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GATE-2016
: 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com
IInnddeexx
11.. QQuueessttiioonn PPaappeerr AAnnaallyyssiiss
22.. QQuueessttiioonn PPaappeerr && AAnnsswweerr kkeeyyss
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GATE-2016 IN-SET-1
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NETWORK 10%
CONTROL SYSTEM 7%
SIGNAL & SYSTEM 8%
ANALOG CIRCUITS 10%
COMMUNICATION6%
Measurements 8%
Transducer 7%
Optical Instrumentation6%
GA 15%
ANALYSIS OF GATE 2016
Instrumentation Engineering
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IN ANALYSIS-2016_31-JAN_Afternoon Session
SUBJECT No. OF
QUESTIONS Topics Asked in Paper
Level of Toughness
Total Marks
Networks 1 M: 4 2 M: 3
Basic Networks; Dependent Source; Sinusoidal Analysis
EASY 10
Control System 1 M: 1 2 M: 3
Nyquist Plot; Initial Value Time Response;
EASY 7
Signals & Systems
1 M: 2 2 M: 3
Convolutions ;Digital filters; Energy and Power; DFT; Periodic signal, Fourier series,
EASY 8
Digital & µP 1 M: 3 2 M: 4
Combinational Circuit ; Flip-flop; Finite State Machines* ;Memory; Converter; Counter.
MODERATE 11
Analog Circuits 1 M: 2 2 M: 4
Op-Amp; Elber-Moll Model; Current Mirror; Diode based
TOUGH 10
Com. Systems 1 M: 2 2 M: 2
SSB; F.M; Noise TOUGH 6
Transducer 1 M: 3 2 M: 2
Piezoelectric; strain gauge; Pressure measurement
MODERATE 7
Measurements 1 M: 2 2 M: 3
Flow Measurements; wheatstone bridge ; Potentiometer ; Power Measurements.
MODERATE 8
Optical Instrumentation
1 M: 2 2 M: 2
LED. MODERATE 6
Engineering Mathematics
1 M: 4 2 M: 4
Calculus ;Linear Algebra, Probability & Distributions ;Limit & Continuity ; Laplace
MODERATE 12
GA 1 M: 5 2 M: 5
Verbal Ability; Time and Work; Directions; Venn Diagrams; Mensuration and Area; Clock
EASY 15
Total 65 MODERATE 100
* Indicates Questions from New Syllabus
Faculty Feedback: Few questions came from new syllabus; General Ability was pretty easy;
question required logic for solving , qualifying is easy but scoring is tough ,
Time Management very critical. Practice previous year question papers will be
beneficial and practice online test series.
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GATE 2016 Examination
Instrumentation Engineering
Test Date: 31/01/2016
Test Time: 2:00 PM to 5:00 PM
Subject Name: INSTRUMENTATION ENGINEERING
Section: General Aptitude
Q No. 1.
Q No. 2.
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Q No. 3.
Q No. 4.
Q No. 5.
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Q No. 6.
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Q No. 7.
Q No. 8.
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Q No. 9.
Q No. 10.
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Section: Technical
Q No. 1
[Ans. *] Range: 3 to 3
Slope =y2 − y1
x2 − x1=
6 − 0
2 − 0= 3
Q No. 2
[Ans. *] Range: 0.5 to 0.5 Here ‘n’ tends to infinity,
Let us try n = 100, then √100 × 100 + 100 − √100 × 100 + 1 = 0.493
try n = 1000, then √1000 × 1000 + 1000 − √1000 × 1000 + 1 = 0.4993
try n = 10000, then √10000 × 10000 + 10000 − √10000 × 10000 + 1 = 0.49993 ∴ We observe that as n tends larger and larger the value of the limit approaches 0.5 asymptotically ∴ The value of the limit is 0.5
Q No. 3
[Ans. *] Range: 2.5 to 2.5 V1 = 100 ± 1.5 V V2 = 150 ± 2 V V3 = V1 + V2 Standard deviation in V3 will be
σV3= √(
∂V3
∂V1)2
σV12 + (
∂V3
∂V2)2
∂V3
∂V1=
∂
∂V2 (V1 + V2) = 1
∂V3
∂V1=
∂
∂V1
(V1 + V2) = 1
σV3 = √12 × (1.5)2 + 12 × (2)2
= √6.25 = 2.5 V
.
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Q No. 4
[Ans. D]
We know that if a⃗ and b⃗ are perpendicular
Then a⃗ ⋅ b⃗ = 0 Options (A), (B), (C) are perpendicular. Option (D) is not perpendicular.
Q No. 5
[Ans. A] f(z) is of the from 1 + x + x2 + ⋯ = (1 + x)−1 Please note f(z) four is expanded in neighborhood of z = 1, so the series will converge
f(z) =1
1 − (1 − z)=
1
z
Q No. 6
[Ans. *] Range: 5 to 5 By KCL clearly i1 = i2 + i3 put here i1, i2, i3 have different phase so, the phases diagram will look like
∴ I3 = √32 + 42 = 5A
i2 = 3 sinωt −i2 = −3 sinωt
i1 = 4 cosωt i3
.
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Q No. 7
[Ans. *] Range: 1 to 1
Q =ωL
R= 5 at 100 kHz (given)
⇒L
R=
5
ω=
5
2π × 100 kHz……… .①
At 20 kHzQ = ω ×L
R= 2π × 20 kHz ×
5
2π × 100 kHz [Using
L
R from ①]
∴ Q = 1
Q No. 8
[Ans. *] Range: 8 to 8
V =1
c∫ i. dt = V =
1
1 ∫8dt
2
1
= 8[t]12
= 8 Volts
Q No. 9
8A
i(t) =
1s 2s
x
t
.
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[Ans. *] Range: 2 to 2 1 1 −1 0 1 1 1 −1 0 1 1 ① −1 0
−1 −1 −1 +1 0 0 0 0 0 0
∴ y(n) ={1,2, −1,−2,1,0,0}
↑
∴ y(−1) = 2
Q No. 10
[Ans. *] Range: 0.28 to 0.283
V1 = 1 sin (2π 10000 t) V V2 = 1 sin(2π 30000 t) V ∴ f1 = 10000 Hz f2 = 30000 Hz = 3 f1 For LC – 2
ω0 =1
√LC⇒ f0 =
1
2π√LC
So, for f1 = f0|LC=2 ⇒ Resonance It act as open so I = 0 For LC − 1 to be open circuit i. e. , for resonance
(3 f1)30,000 =1
2π√LC
(3f1)30,000 =1
(4π)2 × 9 × 108 × 100 × 10−6= 0.28μF
~
C 2.53 μF
LC − 1 LC − 2
R
100 μH 100 μH
V1 + V2
I
.
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Q No. 11
[Ans. *] Range: 0 to 0
limt→∞
f(t) = lims→0
sF(s) =S × 20(S + 2)
(S + 1)(S + 3)2= 0
Q No. 12
[Ans. *] Range: 1 to 1
G(s) =s − 1
s + 1 G(jω) =
jω − 1
jω + 1
= 1∠180° − 2 tan−1 ω On varying ω = 0+to + ∞ Now on varying ω = 0−to − ∞ G(jω)|ω=0 = 1∠180° G(jω)|ω=0− = 1∠180° G(jω)|ω=1 = 1∠90° G(jω)|ω=−1 = 1∠270° G(jω)|ω=∞ = 1∠0° G(jω)|ω=−∞ = 1∠360° On plotting
∴ No. of times the Nyquist plot encircle the origin clockwise = 1
Q No. 13
[Ans. D]
Routh array
s3 1 2 s2 3 90
s1 6 − 0
3 0
s0 90 0 6 − 90
3> 0
⇒ 6 > 90
ω = +∞
ω = −∞
ω = 0+
ω = 0−
Im [G(jω)]
Re [G(jω)]
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Q No. 14
[Ans. *] Range: 0.14 to 0.15 Here x(t) is a sinusoid (Eigen function of a system) ∴ |y(t)| = |G(jω|. |x(t)| ω = 3r/s
=8
(√ω2 + 102)2 × 2 =
16
109= 0.1467
Q No. 15
[Ans. *] Range: 0 to 0
Let the diode is ON The diode is replaced by short circuit KCL at node B: VB − 3
110+
VB
5− 1 × 10−3 = 0
⇒ VB =10
3= 3.33 V
⇒ ID =3 − 3.33
1 × 103= −ve not possible
⇒ Diode is off So the current passing through diode = 0 ⇒ Vab = V1K = 0 Volts
1 k b VB
1 mA 5 k 3 V
a ID
.
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Q No. 16
[Ans. *] Range: −𝟏 𝐭𝐨 − 𝟏 Thevenin’s circuit of input is [This is the important point here]
∴ The overall circuit looks like
V0 = −1V
Q No. 17
20 k
10 k Vo
10 k
−
+ 10 k
1V
10 k
.
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[Ans. *] Range: 1.55 to 1.65 During positive half cycle When vi > 0.6 V ⇒ D1 − ON and D2 − OFF Then equivalent circuit
V0 = −vi [16
10] = −1.6 vi
For negative half cycle when vi < −0.6 V ⇒ D1 − OFF and D2 − ON
Then magnitude of the negative peak value of the output Vo is 1.6 V
Q No. 18
[Ans. C] XY + (X̅ + Y̅)Z = XY + XY̅̅̅̅ Z
= (XY + XY̅̅̅̅ )(XY + Z) = XY + Z = (X + Z) (Y + Z)
Vo = 0
−
+ 10 kΩ
16 k
0.6 V
Vi
Vo
−
+ 10 kΩ
16 k
0.6 V
Vi
.
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Q No. 19
[Ans. B]
f = (x. y.̅̅ ̅̅ ̅ ). (y̅. z) = (x. y̿̿ ̿̿ ) + (y̅z̿̿̿)
By Demorgan’s law f = xy + y̅z
Q No. 20
[Ans. *] Range: 2 to 2
Accuracy = ±1% of Range =±1
100× 200 = ±2 mA
∴Reading can be 100 mA ± 2 mA
x
y
𝑧
(x. y̅̅ ̅̅ ). (y̅. z̅̅ ̅̅ )̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅
y̅
x. y̅̅ ̅̅
y̅. z̅̅ ̅̅
.
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Q No. 21
[Ans. *] Range: 1.1234 to 1.1234 The unknown voltage Vx will be Vx = IP [11 resistors of 10Ω + Slide wire resistance]
= 10 × 10−3 (11 × 10 +234
1000× 10)
= 1.1234 V
Q No. 22
[Ans. *] Range: 20 to 20 R = RF ∥ Ri to cancel Bias current ⇒ R = 60 k ∥ 30 k = 20 kΩ
.
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Q No. 23
[Ans. *] Range: 0 to 0 0 Volts, PIEZO CANNOT SENSE STATIC LOAD
Q No. 24
[Ans. D] SSB signal is given by = ACm(t)cos(ωct) ± ACm̂ sin(ωct) In phase component = ACm(t) Quadrature phase component = ACm̂(t) Now, the output of envelop detector can be given by
|S(t)| = AC√m2(t) + m̂2(t)
= √cos2 θ + sin2 θ = 1
Q No. 25
[Ans. B]
Q No. 26
[Ans. *] Range: −𝟏𝟑 𝒕𝒐 − 𝟏𝟑 f ′(x) = 6x2 − 4x3 = 0 formin or max point
= x2(6 − 4x) = 0
⇒ x = 0 or 6 − 4x = 0 ⇒ x =3
2not possible
.
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x is in (−1 , 1)as given f ′′(x) = 12x − 12x2
= 0@x = 0 f(x) = 0 & f ′′(x) = 0 @ x = 0 ∴ x is not amin point f(x)@ − 1 = −2 − 1 − 10 = −13 f(x)@1 = 2 − 1 − 10 = −9 f(x)@0 = 0 − 0 − 10 = −10 ∴ f(x)has amin value @ x = −1; f(x) = −13
Q No. 27
[Ans. A] P(red ball) = P(green in 1 trial followed by red) or P(red in 1 trail followed by red) Here, we get to pick 12 balls in TRIAL 1 and 13 balls in TRIAL 2
=7
12×
5
13+
5
12×
6
13=
35 + 30
156=
65
156
Q No. 28
[Ans. *] Range: −𝟔 𝒕𝒐 − 𝟔 If E value of ‘A’ is λ1, λ2, λ3
Then by property E value of Ak = λ1k, λ2
k, λ3k
E Value of μA = μλ1, μλ2, μλ3 (μ is ‘n’ constant) ∴ E Value of A3 = 13, 23, 33 = 1, 8, 27 E Value of A2 = 12, 22, 32 = 1, 4, 9 E Value of 3A2 = 3, 12, 27
TRACE OF A MATRIX = Σ (E Value) ∴ TRACE OF A3 = 36 TRACE OF 3A2 = 42 ∴ TRACE OF A3 − 3A2 = −6
.
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Q No. 29
[Ans. A] Taking L.T on both sides, (M2s + Bs + K)X(s) = F(s)
∴X(s)
F(s)=
1
M2s + Bs + K=
1
0.1s2 + 2s + 10=
10
s2 + 20s + 100
Q No. 30
[Ans. *] Range: 1 to 1 1
2πj ∫
z2 + 1
(z2 − 1)c
dz =1
2πj ∫
z2 + 1
(z − 1)(z + 1)c
Given circle is |z − 1| = 1 pole z = 1 lies inside C pole z = −1 lies outside C
Res f(z)at z = 1 is = limz→1
(z − 1)z2 + 1
(z − 1)(z + 1)=
2
2= 1
Res f(z)at z = – 1 is = 0 By Cauchy’s residue theorem 1
2πj ∫
z2 + 1
z2 − 1 C
dz =1
2πj× 2πj(1 + 0) = 1
.
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Q No. 31
[Ans. *] Range: 10 to 10 By nodal analysis at ‘x’ we get Vx − 1
100+
Vx
100− 10 mA = 0
∴2Vx
100= 10 mA + 10 mA
∴ Vx = 50(20 mA) = 1V
∴ Ix =1V
100 Ω= 10 mA
Q No. 32
[Ans. *] Range: 100 to 100 According to statement given circuit is at resonance
Y =1
(R + jXL)+
1
−jXC
Y =R − jXL
R2 + XL2 +
j
XC
Imaginary part is zero Is = Vs ⋅ Yreal part
= 101 ⋅ [10
102 + 1002]
=101 × 10
10100= 0.1A
= 100 mA
.
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Q No. 33
[Ans. *] Range: 3463 to 3465
Q No. 34
[Ans. *] Range:2.9 to 3.1
f =1
2πRC= 150 Hz (given)
∴ C =1
2πRF
Thevenin’s equation of bridge. Here, we need RTH only of bridge to find C RTH of bridge Seen pay cap is
R1 R2
R4 R3
V
R
C
.
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R1 ∥ R4 + R2 ∥ R3
=R(1 − X)
2+
R(1 + X)
2
= R = 350 Ω
∴ C =1
2πRf= 3.03 μF
Q No. 35
[Ans. *] Range: 1.62 to 1.64 For t < 0; ckt is in steady state with V(t) = 3V ∴ iL(θ) = 1A
After t = 0
Here the final inductor current would be 2A
∴ iL(t) = ifinal + (iinitial − ifinal)e−t τ⁄
∴ iL(t) = 2 + (1 − 2)e−t τ⁄ = 2 − e−t τ⁄
τ =L
Req where Req is Rthevein seen by τ
+ 1 Ω
1 Ω
1 Ω 6V −
2 A 2 A
4 A i(∞)
+ 1 Ω
1 Ω
1 Ω 6V −
3V
(L is like short)
1 Ω
1 Ω
1 Ω
2A
iL(0−)
1 A 1 A
.
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Here Req is
∴ τ = 1s ∴ iL(t) = 2 − e−t ∴ iL(1s) = 2 − e−1 = 1.632 A
Q No. 36
[Ans. A] Given signal satisfies odd and half wave symmetry so it contains only odd harmonics. ∴ Power in 10th harmonic will be zero.
Q No. 37
[Ans. *] Range: 8 to 8
x[n] = sin [301
4 πn]
N0 =2π
ω0=
2π
301π/4× m
=8
301× 301 = 8
1 Ω
1 Ω
1 Ω = 1.5 Ω
Req
.
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Q No. 38
[Ans. B] Since the initial slope is 0 dB/decade, thus it a type zero system ∴ 20 log K = 0 K = 1 now at ω1 = 10° Slope goes to 40 dB/decade ∴ Two zeros added at ω1 = 10° and at ω2 = 102 Slope goes to 40 dB/dec ∴ Two poles added at ω2 = 102
Thus T. F =K (
sω1
+ 1)2
(s
ω2+ 1)
2 =K(
s1
+ 1)2
(s
100+ 1)
2
= 104(s + 1)2
(s + 100)2
.
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Q No. 39
[Ans. C]
For C(s) = 3 +9
2
Characteristic equation 1 + C(s) G(s) = 0
1 + (3 +9
2) (
1
(s + 1)2) = 0
1 + (3s + 9
s) (
1
s2 + 2s + 1) = 0
s3 + 2s2 + 4s + 9 = 0 Using RH criteria
s3 1 4 s2 2 9
s1 5 − 9
2 0
s0 9 0 ∵ 1st column of RH table does not contains positive sign only ∴ System become unstable.
.
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Q No. 40
[Ans. B]
Q No. 39
[Ans. C]
For C(s) = 3 +9
2
Characteristic equation 1 + C(s) G(s) = 0
.
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1 + (3 +9
2) (
1
(s + 1)2) = 0
1 + (3s + 9
s) (
1
s2 + 2s + 1) = 0
s3 + 2s2 + 4s + 9 = 0 Using RH criteria
s3 1 4 s2 2 9
s1 5 − 9
2 0
s0 9 0 ∵ 1st column of RH table does not contains positive sign only ∴ System become unstable.
Q No. 40
[Ans. B]
P(s) =−1
s + 1
Response C(s) = P(s) ⋅ u(s) =−1
s + 1⋅1
s
c(∞) = lims→0
s. c(s) = lim s→0
s [−1
s + 1⋅1
s] = −1
∴ P(s) ⇒ 1 Match
Q(s) =2(s − 1)
(s + 10)(s + 2)
C(s) = Q(s). u(s) =2(s − 1)
(s + 10)(s + 2)⋅1
s
.
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c(∞) = lims→0
SC(s) = lims→0
s [2(s − 1)
(s + 10)(s + 2)⋅1
s] =
−1
2= −0.1
Q(s) ⇒ 2 Match
R(s) =1
(s + 1)2; C(s) =
1
(S + 1)2⋅1
s
C(∞) = lims→0
s C(s) = lims→0
1
(s + 1)2⋅1
s= 1
R(s) ⇒ 3 Match; P(s) → 1, Q(s) → 2, R(s)
Q No. 41
[Ans. *] Range: 40 to 41 Q No. 42
[Ans. *] Range: 100 to 100
.
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Apply KCL at node A 0 − Va
1k=
Va − V0
10k
⇒ Va =V0
11
Using virtual ground concept
VA = VB = Vy =V0
11
Apply KCL at Node B Vy − Vx
1K+
Vy
1/CS+
Vy − V0
1/CNS= 0
Vx = Vy [Given]
Vy(CS) + (Vy − V0)CN. S = 0 V0
11(CS) + (
−10
11)V0CNS = 0
CN = 0.1C = 0.1 nF = 100 pF
Q No. 43
+
−
A B
C
1 kΩ
1 kΩ
10 kΩ
CN
Vo
Vx
.
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[Ans. *] Range: 100 to 100
Apply KCL at Node A 0 − Va
100 k=
Va − V0
100 k
⇒ Va =V0
2
Using virtual ground concept
Va = VB =V0
2
Apply KCL at Node B VB − 1
10k+
VB − V0
10K+ IL = 0
⇒ IL = 0.1 mA = 100μA
Q No. 44
[Ans. A] F(x, y, z) = y z(x) + yz(0) + y z(x) + yz(1)
= xy z + xy z + yz Σm(2, 3, 4, 7)
+
−
10 k
100 k 100 k
10 k
RL IL
VB
VA
1 V
Vo
A
.
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Q No. 45
[Ans. B] Note: OBSERVE QA is toggling on every clock ∴ JA = 1 KA = 1 JB KB can be both connected QA This matches with the option given
Q No. 46
[Ans. D] Y5 is connect to CS of memory will be active only when A11 is ‘1’ & A10 = ′0′ & A14, A13, A12 = 101
.
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A14 A13 A12
1 0 1|A11 A10 A9 …… . . A0
1 0 |
μP has 16 address Bus and A15 is not used for decoding when A15 is “0” as well as A15 = 1 with above condition met CS of memory will be selected
∴ Address Dec is 5800
(A15 = 0)−5BFF &
D800(A15 = 1)
+ DBFF
Q No. 47
[Ans. *] Range: 10 to 10 C1 = 110 pF, C2 = pF
n =f2f1
=2f
f= 2
Cd =C1 − n2C2
n2 − 1=
110 − 4 × 20
3= 10 pF
Q No. 48
[Ans. *] Range: 1 to 1
−
+
−
+
Low-Pass
Filter Vo
VXOR
V01
V02
V2
V1
.
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Output of low pass filter is the average value of output of XOR GATE
Vdc =1
2π[∫ 5dt
π/5
0
+ ∫ 5dt6π/5
0
] =1
2π[5 (
π
5) + 5 (
6π
5− π)]
=1
2π[π + π] = 1 Volt
Q No. 49
[Ans. *] Range: 199 to 201
Q No. 50
[Ans. *] Range: 10 to 10
Output Waveform
π 2π 3π 4π 5π V01
VXOR π/5 6π/5 11π/5 16π/5 21π/5 26π/5
π/5 6π/5 11π/5 16π/5 21π/5 26π/5 π 2π 3π 4π 5π
V02
.
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𝑅𝜃 = 1000[1 + 0.004 𝜃] For 1° C rise in temperature 𝑅𝜃 = 1000[1 + 0.004]
= 1004 𝛺 Redraw the CKT Voltage at Node A by potential divider
𝑉𝐴 =(−5)10
10 + 10= −2.5 𝑉
Using virtual ground 𝑉𝐴 = 𝑉𝐵 = −2.5 𝑉 Apply KCL at Node B 𝐼1 = 𝐼2 𝑉𝐵 − (−5)
1000=
𝑉0 − 𝑉𝐵
1004
𝑉0 = 10 𝑚𝑉 Sensitivity = 10 mV/°C
Q No. 51
[Ans. *] Range: 2.0 to 2.1 OUTPUT current through photo diode (V0) = Intensity × Sensitivity × Area
V0 =4W
m2× 0.5
A
W× 10 × 10−6 m2 = 20 μA
Total current flow through photo diode = IP + ID (Dark current) OUTPUT voltage i to v converter
V0 = (920 + 1)μA ×100 mV
μA= 2100 mV = 2.1 V
−
+
+ −
5 V 10 kΩ
1 kΩ
10 kΩ
I1
I2 1004 Ω
Vo
.
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: 080-617 66 222, [email protected] ©Copyright reserved. Web:www.thegateacademy.com 36
Q No. 52
[Ans. *] Range: 4.9 to 5.1
V = √2gΔP ; ΔP = [Sm
Sw− 1] × h
V = √2 × 9.81 × 1.176 ; ΔP = [13.5 − 1] ×94.1
1000
≃ 5 m/s
Q No. 53
[Ans. *] Range: 1.9 to 2.1
Eg =hc
λ
=4.13567 × 10−15 × 2.998 × 108
620 × 10−9= 2 eV
Q No. 54
[Ans. *] Range: 60.0 to 60.2
.
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m(t) =sin(100πt)
100 πt= sin c(100t)
Kf = 30 kHz Volt⁄
β =KfAm
fm=
30 × 1
50
β =3
5K = 600
BW = (β + 1)2f1 = (600 + 1) × 100 BW = 60.1 kHz
Q No. 55
[Ans. *] Range: 15 to 16 Signal power = 3 W
Resolution (Δ) =3.5 − (−3.5)
23=
7
8
Quantization noise power
N =Δ2
12= 0.0638
Signal to noise ratio S
N=
3
0.0638
(S
N)dB
= 10 log10(47.02)
= 16.72 dB
f
M(f)
−50 0 50