2 ]3 2
[13 + ( y - 4) 2 ]3 2 úû
To obtain Ex = 0,
0.48 y 2 + 1392
. y + 7312
. =0
which yields the two value y = -6.89, - 22.11
6. (B) Ep =
2 ´ 10 -6 R AP
4 pe o |R AP|3
20 ´ 10 -8
4 pe o
é R1
R2 ù
ê-|R |3 - |R |3 ú
ë
1
2 û
R1 = (0, y, 0) - (3, 0, 0) =(-3, y, 0)
R 2 = (0, y 0) - (-3, 0, 0) = (3, y, 0),
|R1 | = |R 2 | =
9 + y2
é -3u x + yu y 3u x + yu y ù
+
Ep = 20 ´ 10 -9 ´ 9 ´ 10 9 ê
ú
2 3
( 9 + y 2 ) 3 úû
êë ( 9 + y )
-1080 u x
1080
, |E| = =
=
2 32
(9 + y )
(9 + y 2 ) 3 2
9. (B) The field at P will be
Q é -2 u x + 2 u y - u z ù
Ep = 0 ê
ú, Ez = 1 kV m
4 pe o ë
93 2
û
Q0 = -4 pe o ´ 9 3 2 ´ 10 3 = -3 m C
10. (C) This will be 8 times the integral of r n over the
first octant
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UNIT 8
¥
Q = 8ò
¥
¥
0
0
ò ò
0
é 6 u x - 3u y 6 u x + 3u y ù
EP = 20nC ´ 2 ´ 9 ´ 10 9 ê
36 + 9 úû
ë 36 + 9
r oe - x - y- z dxdydz = 8ro
11. (C) Q =
2p
p
0
0
= -48u y V m
0 .05
ò ò ò
0.2 r sin q drd qdf
2
0 .03
é
r3 ù
= ê4 p(0.2) ú = 82.1 pC
3 û 0 .03
ë
12. (A) Q =
òòò
¥ -p 0
ée
=5 ê
ë
R xp = ( -3, 0, - 1, ) - ( -3, 0, 0, ) = (0, 2, -1)
Similarly R yp = (-3, 0, -1), R zp = (-3, 2, 0)
5 e -0 .1 r( p-|f|)
z 2 + 10
10
-0 .1 r
é R xp
R yp
R zp ù
+
+
ê
3
3
3ú
ë|R xp| |R yp| |R zp| û
rL
2 pe o
17. (B) EP =
0 .05
-¥ p 10
Electromagnetics
ù
( -0.1 - 1)
( -0.101) ú ´
(0.1) 2
û0
Ep = 100 ´ 10 -9 ´ 2 ´ 9 ´ 10 -9
¥
é 2u y - u z -3u x - u z -3ux + 2u y ù
+
+
E p = 100 ´ 10 -9 ´ 2 ´ 9 ´ 10 -9 ´ ê
ú
5
10
13
ë
û
2p
( p - f) dz
2
+ 10
0 z
ò 2ò
-¥
= -0.96 u x + u y - 0.54 u z kV m
¥
p2 dz
-¥ z + 10
Q = 5 ´ 26.4 ò
18. (C) At y = 4, E =
2
¥
z ù
é 1
= 5 ´ 26.4 ´ p2 ê
tan -1
= 129
. mC
10 úû -¥
ë 10
dF = dqE = r L dzE
r 2L dzu y
1
F=
ò
2 pe o
0
13. (A)
f = 10
1 .1 p 25° 5
f
ò ò ò ( r - 4)( r - 5) sin q × sin 2 r
0 .9p 0 4
25°
5
1 .1 p
é r 5 9 r 4 20 r 3 ù é 1
1
fù
ù
é
= 10 ê +
ú ê2 q - 4 sin 2 q ú ´ ê-2 cos 2 ú
5
4
3
ë
û 0 .9p
ë
û
ë
û4
0
= 10 [ -3.39 ][0.0266 ][0.626 ] = 0.57 C
( 6) = 18.75 u y mN
r s dS R - R¢
4 pe o |R - R¢|3
ò ò
19. (A) E =
sin q drdqdf
2
rL
uy ,
2 pe o
where R = 3u x and R¢ = yu y + zu z ,
¥ 2
3u x - yu y - zuz
r
dydz
EPA = L ò ò
4 pe o -¥ -2 (9 + y 2 + z 2 ) 3 2
Due to odd function
Rp
r
,
14. (D) Ep = L
2 pe o |R p|2
EPA =
R p = (1, 2, 3) - (1, -2, 5) = (0, 4, -2)
So there is only x component.
|R |
p
2
= 20,
5 ´ 10 -9
Ep =
2 pe o
é4u y - 2u z ù
ê
ú = 18 u y - 9 u z
20
ë
û
15. (C) With z = 0, the general field is
Ez = 0 =
r L é( y + 2) u y - 5 u z ù
2 pe o êë ( y + 2) 2 + 25 úû
we require |E2 | = |2E y|
So 2 ( y + 2) = 5
E=
Þ
y=
1
2
6.25 é2.5 u y - 5 u z ù
= 9 u y - 18 u z
2 pe o êë 6.25 + 25 úû
rL
4 pe o
¥
2
òò
-¥ -2
3u x dydz
(9 + y 2 + z 2 ) 3 2
20. (D) There will be z component of E only
R = zu z , R¢ = ru r ,
Ez , Pa =
rs
4 pe o
2 p 0 .2
R - R¢ = zu z - ru z
zrdrdf
2
+ r 2)3 2
ò ò (z
0
0
0 .2
ù
ù
2pr s z é
rzé 1
1
1
=
ú = s ê 2 ê 2
ú
2
2
eo ê z
4 pe o ê z + r ú
z + 0.04 úû
û0
ë
ë
Ez =
rs
eo
é
z
ê1 2
z + 0.04
êë
ù
ú, at z = 0.5, Ez = 20.2 V m
úû
21. (A) Since charge sheet are infinite, the field
magnitude associated with each one will be r s 2 e o ,
which is position independent. The field direction will
r
16. (C) Ep = L
2 pe o
é R+ Q
R- Q ù
ê
2
2ú
ë|R + Q| |R - Q| û
R+ Q = (6, 0, 6) - (0, 3, 6) = (6, -3, 0)
R- Q = (6, 0, 6) - (0, -3, 6) = (6, 3, 0)
Page
476
depend on which side of a given sheet one is positioned.
é10 ´ 10 -9
40 ´ 10 -9
50 ´ 10 -9 ù
EA = ê
ux uy uzú
2 eo
2 eo
ë 2 eo
û
= 0.56 u x + 2.23u y - 2.8 u z
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Electrostatics
22. (A)
dy E y -15 x 2 y -3 y
=
=
=
dx Ex
5 x3
x
dy
3dx
=y
3
Þ
At P, 2 =
C1
43
Þ
Þ
a a
ln y = -3 ln x ln C y =
C1 = 128
Chap 8.2
Þ
y =
C1
,
x3
Front
a a
+
128
x3
ò ò -x
a a
= a4 + 0 -
¶y
= 20 x 4 y 3z 4
cube
=(3.05
1.05
2.05)
and
Volume
Ñ V = (0.1) 3 = 0.001
31. (C) F = (Ñ × D ) Dv
h2 = 2 4 2 - 1 = 6.93
=
QT = 2 ´ 7.75 ´ 40n + 2 ´ 6.93 ´ 50 + 20n =1.33 mC
ò ò
of
¶D y
f = 20 ( 305
. ) 4 (105
. ) 3 (2.05) 3 (2.05) 4 0.001 =35.4
24. (C) h1 = 2 4 2 - 1 = 7.75,
0 .05 90 °
Top
a
a
+
+ 0 + 6a5 = a4 + 6a5
3
3
30. (C) Ñ × D =
Center
Botoom
Right
4
arrangement of line charges, whose field will all cancel
D = -19.1u y + 25.5 u z pC m 2
a a
dxdz + ò ò x 2 dxdz + ò ò -6(0) 3 dxdy + ò ò 6a 3dxdy
0 0
0 0
0 0
0 0
1
4
4244
3 1
4243 1
442443 1
4
4244
3
4
at that point. Thus D arises from the point charge alone
10 ´ 10 -9 ( -3u y + 4 u z )
,
D=
( 32 + 4 2 )1 .5
4p
Back
a a
2
Left
23. (A) This point lies in the center of a symmetric
25. (D) Q =
a a
29. (C) Q = ò D × n dS = ò ò 2 aydydz + ò ò -2(0) dydz
S
0 0
0 0
1
4 24 3 1
44244
3
¶( x) æ 4
3ö
. nC
ç p(0.003) ÷ = 1131
¶x è 3
ø
é10 y
æ 2 + 10 x 2 y öù
=8.96
32. (B) r v = Ñ × D = ê
÷÷ú
+ 0 + çç
z3
øû ( -2 , 3, 5)
è
ë z
5 e -20 z (0.08) dfdz nC
0 .01 30
0 .05
æp pö
æ 1 ö -20 z
= ç - ÷(5)(0.08)ç ÷e
è2 6 ø
è 20 ø
0 .01
33. (C) Ñ × D =
= 9.45 ´ 10 -3 nC = 9.45 pC
ò Ñ × D dv = 6 ´ (0.4)
26. (A) Out of the 6 surface only 2 will contribute to the
net outward flux. The y component of D will penetrate
the surface y = 0 and y = z and net flux will be zero. At
x = 0 plane Dx = 0 and at z = 0 plane Dz = 0.
This leaves the 2 remaining surfaces at x = 2 and z = 5.
The net outward flux become
5 3
f = ò ò D x = 2 × u x dydz +
0 0
3
3
0
0
2p p
ò òD
x =2
× u z dzdy
0 0
é -r e
Q = 8 pê
êë 1000
2 -1000 r
-1000 r
0
2 e -1000 r ( -1000 r - 1)
1000(1000) 2
0
Q = 4 ´ 10 -9nC,
Dr =
4.0 ´ 10 -9
Q
= 0.32 C m 2
=
2
4 pr
4 p ´ (0.001) 2
28. (C) 4 pr Dr = 4 p(2 ) 20 ´ 10
2
Dr =
80
nC m 2
r2
2
= -5( 4 u x - 3u y + 5 u z ) ×
=-
10
3
( u x + u y + u z )(2)
3
( 4 - 3 + 5) = -34.64 J
æ
( u x - u y + u z )(2 ´ 10 -3) ö
÷
= -( 60 ´ 10 -6 )ç 100 u r ×
ç
÷
3
è
ø
r 2 sin q drdqdf
0 .001
+
34. (C) dW = - qE × dL
(2, 1, 4) - (7, 2, 3) = (1, -1, 1)
ux - u y + uz
, dW = - qE × dL
u PQ =
3
0
0 .001
= 0.38
V
35. (B) The vector in this direction is
0 0
0 .001
ò ò ò 2e
3
3 2
= 5 ò 4(2) ydy + 2 ò 4(5) ydy = 360 C.
27. (D) Q =
1 d 2
( r ´ 2 r) = 6 ,
r 2 dr
-9
2
Cm ,
ù
ú
úû
= - 12 ´
10 -6
3
( u r × u x - u r × u y)
æ2 ö
At r , f = tan -1 ç ÷ = 63.4 °
è1ø
u r × u x = cos 63.4 ° = 0.447,
u r × u y = sin 63.4 ° = 0.894
dW = 31
. mJ
36. (A) W = - q ò E × dL
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UNIT 8
Electromagnetics
= -10 ò (2 xu x - 3 y 2 u y + 4 u z ) ( dxu x + dyu y + dzu z )
3
1
-1
0
0
0
é
ù
18 z
-ê6 xy 2 z 2 + 2
uz
2
2 ú
x
+
y
+
z
2
3
ë
û
= -10 ò 2 xdx + 10 ò 3 y 2 dy - 10 ò 4 dz = -40 J
. u z V m,
Ep = 7.1u x + 22.8 u y - 711
|E|= 75 V m
37. (D) rA = 2 cm, rB = 3 cm
V ( r) =
43. (D) E = -ÑV = -
Q
4 pa 2r s ( 6 ´ 10 -3) 2 ´ 30 ´ 10 -9
=
=
4 pe o r
4 pe o r
8.85 ´ 10 -12 ´ r
= -42 r -0 .4 u r V m,
0.122
0.122 0.122
, V AB = V A - VB =
V ( r) =
=
= 2.03
r
0.02
0.03
38. (D) W = -
Q1Q2
4 pe o
rB
ò
rA
dr Q1Q2
=
r 2 4 pe o
D = e oE = -42 r -0 .4 e o u r C m 2
1 d 2
1 d
e
rv = Ñ × D = 2
( r Dr ) = 2
( -42 e o r 1 .6 )= - 67.2 1o.4
r dr
r dr
r
é1
1ù
êr - r ú
ë B
Aû
At r = 0.6 m,
rv = -
rA = rB , W = 0
39. (B) VP ( s) =
Q
,
4pe o r
r L dr
r
+ C1 = - L ln r + C1
2 pe or
2 pe o
Vs ( z) = - ò
rs
rz
dz + C2 = s + C2
2 eo
2 eo
D = e oE = -80 e o (2 r cos q u r - r sin q u q)
é1 2 2
ù
1
¶
rv = Ñ × D = ê 2
( r Dr ) +
( Dr sin q) ú
r sin q ¶q
ë r ¶r
û
Here r , r , z are the scalar distance from the charge.
r = 2 2 + 5 2 = 29 , r = (5 - 4) 2 = 1 , z = 5
By putting these value. C = -193
. ´ 10 3
2
r = 12 ` + 12 = 2 , z = 3,
40. (D) Vr =
¥
-¥
2
é1 ¶
ù
1 ¶
r v = eo ê
(rEr) +
Ef ú
r ¶f û
ër ¶r
é (50 + 150 sin f)
150 sin f ù
50e o
= e o ê+
=C m2
ú
r
r
r
ë
û
1 2p 2
Q =ò
R = zu z , R¢ = ru r, dS = rdrdf,
Vr =
ò ò
0
0 .01
(5 ´ 10 -9)r 2 drdf
4 pe o r 2 + z 2
0
ù
z2
5 ´ 10 -9 ér
2
2
r
+
z
ln (r + r 2 + z 2 ) ú
ê
2 eo
2
ë2
û 0 .01
At z = 0.02, Vp = 0.081 V
0 0
50 e o
rdrdfdz = - 2 p(50) e o 2 = -5.56 nC
r
46. (A) V =
42. (C) E = -ÑV
é
ù
6x
= - ê2 y 2 z 2 + 2
ux
2
2ú
x + 2 y + 3z û
ë
é
ù
12 y
- ê4 xyz 3 + 2
uy
2
2 ú
x + 2 y + 3z û
ë
Qd cos q 100 cos q
,
=
4 pe o r 2
r2
1 ¶V
æ ¶V
ö
E = -ÑV = -ç
ur +
uq ÷
¶
r
r
¶q
è
ø
=
Page
478
òò
,
0 .03
Vp =
1 ¶V
¶V
ur uf
¶r
r ¶f
D = e o E, r v = Ñ × D = e oÑ × E
¥
r s dS
41. (A) Vr = òò
,
4 pe o|R - R¢|
2 p 0 .03
r V = -320 e o cos q = -2.45 nC m 3
= -(50 + 150 sin f) u r - (150 cos f) u f ,
VB = 198
. kV
6 ´ 10 -9 dz
r L dz
= ò
= 108 V
2
32
oR
-¥ 4 pe o ( z + 1)
ò 4 pe
é1
12 sin q cos q ù
r v = -80 e o ê 2 ´ 3r 2 cos q ú
r sin q
ër
û
45. (C) E = -ÑV = -
At point N, r = (2 - 1) + 2 + z = 14
2
¶V
1 ¶V
ur uq
¶r
r ¶q
= -160 r cos q u r + 80 r sin q u q V m
Q
r
r
- L ln r - s z + C
4 pe o r 2 pe o
2 pe o
V =
67.2 ´ 8.85 ´ 10 -12
= -1.22 nC m 3
(0.6)1 .4
44. (A) E = -ÑV = -
Vl (r) = - ò
dV
u r = -(0.6)(70) r -0 .4 u r
dr
100
(2 cos q u r + sin q u q),
r3
|E| = 100( 4 cos 2 q + sin 2 q)1 2
47. (A) E =
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= 100 ´
5
= 158.11 V m
2
2 ´ 10 -9
[2 cos q u r + sin q u q ],
4 pe o r 3
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Electrostatics
55. (A) I = òò J × n z = 0 .2 dS
y = z lies at q = 45°
2 ´ 10 -9 1
= 10 -3
Eq =
4 pe o r 3 2
S
(required)
=
r 3 = 12.73 ´ 10 3, r = 23.35
0
56. (B) So Eal = Est =
Solving
4 pe o (1 + 1 + 8 2 )1 .5
é 1
ù
1
1
49. (A) E = -ÑV = 40 ê 2 u x +
uy +
uzú
2
2
x
yz
xy
z
xyz
ë
û
òòò E × Edv
2
2
2
ò ò
1
1
1
Þ
J al =
sac
J st
sst
J st = 3.2 ´ 10 5 A m 2
57. (B) J =
We = 800 e o ò
J al J st
=
sal sst
I = p(2 ´ 10 -3) J st + p[( 4 ´ 10 -3) 2 - (2 ´ 10 -3) 2 ]J al = 80
( 3u x - 5 u y + 10 u z ) × ( u x + u y + 8 u z ) ´ 10 -9
=1.31 V
eo
2
ò ò
0 .4
40
40
rdrdf =
log[r 2 + (0.1) 2 ] (2 p)
2
2
r + (0.1)
2
0
= 40 p log 17 = 356 A
where R - R¢ = Q - P = (1, 1, 8)
We =
2 p 0 .4
0
P × (R - R¢)
48. (A) V =
4 pe o|R - R¢|3
So VP =
Chap 8.2
E=
4
u r A m2,
2prl
J
4
12.73
=
=
ur V m
s 2 prls
rl
3
5
12.73
12.73 5
6.51
V
u r × u rdr =
ln
=
r
l
l
3
l
3
V = - ò E × dL = ò
é 1
1
1 ù
ê x 4 y 2 z 2 + x 2 y 4 z 2 + x 2 y 2 z 4 ú dxdydz
ë
û
5
V 6.51 1.63
R=
=
=
W
I
4l
r
= 1548 pJ
æ ¶V
ö
1 ¶V
¶V
58. (D) E = -ÑV = -çç
ur +
uf +
u z ÷÷
r ¶f
¶z
è ¶r
ø
(r + 1) 2
= -z 2 cos f u r +
z sin f u f - 2(r + 1) z cos f u z
r
1 4
50. (A) W = å qn Vn
2 n =1
V1 = V21 + V31 + V41 =
q é 1
1
1
ù
+
+
ê
4 pe o ë0.04 0.04 0.04 2 úû
E = -1.82 u f + 14.5 u f - 2.67 u z V m
E ×E
r s = eo E × n s = eo
|E|
V1 = V2 = V3 = V4
W =
1
2 ´ (1.2 ´ 10 -9) 2
( 4) q1 V1 =
2
4 pe o (0.04)
51. (B) I = òò J × n
=
ò ò 10
5
r s = e o 1.82 2 + 14.5 2 + 2.67 2 = 1315
. pC m 2
2 1
y =1
S
2 1
1 ù
é
. mJ
êë2 + 2 úû = 175
dS = ò ò J × u y
0 0
y =1
dxdz
59. (C) V =
cos (2 x) 2 -2 y dxdz = 12307 kA = 12.3 MA
p
p
sin
3
2 = 2.5 V
23
40 cos
So the equation of the surface is
40 cos q sin f
= 2.5, 16 cos q sin f = r 3
r2
0 0
52. (C) I = òò J × n dS
S
=
60. (A) E = -ÑV, D = eE = - e oÑV
2 p 0 .3p
ò
0
800 sin q
(0.80) 2 sin q dqdf =154.8 A
ò
2
(
0
.
80
)
+
4
0 .1 p
53. (D) F = ma = qE,
a=
qE ( -1.602 ´ 10 -19)( -4 ´ 106 )
=
u z = 7.0 ´ 1017 u z m s 2 ,
m
9.11 ´ 10 -3
v = at = 7.0 ´ 10 tu z m s
17
54. (D)
=
¶r V
1 ¶
¶J z
= -Ñ × J =
(rJ r) +
¶t
r ¶r
¶z
1 ¶
¶ æ -20 ö
(25) +
ç
÷ =0
r ¶r
¶z çè r 2 + 0.01 ÷ø
é
ù
x
200 x
¶
ux + 2
u z ú C m2
= -e o ê200 z
2
(
)
x
x
+
4
x
+
4
¶
ë
û
200 e o x
D (z= 0 ) = - 2
u z C m2,
x +4
r s = D × u z z= 0 =
Q=
0
2
-3
0
ò ò
- 200 e o x
C m2,
x2 + 4
2
1
- 200 e o x
dxdy = -( 3)(200) e o ln[ x 2 + 4 ]
2
x2 + 4
0
= - 300 ln 2 = - 1.84 nC
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UNIT 8
61. (B) E = -ÑV ,
-3 xy 2 z 3 satisfy this equation.
62. (D) The plane can be replaced by -60 nC at Q (2, 5,
-6).
Electromagnetics
Since E is normal to the surface
e
2 1000
400
En 2 = r1 En 1 = ´
ur = 2 ur V m
r2
r
er 2
5
R = (5, 3, 1) - (2, 4, 6) = (3, -1, -5)
68. (D) D n 1 = D n 2
R¢ = (5, 3, 1)-(2, 4, -6) = (3, -1, 7),
D n 1 = 80 ´ 3e o = D n 2 = 5 e oEn 1 ,
|R |= 35 , |R¢|= 59
E2 = 48 u x - 60 u y - 40 u z
VP =
q
q
q é 1
1 ù
=
ê
4 pe o R 4 pe o R¢ 4 pe o ë 35
59 úû
-9
Vp = 60 ´ 10 ´ 9 ´ 10
-9
rL
2p eo
0 = Ñ × ( - f ( x)ÑV )
é dF ¶V
¶2V
¶2V
¶2V ù
=-ê
+ f ( x) 2 + f ( x) 2 + f ( x) 2 ú
¶x
¶y
¶z û
ë dx ¶x
é Final Distance from the charge ù
rL
ln ê
2 pe o ëInitial Distance from the charge úû
é 1
12 + (2 - 1) 2
12 + (2 - (-2)) 2
12 + 32 ù
- ln
- ln
êln + ln
ú
1
1
1
êë 2
úû
= 2.40 kV
64. (C) P = D - e oE = D -
Þ
. e o = 2.5 e o En 2
= 2 ´ 15
Et1 = Et 2 ,
Þ
D D
= ( e r - 1)
er er
e1E N1 = e 2E N 2
En 1 = - 817
.
1
6
6
f ( x1 ) = -4 e 2 x + 3 x 2 + C, f (0, 1) = -4 + C2
V (0, 0) = 0 = 4 + f (0)
f ( x) = -4 e
C2 = 0. f (0) = -4
+ 3x ,
2x
- 4 e2 x + 3 x 2 - 3 y 2 = 3 ( x 2 - y 2 )
[u x - u y + 2u z ]
*********
Et1 = Et 2 and D n 1 = D n 2
e r1 e oEn 1 = e r 2 e oEn 2
e
1
1
= r1 En 1 = E n 1 , E2 = Et 2 + En 1
er 2
4
4
= 133.3 u x + 166.7 u y + 16.67 u z - 8.33 u x + 8.33 u y - 16.67 u z
= 125 u x + 175 u y V m
67. (B) D n 1 = D n 2 ,
Page
480
Þ
2
[100 - 200 - 100 ] = -817
. Vm
Et1 = E1 - En 1 = 133.3u x + 166.7 u z + 16.67 u z
En 2
2x
V ( x, y) = 4 e
= -33.33u x + 33.33u y - 66.67 u z V m
Þ
Ñ 2 V = 0,
Integrating again
66. (D) The unit vector that is normal to the surface is
u x - u y + 2u z
ÑF
,
uN =
=
|ÑF|
6
1
Þ
It follow that C1 = 0
En 2 = 12
.
E2 = 1.2 u x - 3u y + 1u z
En 1 = E1 × u N =
70. (A) r v = 0
¶2 f
- 6 =0
¶x 2
¶f
¶f
= -16 e 2 x + 6 Þ
= - 8 e 2 x + 6 x + C1
¶x
¶x
¶V
¶f
Ex =
= 8 e2 x +
¶x
¶x
¶f
¶f
Ex (0) = 8 +
=0 Þ
= -8
¶x x = 0
¶x x = 0
P = 12
. u x - 2.3u y + 2.9 u z nC m 2
65. (C) D n 1 = D n 2
é dF ¶V
ù
=-ê
+ f ( x)Ñ 2 V ú
ë dx ¶x
û
1 dF ¶V
Þ
Ñ 2V = f ( x) dx ¶x
Ñ 2 V = 16 e2 x +
(2.4 - 1) ´ 10 -9
2.4
P = (2 u x - 4 u y + 5 u z )
e r1 e oEn 1 = e r 2 e oEn 2
En1 = 48
Ñ × D = r v = 0 = Ñ × ( - f ( x)ÑV )
1 ù
é 1
êë 35 - 59 úû = 21 V
V0 = 0,
Vp = -
D = eE
69. (A) D = eE = - f ( x)ÑV ,
63. (A) Using method of images
Vp - V0 = -
and Et1 = Et 2 ,
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UNIT 8
Electromagnetics
Statement for Q.15–16:
Statement for Q.9–11:
An infinite filament on the z-axis carries 10 mA in
In the cylindrical region
the u z direction. Three uniform cylindrical current
2 r
+
for r £ 0.6
r 2
3
for r > 0.6
Hf =
r
Hf =
sheets are also present at 400 mA m at r = 1 cm, -250
mA m at r = 2 cm and 300 mA m at r = 3 cm.
9. The magnetic field H f at r = 0.5 cm is
(A) 0.32 A m
(B) 0.64 A m
15. The current density J for r < 0.6 mm is
(C) 1.36 mA m
(D) 0
(A) 2u z A m
(B) -2u z A m
(C) u z A m
(D) 0
10. The magnetic field H f at r = 15
. cm is
(A) 1.63 A m
(B) 0.37 A m
16. The current density J for r > 0.6 mm is
(C) 2.64 A m
(D) 0
(A) 2u z A m
(B) -3u z A m
(C) 3u z A m
(D) 0
11. The magnetic field H f at r = 3.5 cm is
(A) 0.14 A m
(C) 0.27 A m
(B) 0.56 A m
17.
(D) 0.96 A m
v = ( 3u x + 12 u y - 4 u z ) ´ 10
An
electron
5
with
velocity
m s experiences no net
forces at a point in a magnetic field B = u x + 2 u y + 3u z
Statement for Q.12–14:
mWb m 2 . The electric field E at that point is
In the fig. P8.3.12–14 The region 0 £ z £ 2 is filled
(A) -4.4 u x + 1.3u y + 0.6 u z kV m
with an infinite slab of magnetic material (m r = 2.5). The
(B) 4.4 u x - 1.3u y - 0.6 u z kV m
surface of the slab at z = 0 and z = 2, respectively, carry
(C) -4.4 u x + 1.3u y + 0.6 u z kV m
surface current 30u x A m and -40u x as shown in fig.
(D) 4.4 u x - 1.3u y - 0.6 u z kV m
z
18. A point charge of 2 ´ 10 -16 C and 5 ´ 10 -26 kg is
mo
z=2
-40ux A/m
moving in the combined fields B = - 3u x + 2 u y - u z mT
and E = 100 u x - 200 u y + 300 u z V m. If the charge
velocity at t = 0 is v(0) = (2 u x - 3u y - 4 u z ) 10 5 m s, the
mr = 2.5
mo
z=0
30ux A/m
Fig. P8.3.12–14
12. In the region 0 < z < 2 the H is
(A) -35u y A m
(B) 35u y A m
(C) -5u y A m
(D) 5u y A m
x
acceleration of charge at t = 0 is
(A) 600[ 3u x + 2 u y - 3u z ]10 9 m s 2
(B) 400[ 6 u x + 6 u y - 3u z ]10 9 m s 2
(C) 400[ 6 u x - 6 u y + 3u z ]10 9 m s 2
(D) 800[ 6 u x + 6 u y - u z ]10 9 m s 2
19. An electron is moving at velocity v = 4.5 ´ 10 7 u y
13. In the region z < 0 the H is
(A) 5u y A m
(B) -5u y A m
(C) 10u y A m
(D) -10u y A m
14. In the region z > 2 the H is
(A) 5u y A m
(B) -5u y A m
(C) 35u y A m
(D) -35u y A m
Page
482
m s along the negative y-axis. At the origin, it
encounters the uniform magnetic field B = 2 .5 u z mT,
and remains in it up to y = 2.5 cm. If we assume that the
electron remains on the y –axis while it is in the
magnetic field, at y = 50 cm the x and z coordinate are
respectively
(A) 1.23 m, 0.23 m
(B) -1.23 m, -0.23 m
(C) -11.7 cm, 0
(D) 11.7 cm, 0
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Magnetostatics
Statement for Q.20–22:
Chap 8.3
25. If the current filament is located at y = 0.5, z = 0,
A rectangular loop of wire in free space joins points
and u L = u x , then F is
A(1, 0, 1) to B(3, 0, 1) to C(3, 0, 4) to D(1, 0, 4) to A. The
(A) 35 .2 u y nN m
(B) 68.3u x nN m
wire carries a current of 6 mA flowing in the u z
(C) 105.6 u z nN m
(D) 0
direction from B to C. A filamentary current of 15 A
flows along the entire z, axis in the u z directions.
20. The force on side BC is
(A) -18u x nN
(B) 18u x nN
(C) 3.6 u x nN
(D) -3.6 u x nN
21. The force on side AB is
(A) 23.4 u z mN
(B) 16.4 u z mN
(C) 19.8 u z nN
(D) 26.3u z nN
(B) -36u x nN
(C) 54u x nN
(D) -54u x nN
fig.
P8.3.23.
The
magnetic
B = 6 xu x - 9 yu y + 3zu z Wb m 2 .
flux
The
(C) 4u x mN m
(D) -4u x mN m
27. The force exerted on the filament by the current
density
total
(B) -0.4 u x N m
A conducting current strip carrying K = 6 u z A m
lies in the x = 0 plane between y = 0.5 and y = 15
. m.
There is also a current filament of I = 5 A in the u z
direction on the z –axis.
23. Consider the rectangular loop on z = 0 plane shown
in
(A) 0.4 u x N m
Statement for Q.27–28:
22. The total force on the loop is
(A) 36u x nN
26. Two infinitely long parallel filaments each carry 100
A in the u z direction. If the filaments lie in the plane
y = 0 at x = 0 and x = 5 mm, the force on the filament
passing through the origin is
is
force
experienced by the rectangular loop is
strip is
(A) 12.2 u y mN m
(B) 6.6 u y mN m
(C) -12.2 u y mN m
(D) -6.6 u y mN m
28. The force exerted on the strip by the filament is
y
2
5A
(A) -6.6 u y mN m
(B) 6.6 u y mN m
(C) 2.4 u x mN m
(D) -2.4 u x mN m
Statement for Q.29–32:
1
In a certain material for which m r = 6.5,
0
1
2
3
x
H = 10 u x + 25 u y - 40 u z A m
Fig. P8.3.23
(A) 30u z N
(B) -30u z N
(C) 36u z N
(D) -36u z N
29. The magnetic susceptibility c m of the material is
(A) 5.5
(C) 7.5
(B) 6.5
(D) None of the above
30. The magnetic flux density B is
Statement for Q.24–25:
Three uniform current sheets are located in free
(A) 82 u x + 204 u y - 327 u z mWb m 2
space as follows: 8u z A m at y = 0, -4u z A m at y = 1
(B) 82 u x + 204 u y - 327 u z mA m
and -4u z A m at y = -1. Let F be the vector force per
(C) 82 u x + 204 u y - 327 u z mT
meter length exerted on a current filament carrying 7
(D) 82 u x + 204 u y - 327 u z mA m
mA in the u L direction.
31. The magnetization M is
24. If the current filament is located at x = 0, y = 0.5
(A) 75 u x + 187.5 u y - 300 u z A m 2
and u L = u z , then F is
(B) 75 u x + 187.5 u y - 300 u z A m 2
(A) 35 .2 u y nN m
(B) -35 .2 u y nN m
(C) 55 u x + 137.5 u y - 220 u z A m 2
(C) 105.6 u y nN m
(D) 0
(D) 55 u x + 137.5 u y - 220 u z A m 2
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UNIT 8
32. The magnetic energy density is
(A) 19 mJ m
2
If H increases from 0 to 210 A m, the energy
(B) 9.5 mJ m
(C) 16.3 mJ m
2
Electromagnetics
stored per unit volume in the alloy is
2
(D) 32.6 mJ m 2
(A) 6.2 MJ m 3
(B) 1.3 MJ m 3
(C) 2.3 kJ m 3
(D) 2.9 kJ m 3
Statement for Q.33–34:
For a given material magnetic susceptibility
c m = 31
. and within which B = 0.4 yu z T.
(B) 151.6 yu z kA m
(C) 102.7 yu z kA m
(D) 77.6 yu z kA m
cube of size a, the magnetization volume current
density is
12
(A)
uz
a
33. The magnetic field H is
(A) 986.8 yu z kA m
40. If magnetization is given by H = 6a ( - yu x + xu y) in a
(C)
34. The magnetization M is
6
uz
a
(B)
6
( x - y)
a
(D)
3
( x - y)
a
41. The point P(2, 3, 1) lies on the planner boundary
(A) 241yu z kA m
(B) 318.2 yu z kA m
separating region 1 from region 2. The unit vector
(C) 163yu z kA m
(D) None of the above
u N12 = 0.6 u x + 0.48 u y + 0.64 u z is directed from region 1
to
35. In a material the magnetic field intensity is
H = 1200 A m when B = 2 Wb m 2 . When H is reduced to
400
A m,
B = 1.4
Wb m 2 .
The
change
in
the
magnetization M is
(A) 164 kA m
(B) 326 kA m
(C) 476 kA m
(D) 238 kA m
region
2.
If
m r1 = 2,
H1 = 100 u x - 300 u y + 200 u z A m, then H 2 is
(A) 40.3u x + 48.3u y - 178.9 u z A m
(B) 80.2 u x - 315.8 u y + 178.9 u z A m
(D) 80.2 u x + 48.3u y + 178.9 u z A m
each atom has a dipole moment of 2.6 ´ 10 -30 u y A × m 2 .
The H in material is (m r = 4.2)
(A) 2.94 u y A m
(B) 0.22 u y A m
(C) 0.17 u y A m
(D) 2.24 u y A m
42. The plane separates air ( z > 0, m r = 1) from iron
( z £ 0, m r = 20). In air magnetic field intensity is
H = 10 u x + 15 u y - 3u z A m. The magnetic flux density
in iron is
(A) 5.02 u x + 7.5 u y - 0.076 u z mWb m 2
(B) 12.6 u x + 18.9 u y - 75.4 u z mWb m 2
37. In a material magnetic flux density is 0.02 Wb m 2
(C) 251u x + 377 u y - 377
. u z mWb m 2
and the magnetic susceptibility is 0.003. The magnitude
(D) 251u x + 377 u y - 1508 u z mWb m 2
of the magnetization is
(B) 23.4 A m
(C) 16.3 A m
(D) 8.4 A m
43. The plane 2 x + 3 y - 4 z = 1 separates two regions.
Let m r1 = 2 in region 1 defined by 2 x + 3 y - 4 z > 1, while
m r 2 = 5 in region 2 where 2 x + 3 y - 4 z < 1. In region
38. A uniform field H = - 600 u y A m exist in free space.
H1 = 50 u x - 30 u y + 20 u z A m. In region 2, H 2 will be
The total energy stored in spherical region 1 cm in
(A) 63.4 u x + 4318
. u y - 19.4 u z A m
radius centered at the origin in free space is
(B) 52.9 u x - 25.66 u y + 14.2 u z A m
(A) 0.226 J m
3
(B) 1.452 J m
(C) 1.68 J m 3
(C) 48.6 u x - 16.4 u y - 46.3u z A m
3
(D) None of the above
(D) 0.84 J m 3
39. The magnetization curve for an iron alloy is
approximately given by
B=
Page
484
1
H + H 2 mWb m 2
3
and
(C) 40.3u x - 315.8 u y - 178.9 u z A m
36. A particular material has 2.7 ´ 10 29 atoms m 3 and
(A) 47.6 A m
m r2 = 8
********
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Magnetostatics
SOLUTIONS
1. (C) H =
=
I
4p
IdL ´ u R
4 pR 2
-¥
ò
[2 + ( 3 - y) ]
2
0
2 32
R
R = (1 + x) + 32 + 2 2 = x 2 + 2 x + 14
2
H=
-u ydy[2 u x + ( 3 - y) u y ]
¥
-(1 + x) u x + 3u y + 2 u z
uR =
¥
ò
Chap 8.3
4 dxu x ´ [ -(1 + x) u x + 3u y + 2 u z ]
¥
ò
4 p ( x 2 + 2 x + 14) 3 2
-¥
=
(12 u z - 8 u y) dx
¥
ò 4 p( x
=
+ 2 x + 14)
2
-¥
¥
3/2
=
2 u z dy
I
=
ò
2
4 p 0 [2 + ( 3 - y) 2 ]3 2
= 0.147 u z - 0.098 u y A m
3 - y = 2 tan q, -dy = 2 sec 2 q,
5. (B) H =
q1 = 56.31°, q 2 = -90 °
H=
I
4p
56 .31 °
ò
-90 °
2 u z dq
2 sec q
=
I
u z [sin q ] -5690.31° °= 145.8 u z mA m
=
4p
2. (A) H = H y + H z , H z =
Iz
uf
2pr
r = ( -3) 2 + ( 4) 2 = 5
3u y + 4 u x
=
5
5
24 ( 4 u x + 3u y)
Hz =
= 0.611u x + 0.458 u y mA m
2 p(5)
5
uf =
Hy =
- u z ´ ( -3u x + 4 u y)
Iy
2pr
2
I
uf 2 pr
( -3u x + 5 u z )
34
=
3u z - 5 u x
34
( -5 u x + 3u z )
12
2 p 34
-a
3. (A) H =
I
uf 2 pr
4
ò
-4
Idzu z ´ (ru r - zu z )
4 p(r 2 + z 2 ) 3 2
I æç
a
12
2 pr ç
(r + a 2 )
è
I
At r = 1, H =
2pr
Þ
a
1Þ
1+ a
1
a=
3
=
2p
ò
-a
Iau f
2 pr(r 2 + z 2 ) 3 2
ö
÷u A m
÷ f
ø
1
2
= 0.577 m
IdL ´ u R
4 pR 2
Idfu f ´ ( - ur )
I
uz A m
=
2a
4 pa
K ´ u R dxdy
4 pR 2
¥
4 u x ´ ( - xu x - yu y - 3u z ) dx dy
7. (D) H = òò
2
=
ò ò
4 p( x 2 + y 2 + 9) 3 2
-2 -¥
=
ö
I æç
4
÷u
1f
2 pr ç
(r 2 + 16) ÷ø
è
=-
IdL ´ u R
, IdL = 4 dxu x
4 pR 2
=
=
I = 3 A, a = 0.5 m, H = 3u z A m
I
I
8
uf uf
2 pr
4 p r (r 2 + 16)
4. (B) H = ò
2
a
6. (A) H = ò
=
H = 0.1u f A m
rIdzu f
4 p(r 2 + z 2 ) 3 2
=
4
f = 60 °, I = 3p,
-a
Idzu z ´ (ru r - zu z )
4 p(r 2 + z 2 ) 3 2
rIdzu f
rIu f
z
ò- a 4 p(r 2 + z 2 ) 3 2 = 4 p r 2 (r 2 + z 2 ) 3 2
rdz
I
I
uf uf
ò
2
2 pr
4 p -4 4 p(r + z 2 ) 3 2
At r = - 3,
a
ò
4 p13
a
0
34
H = H y + H z = 0.331u x + 0.458 u y + 0.168 u z mA m
=
ò
2
= - 0.281u x + 0.168 u z mA m
=
a
u f , r = ( -3) + (5) = 34
uf = u y ´
Hy =
I
uf 2 pr
2(12 u z - 8 u y)
2
¥
ò ò
4 ( - yu z - 3u y) dx dy
-2 -¥
2
¥
4 p( x 2 + y 2 + 9) 3 2
12 u ydx dy
ò ò 4 p( x
-2 -¥
3
uy
p
2
ò
-2
2
+ y 2 + 9) 3 2
2
dy
y2 + 9
2
-
6
1
æ yö
u y tan -1 ç ÷ = -0.75 u y A/m
p
3
è 3 ø -2
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UNIT 8
IdL ´ u R
4 pR 2
-Idzu z ´ ( -zu z + u y)
8. (D) H = ò
¥
=ò
0
¥
=ò
0
4 p(1 + z )
2 32
Idzu x
+
4 p(1 + z 2 ) 3 2
æ
I ç zu x
=
4 p çç (1 + z 2 )
è
=
0
¥
+
Idxu x ´ ( - xu x + u y)
¥
ò
4 p(1 + x )
2 32
0
¥
ò
0
Idxu z
4 p(1 + x 2 ) 3 2
uy
2
12
uzù
3 ú ´ 10 5 ´ 10 -3
ú
-4 úû
= [ u x ( -8 - 36) - u y( -4 - 9) + u z (12 - 6)] ´ 10 2 V m
= -4.4 u x + 1.3u y + 0.6 u z kV m
18. (D) v(0) ´ B = (2 u x - 3u y - 4 u z )10 5
ö
÷
+
÷
(1 + x 2 ) ¥ ÷
ø
xu z
éu x
ê1
ê
êë 3
Electromagnetics
0
´ ( -3u x + 2 u y - u z )10 -3
I
( u x + u z ) = 0.8 ( u x + u z ) mA m
4p
= 1100 u x + 1400 u y - 500 u z
F(0) = Q [E + v ´ B]
= 2 ´ 10 -16 [1200 u x + 1200 u y - 200 u z ]
9. (A) Using Ampere’s circuital law
= 4 ´ 10 -14 [ 6 u x + 6 u y - u z ]
ò H × dL = 2prHf = I encl
F = ma
At r = 0.5 cm,
I encl = 10 mA
-3
2 p(5 ´ 10 ) H f = 10, H f = 0.32 A m
10. (B) At r = 15
. cm enclosed current
I encl = 10 + 2 p(0.01)( 400) = 35.13 mA
2 p(0.015) H f = 35.13 ´ 10
-3
Þ
H f = 0.37 A m
11. (C) The enclosed current is
I encl = 10 + 2 p(0.01) 400 - 2 p(0.02)250 + 2 p(0.03) 300
= 60.3 mA m
2 p(0.035) H f = 60.3 M
12. (A) H =
Þ
H f = 0.27 A m
1
( -30 - 40) u x ´ ( -u z ) = -35 u y A m
2
Þ
a=
F 4 ´ 10 -14
=
[6u x + 6u y - u z ]
m 5 ´ 10 -26
= 800[ 6 u x + 6 u y - u z ] 10 9 m s 2
19. (C) F = e v ´ B
= - (1.6 ´ 10 -19)( 4.5 ´ 10 7 u y)(2.5 ´ 10 -3 u z )
= -1.8 ´ 10 -14 u x N
This force will be constant during the time the electron
travels the field. It establishes a negative x –directed
velocity as it leaves the field, given by the acceleration
times the transit time tt ,
Ftt æ -1.8 ´ 10 -14 öæ 2.5 ´ 10 -2
֍
=ç
m çè 9.1 ´ 10 -31 ÷øçè 4.5 ´ 10 7
0.5 - 0.025
=
= 106
. ´ 10 -8 s
4.5 ´ 10 7
vx =
t50
ö
÷÷ = -11
. ´ 10 7 m s
ø
In that time, the electron moves to an x coordinate
1
13. (B) H = K ´ u n
2
1
= ( 30 - 40) u x ´ ( -u z ) = -5 u y A m
2
given by
. ´ 10 7)(106
. ´ 10 -8 ) = -0.117 m
x = vx t50 = -(11
x = -117
. cm, z = 0
C
1
14. (A) H = ( -30 + 40) u x + ( -u z ) = 5 u y A m
2
15. (C) J = Ñ ´ H =
=
1 ¶ æ
r2
çç 2 +
2
r ¶r è
1 ¶(rH f)
uz
r ¶r
ö
÷÷u z = u z A m
ø
20. (A) FBC = ò I loop dL ´ Bfrom
4
= ò ( 6 ´ 10 -3) dzu z ´
1
1 ¶ æ 3ö
çr ÷ = 0
r ¶r çè r ÷ø
position along the loop segment.
3
ò ( 6 ´ 10
1
=
-3
) dxu x ´
15m o
uy
2 px
45 ´ 10 -3
m o ln 3u z = 19.8 u z nN
p
17. (A) F = e(E + v ´ B)
If F = 0, E = -v ´ B = B ´ v
Page
486
15m o
u y = -1.8 ´ 10 -8 u x = -18 u x nN
2 p( 3)
21. (C) The field from the long wire now varies with
FAB =
16. (D) J =
wire at BC
B
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Magnetostatics
22. (A) This will be the vector sum of the forces on the
four sides. By symmetry, the forces on sides AB and CD
will be equal and opposite, and so will cancel. This
leaves the sum of forces on side BC and DA
4
FDA =
ò -( 6 ´ 10
-3
) dxu z ´
1
15m o
u y = 54 u x nN
2 p(1)
Chap 8.3
28. (A) F =
ò K ´ BdS =
area
1 1 .5
ò ò 6u
z
´
0 0 .5
-5m o u x
dy
2 py
15m o æ 15
. ö
=ln ç
÷u y = -6.6 u y mN m
p
è 0.5 ø
29. (A) c m + 1 = m r , c m + 1 = 6.5 , c m = 5.5
Ftotal = FDA + FBC = (54 - 18) u x = 36 u x nN
30. (A) B = mH = m om rH
23. (A) F = ò IdL ´ R
= 4 p ´ 10 -7 ´ 6.5(10 u x + 25 u y - 40 u z )
2
2
3
1
1
1
1
2
= I ò dxu x ´ B + I ò dyu y ´ B + I ò dxu x ´ B + I ò dyu y ´ B
= 82 u x + 204 u y - 327 u z mWb m 2
u x ´ B = u x [ 6 xu x - 9 yu y + 3zu z ] = 3zu y - 9 yu z
31. (C) M = c m H = 55 u x + 137.5 u y - 220 u z A m
u y ´ B = u y[ 6 xu x - 9 yu y + 3zu z ] = 3zu x - 6 xu z
z = 0 for all element
3
2
1
1
1
1
3
2
32. (B) W =
F = I ò dx (-9 yu z) y=1 + I ò dy(-6 xu z) x=3 + I ò dx (-9 yu z) y=2 + I ò dy(-6 xu z) x=1
=
= I ( -18 - 18 + 36 + 6) u z = 5 ´ 6 u z = 30 u z N
24. (B) Within the region -1 < y < 1, the magnetic fields
from the two outer sheets (carrying -4u z A m) cancel,
leaving only the field from the center sheet. Therefore
1
1
H × B = mH 2
2
2
1
´ 6.5 ´ 4 p ´ 10 -7 (100 + 625 + 1600) = 9.5 mJ m 2
2
33. (D) m r = c m + 1 = 31
. + 1 = 4.1, m = m om r = 4.1m o
H=
0.4 yu z
B
=
= 77.6 yu z kA m
m 4.1 ´ 4 p ´ 10 -7
H = -4 u x A m (0 < y < 1) and H = 4 u x A m ( -1 < y < 0).
Outside (y > 1 and y < - 1) the fields from all three sheet
cancel, leaving H = 0 ( y > 1, y < - 1). So at x = 0, y = 0.5
F
= Iu z ´ B = (7 ´ 10 -3) u z ´ - 4m o u x = -35.2 u y nN m
m
0
0
-100 m o u y
2 p (5 ´ 10 -3)
= 0.4 u x N m
m r1 =
27. (B) The field from the current strip at the filament
location
6m o u x
3m o
. ö
æ 15
dy =
ln ç
÷u x
y
2
p
p
0
è .5 ø
0 .5
1 .5
B=
ò
= 1.32 ´ 10 -6 u x Wb m 2
1
F = ò IdL ´ B
0
1
m
1
1
=
´
= 1326.3
m o 600 4 p ´ 10 -7
M1 = c m H1 = 1590
.
´ 106 A m
B
1.4
For case 2, m = 2 =
H 2 400
ò IdL ´ B
= ò 100 dzu z ´
B1
2
=
H 1200
c m = m r - 1 = 1325.3
1
1
35. (C) For case 1, m =
m r1 =
F
25. (D)
= Iu x ´ ( -4m o u x ) = 0
m
26. (A) F =
34. (A) M = c m H = ( 31
. )(77.6) yu z = 241 yu z kA m
= ò 5 dzu z ´ 1.32 ´ 10 -6 u x dz = 6.6 u y mN m
1.4
= 2785.2
400 ´ 4 p ´ 10 -7
c m = 2784.2 2784.2
M 2 = c m H 2 = 1114
.
´ 106 A m
DM = (1590
.
- 1114
.
) ´ 106 = 476 kA m
36. (B) M = Nm = (2.7 ´ 10 29)(2.6 ´ 10 -30 u y)
= 0.7 u y A m
H=
0.7 u y
M
=
= 0.22 u y A m
m r - 1 4.2 - 1
0
37. (A) M =
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B
mo
ö
æ 1
çç
+ 1 ÷÷
ø
è cm
-1
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487
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GATE EC BY RK Kanodia
UNIT 8
-1
=
Electromagnetics
The normal component of H1 is
0.02 æ 1
ö
+ 1 ÷ = 47.6 A m
ç
4 p ´ 10 -7 è 0.003
ø
H N1 = (H1 × u N 21 ) u N 21
H1 × u N 21 = (50 u x - 30 u y + 20 u z ) × (0.37 u x + 0.56 u y - 0.74u z )
= 18.5 - 16.8 - 14.8 = -131
.
1
1
38. (A) W = H × B = m o H 2
2
2
1
= ( 4 p ´ 10 -7)( 600) 2 = 0.226 J m 3
2
=
Ho
0
0
= -4.83u x - 7.24 u y + 9.66 u z A m
Tangential component of H1 at the boundary
æ1
HT1 = H1 - H N1
ö
ò H. dB = ò Hçè 3 + 2 H ÷ødH
39. (A) W =
2
o
Ho
(H1 × u N 21 ) u N 21 = ( -131
. )(0.37 u x + 0.56 u y - 0.74 u z )
= (50 u x - 30 u y + 20 u z ) - ( -4.83u x - 7.24 u y + 9.66 u z )
= 54.83u x - 22.76 u y + 10.34 u z A m
3
o
2H
H
+
= 6.2 MJ m 3
6
3
40. (A) J b = Ñ ´ M =
H T 2 = H T1
m
2
H N 2 = r1 H N1 = ( -4.83u x - 7.24 u y + 9.66 u z )
m r2
5
12
uz
a
= -193
. u x - 2.90 u y + 3.86 u z A m
H 2 = H T 2 + H N 2 = (54.83u x - 22.76 u y + 10.34 u z )
41. (B) B1 = 200m o u x - 600m o u y + 400m o u z
+ ( -193
. u x - 2.9 u y + 3.86 u z )
= 52.9 u x - 25.66 u y + 14.2 u z
Its normal component at the boundary is
B1 N = (B1 × u N12 ) u N12
********
= (52.8 u x + 42.24 u y + 56.32 u z ) m o = B2 N
B
Þ H 2 N = 2 N = 6.60 u x + 5.28 u y + 7.04 u z
8m o
H1 N =
B1 N
= 26.40 u x + 2112
. u y + 28.16 u z
12m o
H1 T = H1 - H1 N = (100 u x - 300 u y + 200 u z )
-(26.4 u x + 2112
. u y + 28.16 u z )
= 73.6 u x - 32112
. u y + 171.84 u z
H1 T = H 2 T
H 2 = H 2 N + H 2 T = 80.2 u x - 315.8 u y + 178.9 u z A m
42. (C) H N1 = - 3u z , H T1 = 10 u x + 15 u y
H T 2 = H T1 = 10 u x + 15 u y
m
1
H N 2 = 1 H N1 =
( -3u z ) = 0.15 u z
m2
20
H 2 = H N 2 + H T 2 = 10 u x + 15 u y - 0.15 u z
B2 = m 2H 2 = 20 ´ 4 p ´ 10 -7(10 u x + 15 u y - 0.15 u z )
= 251u x + 377 u y - 377
. u z mWb m 2
43. (B) At the boundary normal unit vector
2 u x + 3u y - 4 u z
Ñ (2 x + 3 y - 4 z)
un =
=
|Ñ (2 x + 3 y - 4 z)|
29
= 0.37 u x + 0.56 u y - 0.74 u z
Since this vector is found through the gradient, it will
point
in
the
direction
of
increasing
values
of
2 x + 3 y - 4 z, and so will be directed into region 1. Thus
u n = u n 21 .
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488
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UNIT 8
Statement for Q.8–9:
Electromagnetics
Statement for Q.12–13:
The location of the sliding bar in fig. P8.4.8–9 is
Consider the fig. P8.4.12–13. The rails have a
given by x = 5 t + 4 t . The separation of the two rails is
resistance of 2 W m. The bar moves to the right at a
30 cm. Let B = x u z T.
constant speed of 9 m s in a uniform magnetic field of
3
2
y
0.8 T. The bar is at x = 2 m at t = 0.
B
B
z
B
a
VM
v
0.2 cm
I
v
x
b
Fig. P8.4.8–9.
8. The voltmeter reading at t = 0.5 s is
16 cm
(A) -21.6 V
(B) 21.6 V
Fig. P8.4–12–14
(C) -6.3 V
(D) 6.3 V
12. If 6 W resistor is present across the left-end with the
9. The voltmeter reading at x = 0.6 m is
right end open-circuited, then at t = 0.5 sec the current
(A) -1.68 V
(B) 1.68 V
I is
(C) -0.933 V
(D) 0.933 V
(A) -45 mA
(B) 45 mA
(C) -60 mA
(D)60 mA
Statement for Q.10–11:
A perfectly conducting filament containing a 250W
13. If 6 W resistor is present across each end, then I at
resistor is formed into a square as shown in fig.
0.5 sec is
P8.4.10-11.
(A) -12.3 mA
(B) 12.3 mA
(C) -7.77 mA
(D) 77.7 mA
y
I(t)
Statement for Q.14–15:
0.5 cm
B
The internal dimension of a coaxial capacitor is
250 W
a = 1.2 cm, b = 4 cm and c = 40 cm. The homogeneous
material inside the capacitor has the parameter
x
e = 10 -11 F m, m = 10 -5 H m and s = 10 -5 S m.The electric
Fig. P8.4.10–11
field intensity is E = 10r cos (10 5 t) u p V m.
7
10. If B = 6 cos (120 pt - 30 ° ) u z T, then the value of I ( t)
is
14. The current density J is
(A) 2.26 sin (120 pt - 30 ° ) A
(B) 2.26 cos (120 pt - 30 ° ) A
(C) -2.26 sin (120 pt - 30 ° ) A
(D) -2.26 cos (120 pt - 30 ° ) A
11. If B = 2 cos p( ct - y) u z mT, where c is the velocity of
light, then I ( t) is
200
sin (10 5 t) u r A m 2
r
(B)
400
sin (10 5 t) u r A m 2
r
(C)
100
cos (10 5 t) u r A m 2
r
(A) 1.2(cos pct - sin pct) mA
(D) None of the above
(B) 1.2(cos pct - sin pct) mA
15. The quality factor of the capacitor is
(C) 1.2(sin pct - sin pct) mA
(A) 0.1
(B) 10
(C) 0.2
(D) 20
(D) 1.2(sin pct - sin pct) mA
Page
490
(A)
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Maxwell’s Equations
16. The following fields exist in charge free regions
(B) -4 a sin (15
. ´ 108 t - ax) u z mA m
P = 60 sin ( wt + 10 x) u z
(C) 4 a sin (15
. ´ 108 t - ax) u z m A m
Q = 10r cos ( wt - 2r) u f
(D) 4 a sin (15
. ´ 108 t - ax) u z mA m
R = 3r 2 cot f u r + 1r cos fu f
21. The value of a is
S=
(A) 4.3
(B) 2.25
(C) 5
(D) 6
1
r
sin q sin ( wt - 6 r) u q
The possible electromagnetic fields are
(A) P, Q
(B) R, S
(C) P, R
(D) Q, S
Statement for Q.22–23:
Let H = 2 cos (1010 t - bx) u z A m, m = 3 ´ 10 -5 H m,
17. A parallel-plate capacitor with plate area of 5 cm 2
and plate separation of 3 mm has a voltage 50 sin (10 3 t)
V
Chap 8.4
applied to its plates. If e r = 2,
the displacement
current is
e = 1.2 ´ 10 -10 F m and s = 0 everywhere.
22. The electric flux density D is
(A) 120 cos (1010 t - bx) nC m 2
(A) 148 cos (1010 t) nA
(B) 261 cos (1010 t) mA
(B) -120 cos (1010 t - bx) nC m 2
(C) 261 cos (1010 t) nA
(D) 148 cos (1010 t) mA
(C) 120 cos (1010 t + bx) nC m 2
(D) None of the above
18. In a coaxial transmission line ( e r = 1), the electric
23. The magnetic flux density B is
field intensity is given by
E=
(A) 6.67 ´ 10 4 cos (1010 t + bx) T
100
cos (10 9 t - 6 z) u r V m.
r
(B) 6.67 ´ 10 4 cos (1010 t - bx)
(C) 6 ´ 10 -5 cos (1010 t + bx) T
The displacement current density is
100
(A) sin (10 9 t - 6 z) u r A m 2
r
(B)
(D) 6 ´ 10 -5 cos (1010 t - bx) T
Statement for Q.24–25:
116
sin (10 9 t - 6 z) u r A m 2
r
(C) -
0.9
sin (10 9 t - 6 z) u r A m 2
r
(D) -
216
cos (10 9 t - 6 z) u r A m 2
r
A material has s = 0 and e r = 1. The magnetic field
intensity is H = 4 cos (106 t - 0.01z) u y A m.
24. The electric field intensity E is
(A) 4.52 sin (106 t - 0.01z) kV m
(B) 4.52 sin (106 t - 0.01z) V m
Statement for Q.19–21:
(C) 4.52 cos (106 t - 0.01z) V m
Consider the region defined by |x|,|y| and |z|< 1.
Let e = 5 e o , m = 4m o , and s = 0 the displacement current
(D) 4.52 cos (106 t - 0.01z) kV m
density J d = 20 cos (15
. ´ 108 t - ax) u y m A m 2 . Assume
25. The value of m r is
no DC fields are present.
(A) 2
(B) 3
(C) 4
(D) 16
19. The electric field intensity E is
(A) 6 sin (15
. ´ 108 t - ax) u y mV m
26. The surface r = 3 and 10 mm, and z = 0 and 25 cm
(B) 6 cos (15
. ´ 10 t - ax) u y mV m
are perfect conductors. The region enclosed by these
(C) 3 cos (15
. ´ 108 t - ax) u y mV m
surface has m = 2.5 ´ 10 -6 H m, e = 4 ´ 10 -11 F m and
8
(D) 3 sin (15
. ´ 108 t - ax) u y mV m
s = 0. If H = 2r cos 8 pz cos wt u f A m, then the value of w
is
20. The magnetic field intensity is
(A) 2 p ´ 106 rad s
(B) 8 p ´ 106 rad s
(A) -4 a sin (15
. ´ 108 t - ax) u z m A m
(C) 2 p ´ 108 rad s
(D) 8 p ´ 108 rad s
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UNIT 8
Electromagnetics
27. For distilled water m = m o , e = 81e o , and s = 2 ´ 10 -3
SOLUTIONS
S m, the ratio of conduction current density to
displacement current density at 1 GHz is
(A) 111
. ´ 10 -5
(B) 4.44 ´ 10 -4
(C) 2.68 ´ 10 -6
(D) 1.68 ´ 10 -7
1. (B) emf = -
= 2 p(0.2) 2 (20)( 377) sin 377 t mV = 0.95 cos 377 t V
28. A conductor with cross-sectional area of 10 cm 2
carrier
a
conductor
current
2 sin (10 9 t)
mA.
If
2. (A) emf =
s = 2.5 ´ 106 S m and e r = 4.6, the magnitude of the
(A) 48.4 m A m
(B) 8.11 nA m
2
F = p(0.1) 2 B = 0.31 cos (120 pt)
(D) 16.4 m A m 2
(C) 32.6 nA m 2
emf = Vba ( t) = -
29. In a certain region
dF
dt
= 0.31(120 p) sin (120 pt) = 118.43 sin(120 pt)
J = ( 4 yu x + 2 xzu y + z u z ) sin (10 t) A m
3
4
Vab = - 118.43 sin (120 pt)
If volume charge density r v in z = 0 plane is zero,
4. (D) I =
then r v is
(A) 3z 2 cos (10 4 t) mC m 3
(B) 0.3z 2 cos (10 4 t) mC m 3
(C) -3z cos (10 t) mC m
2
1
1
Bo wL2 = ( 4)(2)(2) 2 = 16 V
2
2
3. (C) Since B is constant over the loop area, the flux is
displacement current density is
2
dF
d
=B × dS
dt
dt ò
4
Vab 118.43 sin (120 pt)
=
= 0.47 sin (120 pt)
R
250
5. (A) emf = -
3
(D) -0.3z 2 cos (10 4 t) mC m 3
=
30. In a charge-free region (s = 0, e = e o e r , m = m o )
magnetic field intensity is
H = 10 cos (10 t - 4 y) u z
11
dF
d
=dt
dt
òò B × u dz
z
loop area
d
( 3)( 4)( 6) cos 5000 t = -360000 sin 5000 t
dt
I=
emf
360000 sin 5000 t
== - 0.4 sin 5000 t A
R
900 ´ 10 3
A m. The displacement current density is
1 1
(A) -40 sin (10 9 t - 4 y) u y A m
6. (C) F =
(B) 40 sin (10 9 t - 4 y) u y A m
ò ò 20m
o
cos ( 3 ´ 108 t - y) dx dy
0 0
= [20m o sin ( 3 ´ 108 t - y)]10
(C) -40 sin(10 3 t - 4 y) u x A m
= 20 m o [sin ( 3 ´ 108 t - 1) - sin ( 3 ´ 108 t)] Wb
(D) 40 sin (10 9 t - 4 y) u x A m
31. In a nonmagnetic medium ( e r = 6.25) the magnetic
field of an EM wave is H = 6 cos bx cos (108 t) u z A m.
The corresponding electric field is
(A) 903 sin (0.83 x) sin (108 t) V m
Emf = -
dF
dt
= - 20 ´ (4p ´ 10 -7)(3 ´ 108 ) ´ [cos (3 ´ 108 t - 1) - cos (3 ´ 108 t)]
= 7540[cos ( 3 ´ 108 t) - cos ( 3 ´ 108 t - 1)] V
7. (D) In this case F = [20m o (2 p) sin ( 3 ´ 108 t - y)]20 p = 0
8
(B) 903 sin (12
. x) sin (10 t) V m
(C) 903 sin (0.83 x) cos (108 t) V m
8. (A) F =
(D) 903 sin (12
. x) cos (108 t) V m
òò B × dS =
area
E = 5 cos(10 9 t - 8 x) u x + 4 sin(10 9 t - 8 x) u z V m.
(A) 3.39
(B) 1.84
(C) 5.76
(D) 2.4
***********
Page
492
ò ò t dtdy
2
0
0
= 0.1 x 3 = 0.1(5 t + 4 t 3) 3 Wb
32. In a nonmagnetic medium
The dielectric constant of the medium is
0 .3 x
emf = -
dF
= -0.1( 3)(5 t + 4 t 3) 2 (5 + 12 t 2 )
dt
At t = 0.5 s, emf = -0.1( 3)(2.5 + 0.5) 2 (5 + 3) = -21.6 V
9. (C) At x = 0.6 m, 0.6 = 5 t + 4 t 3
At t = 0.119 s,
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emf = -0.933 V
Þ
t = 0.119 s
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Maxwell’s Equations
10. (A) F =
òò B × dS = 6(0.5)
2
Chap 8.4
I d = 2 prlJ d = - 2 pl(10) sin (10 5 t) = - 8 psin (10 5 t) A
cos (120 pt - 30 ° ) Wb
area
Quality factor
dF
emf = = 6(0.5) 2 (120 p) sin (120 pt - 30 ° )
dt
The current is
emf 6(0.5) 2 (120 p)
=
sin (120 pt - 30 ° ) A
R
250
= 2.26 sin (120 pt - 30 ° ) A
òò B × dS = (0.5)(2) ò cos( py - pct) dy
area
0
1é æ
pö
ù 1
sinç pct - ÷ - sin pct ú = [ - cos pct - sin pct ] mWb
p êë è
2ø
û p
dF
emf = = c [cos pct - sin pct ] mV
dt
=
I ( t) =
=
Ic
8
= 0.1
80
16. (A) Ñ × P = 0, Ñ ´ P = -
emf 3 ´ 108
=
[cos pct - sin pct ] mA
250
R
Q
is a possible EM field
sin f
1 ¶
( 3r 2 cot f) ×
Ñ ×R =
¹ 0, R is not an EM field.
r ¶r
r
Ñ ×S =
¶(sin 2 f)
1
sin ( wt - 6 r)
¹ 0
r sin q
¶r
2
S is not an EM field. Hence (A) is correct.
= 12
. [cos pct - sin pct ] A
17. (A) D = eE = e
12. (A) The flux in the left-hand closed loop is
Id = J × S =
Fl = B ´ area = (0.8)(0.2)(2 + 9t)
dF L
= - (0.16)(9) = -1.44 V
emfl = dt
V
d
Þ
increasing with time,
Rl = 6 + 2[2 (2 + 9 t)]W, At t = 0.5, Rl = 32 W
1.44
Il = = - 45 mA
32
18. (C)
¶D
¶E
100
=e
Jd =
= eo
[ - sin (10 9 t - 6 z)] 10 9 u r A m 2
¶t
¶t
r
=-
current from the right loop, which is now closed. The
flux in the right loop, whose area decreases with time,
is Fr = (0.8)(0.2)(16 - 2 - 9 t)
dF R
emfR = = 1.44 V
dt
0.9
sin (10 9 t - 6 z) u r A m 2
r
19. (D) D = ò J d dt + C1 =
20 ´ 10 -6
sin (15
. ´ 108 - ax) u y
15
. ´ 108
. ´ 108 t - ax) u y C m 3
= 1.33 ´ 10 -13 sin (15
C1 is set to zero since no DC fields are present.
E=
Rr = 44 W
The contribution to the current from the right loop
-144
Ir =
= 032.7 mA
44
The total current = -32.7 - 45 = -77.7 mA
14. (C) J = s E =
dD e dV
=
dt d dt
eS dV 2 e o 5 ´ 10 -4
=
10 3 ´ 50 cos (10 3 t)
d dt
3 ´ 10 -3
13. (C) In this case, there will be contribution to the
At 0.5 s,
Jd =
= 148 cos (1010 t) nA
While the bar in motion, the loop resistance is
Rr = 6 + 2 (2 (14 - 9 t)),
¶Pz
uy ¹ 0
¶x
P is a possible EM field
1 ¶
Ñ × Q = 0, Ñ ´ Q =
[10 cos ( wt - 2r)]u z ¹ 0
r ¶r
0 .5
11. (D) F =
Id
100
cos (10 5 t) u r A m 2
r
15. (A) Total conduction current
100
cos (10 5 t) u r A m 2
I C = òò J × dS = 2prlJ = 2 prl
r
= 80 pcos (10 5 t) A
¶D ¶eE
10
=
sin (10 5 t) A m 2
Jd =
=¶t
¶t
r
D 1.33 ´ 10 -13
=
sin (15
. ´ 108 - ax) u y
e
5 eo
= 3 ´ 10 -3 sin (15
. ´ 108 t - ax) V m
20. (D) Ñ ´ E =
¶E y
¶x
uz = -
¶B
¶t
= -a( 3 ´ 10 -3) cos (15
. ´ 108 t - ax) u z = -
¶B
¶t
B=
a( 3 ´ 10 -3)
sin (15
. ´ 108 t - ax) u z
15
. ´ 108
H=
2 ´ 10 -11
B
=
sin (15
. ´ 108 t - ax) u z A m
m 4 ´ 4 p ´ 10 -7
. ´ 108 t - ax) u z mA m
= 4 ´ 10 -6 a sin (15
21. (B) Ñ ´ H = www.gatehelp.com
¶H z
u y = Jd
¶x
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UNIT 8
= a 2 ( 4 ´ 10 -6 ) cos (15
. ´ 108 t - ax) = J D
Electromagnetics
27.(B) At high frequency
Comparing the result
a 2 4 ´ 10 -6 = 20 ´ 10 -6 , a = 5 = 2.25
22. (B) Ñ ´ H = -
=
¶H z
¶D
uy =
¶x
¶t
Þ
2b
cos (1010 t - bx) u y C m 2
D =1010
Þ
b=
w
. ´ 10 -10 = 600
= 1010 me = 1010 3 ´ 10 -5 ´ 12
v
Þ
¶z
Jd =
E=
Ic
sS
4.6 e o (10 9)
2 cos (10 9 t) 10 -3
2.5 ´ 106 ´ 10 ´ 10 -4
29. (B) Ñ × J = (0 + 0 + 3z 2 ) sin (10 4 t) = -
23. (D) B = mH = 6 ´ 10 -5 cos (1010 t - bx) u z T
¶H y
Ic
= sE Þ
S
¶E
e ¶I c
Jd = e
=
¶t sS ¶t
|J d | = 32.6 nA m 2
D = -120 cos (1010 t - bx) u y nC m 2
24. (A) Ñ ´ H = -
2 ´ 10 -3
= 4.44 ´ 10 -4
2 p ´ 10 9 ´ 81 ´ e o
28. (C) J c =
¶D
= 2b sin (1010 t - bx) u y
¶t
Jc
sE
s
=
=
J d weE we
3z 2 cos (10 4 t)
+ C1
10 4
rv =
ux
8r
¶t
At z = 0, r v = 0, C1 = 0
r v = 0.3z 2 cos (10 4 t) mC m 3
¶E
Ñ ´ H = 0.04 cos (10 t - 0.01z) u x = e o
¶t
6
30. (D) J d = Ñ ´ H =
0.04 sin (106 t - 0.01z) u x
E=
106 e o
¶H z
ux
¶y
= 40 sin (10 9 t - 4 y) u y A m
= 4.52 sin (10 t - 0.01z) u x kV m
6
31. (A) Ñ ´ H = -
¶Ex
¶H
25. (B) Ñ ´ E =
uy = -m
¶z
¶t
= 6b sin (bx) cos (108 t) u y
1
E = ò 6b sin (bx) cos (108 t) u y dt
e
6b
=
sin (bx) sin (108 t) u y
e1010
-0.04(0.01)
¶H
cos (106 t - 0.01z) u y = - m rm o
106 e o
¶t
H=
0.04(0.01)
sin (106 t - 0.013) u y
(106 )(106 )m rm o e o
0.04(0.01)
=4
1012 m rm o e o
Þ
mr =
(0.04)(0.01)
´ ( 3 ´ 108 ) 2 = 9
4(1012 )
H=
¶Er
(16 p)( 8 p)
¶H
uf =
cos ( 8 pz) sin ( wt) u f = -m
¶z
rew
¶t
6(0.833)
sin (bx) sin (108 t) u y V m
6.25 e o ´ 108
b=
w w
=
er ,
v c
8=
10 9
´ 108 e r
3
128 p2
cos ( 8 pz) cos ( wt) u f
rew2
w=
8p
me
=
8p
4 ´ 10
-11
´ 2.5 ´ 10 -6
= 8 p ´ 108 rad s
Page
494
w = 10 9, b = 8,
Þ
e r = 5.76
**********
This result must be equal to the given H field. Thus
Þ
6.25 = 0.833
32. (C) For nonmagnetic medium m r = 1
16 p
sin ( 8 pz) sin ( wt) u r
rew
128 p2 2
=
remw2 r
w w
108
=
er =
v c
3 ´ 108
= 903 sin (0.83 x) sin (108 t) u y V m
16 p
¶E
=
sin ( 8 pz) cos ( wt) u r = e
r
¶t
Ñ ´E=
e = 6.25 e o , b =
E=
¶H f
26. (D) Ñ ´ H = ur
¶z
E=
¶H z
¶E
uz = e
¶y
¶t
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CHAPTER
8.5
ELECTROMAGNETIC WAVE PROPAGATION
Statement for Q.1–3:
A y-polarized uniform plane wave with a frequency
(A) 6.43 ´ 106 m s
(B) 2.2 ´ 10 7 m s
(C) 1.4 ´ 108 m s
(D) None of the above
of 100 MHz propagates in air in the + x direction and
impinges normally on a perfectly conducting plane at
Statement for Q.5–6:
x = 0. The amplitude of incident E-field is 6 mV m.
1. The phasor H s of the incident wave in air is
(A) 16 e
(C) 16 e
-j
2p
x
3
-j
2p
x
3
uz m A m
(B) -16 e
ux m A m
-j
2p
x
3
-j
2p
x
3
(D) -16 e
A uniform plane wave in free space has electric
field Es = (2 u z + 3u y) e - jbx V m.
uz m A m
5. The magnetic field phasor H s is
ux m A m
(B) (5.3u y - 8 u z ) e - jbx m A m
(A) ( -5.3u y - 8 u z ) e - jbx m A m
(C) ( -5.3u y + 8 u z ) e - jbx m A m
2. The E-field of total wave in air is
(D) (5.3u y + 8 u z ) e - jbx m A m
æ 2p ö
(A) j12 sin ç
x ÷ u y mV m
è 3 ø
6. The average power density in the wave is
æ 2p ö
(B) - j12 sin ç
x ÷ u y mV m
è 3 ø
(A) 34 mW m 2
(B) 17 mW m 2
(C) 22 mW m 2
(D) 44 mW m 2
æ 2p ö
(C) 12 cos ç
x ÷ u y mV m
è 3 ø
7. The electric field of a uniform plane wave in free
æ 2p ö
(D) -12 cos ç
x ÷ u y mV m
è 3 ø
field phasor H s is
space is given by Es = 12 p( u y + ju z ) e - j15x . The magnetic
3. The location in air nearest to the conducting plane,
where total E-field is zero, is
(A) x = 15
. m
(B) x = -15
. m
(C) x = 3 m
(D) x = - 3 m
vp is
( -u z + ju y) e - j15x
(B)
12
ho
( u z + ju y) e - j15x
(C)
12
ho
( -u z - ju y) e - j15x
(D)
12
ho
( u z - ju y) e - j15x
A lossy material has m = 5m o , e = 2 e o . The phase
uniform plane wave propagating in a certain lossless
material is ( 6 u y - j5 u z ) e
12
ho
Statement for Q.8–9:
4. The phasor magnetic field intensity for a 400 MHz
- j18 x
(A)
A m . The phase velocity
constant is 10 rad m at 5 MHz.
8. The loss tangent is
(A) 2913
(B) 1823
(C) 2468
(D) 1374
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UNIT 8
Electromagnetics
9. The attenuation constant a is
16.
(A) 4.43
(B) 9.99
m r = 1, e 2 = 1) pipe with inner and outer radii 9 mm and
(C) 5.57
(D) None of the above
12 mm carries a total current of 16 sin (106 pt) A. The
A
60
m
long
aluminium
( s = 35
. ´ 10 7 S m,
effective resistance of the pipe is
Statement for Q.10–11:
At
50
MHz
a
lossy
dielectric
material
is
characterized by m = 2.1m o , e = 3.6 e o and s = 0.08 S m.
(A) 0.19 W
(B) 3.48 W
(C) 1.46 W
(D) 2.43 W
The electric field is Es = 6 e - jgx u z V m.
17. Silver plated brass wave guide is operating at 12
10. The propagation constant g is
m r = e r = 1) is 5d, the minimum thickness required for
(A) 7.43 + j2.46 per meter
wave-guide is
(B) 2.46 + j7.43 per meter
(A) 6.41 mm
(B) 3.86 mm
(C) 6.13 + j5.41 per meter
(C) 5.21 mm
(D) 2.94 mm
GHz. If at least the thickness of silver ( s = 6.1 ´ 10 7 S m,
(D) 5.41 + j 6.13 per meter
Statement for Q.18–19:
11. The impedance h is
A uniform plane wave in a lossy nonmagnetic
(A) 101.4 W
(B) 167.4 W
(C) 98.3 W
(D) 67.3 W
media has
Es = (5 u x + 12 u y) e - gz , g = 0.2 + j 3.4 m -1
Statement for Q.12–13:
A
non
magnetic
18. The magnitude of the wave at z = 4 m and t = T 8
medium
has
an
intrinsic
impedance 360 Ð30 ° W.
is
(A) 10.34
(B) 5.66
(C) 4.36
(D) 12.60
12. The loss tangent is
(A) 0.866
(B) 0.5
19. The loss suffered by the wave in the interval
(C) 1.732
(D) 0.577
0 < z < 3 m is
13. The Dielectric constant is
(A) 1.634
(B) 1.234
(C) 0.936
(D) 0.548
(A) 4.12 dB
(B) 8.24 dB
(C) 10.42 dB
(D) 5.21 dB
Statement for Q.20–22:
The plane wave E = 42 cos ( wt - z) u x V m in air
Statement for Q.14–15:
normally hits a lossless medium (m r = 1, e r = 4) at z = 0.
The amplitude of a wave traveling through a lossy
nonmagnetic medium reduces by 18% every meter. The
20. The SWR s is
wave operates at 10 MHz and the electric field leads the
(A) 2
magnetic field by 24°.
(C)
14. The propagation constant is
(A) 0.198 + j0.448 per meter
(B) 0.346 + j0.713 per meter
(C) 0.448 + j0.198 per meter
1
3
(D) 3
22. The reflected electric field is
15. The skin depth is
(A) 2.52 m
(B) 5.05 m
(C) 8.46 m
(D) 4.23 m
Page
496
(D) None of the above
21. The transmission coefficient t is
2
4
(A)
(B)
3
3
(C)
(D) 0.713 + j0.346 per meter
1
2
(B) 1
(A) -14 cos ( wt - z) u x V m
(B) -14 cos ( wt + z) u x V m
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UNIT 8
Electromagnetics
33. The region z < 0 is characterized by e r = m r = 1 and
Region 2 (0 < z < 6 cm): m 2 = 2 mH m, e 2 = 25 pF m
s = 0.
Region 3 (z > 6 cm):
The
total
electric
field
E s = 150 e - j10 z u x + 50 Ð20 ° e j10 z u x
here
is
given
39. The lowest frequency, at which a uniform plane
impedance of the region z > 0 is
(A) 692 + j176 W
(C) 176 + j 692 W
(B) 193 - j 49 W
wave incident from region 1 onto the boundary at z = 0
(D) 49 - j193 W
will have no reflection, is
Statement for Q.34–35:
Region 1, z < 0 and region 2, z > 0, are both perfect
dielectrics. A uniform plane wave traveling in the u z
direction
has
a
frequency
m 3 = 4 mH m, e 3 = 10 pF m
V m. The intrinsic
of
3 ´ 1010
rad s.
Its
wavelength in the two region are l1 = 5 cm and l2 = 3
cm.
(A) 2.96 GHz
(B) 4.38 GHz
(C) 1.18 GHz
(D) 590 MHz
40. If frequency is 50 MHz, the SWR in region 1 is
(A) 0.64
(B) 1.27
(C) 2.38
(D) 4.16
41. A uniform plane wave in air is normally incident
34. On the boundary the reflected energy is
onto a lossless dielectric plate of thickness l 8 , and of
(A) 6.25%
(B) 12.5%
intrinsic impedance h = 260 W. The SWR in front of the
(C) 25%
(D) 50%
plate is
35. The SWR is
(A) 1.67
(B) 0.6
(C) 2
(D) 1.16
(A) 1.12
(B) 1.34
(C) 1.70
(D) 1.93
42. The E-field of a uniform plane wave propagating in
a dielectric medium is given by
36. A uniform plane wave is incident from region 1
(m r = 1, s = 0) to free space. If the amplitude of incident
z ö
z ö
æ
æ 8
E = 2 cos ç 108 t ÷ u x - sin ç 10 t ÷ uy V m
3ø
3ø
è
è
wave is one-half that of reflected wave in region, then
the value of e r is
(A) 4
(B) 3
(C) 16
(D) 9
37. A 150 MHz uniform plane wave is normally incident
from air onto a material. Measurements yield a SWR of
3 and the appearance of an electric field minimum at
0.3l in front of the interface. The impedance of material
is
(A) 502 - j 641 W
(B) 641 - j502 W
(C) 641 + j502 W
(D) 502 + j 641 W
The dielectric constant of medium is
(A) 3
(B) 9
(C) 6
(D)
43. An electromagnetic wave from an under water
source with perpendicular polarization is incident on a
water-air interface at angle 20° with normal to surface.
For water assume e r = 81, m r = 1. The critical angle q c is
(A) 83.62°
(B) 6.38°
(C) 42.6°
(D) None of the above
***********
38. A plane wave is normally incident from air onto a
semi-infinite slab of perfect dielectric ( e r = 3.45). The
fraction of transmitted power is
(A) 0.91
(B) 0.3
(C) 0.7
(D) 0.49
Statement for Q.39–40:
Consider three lossless region :
Region 1 (z < 0):
Page
498
6
m 1 = 4 mH m, e1 = 10 pF m
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Electromagnetic Wave Propagation
SOLUTIONS
1. (A) w = 2 p ´ 10
8
8. (B) Loss tangent
rad s
w
2 p ´ 10
2p
=
=
rad m
c
3 ´ 108
3
Es = 6 e
2p
-j x
3
u y mV m
uE ´ u H = ux ,
Hs =
6
e
120 p
-j
2p
x
3
u y ´ u H = ux ,
uz
= 16 e
-j
2p
x
3
u H = uz
Þ
10 =
Þ
x=
uz m A m
Er = -6 e
j
2p
x
3
5´2
2
[ 1+ x
s
= 1823
we
1 + x2 - 1
1 + x2 + 1
2
ù
me é
æ s ö
ê 1+ç
÷ - 1ú
2 ê
è we ø
úû
ë
æ 2p ö
= - j12 sin ç
x ÷ u y mV m
è 3 ø
10. (D) a = w
3. (B) The electric field vanish at the surface of the
s
0.08
=
=8
we
3.6 ´ 50 ´ 106 ´ 2 pe o
conducting plane at x = 0. In air the first null occur at
3
l
p
x =- 1 ==- m
2
2
b1
w 2 p ´ 400 ´ 10
=
= 1.4 ´ 108 m s
b
18
5. (C) The wave is propagating in forward x direction.
2 p ´ 50 ´ 106
3 ´ 108
=
(2.1)( 3.6)
( 65 - 1) = 5.41
2
2
ù
me é
æ s ö
ê 1+ç
÷ + 1ú
2 ê
è we ø
úû
ë
2 p ´ 50 ´ 106
3 ´ 108
(2.1)( 3.6)
( 65 + 1) = 6.13
2
g = a + jb = 5.41 + j 6.13 per meter.
Therefore u E ´ u H = u x .
Þ
uH = - uy
For u E = u y , u y ´ u H = u x Þ u H = u z
1
Hs =
( -2 u y + 3u z ) e - jbx = ( -5.3u y + 8 u z ) e - jbx mA m
120 p
6. (B) Pavg =
a=
b=w
6
For u E = u z , u z ´ u H = u x
]
+1
a = 10 ´ 0.999 = 9.99
2p
-j x
æ - j 2 px
ö
E = Ei + Er = çç 6 e 3 u y - 6 e 3 u y ÷÷ mV m
è
ø
4. (C) vp =
2
a
1822
=
b
1824
Þ
u y mV m,
2 p ´ 5 ´ 106
3 ´ 108
a
=
b
9. (B)
2. (B) For conducting plane G = -1,
s
=x
we
2
ù
me é
æ s ö
ê 1+ç
÷ + 1ú
2 ê
è we ø
úû
ë
b=w
8
b=
Chap 8.5
11. (A) |h| =
1
Re {Es ´ H s*}
2
12. (C)
1
{(5.3) u x + 3( 8) u x } ´ 10 -3 = 17.3u x mW m 2
2
7. (D) Since Pointing vector is in the positive x
direction, therefore u E ´ u H = u x .
For u E = u y , u y ´ u H = u x
Þ
u H = uz
For u E = u z , u z ´ u H = u x
12
Hs =
( u z - ju y) e - j 15x
ho
Þ
u H = -u y ,
m
e
æ
ç 1 + æç s ö÷
ç
è we ø
è
ö
÷
÷
ø
=
1
4
2.1
3.6 = 101.4
1
64 4
s
= tan 2 q n = tan 60 ° = 1732
.
we
13. (D) |h| =
Þ
2
120 p
360 =
m
e
1
2 4
æ
ö
ç 1 + æç s ö÷ ÷
ç
è we ø ÷ø
è
120p
er
Þ
1
e r = 0.548
2 4
(1 + 1732
.
)
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UNIT 8
14. (A) |E| = Eo e - az
ho
- ho
1
h2 - h1
G=
= 2
=ho
3
h2 + h1
+ ho
2
1
1 + |G| 1 + 3
s=
=
=2
1 - |G| 1 - 1
3
Eo e - a1 = (1 - 0.18) Eo
e - a1 = 0.82
q n = 24 °
Þ
Þ
1
= 0.198
0.82
s
tan 2 q n =
= 1111
.
we
a = ln
2
æ s ö
1+ç
÷ -1
è we ø
a
=
b
h
2 o
2h2
2 =2
21. (A) t =
=
h2 + h1 ho + 2
3
2
2
æ s ö
1+ç
÷ +1
è we ø
0.198
=
b
234 - 1
Þ
234 + 1
b = 0.448
22. (A) Eor = GEoi = -
g = a + jb = 0.198 + j0.448
15. (B) d =
d=
1
pfsm
Rac =
=
Þ
23. (C) h1 = ho , h2 =
1
p ´ 5 ´ 10 ´ 35
. ´ 10 7 ´ m o
= 120 mm
l
sdw
Since d is very small, w = 2pr outer
60
= 0.19 W
Rac =
7
. ´ 10 ´ 120 ´ 10 -6 ´ 2 p ´ 12 ´ 10 -3
35
=
5
p ´ 12 ´ 10 ´ m o ´ 6.1 ´ 10 7
= 2.94 mm
h1 = ho
18. (B) E = Re{E s e jwt } = (5 u x + 12 u y) e -0 .2 z cos ( wt - 3.4 z)
T
8
Þ
377
h - 377
= 2
3000 h 2 + 377
Þ
e r = 12.5,
Þ
æp
ö
|E|= 13e -0 .8 cos ç - 13.6 ÷ = 5.66
è4
ø
G=
19. (D) Loss = aDz = 0.2 ´ 3 = 0.6 Np
Eor = -
20. (A) h1 = ho , h2 = ho
Page
500
mr
h
= o
2
er
æ h - h1 ö 18
÷÷
h1 = çç 2
-3
è h2 + h1 ø 6 ´ 10
h2 = 485.37 = ho
mr
er
m r = 20.75
25. (A) h1 = ho , h2 = ho
0.6 Np = 5. 21 dB.
Þ
æ h - ho ö
÷÷3000
ho = çç 2
è h2 + ho ø
æp
ö
E = (5 u x + 12 u y) e -0 .8 cos ç - 13.6 ÷
è4
ø
1 Np = 8.686 DB,
mr
mr
= ho
12.5
er
Eor
h - h1
=G = 2
Eoi
h2 + h1
æ h - h1 ö
÷÷ Eoi
h1 H or = çç 2
è h2 + h1 ø
pfms
At z = 4 m, t =
24. (D) h1 = ho , h2 = ho
But Eor = h1 H or = GEoi
5
9
m ho ho
=
=
2
e er
ho
- ho
1
h2 - h1
G=
= 2
=ho
3
h2 + h1 1
+ ho
2
f = 5 ´ 10 5 Hz,
5
17. (D) t = 5 d =
1
( 42) = -14
3
Er = - 14 cos ( wt - z) u x V m
1
1
=
= 5.05
a
0.198
16. (A) w = p106
Electromagnetics
m r ho
=
2
er
h2 - h1
1
=h2 + h1
3
1
10
(10) = 3
3
10
Eor
H or =
=
= 8.8 ´ 10 -3
ho
3 ´ 377
uE ´ u H = uk ,
-u y ´ u H = -u z
H r = -8.8 cos ( wt - z) u x mA m
w w
b =1 = =
m r er
v c
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u H = -u x
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Electromagnetic Wave Propagation
w=
3 ´ 108
= 0.5 ´ 108 rad s.
12 ´ 3
26. (D) h1 = ho , h2 = ho
33.(A) G =
G=
m r ho
=
= 0.58ho
er
3
h - h1 0.58ho - ho
G= 2
=
= -0. 266
h2 + h1 0.58h o + h o
27. (B) ETotal = Ei + Er ,
Eor = GEoi = -2.66
ho
G=
Þ
1
sin 45 ° = 0.333
4.5
sin q t1 =
q t1 = 19.47 °
q t2 = sin -1 0.47 = 28 °
But cos q t =
Þ
G=
h1
h
cos q B = o cos 58.2 ° = 2.6 cos 58.2 °
ho
h2
2.6
q t = 31.8 °
mr
h
= o = 0.447ho
er
5
Ei
Er
=
ho -
Þ
h - h1
G= 2
= - 0.38, t = 1 + G = 0.62
h2 + h1
G=
Þ
Þ
h2 - h1
=
h2 + h1
m r2
m r1
+ ho
er 2
e r1
ho
=
m r 2 1 + 0.447
=
= 0.382, 2.62
m r1 1 ± 0.447
e r1 æ m r 2
=ç
e r 2 çè m r1
3
ö
÷÷ = 0.056, 17.9
ø
m r1
m r2
m r32
m r31
m r2
m r1
+
m r32
m r31
2
e r1
-1
er 2
e r1
er 2
l2
-1
l
= 1
l2
+1
+1
l1
mr
h
= o
er
er
1 h2 - h1
=
2 h2 + h1
ho
er 1
=
ho
2
ho +
er
Þ
37. (C) At minimum
G = ± 0.447
m r2
m r1
- ho
er 2
e r1
ho
ö
÷÷
ø
1+
Et = tEi = 92.7 cos ( wt - 8 y) u z V m
32. (B) |G| = 0.2 ,
e r1 æ l2
=ç
e r 2 çè l1
l2 - l1 3 - 5
1
=
=4
l2 + l1 3 + 5
36. (D) h2 = ho , h1 = ho
G=
2
Þ
1
4 = 5
35. (A) s =
=
1
1 - |G|
3
14
2.6 e o
= 2.6
eo
31. (A) h1 = ho , h2 = ho
ho
er 2
e r1
h2 - h1
=
=
ho
ho
h2 + h1
+
er 2
e r1
1 + |G|
30. (A) Since both media are non magnetic
e1
=
e2
Þ
-
2
The fraction of the incident energy that is reflected is
1
G2 =
= 6.25%.
16
e
4.5
29. (B) sin q t 2 = 1 sin q t1 =
(0.333) = 0.47
e2
2.25
tan q B =
ö
÷
÷ = 692 + j176 W
÷
÷
ø
e j 20
3
e j 20
3
2
28. (B) m o = m 1 = m 2
eo
sin q i
e1
æ
ç1 +
ö
÷÷ = 377ç
ç
ø
ç1è
æ 2 pc ö
æ 2 pc ö
34. (A) e r1 = çç
÷÷ , e r 2 = çç
÷÷
è l1 w ø
è l2 w ø
ETotal = 10 cos ( wt - z) u y - 2.66 cos ( wt + z) u y V m
Þ
h2 - h1
, h1 = ho ,
h2 + h1
æ1 + G
h2 = ho çç
è1-G
Et = 7.34 cos ( wt - z) u y V m
Þ
Er 50 Ð20 ° e j 20
=
=
Ei
150
3
Eot = tEoi = 7.34
t = 1 + G = 0.734,
sin q t1 =
Chap 8.5
er = 9
( f + p)
= 0.3l ,
2b
2p
Þ
f = 0.2 p
l
s -1
3 -1 1
=
=
|G| =
s+1
3+1 2
h - ho
G = 0.5 e j 0 .2 p = 2
h2 + ho
b=
=
m r1 - m r 2
m r1 + m r 2
Þ
æ 1 + 0.5 e j 0 .2 p ö
÷ = 641 + j502 W
h2 = ho çç
j 0 .2 p ÷
è 1 - 0.5 e
ø
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UNIT 8
ho
3.45
ho
3.45
- ho
+ ho
= -0.3
43. (B) q c = sin -1
39. (C) This frequency gives the condition b2 d = p
Where d = 6 cm, b2 = w m 2 e 2
Þ
f =
p
0.06
1
2 ´ 0.06 2 ´ 10 -6 ´ 25 ´ 10 -12
= 118
. GHz
40. (B) At 50 MHz,
b2 = w m 2 e2 = 2 p ´ 50 ´ 10 6 2 ´ 10 -6 ´ 25 ´ 10 -12 = 2. 2
b2 d = 2.22(0.06) = 0.133
h1 =
m1
4 ´ 10 -6
=
= 632 W
e1
10 -11
h3 = 632 W
h2 =
m2
2 ´ 10 -6
=
= 283 W
e2
25 ´ 10 -12
The input impedance at the first interface is
æ h + jh2 tan (b2 d) ö
æ 632 + j283(0.134) ö
÷÷ = 283çç
hin = h2 çç 3
÷÷
è 283 + j 632(0.134) ø
è h2 + jh3 tan (b2 d) ø
= 590 - j138
h - h1 590 - j138 - 632
G = in
=
= 0.12 Ð - 100.5°
h in + h1 590 - j138 + 632
s=
1 + |G|
1 - |G|
=
1 + 0.12
= 1.27
1 - 0.12
41. (C) bd =
2p l p
p
× = , tan
=1
l 8 4
4
h2 = 260, h1 = h3 = ho
æ h + jh2 tan (b2 d) ö
æ 377 + j260 ö
÷÷ = 260çç
hin = h2 çç 3
÷÷
h
+
j
h
tan
(
b
d
)
è 260 + j 377 ø
ø
è 2
3
2
= 243 - j92 W
h - ho 243 - j92 - 377
G = in
=
= 0.26 Ð - 137 °
hin + ho 243 - j92 + 377
s=
1 + |G|
1 - |G|
=
126
.
= 170
.
0.74
42. (A) w = 108 rad s, b =
Page
502
er
=
1
3
Þ
e r = 3.
er 2
= sin -1
e r1
1
= 6.38 °
81
********
2
w m 2 e2 =
108
Þ
The transmitted fraction is 1 - |G| = 1 - 0.09 = 0.91.
Þ
3 ´ 108
mr
ho
=
er
3.45
38. (A) h1 = ho , h2 = ho
h - h1
G= 2
=
h2 + h1
Electromagnetics
1
3
rad m, v =
c
er
=
w
b
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GATE EC BY RK Kanodia
CHAPTER
8.7
WAVEGUIDES
Statement for Q.1–3:
Statement for Q.6–7:
A 2 cm by 3 cm rectangular waveguide is filled
In an air-filled rectangular waveguide the cutoff
with a dielectric material with e r = 6. The waveguide is
frequencies for TM11 and TE03 modes are both equal to
operating at 20 GHz with TM11 mode.
12 GHz.
1. The cutoff frequency is
6. The dominant mode is
(A) 3.68 GHz
(B) 22.09 GHz
(A) TM10
(B) TM 01
(C) 9.02 GHz
(D) 16.04 GHz
(C) TE01
(D) TE10
2. The phase constant is
(A) 816 rad m
(B) 412 rad m
(C) 1009 rad m
(D) 168 rad m
7. At dominant mode the cutoff frequency is
(A) 11.4 GHz
(B) 4 GHz
(C) 5 GHz
(D) 8 GHz
3. The phase velocity is
(A) 1.24 ´ 108 m s
(B) 154
. ´ 106 m s
(C) 305
. ´ 10 m s
(D) 7.48 ´ 10 m s
8
8. For an air-filled rectangular waveguide given that
8
4. In an an-filled rectangular wave guide, the cutoff
frequency of a TE10 mode is 5 GHz where as that of TE01
æ 2 px ö
æ 3py ö
12
Ez = 10 sin ç
÷ sin ç
÷ cos (10 t - bz) V m
è a ø
è b ø
If the waveguide has cross-sectional dimension
mode is 12 GHz. The dimensions of the guide is
a = 6 cm and b = 3 cm, then the intrinsic impedance of
(A) 3 cm by 1.25 cm
(B) 1.25 cm by 3 cm
this mode is
(C) 6 cm by 2.5 cm
(D) 2.5 cm by 6 cm
(A) 373.2 W
(B) 378.9 W
(C) 375.1 W
(D) 380.0 W
5. Consider a 150 m long air-filled hollow rectangular
waveguide with cutoff frequency 6.5 GHz. If a short
pulse of 7.2 GHz is introduced into the input end of the
guide, the time taken by the pulse to return the input
end is
Statement for Q.9–10:
In an air-filled waveguide, a TE mode operating at
6 GHz has
(A) 920 ns
(B) 460 ns
(C) 230 ns
(D) 430 ns
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æ 2 px ö
æ py ö
E y = 15 sin ç
÷ cos ç ÷ sin ( wt - 12 z) V m
è a ø
è b ø
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511
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GATE EC BY RK Kanodia
UNIT 8
9. The cutoff frequency is
Electromagnetics
16. The cross section of a waveguide is shown in fig.
(A) 4.189 GHz
(B) 5.973 GHz
P8.7.16. It has dielectric discontinuity as shown in fig.
(C) 8.438 GHz
(D) 7.946 GHz
P8.7.16. If the guide operate at 8 GHz in the dominant
mode, the standing wave ratio is
y
10. The intrinsic impedance is
(A) 35.72 W
(B) 3978 W
(C) 1989 W
(D) 71.44 W
2.5 cm
mo, eo
Statement for Q.11–12.
x
mo, 2.25eo
z
5 cm
Fig. P8.7.16
Consider an air-filled rectangular wave guide with
cm. The y-component of the
a = 2.286 cm and b = 1016
.
(A) -3.911
(B) 2.468
TE mode is
(C) 1.564
(D) 4.389
æ 2 px ö
æ 3py ö
10
E y = 12 sin ç
÷ cos ç
÷ sin (10 p ´ 10 t - bz) V m
è a ø
è b ø
Statement for Q.17–19:
Consider the rectangular cavity as shown in fig.
11. The propagation constant g is
P8.7.17–19.
(A) j4094.2
(B) j400.7
(C) j2733.3
(D) j276.4
y
z
12. The intrinsic impedance is
(A) 743 W
(B) 168 W
(C) 986 W
(D) 144 W
b
x
Consider a air-filled waveguide operating in the
TE12 mode at a frequency 20% higher than the cutoff
frequency.
13. The phase velocity is
(A) 1.66 ´ 108 m s
(B) 5.42 ´ 108 m s
(C) 2.46 ´ 108 m s
(D) 9.43 ´ 108 m s
(A) 1.66 ´ 108 m s
(B) 4.42 ´ 108 m s
(C) 2.46 ´ 108 m s
(D) 9.43 ´ 108 m s
rectangular
waveguide
is
filled
17. If a < b < c, the dominant mode is
(A) TE011
(B) TM110
(C) TE101
(D) TM101
18. If a > b > c, then the dominant mode is
14. The group velocity is
A
0
Fig. P8.7.17–19
Statement for Q.13–14:
15.
a
c
(A) TE011
(B) TM110
(C) TE101
(D) TM101
19. If a = c > b, then the dominant mode is
with
(A) TE011
(B) TM110
(C) TE101
(D) TM101
a
polyethylene ( e r = 2.25) and operates at 24 GHz. The
20. The air filled cavity resonator has dimension a = 3
cutoff frequency of a certain mode is 16 GHz. The
cm, b = 2 cm, c = 4 cm. The resonant frequency for the
intrinsic impedance of this mode is
TM110 mode is
(A) 2248 W
(B) 337.2 W
(A) 5 GHz
(B) 6.4 GHz
(C) 421.4 W
(D) 632.2 W
(C) 16.2 GHz
(D) 9 GHz
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GATE EC BY RK Kanodia
UNIT 8
Electromagnetics
SOLUTIONS
frequency, the TM1 mode propagates through the guide
without suffering any reflective loss at the dielectric
interface. This frequency is
1. (A) fc =
er2 = 2.1
er1 = 4 Incident wave
1 cm
z
=
Fig. P8.7.34
(A) 8.6 GHz
(B) 12.8 GHz
(C) 4.3 GHz
(D) 7.5 GHz
2
2
2
2
v æmö
æ nö
ç ÷ +ç ÷
2 è aø
è bø
3 ´ 108
2 6 ´ 10 -2
æ1ö
æ1ö
ç ÷ + ç ÷ = 3.68 GHz
è2 ø
è 3ø
2
æf ö
æf ö
w
2. (C) bp = b 1 - çç c ÷÷ =
1 - çç c ÷÷
f
v
è f ø
è ø
Statement for Q.35–36:
2
2
A 6 cm ´ 4 cm rectangular wave guide is filled with
Þ
bp =
2 p ´ 20 ´ 10 9 6
æ 3.68 ö
1 -ç
÷ = 1009 rad m
3 ´ 108
è 20 ø
dielectric of refractive index 1.25.
35. The range of frequencies over which single mode
3. (A) vp =
w
2 p ´ 20 ´ 10 9
=
= 1.24 ´ 108 m s
bp
1009
operation will occur is
(A) 2.24 GHz < f < 3.33 GHz
4. (A) For TE10 mode fc =
(B) 2 GHz < f < 3 GHz
(C) 4.48 GHz < f <7.70 GHz
v
3 ´ 108
=
= 3 cm
2 fc 2 ´ 5 ´ 10 9
a=
(D) 4 GHz < f < 6 GHz
For TE01 mode fc =
36. The range of frequencies, over which guide support
b=
both TE10 and TE01 modes and no other, is
(A) 3.35 GHz < f < GHz
(B) 2.5 GHz < f < 3.6 GHz
c
æf ö
1 - çç c ÷÷
è f ø
(C) 3 GHz < f < 3.6 GHz
37. Two identical rectangular waveguide are joined end
t=
v
,
2b
v
3 ´ 108
=
= 1.25 cm
2 fc 2 ´ 12 ´ 10 9
5. (D) v =
(D) 2.5 GHz < f < 4.02 GHz
v
,
2a
2
3 ´ 108
=
æ 6.5 ö
1 -ç
÷
è 7.2 ø
2
= 6.975 ´ 108 ms
2l
2 ´ 150
=
= 430 ns
v 6.975 ´ 108
to end where a = 2 b. One guide is air filled and other is
filled with a lossless dielectric of e r . it is found that up
6. (C) 12 ´ 10 9 =
to a certain frequency single mode operation can be
simultaneously
ensured
in
both
guide.
For
this
frequency range, the maximum allowable value of e r is
(A) 4
(B) 2
(C) 1
(D) 6
3 ´ 108
æ 3ö
0 +ç ÷
2
è bø
2
12
. ´ 10 9 =
2
Þ
b = 375
. cm
2
3 ´ 108 æ 1 ö
1
æ
ö
ç ÷ +ç
÷ Þ a = 1.32 cm
2
. ´ 10 -2 ø
èaø
è 375
Since a < b, the dominant mode is TE01 .
38. A parallel-plate guide operates in the TEM mode
7. (B) fc 01 =
3 ´ 108
v
=
= 4 GHz
2 b 2 ´ 375
. ´ 10 -2
only over the frequency range 0 < f < 3 GHz. The
dielectric between the plates is teflon ( e r = 2.1). The
8. (C) Ez ¹ 0, this must be TM 23 mode ( m = 2, n = 3)
maximum allowable plate separation b is
(A) 3.4 cm
(B) 6.8 cm
(C) 4.3 cm
(D) 8.6 cm
*************
Page
514
2
2
fc =
3 ´ 108
2 ´ 10 -2
f =
w 1012
=
= 159.2 GHz
2p 2p
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æ2 ö
æ 3ö
ç ÷ + ç ÷ = 15.81 GHz
è6ø
è 3ø
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Waveguides
2
2
Chap 8.7
ho
377
æf ö
æ 15.81 ö
hTM = 377 1 - çç c ÷÷ = 377 1 - ç
÷ = 375.1 W
f
è 159.2 ø
è ø
h1 =
9. (B) m = 2, n = 1, bp = 12, f = 6 GHz
In dielectric medium
æf ö
w
1 - çç c ÷÷
bp =
v
è f ø
Þ
2
2 p ´ 6 ´ 10 9
æf ö
12 =
1 -ç c ÷
8
3 ´ 10
è6ø
Þ
2
fc =
h=
fc = 5.973 GHz
377
10. (B) hTE =
æf ö
1 - çç c ÷÷
è f ø
377
=
2
æ 5.973 ö
1 -ç
÷
è 6 ø
= 3978 W
2
11. (B) m = 2, n = 3,
2
fc =
2
2
æ 2 ö
æ 3 ö
çç
÷÷ + ç
÷
2
.
286
.
è 1016
ø
è
ø
3 ´ 108
c æmö
æ nö
ç ÷ +ç ÷ =
2 è aø
2 ´ 10 -2
è bø
2
æf ö
1 - çç c ÷÷
è f ø
2
er
=
3 ´ 108
377
2.25
= 251.33 W, h2 =
æf ö
1 - çç c ÷÷
è f ø
h2 - h1
259.23 - 406.7
=
= -0.22
259.23 + 406.7
h2 + h1
s=
1 + |G| 1 + 0.22
=
= 1564
.
1 -|G| 1 - 0.22
æ 46.2 ö
1 -ç
÷
è 50 ø
377
=
æ 46.2 ö
1 -ç
÷
è 50 ø
2
hTE =
h
æf ö
1 - çç c ÷÷
è f ø
2
æ f ö
1 - çç c ÷÷
è 1.2 f c ø
=
n = 1, 2, 3.....
= 986 W
2
p = 1, 2, 3...... ,
1
1 1
>
>
a
b c
if a < b < c, then
fr1 =
2
2
377
= 251.33 W
.
15
251.33
æ 16 ö
1 -ç
÷
è 24 ø
2
v æ1ö
æ1ö
ç ÷ +ç ÷
2 èaø
è bø
2
The lowest TE mode is TE011 with
= 337.2 W
16. (C) Since a > b, the dominant mode is TE10 .
In free space fc =
2
For TE mode to z
= 5. 42 ´ 108 m s
2
er
2
v æmö
æ nö
æ pö
ç ÷ +ç ÷ +ç ÷
2 è aø
b
è ø
ècø
The lowest TM mode is TM110 with
3 ´ 10
=
= 259.23 W
p = 0, 1, 2 ......
æ f ö
æf ö
14. (A) v g = v 1 - çç c ÷÷ = c 1 - çç c ÷÷ = 1.66 ´ 108 m s
. fc ø
è f ø
è 12
377
2
n = 1, 2, 3...... ,
2
fr 2 =
15. (B) h =
æ2 ö
1 -ç ÷
è8ø
m = 1, 2, 3.....,
8
=
2
251.33
G=
2
v
= 2 GHz
2 ´ 0.05 2.25
m = 1, 2, 3...... ,
13. (A) v = c, f = 1.2 fc
vp =
= 406.7 W
2
where for TM mode to z
2 p ´ 50 ´ 10 9
=
3 ´ 108
377
ho
æ 3ö
1 -ç ÷
è8ø
2
= 400.7 m -1 , g = jbp = j 400.7
12. (C) hTE =
2 a er
=
17. (A) fr =
10 p ´ 1010
= 50 GHz
2p
æf ö
w
1 - çç c ÷÷
bp =
v
è f ø
c
=
2
= 46.2 GHz
f =
æf ö
1 - çç c ÷÷
è f ø
2
c
3 ´ 108
=
= 3 GHz
2 a 2 ´ 0.05
v æ1ö
æ1ö
ç ÷ +ç ÷
2 è bø
è cø
2
fr 2 > fr1 , Hence the dominant mode is TE011
18. (B) If a > b > c then
1 1 1
< <
a b c
The lowest TM mode is TM110 with
2
fr1 =
v æ1ö
æ1ö
ç ÷ +ç ÷
2 èaø
è bø
2
The lowest TE mode is TE101 with
2
fr 2 =
v æ1ö
æ1ö
ç ÷ +ç ÷
2 èaø
è cø
fr 2 > fr1
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2
Hence the dominant mode is TM110 .
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UNIT 8
1 1 1
= <
a c b
19. (C) If a = c > b, then
The lowest TM mode is TM110 with
2
fr1 =
v æ1ö
æ1ö
ç ÷ +ç ÷
2 èaø
è bø
fr 2 =
v æ1ö
æ1ö
ç ÷ +ç ÷
2 èaø
è cø
2
2
2
v æmö
æ nö
æ pö
ç ÷ +ç ÷ +ç ÷
2 è aø
è bø
ècø
3 ´ 108 2
2
a
m<
mc
Þ
=
2 b er
f >
,
Þ
a = 10.6 cm
2 ´ 3 ´ 108
2 ´ 0.01 e r
mc
2 b er
Þ
15
. ´ 108
a
12
. ´ 15
. ´ 108
a
Þ
3 ´ 10 9 <
9
Þ
er = 9
m<
15
. ´ 108
,
b
0.8 ´ 15
. ´ 108
b
b < 4 cm, Thus (C) is correct option.
27. (C) fc =
= 10 ´ 10
Þ
c
3 ´ 108
=
= 2.3 GHz
2 a 2 ´ 0.065
c
vp =
æf ö
1 - çç c ÷÷
è f ø
2
=
3 ´ 108
æ 2.3 ö
1 -ç
÷
è 3 ø
2
= 4.7 ´ 108 m s
28. (B) For TE10 mode
2 fb e r
c
Rs =
m < 316
.
p ´ 9 ´ 10 9 ´ 4 p ´ 10 -7
= 0.0568
. ´ 10 7
11
1
pfm
=
=
sc d
sc
2
2
æ
æ
2 b æ fc ö ö÷
2 ´ 15
. æ 3.876 ö ö÷
Rs ç 1 +
çç ÷÷
0.0568ç 1 +
ç
÷
ç
ç
a è f ø ÷
2.4 è 9 ø ÷ø
ø=
è
ac = è
2
2
æf ö
æ 3.876 ö
15
. ´ 10 -2 ´ 233.8 1 - ç
bh 1 - çç c ÷÷
÷
è 9 ø
è f ø
= 0.022
,
fc 2 = 2 fc1 = 15 GHz
29. (B)
f =
c
3 ´ 108
=
= 20 GHz
l
0.015
vg2
æf ö
æ 15 ö
8
= c 1 - çç c ÷÷ = 3 ´ 108 1 - ç
÷ = 2 ´ 10 m s
è 20 ø
è f ø
2
2
s
2
Page
516
3 ´ 10 9 >
m 2 + k2 n2
a > 6 cm
Þ
Thus total 7 modes.
25. (A) fc =
Þ
3 GHz < 0.8 fc 01
The maximum allowed m is 3. The propagating mode
2 b er
2
c æmö
15
. ´ 108
æ nö
ç ÷ +ç ÷ =
2 è aø
a
è bø
The next higher mode is TE01 , fc 01 =
will be TM1 , TE1 , TM 2 , TE2 , TM 3 , TE3 and TEM
24. (B) fcm =
7.21
2
2 ´ 30 ´ 10 9 ´ 0.01 2.5
3 ´ 108
mc
6.15
3 GHz > 12
. fc
23. (A) For a propagating mode f > fcm ,
fcm =
6.8
2
2
2 b er
14.4
Dominant mode is TE10 , fc10 =
21. (A) m = n = 1, p = 0, a = b = c, fr = 2 Ghz,
22. (A) fc =
lc (cm)
2
v æmö
æ nö
æ pö
ç ÷ +ç ÷ +ç ÷
2 è aø
è bø
ècø
mc
TE20
fcmn =
2
2 ´ 10 9 =
TE11
26. (C) Let a = kb , 1 < k < 2
3 ´ 108 æ 1 ö
æ1ö
ç ÷ + ç ÷ = 9 GHz
2 ´ 0.01 è 3 ø
è2 ø
fr =
TE01
2
2
=
TE10
l > lc . Hence TE10 mode can be used.
fr2 < fr1 Hence the dominant mode is TE101 .
20. (D) fr =
Mode
2
The lowest TE mode is TE101 with
2
Electromagnetics
sd
10 -15
10 -15
=
=
we 2 p ´ 9 ´ 10 9 ´ 2.6 ´ 8.85 ´ 10 -12
1.3
sd
<< 1, hence v »
we
fc =
1
me
3 ´ 108
2 ´ 2.4 ´ 10 -2 2.6
=
c
2.6
, h»
377
2.6
= 233.8
= 3.876 GHz
2
2
c æmö
æ nö
ç ÷ + ç ÷ , lc =
2
2
2 è aø
b
è ø
æmö
æ nö
ç ÷ +ç ÷
è aø
è bø
ad =
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sdh
æf ö
1 - çç c ÷÷
è f ø
2
=
10 -15 ´ 233.8
æ 3.876 ö
2 1 -ç
÷
è 9 ø
2
= 1.3 ´ 10 -13 Np m
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Waveguides
30. (D) Dominant mode is TE10 mode
fc =
2
9
-7
= 1.429 ´ 10 -2 W
2
æ
2 b æ fc ö ö÷
2 ´ 3.4 æ 2.08 ö ö
-2 æ
Rs ç 1 +
çç ÷÷
ç
a è f ø ÷ 1.429 ´ 10 ç 1 + 7.2 çè 3 ÷ø ÷
ø=
è
ø
ac = è
2
2
æf ö
æ 2.08 ö
377 ´ 0.034 1 - ç
bh 1 - çç c ÷÷
÷
è 3 ø
è f ø
e- a c z
1
=
2
-3
Np m
Þ
z=
1
ln 2 = 308 m
ac
2
2
2
c æmö
3 ´ 108 æ m ö
æ nö
æ n ö
ç ÷ +ç ÷ =
ç ÷ +ç
÷
2 è aø
2 ´ 0.01 è 8 ø
è bø
è 10 ø
2
34. (B) The ray angle is such that the wave is interface
fc1 =
f =
c
2
=
2 b e r1
3 ´ 1010
= 7.5 GHz
2 ´ 1´ 2
fc1
7.5
=
= 12.8 GHz
cos q cos 54.1°
fc10 =
c
a2 e r
fc 01 =
2
æ n ö
æmö
= 15 ç ÷ + ç
÷ GHz
è 8ø
è 10 ø
c
b2 e r
2
c
2 er
æmö
æ nö
ç ÷ +ç ÷
è aø
è bø
3 ´ 108
= 2 GHz
. ´ 0.06
2 ´ 125
=
3 ´ 108
= 3 GHz
. ´ 0.04
2 ´ 125
2 GHz < f < 3 GHz
fc 01 = 15
. GHz,
fc11 = 2.4 GHz
fc 20 = 375
. GHz ,
fc 02 = 3 GHz,
fc 21 = 4.04 GHz,
fc12 = 354
. GHz,
36. (C) fc11 =
2
3 ´ 108
. ´ 10 -2
2 ´ 125
If fc < f , then mode will be transmit. Hence six mode
37. (A) fc =
will be transmitted.
2
c
2 er
2
æmö
æ nö
ç
÷ + ç ÷ , In guide 1 e r = 1
è 2b ø
è bø
lowest cutoff frequency fc10 =
32. (C) For dominant mode (m = 1, n = 0)
Since given frequency is below the cutoff frequency, 3
GHz will not be propagated and get attenuated
2
æ mp ö
æ np ö æ w ö
g = a + jb = ç
÷ +ç
÷ -ç ÷
è a ø
è b ø ènø
2
æ mp ö æ w ö
æ p ö æ 2 p ´ 3 ´ 10
a= ç
÷ -ç ÷ = ç
÷ - çç
8
a
c
è
ø è ø
è 0.04 ø è 3 ´ 10
2
33. (B) fc =
fc10 =
c æmö
æ nö
ç ÷ +ç ÷
2 è aø
è bø
c
2b
In guide 2 lowest cutoff frequency fc¢10 =
Next lowest cutoff frequency fc¢20 =
2
c
2 e r 2 (2 b)
c
2 e r 2 ( b)
For single mode fc¢10 < f < fc¢10
b = 0, Since wave is attenuated,
2
c
2(2 b)
Next lowest cutoff frequency fc 20 =
2
c æmö
3 ´ 108
æ nö
GHz
.
fc =
= 375
ç ÷ +ç ÷ =
2 è aø
2 ´ 0.04
è bø
2
2
æ1ö
æ1ö
ç ÷ + ç ÷ = 3.6 GHz
6
è4ø
è ø
3 GHz < f < 3.6 GHz
fc 30 = 5.625 GHz , fc 03 = 4.5 GHz
2
2
=
fc10 = 1.875 GHz
2
2.1
= 35.9 °.
4
The ray angle q = 90 °-35.9 ° = 54.1°
35. (A) fc =
31. (B)
fcmn =
2
æ 2p ´ 1.5 ´ 1.875 ö
æ p ö
÷
ç
÷ -ç
3 ´ 108
è
ø
è 0.08 ø
at Brewster’s angle q B = tan -1
For TE10 mode
= 2.25 ´ 10
2
= j439
.
p ´ 3 ´ 10 ´ 4 p ´ 10
5.8 ´ 10 7
pfm o
=
sc
2
æ mp ö
æ np ö
æwö
ç
÷ + ç
÷ -ç ÷ =
è a ø
è b ø
ènø
g=
c
3 ´ 108
=
= 2.08 GHz
2 a 2 ´ 0.072
Rs =
Chap 8.7
Þ
9
2
ö
÷÷ = 47.1
ø
c
c
<f <
2(2 b)
2 e r ( b)
38. (A) f < fc
Þ
f <
2
Þ
3 ´ 10 9 <
3 ´ 108
c
=
= 1.875 GHz
2 a 2 ´ 0.08
Þ
er < 4
v
3 ´ 108
=
2 b 2 ´ b ´ 2.1
3 ´ 108
2 ´ b ´ 2.1
Þ
b < 3.4 cm
*******************
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CHAPTER
8.8
ANTENNAS
1. A Hertizian dipole at the origin in free space has
5. The time-average poynting vector at 50 km is
dl = 10 cm and I = 20 cos (2 p ´ 10 t) A. The | E| at the
(A) 6.36u r mW m 2
(B) 4.78u r mW m 2
distant point (100, 0, 0) is
(C) 9.55u r mW m 2
(D) 12.73u r mW m 2
7
(A) 0.252 V m
(B) 0.126 V m
(C) 0.04 V m
(D) 0.08 V m
6. The maximum electric field at that location is
Statement for Q.2–3:
A 25 A source operating at 300 MHz feeds a
Hertizian dipole of length 4 mm situated at the origin.
Consider the point P(10, 30°, 90°).
2. The H at point P is
(A) j0.25 mA m
(B) 94.25 mA m
(C) j0.5 mA m
(D) 188.5 mA m
(A) 24 mV m
(B) 85 mV m
(C) 109 mV m
(D) 12 mV m
7. In free space, an antenna has a far-zone field given
by E =
1
r
10 sin 2 q e - jbr u q V m. The radiated power is
(A) 0.23 W
(B) 0.89 W
(C) 1.68 W
(D) 1.23 W
8.
At
Pave =
the
far
field,
an
antenna
cos q cos f u r W m , where
1
r2
p
2
2
0<q<p
and
0 < f < . The directive gain of the antenna is
3. The E at point P is
(A) j0.25 mV m
(B) j0.5 mV m
(A) cos q cos f
(B) 2 sin q cos f
(C) j94.25 mV m
(D) j188.5 mV m
(C) 8 cos q sin f
(D) 8 sin q cos f
4. An antenna can be modeled as an electric dipole of
Statement for Q.9–10:
length 4 m at 3 MHz. If current is uniform over its
length, then radiation resistance of the antenna is
(A) 1.974 W
(B) 1.263 W
(C) 2.186 W
(D) 2.693 W
Statement for Q.5–6:
A antenna located on the surface of a flat earth
Page
518
produces
The radiation intensity of antennas has been
given. Determine the directivity of antenna.
9. U( q, f) = sin 2 q, 0 < q < p ,
0 < f < 2p
(A) 1.875
(B) 2.468
(C) 3.943
(D) 6.743
transmit an average power of 150 kW. Assume that all
10. U( q, f) = 4 sin 2 q sin 2 f ,
the power is radiated uniformly over the surface of
(A) 15
(B) 12
hemisphere with the antenna at the center.
(C) 3
(D) 6
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0 < q < p,
0 <f<p
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Antennas
11. The radiation intensity of a antenna is given by
U( q, f) = 8 sin q cos f , where 0 < q < p and 0 < f < p. The
2
2
directive gain is
(A) 6 sin 2 q cos 2 f
(C) 3 sin 2 f cos 2 q
(B) 3 sin 2 q cos 2 f
(D) 6 sin 2 f cos 2 q
Chap 8.8
19. Two identical antenna separated by 12 m are
oriented for maximum directive gain. At a frequency of
5 GHz, the power received by one is 30 dB down from
the transmitted by the other. The gain of antenna is
(A) 22 dB
(C) 19 dB
Statement for Q.12–13:
(B) 16 dB
(D) 13 dB
Statement for Q.20–21:
At the far field, an antenna radiates a field
12. The total radiated power is
An L-band pulse radar has common transmitting
and receiving antenna. The antenna having directive
gain of 36 dB operates at 1.5 GHz and transmits 200
kW. The object is 120 km from the radar and its
scattering cross section is 8 m 2 .
(A) 1.36 W
(C) 0.844 W
20. The magnitude of the incident electric field
0.4 cos 2 q - jbr
Ef =
e
kV m
4 pr
(B) 2.14 W
(D) 3.38 W
intensity of the object is
13. The directive gain at q = p 3 is
(A) 0.3125
(C) 1.963
14. An antenna has directivity of 100 and operates at
150 MHz. The maximum effective aperture is
(A) 31.8 m 2
(C) 26.4 m 2
21. The magnitude of the scattered electric field at the
radar is
(A) 18 mW
(C) 17 mW
(B) 62.4 m 2
(D) 13.2 m 2
15. Two half wave dipole antenna are operated at 100
MHz and separated by 1 km. If 100 W is transmitted by
one, the power received by the other is (D = 1.68)
(A) 12 mW
(C) 18 mW
(B) 2.46 V m
(D) 0.17 V m
(A) 1.82 V m
(C) 0.34 V m
(B) 0.625
(D) 3.927
(B) 12 mW
(D) 126 mW
22. A transmitting antenna with a 300 MHz carrier
frequency produces 2 kW of power. If both antennas has
unity power gain, the power received by another
antenna at a distance of 1 km is
(A) 11.8 mW
(C) 18.4 mW
(B) 10 mW
(D) 16 mW
16. The electric field strength impressed on a half wave
(B) 18.4 mW
(D) 12.7 mW
23. A bistatic radar system shown in fig. P8.7.23 has
dipole is 6 mV m at 60 MHz. The maximum power
following parameters: f = 5 GHz,
received by the antenna is (D = 1.68)
22 dB. To obtain a return power of 8 pW the minimum
(A) 159 nW
(C) 196 mW
necessary radiated power is
(B) 230 nW
(D) 318 mW
Gdt = 34 dB,
Gdr =
Target s = 2.4m2
Scattered wave
17. The power transmitted by a synchronous orbit
Incident
wave
satellite antenna is 480 W. The antenna has a gain of
4 km
40 dB at 15 GHz. The earth station is located at
distance of 24, 567 km. If the antenna of earth station
has a gain of 32 dB, the power received is
(A) 32 pW
(C) 10.2 pW
Receiving
antenna
Transmitting
antenna
3 km
Fig. P8.7.23
(B) 3.2 fW
(D) 1.3 fW
(A) 1.394 kW
(C) 1.038 kW
18. The directive gain of an antenna is 36 dB. If the
(B) 2.046 kW
(D) 3.46 kW
time average power density at that distance is
24. The radiation resistance of an antenna is 63 W and
loss resistance 7 W. If antenna has power gain of 16,
then directivity is
(A) 9.42 mW m 2
(C) 1.32 mW m 2
(A) 48.26 dB
(C) 38.96 dB
antenna radiates 15 kW at a distance of 60 km, the
(B) 6.83 mW m 2
(D) 10.46 mW m 2
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(B) 12.5 dB
(D) 24.7 dB
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GATE EC BY RK Kanodia
UNIT 8
Electromagnetics
25. An antenna is desired to operate on a frequency of
SOLUTIONS
40 MHz whose quality factor is 50. The bandwidth of
antenna is
(A) 5.03 MHz
(C) 127 kHz
(B) 800 kHz
(D) None of the above
26. A thin dipole antenna is l 15 long. If its loss
(B) 59%
(C) 74.5%
(D) 25.5%
Statement for Q.27–29:
An array comprises of two dipoles that are
separated by the wavelength. The dipoles are fed by
currents of the same magnitude and phase.
27. The array factor is
(A) 2 cos ( p cos q + 45 ° )
(C) 2 cos ( p sin q + 45 ° )
(B) 2 cos ( p sin q )
(D) 2 cos ( p cos q )
(B) 60°, 120°
(C) 45°, 135°
(D) 0, 180°
At far field | Eq| =
2. (A) b =
Hf =
29. The maximum of the pattern occur at
(A) q = 45 ° , 135 °
(B) q = 0, 90 ° , 180 °
(C) q = 30 ° , 150 °
(D) q = 60 ° , 150 °
)
(C) cos ( 2p cos q + 2p )
(B) cos ( 4p cos q +
)
(D) sin ( 2p cos q + 2p )
p
2
(B) 4 cos ( 4p cos q) cos (8p cos q)
(C) 4 cos ( 4p cos q) sin ( 2p cos q)
(D) 4 cos ( 4p cos q) sin (8p cos q)
***********
Page
520
c 3 ´ 108
=
= 100
f 3 ´ 106
dl
4
1
1
=
=
<
l 100 25 10
80 p2
æ dl ö
Rrad = 80 p2 ç ÷ =
= 1263
.
W
625
è lø
5. (C) Prad = ò Pave × dS = Pave × 2 pr 2
Pave =
Prad
150 ´ 10 3
=
= 9.55 mW m 2
2 pr 2 2 p(50 ´ 10 3) 2
P = 9.55 u r mW m 2
31. An antenna consists of 4 identical Hertizian dipoles
uniformly located along the z–axis and polarized in the
z-direction. The spacing between the dipole is l / 4 . The
group pattern function is
(A) 4 cos ( 4p cos q) cos ( 2p cos q)
j(2.5)(2 p) ( 4 ´ 10 -3) - j 2 p10
e
= j0.25 mA m
4 p(10)
2
30. An array comprises two dipoles that are separated
by half wavelength. If the dipoles are fed by currents,
that are 180° out of phase with each other, then array
factor is
p
4
w 2 p ´ 300 ´ 106
=
= 2p
c
3 ´ 108
3. (C) E = Eq = hH f = 377 H f = j94.25 mV m
4. (B) l =
(A) sin ( 4p cos q +
hI obdl
sin q
4 pr
r = 10 m, q = 30 ° , f = 90 °
jI bdl
At far field H = H f = o
sin q e - jbr
4 pr
Þ
28. The nulls of the pattern occur when q is
(A) 30°, 150°
w 2 p ´ 10 7 2 p
=
=
c
3 ´ 108
30
h = 120 p = 377, I o = 20, dl = 10 cm
p
At (100 cm, 0, 0), q =
2
120 p ´ 20 ´ 0.1 2 p
´
= 0.126 V m
| Eq| =
4 p ´ 100
30
resistance is 1.2 W, the efficiency is
(A) 41.1%
1. (B) b =
6. (B) Pave =
( Emax ) 2
2h
Þ
Emax = 2hPave
Emax = 2 ´ 377 ´ 9.55 ´ 10 -6 = 85 mV m
Þ
7. (B) Pave =
Prad = òò
|E|2
2h
100 sin 2 2 q
sin q dq df
2h
p
=
100
(2 p)(2 sin q cos q) 2 sin q dq
2 ´ 120 p ò0
=
10
sin 3 q cos 2 qdq = 0.89 W
3 ò0
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UNIT 8
Gd =
2p
l = 2p , a = 0
l
æ bd cos q + a ö
AF = 2 cos ç
÷ = 2 cos ( p cos q)= 2 cos ( p cos q)
2
è
ø
4 p(12) 1
= 79.48 = 19 dB
0.06
10 3
27. (D) bd =
20. (A) Gd = 36 dB = 3981
Gd =
4 pr 2 Pi
Prad
Þ
Pi =
Gd Prad |E|2
=
4 pr 2
2h
28. (B) cos ( p cos q) = 0
(240 p) Gd Prad
Þ |E|=
4 pr 2
( 60)( 3981)(200 ´ 10 3)
(120 ´ 10 3) 2
=
= 1.82 V m
21. (B) |Es|=
1.82
=
120 ´ 10 3
22. (D) l =
|Er|2 s |Ei| s
=
4 pr 2
4p
r
8
4p
2
æ l ö
æ 1 ö
Pr = Gdr Gdt ç
2000 = 12.7 mW
÷ P = (1)(1) ç
3÷
4
p
r
è
ø
è 4 p10 ø
Gdr Gdt æ l
ç
4 p çè 4 pr1 r2
p
3p
, ±
2
2
cos q = ±
1
2
q = 60 ° , 120 °
Þ
29. (B) Maxima occur when
d( AF )
=0
dq
sin ( p cos q) p sin q = 0
q = 0, 90 ° , 180 °
Þ
31. (A) ( AF ) N
æ Nf ö
sinç
÷
è 2 ø
=
æyö
sinç ÷
è2ø
y = bd cos q + a , N = 4
sin 4 x 2 sin 2 x cos 2 x
=
= 4 cos x cos 2 x
sin x
sin x
2p l p
y p
bd =
= , a = 0,
= cos q
l 4 2
2 4
æp
ö
æp
ö
AF = 4 cos ç cos q ÷ cos ç cos q ÷ .
4
2
è
ø
è
ø
2
ö
÷÷ sPrad
ø
Gdt = 34 dB = 2512, Gdr = 22 dB = 158.5
r1 = 3 km, r2 = 32 + 4 2 = 5 km
l=
p cos q = ±
2p l
=p , a =p
l 2
pö
æ bd cos q + a ö
æp
AF = 2 cos ç
÷ = cos ç cos q + ÷
2
2ø
è
ø
è4
c
3 ´ 108
=
=1 m
f 300 ´ 106
23. (C) Pr =
Þ
30. (B) bd =
= 12 mW
2
Electromagnetics
3 ´ 108
= 0.06 m, Pr = 8 pW
5 ´ 10 9
2
8 ´ 10
Þ
-12
0.06
(2512)(158.5) æ
ö
=
ç
÷ (2.4) Prad
4p
4
p
(
3
k
)(
5
k
)
è
ø
kW
Prad = 1038
.
24. (B) Efficiency =
D=
************
63
= 0.9
63 + 7
Gain
16
=
= 17.78 = 12.5 dB
Efficiency 0.9
25. (B) BW =
f 40 ´ 106
=
= 800 kHz
Q
50
æ dl ö
26. (C) Radiation resistance Rr = 80 p2 ç ÷
è l ø
2
2
æ 1 ö
= 80 ´ p2 ´ ç
. W
÷ = 351
è 15 ø
Efficiency =
Page
522
351
.
Rr
=
= 74.5 %
. + 12
.
Rr + RL 351
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CHAPTER
9.1
LINEAR ALGEBRA
é 0
1. If A = ê - 1
ê
êë 2
1
0
-2
-2 ù
3 ú is a singular matrix, then l is
ú
l úû
(A) 0
(B) -2
(C) 2
(D) -1
7. Every diagonal elements of a Hermitian matrix is
(A) Purely real
(B) 0
(C) Purely imaginary
(D) 1
8. Every diagonal element of a Skew–Hermitian matrix
is
2. If A and B are square matrices of order 4 ´ 4 such
(A) Purely real
(B) 0
that A = 5B and A = a × B , then a is
(C) Purely imaginary
(D) 1
(A) 5
(B) 25
(C) 625
(D) None of these
9. If A is Hermitian, then iA is
3. If A and B are square matrices of the same order
such that AB = A and BA = A , then A and B are both
(A) Symmetric
(B) Skew–symmetric
(C) Hermitian
(D) Skew–Hermitian
(A) Singular
(B) Idempotent
10. If A is Skew–Hermitian, then iA is
(C) Involutory
(D) None of these
(A) Symmetric
(B) Skew–symmetric
(C) Hermitian
(D) Skew–Hermitian.
é-5 -8 0 ù
4. The matrix, A = ê 3 5 0 ú is
ê
ú
êë 1 2 -1ûú
(A) Idempotent
(C) Singular
é- 1 - 2
11. If A = ê 2
1
ê
êë 2 - 2
(B) Involutory
(D) None of these
5. Every diagonal element of a skew–symmetric matrix
-2ù
- 2 ú, then adj. A is equal to
ú
1úû
(A) A
(B) c t
(C) 3A t
(D) 3A
is
(A) 1
(C) Purely real
é
6. The matrix, A = ê
êë-
(B) 0
(D) None of these
1
2
i
2
i
2
-
1
2
ù
ú is
úû
(A) Orthogonal
(B) Idempotent
(C) Unitary
(D) None of these
é-1 2 ù
12. The inverse of the matrix ê
úis
ë 3 -5 û
é5
(A) ê
ë3
2ù
1 úû
é-5 -2 ù
(C) ê
ú
ë -3 -1û
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é5
(B) ê
ë2
3ù
1 úû
(D) None of these
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UNIT 9
é1
13. Let A = ê5
ê
êë3
0
é 4 0
1
(A) ê-10 2
4ê
êë -1 -1
0ù
0ú
ú
2 úû
é 1
(C) ê-10
ê
êë -1
0
2
-1
0ù
0 ú, then A -1 is equal to
ú
2 úû
2
1
0
é 2
1
(B) ê-5
1
2ê
êë -1 -1
0ù
0ú
ú
2 úû
Engineering Mathematics
19. The system equationsx + y + z = 6, x + 2 y + 3z = 10,
x + 2 y + lz = 12 is inconsistent, if l is
0ù
0ú
ú
2 úû
(A) 3
(B) -3
(C) 0
(D) None of these.
20.
The
system
of
equations
5 x + 3 y + 7 z = 4,
3 x + 26 y + 2 z = 9, 7 x + 2 y + 10 z = 5 has
(A) a unique solution
(D) None of these
(B) no solution
(C) an infinite number of solutions
é 2 -1
14. If the rank of the matrix, A = ê 4 7
ê
êë 1 4
3ù
lú is 2, then
ú
5 úû
the value of l is
(D) none of these
21. If A is an n–row square matrix of rank (n - 1), then
(A) adj A = 0
(B) adj A ¹ 0
(D) None of these
(A) -13
(B) 13
(C) adj A = I n
(C) 3
(D) None of these
22.
The
system
of
equations
x - 4 y + 7 z = 14,
15. Let A and B be non–singular square matrices of the
3 x + 8 y - 2 z = 13, 7 x - 8 y + 26 z = 5 has
same order. Consider the following statements.
(A) a unique solution
(I) ( AB) T = A TBT
(II) ( AB) -1 = B-1 A -1
(B) no solution
(III) adj( AB) = (adj. A)(adj. B)
(IV) r( AB) = r( A)r(B)
(C) an infinite number of solution
(D) none of these
(V) AB = A × B
(A) I, III & IV
(B) IV & V
4ù
é3
23. The eigen values of A = ê
ú are
9
5
ë
û
(C) I & II
(D) All the above
(A) ± 1
(B) 1, 1
(C) -1, - 1
(D) None of these
Which of the above statements are false ?
1 -1ù
3 -2 ú is
ú
4 -3úû
é 2
16. The rank of the matrix A = ê 0
ê
êë 2
(A) 3
(B) 2
2ù
é 8 -6
ê
24. The eigen values of A = -6
7 - 4 ú are
ê
ú
3 úû
êë 2 - 4
(C) 1
(D) None of these
(A) 0, 3, -15
(B) 0, - 3 , - 15
(C) 0, 3, 15
(D) 0, - 3, 15
17.
The
system
15 x - 6 y + 5 z = 0,
of
equations
lx - 2 y + 2 z = 0
has
3 x - y + z = 0,
a
non–zero
solution, if l is
(A) 6
(B) -6
(C) 2
(D) -2
18.
The
system
of
then the eigen values of the matrix 2A are
1
3
(B) 2 , - 4 , 6
(A) , - 1 ,
2
2
(C) 1 , - 2, 3
equation
x - 2 y + z = 0,
2 x - y + 3z = 0, lx + y - z = 0 has the trivial solution as
the only solution, if l is
(A) l ¹ (C) l ¹ 2
Page
526
4
5
(B) l =
25. If the eigen values of a square matrix be 1, - 2 and 3,
4
3
(D) None of these
(D) None of these.
26. If A is a non–singular matrix and the eigen values
of A are 2 , 3 , - 3 then the eigen values of A -1 are
1 1 -1
(A) 2 , 3 , - 3
(B) , ,
2 3 3
(C) 2 A , 3 A , - 3 A
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(D) None of these
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Linear Algebra
é cos 2 f
cos f sin fù
B=ê
ú
cos
f
sin
f
sin 2 f û
ë
27. If -1 , 2 , 3 are the eigen values of a square matrix A
then the eigen values of A 2 are
(A) -1 , 2 , 3
(B) 1, 4, 9
(C) 1, 2, 3
(D) None of these
is a null matrix, then q and f differ by
(A) an odd multiple of p
28. If 2 , - 4 are the eigen values of a non–singular
matrix A and A = 4, then the eigen values of adj A are
(A)
1
2
, -1
(C) 2 , - 4
,
1
4
(C) 4, 16
(C) an odd multiple of
p
2
p
2
(D) an even multiple
(D) 8 , - 16
35. If A and B are two matrices such that A + B and AB
eigenvalues of A T are
1
2
(B) an even multiple of p
(B) 2 , - 1
29. If 2 and 4 are the eigen values of A then the
(A)
Chap 9.1
are both defined, then A and B are
(A) both null matrices
(B) 2, 4
(B) both identity matrices
(D) None of these
(C) both square matrices of the same order
(D) None of these
30. If 1 and 3 are the eigenvalues of a square matrix A
é 0
36. If A = ê
ëtan
then A 3 is equal to
(A) 13( A - I 2 )
(B) 13A - 12I 2
(C) 12( A - I 2 )
(D) None of these
31. If A is a square matrix of order 3 and A = 2 then
A (adj A) is equal to
é2
(A) ê0
ê
êë0
0
é1
(C) ê0
ê
êë0
0
2
0
1
0
0ù
ú
0ú
1 ú
2 û
0ù
0ú
ú
2 úû
é 12
ê
(B) ê0
êë0
0ù
0ú
ú
1 úû
(D) None of these
0
1
2
0
é8
32. The sum of the eigenvalues of A = ê4
ê
êë2
écos a - sin a2 ù
then (I - A ) × ê
ú is equal to
cos a û
ësin a
(A) I + A
(B) I - A
(C) I + 2 A
(D) I - 2 A
- 4ù
, then for every positive integer
- 1 úû
é3
37. If A = ê
ë1
2
5
0
- tan a2 ù
ú
0
û
a
2
n, A n is equal to
3ù
9 ú is
ú
5 úû
é1 + 2 n
(A) ê
ë n
4n ù
1 + 2 núû
é1 + 2 n
(B) ê
ë n
é1 - 2 n
(C) ê
ë n
4n ù
1 + 2 núû
(D) None of these
(A) 18
(B) 15
é cos a
38. If A a = ê
ë- sin a
(C) 10
(D) None of these
statements :
equal to
sin a ù
, then consider the following
cos a úû
I. A a × A b = A ab
33. If 1, 2 and 5 are the eigen values of the matrix A
then A is equal to
(A) 8
(B) 10
(C) 9
(D) None of these
34. If the product of matrices
écos q
cos q sin q ù
A =ê
ú and
2
ëcos q sin q sin q û
2
- 4n ù
1 - 2 núû
II. A a × A b = A ( a + b)
é cos n a
III. ( A a ) n = ê
n
ë- sin a
sin n a ù
ú
cos n a û
é cos na
IV. ( A a ) n = ê
ë- sin na
sin na ù
cos na úû
Which of the above statements are true ?
(A) I and II
(B) I and IV
(C) II and III
(D) II and IV
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UNIT 9
39. If A is a 3-rowed square matrix such that A = 3,
then adj (adj A) is equal to :
(A) 3A
(B) 9A
(C) 27A
(D) none of these
40. If A is a 3-rowed square matrix, then adj (adj A) is
Engineering Mathematics
é1 2 0 ù
T
45. If A = ê
ú, then AA is
ë3 -1 4 û
é 1 3ù
(A) ê
ú
ë-1 4 û
é 1 0 1ù
(B) ê
ú
ë-1 2 3û
1ù
é2
(C) ê
ú
ë1 26 û
(D) Undefined
equal to
(A) A
6
(B) A
3
46. The matrix, that has an inverse is
(C) A
4
(D) A
2
é 3 1ù
(A) ê
ú
ë6 2 û
é5 2 ù
(B) ê
ú
ë2 1 û
é6 2 ù
(C) ê
ú
ë9 3û
é8 2 ù
(D) ê
ú
ë4 1 û
41. If A is a 3-rowed square matrix such that A = 2,
then adj (adj A 2 ) is equal to
(A) 2 4
(B) 28
47. The skew symmetric matrix is
(C) 216
(D) None of these
é 0 -2 5 ù
(A) ê 2
0 6ú
ê
ú
êë-5 -6 0 úû
é1 5 2ù
(B) ê6 3 1ú
ê
ú
êë2 4 0 úû
é0 1 3ù
(C) ê1 0 5 ú
ê
ú
êë3 5 0 úû
é0 3 3ù
(D) ê2 0 2 ú
ê
ú
êë1 1 0 úû
é2 x
42. If A = ê
ë x
0ù
é 1
and A -1 = ê
x úû
ë-1
0ù
, then the value
2 úû
of x is
(A) 1
(C)
1
2
(B) 2
(D) None of these
é1 ù
é1 1 0 ù
ê ú
48. If A = ê
ú and B = ê0 ú, the product of A and B
ë1 0 1û
êë1 úû
é1 2 ù
43. If A = ê2 1ú then A -1 is
ê
ú
êë1 1úû
is
é1 4 ù
(A) ê3 2 ú
ê
ú
êë2 5 úû
é 1 -2 ù
(B) ê-2
1ú
ê
ú
êë 1 2 úû
é2 3ù
(C) ê3 1ú
ê
ú
êë2 7 úû
(D) Undefined
é-1 -8 -10 ù
(C) ê 1 -2
-5 ú
ê
ú
15 úû
êë 9 22
Page
528
é1 0 ù
(B) ê
ú
ë0 1 û
é1 ù
(C) ê ú
ë2 û
é1 0 ù
(D) ê
ú
ë0 2 û
éA
49. Matrix D is an orthogonal matrix D = ê
ëC
é 2 -1ù
é1 -2 -5 ù
44. If A = ê 1 0 ú and B = ê
then AB is
ê
ú
0 úû
ë3 4
3
4
êë
úû
é-1 -8 -10 ù
(A) ê-1 -2
5ú
ê
ú
15 úû
êë 9 22
é1 ù
(A) ê ú
ë0 û
0 -10 ù
é 0
ê
(B) -1 -2
-5 ú
ê
ú
êë 0 21 -15 úû
é0 -8 -10 ù
(D) ê1 -2
-5 ú
ê
ú
êë9 21 15 úû
value of B is
1
(A)
2
(C) 1
(B)
Bù
. The
0 úû
1
2
(D) 0
50. If A n ´n is a triangular matrix then det A is
n
(A)
Õ ( -1) a
i =1
n
ii
(B)
ii
(D)
i =1
n
(C)
å ( -1) a
i =1
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ii
n
åa
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ii
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Linear Algebra
ét 2 cos t ù
dA
51. If A = ê t
ú, then dt will be
sin
e
t
ë
û
Chap 9.1
SOLUTIONS
ét 2 sin t ù
(A)ê t
ú
ë e sin t û
é2t cos t ù
(B) ê t
ú
ë e sin t û
1. (B) A is singular if A = 0
é2t - sin t ù
(C) ê t
cos t úû
ëe
(D) Undefined
é 0
Þ ê -1
ê
êë 2
1
0
-2
-2 ù
3 ú =0
ú
l úû
-2 ½
½ + 2½
½1
l½ ½ 0
½ 1
- ( -1)½
½-2
- 2½
½
½ 0
+ 0½
3½ ½ - 2
3½
½= 0
l½
52. If A Î R n ´n , det A ¹ 0, then
Þ
(A) A is non singular and the rows and columns of A
are linearly independent.
Þ ( l - 4) + 2( 3) = 0
(B) A is non singular and the rows A are linearly
dependent.
2. (C) If k is a constant and A is a square matrix of
(C) A is non singular and the A has one zero rows.
order n ´ n then kA = kn A .
(D) A is singular.
A = 5B Þ A = 5B = 5 4 B = 625 B
Þ l - 4 + 6 =0 Þ l = -2
Þ a = 625
************
3. (B) A is singular, if A = 0,
A is Idempotent, if A 2 = A
A is Involutory, if A 2 = I
Now, A 2 = AA = ( AB) A = A( BA) = AB = A
and B2 = BB = (BA)B = B( AB) = BA = B
Þ
A 2 = A and B2 = B,
Thus A & B both are Idempotent.
é-5 -8
4. (B) Since, A = ê 3 5
ê
êë 1 2
0 ù é-5 -8
0úê 3 5
úê
-1úû êë 1 2
2
é1 0 0 ù
= ê0 1 0 ú = I,
ê
ú
êë0 0 1úû
A2 = I
Þ
0ù
0ú
ú
-1úû
A is involutory.
5. (B) Let A = [ aij ] be a skew–symmetric matrix, then
AT = - A ,
Þ
aij = - aij ,
if i = j then aii = - aii
Þ
2 aii = 0
Þ
aii = 0
Thus diagonal elements are zero.
6. (C) A is orthogonal if AA T = I
A is unitary if AA Q = I , where A Q is the conjugate
transpose of A i.e., A Q = ( A) T .
Here,
é
ê
AA Q = ê
êêë
1
2
i
2
ùé
2 úê
úê
1 úê
2 ûú ëê
i
1
2
i
2
ù
2 ú é1
ú=
1 ú êë0
2 ûú
i
0ù
= I2
1úû
Thus A is unitary.
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UNIT 9
7. (A) A square matrix A is said to be Hermitian if
A = A. So aij = a ji . If i = j then aii = aii i.e. conjugate of
Q
an element is the element itself and aii is purely real.
8. (C) A square matrix A is said to be Skew-Hermitian
if A = - A.
If A is Skew–Hermitian then A = - A
Q
Þ
Q
a ji = - aij ,
if i = j then aii = - aii Þ
Now, ( iA) Q = i A Q = - iA Q = - iA, Þ
( iA ) Q = - ( iA )
Thus iA is Skew–Hermitian.
Now, ( iA) Q = i A Q = - ( -A ) = iA then iA is Hermitian.
11. (C) If A = [ aij ]n ´n then det A = [ cij ]nT ´n
Where cij is the cofactor of aij
Also cij = ( -1) i + j M ij , where M ij is the minor of aij ,
the
row
and
the
the remaining matrix.
½ 1
Now, M11 = minor of a11 i.e. -1 =½
½-2
-2 ½
½
1½
= -3
Similarly
1½
½
½2 - 2 ½
½2
½= 6 ; M13 =½
=-6
M12 = ½
1½
½2 - 2 ½
½2
-2½
½
½-1
= - 6 ; M 22 =½
1½
½ 2
½-1
M 23 =½
½ 2
- 2½
½= 6 ; M 31
- 2½
½-2
=½
½ 1
½-1 - 2 ½
½= 6 ; M 33
M 32 =½
½ 2 - 2½
-2 ½
½
=3 ;
1½
- 2½
½= 6 ;
-2½
½-1 - 2
=½
½ 2
½
½= 3
1½
C11 = ( -1)1 + 1 M11 = -3; C12 = ( -1)1 + 2 M12 = -6 ;
1+ 3
C13 = ( -1)
C22 = ( -1)
2+ 2
M13 = -6; C21 = ( -1)
M 22 = 3; C23 = ( -1)
2+1
2+ 3
M 21 = 6;
M 23 = -6;
C31 = ( -1) 3+ 1 M 31 = 6; C32 = ( -1) 3+ 2 M 32 = -6 ;
C33 = ( -1) 3+ 3 M 33 = 3
éC11
det A = êC21
ê
êëC31
C12
C22
C32
T
C13 ù
C23 ú
ú
C33 úû
T
é-5
ê-3
ë
Þ
é-5
adj A = ê
ë-3
3
1
2
2ù
1 úû
T
10 -10 ù
é 4 0
ú
2
-1 = ê10 2
ú
ê
0
2 úû
êë-1 -1
4
0
0
é 4 0
1ê
10 2
4ê
êë-1 -1
0ù
0ú
ú
2 úû
0ù
0ú
ú
2 úû
14. (B) A matrix A ( m ´n ) is said to be of rank r if
(i) it has at least one non–zero minor of order r, and
(ii) all other minors of order greater than r, if any; are
zero. The rank of A is denoted by r( A). Now, given that
r( A) = 2 ® minor of order greater than 2 i.e., 3 is zero.
½2 - 1
Thus A =½4
7
½
1
4
½
3½
l½ = 0
½
5½
Þ
2( 35 - 4 l) + 1(20 - l) + 3(16 - 7) = 0,
Þ
70 - 8 l + 20 - l + 27 = 0,
Þ
9 l = 117
Þ
l = 13
15. (A) The correct statements are
( AB) T = BT A T , ( AB) -1 = B-1 A -1 ,
adj ( AB) = adj (B) adj ( A)
r( AB) ¹ r( A) r(B), AB = A × B
Thus statements I, II, and IV are wrong.
A = 2( -9 + 8) + 2( -2 + 3) = - 2 + 2 = 0
Þ
T
- 2ù
- 1 úû
1
adj A
A
0
0 = 4 ¹ 0,
A -1 =
T
- 2 ù é5
=
- 1 úû êë3
0
2
é
adj A == ê
ê
êë
= -1
16. (B) Since
é -3 -6 -6 ù
é -1 -2 -2 ù
=ê 6
3 -6 ú = 3ê 2
1 -2 ú = 3A T
ê
ú
ê
ú
3úû
1ûú
êë 6 -6
êë 2 -2
Page
530
1
-1
- 3ù
- 1 úû
1
A =5
column
corresponding to aij and then take the determinant of
½-2
M 21 =½
½-2
3 -5
é-5
Also, adj A = ê
ë-2
10. (C) A is Skew–Hermitian then A Q = - A
leaving
2
13. (A) Since, A -1 =
9. (D) A is Hermitian then A Q = A
by
-1
Now, Here A =
it is only possible when aii is purely imaginary.
obtained
1
adj A
A
12. (A) Since A -1 =
A -1 =
aii + aii = 0
Engineering Mathematics
r( A ) < 3
½2 1½
½= 6 ¹ 0
Again, one minor of order 2 is ½
½0 3½
Þ
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UNIT 9
5 ½
½3-l
½= 0
Þ ½
- 5 - l½
½-4
31. (A) Since A(adj A) = A I 3
Þ
( 3 - l)( - 5 - l) + 16 = 0 Þ
Þ
l + 2l + 1 = 0
2
Engineering Mathematics
Þ
- 15 + l + 2 l + 16 = 0
2
( l + 1) = 0
2
Þ
Þ
l = - 1, - 1
é1
A(adj A) = 2 ê0
ê
êë0
0
1
0
0 ù é2
0 ú = ê0
ú ê
1 úû êë0
0
2
0
0ù
0ú
ú
2 úû
Thus eigen values are -1 , - 1
32. (A) Since the sum of the eigenvalues of an n–square
24. (C) Characteristic equation is A - lI = 0
matrix is equal to the trace of the matrix (i.e. sum of the
-6
2½
½8 -l
Þ ½-6
7-l
- 4½= 0
½
½
- 4 3 - l½
½ 2
Þ
l2 - 18 l2 + 45 l = 0
Þ
l( l - 3)( l - 15) = 0
diagonal elements)
so, required sum = 8 + 5 + 5 = 18
33. (B) Since the product of the eigenvalues is equal to
Þ
the determinant of the matrix so A = 1 ´ 2 ´ 5 = 10
l = 0 , 3 , 15
25. (B) If eigen values of A are l1 , l2 , l3 then the eigen
values of kA are kl1 , kl2 , kl3. So the eigen values of 2A
are 2 , - 4 and 6
34. (C)
écos q cos f cos ( q - f) cos q sin f cos ( q - f) ù
AB = ê
ú =A
ëcos f sin q cos ( q - f) sin q sin f cos ( q - f) û
null matrix when cos ( q - f) = 0
26. (B) If l1 , l2 ,........, l
n
are the eigen values of a
non–singular matrix A, then A -1 has the eigen values
1 1
1
1 1
,
, ........,
. Thus eigen values of A -1 are , ,
l1 l2
ln
2 3
-1
.
3
This happens when ( q - f) is an odd multiple of
p
.
2
35. (C) Since A + B is defined, A and B are matrices of
the same type, say m ´ n. Also, AB is defined. So, the
number of columns in A must be equal to the number of
rows in B i.e. n = m. Hence, A and B are square matrices
27. (B) If l1 , l2 , ......, l
n
are the eigen values of a matrix
A, then A has the eigen values l12 , l22 , ........, l 2n . So,
of the same order.
2
eigen values of A 2 are 1, 4, 9.
28. (B) If l1 , l2 ,...., l
n
are the eigen values of A then
the eigen values adj A are
eigenvalues of adj A are
A
l1
,
A
l2
,......,
A
l
; A ¹ 0. Thus
n
4 -4
,
i.e. 2 and-1.
2
4
29. (B) Since, the eigenvalues of A and A are square so
T
the eigenvalues of A T are 2 and 4.
30. (B) Since 1 and 3 are the eigenvalues of A so the
characteristic equation of A is
( l - 1) ( l - 3) = 0 Þ l2 - 4 l + 3 = 0
Also, by Cayley–Hamilton theorem, every square
matrix satisfies its own characteristic equation so
A 2 - 4 A + 3I 2 = 0
Þ
A 2 = 4 A - 3I 2
Þ A 3 = 4 A 2 - 3A = 4( 4 A - 3I) - 3A
Þ A 3 = 13A - 12I 2
Page
532
a
1 - tan 2
2
a
2 = 1-t
36. (A) Let tan = t, then, cos a =
a
2
t + t2
1 + tan 2
2
2 tan
and sin a =
a
1 + tan 2
2
écos a
(I - A ) × ê
ësin a
é
ê 1
=ê
ê- tan a
ë
2
a
2
=
2t
1 + t2
- sin a ù
cos a úû
aù
2ú´
ú
1 ú
û
tan
é1 - t2
t ù ê1 + t 2
é 1
=ê
ú´ê
ë -t 1 û ê 2 t
2
ëê(1 + t )
écos a
êsin a
ë
- sin a ù
cos a úû
-2 t ù
(1 + t 2 ) ú
ú
1 - t2 ú
1 + t 2 úû
aù
é
1 - tan ú
é 1 - tù ê
2 = (I + A )
=ê
ú
ú =ê
t
1
a
ë
û êtan
1 ú
ë
û
2
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Linear Algebra
- 4 ù é3
- 1 úû êë1
é3
37. (B) A 2 = ê
ë1
é1 + 2 n
=ê
ë n
- 4 ù é5
=
- 1 úû êë2
- 8ù
- 3 úû
é1 3ù
é1 2 0 ù ê
ú
45. (C) AA = ê
ú 2 -1
ë3 -1 4 û ê0 4 ú
êë
úû
T
- 4n
ù
, where n = 2.
1 - 2 núû
é cos a
38. (D) A a × A b = ê
ë- sin a
é cos ( a + b)
=ê
ë- sin ( a + b)
sin a ù é cos b
cos a úû êë- sin b
(1)( 3) + (2)( -1) + (0)( 4) ù
é (1)(1) + (2)(2) + (0)(0)
=ê
(
3
)(
1
)
+
(
1
)(
2
)
+
(
4
)(
0
)
(
3
)( 3) + ( -1)( -1) + ( 4)( 4) úû
ë
sin bù
cos b úû
1ù
é5
=ê
ú
1
26
ë
û
sin ( a + b) ù
= Aa+ b
cos ( a + b) úû
46. (B) if A is zero, A -1 does not exist and the matrix A
Also, it is easy to prove by induction that
sin na ù
é cos na
(A a )n = ê
ú
ë- sin na cos na û
39. (A) We know that adj (adj A) = A
n -2
is said to be singular. Only (B) satisfy this condition.
5 2
A =
= (5)(1) - (2)(2) = 1
2 1
× A.
47. (A) A skew symmetric matrix A n ´n is a matrix with
Here n = 3 and A = 3.
A T = -A . The matrix of (A) satisfy this condition.
So, adj (adj A) = 3( 3- 2 ) × A = 3A.
40. (C) We have adj (adj A) = A
é1ù
é1 1 0 ù ê ú é(1)(1) + (1)(0) + (0)(1) ù é1 ù
48. (C) AB = ê
ú 0 =ê
ú =ê ú
ë1 0 1û ê1ú ë(1)(1) + (0)(0) + (1)(1) û ë2 û
êë úû
( n -1 ) 2
4
Putting n = 3, we get adj (adj A) = A .
49. (C) For orthogonal matrix
41. (C) Let B = adj (adj A 2 ).
Then, B is also a 3 ´ 3
2
[K
A2 = A
é2 x
42. (C) ê
ë x
Þ
é2 x
ê0
ë
2
( 3-1 ) 2
]
0ù é 1
x úû êë-1
0 ù é1
=
2 x úû êë0
det M = 1 And M -1 = M T , therefore Hence D -1 = D T
é A Cù
1 é 0 -B ù
DT = ê
= D -1 =
ú
A úû
-BC êë-C
ëB 0 û
-C
1
This implies B =
Þ B=
Þ B = ±1
-BC
B
matrix.
adj {adj (adj A 2 )} = adj B = B3
= adj (adj A 2 ) = é A 2
ëê
Chap 9.1
3-1
2
2
ù = A 16 = 216
ûú
0 ù é1
=
2 úû êë0
0ù
,
1úû
=B
Hence B = 1
50. (B) From linear algebra for A n ´n triangular matrix
0ù
1úû
So, 2 x = 1
n
det A = Õ aii , The product of the diagonal entries of A
i =1
Þ
1
x= .
2
43. (D) Inverse matrix is defined for square matrix only.
é 2 -1ù
é1 -2 -5 ù
44. (C) AB = ê 1 0 ú ê
ê
ú ë3 4
0 úû
êë-3 4 úû
é d( t 2 )
dA ê dt
51. (C )
=ê
t
dt ê d( e )
ë dt
d(cos t) ù
dt ú = é2 t - sin t ù
d(sin t) ú êë e t
cos t úû
ú
dt û
52. (A) If det A ¹ 0, then A n ´n is non-singular, but if
A n ´n is non-singular, then no row can be expressed as a
é(2)(1) + ( -1)( 3) (2)( -2) + ( -1)( 4) (2)( -5) + ( -1)(0) ù
= ê (1)(1) + (0)( 3)
(1)( -2) + (0)( 4)
(1)( -5) + (0)(0) ú
ê
ú
êë( -3)(1) + ( 4)( 3) ( -3)( -2) + ( 4)( 4) ( -3)( -5) + ( 4)(0) úû
linear combination of any other. Otherwise det A = 0
é-1 -8 -10 ù
= ê 1 -2
-5 ú
ê
ú
15 úû
êë 9 -22
************
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CHAPTER
9.2
DIFFERENTIAL CALCULUS
1. If f ( x) = x 3 - 6 x 2 + 11 x - 6 is on [1, 3], then the point
6. A point on the curve y = x - 2 on [2, 3], where the
c « ] 1, 3 [ such that f ¢ ( c) = 0 is given by
1
1
(A) c = 2 ±
(B) c = 2 ±
2
3
tangent is parallel to the chord joining the end points of
(C) c = 2 ±
1
2
(D) None of these
2. Let f ( x) = sin 2 x, 0 £ x £
Then, c is equal to
p
(A)
4
(C)
the curve is
p
2
and f ¢ ( c) = 0 for c « ] 0, 2p [.
x
2
3. Let f ( x) = x( x + 3) e , -3 £ x £ 0. Let c « ] - 3, 0 [ such
that f ¢ ( c) = 0. Then, the value of c is
(B) -3
(C) -2
1
(D) 2
æ7 1ö
(C) ç , ÷
è4 2ø
æ9 1 ö
(D) ç , ÷
è2 4 ø
value of c of the mean value theorem is
(D) None
(A) 3
æ7 1 ö
(B) ç , ÷
è2 4 ø
7. Let f ( x) = x( x - 1)( x - 2) be defined in [0, 12 ]. Then, the
p
(B)
3
p
6
æ9 1ö
(A) ç , ÷
è4 2ø
(A) 0.16
(B) 0.20
(C) 0.24
(D) None
8. Let f ( x) = x 2 - 4 be defined in [2, 4]. Then, the value
of c of the mean value theorem is
(A) - 6
(B)
(C)
(D) 2 3
3
6
4. If Rolle’s theorem holds for f ( x) = x 3 - 6 x 2 + kx + 5 on
1
[1, 3] with c = 2 +
, the value of k is
3
(A) -3
(B) 3
mean-value theorem is
(A) 0.5
(B) ( e - 1)
(C) 7
(C) log ( e - 1)
(D) None
(D) 11
5. A point on the parabola y = ( x - 3) 2 , where the
Page
534
9. Let f ( x) = e x in [0, 1]. Then, the value of c of the
tangent is parallel to the chord joining A (3, 0) and B (4,
10. At what point on the curve y = (cos x - 1) in ]0, 2p[ ,
1) is
is the tangent parallel to x –axis ?
(A) (7, 1)
æ3 1ö
(B) ç , ÷
è2 4 ø
æp
ö
(A) ç , - 1 ÷
è2
ø
(B) ( p, - 2)
æ7 1 ö
(C) ç , ÷
è2 4 ø
æ 1 1ö
(D) ç - , ÷
è 2 2ø
æ 2 p -3 ö
(C) ç
,
÷
è 3 2 ø
(D) None of these
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Differential Calculus
11. log sin ( x + h) when expanded in Taylor’s series, is
equal to
Chap 9.2
æ
16. If u = tan -1 ç
ç
è
(A) log sin x + h cot x -
1 2
h cosec2 x + K
2
(A) 2 cos 2 u
(B) log sin x + h cot x +
1 2
h sec 2 x + K
2
(C)
(C) log sin x - h cot x +
1 2
h cosec2 x + K
2
(B)
1
tan u
4
(D) None of these
x
2
1
sin 2 u
2
(B) sin 2u
(C) sin u
2
3
x 3 + y 3 + x 2 y - xy 2
, then the value of
x 2 - xy + y 2
¶u
¶u
is
+y
¶x
¶y
(A)
pö
pö
æ
æ
çx- ÷
çx- ÷
2
2
è
ø
è
ø
(B) 1 +
-K
2!
4!
1
sin 2 u
4
(D) 2 tan 2 u
17. If u = tan -1
pö
æ
12. sin x when expanded in powers of ç x - ÷ is
2
è
ø
2
3
2
pö
pö
pö
æ
æ
æ
çx- ÷
çx- ÷
çx- ÷
2ø
2ø
2ø
è
è
è
(A) 1 +
+
+
+K
2!
3!
4!
x + y ö÷
¶u
¶u
equals
, then x
+y
÷
¶
x
¶y
x+ yø
(D) 0
æ yö
æ yö
18. If u = fç ÷ + xyç ÷, then the value of
è xø
è xø
x2
5
pö
pö
æ
æ
çx- ÷
çx- ÷
2ø
2ø
pö
è
è
æ
(C) ç x - ÷ +
+
+K
!
!
2
3
5
è
ø
¶2u
¶2u
¶2u
+ 2 xy
+ y 2 2 , is
2
dx
dx dy
¶y
2
(A) 0
(B) u
(C) 2u
(D) -u
(D) None of these
æp
ö
13. tanç + x ÷ when expanded in Taylor’s series, gives
è4
ø
4
2
(A) 1 + x + x + x 3 + K
3
(B) 1 + 2 x + 2 x 2 +
(C) 1 +
8 3
x +...
3
x2 x4
+
+K
2! 4!
19. If z = e x sin y, x = log e t and y = t 2 , then
by the expression
dz
is given
dt
(A)
ex
(sin y - 2 t 2 cos y)
t
(B)
ex
(sin y + 2 t 2 cos y)
t
(C)
ex
(cos y + 2 t 2 sin y)
t
(D)
ex
(cos y - 2 t 2 sin y)
t
20. If z = z( u, v) , u = x 2 - 2 xy - y 2 , v = a, then
(A) ( x + y)
¶z
¶z
= ( x - y)
¶x
¶y
(B) ( x - y)
¶z
¶z
= ( x + y)
¶x
¶y
¶ 3u
14. If u = e , then
is equal to
¶ x¶ y¶ z
(C) ( x + y)
¶z
¶z
= ( y - x)
¶x
¶y
(D) ( y - x)
¶z
¶z
= ( x + y)
¶x
¶y
(A) e xyz [1 + xyz + 3 x 2 y 2 z 2 ]
21. If f ( x, y) = 0, f( y, z) = 0, then
(D) None of these
xyz
(B) e
xyz
[1 + xyz + x y z ]
(C) e
xyz
[1 + 3 xyz + x y z ]
3
2
2
¶f ¶f ¶f ¶f dz
×
=
×
×
¶y ¶z ¶x ¶y dx
(B)
(C)
¶f ¶f dz ¶f ¶f
×
×
=
×
¶y ¶z dx ¶x ¶y
(D) None of these
2
(D) e xyz [1 + 3 xyz + x 3 y 3z 3 ]
15. If z = f ( x + ay) + f( x - ay), then
(A)
¶ z
¶ z
= a2 2
¶x 2
¶y
(B)
¶ z
¶ z
= a2 2
¶y 2
¶x
(C)
¶2z
1 ¶2z
=- 2
2
¶y
a ¶x 2
(D)
¶2z
¶2z
= - a2 2
2
¶x
¶y
2
(A)
3 3
2
2
¶f ¶f ¶f ¶f dz
×
×
=
×
¶y ¶z ¶x ¶x dx
22. If z = x 2 + y 2 and x 3 + y 3 + 3 axy = 5 a 2 , then at
2
x = a, y = a,
dz
is equal to
dx
(A) 2a
(B) 0
(C) 2 a 2
(D) a 3
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UNIT 9
23. If x = r cos q, y = r sin q where r and q are the
functions of x, then
dx
is equal to
dt
Engineering Mathematics
¶2u
¶2 y
æ yö
30. If u = x n -1 yf ç ÷, then x 2 + y
is equal to
¶x
¶y ¶x
è xø
(B) n( n - 1) u
(A) nu
(A) r cos q
dr
dq
- r sin q
dt
dt
(B) cos q
dr
dq
- r sin q
dt
dt
(C) r cos q
dr
dq
+ sin q
dt
dt
(D) r cos q
dr
dq
- sin q
dt
dt
¶u
¶x
(C) ( n - 1)
(D) ( n - 1)
¶u
¶y
31. Match the List–I with List–II.
List–I
¶ r ¶ r
is equal to
+
dx 2 ¶y 2
2
24. If r 2 = x 2 + y 2 , then
2
2
æ ¶r ö ïü
ïì æ ¶r ö
(A) r 2 í ç ÷ + çç ÷÷ ý
ïî è ¶x ø
è ¶y ø ïþ
æ ¶r ö
1 ìï æ ¶r ö
÷ + çç ÷÷
2 íç
r ï è ¶x ø
è ¶y ø
î
2
(C)
2
ïü
ý
ïþ
2
2
2
æ ¶r ö ïü
ïì æ ¶r ö
(B) 2 r 2 í ç ÷ + çç ÷÷ ý
ïî è ¶x ø
è ¶y ø ïþ
(D) None of these
25. If x = r cos q, y = r sin q, then the value of
is
(A) 0
(C)
(i) If u =
(D)
¶2q ¶2q
+
¶x 2 ¶y 2
¶r
¶y
1
1
then x 2
x4 + y4
1
3
u
16
(C) udu = mxdx + nydy
(D)
2
27. If y 3 - 3ax 2 + x 3 = 0, then the value of
to
(B)
(C) -
2 a2 x4
y5
(D) -
d y
is equal
dx 2
2 a2 x2
y5
(4) -
(B) dz =
xdy + ydx
x2 + y2
(C) dz =
xdx - ydy
x2 + y2
(D) dz =
xdx - ydy
x2 + y2
(B) 1
(C) u
(D) eu
(II)
(III)
(IV)
(A)
1
2
3
4
(B)
2
1
4
3
(C)
2
1
3
4
(D)
1
2
4
3
in the area is approximately equal to
(A) 1%
(B) 2%
(C) p%
(D) 4%
33. Consider the Assertion (A) and Reason (R) given
below:
x2 + y2
¶u
¶u
, then x
is equal to
+y
x+ y
¶x
¶y
(A) 0
(I)
and minor axes of an ellipse, then the percentage error
y
, then
x
xdy - ydx
x2 + y2
1
u
4
32. If an error of 1% is made in measuring the major
2 a2 x2
y5
(A) dz =
¶u
¶x
Correct match is—
du
dx
dy
=m
+n
u
x
y
a2 x2
y5
¶2u
¶2u
¶2u
+ 2 xy
+ y2
2
¶x
¶x ¶y
¶y 2
(2)
(3) 0
(A) -
2
¶2u
¶2u
2 ¶ u
+
2
xy
+
y
¶x 2
¶x ¶y
¶y 2
¶u
¶u
æ yö
(iv) If u = f ç ÷ then x
+y
¶x
¶y
è xø
(1) -
(B) du = mdx + ndy
Page
536
1
List–II
(A) du = mx m -1 y n + nx m y n -1
29. If u = log
(ii) If u =
1
26. If u = x m y n , then
28. z = tan -1
1
x2 - y2
(iii) If u = x 2 + y 2 then x 2
(B) 1
¶r
¶x
x2 y
¶2u
¶2u
then x 2 + y
x+ y
¶x
¶x ¶y
¶u
¶u
æ yö
Assertion (A): If u = xyf ç ÷, then x
+y
= 2u
¶x
¶y
è xø
Reason (R): Given function u is homogeneous of
degree 2 in x and y.
Of these statements
(A) Both A and R are true and R is the correct
explanation of A
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Differential Calculus
Chap 9.2
(B) Both A and R are true and R is not a correct
explanation of A
41. If a < 0, then f ( x) = e ax + e - ax is decreasing for
(A) x > 0
(B) x < 0
(C) A is true but R is false
(C) x > 1
(D) x < 1
(D) A is false but R is true
42. f ( x) = x 2 e - x is increasing in the interval
34. If u = x log xy, where x 3 + y 3 + 3 xy = 1, then
equal to
(A) (1 + log xy) -
x æ x2 + y ö
÷
ç
y çè y 2 + x ÷ø
(B) (1 + log xy) -
yæ y + xö
÷
ç
x çè x 2 + y ÷ø
(C) (1 - log xy) -
xæ x + yö
÷
ç
y çè y 2 + x ÷ø
du
is
dx
(A) ] -¥, ¥ [
(B) ] -2, 0 [
(C) ] 2, ¥ [
(D) ] 0, 2 [
43. The least value of a for which f ( x) = x 2 + ax + 1 is
increasing on ] 1, 2, [ is
2
(A) 2
(B) -2
(C) 1
(D) -1
2
44. The minimum distance from the point (4, 2) to the
parabola y 2 = 8 x, is
y æ y2 + x ö
(D) (1 - log xy) - çç 2
÷
x è x + y ÷ø
(A)
(B) 2 2
2
(C) 2
(D) 3 2
¶z
¶z
æ yö
35. If z = xyf ç ÷, then x
is equal to
+y
¶x
¶y
è xø
(A) z
(B) 2z
45. The co-ordinates of the point on the parabola
(C) xz
y = 3 x - 3, are
y = x 2 + 7 x + 2 which is closest to the straight line
(D) yz
36. f ( x) = 2 x 3 - 15 x 2 + 36 x + 1 is increasing in the
interval
(A) (-2, -8)
(B) (2, -8)
(C) (-2, 0)
(D) None of these
(A) ] 2, 3 [
(B) ] -¥, 3 [
46. The shortest distance of the point (0, c), where
(C) ] -¥, 2 [È ] 3, ¥
(D) None of these
0 £ c < 5, from the parabola y = x 2 is
37. f ( x) =
x
is increasing in the interval
( x 2 + 1)
(A) ] -¥, - 1 [ È ] 1, ¥ [
(B) ] -1, 1 [
(C) ] -1, ¥ [
(D) None of these
4c + 1
2
(A)
4c + 1
(B)
(C)
4c - 1
2
(D) None of these
x
æ1ö
47. The maximum value of ç ÷ is
è xø
38. f ( x) = x 4 - 2 x 2 is decreasing in the interval
(A) ] -¥, -1 [ È ] 0, 1 [
(B) ] -1, 1 [
(C) ] -¥, -1 [ È ] 1, ¥ [
(D) None of these
(A) e
æ1ö
(C) ç ÷
è eø
39. f ( x) = x 9 + 3 x 7 + 6 is increasing for
(A) all positive real values of x
(B) e
-
1
e
e
(D) None of these
(B) all negative real values of x
250 ö
æ
48. The minimum value of ç x 2 +
÷ is
x ø
è
(C) all non-zero real values of x
(A) 75
(B) 50
(C) 25
(D) 0
(D) None of these
40. If f ( x) = kx 3 - 9 x 2 + 9 x + 3 is increasing in each
49. The maximum value of f ( x) = (1 + cos x) sin x is
interval, then
(A) 3
(B) 3 3
(C) 4
(D)
(A) k < 3
(B) k £ 3
(C) k > 3
(D) k ³ 3
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UNIT 9
50. The greatest value of
f ( x) =
(A)
SOLUTIONS
sin 2 x
pö
æ
sinç x + ÷
4ø
è
on the interval [0, 2p ] is
1
2
1. (B) A polynomial function is continuous as well as
differentiable. So, the given function is continuous and
differentiable.
(B)
2
f (1) = 0 and f ( 3) = 0. So, f (1) = f ( 3).
By Rolle’s theorem Ec such that f ¢( c) = 0.
(D) - 2
(C) 1
Engineering Mathematics
Now, f ¢ ( x) = 3 x 2 - 12 x + 11
51. If y = a log x + bx 2 + x has its extremum values at
Þ
x = -1 and x = 2, then
1
(A) a = - , b = 2
2
Now, f ¢ ( c) = 0
(C) a = 2, b = 52.
The
(B) a = 2, b = -1
1
2
Þ
f ¢ ( c) = 3c 2 - 12 c + 11.
Þ
3c 2 - 12 c + 11 = 0
1 ö
æ
c = ç2 ±
÷.
3 ø
è
(D) None of these
2. (A) Since the sine function is continuous at each
co-ordinates
of
the
point
on
the
curve
4 x 2 + 5 y 2 = 20 that is farthest from the point (0, -2) are
(A) ( 5 , 0)
(B) ( 6 , 0)
(C) (0, 2)
(D) None of these
pö
æ
53. For what value of xç 0 £ x £ ÷, the function
2
è
ø
x
has a maxima ?
y=
(1 + tan x)
(A) tan x
(B) 0
(C) cot x
(D) cos x
é pù
x « R, so f ( x) = sin 2 x is continuous in ê0, ú.
ë 2û
f ¢ ( x) = 2 cos 2 x, which clearly exists for all
p
p
x « ]0, [ .So, f ( x) is differentiable in x « ]0, [.
2
2
Also,
æ pö
Also, f (0) = f ç ÷ = 0. By Rolle’s theorem, there exists
è2 ø
p
c « ]0, [ such that f ¢( c) = 0.
2
2 cos 2 c = 0
2c =
Þ
p
2
Þ
c=
p
.
4
3. (C) Since a polynomial function as well as an
exponential function is continuous and the product of
two continuous functions is continuous, so f ( x) is
*************
continuous in [-3, 0].
f ¢ ( x) = (2 x + 3) × e
-
x
2
- é x + 6 - x2 ù
1 -2 2
e ( x + 3 x) = e 2 ê
ú
2
2
ë
û
x
-
x
which clearly exists for all x « ] - 3, 0 [.
f ( x) is differentiable in ] -3, 0 [.
Also, f ( -3) = f (0) = 0.
By Rolle’s theorem c « ] -3, 0 [ such that f ¢ ( c) = 0.
- é c + 6 - c2 ù
e 2ê
ú =0
2
ë
û
c
Now, f ¢ ( c) = 0
Þ
c + 6 - c 2 = 0 i.e. c 2 - c - 6 = 0
Þ
( c + 2) ( 3 - c) = 0
Þ
c = -2, c = 3.
Hence, c = -2 « ] -3, 0 [ .
4. (D) f ¢ ( x) = 3c 2 - 12 x + k
f ¢ ( c) = 0
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538
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3c 2 - 12 c + k = 0
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UNIT 9
¶u
¶u 1
+ y sec 2 u
= tan u
¶x
¶y 2
æ pö
æ pö
æ pö
f ç ÷ = 1, f ¢ç ÷ = 0, f ¢¢ç ÷ = -1,
è2 ø
è2 ø
è2 ø
x sec 2 u
æ pö
æ pö
f ¢¢¢ç ÷ = 0, f ¢¢¢¢ç ÷ = 1, ....
2
è ø
è2 ø
Þ
13. (B) Let f ( x) = tan x Then,
17. (A) Here tan u =
2
3
æp
ö
æ pö
æ pö x
æ p ö x ¢¢¢æ p ö
f ç + x ÷ = f ç ÷ + xf ¢ç ÷ +
× f ¢¢ç ÷ +
f ç ÷+...
è4
ø
è4ø
è 4 ø 2!
è 4 ø 3! è 4 ø
f ¢( x) = sec , f ¢¢( x) = 2sec x tan x,
2
2
f ¢¢¢( x) = 2sec 4 x + 4sec 2 x tan 2 x etc.
æ pö
æ pö
æ pö
æ pö
f ç ÷ = 1, f ¢ç ÷ = 2, f ¢¢ç ÷ = 4, f ¢¢¢ç ÷ = 16, ...
è4ø
è4ø
è4ø
è4ø
x2
x3
æp
ö
Thus tanç + x ÷ = 1 + 2 x +
×4 +
× 16 + K
2
6
è4
ø
8
= 1 + 2 x + 2 x + x3 + K
3
2
14. (C) Here u = e xyz
Þ
¶u
= exyz × yz
¶x
= e xyz (1 + 3 xyz + x 2 y 2 z 2 )
Which is homogeneous of degree 1
¶f
¶f
Thus x
+y
= f
¶x
¶y
¶z
= f ¢ ( x + ay) + f¢ ( x - ay)
¶x
¶ z
= a 2 f ¢¢( x + ay) + a 2 f¢¢( x - ay)....(2)
¶y 2
2
Hence from (1) and (2), we get
tan u =
x+
x+ y
x+
y
Þ
x2
and x 2
2
¶2v
¶2v
2 ¶ v
+
2
xy
+
y
= 0....(1)
¶x 2
¶x ¶y
¶y 2
¶2w
¶2w
¶2w
+ 2 xy
+ y2
= 0....(2)
2
¶x
¶x ¶y
¶y 2
Adding (1) and (2), we get
x2
2
¶2
¶2
2 ¶
(
v
+
w
)
+
2
xy
(
v
+
w
)
+
y
( v + w) = 0
¶x 2
¶x ¶y
¶y 2
x2
¶2u
¶2u
¶2u
+ 2 xy
+ y2 2 = 0
2
¶x
¶x ¶y
¶y
¶z
= ex sin y
¶x
¶z
dx 1
And
= e x cos y, x = log e t Þ
=
¶y
dt t
¶ z
= f ¢¢( x + ay) + f¢¢( x - ay)....(1)
dx 2
¶z
= af ¢( x + ay) - af¢ ( x - ay)
¶y
x+ y
Then u = v + w
19. (B) z = e x sin y
2
æ
16. (B) u = tan -1 ç
ç
è
¶f
¶u 1
+y
= sin 2 u
¶x
¶y 2
æ yö
æ yö
18. (A) Let v = fç ÷ and w = xYç ÷
è xø
è xø
Þ
15. (B) z = f ( x + ay) + f( x - ay)
¶2z
¶2z
= a2 2
2
¶y
¶x
ö
÷
y ÷ø
= f (say)
Which is a homogeneous equation of degree 1/2
¶f
¶f 1
By Euler’s theorem. x
+y
= f
¶x
¶y 2
x
x 3 + y 3 + x 2 y - xy 2
= f (say)
x 2 - xy + y 2
homogeneous of degree one
¶ 3u
= e xyz × (1 + 2 xyz) + ( z + xyz 2 ) e xyz × xy
¶ x¶ y¶ z
Þ
¶u
¶u 1
1
+y
= sin u cos u = sin 2 u
¶x
¶y 2
4
Now v is homogeneous of degree zero and w is
¶2u
= ze xyz + yzexyz × xz = e xyz ( z + xyz 2 )
¶x¶y
Þ
x
As above question number 16 x
Now,
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540
Engineering Mathematics
¶(tan u)
¶(tan u) 1
+y
= tan u
¶x
¶y
2
Þ
dy
= 2t
dt
dz ¶z dx ¶z dy
=
×
+
×
dt ¶x dt ¶y dt
And y = t 2
= e x sin y ×
Þ
ex
1
+ e x cos y × 2 t = (sin y + 2 t 2 cos y)
t
t
20. (C) Given that
z = z( u, v), u = x 2 - 2 xy - y 2 , v = a....(i)
¶z ¶z ¶u ¶z ¶v
=
×
+
× ....(ii)
¶x ¶u ¶x ¶v ¶x
¶z ¶z ¶u ¶z ¶v
and
=
×
+
× ....(iii)
¶y ¶u ¶y ¶v ¶y
From (i),
¶u
¶u
¶v
¶v
=2x -2y ,
= -2 x - 2 y,
= 0,
=0
¶x
¶y
¶x
¶y
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Substituting these values in (ii) and (iii)
24. (C) r 2 = x 2 + y 2
¶z ¶z
¶z
=
(2 x - 2 y) +
× 0....(iv)
¶x ¶u
¶v
and
and
¶z ¶z
¶z
=
× ( -2 x - 2 y) +
× 0....(v)
¶y ¶u
¶v
Þ
¶2r ¶2r
+
=2 + 2 + 4
¶x 2 ¶y 2
2
2
2
æ ¶r ö üï
¶ 2 r ¶ 2 y 1 ìï æ ¶r ö
+
=
+
ç
÷
ç
÷
í
ç ¶y ÷ ý
¶x 2 ¶ y 2 r 2 ï è ¶x ø
è ø þï
î
Þ
21. (C) Given that f ( x, y) = 0, f( y, z) = 0
25. (A) x = r cos q ,
These are implicit functions
¶f
dz
¶y
=¶f
dy
¶z
æ ¶f ö æ ¶f ö
ç÷ ç÷
dy dz ç ¶x ÷ ç ¶y ÷
×
=
´
dx dy ç ¶f ÷ ç ¶f ÷
ç ¶y ÷ ç ¶z ÷
è
ø è
ø
or,
¶r
¶r
= 2 x and
=2y
¶x
¶y
¶2r
¶2r
= 2 and
=2
2
¶x
¶y 2
2
¶z
¶z
= ( y - x)
¶x
¶y
¶f
dy
= - ¶x ,
¶f
dx
¶y
Þ
æ ¶r ö
æ ¶r ö
and ç ÷ + çç ÷÷ = 4 x 2 + 4 y 2 = 4 r 2
¶
x
è ø
è ¶y ø
From (iv) and (v), we get
( x + y)
Chap 9.2
Þ
æ yö
q = tan -1 ç ÷
è xø
Þ
tan q =
Þ
¶q
1
-y
æ -y ö
=
= 2
2 ç
2 ÷
¶x 1 + ( y x) è x ø x + y 2
and
¶2q
-2 xy
=
¶x 2 ( x 2 + y 2 ) 2
Similarly
¶f ¶f dz ¶f ¶f
×
×
=
×
¶y ¶z dx ¶x ¶y
y
x
y = r sin q
¶2q
2 xy
¶2q ¶2q
and
= 2
+
=0
2
2 2
¶y
(x + y )
¶x 2 ¶y 2
26. (D) Given that u = x m y n
Taking logarithm of both sides, we get
22. (B) Given that z = x + y
2
2
log u = m log x + n log y
and x + y + 3axy = 5 a ...(i)
dz ¶z ¶z dy
=
+
× ....(ii)
dx ¶x ¶y dx
3
3
from (i),
Differentiating with respect to x, we get
1 du
1
1 dy
du
dx
dy
or,
= m× + n×
=m
+ n×
u dx
x
y dx
u
x
y
2
¶z
1
¶z
=
×2x ,
=
2
2
¶x 2 x + y
¶y 2
1
x + y2
2
×2y
dy
dy
and 3 x + 3 y
+ 3ax
+ 3ay.1 = 0
dx
dx
2
Þ
2
æ x 2 + ay ö
dy
÷÷
= - çç 2
dx
è y + ax ø
Substituting these value in (ii), we get
dz
=
dx
x
x2 + y2
+
æ x 2 + ay ö
÷÷
çç - 2
x 2 + y 2 è y + ax ø
y
a
æ dz ö
=
+
ç
÷
2
dx
è
ø( a , a )
a + a2
æ a 2 + aa ö
÷÷ = 0
çç - 2
a 2 + a 2 è a + a. a ø
a
27. (D) Given that f ( x, y) = y 3 - 3ax 2 + x 3 = 0
fx = - 6 ax + 3 x 2 , f y = 3 y 2 , fxx = - 6 a + 6 x ,
f yy = 6 y , fxy = 0
é fxx ( f y) 2 - 2 fx f y fxy + f yy( fx ) 2 ù
d2 y
=-ê
ú
2
( f y) 3
dx
ë
û
é( 6 x - 6 a( 3 y 2 ) 2 - 0 + 6 y( 3 x 2 - 6 ax) 2 ù
= -ê
ú
(3 y2 )3
ë
û
2
= - 5 ( -ax 3 - ay 3 + 4 a 2 x 2 )
y
=-
2
[ -a( a 3 + y 3) + 4 a 2 x 2 ]
y5
23. (B) Given that x = r cos q, y = r sin q....(i)
=-
2
[ -a( 3ax 2 ) + 4 a 2 x 2 ] [ \ x 3 + y 3 - 3ax 2 = 0]
y5
dx ¶x dr ¶x dq
=
×
+
× ....(ii)
dt ¶r dt ¶q dt
=-
2 a2 x2
y5
From (i),
¶x
¶x
= cos q,
= - r sin q
¶r
¶q
Substituting these values in (ii), we get
dx
dr
dq
= cos q
- r sin q ×
dt
dt
dt
28. (A) Given that z = tan -1
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....(i)
x
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UNIT 9
dz ¶z ¶z dy
=
+
× ....(ii)
dx ¶x ¶y dx
From (i)
¶z
=
¶y
¶z
=
¶x
In (c) u = x1 2 + y1 2 It is a homogeneous function of
1
degree .
2
-y
æ -y ö
×ç 2 ÷ = 2
+ y2
x
x
ø
æ yö è
1+ç ÷
è xø
1
2
x2
x
æ1ö
×ç ÷ = 2
2
æ yö è xø x + y
1+ç ÷
è xø
1
=
2
Hence correct match is
f is a homogeneous function of degree one
Þ
¶e u
¶e u
x
+y
= eu
¶x
¶y
¶u
¶u
+y
=1
¶x
¶y
d
2
1
3
4
Þ
log A = log p + log a + log b
Þ
¶(log A) = ¶(log p) + ¶(log a) + ¶(log b)
¶A
¶a ¶b
Þ
=0 +
+
A
a
b
100
100
100
Þ
¶A =
¶a +
× ¶b
A
a
b
100
100
But it is given that
¶a = 1, and
¶b = 1
a
b
100
¶A = 1 + 1 = 2
A
Differentiating partially w.r.t. x, we get
¶ 2 u ¶u
¶2u
n ¶u
+
+
y
=
2
¶x
¶x
¶y ¶x
¶x
x
c
Area A = pab
It is a homogeneous function of degree n
¶u
¶u
Euler’s theorem x
+y
= nu
¶x
¶y
Þ
b
ellipse
æ yö
30. (C) Given that u = x n -1 yf ç ÷.
è xø
x
a
32. (B) Let 2a and 2b be the major and minor axes of the
¶u
¶u
or xe u
+ ye u
= eu
¶x
¶y
or, x
1 æ1
1
ö
ç - 1 ÷u = - u
2 è2
4
ø
zero.
¶u
¶u
x
+y
= 0. u = 0
¶x
¶y
x2 + y2
x2 + y2
, eu =
= f (say)
x+ y
x+ y
¶f
¶f
x
+y
= f
¶x
¶y
¶2u
¶2u
¶2u
+ 2 xy
+ y2
= n( n - 1) u
2
¶x
¶x dy
¶y 2
æ yö
In (d)u = f ç ÷ It is a homogeneous function of degree
è xø
Substituting these in (ii), we get
dz
-y
x
dy
xdy - ydx
, dz =
=
+
×
dx x 2 + y 2 x 2 + y 2 dx
x2 + y2
29. (B) u = log
Thus percentage error in A =2%
¶2u
¶2u
¶u
+y
= ( n - 1)
2
¶x
¶y ¶x
¶x
æ yö
33. (A) Given that u = xyf ç ÷. Since it is a homogeneous
è xø
x2 y
It is a homogeneous function of
31. (B) In (a) u =
x+ y
function of degree 2.
By Euler’s theorem x
degree 2.
¶2u
¶2u
¶u ¶u
(as in question 30)
x 2 +y
= ( n - 1)
=
¶x
¶x ¶y
¶x ¶x
In (b) u =
x1 2 - y1 2
. It is a homogeneous function of
x1 4 + y1 4
æ1 1ö 1
degree ç - ÷ =
è2 4 ø 4
x2
=
Page
542
¶2u
¶2u
¶2u
+ 2 xy
+ y 2 2 = n( n - 1) u
2
¶x
¶x ¶y
¶y
1 æ1
3
ö
u
ç - 1 ÷u = 4 è4
16
ø
Engineering Mathematics
Thus x
¶u
¶u
+y
= nu (where n = 2)
¶x
¶y
¶u
¶u
+y
= 2u
¶x
¶y
34. (A) Given that u = x log xy....(i)
x 3 + y 3 + 3 xy = 1....(ii)
¶u ¶u ¶u dy
we know that
....(ii)
=
+
¶x ¶x ¶y dx
From (i)
and
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¶u
1
= x×
× y + log xy = 1 + log xy
¶x
xy
¶u
1
x
= x×
×x=
¶y
xy
y
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Now, D = (2 t 2 - 4) 2 + ( 4 t - 2) 2 is minimum when
From (ii), we get
3 x2 + 3 y2
Chap 9.2
dy
æ dy
ö
+ 3ç x
+ y × 1÷ = 0
dx
è dx
ø
æ x + yö
dy
÷÷
= - çç 2
dx
è y + xø
2
Þ
Substituting these in (A), we get
du
x ì æ x 2 + y öü
÷ý
= (1 + log xy) + í -çç 2
dx
y î è y + x ÷øþ
35. (B) The given function is homogeneous of degree 2.
¶z
¶z
Euler’s theorem x
+y
= 2z
¶x
¶y
E = (2 t 2 - 4) 2 + ( 4 t - 2) 2 is minimum.
Now, E = 4 t 4 - 16 t + 20
dE
Þ
= 16 t 3 - 16 = 16 ( t - 1) ( t 2 + t + 1)
dt
dE
=0 Þ
t =1
dt
éd2 E ù
d2 E
2
.
So,
=
48
t
= 48 > 0.
ê dt 2 ú
dt 2
ë
û ( t =1 )
So, t = 1 is a point of minima.
Thus Minimum distance = (2 - 4) 2 + ( 4 - 2) 2 = 2 2 .
36. (C) f ¢( x) = 6 x - 30 x + 36 = 6( x - 2)( x - 3)
2
Clearly, f ¢( x) > 0 when x < 2 and also when x > 3.
45. (A) Let the required point be P ( x, y). Then,
f ( x) is increasing in ] -¥, 2 [ È ] 3, ¥ [.
perpendicular distance of P ( x, y) from y - 3 x + 3 = 0 is
37. (B) f ¢ ( x) =
p=
( x + 1) - 2 x
1- x
= 2
( x 2 + 1) 2
( x + 1) 2
2
2
2
Clearly, ( x 2 + 1) 2 > 0 for all x.
So, f ¢ ( x) > 0
Þ
Þ
=
y - 3x + 3
10
x2 + 4 x + 5
(1 - x 2 ) > 0
10
=
=
x2 + 7 x + 2 - 3 x + 3
10
( x + 2) 2 + 1
10
or p =
( x + 2) 2 + 1
10
dp 2 ( x + 2)
d p
2
and
=
=
2
dx
dx
10
10
2
(1 - x) (1 + x) > 0
So,
This happens when -1 < x < 1.
æ d2 p ö
÷
> 0.
x = -2, Also, çç
2 ÷
è dx øx = -2
So, f ( x) is increasing in ] -1, 1 [.
dp
=0
dx
38. (A) f ¢ ( x) = 4 x 3 - 4 x = 4 x( x - 1)( x + 1).
So, x = -2 is a point of minima.
Clearly, f ¢ ( x) < 0 when x < - 1 and also when x > 1.
When x = -2, we get y = ( -2) 2 + 7 ´ ( -2) + 2 = -8.
Sol. f ( x) is decreasing in ] -¥, -1 [ È ] 1, ¥ [.
The required point is ( -2, - 8).
39.(C) f ¢ ( x) = 9 x8 + 21 x6 > 0 for all non-zero real values
46. (C) Let A (0, c) be the given point and P ( x, y) be any
of x.
point on y = x 2 .
Þ
D = x 2 + ( y - c) 2 is shortest when E = x 2 + ( y - c) 2 is
40. (C) f ¢ ( x) = 3kx - 18 x + 9 = 3 [ kx - 6 x + 3]
shortest.
This is positive when k > 0 and 36 - 12 k < 0 or k > 3.
Now,
2
2
41. (A) f ( x) = ( e ax + e - ax ) = 2 cosh ax.
f ¢( x) = 2 a sinh ax < 0 When x > 0 because a < 0
42. (D) f ¢( x) = - x 2 e - x + 2 xe - x = e - x x(2 - x).
Clearly, f ¢( x) > 0 when x > 0 and x < 2.
43. (B) f ¢( x) = (2 x + a)
1 < x <2 Þ
Þ
2 <2x < 4 Þ 2 + a <2x + a < 4 + a
(2 + a) < f ¢( x) < ( 4 + a).
For f ( x) increasing, we have f ¢( x) > 0.
E = x 2 + ( y - c) 2 = y + ( y - c) 2 Þ E = y 2 + y - 2 cy + c2
dE
d2 E
= 2 y + 1 - 2 c and
= 2 > 0.
dy
dy 2
dE
=0
dy
Þ
1ö
æ
y =çc - ÷
2
è
ø
1ö
æ
Thus E minimum, when y = ç c - ÷
2ø
è
1ö æ
1
æ
ö
Also, D = ç c - ÷ + ç c - - c ÷
2
2
è
ø è
ø
= c-
\2 + a ³ 0 or a ³ - 2. So, least value of a is -2.
2
é
1 öù
æ
2
ê . .. x = y = ç c - 2 ÷ú
è
øû
ë
1
4c - 1
=
4
2
44. (B) Let the point closest to (4, 2) be (2 t 2 , 4).
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543
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GATE EC BY RK Kanodia
UNIT 9
x
æ1ö
47. (B) Let y = ç ÷ then, y = x - x
è xø
51. (C)
d2 y
1
= x - x (1 + log x) 2 - x - x ×
2
dx
x
dy
=0
Þ
1 + log x = 0 Þ
dx
éd y ù
æ1ö
ê dx 2 ú æ 1 ö = -ç e ÷
è ø
ë
û çç x = ÷÷
2
è
So, x =
-
1
-1
e
é dy ù
=0
êë dx úû
(x = 2 )
x=
1
e
Þ
48. (A) f ¢( x) = 2 x -
250
500 ö
æ
and f ¢¢( x) = ç 2 + 3 ÷
x2
x ø
è
f ¢( x) = 0
250
=0
x2
Þ
a
+ 4b + 1 = 0
2
a + 8 b = -2....(ii)
52. (C) The given curve is
ellipse.
dz
=0
df
Þ
x = p 3 or
x = p.
æ p ö -3 3
f ¢¢ç ÷ =
< 0. So, x = p 3 is a point of maxima.
2
è 3ø
p öæ
pö 3 3
æ
.
Maximum value = ç sin ÷ç 1 + cos ÷ =
3 øè
3ø
4
è
2 sin x cos x
sin x + cos x
Þ
cos f = 0
2 cos f ( 4 - sin f) = 0
Þ
when f =
d 2z
= -2 cos 2 f - 8 sin f
df2
p d 2z
,
< 0.
2 df2
p
z is maximum when f = . So, the required point is
2
p
pö
æ
ç 5 cos , sin ÷ i.e. (0, 2).
2
2ø
è
53. (D) Let z =
1 + tan x 1 tan x
= +
x
x
x
dz
1
d 2z 2
= - 2 + sec 2 x and
=
+ 2sec2 x tan x
dx
x
dx 2 x 3
dz
1
Þ
x = cos x.
=0 Þ
- 2 + sec 2 x = 0
x
dx
2 2
2 2
(say),
=
(sec x + cosec x)
z
Then,
where z = (sec x + cosec x).
dz
cos x
= sec x tan x - cosec x cot x =
(tan 3 x - 1).
dx
sin 2 x
éd 2z ù
= 2 cos 3 x + 2sec2 x tan x > 0.
ê dx 2 ú
ë
û x = cos x
p
é pù
in ê0, ú.
4
ë 2û
dz
=0
dx
Þ
Sign of
dz
changes from -ve to +ve when x passes
dx
x=
p
f= .
2
dz
= - sin 2 f + 8 cos f Þ
df
2
Þ
x2 y2
+
= 1 which is an
5
4
z = 5 cos 2 f + 4 (1 + sin f) 2
dz
Þ
= -10 cos f sin f + 8(1 + sin f) cos f
df
49. (D) f ¢( x) = (2 cos x - 1)(cos x + 1) and
tan x = 1
1
and a = 2.
2
when z = D 2 is maximum
250 ö
æ
Thus minimum value = ç 25 +
÷ = 75.
5 ø
è
=
Thus z has a minima and therefore y has a maxima at
x = cos x.
through the point p 4. So, z is minimum at x = p 4 and
************
therefore, f ( x) is maximum at x = p 4.
Maximum value =
Page
544
a + 2 b = 1....(i)
D = ( 5 cos f - 0) 2 + (2 sin f + 2) 2 is maximum
x = 5.
f ¢¢( x) = - sin x(1 + 4 cos x).
1
or cos x = -1 Þ
f ¢( x) = 0 Þ cos x =
2
Þ
Let the required point be ( 5 cos f , 2 sin f). Then,
f ¢¢(5) = 6 > 0. So, x = 5 is a point of minima.
50. (C) f ( x) =
Þ
- a - 2b + 1 = 0
< 0.
1
is a point of maxima. Maximum value = e1 e .
e
2x -
Þ
Solving (i) and (ii) we get b = -
eø
Þ
dy a
= + 2 bx + 1
dx x
é dy ù
=0
êë dx úû
( x = -1 )
dy
= - x - x (1 + log x)
dx
Þ
Engineering Mathematics
2 2
= 1.
[sec( p 4) + cosec ( p 4)]
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GATE EC BY RK Kanodia
CHAPTER
9.3
INTEGRAL CALCULUS
1.
òx
(A)
2
x
dx is equal to
+1
1
log ( x 2 + 1)
2
(C) tan -1
(B) log ( x 2 + 1)
x
2
2. If F ( a) =
5.
(D) 2 tan -1 x
1
,
log a
to
1
(A)
( a x - a a + 1)
log a
(C)
3.
1
( a x + a a + 1)
log a
1
(B)
(ax - aa)
log a
(D)
1
( a x + a a - 1)
log a
(B) cot x + cosec x + c
(C) tan x - sec x + c
(D) tan x + sec x + c
4.
( 3 x + 1)
dx is equal to
2
-2x + 3
ò 2x
æ2x -1ö
3
5
÷
(A) log (2 x 2 - 2 x + 3) +
tan -1 çç
÷
4
2
è 5 ø
(B)
æ2x -1ö
4
÷
log (2 x 2 - 2 x + 3) + 5 tan -1 çç
÷
3
è 5 ø
æ2x -1ö
4
2
÷
(C) log (2 x 2 - 2 x + 3) +
tan -1 çç
÷
3
5
è 5 ø
æ2x -1ö
3
2
÷
(D) log (2 x 2 - 2 x + 3) +
tan -1 çç
÷
4
5
è 5 ø
is equal to
tan -1 (tan x)
(B) 2 tan -1 (tan x)
(C)
1
2
tan -1 (2 tan x)
(D) 2 tan -1 ( 12 tan x)
2 sin x + 3 cos x
ò 3 sin x + 4 cos x dx is equal to
(A)
9
1
x+
log( 3 sin x + 4 cos x)
25
25
(B)
18
2
x+
log( 3 sin x + 4 cos x)
25
25
(C)
18
1
x+
log( 3 sin x + 4 cos x)
25
25
(D) None of these
ò
(A)
(A) - cot x + cosec x + c
x
1
2
7.
dx
ò 1 + sin x is equal to
2
(A)
6.
a > 1 and F ( x) = ò a 2 dx + K is equal
dx
ò 1 + 3 sin
(B)
(C)
3 + 8 x - 3 x 2 dx is equal to
3x - 4
3 3
3 + 8 x - 3 x2 -
æ 3x - 4 ö
sin -1 ç
÷
18 3
è 5 ø
25
3x - 4
25 3
æ 3x - 4 ö
3 + 8 x - 3 x2 +
sin -1 ç
÷
6
18
è 5 ø
3x - 4
6 3
3 + 8 x - 3 x2 -
æ 3x - 4 ö
sin -1 ç
÷
18 3
è 5 ø
25
(D) None of these
8.
dx
ò
(A)
(C)
2 x + 3x + 4
2
1
2
1
2
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sin -1
is equal to
4x + 3
cosh -1
23
4x + 3
23
(B)
1
2
sinh -1
4x + 3
23
(D) None of these
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UNIT 9
9.
2x + 3
ò
x + x+1
2
dx is equal to
(A) 2 x 2 + x + 1 + 2 sinh -1
(C) 2 x 2 + x + 1 + sinh -1
(D) 2 x 2 + x + 1 - sinh -1
10.
(A)
ò1+ x + x
(A)
1
2
é
( x + 1) 2
-1
êlog x 2 + 1 + tan
ë
(B)
1
4
é
( x + 1) 2
log
+ 2 tan -1
ê
x2 + 1
ë
ù
xú
û
2x + 1
(C)
1
2
é
( x + 1) 2
-1
êlog x 2 + 1 - 2 tan
ë
3
ù
xú
û
2x + 1
+ x3
is equal to
ù
xú
û
(D) None of these
is equal to
x - x2
2
3
16.
dx
ò
3
dx
15.
3
2x + 1
x 2 + x + 1 + 2 sinh -1
(B)
2x + 1
Engineering Mathematics
x - x2 + c
(C) log (2 x - 1) + c
sin x
ò 1 - sin x dx
is equal to
(A) - x + sec x + tan x + k
(B) - x + sec x + tan x
(B) sin -1 (2 x - 1) + c
(C) - x + sec x - tan x
(D) - x - sec x - tan x
(D) tan -1 (2 x - 1) + c
17.
ò e { f ( x) + f ¢( x)} dx is equal to
x
(A) e f ¢( x)
(B) e x f ( x)
(C) e x + f ( x)
(D) None of these
x
1
11.
ò ( x + 1)
(A)
æ 2 ö
÷
2 cosh -1 çç
÷
è1 + xø
1 - 2 x - x2
dx is equal to
(B)
æ 2 ö
÷
(C) - 2 cosh -1 çç
÷
è1 + xø
12.
(A)
(C)
13.
dx
ò sin x + cos x
1
2
æ 2 ö
÷
cosh -1 çç
÷
2
è1 + xø
1
æ 2 ö
÷
cosh -1 çç
÷
2
è1 + xø
1
(D) -
is equal to
pö
æ
log tanç x + ÷
4
è
ø
æ x pö
log tanç + ÷
2
è2 8ø
dx
ò sin( x - a) sin( x - b)
(D)
1
2
æ x pö
log tanç + ÷
è2 6ø
æ x pö
log tanç + ÷
2
è4 4ø
1
is equal to
(A) sin( x - a) log sin( x - b)
ì sin( x - a) ü
(C) sin( a - b) log í
ý
î sin( x - b) þ
dx
ò ex - 1 is equal to
(C) log ( e
Page
546
-x
- 1)
x
x
+c
2
(B) e x cot
(C) e x tan x + c
òx
x
+c
2
(D) e x cot x + c
x3
dx is equal to
+1
2
(A) x 2 + log ( x 2 + 1) + c
(B) log ( x 2 + 1) - x 2 + c
(C)
1 2 1
x - log ( x 2 + 1) + c
2
2
(D)
1 2 1
x + log( x 2 + 1) + c
2
2
(A) x sin -1 x + 1 - x 2 + c
(B) x sin -1 x - 1 - x 2 + c
(C) x sin -1 x + 1 + x 2 + c
(D) x sin -1 x - 1 - x 2 + c
21.
ì sin( x - a) ü
1
(D)
log í
ý
sin( a - b)
î sin( x - b) þ
(A) log ( e x - 1)
æ 1 + sin x ö
ò e ççè 1 + cos x ÷÷ødx is
20. ò sin -1 x dx is equal to
æ x -aö
(B) log sinç
÷
è x -bø
14.
(A) e x tan
19.
(B)
1
18. The value of
ò
sin x + cos x
1 + sin 2 x
dx is equal to
(A) sin x
(B) x
(C) cos x
(D) tan x
1
22. The value of
(B) log (1 - e x )
(D) log (1 - e )
x
(A) -1/2
(C) 1/2
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ò 5 x - 3 dx is
0
(B) 13/10
(D) 23/10
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GATE EC BY RK Kanodia
UNIT 9
1
39.
x
ò ò(x
46. The area bounded by the curve r = q cos q and the
p
lines q = 0 and q = is given by
2
ö
ö
p æ p2
p æ p2
(B)
(A) çç
çç
- 1 ÷÷
- 1 ÷÷
4 è 16
16 è 6
ø
ø
+ y 2 ) dydx is equal to
2
0 x
(A)
7
60
(B)
(C)
4
49
(D) None of these
3
35
(C)
1
1 + x2
0
0
ò ò dy dx is
40. The value of
p
log ( 2 + 1)
4
(B)
(C)
p
log ( 2 + 1)
2
(D) None of these
p
log ( 2 - 1)
4
(C)
A
(B)
36
5
(D) None of these
x + y = a and x + y = a in the first quadrant is given
2
2
by
a2 - x 2
a
(A)
(C) 4 ò
p2
ò
ò
a cos 2 q
0
rdrdq
rdrdq
(B) 2 ò
p2
ò
0
(D) 2 ò
cos 2 q
a
0
p a cos 2 q
ò
0 0
rdrdq
rdrdq
48. The area of the region bounded by the curve
1
x2 4
0
y= 0
(A)
ò ò
(C)
ò ò
2
0
dxdy
3x ( x 2 + 2)
y= x 2 4
dydx
x2 4
y= 0
1
ò
ò ò
(D)
ò
y= 0
dydx
3x ( x 2 + 2)
y= x 2 4
dxdy
below by z = 0 and bounded above by z = h is given by
(B) pa 2 h
1
pa 3h
3
(D) None of these
50.
òòò
0
1
0
(B)
49. The volume of the cylinder x 2 + y 2 = a 2 bounded
(C)
0
cos 2 q
a
0
ò ò dxdy
(B)
a-x
0
p4
(A) pah
a2 - x 2
a
ò ò dxdy
(D) None of these
y ( x 2 + 2) = 3 x and 4 y = x 2 is given by
42. The area of the region bounded by the curves
2
(A) 4 ò
0
òò ydxdy is equal to
32
5
ö
æ p2
çç
- 1 ÷÷
ø
è 16
0
41. If A is the region bounded by the parabolas y 2 = 4 x
and x = 4 y, then
48
(A)
5
p
16
47. The area of the lemniscate r 2 = a 2 cos 2 q is given by
(A)
2
Engineering Mathematics
(D) None of these
a 2 - y2 a
ò ò dxdy
(C)
a-x
0
43. The area bounded by the curves y = 2 x , y = - x,
x = 1 and x = 4 is given by
33
(B)
2
(A) 25
47
(C)
4
44. The area bounded by the curves y 2 = 9 x, x - y + 2 = 0
e x + y+ z dxdydz is equal to
(B)
(C) ( e - 1) 2
(D) None of these
1
z
x+ z
-1
0
x -z
ò ò ò
( x + y + z) dy dx dz is equal to
(A) 4
(B) -4
(C) 0
(D) None of these
is given by
(A) 1
(B)
1
2
3
2
(D)
5
4
(C)
*************
45. The area of the cardioid r = a (1 + cos q) is given by
(A) 2 ò
p
ò
a (1 + cos q)
ò
a (1 + cos q)
q= 0 r = 0
(C) 2 ò
p2
0
Page
548
r=0
rdrdq
(B) 2 ò
rdrdq
(D) 2 ò
p
0
p4
0
ò
a (1 + cos q)
r=a
ò
a (1 + cos q)
r=0
3
( e - 1)
2
(A) ( e - 1) 3
51.
101
(D)
6
1 1 1
0 0 0
rdrdq
rdrdq
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GATE EC BY RK Kanodia
Integral calculus
SOLUTIONS
1. (A)
ò
Þ
F ( a) =
ax
+K
log a
aa
+K
log a
1
aa
1 - aa
K =
=
log a log a log a
F ( x) =
3. (C)
a
1-a
1
+
=
[ a x - a a + 1]
log a log a log a
x
a
dx
ò 1 + sin x
dx
x
x
x
x
ö
2
+ cos 2 ÷ + 2 sin cos
ç sin
2
2ø
2
2
è
x
sec 2
dx
2
=ò
=ò
dx
2
2
x
xö
xö
æ
æ
ç cos + sin ÷
ç 1 + tan ÷
2
2ø
2ø
è
è
x
Put 1 + tan = t
2
x
2 dt
2
Þ sec2 dx = 2 dt Þ ò 2 dt = - + K
2
t
t
x
-2 cos
-2
2
=
+K =
+K
x
x
x
1 + tan
cos + sin
2
2
2
x
x
x
-2 cos
cos - sin
2
2
2
=
´
+K
x
x
x
x
cos + sin
cos - sin
2
2
2
2
x
x
2 x
-2 cos
+ 2 sin cos
2
2
2 +K
=
2 x
2 x
- sin
cos
2
2
-(1 + cos x) + sin x
=
+ k = tan x - sec x - 1 + K
cos x
òæ
= tan x - sec x + c
3x + 1
dx
2x -2x + 3
2
=
p=
Þ
3
5
dx
log (2 x 2 - 2 x + 3) + ò
2
2
4
4 æ
æ 5ö
1ö
÷
ç x - ÷ + çç
÷
2ø
è
è 2 ø
3
5 1
= log (2 x 2 - 2 x + 3) +
tan -1
4
4æ 5ö
ç
÷
ç 2 ÷
è
ø
5. (C) Let I = ò
=ò
1
2
5
2
x-
dx
1 + 3 sin 2 x
cosec2 x dx
cosec2 x dx
=
cosec 2 x + 3 ò (1 + cot2 x) + 3
Put cot x = t Þ
I =ò
=
3
5
, q =
4
2
3
4x -2
5
dx
dx + ò 2
4 ò 2 x2 - 2 x + 3
2 2x -2x + 3
I=
1
1
log t = log ( x 2 + 1)
2
2
2. (A) F ( x) = ò a x dx + K =
=
4. (A) Let I = ò
Let 3 x + 1 = p( 4 x - 2) + q
x
dx
2
x +1
Put x 2 + 1 = t Þ 2 xdx = dt
1 1
x
ò x 2 + 1 dx = ò 2 × t dt
=
Chap 9.3
- cosec2 x dx = dt
1
-dt
t 1
æ cot x ö
= cot-1 = cot-1 ç
÷
4 + t2 2
2 2
è 2 ø
1
tan -1 (2 tan x)
2
6. (C) Let I = ò
2 sin x + 3 cos x
dx
3 sin x + 4 cos x
Let (2 sin x + 3 cos x) = p( 3 cos x - 4 sin x)
+ q( 3 sin x + 4 cos x)
p=
1
18
, q=
25
25
I=
1
25
=
3 cos x - 4 sin x
3 sin x + 4 cos x
18
ò 3 sin x + 4 cos x dx + 25 ò 3 sin x + 4 cos x dx
1
18
log ( 3 sin x + 4 cos x) +
x
25
25
2
ò
1
= 3
2
ì
4 öü
æ
2
2
2
ç x - ÷ï
ïæ
4 ö æ5 ö æ
4ö
æ5 ö
-1
3
í ç x - ÷ ç ÷ - ç x - ÷ + ç ÷ sin ç 5 ÷ý
3
3
3
3
è
ø
è
ø
è
ø
è
ø
çç
÷÷ï
ï
è 3 øþ
î
=
3 + 8 x - 3 x 2 dx = 3
ò
2
4ö
æ5 ö æ
ç ÷ - ç x - ÷ dx
3
3ø
è ø è
7. (B)
3x - 4
25 3
3x - 4
3 + 8 x - 3 x2 +
sin -1
6
18
5
8. (B)
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ò
dx
2 x + 3x + 4
2
=
1
2
ò
dx
æ 23 ö
3ö
æ
÷
ç x + ÷ + çç
÷
4
è
ø
è 4 ø
2
2
Page
549
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GATE EC BY RK Kanodia
UNIT 9
1
=
sinh
2
æ
ç
ç
è
2x + 3
ò
9. (B)
3
4 = 1 sinh -1 4 x + 3
2
23
23 ö
÷
÷
4 ø
x+
-1
=ò
x2 + x + 1
=ò
x2 + x + 1
=
dx +
2x + 1
ò
dx + 2 ò
= 2 x 2 + x + 1 + 2 sinh -1
dx
ò
x 1- x
13. (D)
x2 + x + 1
æ 3ö
1ö
æ
÷
ç x + ÷ + çç
÷
2ø
è
è 2 ø
2
2
1
x+
2
3
2
2x + 1
I =ò
( x + 1) 1 - 2 x - x 2
Þ
dx = -
1
dt
t2
ò
dt
æ 1
t 2 - çç
è 2
ö
÷
÷
ø
2
=-
2
2
ò
1
sin( a - b)
ò sin( x - a) sin( x - b)
=
1
sin ( a - b)
sin [( x - b) - ( x - a)]
dx
sin( x - b) cos( x - a) - cos( x - b) sin( x - a)
dx
sin( x - a) sin( x - b)
=
1
[cot( x - a) - cot( x - b)]dx
sin( a - b) ò
=
1
[log sin ( x - a) - log sin ( x - b)] dx
sin ( a - b)
=
ì sin( x - a) ü
1
log í
ý
sin ( a - b)
î sin( x - b) þ
14. (D) Let I =
òe
dx
e - x dx
=ò
1 - e- x
-1
x
dx
15. (B) Let I = ò
=ò
dt
2 t2 - 1
dx
1 + x + x2 + x3
dx
(1 + x) (1 + x 2 )
A
Bx + C
1
=
+
2
(1 + x)(1 + x ) 1 + x 1 + x 2
Let
1 = A(1 + x 2 ) + ( Bx + C)(1 + x)
1
2
æ 2 ö
÷
cosh -1 çç
÷
2
è x + 1ø
1
=-ò
sin( a - b) dx
Put 1 - e - x = t Þ e - x dx = dt
dt
I =ò
= log t = log (1 - e - x )
t
1
dt
t2
1
12. (C)
Page
550
1
1
æ1
ö æ1
ö
1 - 2ç - 1 ÷ - ç - 1 ÷
t
t
t
è
ø è
ø
2
=-
=
1
t
-
1
=-
x+c
dx
ò sin( x - a) sin( x - b)
=
=I
(2 x - 1) + c
Put x + 1 =
2
pö
æ
ò cosecçè x + 4 ÷ødx
ò sin( x - a) sin( x - b)
3
I = ò 2 dq = 2 q + c = 2 sin -1
11. (D) Let I = ò
1
1
sin( a - b)
´ò
Put x = sin q Þ dx = 2 sin q cos q dq
2 sin q cos q
2 sin q cos q
I =ò
dq = ò
dq
2
sin q cos q
sin q 1 - sin q
I = sin
pö
æ
sinç x + ÷
4ø
è
=
=
dx
2
-1
dx
1 é
1æ
p öù
1
æ x pö
log tan ç + ÷
- log cot ç x + ÷ú =
ê
2è
4 øû
2 ë
2
è2 8ø
=
2 dx
( x 2 + x + 1)1 2
+ 2 sinh -1
1
2
10. (B)
2
ò
dx
x2 + x + 1
2x + 1
1
=
Engineering Mathematics
dx
ò sin x + cos x
dx
p
p
sin x cos + cos x sin
4
4
t
cosh -1
1
Comparing the coefficients of x 2 , x and constant terms,
2
A + B = 0, B + C = 0, C + A = 1
Solving these equations, we get
1
1
1
A = , B=- , C=
2
2
2
1
1
1 x -1
I =
dx - ò 2
dx
2 ò1+ x
2 x +1
=
1
1
1
log (1 + x) - log ( x 2 + 1) + tan -1 x
2
2
2
=
1
4
é
( x + 1) 2
log
+ 2 tan -1
ê
x2 + 1
ë
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û
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Integral calculus
16. (B) Let I = ò
sin x
dx
1 - sin x
ò
=
=ò
1 - (1 - sin x)
dx
1 - sin x
=ò
=ò
1
1 + sin x
dx - ò dx = ò
dx - x
1 - sin x
1 - sin 2 x
=ò
=ò
1 + sin x
dx - x = ò (sec2 x + sec x tan x) dx - x
cos 2 x
= tan x + sec x - x
Chap 9.3
sin x + cos x
(sin x + cos 2 x) + 2 sin x cos x
2
sin x + cos x
(cos x + cos x) 2
dx
sin x + cos x
dx = ò dx = x
sin x + cos x
35
35
ò 5 x - 3 dx = - ò 5 x - 3 dx +
22. (D)
0
0
= ò e x f ( x) dx + ò e x f ¢( x) dx
= { f ( x) e x - ò f ¢( x) ex dx} + ò ex f ¢( x) dx = f ( x) × ex
9ö
æ 9
=ç+ ÷+
è 10 5 ø
=
æ 1 + sin x ö
18. (A) Let I = ò e çç
÷÷ dx
è 1 + cos x ø
x
23. (B)
24. (D)
c
dt
= [tan -1 t ]1e
+1
2
p
4
c
ò x(1 - x) dx = ò ( x - x ) dx
2
0
1 ö
1
æ1
= ç x 2 - x 3 ÷ = c 2 ( 3 - 2 c)
2
3
è
ø0 6
x3
x × x2
dx = ò 2
dx
x +1
x +1
c
x( x + 1 - 1)
x
dx = ò xdx - ò 2
dx
2
x +1
x +1
2
20. (A) Let I = ò sin
= sin -1 x × x - ò
-1
1
1 - x2
x
1 - x2
x dx = ò sin
-1
x × 1 × dx
× x dx
dx
xdx = - tdt = x sin -1 x + ò dt
= x sin -1 x + t = x sin -1 x + 1 - x 2 + c
sin x + cos x
1 + sin 2 x
dx
ò
x(1 - x) dx = 0
1 2
c ( 3 - 2 c) = 0
6
Þ
0
Þ
c=
3
2
25. (D) Put x 2 + x = t
In second part put 1 - x 2 = t 2
ò
e
òt
e x dx = dt =
c
1 2 1
x - log ( x 2 + 1) + c
2
2
21.
e x dx
2x
+1
0
0
2
ò
Þ
1
òe
= tan -1 e - tan -1 1 = tan -1 e -
x
= e tan + c
2
x-
dx
=
+ e- x
1
x
= x sin
x
Put e x = t
1ì x
x
x
x
ü
x
x
í e × 2 tan - ò e × 2 tan dx ý + ò e tan dx
2î
2
2
2
þ
-1
òe
0
1
x
x
= ò e x sec 2 dx + ò e x tan dx
2
2
2
=
éæ 5
ö æ 9 9 öù
êç 2 - 3 ÷ - ç 10 - 5 ÷ú
è
ø è
øû
ë
9 æ 1 9 ö 13
+ ç- +
÷=
10 è 2 10 ø 10
1
x
xö
æ
ç 1 + 2 sin cos ÷
x
2
2
÷ dx
=òe ç
x
÷÷
çç
2 cos 2
è
ø
2
=ò
ò 5 x - 3 dx
35
1
35
19. (C) I = ò
1
ö
æ 5 x2
æ 5
ö
= ç - x 2 + 3 x ÷ + çç
- 3 x ÷÷
è 2
ø0
ø3 5
è 2
17. (B) Let I = ò e { f ( x) + f ¢( x)} dx
x
=
dx
1
ò
2x + 1
0
x+ x
26. (A)
2
2
dx = ò
0
dt
t
Þ
(2 x + 1) dx = dt
= 2( t1 2 ) 20 = 2 2
p
òx
4
sin 5 xdx
p
Since, f ( - x) = ( - x) 4 sin 5 ( - x) = -x 4 sin 5 x
f ( x) is odd function thus
p
òx
4
sin 5 x dx = 0
p
27. (A)
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p2
p2
0
0
2
ò cos x dx =
ò
1
(cos 2 x + 1) dx
2
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UNIT 9
Engineering Mathematics
1 æ1
ö
ç sin 2 x + x ÷
2 è2
ø0
pö
æp
= ò 2 sinç (1 - t) - ÷dt =
2
4
è
ø
0
=
1 é1
æp
öù
(sin p - sin 0) + ç - 0 ÷ú
2 êë2
è2
øû
pö
æp
= - ò 2 sin ç t - ÷dt = - 1
2
4
è
ø
0
=
1 é1
pù p
(0 - 0) - 0 + ú =
ê
2 ë2
2û 4
2I = 0
1
p2
=
p ö
ò 2 sinçè 4 - 2 t ÷ødt
0
1
æ 3ö æ 1ö
Gç ÷ Gç ÷ 1 p
p2
p
2
2
Aliter 1. ò cos 2 x dx = è ø è ø = 2 =
2
4
æ4ö
0
2 Gç ÷
è2 ø
p2
Aliter 2. Use Walli’s Rule
æp
1
ò cos
2
x =
0
Þ
I =0
31. (C) Let I =
2a
f ( x)
ò f ( x) + f (2 a - x) dx ....(1)
0
I=
f (2 a - x)
2a
ò f (2 a - x) + f ( x) dx....(2)
0
1 p p
× =
2 2 4
Adding (1) and (2), we get
2I =
2a
f ( x) + f (2 a - x)
ò f ( x) + f (2 a - x) dx
=
0
a
28. (B) Let I = ò a 2 - x 2 dx
Þ
Put x = a sin q Þ dx = a cos q dq when x = 0, q = 0,
p
when x = a, q =
2
32. (C) Let I =
2a
ò 1 × dx = [ x ]
2a
0
= 2a
0
I = a
0
Put
0
Þ
p2
= a 2 ò cos 2 q dq = a 2 ×
0
1 p
(By Walli’s Formula)
×
2 2
1 - x2
1 - x2
xdx
1 - x2 = t
1
2 1 - x2
( -2 x) dx = dt
when x = 0, t = 1, when x = 1, t = 0
0
pa 2
4
I = ò -e t dt = -[ e t ]10 = -[ e 0 - e1 ] = e - 1
1
a
Aliter:
ò
e
0
p2
I = ò a 2 - a 2 sin 2 q a cos q dq
=
1
ò
a 2 - x 2 dx
1
0
a
é
1
xù
pa 2 ù pa 2
é1
= ê x a 2 - x 2 + a 2 sin -1 ú = ê0 +
=
4
2
a û0 ë
4 úû
ë2
p2
dx
1
x
+ x2
0
33. (B) Let I = ò
1
1
=ò
0
29. (D) Let I = ò log (tan x) dx ....(1)
0
dx
æ 3ö
1ö
æ
÷
ç x - ÷ + çç
÷
2ø
è
è 2 ø
2
p2
æp
ö
I = ò log tan ç - x ÷ dx
2
è
ø
0
=
0
=
æ
2 é
1
1 öù 2 æ p p ö
-1
÷ú=
- tan -1 çç ç + ÷
êtan
3 ë
3
3 ÷øû
3 è6 6ø
è
=
p2
I = ò log (cot x)....(2)
2
é
1ù
x- ú
1 ê
-1
2ú
êtan
3 ê
3 ú
2 êë
2 úû 0
2p
3 3
=
2p 3
9
Adding (1) and (2), we get
p2
2 I = ò [log (tan x) + log (cot x)]dx
1
34. (B) Let I =
-1
0
0
p2
= ò log (tan x × cot x) dx
0
p2
= ò log 1 dx = 0
Þ
I =0
=
1
Page
552
x
-x
ò-1 x dx +
0
dx =
1
ò -1dx + ò 1 × dx = -[ x ]
0
-1
-1
+ [ x ]10
0
= - [0 - ( -1)] + [1 - 0 ] = 0
0
æ pt p ö
30. (D) Let I = ò 2 sin ç - ÷dt ....(i)
è 2 4ø
0
ò
x
35. (C)
100 p
p
0
0
ò |sin x|dx = 100 ò |sin x|dx
[ . .. sin x is periodic with period p]
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0
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Integral calculus
p
1
= 100 ò sin x dx = 100( - cos x) 0p
40. (D)
0
1 + x2
ò ò
0
= 100( - cos p + cos 0) = 100(1 + 1) = 200.
p
p
0
0
= ( - cos x) m (sin x) n
1
[ x 1 + x 2 + log( x + 1 + x 2 )]10
2
1
= [ 2 + log (1 + 2 )]
2
=
41. (A) Let I = òò ydxdy,
= -cos m x sin n x, if m is odd
I = ò cos
dx
0
f ( p - x) = cos m ( p - x) sin n ( p - x)
m
x2
0
1
Where f ( x) = cos m x sin n x
p
0
1
dydx = ò [ y ]01 +
= ò 1 + x 2 dx
36. (C) Let I = ò cos m x sin nx dx = ò f ( x) dx
A
x sin x dx = 0, if m is odd
Solving the given equations y 2 = 4 x and x 2 = 4 y , we get
n
x = 0, x = 4 . The region of integration A is given by
0
2 x
4
é y2 ù
=
ydydx
ò
ò
ê 2 ú dx
û x2 4
0 x2 4
0 ë
4 2 x
p
A =ò
37. (A) Let I = ò xF (sin x) dx ....(1)
0
p
4
é
48
1æ
x4 ö
x5 ù
÷÷dx = ê x 2 çç 4 x ú = 5
160
2
10
ø
è
ë
û0
0
4
= ò ( x - p) F [sin ( p - x)]dx
=ò
0
I =
Chap 9.3
p
ò ( p - x) F(sin x) dx ....(2)
42. (A) The curves are
0
Adding (1) and (2), we get
p
2 I = ò pF (sin x) dx
x 2 + y 2 = a 2 ...
....(i)
x + y = a...
....(ii)
0
Þ
I=
The curves (i) and (ii) intersect at A (a, 0) and B (0,a)
p
1
pF (sin x) dx
2 ò0
p2
38. (B) Let I = ò
0
The required area A =
ex æ 2 x
xö
+ 2 tan ÷dx
ç sec
2 è
2
2ø
p2
1
x
= ò e xsec 2 dx +
2
2
0
4 2 x
A =ò
1
x
Here, I1 = ò e x sec 2 dx
2
2
0
òe
0
= e - I 2 , I1 + I 2 = e
p2
I = I1 + I 2 = e
x
ò0 e × 2 tan 2 dx
x
tan
ò dydx
y = - x....(ii)
2 x
-x
1
4
= ò [2 x + x ]dx
1
æ 32
ö æ 4 1 ö 101
=ç
+ 8÷ -ç + ÷ =
6
è 3
ø è3 2ø
p2
x
4
ò dydx = ò [ y ]
1 -x
p2
y= a - x
ò
If a figure is drawn then from fig. the required area is
p2
p
æ
ö
= ç e p 2 tan - 0 ÷ 4
è
ø
x =0
y = 2 x i.e., y 2 = 4 x....(i)
x
ò0 e tan 2 dx = I1 + I 2
x
p2
a2 - x 2
43. (D) The given equations of the curves are
p2
xù
1
é1
= ê e x × 2 tan ú 2 û0 2
ë2
a
44. (B) The equations of the given curves are
x
dx
2
y 2 = 9 x....(i)
x - y + 2 = 0....(ii)
The curves (i) and (ii) intersect at
p2
A(1, 3) and B(4, 6)
p2
If a figure is drawn then from fig. the required area is
1
39. (B)
ò
0
1
x
x
1 3ù
é 2
2
2
òx ( x + y ) dy dx = ò0 êë x y + 3 y úû x dx
1
1
1 ù
é
= ò ê x 5 2 + x 3 2 - x 3 - x 3 ú dx
3
3 û
0 ë
1
2 52 1 4 ù
3
é2
= ê x7 2 +
x - x ú =
15
3 û 0 35
ë7
4 3 x
A =ò
4
ò dydx = ò [ y ]
1 x+ 2
3 x
x+ 2
dx
1
4
4
1
é
ù
= ò [ 3 x - ( x + 2)]dx = ê2 x 3 2 - x 2 - 2 x ú
2
ë
û1
1
1
æ
ö 1
= (16 - 8 - 8) - ç 2 - - 2 ÷ =
2
è
ø 2
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UNIT 9
Thus the equation volume is V = 4 ò zdxdy
45. (A) The equation of the cardioid is
r = a (1 + cos q)
A
....(i)
If a figure is drawn then from fig. the required area is
p a (1 + cos q)
ò
Required area A = 2
=
=
1
2
1
4
cos 2 qdq =
0
2
ò q dq +
0
1
4
1
4
dq
1 1 1
50. (A)
p2
ò q (1 + cos 2 q)dq
2
òòòe
1 1
= ò ò [ ex +
0
2
ò q cos 2 q dq
dxdydz
1 1
] dydz = ò ò [ e1 +
0 0
0
1
= ò [( e 2 + z - e1 + z ) - ( e1 + z - e z )]dz
0
1
= ò ( e 2 + z - 2 e1 + z + e z ) dz = [ e 2 + z - 2 e1 + z + e z ]10
0
=
p3 1 æ -p
ö 1
+ ç
-0÷ 96 4 è 4
ø 8
= ( e 3 - 2 e 2 + e) - ( e 2 - 2 e + 1)
p2
= e 3 - 3e 2 + 3e - 1 = ( e - 1) 3
p2
1
ò cos 2 q dq
51. (C)
0
z x+ z
ò ò ò ( x + y + z) dydxdz
-1 0 x - z
p2
p
p3
p 1æ1
ö
- ç sin 2 q ÷ =
96 16 8 è 2
ø0 16
ö
æ p2
çç
- 1 ÷÷
16
ø
è
47. (A) The curve is r 2 = a 2 cos 2 q
If a figure is drawn then from fig. the required area is
p4
a
é1 2 ù
òr = 0rdrdq = 4 ò0 êë2 r úû 0
p4
x+ z
=
é( x + y + z) 2 ù
ò-1 ò0 êë
ú dxdz
2
ûx- y
=
é(2 x + 2 z) 2 æ 2 x ö2 ù
-ç
÷ ú dxdz
òòê 2
è 2 ø úû
-1 0 ê
ë
1 z
1 z
z
1
1 3
é
ù
é( x + z) 3 x 3 ù
= 2 ò êò (( x + z) 2 - x 2 ) dx ú dz = 2 ò ê
- ú dz
3
3 û0
-1 ë 0
-1 ë
û
cos 2 q
dq
1
ésin 2 q ù
= 2 ò a cos 2 q dq= 2 a ê
= a2
ú
2
ë
û0
0
2
2
= ò [(2 z) 3 - z 3 - z ]dz =
3 -1
3
48. (C) The equations of given curves are
æ1 1ö
= 4ç - ÷ = 0
è4 4ø
p4
2
y( x 2 + 2) = 3 x....(i)
1
éz4 ù
6
z
dz
=
4
ò
ê4ú
ë û -1
-1
1
3
and 4 y = x 2 ....(ii)
The curve (i) and (ii) intersect at A (2, 1).
If a figure is drawn then from fig. the required area is
The required area A =
2
3x ( x 2 + 2 )
x =0
y= x 2 4
ò
ò dxdy
49. (B) The equation of the cylinder is x 2 + y 2 = a 2
The equation of surface CDE is z = h.
If a figure is drawn then from fig. the required area is
Page
554
- e y + z ]dydz
= ò [ e1 + y + z - e y + z ]10 dz
p3 1 éæ
cos 2 q ö
æ cos 2 q ö ù
- êç -q
÷ - òç÷d q ú
96 4 ëêè
2 ø0
2 ø úû
0è
2
y+ z
1
0
p 4 a cos 2 q
x + y+ z
y+ z 1
0
0 0
p2
=
ò
1 p
× = pa 2 h.
2 2
0 0 0
p2
q= 0
dx = a cos q dq,
0
p2
ù
p3 1 é
=
+ ê- ò q sin 2 q dq ú
96 4 ë 0
û
A =4
0
0
p2
p2
p2
ù
1 é1 ù
1 éæ sin 2 q ö
sin 2 q
= ê q 3 ú + êç q 2
dq ú
÷ - ò 2q
4 ë3 û0
4 êëè
2 ø0
2
úû
0
=
a
dx = 4 hò a 2 - x 2 dx
q cos q
p2
p2
p2
Þ
= 4 ha 2 ò cos 2 q dq = 4 ha 2 ×
é1 2 ù
òr = 0rdrdq = ò0 êë2 r úû o
2
0
0
p2
p 2 q cos q
òq
a2 - x 2
0
Volume V = 4 h ò a 2 - a 2 sin 2 q × a cos q dq
The required area
ò
a
p2
r = q cos q....(i)
q= 0
a
ò ò hdydx = 4 hò [ y ]
Let x = a sin q,
r=0
46. (C) The equation of the given curve is
A=
=4
a2 - x 2
0
ò rdrdq
q= 0
Engineering Mathematics
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Complex Variables
Chap 9.5
10. The integration of f ( z) = x 2 + ixy from A(1, 1) to B(2,
17. The value of f ( 3) is
4) along the straight line AB joining the two points is
(A) 6
(B) 4i
-29
(A)
+ i11
3
(C) -4i
(D) 0
(C)
29
(B)
- i11
3
23
+ i6
5
(D)
18. The value of f ¢(1 - i) is
23
- i6
5
11.
e2 z
òc ( z + 1) 4 dz = ? where c is the circle of z = 3
(A)
4 pi -3
e
9
(B)
4 pi 3
e
9
(C)
4 pi -1
e
3
(D)
8 pi -2
e
3
12.
1 - 2z
òc z( z - 1)( z - 2) dz = ? where c is the circle z = 15.
(A) 2 + i 6 p
(B) 4 + i 3p
(C) 1 + ip
(D) i3p
(A) 7 ( p + i2)
(B) 6 (2 + ip)
(C) 2 p (5 + i13)
(D) 0
Statement for 19–21:
Expand the given function in Taylor’s series.
19. f ( z) =
(A) 1 + 2( z + z 2 + z 3......)
(B) -1 - 2( z - z 2 + z 3......)
(C) -1 + 2( z - z 2 + z 3......)
(D) None of the above
20. f ( z) =
13. ò ( z - z 2 ) dz = ? where c is the upper half of the circle
(B)
-3
(D)
2
3
(C)
2
14.
2
3
-1 é
1
1
ù
1 - ( z - 1) + 2 ( z - 1) 2 .......ú
ê
2 ë
2
2
û
(B)
1é
1
1
ù
1 - ( z - 1) + 2 ( z - 1) 2 .......ú
2 êë
2
2
û
(C)
1é
1
1
ù
1 + ( z - 1) + 2 ( z - 1) 2 .......ú
2 êë
2
2
û
cos pz
dz = ? where c is the circle z = 3
c z -1
(B) - i2 p
(C) i6 p2
(D) - i6 p2
15.
(D) None of the above
ò
(A) i2 p
21. f ( z) = sin z about z =
sin pz 2
òc ( z - 2)( z - 1) dz = ? where c is the circle z = 3
(A) i6p
(B) i2p
(C) i4p
(D) 0
16. The value of
1 cos pz
dz around a rectangle with
2 pi òc z 2 - 1
vertices at 2 ± i , -2 ± i is
(A) 6
(B) i2 e
(C) 8
(D) 0
2
ù
1 é
pö 1 æ
pö
æ
1
+
z
z
ç
÷
ç
÷ - .......ú
ê
4 ø 2 !è
4ø
2 êë
è
úû
(B)
2
ù
1 é
pö 1 æ
pö
æ
ê1 + ç z - ÷ + ç z - ÷ + .......ú
4 ø 2 !è
4ø
2 êë
è
úû
(C)
2
ù
1 é æ
pö 1 æ
pö
ê1 - ç z - ÷ - ç z - ÷ - .......ú
4 ø 2 !è
4ø
2 êë è
úû
(D) None of the above
22. If z + 1 < 1, then z -2 is equal to
(A) 1 +
x + y = 4.
¥
å ( n + 1)( z + 1)
n -1
n =1
(B) 1 +
¥
å ( n + 1)( z + 1)
n +1
n =1
3z 2 + 7 z + 1
f ( z0 ) = ò
dz , where c is the circle
( z - z0 )
c
2
p
4
(A)
Statement for Q. 17–18:
2
1
about z = 1
z +1
(A)
c
z =1
-2
(A)
3
z -1
about the points z = 0
z +1
(C) 1 +
¥
å n( z + 1)
n
n =1
(D) 1 +
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¥
å ( n + 1)( z + 1)
n
n =1
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UNIT 9
Statement for Q. 23–25.
28. The Laurent’s series of f ( z) =
1
in Laurent’s
( z - 1)( z - 2)
Expand the function
series for the condition given in question.
23. 1 < z < 2
1 2
3
(A) + 2 + 3 + .......
z z
z
(B) K - z -3 - z -2 - z -1 (C)
Engineering Mathematics
1 1
1
1 3
- z - z2 z -K
2 4
8
18
where z < 1
1
5 3 21 5
(A) z z +
z ..........
4
16
64
(B)
1 1 2
5 4 21 6
+ z +
z +
z ..........
2 4
16
64
(C)
1
3
15 5
z - z3 +
z ..........
2
4
8
(D)
1 1 2 3 4 15 6
+ z + z +
z ..........
2 2
4
8
1
3
7
+ 2 + 4 ...........
2
z
z
z
1 - e Zz
at its pole is
z4
-4
(B)
3
(D) None of the above
29. The residue of the function
24. z > 2
(A)
4
3
(C)
-2
3
(A)
6 13 20
+
+ 3 + ........
z z2
z
(B)
1 8 13
+
+
+ .........
z z2 z3
(C)
1
3
7
+ 3 + 4 + .........
2
z
z
z
(D)
2
3
4
- 3 + 4 - ........
2
z
z
z
25. z < 1
(A) 1 + 3z
+7 2 15 2
z +
z .....
2
4
(B)
1 3
7
15 3
+ z + z2 +
z ...
2 4
8
16
(C)
1 3 z2 z3
+ +
+
.......
4 4
8 16
26. If z - 1 < 1 , the Laurent’s series for
30. The residue of z cos
1
is
z( z - 1)( z - 2)
( z - 1) 3 ( z - 1) 5
- .........
2!
5!
(D) - ( z - 1)
-1
1
at z = 0 is
z
-1
(B)
2
1
2
(C)
1
3
31.
ò z(1 - z)( z - 2) dz = ? where c is
(D)
1 - 2z
-1
3
z = 15
.
(A) - i3p
(B) i3 p
(C) 2
(D) -2
32.
z cos z
dz = ? where c is z - 1 = 1
pö
c z ç
÷
2ø
è
òæ
(A) 6 p
(B) - 6p
(C) i2p
(D) None of the above
(C) - ( z - 1) - ( z - 1) - ( z - 1) - ..........
3
2
3
(A)
( z - 1) 3 ( z - 1) 5
- ...........
2!
5!
(B) - ( z - 1) -1 -
(D)
c
(D) None of the above
(A) - ( z - 1) -
z
is,
( z + 1)( z 2 + 4)
2
5
1
- ( z - 1) - ( z - 1) - ( z - 1) - .........
3
5
33.
2
ò z e z dz = ? where c is z = 1
c
27. The Laurent’s series of
1
for z < 2 is
z( e z - 1)
1
1
1
1 2
(A) 2 +
+
+ 6z +
z + ..........
z
2 z 12
720
1
1
1
1 2
(B) 2 +
z + ..........
z
2 z 12 720
(C)
1
1
1
1 2
+
+
z2 +
z + ..........
z 12 634
720
(D) None of the above
Page
566
(B) - i3p
(A) i3p
C)
34.
ip
3
(D) None of the above
2p
dq
ò 2 + cos q = ?
0
(A)
-2 p
2
(C) 2 p 2
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(B)
2p
3
(D) -2 p 3
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Complex Variables
¥
35.
x2
ò 2 2 2 2 dx = ?
-¥ ( x + a )( x + b )
(A)
p ab
a+b
(B)
(C)
p
a+b
(D) p ( a + b)
36.
¥
dx
ò1+ x
6
p ( a + b)
ab
=?
0
(A)
(C)
p
6
(B)
2p
3
(D)
p
2
***************
p
3
Chap 9.5
SOLUTIONS
1. (C) Since, f ( z) = u + iv =
Þ
u=
x3 - y3
;
x2 + y2
v=
x 3(1 + i) - y 3(1 - i)
; z ¹0
x2 + y2
x3 + y3
x2 + y2
Cauchy Riemann equations are
¶u ¶v
¶u
¶v
and
=
=¶x ¶y
¶y
¶x
By differentiation the value of
we get
¶u ¶y ¶v ¶v
at(0, 0)
,
,
,
¶x ¶y ¶x ¶y
0
, so we apply first principle method.
0
At the origin,
¶u
u(0 + h, 0) - u(0, 0)
h3 h2
= lim
= lim
= 1
h ®0
h ®0
¶x
h
h
¶u
u(0, 0 + k) - u(0, 0)
- k3 k2
= lim
= lim
= -1
k ®0
¶v h ®0
k
k
¶v
v(0 + h, 0) - v(0, 0)
h3 h2
= lim
= lim
=1
h ®0
¶x h ®0
h
h
¶v
v(0, 0 + k), v(0, 0)
k3 k2
= lim
= lim
=1
k ®0
¶y k ®0
k
k
Thus, we see that
¶u ¶v
¶u
¶v
and
=
=¶x ¶y
¶y
¶x
Hence, Cauchy-Riemann equations are satisfied at
z = 0.
Again, f ¢(0) = lim
z ®0
f ( z) - f (0)
z
é( x 3 - y 3) + i( x 3 + y 3)
1 ù
= lim ê
2
2
z ®0
(x + y )
( x + iy) úû
ë
Now let z ® 0 along y = x, then
é( x 3 - y 3) + i( x 3 + y 3)
1 ù
2i
1+ i
f ¢ (0) = lim ê
=
=
2
2
ú
z ®0
(x + y )
( x + iy) û 2(1 + i)
2
ë
Again let z ® 0 along y = 0, then
é x 3 + i( x 3) 1 ù
f ¢ (0) = lim ê
=1 + i
2
x ®0
x úû
ë (x )
So we see that f ¢(0) is not unique. Hence f ¢(0) does not
exist.
df
Df
= lim
dz Dz ®0 Dz
Du + iDv
or f ¢( z) = lim
Dz ®0 Dx + iDy
2. (A) Since, f ¢( z) =
....(1)
Now, the derivative f ¢( z) exits of the limit in equation
(1) is unique i.e. it does not depends on the path along
which Dz ® 0.
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UNIT 9
Let Dz ® 0 along a path parallel to real axis
Þ
Dy = 0 \ Dz ® 0
Þ
Now let v be the conjugate of u then
¶v
¶v
¶u
¶u
dv =
dx +
dy = dx +
dy
¶x
¶y
¶y
¶x
Dx ® 0
Now equation (1)
Du + iDv
Du
Dv
f ¢( z) = lim
= lim
+ i lim
Dx ®0
Dx ®0 Dx
Dx ®0 Dx
Dx
¶u
¶v
f ¢( z) =
+i
¶x
¶x
Engineering Mathematics
(by Cauchy-Riemann equation)
Þ
....(2)
dv = 2 x dx + 2(1 - y) dy
On integrating v = x 2 - y 2 + 2 y + C
Again, let Dz ® 0 along a path parallel to imaginary
5. (C) Given f ( z) = u + i v
....(1)
axis, then Dx ® 0 and Dz ® 0 ® Dy ® 0
Þ
....(2)
Thus from equation (1)
Dz + iDv
Du
Dv
¶u ¶v
= lim
+ i lim
=
+
f¢( z) = lim
Dy ®0
Dy ®0 iDy
Dy ®0 iDz
iD y
i ¶y ¶y
add equation (1) and (2)
f ¢( z) =
-i ¶ u ¶ v
+
¶y
¶y
....(3)
Now, for existence of f ¢( z) R.H.S. of equation (2) and (3)
must be same i.e.,
¶u
¶v ¶v
¶u
+i
=
-i
¶x
¶x ¶y
¶y
if ( z) = -v + iu
Þ
(1 + i) f ( z) = ( u - v) + i( u + v)
Þ
F ( z) = U + iV
where, F ( z) = (1 + i) f ( z); U = ( u - v); V = u + v
Let F ( z) be an analytic function.
Now, U = u - v = e x (cos y - sin y)
¶U
¶U
and
= e x (cos y - sin y)
= e x ( - sin y - cos y)
¶x
¶y
Now, dV =
¶u ¶v
¶v -¶u
and
=
=
¶x ¶y
¶x
¶y
-¶U
¶U
dx +
dy....(3)
¶y
¶x
= e x (sin y + cos y) dx + e x (cos y - sin y) dy
¶u
¶u ¶v
¶v
f ¢( z) =
-i
=
+i
¶x
¶y ¶y
¶x
= d[ e x (sin y + cos y)]
on integrating V = e x (sin y + cos y) + c1
3. (A) Given f ( z) = x 2 + iy 2 since, f ( z) = u + iv
F ( z) = U + iV = e x (cos y - sin y) + ie x (sin y + cos y) + ic1
Here u = x 2 and v = y 2
¶u
¶u
Now, u = x 2
Þ
= 2 x and
=0
¶x
¶y
= e x (cos y + i sin y) + ie x (cos y + i sin y) + ic1
and v = y 2
and f ¢( z) =
(1 + i) f ( z) = (1 + i) e z + ic1
i
i(1 - i)
( i + 1)
Þ f ( z) = e z +
c1 = e z + c1
= ez +
c1
1+ i
(1 + i)(1 - i)
2
¶v
¶v
= 0 and
=2y
¶x
¶y
Þ
we know that
F ( z) = (1 + i) e x + iy + ic1 = (1 + i) ez + ic1
f ¢( z) =
¶u
¶u
-i
¶x
¶y
....(1)
Þ
f ( z) = e z + (1 + i) c
6. (C) u = sinh x cos y
¶v
¶v
+ i ....(2)
¶y
¶x
Now, equation (1) gives f ¢( z) = 2 x
....(3)
and equation (2) gives f ¢( z) = 2 y
....(4)
Now, for existence of f ¢( z) at any point is necessary that
the value of f ¢( z) most be unique at that point, whatever
be the path of reaching at that point
¶u
= cosh x cos y = f( x, y)
¶x
¶u
and
= - sinh x sin y = y( x, y)
¶y
by Milne’s Method
f ¢( z) = f( z, 0) - iy( z, 0) = cosh z - i × 0 = cosh z
On integrating f ( z) = sinh z + constant
From equation (3) and (4) 2 x = 2 y
Hence, f ¢( z) exists for all points lie on the line x = y.
Þ
f ( z) = w = sinh z + ic
¶u
¶2u
4. (B)
= 2(1 - y) ;
=0
¶x
¶x 2
....(1)
the function x and hence i.e. in w).
¶u
¶2u
= -2 x ;
=0
¶y
¶y 2
....(2)
7. (A)
(As u does not contain any constant, the constant c is in
Þ
Page
568
¶2u ¶2u
+
= 0, Thus u is harmonic.
¶x 2 ¶y 2
¶v
¶v
= 2 y = h( x, y),
= 2 x = g( x, y)
¶x
¶y
by Milne’s Method f ¢( z) = g( z, 0) + ih( z, 0) = 2 z + i 0 = 2 z
On integrating f ( z) = z 2 + c
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Complex Variables
8. (D)
¶v -( x 2 + y 2 ) - ( x - y)2 y
=
¶y
( x2 + y2 )2
f ¢¢( z o) =
y - x - 2 xy
= g( x, y)
( x2 + y2 )2
2
=
2
Chap 9.5
n!
2 pi
f ( z) dz
ò (z - z )
o
c
Taking n = 3,
or
n +1
c
f ( z) dz
ò (z - z )
4
=
o
c
f ( z) dz
ò (z - z )
pi
f ¢¢( z o)
3
n +1
=
o
2pi n
f ( z o)
n!
....(1)
e 2 z dz
e 2 z dz
=
( z + 1) 4 òc [ z - ( -1)]4
¶v ( x 2 + y 2 ) - ( x - y)2 x y 2 - x 2 + 2 xy
=
=
= h( x, y)
¶x
( x2 + y2 )2
( x2 + y2 )2
Given fc
By Milne’s Method
Taking f ( z) = e 2 z , and z o = -1 in (1), we have
f ¢( z) = g( z, 0) + ih( z, 0) = -
e 2 z dz
1
1
æ 1 ö
+ iç - 2 ÷ = - (1 + i) 2
z2
z
z
è
ø
ò ( z + 1)
=
c
pi
f ¢¢¢( -1)....(2)
3
Þ
f ¢¢¢( -1) = 8 e
Þ
e 2 z dz
ò ( z + 1)
4
Since, f ( z) is analytic within and on z = 3
8 pi - z
e 2 z dz
=
e
ò
4
3
|z |= 3 ( z + 1)
By Milne’s Method
12. (D) Since,
f ¢( z) = f( z, 0) - iy( z, 0)
2 cos 2 z - 2
-2
=
- i(0) =
= - cosec2 z
(1 - cos 2 z) 2
1 - cos 2 z
1 - 2z
c
Formula
1
I1 = ò dz = 2 pi
c z
On A, z = 1 + i and On B, z = 2 + 4 i
Let z = 1 + i corresponds to t = 0
é
1
ê f ( z o) = 2pi
ë
x = b, y = d
b = 1, d = 1
ò f ( z) dz = ò ( x
c
2
x = a + b, y = c + d
a = 1, c = 3
+ ixy)( dx + idy)
ò [( t + 1)
everywhere in c i.e. z = 15
. , hence by Cauchy’s integral
+ i( t + 1)( 3t + 1)][ dt + 3i dt ]
t= 0
theorem
1
I3 = ò
dz = 0....(4)
z
-2
c
1
= ò [( t 2 + 2 t + 1) + i( 3t 2 + 4 t + 1)](1 + 3i) dt
0
1
é t3
ù
29
= (1 + 3i) ê + t 2 + t + i( t 3 + 2 t 2 + t) ú = + 1 1i
3
3
ë
û0
using equations (2), (3), (4) in (1), we get
1 - 2z
1
3
òc z( z - 1)( z - 2) dz = 2 (2 pi) + 2 pi - 2 (0) = 3pi
11. (D) We know by the derivative of an analytic
function that
ò
the circle z = 15
. , so the function f ( z) is analytic
1
=
....(2)
inside z = 15
. , therefore I 2 = 2 pi....(3)
1
For I 3 = ò
dz, the singular point z = 2 lies outside
-2
z
c
dx = dt ; dy = 3 dt
c
2
and it
f ( z) dz ù
ú [Here f ( z) = 1 = f ( z o) and z o = 0]
c z - zo
û
1
Similarly, for I 2 = ò
dz, the singular point z = 1 lies
z
-1
c
and z = 2 + 4 i corresponding to t = 1
2 = a + 1, 4 = c + 1 Þ
1
ò z dz
z = 15
. , therefore by Cauchy’s integral
lies inside
AB is , y = 3t + 1 Þ
1
3
I1 + I 2 - I 3....(1)
2
2
c
10. x = at + b, y = ct + d
Þ
=
Since, z = 0 is the only singularity for I1 =
f ( z) = - ò cosec2 z dz + ic = cot z + ic
Þ
1 - 2z
1
1
3
=
+
z( z - 1)( z - 2) 2 z z - 1 2( z - 2)
ò z( z - 1)( z - 2) dz
On integrating
and t = 1
....(3)
If is the circle z = 3
¶u
2 sin 2 x sinh 2 y
=
= y( x, y)
¶y (cosh 2 y - cos 2 x) 2
Þ
8 pi -2
e
3
=
c
2 cos 2 x cosh 2 y - 2
= f( x, y)
(cosh 2 y - cos 2 y) 2
Þ
-2
equation (2) have
¶u 2 cos 2 x (cosh 2 y - cos 2 x) - 2 sin 2 2 x
9. (A)
=
¶x
(cosh 2 y - cos 2 x) 2
then, t = 0
f ¢¢¢( z) = 8 e 2 z
Þ
Now, f ( z) = e 2 z
On integrating
1
1
f ( z) = (1 + i) ò 2 dz + c = (1 + i) + c
z
z
=
4
13. (B) Given contour c is the circle z = 1
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UNIT 9
Þ
z = e iq
3z 2 + 7 z + 1
òc z - zo dz = 2 pif( zo)
dz = ieiqdq
Þ
Now, for upper half of the circle, 0 £ q £ p
2
ò ( z - z ) dz =
c
p
ò (e
iq
Þ
- e 2 iq) ie iqdq
Þ
f ¢( z o) = 2pif¢( z o)
f¢ ( z) = 6 z + 7 and f¢¢( z) = 6
f ¢(1 - i) = 2 pi[ 6(1 - i) + 7 ] = 2 p (5 + 13i)
1 é1
1
ù 2
× ( e 2 pi - 1) - ( e 3px - 1) ú =
i êë2
3
û 3
19. (C) f ( z) =
14. (B) Let f ( z) = cos pz then f ( z) is analytic within and
on z = 3, now by Cauchy’s integral formula
1
f ( z)
f ( z) dz
f ( z o) =
dz Þ ò
= 2pif ( z o)
2pi òc z - z o
c z - zo
Þ
Þ
take f ( z) = cos pz, z o = 1, we have
cos pz
òz = 3 z - 1 dz = 2 pif (1) = 2pi cos p = -2pi
z -1
2
=1 z +1
z +1
f (0) = -1, f (1) = 0
2
f ¢( z) =
Þ
( z + 1) 2
f ¢¢( z) =
-4
( z - 1) 3
f ¢¢¢( z) =
12
( z + 1) 4
Þ
Þ
f ¢(0) = 2;
f ¢¢(0) = -4;
f ¢¢¢(0) = 12; and so on.
Now, Taylor series is given by
sin pz 2
15. (D) ò
dz
- 1)( z - 2)
c (z
f ( z) = f ( z 0 ) + ( z - z 0 ) f ¢( z 0 ) +
( z - z0 ) 2
f ¢¢( z 0 ) +
2!
sin pz 2
sin pz 2
dz - ò
dz
z -2
z -1
c
c
( z - z0 ) 3
f ¢¢¢( z 0 ) + .....
3!
ò
= 2 pif (2) - 2 pif (1) since, f ( z) = sin pz 2
about z = 0
Þ
f ( z) = -1 + z(2) +
f (2) = sin 4 p = 0 and f (1) = sin p = 0
16. (D) Let, I =
=
Þ
since, f( z) = 3z 2 + 7 z + 1
p
é e 2 iq e 3iq ù
= i ò ( e 2 iq - e 3iq)dq = i ê
3i úû 0
ë 2i
0
=
f ( z o) = 2pif( z o)
and f ¢¢( z o) = 2pi f¢¢( z o)
q= 0
p
=i×
Engineering Mathematics
1
2 pi
òz
c
2
1
cos pz dz
-1
z2
z3
( -4) +
(12) + ....
2!
3!
= -1 + 2 z - 2 z 2 + 2 z 3....
f ( z) = -1 + 2( z - z 2 + z 3 ....)
1
1 ö
æ 1
ç
÷ cos pz dz
ò
2 × 2 pi c è z - 1 z + 1 ø
20. (B) f ( z) =
1
z +1
f ¢( z) =
-1
( z + 1) 2
Þ
f ¢(1) =
-1
4
3z 2 + 7 z + 1
òc z - 3 dz , since zo = 3 is the only
f ¢¢( z) =
2
( z + 1) 3
Þ
f ¢¢(1) =
1
4
3z 2 + 7 z + 1
singular point of
and it lies outside the
z -3
f ¢¢¢( z) =
-6
( z + 1) 4
Þ
f ¢¢¢(1) = -
Or I =
1 æ cos nz cos nz ö
ç
÷dz
4 pi òc è z - 1
z +1 ø
17. (D) f ( 3) =
circle x 2 + y 2 = 4 i.e., z = 2, therefore
3z 2 + 7 z + 1
is
z -3
Þ
f (1) =
f ( z) = f ( z 0 ) + ( z - z 0 ) f ¢( z 0 ) +
( z - z0 ) 2
f ¢¢( z 0 )
2!
+
Hence by Cauchy’s theorem—
c
3z 2 + 7 z + 1
dz = 0
z -3
( z - z0 ) 3
f ¢¢¢( z 0 ) + K
3!
about z = 1
f ( z) =
18. (C) The point (1 - i) lies within circle z = 2 ( . .. the
distance of 1 - i i.e., (1, 1) from the origin is 2 which is
less than 2, the radius of the circle).
Let f( z) = 3z 2 + 7 z + 1 then by Cauchy’s integral formula
Page
570
3
and so on.
8
Taylor series is
analytic everywhere within c.
f ( 3) = ò
1
2
=
2
1
æ -1 ö ( z - 1)
+ ( z - 1)ç
÷+
2
2!
è 4 ø
3
æ 1 ö ( z - 1) æ 3 ö
ç - ÷+K
ç ÷+
3! è 8 ø
è4ø
1 1
1
1
( z - 1) + 3 ( z - 1) 2 - 4 ( z - 1) 3 +....
2 22
2
2
or f ( z) =
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1é
1
1
1
ù
1 - ( z - 1) + 2 ( z - 1) 2 - 3 ( z - 1) 3 + ....ú
2 êë
2
2
2
û
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Complex Variables
21. (A) f ( z) = sin z
f ¢( z) = cos z
p
1
æ pö
f ç ÷ = sin =
4
2
è4ø
Þ
ö 1æ
1æ
z z2 z3
1 1
1
ö
çç 1 + +
+
+ .. ÷÷ - ç 1 + + 2 + 3 + K÷
2è
2
4
9
z
z
z
z
è
ø
ø
1 1
1
1 3
or f ( z) = K-z -4 - z -2 - z -1 - - z - z 2 z -K
2 4
8
18
f ( z) = -
1
æ pö
f ¢ç ÷ =
2
è4ø
Þ
f ¢¢( z) = - sin z
Þ
1
æ pö
f ¢¢ç ÷ = 4
2
è ø
f ¢¢¢( z) = - cos z
Þ
1
æ pö
and so on.
f ¢¢¢ç ÷ = 4
2
è ø
24. (C)
( z - z0 )
f ¢¢( z 0 )
2!
about z =
( z - z0 ) 3
f ¢¢¢( z 0 ) + ....
3!
p
4
2
æ
1 ö
÷
çç
2 ÷ø
è
pö
æ
çz - ÷
4ø
+è
3!
22. (D) Let f ( z) = z -2 =
1
1
=
2
z
[1 - (1 + z)]2
=
-1
=
1æ
2
4
8
ö
ç 1 + + 2 + 3 + .... ÷
zè
z z
z
ø
f ( z) =
1
3
7
+
+
+K
z2 z3 z4
25. (B) z < 1,
zö
1
1
1æ
= - ç1 - ÷
z -2 z -1
2è
2ø
1
1
1
1
=
+
z( z - 1)( z - 2) 2 z z - 1 2( z - 2)
Since, 1 + z < 1, so by expanding R.H.S. by binomial
For z - 1 < 1 Let z - 1 = u
theorem, we get
Þ
+ ( n + 1)(1 + z) n + K
or f ( z) = z -2 = 1 +
å ( n + 1)( z + 1)
z = u + 1 and u < 1
1
1
1
1
=
+
z( z - 1)( z - 2) 2 z z - 1 2( z - 2)
=
n
n =1
1
1
1
23. (B) Here f ( z) =
....(1)
=
( z - 1)( z - 2) z - 2 z - 1
Since, z > 1
Þ
1
< 1 and z < 2
z
1
1
1æ
1ö
=
= ç1 - ÷
1ö zè
z -1
zø
æ
zç 1 - ÷
zø
è
=
Þ
z
2
<1
and
-1 æ
zö
1
1
=
ç1 - ÷ = z -2 2 è
2ø
2
equation (1) gives—
1
1
1
1
1
- +
= (1 + u) -1 - u-1 - (1 - u) -1
2( u + 1) u 2( u - 1) 2
2
1
1
[1 - u + u2 - u3 + ... ] - u-1 - (1 + u + u2 + u3 + ...)
2
2
1
3
-1
= ( -2 u - 2 u - ...) - u = -u - u3 - u5 - K - u-1
2
=
Required Laurent’s series is
f ( z) = -( z - 1) -1 - ( z - 1) - ( z - 1) 3 - ( z - 1) 5 - K
-1
27. (B) Let f ( z) =
1æ
1 1
1
ö
ç 1 + + 2 + 3 + K÷
zè
z z
z
ø
-1
+ (1 - z) -1
1
2
26. (D) Since,
¥
-1
é
ù
z z2 z3
1
+
+
+
+ Kú + (1 + z + z 2 + z 3 + ...)
ê
2
4
8
ë
û
1 3
7
15 3
f ( z) = + z + z 2 +
z +K
2 4
8
16
=-
f ( z) = [1 - (1 + z)]-2
f ( z) = 1 + 2(1 + z) + 3(1 + z) 2 + 4(1 + z) 3 + K
1
<1
z
1æ
1 1
1
ö
ç 1 + + 2 + 3 + K÷
z z
z
2è
ø
1
1æ
2ö
= ç1 - ÷
z -2 z è
zø
3
2
3
ù
1 é
1æ
pö 1 æ
pö
pö
æ
ê1 + ç z - ÷ - ç z - ÷ - ç z - ÷ -...ú
4 ø 2 !è
4ø
3! è
4ø
2 êë
è
úû
-1
Þ
1æ1
3
7
ö
ç + 2 + 3 + K÷
zèz z
z
ø
Þ
æ
1 ö
÷ +K
çç
2 ÷ø
è
1 1
< <1
z 2
Þ
Laurent’s series is given by
1æ
2
4 98
1 1
1
ö 1æ
ö
f ( z) = ç 1 + + 2 + 3 + .. ÷ - ç 1 + + 2 + 3 + .. ÷
zè
z z
z
z
z
z
z
ø
ø
è
=
pö
æ
çz - ÷
1
pö 1
4ø
æ
f ( z) =
+ çz - ÷
+è
2!
4ø 2
2
è
f ( z) =
and
2
+
2
<1
z
1
1æ
1ö
= ç1 - ÷
z -1 z è
zø
Taylor series is given by
f ( z) = f ( z 0 ) + ( z - z 0 ) f ¢( z 0 ) +
Chap 9.5
=
é
ù
z z
z
ê1 + 2 + 4 + 9 + Kú
ë
û
2
3
1
z( e z - 1)
1
é
ù
z
z3 z4
+
+
+ K - 1ú
z ê1 + z +
2 ! 3! 4 !
ë
û
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Complex Variables
1
p
1
ò f ( z) dz = 2 pi ´ 6 = 3 pi
r
-idz
dq =
; z £ q £ 2p
z
Þ
0
çç e 2 iq
è
ò b( z) dz = 0
r
¥
c: z =1
ò (x
dz
z + 4z + 1
x
p
dz =
2
2
+ a )( x + b )
a+b
2
2
-¥
2
dz
= ò f ( z) dz
1
+ z6
c
c
36. (C) Let I = ò
2
Let f ( z) =
òæ
e 3iq
dq
R
a 2 öæ
b2 ö
+ 2 ÷÷çç e 2 iq + 2 ÷÷
R øè
R ø
Now when R ® ¥,
-idz
2p
dq
ò0 2 + cos q = òc 1 æz 1 ö;
2 + çz + ÷
2è
zø
c
p
=
1æ
1ö
and cos q = ç z + ÷
2è
zø
= - 2iò
ie 2 iqiRe iqdq
2 2 iq
+ a 2 )( R 2 e2 iq + b2 )
0 (R e
ò f ( z) dz = ò
Now
c
34. (B) Let z = eiq
Chap 9.5
1
z2 + 4z + 1
c is the contour containing semi circle r of radius R and
f ( z) has poles at z = - 2 + 3, -2 - 3 out of these only
segment from -R to R.
For poles of f ( z), 1 + z6 = 0
z = -2 + 3 lies inside the circle c : z = 1
Þ
ò f ( z) dz = 2pi(Residue at z = -2 +
where n = 0, 1, 2, 3, 4, 5, 6
3)
z = ( -1) p 6 = e i ( 2 n + 1 ) p 6
c
=
lim
z ®-2 +
3
( z + 2 - 3) f ( z) = lim
z ®-2 +
ò f ( z) dz = 2 pi ´ 2
1
c
2p
dq
ò 2 + cos q = -2 i ´
0
=
3
pi
3
1
3
( z + 2 + 3)
=
1
2 3
pi
3
=
- 3+i
, i,
2
Only poles z =
Now, residue at z = -2 + 3
2p
+ 3+i
2
1
=
( z1 - z 2 )( z1 - z 3)( z1 - z 4 )( z1 - z 5)( z1 - z6 )
Residue at z =
1
=
3
3i(1 + 3 i)
=
where c is be semi circle r with segment on real axis
from -R to R.
The poles are z = ± ia, z = ± ib. Here only z = ia and
z = ib lie within the contour c
ò f ( z) dz = 2pi
c
Residue at z =
1 + 3i
is
12 i
ò f ( z) dz +
Now
Residue at z = ib
z2
-b
= lim ( z - ib)
=
z ®ib
( z - ia)( z + ia)( z + ib)( z - ib) 2 i( a 2 - b2 )
R
ò f ( z) dz = ò f ( z) dz + ò f ( z) dz
c
=
r
ò f ( z) dz
ò f ( z) dz =
-R
2p
....(1)
3
iq
iRe dq
ò f ( z) dz = ò 1 + R e
6 6 iq
=
0
where R ® ¥,
1 + 3i
12 i
-R
R
c
=
R
2 pi
2p
(1 - 3i + 1 + 3i + 2 i) =
12 i
3
r
p
z
a
=
( z - ia)( z - ia)( z 2 + b2 ) 2 i( a 2 - b2 )
1
3i(1 - 3i)
=
c
Residue at z = ia,
z ®ia
=
ò
r
2
1
6i
f ( z) dz = ò f ( z) dz +
or
(sum of residues at z = ia and z = ib)
= lim ( z - ia)
1 - 3i
12 i
Residue at z = i is
z2
35. (C) I = ò 2
dz = ò f ( z) dz
2
2
2
c ( z + a )( z + b )
c
3+i
lie in the contour
2
p
ò
0
ie iqdq
R5
1
+ e6 iq
R6
ò f ( z) dz ® 0
r
(1) ®
¥
ò
0
ax
2p
=
1 + x6
3
-R
p
2 pi
( a - b) =
a+b
2 i ( a 2 - b2 )
********
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CHAPTER
9.6
PROBABILITY AND STATISTICS
1. In a frequency distribution, the mid value of a class is
6. A distribution consists of three components with
15 and the class interval is 4. The lower limit of the
frequencies 45, 40 and 15 having their means 2, 2.5 and
class is
2 respectively. The mean of the combined distribution is
(A) 14
(C) 12
(A) 2.1
(C) 2.3
(B) 13
(D) 10
2. The mid value of a class interval is 42. If the class
7. Consider the table given below
size is 10, then the upper and lower limits of the class
are
(A) 47 and 37
(C) 37.5 and 47.5
(B) 37 and 47
(D) 47.5 and 37.5
3. The following marks were obtained by the students
in a test: 81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95,
85,79, 62. The range of the marks is
(A) 9
(B) 17
(C) 27
(D) 33
(B) 2.2
(D) 2.4
Marks
Number of Students
0 – 10
12
10 – 20
18
20 – 30
27
30 – 40
20
40 – 50
17
50 – 60
6
The arithmetic mean of the marks given above, is
4. The width of each of nine classes in a frequency
(A) 18
(B) 28
distribution is 2.5 and the lower class boundary of the
(C) 27
(D) 6
lowest class is 10.6. The upper class boundary of the
highest class is
(A) 35.6
(B) 33.1
(C) 30.6
(D) 28.1
5.
In
a
monthly
test,
the
marks
8. The following is the data of wages per day: 5, 4, 7, 5,
8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8 The mode of the data is
(A) 5
(C) 8
obtained
in
(B) 7
(D) 10
9. The mode of the given distribution is
mathematics by 16 students of a class are as follows:
0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8
Weight (in kg)
40
43
46
49
52
55
Number of Children
5
8
16
9
7
3
The arithmetic mean of the marks obtained is
(A) 3
(B) 4
(C) 5
(D) 6
Page
574
(A) 55
(B) 46
(C) 40
(D) None
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Probability and Statistics
10. If the geometric mean of x, 16, 50, be 20, then the
16. The mean deviation of the following distribution is
value of x is
(A) 4
(C) 20
(B) 10
(D) 40
11. If the arithmetic mean of two numbers is 10 and
Chap 9.6
x
10
11
12
13
14
f
3
12
18
12
3
(A) 12
(C) 1.25
(B) 0.75
(D) 26
their geometric mean is 8, the numbers are
(A) 12, 18
(C) 15, 5
17. The standard deviation for the data 7, 9, 11, 13,
(B) 16, 4
(D) 20, 5
15 is
12. The median of
0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5, 6 is
(B) -1.5
(D) 3.5
(A) 0
(C) 2
(A) 2.4
(B) 2.5
(C) 2.7
(D) 2.8
18. The standard deviation of 6, 8, 10, 12, 14 is
13. Consider the following table
(A) 1
(B) 0
(C) 2.83
(D) 2.73
Diameter of heart (in mm)
Number of persons
19. The probability that an event A occurs in one trial of
120
5
an experiment is 0.4. Three independent trials of
121
9
122
14
123
8
124
5
125
9
experiment are performed. The probability that A
occurs at least once is
(A) 0.936
(B) 0.784
(C) 0.964
(D) None
20.
Eight
coins
are
tossed
simultaneously.
The
probability of getting at least 6 heads is
The median of the above frequency distribution is
(A) 122 mm
(B) 123 mm
(C) 122.5 mm
(D) 122.75 mm
(A)
7
64
(B)
37
256
(C)
57
64
(D)
249
256
21. A can solve 90% of the problems given in a book and
14. The mode of the following frequency distribution, is
Class interval
Frequency
3–6
2
6–9
5
9–12
21
12–15
23
15–18
10
18–21
12
21–24
3
(A) 11.5
(B) 11.8
(C) 12
(D) 12.4
B can solve 70%. What is the probability that at least
one of them will solve a problem, selected at random
from the book?
(A) 0.16
(B) 0.63
(C) 0.97
(D) 0.20
22. A speaks truth in 75% and B in 80% of the cases. In
what percentage of cases are they likely to contradict
each other narrating the same incident ?
(A) 5%
(B) 45%
(C) 35%
(D) 15%
23. The odds against a husband who is 45 years old,
living till he is 70 are 7:5 and the odds against his wife
15. The mean-deviation of the data 3, 5, 6, 7, 8, 10,
who is 36, living till she is 61 are 5:3. The probability
11, 14 is
that at least one of them will be alive 25 years hence, is
(A) 4
(B) 3.25
(A)
61
96
(B)
(C) 2.75
(D) 2.4
(C)
13
64
(D) None
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UNIT 9
Engineering Mathematics
24. The probability that a man who is x years old will
30. If 3 is the mean and (3/2) is the standard deviation
die in a year is
of a binomial distribution, then the distribution is
A1 , A2 , K, An each x
p. Then amongst n
persons
years old now, the probability
that A1 will die in one year is
(A)
1
n2
(B) 1 - (1 - p) n
(C)
1
[1 - (1 - p) n ]
n2
(D)
1
[1 - (1 - p) n ]
n
drawn from each bag, the probability that both are
white is
5
24
60
æ1 4ö
(D) ç + ÷
è5 5 ø
the distribution is
12
æ1 3ö
(B) ç + ÷
è4 4ø
24
æ1 1ö
(D) ç + ÷
è2 2 ø
32
1
(B)
4
æ1 5ö
(C) ç + ÷
è6 6ø
(D) None
32. A die is thrown 100 times. Getting an even number
is considered a success. The variance of the number of
contains 4 white and 2 red balls. If one ball is drawn
(A) 50
(C) 10
from each bag, the probability that one is white and one
is red, is
8
27
16
æ1 1ö
(A) ç + ÷
è7 8 ø
successes is
(C)
5
binomial distribution are 24 and 18 respectively. Then,
26. A bag contains 5 white and 4 red balls. Another bag
13
(A)
27
12
31. The sum and product of the mean and variance of a
bag contains 3 white and 5 black balls. If one ball is
(C)
æ 1 3ö
(B) ç + ÷
è2 2 ø
æ4 1ö
(C) ç + ÷
è5 5ø
25. A bag contains 4 white and 2 black balls. Another
1
(A)
24
12
æ3 1ö
(A) ç + ÷
è4 4ø
(B) 25
(D) None
33. A die is thrown thrice. Getting 1 or 6 is taken as a
5
(B)
27
success. The mean of the number of successes is
3
2
(A)
(B)
2
3
(D) None
(C) 1
(D) None
27. An anti-aircraft gun can take a maximum of 4 shots
34. If the sum of mean and variance of a binomial
at an enemy plane moving away from it. The
distribution is 4.8 for five trials, the distribution is
probabilities of hitting the plane at the first, second,
æ1 4ö
(A) ç + ÷
è5 5 ø
third and fourth shot are 0.4, 0.3, 0.2 and 0.1
respectively. The probability that the gun hits the plane
is
(A) 0.76
(B) 0.4096
(C) 0.6976
(D) None of these
æ2 3ö
(C) ç + ÷
è5 5 ø
5
æ1 2ö
(B) ç + ÷
è 3 3ø
5
5
(D) None of these
35. A variable has Poission distribution with mean m.
The probability that the variable takes any of the
28. If the probabilities that A and B will die within a
values 0 or 2 is
year are p and q respectively, then the probability that
æ
m2 ö
(A) e - m çç 1 + m +
÷
2 ! ÷ø
è
(B) e m (1 + m) -3 2
(C) e 3 2 (1 + m 2 ) -1 2
æ
m2 ö
(D) e - m çç 1 +
÷
2 ! ÷ø
è
only one of them will be alive at the end of the year is
(A) pq
(B) p(1 - q)
(C) q(1 - p)
(D) p + 1 - 2 pq
29. In a binomial distribution, the mean is 4 and
36.
variance is 3. Then, its mode is
P (2) = 9 P ( 4) + 90 P ( 6), then the mean of X is
(A) 5
(B) 6
(A) ± 1
(B) ± 2
(C) 4
(D) None
(C) ± 3
(D) None
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X
is
a
Poission
variate
such
that
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Probability and Statistics
Chap 9.6
37. When the correlation coefficient r = ± 1, then the
43. If Sxi = 30,
two regression lines
å y = 318 and n = 6, then the regression coefficient bxy
(A)
(B)
(C)
(D)
is
å yi = 42,
å xi yi = 199, å xi2 = 184,
2
i
are perpendicular to each other
coincide
are parallel to each other
do not exist
(A) -0.36
(C) 0.26
(B) -0.46
(D) None
44. Let r be the correlation coefficient between x and y
38. If r = 0, then
(A) there is a perfect correlation between x and y
(B) x and y are not correlated.
and byx , bxy be the regression coefficients of y on x and
x on y respectively then
(A) r = bxy + byx
(B) r = bxy ´ byx
(D) there is a negative correlation between x and y
(C) r = bxy ´ byx
(D) r =
39. If Sxi = 15,
45. Which one of the following is a true statement.
(C) there is a positive correlation between x and y
Syi = 36,
Sxi yi = 110 and n = 5, then
cov ( x, y) is equal to
(A) 0.6
(C) 0.4
(B) 0.5
(D) 0.225
40. If cov ( x, y) = -16.5, var ( x) = 2.89 and var ( y) = 100,
then the coefficient of correlation r is equal to
(A)
1
2
( bxy + byx ) = r
(B)
(C)
1
2
( bxy + byx ) > r
(D) None of these
1
2
( bxy + byx ) < r
46. If byx = 1.6 and bxy = 0.4 and q is the angle between
two regression lines, then tan q is equal to
(A) 0.18
(C) 0.16
(B) -0.64
(D) -0.97
(A) 0.36
(C) 0.97
1
( bxy + byx )
2
(B) 0.24
(D) 0.3
41. The ranks obtained by 10 students in Mathematics
47. The equations of the two lines of regression are :
and Physics in a class test are as follows
4 x + 3y + 7 = 0
and 3 x + 4 y = 8 = 0. The correlation
coefficient between x and y is
Rank in Maths
Rank in Chem.
1
3
2
10
3
5
48. If cov( X , Y ) = 10, var ( X ) = 6.25 and var( Y ) = 31.36,
4
1
then r( X , Y ) is
5
2
6
9
7
4
49. If å x = å y = 15,
8
8
n = 5, then bxy = ?
9
7
(A) - 13
(B) -
10
6
(C) - 14
(D) - 12
The coefficient of correlation between their ranks is
(A) 0.15
(C) 0.625
42. If Sxi = 24,
(B) 0.224
(D) None
å yi = 44,
(A) 1.25
(B) 0.25
(C) -0.75
(D) 0.92
(A)
5
7
(B)
(C)
3
4
(D) 0.256
50. If å x = 125,
4
5
å x 2 = å y 2 = 49, å xy = 44 and
2
3
å y = 100, å x 2 = 1650, å y 2 = 1500,
å xy = 50 and n = 25, then the line of regression of x on
y is
Sxi yi = 306, å xi2 = 164,
å yi2 = 574 and n = 4, then the regression coefficient byx
(A) 22 x + 9 y = 146
(B) 22 x - 9 y = 74
(C) 22 x - 9 y = 146
(D) 22 x + 9 y = 74
is equal to
(A) 2.1
(C) 1.225
(B) 1.6
(D) 1.75
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UNIT 9
SOLUTION
Engineering Mathematics
A.M. = A +
S( fd) æ
300 ö
= ç 25 +
÷ = 28.
Sf
100 ø
è
1. (B) Let the lower limit be x. Then, upper limit is
x + ( x + 4)
x + 4.
= 15 Þ x = 13.
2
8. (C) Since 8 occurs most often, mode =8.
2. (A) Let the lower limit be x. Then, upper limit x + 10.
10. (B) ( x ´ 16 ´ 50)1 3 = 20
x + ( x + 10)
= 42
2
Þ
Þ
x = 37.
9. (B) Clearly, 46 occurs most often. So, mode =46.
Lower limit = 37 and upper limit =47.
3. (D) Range = Difference between the largest value
= (95 - 62) = 33.
4. (B) Upper class boundary = 10.6 + (2.5 ´ 9) = 331
..
x ´ 16 ´ 50 = (20) 3
æ 20 ´ 20 ´ 20 ö
x = çç
÷÷ = 10.
è 16 ´ 50 ø
11. (B) Let the numbers be a and b Then,
a+b
= 10
2
ab = 8
Þ
Þ
( a + b) = 20
and
ab = 64
a - b = ( a + b) 2 - 4 ab = 44 - 256 = 144 = 12.
5. (B)
Solving a + b = 20 and a - b = 12 we get a = 16 and
Marks
Frequency f
f ´1
0
2
0
2
2
4
3
3
9
4
1
4
5
4
20
6
2
12
7
1
7
8
1
8
Diameter of heart
(in mm)
Number of
persons
Cumulative
frequency
å f = 16
å( f ´ x) = 64
120
5
5
121
9
14
122
14
28
123
8
36
124
5
41
125
9
50
A.M. =
b = 4.
12. (D) Observations in ascending order are
-3, -3, -1, 0, 2, 2, 2, 5, 5, 5, 5 6, 6, 6
Number of observations is 14, which is even.
Median =
1
1
[7 the term +8 the term] = (2 + 5) = 35
. .
2
2
13. (A) The given Table may be presented as
å( f ´ x) 64
=
= 4.
åf
16
6. (B) Mean =
45 ´ 2 + 40 ´ 2.5 + 15 ´ 2 220
=
= 2.2.
100
100
7. (B)
n
n
= 25 and
+ 1 = 26.
2
2
Class
Mid
value x
Frequenc
yf
Deviation
d = x- A
f ´d
Here n = 50. So,
0–10
5
12
-20
-240
Medium =
10–20
15
18
-10
-180
20–30
25 = A
27
0
0
30–40
35
20
10
200
14. (B) Maximum frequency is 23. So, modal class is
40–50
45
17
20
320
12–15.
50–60
55
6
30
180
L1 = 12, L2 = 15, f = 23, f1 = 21 and f2 = 10.
Sf = 100
Page
578
Þ
S ( f ´ d) = 390
1
122 + 122
(25th term +26 th term) =
= 122.
2
2
[ . .. Both lie in that column whose c.f. is 28]
Thus Mode = L1 +
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f - f1
( L2 - L1 )
2 f - f1 - f2
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Probability and Statistics
= 12 +
(23 - 21)
(15 - 12) = 12.4.
( 46 - 21 - 10)
Chap 9.6
6
æ 3 + 5 + 6 + 7 + 8 + 10 + 11 + 14 ö
15. (C) Mean = ç
÷ = 8.
8
è
ø
7
2
æ1ö æ1ö
æ1ö 1
= 8 C6 × ç ÷ × ç ÷ + 8 C7 × ç ÷ × + 8 C8
2
2
è ø è ø
è2 ø 2
8´7
1
1
1
37
=
´
+ 8´
+
=
2 ´ 1 256
256 256 256
8
æ1ö
×ç ÷
è2 ø
Sd = 3 - 8 + 5 - 8 + 8 - 8 + 10 - 8 + 11 - 8 + 14 - 8
21. (C) Let E = the event that A solves the problem. and
= 22
F = the event that B solves the problem.
Sd 22
Thus Mean deviation =
=
= 2.75.
n
8
Clearly E and F are independent events.
90
70
P ( E) =
= 0.9, P ( F ) =
= 0.7,
100
100
16. (B)
P ( E º F ) = P ( E) × P ( F ) = 0.9 ´ 0.7 = 0.63
x
f
f ´x
d = x-M
f ´d
10
3
30
2
6
11
12
132
1
12
12
18
216
0
0
13
12
156
1
12
14
3
42
2
6
Sf = 48
Sfx = 576
Thus M =
= P ( E) + P ( F ) - P ( E º F ) = (0.9 +0.7 - 0.63) =0.97.
Sfd = 36
576
= 12.
48
B speaks the truth)]
= P (E and F ) + P (E and F)
2
2
= P ( E) × P ( F ) + P ( E ) × P ( F )
2
2
Sd2 = 7 - 11 + 9 - 11 + 11 - 11 + 13 - 11 + 15 - 11 = 40
s=
Sd 2
40
=
= 8 = 2 2 = 2 ´ 1.41 = 2.8.
n
5
2
2
2
=
3 1 1 4
3
1 7 æ 7
ö
´ + ´ =
+ =
=ç
´ 100 ÷% = 35%.
4 5 4 5 20 5 20 è 20
ø
23. (A) Let E = event that the husband will be alive 25
years hence and F =event that the wife will be alive 25
6 + 8 + 10 + 12 + 14 50
18. (C) M =
=
= 10.
5
5
years hence.
2
Sd2 = 6 - 10 2 + 8 - 10 + 10 - 10 + 12 - 10 + 14 - 10 = 40
6=
F =event that B speaks the truth.
75
3
80 4
Then, P ( E) =
= ,
P( F) =
=
100 4
100 5
3ö 1
4ö 1
æ
æ
P ( E) = ç 1 - ÷ = ,
P( F ) = ç 1 - ÷ =
4ø 4
5ø 5
è
è
= P[(A speaks truth and B tells a lie) or (A tells a lie and
7 + 9 + 11 + 13 + 15 55
=
= 11.
5
5
2
22. (C) Let E =event that A speaks the truth.
P (A and B contradict each other).
Sfd 36
So, Mean deviation =
=
= 0 .75
n
48
17. (D) m =
Required probability = P ( E È F )
Then,
P ( E) =
5
12
and P ( F ) =
3
8
5 ö 7
3ö 5
æ
æ
and P ( F ) = ç 1 - ÷ = .
Thus P ( E) = ç 1 ÷=
12
12
8
è
ø
è
ø 8
Sd2
40
=
n
5
Clearly, E and F are independent events.
= 8 = 2 2 = 2 ´ 1.414 = 2.83 (app.)
So, E and F are independent events.
19. (B) Here p = 0.4, q = 0.6 and n = 3.
P(at least one of them will be alive 25 years hence)
Required probability = P(A occurring at least once)
= 1 - P(none will be alive 24 years hence)
7
5 ö 61
æ
= 1 - P ( E º F ) = 1 - P ( E) × P ( F ) = ç 1 ´ ÷=
12
8 ø 96
è
= 3C1 × (0.4) ´ (0.6) 2 + 3C2 × (0.4) 2 ´ (0.6) + 3C3 × (0.4) 3
4
36
16
6
64 ö 784
æ
=ç3´
´
+ 3´
´
+
= 0.784.
÷=
10 100
100 10 1000 ø 1000
è
20. (B) p =
1
,
2
q=
1
,
2
24. (D) P(none dies)
= (1 - p) (1 - p)....n times = (1 - p) n
n = 8. Required probability
= P (6 heads or 7 heads or 8 heads)
P(at least one dies) = 1 - (1 - p) n .
1
P(A1 dies) = {1 - (1 - p) n }.
n
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Probability and Statistics
Sxi 15
å yi 36
=
= 3, y =
=
= 7.2
n
5
n
5
æ Sx y
ö æ 110
ö
cov( x, y) = ç i i - x y ÷ = ç
- 3 ´ 7.2 ÷ = 0.4
n
5
è
ø è
ø
47. (C) Given lines are : y = -2 -
39. (C) x =
40. (D) r =
cov ( x, y)
var ( x) × var ( y)
=
-16.5
2.89 ´ 100
Chap 9.6
æ 7 3 ö
and x = ç - - y ÷
è 4 4 ø
-3
-3
and
.
byx =
bxy =
4
4
3
æ -3 -3 ö 9
or r = - = -0.75.
So, r 2 = ç
´
÷=
4 ø 16
4
è 4
= -0.97.
41. (B) Di = -2, - 8, - 2, 3, 3, - 3, 3, 0, 2, 4.
[. .. byx and bxy are both negative ® r is negative]
SDi2 = ( 4 + 64 + 4 + 9 + 9 + 9 + 9 + 0 + 4 + 16) = 128.
é
6( SDi2 ) ù æ
6 ´ 128 ö 37
R = ê1 2
ú = çç 1 - 10 ´ 99 ÷÷ = 165 = 0.224.
n
(
n
)
1
ø
ë
û è
48. (A) r( X , Y ) =
( Sxi )( Syi )
n
42. (A) byx =
é 2 ( Sxi ) 2 ù
êSxi - n ú
ë
û
49. (C) byx =
Sxi yi -
50. (B) bxy =
=
( Sxi )( Syi ) ù æ 199 - 30 ´ 42 ö
é
ç
÷
êëSxi yi úû è
6
ø
n
43. (B) byx =
=
é 2 ( Syi ) 2 ù
42 ´ 42 ù
é
êSyi - n ú
êë318 úû
6
ë
û
(199 - 210) -11
=
=
= -0.46.
( 318 - 294) 24
r 2 = bxy ´ byx
45. (C)
cov( X , Y )
var( X ) var( Y )
=
10
6.25 ´ 31.36
=
5
7
nSxy - ( Sx)( Sy)
nSx 2 - ( Sx) 2
æ 5 ´ 44 - 15 ´ 15 ö
1
= çç
÷÷ = 5
´
49
15
´
15
4
è
ø
24 ´ 44 ö
æ
÷
ç 306 4
ø = ( 306 - 264) = 42 = 2.1
=è
2
(164 - 144) 20
é
(24) ù
ê164 - 4 ú
ë
û
44. (C) byx = r ×
3
x
4
nSxy - ( Sx)( Sy)
nSy 2 - ( Sy) 2
25 ´ 50 - 125 ´ 100
9
=
25 ´ 1500 - 100 ´ 100 22
Also,
x=
125
= 5,
25
y=
100
= 4.
25
Required line is x = x + bxy ( y - y)
9
Þ x =5 +
( y - 4) Þ 22 x - 9 y = 74.
22
sy
sx
and bxy = r ×
sx
sy
Þ
r = bxy ´ byx .
1
1 é sy
sx ù
( bxy + byx ) > r is true if êr ×
+r×
>r
2
2 ë sx
sy úû
i.e. if s2y + sx2 > 2 sx s y
i.e. if ( s y - sx ) 2 > 0, which is true.
46. (A) r = 1.6 ´ 0.4 = .64 = 0.8
byx = r ×
m1 =
sy
sx
Þ
sy
sx
=
byx
r
1 sy
1
5
,
×
=
´2=
r sx 0.8
2
æ m - m2
tan q = çç 1
è 1 + m1 m2
=
1.6
=2
0.8
m2 = r ×
sy
sx
= 0.8 ´ 2 = 1.6.
ö æ 2.5 - 1.6 ö 0.9
÷÷ = çç
= 0.18.
÷÷ =
ø è 1 + 2.5 ´ 1.6 ø 5
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(B)
x2
x5
x8
x11
+
+
+
2
20 160 4400
(C)
x2
x5
x8
x11
+
+
+
2
20 160 2400
2
(D)
5
8
Chap 9.7
Statement for Q. 18–19:
dy
For
= 1 + y 2 given that
dx
x:
0
0.2
0.4
0.6
y:
0
0.2027
0.4228
0.6841
11
x
x
x
x
+
+
+
2
40 480 2400
12. For dy dx = xy given that y = 1 at x = 0. Using Euler
method taking the step size 0.1, the y at x = 0.4 is
Using Milne’s method determine the value of y for
x given in question.
(A) 1.0611
(B) 2.4680
18. y (0.8) = ?
(C) 1.6321
(D) 2.4189
(A) 1.0293
(B) 0.4228
(C) 0.6065
(D) 1.4396
Statement for Q. 13–15.
For dy dx = x 2 + y 2 given that y = 1 at x = 0.
19. y (10
. ) =?
Determine the value of y at given x in question using
(A) 1.9428
(B) 1.3428
modified method of Euler. Take the step size 0.02.
(C) 1.5555
(D) 2.168
13. y at x = 0.02 is
Statement for Q.20–22:
(A) 1.0468
(B) 1.0204
(C) 1.0346
(D) 1.0348
Apply Runge Kutta fourth order method to obtain
y (0.2), y (0.4) and y (0.6) from dy dx = 1 + y 2 , with y = 0
at x = 0. Take step size h = 0.2.
14. y at x = 0.04 is
(A) 1.0316
(B) 1.0301
(C) 1.403
(D) 1.0416
20. y (0.2) = ?
15. y at x = 0.06 is
(A) 1.0348
(B) 1.0539
(C) 1.0638
(D) 1.0796
(A) 0.2027
(B) 0.4396
(C) 0.3846
(D) 0.9341
21. y (0.4) = ?
(A) 0.1649
(B) 0.8397
(C) 0.4227
(D) 0.1934
16. For dy dx = x + y given that y = 1 at x = 0. Using
22. y (0.6) = ?
modified Euler’s method taking step size 0.2, the value
(A) 0.9348
(B) 0.2935
(C) 0.6841
(D) 0.563
of y at x = 1 is
(A) 3.401638
(B) 3.405417
(C) 9.164396
(D) 9.168238
23. For dy dx = x + y 2 , given that y = 1 at x = 0. Using
Runge Kutta fourth order method the value of y
17. For the differential equation dy dx = x - y 2 given
x = 0.2 is (h = 0.2)
that
(A) 1.2735
(B) 2.1635
(C) 1.9356
(D) 2.9468
x:
0
0.2
0.4
0.6
y:
0
0.02
0.0795
0.1762
24. For dy dx = x + y given that y = 1 at x = 0. Using
Runge Kutta fourth order method the value of y
Using Milne predictor–correction method, the y at
next value of x is
(A) 0.2498
(B) 0.3046
(C) 0.4648
(D) 0.5114
at
x = 0.2 is
at
(h = 0.2)
(A) 1.1384
(B) 1.9438
(C) 1.2428
(D) 1.6389
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UNIT 9
SOLUTIONS
x2 = x0 = 35
. -
1. (B) Let f ( x) = x 3 - 4 x - 9
Engineering Mathematics
x1 - x0
f ( x0 )
f ( x1 ) - f ( x0 )
0.5
( - 0.5441) = 37888
.
0.3979 + 0.5441
Since f (2) is negative and f ( 3) is positive, a root lies
Since f ( 37888
.
) = - 0.0009 and f ( 4) = 0.3979, therefore
between 2 and 3.
the root lies between 3.7888 and 4.
First approximation to the root is
1
x1 = (2 + 3) = 2.5.
2
Taking x0 = 37888
.
, x1 = 4, we obtain
0.2112
x3 = 37888
.
( - .009) = 37893
.
0.3988
Then f ( x1 ) = 2.5 3 - 4(2.5) - 9 = - 3.375
Hence the required root correct to three places of
i.e. negative\The root lies between x1 and 3. Thus the
decimal is 3.789.
second
approximation
1
x2 = ( x1 + 3) = 2.75.
2
4. (D) Let f ( x) = xe x - 2, Then f (0) = - 2, and
to
the
root
is
Then f ( x2 ) = (2.75) 3 - 4(2.75) - 9 = 0.7969 i.e. positive.
The root lies between x1 and x2 . Thus the third
1
approximation to the root is x3 = ( x1 + x2 ) = 2.625.
2
Then
f ( x3) = (2.625) 3 - 4(2.625) - 9 = - 1.4121
i.e.
negative.
The root lies between x2 and x3 . Thus the fourth
1
approximation to the root is x4 = ( x2 + x3) = 2.6875.
2
Hence the root is 2.6875 approximately.
2. (B) Let f ( x) = x 3 - 2 x - 5
f (1) = e - 2 = 0.7183
So a root of (i ) lies between 0 and 1. It is nearer to 1.
Let us take x0 = 1.
Also f ¢( x) = xe x + e x and f ¢(1) = e + e = 5.4366
By Newton’s rule, the first approximation x1 is
f ( x0 )
0.7183
x1 = x0 =1 = 0.8679
f ¢( x0 )
5.4366
f ( x1 ) = 0.0672, f ¢( x1 ) = 4.4491.
Thus the second approximation x2 is
f ( x1 )
0.0672
x2 = x1 = 0.8679 = 0.8528
f ( x1 )
4.4491
Hence the required root is 0.853 correct to 3 decimal
So that f (2) = - 1 and f ( 3) = 16
places.
i.e. a root lies between 2 and 3.
Taking
x0 = 2, x1 = 3, f ( x0 ) = - 1, f ( x1 ) = 16,
in
the
method of false position, we get
x1 - x0
1
x2 = x0 f ( x0 ) = 2 +
= 2.0588
f ( x1 ) - f ( x0 )
17
5. (B) Let y = x + log10 x - 3.375
To obtain a rough estimate of its root, we draw the
graph of (i ) with the help of the following table :
Now, f ( x2 ) = f (2.0588) = - 0.3908 i.e., that root lies
between 2.0588 and 3.
Taking x0 = 2.0588, x1 = 3, f ( x0 )
x
1
2
3
4
y
-2.375
-1.074
0.102
1.227
= - 0.3908, f ( x1 ) = 16 in (i), we get
0.9412
x3 = 2.0588 ( - 0.3908) = 2.0813
16.3908
Taking 1 unit along either axis = 0.1, The curve crosses
Repeating this process, the successive approxima- tions
approximation to the root.
are
Now let us apply Newton–Raphson method to
x4 = 2.0862, x5 = 2.0915, x6 = 2.0934, x7 = 2.0941,
f ( x) = x + log10 x - 3.375
1
f ¢( x) = 1 + log10 e
x
the x–axis at x0 = 2.9, which we take as the initial
x8 = 2.0943 etc.
Hence the root is 2.094 correct to 3 decimal places.
Taking x0 = 35
. , x1 = 4, in the method of false position,
f (2.9) = 2.9 + log10 2.9 - 3.375 = - 0.0126
1
f ¢(2.9) = 1 +
log10 e = 11497
.
2.9
we get
The first approximation x1 to the root is given by
3. (C) Let f ( x)2 x - log10 x - 7
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x1 = x0 -
f ( x0 )
0.0126
= 2.9 +
= 2.9109
¢
f ( x0 )
11497
.
at x = 0,
Thus the second approximation x2 is given by
d2 y
=1 + 2 = 3
dx 2
at x = 0, y = 1,
f ( x1 )
0.0001
= 2.9109 +
= 2.91099
f ¢( x1 )
11492
.
2
Hence the desired root, correct to four significant
figures, is 2.911
d 3y
d2 y
æ dy ö
=
2
2
y
ç
÷
dx 3
dx 2
è dx ø
6. (B) Let x = 28 so that x 2 - 28 = 0
Taking f ( x) = x 2 - 28, Newton’s iterative method gives
f ( xn )
x 2 - 28 1 æ
28 ö
÷
= xn - n
= çç xn +
¢
f ( xn )
2 xn
2è
xn ÷ø
é dy d 2 y
d 3y ù
3
+
y
ê dx dx 2
dx 3 úû
ë
d4 y
= -2
dx 4
at x = 0, y = 1
Now since f (5) = - 3, f ( 6) = 8, a root lies between 5 and
6.
d4 y
= 34
dx 4
The Taylor series expression gives
y( x + h) = y( x) + h
Taking x0 = 5.5,
x1 =
1æ
28 ö 1 æ
28 ö
÷÷ = ç 5.5 +
çç x0 +
÷ = 5.29545
2è
x0 ø 2 è
5.5 ø
x2 =
1æ
28 ö 1 æ
28 ö
÷÷ = çç 5.29545 +
çç x1 +
÷ = 5.2915
2è
x1 ø 2 è
5.29545 ÷ø
d 3y
=-8
dx 3
x = 0, y = 1,
at
xn + 1 = xn -
dy
= -1
dx
y = 1,
d2 y
dy
=1 -2y
2
dx
dx
f ( x1 ) = - 0.0001, f ¢( x1 ) = 11492
.
x2 = x1 -
Chap 9.7
dy h2 d 2 y h3 d 3 y h4 d 4 y
+K
+
+
+
3 ! dx 3 4 ! dx 4
dx 2 ! dx 2
y(0.1) = 1 + 0.1( -1) +
(0.1) 2
(0.1) 3
(0.1) 4
3+
( -8) +
34 + ......
2!
3!
4!
= 1 - 0.1 + 0.015 - 0.001333 + 0.0001417 = 0.9138
9. (C) Here f ( x, y) = x 2 + y 2 , x0 = 0
1æ
28 ö 1 æ
28 ö
÷ = ç 5.2915 +
x3 = çç x2 +
÷ = 5.2915
2è
5.2915 ÷ø
x2 ÷ø 2 çè
y0 = 0
We have, by Picard’s method
Since x2 = x3 upto 4 decimal places, so we take
y = y0 +
28 = 5.2915.
x
ò f ( x, y) dx
....(1)
x0
The first approximation to y is given by
7. (B) Let h = 0.1,
dy
= 1 + xy Þ
dx
3
given x0 = 0,
d2 y
dy
=x
+y
dx 2
dx
2
d y
d y
dy
,
=x
+2
dx 3
dx 2
dx
given that
Þ
x1 = x0 + h = 0.1
x = 0,
4
3
2
d y
d y
d y
=x
+3
dx 4
dx 3
dx 2
y =1
The Taylor series expression gives :
dy h2 d 2 y h3 d 3 y
+
+
+
3 ! dx 3
dx 2 ! dx 2
Þ
y (0.1) = 1 + 0.1 ´ 1 +
Þ
y(0.1) = 1 + 0.1 +
(0.1) 2
(0.1) 3
×1 +
2 +K
2!
3!
0.01 0.001
+
+K
2
3
= 1 + 0.1 + 0.005 + 0.000033 .........
8. (B) Let h = 0.1,
given
dy
= x - y2
x1 = x0 + h = 0.1,
dx
x
ò f ( x, y ) dx
0
x0
dy
d2 y
d 3y
d4 y
= 1 ; 2 = 1,
=
2
,
= 3 and so on
dx
dx
dx 3
dx 4
y( x + h) = y( x) + h
y (1 ) = y0 +
x0 = 0,
Where y0 = 0 +
x
x
ò f ( x, 0) dx = ò x dx.
2
0
...(2)
0
The second approximation to y is given by
x
ò f ( x, y
y ( 2 ) = y0 +
(1 )
) dx = 0 +
0
x0
æ
x
= 0 + ò çç x 2 +
9
0è
x
Now,
6
æ
x3 ö
÷dx
3 ÷ø
ö
x
x
÷÷dx =
+
3 63
ø
3
y (0.4) =
10. (C) Here
x
ò f ççè x,
7
(0.4) 3 (0.4) 7
+
= 0.02135
3
63
f ( x, y) = y - x ; x0 = 0, y0 = 2
We have by Picard’s method
= 11053
.
y0 = 1
y = y0 +
ò
x
x0
f ( x, y) dx
The first approximation to y is given by
y (1 ) = y0 +
x
ò f ( x, y0 ) dx = 2 +
x0
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x
ò f ( x, 2) dx
0
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585
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GATE EC BY RK Kanodia
UNIT 9
x
= 2 + ò (2 - x) dx
=2 + 2x -
0
x2
2
....(1)
x
ò f ( x, y
(1 )
=2 +
Here
æ
ò f ççè x, 2 + 2 x -
x0
2
= 2 + ò (2 + 2 x 0
x
2
....(2)
æ
ò f ççè x, 2 + 2 x +
=2 +
x0
y4 = y3 + hf ( x3 , y3) = 10302
.
+ 0.1 f (0.3 , 10302
.
)
x
x ö
÷dx
2
6 ÷ø
2
= 10302
.
+ 0.03090
3
.
y4 = y( 0 .4 ) = 10611
Hence y( 0 .4 ) = 10611
.
ö
æ
x
x
= 2 + ò çç 2 + 2 x +
- ÷÷dx
2
6
ø
0è
x
2
3
13. (B) The Euler’s modified method gives
x2 x3 x4
+
2
6 24
=2 + 2x +
11. (B) Here
f ( x, y) = x + y 2 , x0 = 0
y1* = y0 + hf ( x0 , y0 ),
h
y1 = y0 + [ f ( x0 , y0 ) + f ( x1 , y1*)]
2
y0 = 0
Now, here h = 0.02, y0 = 1, x0 = 0
We have, by Picard’s method
y = y0 +
.
y1* = 1 + 0.02 f (0, 1), y1* = 1 + 0.02 = 102
h
Next y1 = y0 + [ f ( x0 , y0 ) + f ( x , y1*)]
2
0.02
=1 +
[ f (0, 1) + f (0.02, 102
. )]
2
x
ò f ( x, y0 ) dx
x0
The first approximation to y is given by
y (1 ) = y0 +
x
ò f ( x, y ) dx
0
=0 +
x0
x
x
ò f ( x, 0) dx
= 1 + 0.01 [1 + 10204
.
] = 10202
.
0
2
x
= 0 + ò xdx =
2
0
So,
x
ò f ( x, y
(1 )
) dx = 0 +
x0
x
æ
ò f ççè x,
0
x2
2
ö
÷÷dx
ø
= 10202
.
+ 0.02 [ f (0.02, 10202
.
)]
= 10202
.
+ 0.0204
= 10406
.
h
Next y2 = y1 + [ f ( x, y) + f ( x2 , y2* )]
2
0.02
y2 = 10202
.
+
[ f (0.02, 10202
.
) + f (0.04, 10406
.
)]
2
æ
x
x
x ö
÷÷dx =
= ò çç x +
+
4 ø
2
50
0è
x
4
2
5
The third approximation is given by
y ( 3) = y0 +
x
ò f ( x, y
(2 )
= 10202
.
+ 0.01 [10206
.
+ 10422
.
] = 10408
.
) dx
x0
.
y2 = y( 0 .04 ) = 10408
æ
x2
x5 ö
÷dx
= 0 + ò f çç x,
+
2
20 ÷ø
0 è
x
15. (C) y3* = y2 + hf ( x2 , y2 )
x
æ
2 x7 ö
x2
x5
x8
x11
x4
x10
÷÷dx =
+
+
+
= ò çç x +
+
+
2
20 160 4400
4
400
40 ø
0è
12. (A) x: 0
Page
586
0.1
0.2
.
y1 = y (0.02) = 10202
14. (D) y2* = y1 + h f ( x1 , y1 )
The second approximation to y is given by
y ( 2 ) = y0 +
y3 = y2 + hf ( x2 , y2 ) = 101
. + 0.1 f (0.2 , 101
. )
n = 3 in (1) gives
) dx
x0
x
n = 2 in (1) gives
. + 0.0202 = 10302
.
y3 = y( 0 .3) = 101
x
(2 )
h = 0.1
Thus y2 = y( 0 .2 ) = 101
.
3
ò f ( x, y
y0 = 1,
= 1 + 0.1 f (0.1 , 1) = 1 + 0.1 (0.1) = 1 + 0.01
The third approximation to y is given by
y ( 3) = y0 +
x0 = 0,
n = 0 in (1) gives y2 = y1 + h f ( x1 , y1 )
x
x
2
6
=2 + 2x +
....(1)
y1 = 1 + 0.1 f (0, 1) = 1 + 0 = 1
ö
÷÷dx
ø
2
x2
- x) dx
2
2
yn + 1 = yn + h( xn , yn )
y1 = y0 + hf ( x0 , y0 )
) dx
x0
x
Euler’s method gives
n = 0 in (1) gives
The second approximation to y is given by
y ( 2 ) = y0 +
Engineering Mathematics
0.3
0.4
= 10416
.
+ 0.02 f (0.04, 10416
.
)
= 10416
.
+ 0.0217 = 10633
.
h
Next y3 = y2 + [ f ( x2 , y2 ) + f ( x3 , y3*)]
2
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GATE EC BY RK Kanodia
UNIT 9
Engineering Mathematics
1
1 ö
æ
k2 = hf ç x0 + h, y0 + k1 ÷ = (0.2) f (0.1, 0.1) = 0.202
2
2 ø
è
h
k ö
æ
k2 = hf ç x0 + , y0 + 1 ÷
2
2ø
è
1
1 ö
æ
k3 = hf ç x0 + h, y0 + k2 ÷ = (0.2) f (0.1, 0.101) = 0.2020
2
2
è
ø
= (0.2) f (0.1, 11
. ) = 0.2(1.31) = 0.262
h
k ö
æ
k3 = hf ç x0 + , y0 + 2 ÷
2
2 ø
è
k4 = hf ( x0 + h, y0 + k3) = 0.2 f (0.2, 0.2020) = 0.20816
1
k = [ k1 + 2 k2 + 2 k3 + k4 ]
6
1
= [0.2 + 2 (.202) + 2 (.20204) + 0.20816 ],
6
k = 0.2027
such that
y1 = y(0.2) = y0 + k = 0 + 0.2027 = 0.2027
21. (C) We now to find y2 = y(0.4), k1 = hf ( x1 , y1 )
= (0.2) f (0.2, 0.2027) = 0.2 (10410
.
)
= .2082
= 0.2 f (0.1, 1131
. ) = 0.2758
k4 = hf ( x0 + h, y0 + k3)
= (0.2) f (0.2, 12758
.
) = 0.3655
1
k = [ k1 + 2 k2 + 2 k3 + 2 k4 ]
6
1
= [0.2 + 2 (0.262) + 2 (0.2758) + 0.3655 ] = 0.2735
6
Here
y1 = y( 0 .2 ) = y0 + k
= 1 + 0.2735
Þ
12735
.
1
1 ö
æ
k2 = hf ç x1 + h , y1 + k1 ÷
2
2 ø
è
24. (C) Here f ( x, y) = x + y h = 0.2
= (0.2) f (0.3, 0.3068)
k1 = hf ( x0 , y0 ) = 0.2 f (0, 1) = 0.2
h
k ö
æ
. ) = 0.24
k2 = hf ç x0 + , y0 + 1 ÷ = (0.2) f (0.1, 11
2
2ø
è
To find y1 = y( 0 .2 ) ,
= 0.2188
1
1 ö
æ
k3 = hf ç x1 + h , y1 + k2 ÷
2
2
è
ø
h
k ö
æ
. ) = 0.244
k3 = hf ç x0 + , y0 + 2 ÷ = (0.2) f (0.1, 112
2
2 ø
è
= 0.2 f (0.3, 0.3121) = .2194
k4 = hf ( x1 + h, y1 + k3) = 0.2 f (0.4, .4221) = 0.2356
1
k = [ k1 + 2 k2 + 2 k3 + k4 ]
6
1
= [0.2082 + 2(.2188) + 2(.2194) + 0.356 ] = 0.2200
6
k4 = hf ( x0 + h, y0 + k3) = (0.2) f (0.2, 1244
.
) = 0.2888
1
k = [ k1 + 2 k2 + 2 k3 + k4 ]
6
1
= [0.2 + 2(0.24) + 2(0.244) + 0.2888 ] = 0.2428
6
y2 = y( 0 .4 ) = y1 + k = 0.2200 + .2027 = 0.4227
y1 = y( 0 .2 ) = y0 + k = 1 + 0.2428
= 12428
.
22. (C) We now to find y3 = y( 0 .6 ) , k1 = hf ( x2 , y2 )
= (0.2) f (0.4, 0.4228)
= 0.2357
1
1 ö
æ
k2 = hf ç x2 + h, y2 + k1 ÷
2
2
è
ø
***********
= (0.2) f (0.5, 0.5406) = 0.2584
1
1 ö
æ
k3 = hf ç x2 + h, y2 + k2 ÷
2
2 ø
è
= 0.2 f (0.5, .5520) = 0.2609
1
k4 = [ k1 + 2 k2 + 2 k3 + k4 ]
6
1
= [0.2357 + 2(.2584) + 2(0.2609) + 0.2935 ]
6
1
= [0.2357 + 0.5168 + 0.5218 + 0.2935 ] = 0.2613
6
y3 = y( 0 .6 ) = y2 + k = .4228 + 0.2613 = 0.6841
23. (A) Here given
f ( x, y) = x + y
x0 = 0
y0 = 1,
h = 0.2
2
To find y1 = y( 0 .2 ) ,
k1 = hf ( x0 , y0 )
Page
588
= (0.2) f (0, 1) = (0.2) ´ 1 = 0.2
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GATE EC BY RK Kanodia
CHAPTER
10.5
EC-07
1. If E Denotes expectation, the variance of a random
6. For the function e- x , the linear approximation around
variable X is given by
x = 2 is
(A) E[ X 2 ] - E 2 [ X ]
(B) E[ X 2 ] + E 2 [ X ]
2
2
(C) E[ X ]
(D) E [ X ]
2. The following plot shows a function y which varies
2
linearly with X. The value of the integral I = ò y dx is
y
1
(B) 1 - x
(C) [ 3 + 2 2 - 1(1 + 2 x ]e -2
(D) e -2
7. An independent voltage source in series with an
impedance Z s = Rs + jX s delivers a maximum average
power to a load impedance Z L when
3
2
1
-1
(A) ( 3 - x) e -2
1
2
x
3
(A) Z L = RS + jX S
(B) Z L = RS
(C) Z L = jX S
(D) Z L = RS - jX S
8. The RC circuit shown in the figure is
R
(A) 1.0
(B) 2.5
(C) 4.0
(D) 5.0
C
+
Vi
+
R
C
Vo
3. For x << 1, coth( x) can be approximated as
(C)
1
x
4. lim
-
(B) x 2
(A) x
(D)
sin( q/2)
q®0
q
1
x2
is
(C) 2
(D) not defined
(C) a band-pass filter
(D) a band-reject filter
impurities are introduced with a concentration of N A
per cm 3(where N A >> ni ) the electron concentration per
5. Which of the following functions is strictly bounded ?
(C) x 2
(B) a high-pass filter
semiconductor are ni per cm 3 at 300 K. Now, if acceptor
(B) 1
1
x2
(A) a low-pas filter
9. The electron and hole concentrations in an intrinsic
(A) 0.5
(A)
-
(B) e x
(D) e - x
2
cm 3 at 300 K will be
(A) ni
(B) ni + N A
(C) N A - ni
(D)
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ni2
NA
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GATE EC BY RK Kanodia
UNIT 10
10. In a
p+ n junction diode under reverse biased the
Previous year Papers
15. If closed-loop transfer function of a control system is
magnitude of electric field is maximum at
given as T( s) =
(A) the edge of the depletion region on the p -side
(A) an unstable system
(B) the edge of the depletion region on the n -side
(B) an uncontrollable system
(C) the p+ n junction
(C) a minimum phase system
(D) the center of the depletion region on the n-side
(D) a non-minimum phase system
11. The correct full wave rectifier circuit is
16. If the Laplace transform of a signal y( t) is
Output
(B)
Input
Output
(A)
Input
Y ( s) =
1
s ( s -1 )
s-5
( s + 2 )( s + 3)
then It is
, then its final value is
(A) -1
(B) 0
(C) 1
(D) unbounded
17. If R( t) is the auto correlation function of a real,
wide-sense stationary random process, then which of
(A) R( t) = R( -t)
Output
(D)
Input
Output
C
Input
the following is NOT true
(B) R( t) £ R(0)
(C) R( t) = - R( -t)
(D) The mean square value of the process is R(0)
12. In a trans-conductance amplifier, it is desirable to
18. If S( f )is the power spectral density of a real,
have
wide-sense stationary random process, then which of
the following is ALWAYS true?
(A) a large input resistance and a large output
resistance
(B) a large input resistance and a small output
resistance
(A) S(0) £ S( f )
(B) S( f ) ³ 0
(C) S( - f ) = -S( f )
(D)
¥
ò S( f )df
=0
-¥
(C) a small input resistance and a large output
resistance
19. A plane wave of wavelength l is traveling in a
direction making an angle 30 o with positive x -axis. The
(D) a small input resistance and a small output
resistance
®
E field of the plane wave can be represented as (E0 is
constant)
13. X = 01110 and Y = 11001 are two 5-bit binary
numbers represented in two's complement format. The
sum of X and Y represented in two's complement format
using 6 bits is
(A) 100111
(B) 0010000
(C) 000111
(D) 101001
®
æ
3p p ÷ö
j çç wt x z
l l ÷ø
è
®
æ
3p p ö÷
j çç wt +
x z
l l ÷ø
è
(A) E = yE0 e
(C) E = yE0 e
®
æ
p
3p ÷ö
j çç wt - x
z
l l ø÷
è
®
æ
p
3p ö÷
j çç wt - x +
z
l
l ÷ø
è
(B) E = yE0 e
(D) E = yE0 e
20. If C is close curve enclosing a surface S, then the
®
®
the electric flux density D are related by
14. The Boolean function Y = AB + CD is to be realized
using only 2-input NAND gates. The minimum number
of gates required is
(A) 2
(B) 3
(C) 4
(D) 5
Page
640
®
magnetic field intensity H , the current density j and
®
æ®
ö
¶D÷ ®
ç
(A) ò ò H . d s = ò j +
.d l
ç
¶t ÷
s
c
è
ø
®
®
®
æ®
ö
¶D ÷ ®
ç
(B) ò H . d l = òò j +
d .d s
ç
¶t ÷
s
s
è
ø
®
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GATE EC BY RK Kanodia
UNIT 10
Previous year Papers
29. For the circuit shown in the figure, the Thevenin
33. Group I lists four types of p - n junction diodes.
voltage and resistance looking into X - Y are
match each device in Group I with one of the option in
1W
X
Group II to indicate the bias condition of the device in
its normal mode of operation.
Group-I
(P) Zener Diode
(Q) Solar cell
(R) LASER diode
(S) Avalanche Photodiode
2W
2A
1W
2i
Y
(A)
(C)
4
V, 2W
3
2
(B) 4 V, W
3
4
2
V, W
3
3
(D) 4 V, 2W
Group-II
(1) Forward bias
(2) Reverse bias
(A) P - 1 Q - 2 R - 1 S - 2
(B) P - 2 Q - 1 R - 1 S - 2
(C) P - 2 Q - 2 R - 1 S - 2
30. In the circuit shown, Vc is 0 volts at t = 0 sec. for
t > 0, the capacitor current ic ( t), where t is in seconds, is
given by
(D) P - 2 Q - 1 R - 2 S - 2
34. The DC current gain (b) of a BJT is 50. Assuming
that the emitter injection efficiency is 0.995, the base
iC
20 kW
4 mF
20 kW
10 V
transport factor is
+
VC
-
(A) 0.980
(B) 0.985
(C) 0.990
(D) 0.995
35. group I lists four different semiconductor devices.
(A) 0.50 exp( -25 t) mA
match each device in Group I with its characteristic
(B) 0.25 exp( -25 t) mA
property in Group II.
(C) 0.50 exp( -25 t) mA
(D) 0.25 exp( -25 t) mA
31. In the AC network shown in the figure, the phasor
(B) P - 1 Q - 4 R - 3 S - 2
A
~
Group-II
(1) Population inversion
(2)Pinch-off voltage
(3) Early effect
(4) Fat-band voltage
(A) P - 3 Q - 1 R - 4 S - 2
voltage V AB (in volts) is
A
Group-I
(P)BJT
(Q)MOS capacitor
(R) LASER diode
(S) JFET
5W
(C) P - 3 Q - 4 R - 1 S - 2
5W
(D) P - 3 Q - 2 R - 1 S - 4
-j3
j3
36. For the Op-Amp circuit shown in the figure, Vo is
B
(A) 0
(C) 12.5 Ð30 o
2 kW
(B) 5 Ð30 o
(D) 17 Ð30 o
1 kW
vo
1V
32. A p+ n junction has a built-in potential of 0.8 V. The
1 kW
1 kW
depletion layer width at reverse bias of 1.2V is 2 mm.
For a reverse bias of 7.2 V, the depletion layer width
will be
(A) 4 mm
(B) 4.9 mm
(C) 8 mm
(D) 12 mm
(A) -2 V
(B) -1 V
(C) -0.5 V
(D) 0.5 V
37. For the BJT circuit shown, assume that the b of the
transistor is very large and VBE = 0.7 V . The mode of
operation of the BJT is
Page
642
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GATE EC BY RK Kanodia
EC-07
10 kW
2V
10 V
Chap 10.5
(A) 7.00 to 7.29 V
(B) 7.14 to 7.29 V
(C) 7.14 to 7.43 V
(D) 7.29 to 7.43 V
41.
1 kW
The
Boolean
expression
Y = AB CD + ABCD+ ABCD+ ABCD
(A) cut-off
(B) saturation
minimized to
(C) normal active
(D) reverse active
(A) Y = A B CD + A B C + A C D
38. In the Op-Amp circuit shown, assume that the diode
current follows the equation I = I s exp(V/VT ). For
Vi = 2 V , V0 = V01 ,
and
Vi = 4 V , V0 = V02 .
for
The
relationship between V01 and V02 is
can
be
(B) Y = A B CD + B C D + A B C D
(C) Y = AB C D + B C D + AB CD
(D) Y = AB C D + B C D + A B C D
42. The circuit diagram of a standard TTL NOT gate is
D
shown in the figure. Vi = 2.5 V , the modes of operation of
vi
2 kW
the transistors will be
vo
VCC =5 V
(A) V02 = 2 V01
(B) V02 = e V01
(C) V02 = V01 ln 2
(D) V01 - V02 = VT ln 2
100 W
1.4 kW
4 kW
2
Q4
+
39. In the CMOS inverter circuit shown, if the
Q1
Q2
D
+
transconductance parameters of the NMOS and PMOS
transistors are kn = kp = m n Cox
Wn
Ln
= m p Cox
Wp
Lp
Q3
= 40 mA/V 2
1 kW
and their threshold voltages are VTHn = VTHp = 1V , the
-
-
current I is
5V
(A) Q1 : revere active;Q2 : normal active; Q3: saturation;
Q4 :cut-off
PMOS
2.5 V
I
(B) Q1 : revere active;Q2 : saturation; Q3: saturation;
Q4 :cut-off
NMOS
(A) 0 A
(B) 25 mA
(C) 45 mA
(D) 90 mA
(C) Q1 : normal active;Q2 : cut-off; Q3: cut-off;
Q4 :saturation
(D) Q1 : saturation;Q2 : saturation; Q3: saturation;
Q4 :normal active
40. For the Zener diode shown in the figure, the Zener
43. In the following circuit, X is given by
voltage at knee is 7V, the knee current is negligible and
0
the Zener dynamic resistance is 10W. if the input
1
1
0
voltage ( Vi ) range is from 10 to 16V, the output voltage
(V0 ) ranges
from
S1 S0
200 W
+
vi
I0 4-to-1
MUX
I1
Y
I2
I3
A
0
1
1
0
I0 4-to-1
I1 MUX
I2
Y
I3
S1
B
vo
(A) X = A B C + A B C + A BC + ABC
_
(B) X = ABC + A B C + ABC + ABC
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X
S0
C
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UNIT 10
Previous year Papers
(C) X = AB + BC + AC
47. (A) The 3-dB bandwidth of the low-pas signal e- t u( t),
(D) X = A B + B C + AC
44. The following binary values were applied to the X
where u( t) is the unit step function, is given by
1
1
(A)
Hz
(B)
2 - 1 Hz
2p
2p
and Y inputs of NAND latch shown in the figure in the
(C) ¥
sequence indicated below
(D) 1 Hz
48. A Hilbert transformer is a
X = 0, Y = 1; X = 0, Y = 0; X = 1, Y = 1.
(A) non-linear system
(C) time-varying system
X
P
(B) non-causal system
(D) low-pass system
49. The frequency response of a linear, time-invariant
system is given by H ( f ) = 1 + j510 pf . The step response of
Q
Y
the system is
The corresponding stable P , Q outputs will be
(A) P = 1, Q = 0; P = 1, Q = 0; P = 1, Q = 0 or P = 0, Q = 1
(B) P = 1, Q = 0; P = 0, Q = 1; or P = 0 Q = 1; P = 0, Q = 1
(C)
(C) P = 1, Q = 0; P = 1, Q = 1; P = 1, Q = 0 or P = 0, Q = 1
(D)P = 1, Q = 0; P = 1, Q = 1; P = 1, Q = 1
t
- ö
æ
(B) 5çç 1 - e 5 ÷÷u( t)
è
ø
(A) 5( 1 - e -5t ) u( t)
1
(1 - e-5t )u(t)
2
50.
A
5-point
(C)
sequence
t
- ö
1æ
ç 1 - e 5 ÷u( t)
÷
5 çè
ø
x[ n]
is
given
as
45. For the circuit shown, the counter state (Q1Q0 )
x[ -3] = 1, x[ -2 ] = 1, x[ -1] = 0, x[0 ] = 5, x[1] = 1. Let X ( e jw )
follows the sequence
denote the
discrete-time Fourier transform of x[ n]. The
p
value of
ò X (e
jw
)dw is
-p
(A) 5
(B) 10p
(D) 5 + j10 p
(C) 16p
Q0
D0
D1
Q1
51. The z-transform x[ z ] of a sequence x[ n] is given by
X ( z) = 1 -02.z5-1 . It is given that the region of convergence of
x[ n] includes the unit circle. The value of x[0] is
(A) 00, 01, 10, 11, 00
(B) 00, 01, 10, 00, 01
(C) 00, 01, 11, 00, 01
(D) 00, 10, 11, 00, 10
(A) -0.5
(B) 0
(C) 0.25
(D) 0.5
52. A Control system with PD controller is shown in
46.
8085
the figure If the velocity error constant K V = 1000 and
microprocessor system as an I/O mapped I/O as show in
the damping ration z = 0.5, then the value of K P and K D
the figure. The address lines A0 and A1 of the 8085 are
are
An
8255
chip
is
interfaced
to
an
used by the 8255 chip to decode internally its thee ports
R(s)
and the Control register. The address lines A3 to A7 as
C(s)
+
well as the IO/M signal are used for address decoding.
The range of addresses for which the 8255 chip would
get selected is
8255
(A) K P = 100, K D = 0.09
(B) K P = 100, K D = 0.9
(C) K P = 10, K D = 0.09
(D) K P = 10, K D = 0.9
53. The transfer function of a plant is
T( s) =
(A) F8H - FBH
(B) F8H - FCH
(C) F8H - FFH
(D) F0H - F7H
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644
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5
(s + 5)( s2 + s + 1)
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The second-order approximation of T( s) using
dominant pole concept is
1
(A)
(s + 5)(s + 1)
(C)
5
(B)
5
s2 + s + 1
(s + 5)(s + 1)
1
s2 + s + 1
(D)
1
s 2 -1
. If the plant is operated in a unity feedback
stabilize this control system is
10( s - 1)
10( s + 4)
(A)
(B)
s+2
s+2
10( s + 2)
s + 10
(B)
1
s + 11s + 11
(C)
10 s + 10
s + 11s + 11
(D)
1
s + s + 11
2
(A)
(B)
(C)
(D)
decreasing
decreasing
decreasing
increasing
2
the
the
the
the
step size
granular noise
sampling rate
step size
digital communications. The expression for p( t) with
unity roll-off facto is given by
s + 10
p( t) =
55. A unity feedback control system has an open-loop
sin 4 pWt
4 pWt(1 - 16W 2 t 2 )
The value of p( t) at t =
transfer function
G( s) =
2
59. The raised cosine pulse p( t) is used for zero ISI in
2(s + 2)
(D)
10
s + 11s + 11
2
can be reduced by
configuration, then the lead compensator that an
(C)
(A)
58. In delta modulation, the slope overload distortion
54. The open-loop transfer function of a plant is given
as G( s) =
Chap 10.5
K
.
s( s 2 + 7 s + 12)
The gain K for which s = 1 + j1 will lie on the root
1
is
4W
(A) -0.5
(B) 0
(C) 0.5
(D) ¥
60. In the following scheme, if the spectrum M ( f ) of
locus of this system is
(A) 4
(B) 5.5
(C) 6.5
(D) 10
m( t) is as shown, then the spectrum Y ( f ) of y( t) will be
M
m(t)
56. The asymptotic Bode plot of a transfer function is as
S
shown in the figure. The transfer function G( s)
Hilbert
Transform
corresponding to this Bode plot is
0
G(jw)dB
60 dB
-20 dB/dec
(A)
40 dB
(B)
-40 dB/dec
20 dB
w
1
10
20 100
-60 dB/dec
(A)
1
( s + 1)( s + 20)
(B)
1
s( s + 1)( s + 20)
(C)
100
s( s + 1)( s + 20)
(D)
100
s( s + 1)(1 + 0.05 s)
0
0
(D)
C
0
61.
During
0
transmission
over
a
certain
binary
57. The state space representation of a separately
communication channel, bit errors occurs independently
excited DC servo motor dynamics is given as
with probability p. The probability of AT MOST one bit
é
ê
ë
dw
dt
dia
dt
1ù é wù é 0 ù
ù é-1
ú = ê-1 -10 ú êi ú + ê10 ú u
ûë a û ë û
û ë
in error in a block of n bits is given by
(A) pn
(C) np(1 - p) n -1 + (1 - p) n
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(B) 1 - pn
(D) 1 - (1 - p) n
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UNIT 10
Previous year Papers
62. In a GSM system, 8 channels can co-exist in 200
67. A load of 50W is connected in shunt in a 2-wire
KHz bandwidth using TDMA. A GSM based cellular
transmission line of Z 0 = 50W as shown in the figure.
operator is allocated 5 MHz bandwidth. Assuming a
The 2-port scattering parameter matrix (s-matrix) of
1
5
frequency reuse factor of , i.e. a five-cell repeat pattern,
the shunt element is
the maximum number of simultaneous channels that
é- 1
(A) ê 12
ë 2
can exist in one cell is
(A) 200
(C) 25
(B) 40
(D) 5
ù
- úû
é0 1 ù
(B) ê
ú
ë1 0 û
1
2
1
2
2
é- 1
3ù
(C) ê 23
1ú
ë 3 - 3û
63. In a Direct Sequence CDMA system the chip rate is
1.2288 ´ 106 chips per second. If the processing gain is
desired to be at Least 100, the data rate
(A) must be less than or equal to 12.288 ´ 10 3 bits/sec
(B) must be greater than 12.288 ´ 10 3 bits per sec
(C) must be exactly equal to 12.288 ´ 10 3 bits per sec
é 1
(D) ê 43
ë- 4
- 43 ù
1ú
4û
68. The parallel branches of a 2-wire transmission line
are terminated in 100 W and 200 W resistors as shown
in the figure. The characteristic impedance of the line is
Z 0 = 50W and each section has a length of 4l . The voltage
reflection coefficient G at the input is
(D) can take any value less than 122.88 ´10 3 bits/sec
l/4
l/4
200
64. An air-filled rectangular waveguide has inner
W
dimensions of 3 cm ´ 2 cm. The wave impedance of the
TE20 mode of propagation in the waveguide at a
frequency
of
30
GHz
is
(free
space
impedance
h0 = 377 W)
(A) 308W
(B) 355W
(C) 400W
(D) 461W
200 W
l/4
®
65. The H field (in A/m) of a plane wave propagating in
(A) - j
free space is given by
®
H=x
pö
5 3
5
æ
cos( wt - bz) + y sinç wt - bz + ÷
h0
h0
2
è
ø
The time average power flow density in Watts is
h0
100
(B)
(A)
100
h0
(C) 50h20
(D)
7
5
(C) j
5
7
69. A
l
2
(B)
-5
7
(D)
5
7
dipole is kept horizontally at a height of
l0
2
above
a perfectly conducting infinite ground plane. The
®
radiation pattern in the lane of the dipole (E plane)
50
h0
looks approximately as
(A)
®
y
(B)
y
66. The E field in a rectangular waveguide of inner
dimensions a ´ b is given by
®
E=
z
2
wm æ p ö
æ 2 px ö
ç ÷ H 0 sinç
÷ sin( wt - bz) y
h2 è 2 ø
è a ø
y
Where H 0 is a constant, and a and b are the
dimensions
along
the
x -axis
and
the
(C) TM 20
(D) TE10
Page
646
(D)
y -axis
is
(B) TM11
y
C
respectively. The mode of propagation in the waveguide
(A) TE20
z
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EC-07
70. A right circularly polarized (RCP) plane wave is
Common Data for Questions 74, 75 :
incident at an angle of 60 o to the normal, on an
air-dielectric interface. If the reflected wave is linearly
polarized, the relative dielectric constant xr 2 is
Two 4-ray signal constellations are shown. It is
given that f1 and f2 constitute an orthonormal basis for
the two constellations. Assume that the four symbols in
both the constellations are equiprobable. Let N0 / 2
Linearly
Polarized
RCP
Chap 10.5
denote the power spectral density of white Gaussian
air
noise.
Dielectric
(B) 3
(D) 3
(A) 2
(C) 2
0
Common Data for Questions 71, 72, 73:
The
figure
shows
the
high-frequency
capacitance-voltage(C-V) characteristics of a Metal/
SiO2/silicon
(MOS)
capacitor
having
an
area
of
1 ´ 10 -4 cm 2 . Assume that the perimitivities ( e 0 e r ) of
silicon and Si O2 are 1 ´ 10 -12 F/cm and 35
. ´ 10 -13 F/cm
respectively.
C
Constellation 1
74. The ratio of the average energy of constellation 1 to
the average energy of constellation 2 is
(A) 4 a 2
(B) 4
(C) 2
(D) 8
75.
7 pF
Constellation 2
If
these
constellations
are
used
for
digital
communications over an AWGN channel, then which of
the following statements is true ?
(A) Probability of symbol error for Constellation 1 is
lower
1 pF
V
0
71. The gate oxide thickness in the MOS capacitor is
(A) 50 nm
(B) 143 nm
(C) 350 nm
(D) 1 mm
72. The maximum depletion layer width in silicon is
(B) Probability of symbol error for Constellation 1 is
higher
(C) Probability of symbol error is equal for both the
constellations
(D) The value of N 0 will determine which of the two
constellations has a lower probability of symbol error,
(A) 0.143 mm
(B) 0.857 mm
Linked Answer Questions: Q. 76 to Q. 85 Carry
(C) 1 mm
(D) 1.143 mm
Two marks Each.
73. Consider the following statements about the C-V
characteristics plot:
Statement for Linked Answer Questions 76 & 77:
Consider the Op-Amp circuit shown in the figure.
S1: The MOS capacitor has an n-type substrate.
R1
S2: If positive charges are introduced in the oxide, the
R1
C-V plot will shift to the left.
vi
Then which of the following is true?
(A) Both S1 and S2 are true
vo
R
C
(B) S1 is true and Se is false
(C) S1 is false and S2 is true
(D) Both S1 and S2 are false
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UNIT 10
76. The transfer function V0 ( s)/Vi ( s) is
80. The eigenvalue and eigenvector pairs ( li Vi ) for the
(A)
1 - sRC
1 + sRC
(B)
1 + sRC
1 - sRC
(C)
1
1 - sRC
(D)
1
1 + sRC
system are
æ
æ é 1ù ö
é 1ù ö
(A) çç -1, ê ú ÷÷ and çç -2, ê ú ÷÷
ë-1û ø
è
è ë-2 û ø
77. If Vi = V1 sin( wt) and V0 = V2 sin( wt - f), then the
minimum and maximum values of f (in radians) are
respectively
(A) - 2p and
p
2
(C) -p and 0
(B) 0 and
Previous year Papers
p
2
æ
æ é 1ù ö
é 1ù ö
(B) çç -1, ê ú ÷÷ and çç -2, ê ú ÷÷
ë-1û ø
è
è ë-2 û ø
æ é 1ù ö
æ é 1ù ö
(C) çç -1ê ú ÷÷ and çç -2, ê ú ÷÷
1
è ë-2 û ø
è ë ûø
æ
æ é 1ù ö
é 1ù ö
(D) çç -2, ê ú ÷÷ and çç 1, ê ú ÷÷
1
ë ûø
è
è ë-2 û ø
(D) - 2p and 0
Statement for Linked Answer Questions 78 & 79.
An 8085 assembly language program is given
below.
81. The system matrix A is
é 0 1ù
(A) ê
ú
ë-1 1û
1ù
é 1
(B) ê
ú
ë-1 -2 û
1ù
é 2
(C) ê
ú
ë-1 -1û
1ù
é 0
(D) ê
ú
ë-2 -3û
Line 1:
MVI A, B5H
2:
MVI B, OEH
3:
XRI 69H
4:
ADD B
5:
ANI 9BH
6:
CPI 9FH
7:
STA 3010H
density function f ( x) as shown in the figure. Decision
8:
HLT
boundaries of the quantizer are chosen so as t maximize
Statement fo Linked Answer Questions 82 & 83:
An input to a 6-level quantizer has the probability
78. The contents of the accumulator just after execution
of the ADD instruction in line 4 will be
(A) C3H
(B) EAH
(C) DCH
(D) 69H
the entropy of the quantizer output. It is given that 3
consecutive decision boundaries are ' -1' , '0 ' and '1'.
f(x)
a
b
79. After execution of line 7 of the program, the status
-5
of the CY and Z flags will be
(A) CY = 0, Z = 0
(B) CY = 0, Z = 1
(C) CY = A, Z = 0
(D) CY = 1, Z = 1
-1
0
1
5
82. The values of a and b are
(A) a =
1
1
and b =
6
12
(B) a =
1
3
and b =
5
40
(C) a =
1
1
and b =
4
16
(D) a =
1
1
and b =
3
24
Statement for Linked Answer Questions 80 & 81.
Consider a linear system whose state space
representation is x( t) = Ax( t). If the initial state vector of
é 1ù
the system is x(0) = ê ú, then the system response is
ë-2 û
é e -2 x ù
. If the itial state vector of the system
x( t) = ê
-2 t ú
ë-2 e û
83. Assuming that the reconstruction levels of the
quantizer are the mid-points of the decision boundaries,
the ratio of signal power to quantization noise power is
é 1ù
changes to x(0) = ê ú, then the system response
ë-2 û
(A)
152
9
(B)
é e- t ù
becomes x( t) = ê - t ú
ë -e û
(C)
76
3
(D) 28
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648
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EC-07
Statement for Linked Answer Questions 84 & 85.
ANSWER
In the digital-to Analog converter circuit shown in
the figure below, VR = 10 V and R = 10 kW.
R
2R
R
2R
2R
i
R
R
Chap 10.5
1. A
2. B
3. C
4. A
5. D
6. A
7. D
8. C
9. D
10 .C
11. C
12. A
13. C
14. B
15. D
16. A
17. C
18. B
19. A
20. D
21. C
22. A
23. C
24. D
25. B
26. B
27. A
28. D
29. D
30. A
31. D
32. A
33. B
34. B
35. C
2R
R
vo
+
84. The current is
(A) 3125
. mA
(B) 62.5mA
36. C
37. B
38. D
39. D
40. C
(C) 125mA
(D) 250mA
41. D
42. B
43. A
44. C
45. B
46. C
47. A
48. A
49. B
50. B
51. D
52. B
53. C
54. A
55. D
56. D
57. A
58. D
59. C
60. A
61. C
62. B
63. A
64. C
65. D
66. A
67. B
68. D
69. C
70. D
71. A
72. B
73. C
74. B
75. B
76. A
77. C
78. B
79. C
80. A
81. D
82. A
83.
84. B
85. C
85. The voltage V0 is
(A) -0.781 V
(B) -1.562 V
(C) -3.125 V
(D) -6.250 V
************
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CHAPTER
10.1
EC-03
Duration : Three Hours
Maximum Marks : 150
4. The Laplace transform of i( t) is given by
I ( s) =
Q.1—30 carry one mark each
As t ® ¥, The value of i( t) tends to
1. The minimum number of equations required to
analyze the circuit shown in Fig. Q. 1 is
C
C
2
s(1 + s)
(A) 0
(B) 1
(C) 2
(D) ¥
5. The differential equation for the current i( t) in the
R
R
~
C
R
circuit of Fig. Q.5 is
R
i1(t)
Fig. Q1
(A) 3
(B) 4
(C) 6
(D) 7
sin t
2W
2H
~
1F
Fig. Q5
impedance consisting of 1W resistance in series with 1
d 2i
di
(A) 2 2 + 2
+ i( t) = sin t
dt
dt
H inductance. The load that will obtain the maximum
(B) 2
d 2i
di
+2
+ 2 i( t) = cos t
dt 2
dt
(C) 2
d 2i
di
+2
+ i( t) = cos t
2
dt
dt
(D) 2
d 2i
di
+2
+ 2 i( t) = sin t
dt 2
dt
2.. A source of angular frequency 1 rad/sec has a source
power transfer is
(A) 1 W resistance
(B) 1 W resistance in parallel with 1 H inductance
(C) 1 W resistance in series with 1 F capacitor
(D) 1 W resistance in parallel with 1 F capacitor
6. n-type silicon is obtained by doping silicon with
3. A series RLC circuit has a resonance frequency of
(A) Germanium
(B) Aluminium
1 kHz and a quality factor Q = 100. If each of R, L and C
(C) Boron
(D) Phosphorus
is doubled from its original value, the new Q of the
circuit is
7. The bandgap of silicon at 300 K is
(A) 25
(B) 50
(C) 100
(D) 200
(A) 1.36 eV
(B) 1.10 eV
(C) 0.80 eV
(D) 0.67 eV
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EC-03
Chap 10.1
20. A 0 to 6 counter consists of 3 flip flops and a
25. A PD controller is used to compensate a system.
combination circuit of 2 input gate(s). The combination
Compared
circuit consists of
compensated system has
(A) one AND gate
(A) a higher type number
(B) one OR gate
(B) reduced damping
(C) one AND gate and one OR gate
(C) higher noise amplification
(D) two AND gates
(D) larger transient overshoot
21. The Fourier series expansion of a real periodic
26. The input to a coherent detector is DSB-SC signal
signal with fundamental frequency f0 is given by
plus noise. The noise at the detector output is
g p ( t) =
åc
n
n = -¥
e j 2 pf0t . It is given that c3 = 3 + j5. Then c-3 is
to
the
uncompensated
system,
the
(A) the in-phase component
(A) 5 + j 3
(B) -3 - j5
(B) the quadrature component
(C) -5 + j 3
(D) 3 - j5
(C) zero
(D) the envelope
22. Let x( t) be the input to a linear, time-invariant
system. The required output is 4 x( t - 2). The transfer
27. The noise at the input to an ideal frequency detector
function of the system should be
is white. The detector is operating above threshold. The
(A) 4 e
j 4 pf
(C) 4 e
- j 4 pf
23.
A
(B) 2 e
- j8 pf
power spectral density of the noise at the output is
j8 pf
(A) raised-cosine
(B) flat
(C) parabolic
(D) Gaussian
(D) 2 e
sequence
x( n)
with
the
z-transform
X ( z) = z 4 + z 2 - 2 z + 2 - 3z -4 is applied as an input to a
28. At a given probability of error, binary coherent FSK
linear, time-invariant system with the impulse response
is inferior to binary coherent PSK by
h( n) = 2 d( n - 3) where
(A) 6 dB
(B) 3 dB
(C) 2 dB
(D) 0 dB
ì 1, n = 0
d( n) = í
î 0, otherwise
29. The unit of Ñ ´ H is
The output at n = 4 is
(A) Ampere
(A) -6
(B) zero
(C) 2
(D) -4
(C) Ampere/meter
(B) Ampere/meter
2
(D) Ampere-meter
30. The depth of penetration of electromagnetic wave in
24. Fig. Q.24 shows the Nyquist plot of the open-loop
a medium having conductivity s at a frequency of 1
transfer function G( s) H ( s) of a system. If G( s) H ( s) has
MHz is 25 cm. The depth of penetration at a frequency
one right-hand pole, the closed-loop system is
of 4 MHz will be
Im
GH - plane
w=0
(-1, 0)
Re
(A) 6.25 cm
(B) 12.50 cm
(C) 50.00 cm
(D) 100.00 cm
Q.31—90 carry two marks each.
w is positive
31. Twelve 1 W resistance are used as edges to form a
Fig. Q24
cube. The resistance between two diagonally opposite
(A) always stable
(B) unstable with one closed-loop right hand pole
(C) unstable with two closed-loop right hand poles
(D) unstable with three closed-loop right hand poles
corners of the cube is
5
(A) W
6
(C)
6
5
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(B) 1 W
(D)
3
W
2
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