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Exercise‐5
1. A bulb in a staircase has two switches, one switch being at the ground floor
and the other one at the first floor. The bulb can be turned ON and also can
be turned OFF by any one of the switches irrespective of the state of the
other switch. The logic of switching of the bulb resembles
a. an AND gate b. an OR gate
c. an XOR gate d. a NAND gate
2. Consider a vector field ( )A r . The closed loop line integral A dl∫ i can be
expressed as
a. ( )A ds∇ ×∫ i over the closed surface bounded by the loop
b. ( )A dv∇∫∫∫ i over the closed volume bounded by the loop
c. ( )A dv∇∫∫∫ i over the open volume bounded by the loop
d. ( )A ds∇ ×∫ i over the open surface bounded by the loop
3. Two systems with impulse responses h1(t) and h2(t) are connected in
cascade. Then the overall impulse response of the cascaded system is given
by
a. product of h1(t) and h2( t)
b. sum of h1(t) and h2(t)
c. convolution of h1(t) and h2(t)
d. subtraction of h2(t) from h1( t )
4. In a forward biased p‐n junction diode, the sequence of events that best
describes the mechanism of current flow is
a. injection, and subsequent diffusion and recombination of minority
carriers
b. injection, and subsequent drift and generation of minority carriers
c. extraction, and subsequent diffusion and generation of minority carriers
d. extraction, and subsequent drift and recombination of minority carriers
5. In IC technology, dry oxidation (using dry oxygen) as compared to wet
oxidation (using steam or water vapor) produces
a. superior quality oxide with a higher growth rate
b. inferior quality oxide with a higher growth rate
c. inferior quality oxide with a lower growth rate
d. superior quality oxide with a lower growth rate
6. The maximum value of θ until which the approximation sin θ ≈ θ holds to
within 10% error is
a. 10° b. 18° c. 50° d. 90°
7. The divergence of the vector field ˆ ˆ ˆx y zA xa ya za= + + is
a. 0 b. 1/3 c. 1 d. 3
8. The impulse response of a system is h(t) = t u(t). For an input u(t – 1), the
output is
a
c
9. T
T
re
a
10. In
tr
. 2
( )2t u t
. 2( 1) (
2t u−
The Bode p
The gain (
espectively
. 39.8
s
n the circu
ransistor Q
( 1)t −
lot of a tra
20 log |G
y. The pha
b.
uit shown
Q and an id
ansfer func
G(s)|) is 3
se is negat
2
39.8s
below w
deal op‐am
b. ( 1)
2t t −
d. 2 1 (2
t u−
ction G(s) i
2 dB and
tive for all
c. 3
hat is the
mp are used
( 1)u t −
( 1)t −
s shown in
–8 dB at
ω. Then G
32s
e output v
d?
n the figure
t 1 rad/s
(s) is
d. 2
32s
voltage (Vo
e below.
and 10 r
2
2
out) if a sil
rad/s
licon
a
11. C
a
fa
sc
a
12. F
P
. –15 V
Consider a
s shown b
actor k, k
caled by a
. k2
or 8085 m
MVI A
MVI
PTR: ADD B
DCR
JNZ P
b.
delta conn
below. If a
> 0, the e
factor of
b.
icroproces
A, 05H;
B, 05H;
B;
B;
PTR;
–0.7 V
nection of
all elemen
elements o
k
ssor, the fo
f resistors
nts of the
of the corr
c. 1
ollowing pr
c. +0.7
and its eq
delta con
responding
/k
rogram is e
V
uivalent st
nection ar
g star equ
d. k
executed.
d. +15 V
tar connec
re scaled
ivalent wi
k
ction
by a
ll be
A
a
13. T
u
is
a
14. F
th
a
15. In
st
a
b
ADI 0
HLT;
At the end
. 17H
The bit rate
sed is 32‐Q
s
. R/10 Hz
or a perio
he fundam
. 100
n a voltage
tatements
. The input
b. The inpu
03H;
of program
b.
e of a digit
QAM. The
b.
dic signal
mental freq
b.
e‐voltage f
s is TRUE if
t impedan
t impedan
m, accumu
20H
tal commu
minimum
R/10 kHz
v(t) = 30 s
quency in r
300
feedback a
the gain k
ce increas
ce increas
lator conta
c. 2
nication sy
bandwidth
c. R
sin 100t +
ad/s is
c. 5
as shown b
k is increase
es and out
es and out
ains
3H d
ystem is R
h required
R/5 Hz
10 cos 30
00
below, wh
ed?
tput imped
tput imped
d. 05H
kbits/s. Th
for ISI fre
d. R/
0t + 6 sin
d. 15
ich one of
dance decr
dance also
he modula
e transmis
/5 kHz
(500t + π
500
f the follow
reases.
increases.
ation
ssion
/ 4),
wing
.
c. The input impedance decreases and output impedance also decreases.
d. The input impedance decreases and output impedance increases.
16. A band‐limited signal with a maximum frequency of 5 kHz is to be sampled.
According to the sampling theorem, the sampling frequency which is not
valid is
a. 5 kHz b. 12 kHz c. 15 kHz d. 20 kHz
17. In a MOSFET operating in the saturation region, the channel length
modulation effect causes
a. an increase in the gate‐source capacitance
b. a decrease in the transconductance
c. a decrease in the unity‐gain cutoff frequency
d. a decrease in the output resistance
18. Which one of the following statements is NOT TRUE for a continuous time
causal and stable LTI system?
a. All the poles of the system must lie on the left side of the jω axis.
b. Zeros of the system can lie anywhere in the s‐plane.
c. All the poles must lie within |s| = 1.
d. All the roots of the characteristic equation must be located on the left
side of the jω axis.
19. T
a
20. A
h
a
b
c
d
21. A
b
a
22. T
The minimu
. 0
A polynomi
has
. no real ro
b. no negat
. odd num
. at least o
Assuming z
below to a
. u(t)
The transfe
um eigen v
b.
ial f(x) = a4
oots
tive real ro
ber of real
one positiv
zero initia
unit step i
b.
er function
value of the
352
⎡⎢⎢⎢⎣
1
4x4 + a3x
3 +
ot
l roots
ve and one
l conditio
nput u(t) is
tu(t)
2
1
( )( )
V sV s
of t
e following
5 212 7
2 7 5
⎤⎥⎥⎥⎦
c. 2
+ a2x2 + a1
negative r
n, the res
s
c. 2t
the circuit
g matrix is
x – a0 with
real root
sponse y
2
( )2t u t
shown be
d. 3
h all coeffi
(t) of the
d. e–
low is
cients pos
system g
–t u(t)
sitive
given
a. 0.5 1
1s
s+
+ b.
3 62
ss
++
c. 21
ss
++
d. 12
ss
++
23. A source Vs(t) = V cos100πt has an internal impedance of (4 + j3) Ω. If a
purely resistive load connected to this source has to extract the maximum
power out of the source, its value in Ω should be
a. 3 b. 4 c. 5 d. 7
24. The return loss of a device is found to be 20 dB. The voltage standing wave
ratio (VSWR) and magnitude of reflection coefficient are respectively
a. 1.22 and 0.1 b. 0.81 and 0.1
c. –1.22 and 0.1 d. 2.44 and 0.2
25. Let g(t) = e–πt2 , and h(t) is a filter matched to g(t) . If g(t) is applied as input
to h(t) , then the Fourier transform of the output is
a. 2fe−π b.
2 /2fe−π c. | |fe−π d. 22 fe− π
26 Let U and V be two independent zero mean Gaussian random variables of
variances 14 and
19 respectively. The probability P(3V ≥ 2U) is
a. 4/9 b. 1/2 c. 2/3 d. 5/9
27. Let A be an m × n matrix and B an n × m matrix. It is given that determinant
(Im + AB) = determinant (In + BA), where Ik is the k × k determinant of the
matrix given below is identity matrix. Using the above property, the
determinant of the matrix given below is
a
28. In
th
is
a
29. T
W
th
h
a
. 2
n the circu
he Theven
s
. 100 ∠ 90
The open‐lo
When conn
hat will re
hundred tim
. 1
b.
uit shown b
in’s equiva
0°
oop transf
nected in f
educe the
mes as com
b.
2111
⎡⎢⎢⎢⎢⎣
5
below, if t
alent volta
b. 800
fer functio
eedback a
e time co
mpared to
5
1 1 12 1 11 2 11 1 2
⎤⎥⎥⎥⎥⎦
c. 8
he source
age in Volt
0 ∠ 0°
on of a dc
s shown b
nstant of
that of the
c. 1
⎤
⎦
8
voltage V
s as seen b
c. 800 ∠
motor is g
below, the
the close
e open‐loo
0
d. 16
VS = 100 ∠
by the load
∠ 90°
given as Vω
approxima
ed loop sy
op system i
d. 10
6
53.13° V t
d resistanc
d. 100 ∠
( ) 1( ) 1a
sV sω
=+
ate value o
ystem by
s
00
then
ce RL
∠ 60°
010s
.
of Ka
one
30. In
m
m
a
2
31. T
a
a
v
a
n the circu
mA. To ma
minimum p
. 125 and
50
The follow
ttenuator
pplied acr
oltage VYZ2
cross WX.
uit shown b
aintain 5V
power ratin
125
wing arran
which atte
oss WX to
2 = 100V is
Then, VYZ1
below, the
V across R
ng of the Z
b. 125
ngement c
enuates by
get an op
s applied a
1 / VWX1 , VW
e knee curr
L, the min
ener diode
5 and 250
consists o
y a factor o
pen circuit
across YZ t
WX 2 / VYZ2 a
rent of the
nimum val
e in mW, re
c. 250 a
of an idea
of 0.8. An a
voltage VY
to get an o
are respect
e ideal Zen
lue of RL
espectively
and 125
al transfo
ac voltage
YZ1 across Y
open circu
tively,
ner diode i
in Ω and
y, are
d. 250
ormer and
VWX1 = 100
YZ. Next, a
it voltage
is 10
the
and
d an
0V is
an ac
VWX2
a. 125/1
c
32. T
c
W
fr
a
c
33. T
+
a
34. T
M
d
μ
le
100 and 80
. 100/100
Two magne
hosen ope
When conn
requency i
. q1 + q2
. (q1R1 + q2
The impulse
+ δ (t – 3). T
. 0
The small‐s
MOSFET M
ata for M
μA/V2, thre
ength mod
0/100
and 100/1
etically unc
erating fre
nected in s
s
2R2) / (R1 +
e response
The value o
b.
signal resis
shown in
M: device t
eshold vol
dulation eff
b. 100
100
coupled in
equency. T
series, the
R2)
e of a cont
of the step
1
stance (i.e
the figure
transcond
tage VTN =
fects)
0/100 and 8
d. 80/100
ductive co
Their resp
eir effectiv
b. (
d. (q1R2 +
tinuous tim
p response
c. 2
e., dVB/dID
e below, at
uctance p
= 1V, and
80/100
and 80/10
oils have Q
ective res
ve Q facto
1 / q1) + (1
q2R1) / (R1
me system
at t = 2 is
) in kΩ of
t a bias po
parameter
neglect b
00
Q factors q1
sistances a
r at the sa
1 / q2)
+ R2)
is given by
d. 3
ffered by
oint of VB =
kN = μnC’
body effec
1 and q2 at
are R1 and
ame opera
y h(t) = δ(t
the n‐cha
= 2V is (de
’ox (W/L) =
t and cha
t the
d R2.
ating
– 1)
nnel
evice
= 40
nnel
a
35. T
b
F
e
c
a
36. A
L
x
A
. 12.5
The ac sche
below, whe
or the n‐ch
ffect and c
utoff frequ
. 8
A system is
et x(t) be a
1( )
0x t ⎧
= ⎨⎩
Assuming t
b.
ematic of a
ere part o
hannel MO
channel le
uency in H
b.
described
a rectangu
0 totherw
< <
hat y(0) = 0
25
an NMOS c
f the biasi
OSFET M, t
ngth modu
z of the cir
32
d by the dif
lar pulse g
2ise
0 and dydt
=
c. 5
common‐s
ing circuits
the transco
ulation eff
rcuit is app
c. 5
fferential e
given by
0= at t = 0
0
source stag
s has been
onductance
ect are to
proximately
0
equationdd
0, the Lapla
d. 10
ge is show
n omitted
e gm = 1 m
be neglect
y at
d. 20
2
2 5y dydt dy
+ +
ace transfo
00
wn in the fig
for simpli
mA/V, and b
ted. The lo
00
6 ( )y t x+ =
orm of y(t)
gure
icity.
body
ower
( )t .
is
a. 2
( 2)( 3)
ses s s
−
+ + b.
21( 2)( 3)
ses s s
−−+ +
c. 2
( 2)( 3)
ses s
−
+ + d.
21( 2)( 3)
ses s
−−+ +
37. A system described by a linear, constant coefficient, ordinary, first order
differential equation has an exact solution given by y(t) for t > 0, when the
forcing function is x( t) and the initial condition is y(0). If one wishes to
modify the system so that the solution becomes –2y(t) for t > 0, we need
to
a. change the initial condition to –y(0) and the forcing function to 2x(t)
b. change the initial condition to 2y(0) and the forcing function to –x(t)
c. change the initial condition to 2 (0)j y and the forcing function to
2 ( )j x t
d. change the initial condition to –2y(0) and the forcing function to –2x(t)
38. Consider two identically distributed zero‐mean random variables U and V.
Let the cumulative distribution functions of U and 2V be F(x) and G(x)
respectively. Then, for all values of x
a. F(x) – G(x) ≤ 0 b. F(x) – G(x) ≥ 0
c. (F(x) – G(x)) × x ≤ 0 d. (F(x) – G(x))⋅ x ≥ 0
39. The DFT of a vector [ ]a b c d is the vector [ ]α β γ δ . Consider the
product
[ ] [ ]
a b c dd a b c
p q r s a b c dc d a bb c d a
⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦
T
a
c
40. T
YU
a. 25s
s +
41. In
a
The DFT of
. 2 2⎡α β⎣
. [α + β β
The signal
( )( )
Y sU s
for th
16 2
ss
++ +
n the circu
. 4
the vector
2 2 ⎤γ δ ⎦
β + δ δ +
flow grap
his system
b. 2s +
it shown b
b.
r [ p q r
]γ γ + α
h for a sy
is
16 2
ss
++ +
below the o
6
]r s is a s
b. ⎡⎣
d. [
ystem is gi
c. 2 4s
s+
+
op‐amps a
c. 8
scaled vers
⎡ α β⎣
α β γ
iven below
14 2s+
+
re ideal. Th
8
sion of
γ δ
]δ
w. The tra
d. 5s
hen Vout in
d. 10
⎤⎦
nsfer func
2
16 2s+ +
Volts is
0
ction
42. In
v
If
th
a
43. A
v
a
c
44. T
re
th
n the circu
oltage and
f Vcc is + 5 V
hen the Bo
. XY
A voltage 1
oltage me
. sin ωt
. (sin ωt –
Three capa
espectively
he interco
it shown b
d the diode
V, X and Y
oolean exp
b.
000 sin co
asured acr
|sin ωt|) /
acitors C1,
y, have bre
nnection s
below, Q1 h
e drops ne
are digita
pression fo
XY
ot Volts is a
ross WX in
/2
C2 and C
eakdown v
shown bel
has negligi
egligible vo
l signals w
r Z is
c. X
applied acr
Volts, is
b. (sin ωt
d. 0 for all
C3 whose
voltages of
ow, the m
ible collect
oltage acro
with 0 V as
XY
ross YZ. As
+ |sin ωt|)
l t
values a
f 10V, 5V,
maximum s
tor‐to‐emit
oss it under
logic 0 and
d. X
suming ide
) /2
re 10µF,
and 2V re
afe voltag
tter satura
r forward
d Vcc as log
XY
eal diodes,
5µF, and
spectively
e in Volts
ation
bias.
gic 1,
, the
2µF
. For
that
c
in
re
a
c
45. T
a
to
an be app
n µC sto
espectively
. 2.8 and 3
. 2.8 and 3
There are f
s shown in
o addresse
lied across
red in th
y,
36
32
four chips
n the figur
es
s the comb
he effectiv
each of 10
re below. R
bination, a
ve capaci
b. 7
d. 7
024 bytes
RAMs 1, 2,
nd the cor
tance acr
7 and 119
7 and 80
connected
, 3 and 4 r
rrespondin
ross the
d to a 16 b
respective
ng total ch
terminals
bit address
ly are map
arge
are
s bus
pped
a
b
c
d
46. In
v
a
47. L
v
a
. 0C00H‐0F
b. 1800H‐1
. 0500H‐08
. 0800H‐0
n the circu
alue of β. T
. 20
et U and
ariables su
. 3/4
FFFH, 1C00
FFFH, 2800
8FFH, 1500
BFFH, 1800
uit shown
The requir
b.
V be tw
uch that P(
b.
0H‐1FFFH,
0H‐2FFFH,
0H‐18FFH,
0H‐1BFFH,
below, th
red value o
30
wo indepe
U = +1) = P
1
2C00H‐2F
3800H‐3F
3500H‐38
, 2800H‐2B
he silicon
of R2 in kΩ
c. 4
endent an
P(U = –1) =
c. 3
FFH, 3C00
FFH, 4800
8FFH, 5500
BFFH, 3800
npn trans
to produc
40
d identica
= 12. The e
/2
H‐3FFFH
H‐4FFFH
H‐58FFH
0H‐3BFFH
sistor Q ha
e IC = 1 mA
d. 50
ally distrib
ntropy H(U
d. log
as a very
A is
0
buted ran
U + V) in bi
g23
high
dom
its is
Comm
Bits 1 a
respect
48. If
a
49. T
a
Commo
Conside
50. T
th
on Data
nd 0 are tr
ive receive
f the detec
. 12
The optimu
. 12
on Data for
er the follo
The current
he current
Question
ransmitted
ed signals f
ction thres
b.
um thresho
b.
r Question
owing figur
t IS in Amp
t source re
ns: Commo
d with equ
for both bi
hold is 1, t
14
old to achie
45
ns 50 and 5
re
ps in the vo
spectively,
on Data fo
al probabi
its are as s
the BER wi
c. 18
eve minim
c. 1
51:
oltage sou
, are
or Question
lity. At the
hown belo
ll be
18
um bit err
urce, and v
ns 48 and 4
e receiver,
ow.
d. 1
16
ror rate (BE
d. 32
voltage VS
49:
the pdf of
6
ER) is
in Volts ac
f the
cross
a
51. T
a
Linked
Stateme
A mono
directio
and tran
52. T
a
b
c
. 13, –20
The current
. 2
d Answer
ent for Lin
ochromatic
on as show
nsmitted e
The angle o
. 60 and E°
b. 45 and°
. 45 and E°
t in the 1Q
b.
Question
nked Answ
c plane wa
wn in the fig
electric fiel
of incidenc
0 ˆ ˆ(2 x z
E a a−
0 ˆ ˆ(2 x z
E a a+
0 ˆ ˆ(2 x z
E a a−
b. 8, –
Q resistor in
3.33
ns
wer Questio
ave of wa
gure below
d vectors a
e θi and th
410 (3 2)
xj
z eπ× +
−
410 (3)
xj
z eπ× +−
410 (3 2)
xj
z eπ× +
−
–10
n Amps is
c. 1
ons 52 and
velength λ
w. ,i rE E ,
associated
he expressi
)
V/mz+
)
V/mz+
)
V/mz
c. –8, 2
0
d 53:
λ = 600 μ
and tE de
d with the w
ion for iE
0
d. 12
m is prop
enote incid
wave.
are
d. –13, 2
2
pagating in
dent, reflec
20
n the
cted,
d
53. T
a
b
c
d
Stateme
The stat
variable
54. T
a
. 60 and°
The express
. 00.23 (2
E
b. 0 ˆ(2 x
E a−
. 00.44 (2
E
. 0 ˆ(2 x
E a +
ent for Lin
te diagram
e equation
The state‐v
.
[
1X
1
1y
−⎡= ⎢
⎣= −
i
0 ˆ ˆ(2 x z
E a a−
sion rE is
ˆ ˆ( )x za a e−
+
103ˆ )
j
za eπ×
+
ˆ ˆ( )j
x za a e−
+
4103ˆ )
j
za eπ×−
+
nked Answ
m of a syste
s
ariable eq
]
0X
1
1 X u
⎤ ⎡+⎥ ⎢− ⎦ ⎣
− +
4103) V
zj
z eπ×−
410 ( )3 2 V/
x zj π× −
40
V/mz
410 ( )3 2 V/
x zj π× −
( )3 V/m
x z+
wer Questio
em is show
uations of
11
u−⎡ ⎤
⎥⎣ ⎦
V/m
/m
/m
ons 54 and
wn below. A
the system
b. X
y
d 55:
A system i
m shown in
[
1X
1
1y
−⎡= ⎢− −⎣= − −
i
s describe
n the figure
]
0 1X
1 1
1 X u
−⎤ ⎡+⎥ ⎢− ⎦ ⎣
+
d by the st
e above ar
1u⎤
⎥⎦
tate‐
re
c.
[ ]
1 0 1X X
1 1 1
1 1 X
u
y u
− −⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦= − − −
i
d.
[ ]
1 1 1X X
0 1 1
1 1 X
u
y u
− − −⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦= − −
i
55. The state transition matrix eAt of the system shown in the figure above is
a. 0t
t t
ete e
−
− −
⎡ ⎤⎢ ⎥⎣ ⎦
b. 0t
t t
ete e
−
− −
⎡ ⎤⎢ ⎥−⎣ ⎦
c. 0t
t t
ee e
−
− −
⎡ ⎤⎢ ⎥⎣ ⎦
d. 0
t t
t
e tee
− −
−
⎡ ⎤−⎢ ⎥⎣ ⎦
Key for Excerise‐5
Q.NO KEY
1. C
2. D
3. C
4. A
5. D
6. C
7. D
8. C
9. B
10. B
11. B
12. A
13. B
14. A
15. A
16. A
17. D
18. C
19. A
20. D
21. B
22. D
23. C
24. A
25. D
26. B
27. B
28. C
29. C
30. B
31. B
32. C
33. B
34. B
35. A
36. B
37. D
38. D
39. A
40. A
41. C
42. B
43. D
44. C
45. D
46. C
47. C
48. D
49. B
50. D
51. C
52. C
53. A
54. A
55. A