Gate 2014 Ec Morning
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Transcript of Gate 2014 Ec Morning
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8/9/2019 Gate 2014 Ec Morning
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W
hatismyresults.com
W hatismyresults.org
Presents
GATE Answer Key 2014 for
EC
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GATE-2014Exam Solutions (15 Feb)Electronics Engineering(Morning Session)
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Section - I (Technical)
Q.1 In the voltage regulator shown in figure op-amp is ideal. The BJT has VBE
=0.7V
and = 100 and Zener voltage Vzis 4.7 V for a regulated output of 9 V the valueof R in is
+
V = 4.7 Vz
1 k
V = 12 Vi V = 9 Vo
1 K = R1
R = R2
Solution: (1093.0232)
Given circuit is a op-amp series regulator
Vo is given by
Vo = 1
z
2
R1 V
R
+
9 V =2
1k1 4.7
R
+
R2 = 1093.0232
Q.2 A depletion type N-channel MOS is biased in its linear region to use as a voltage
controlled resistor. Assume Vth = 0.5 V, VGS = 20 V, VDS = 5 V, =W
100,L
COX = 108 F/m2, n= 800 cm
2/V-s. Find the resistance of voltage control resistor
in ().
Solution: (641025.641)
Voltage controlled resistor rDSis given by
rDS = n OX GS t
1
W( C ) (V V )
L
rDS = 4 81
800 10 10 100 (20 0.5)
rDS = 641.02 k
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GATE-2014Exam Solutions (15 Feb)Electronics Engineering(Morning Session)
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1 2 3 4 5 6 7 8 9 0
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1 2 3 4 5 6 7 8 9 0
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1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
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Q.3 Capacity of binary symmetric channel with cross-over probability 0.5 ____.
Solution: (0)
Channel capacity of BSC is
C = Plog2P + (1 P) log2(1 P) + 1
C = 0.5log20.5 + 0.5log20.5 + 1
C = 0
It is the case of channel with independent input and output, hence C = 0.
Q.4 In BJT transistor VBE = 0.7 V and VT = 25 mV and reverse saturation current is
1013A. Find the transconductance in mA .V
Solution: (5785.0282)
We know that
gm =c
T
I
V
where Ic =
BE
T
V
VsI e
So, Ic =
0.713 0.02510 e
Ic = 144.6257 mA
Hence, gm =144 mA
5785.02820.025 V
=
Q.5 Find the RMS value of the given pulse
0
1
T/2 T t3T/2
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Corporate Office: 44-A/1, Kalu Sarai, New Delhi-16 | Email : [email protected] | Visit: www.madeeasy.in
GATE-2014Exam Solutions (15 Feb)Electronics Engineering(Morning Session)
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Solution: (0.4082)
RMS value = T
2
0
1 f (t)dtT
where T is time period
= +
2T/ 2 T2
0 T /2
1 2t dt (0) dt
T T
= T/2
2
20
1 4t dt
T T
So, RMS value =1
or 0.4086
Q.6 Let x(n) =
n n1 1
u(n) u( n 1)9 3
ROC of z-transform is
(a) 1 1Z3 9
(d) does not exist
Solution: (c)
x(n) =
> >1 1
Z3 9
Q.7 The amplifier shown in figure. The BJT parameters are, VBE = 0.7 V, = 200 V,
VT = 250 mV. Find the gain =o
i
V_______.
V
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Corporate Office: 44-A/1, Kalu Sarai, New Delhi-16 | Email : [email protected] | Visit: www.madeeasy.in
GATE-2014Exam Solutions (15 Feb)Electronics Engineering(Morning Session)
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
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1 F
Vi
R1 RC = 5 k
R2 11 k
10 kRs
Re 11 k
33 k
1 mF
V = +12 Vcc
Solution: (0.4889)
RB
re
ib
+
Vi= R R1 2
ib
R = 10 ks
R = 5 kC
+
V0
Vi =ibre+ (ib+ ib) RsVi =ibre+ ib(1 + ) RsVi = ib[re+ (1 + ) Rs] ...(i)
V0 = Rcib ...(ii)
0
i
V
V=
c b
b e s
R i
i [ r (1 )R ]
+ +
AV =c
e s
R
r (1 )R
+ + ...(iii)
where re is given by
re =T
E
V
I
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Corporate Office: 44-A/1, Kalu Sarai, New Delhi-16 | Email : [email protected] | Visit: www.madeeasy.in
GATE-2014Exam Solutions (15 Feb)Electronics Engineering(Morning Session)
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
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from dc analysis
V = 3 Vth
R = 8.25 kth
10 k
11 k
5 k
12 V
Rth = 33k 11k = 8.25k
Vth =11 12
3 V44
=
3 = 8.25k IB+ 21k IE
3 =E
E
I8.25 21 k I
1+
+
3 = E8.25
I 21 k201
+
IE = 0.142 mA
re =25 mV
0.142 mA
re = 176.0563
So, A v =5 k 200
200 176.0563 201 10 k
+
= 0.4889
Q.8 A transmission line has characteristic impedance is 50and length l=/8. Ifload ZL = (R + j30), then what is the value of R, if input impedance of transmissionline is real is ______ .
Solution: (40)
Zin = + +
l
l
L oo
o L
Z jZ tanZ
Z jZ tan
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8/9/2019 Gate 2014 Ec Morning
8/24
Corporate Office: 44-A/1, Kalu Sarai, New Delhi-16 | Email : [email protected] | Visit: www.madeeasy.in
GATE-2014Exam Solutions (15 Feb)Electronics Engineering(Morning Session)
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 9 0
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Here =
=
l2
and8
tan l =
=tan 14
Thus, Zin = + + +
= + +
L oo
o L
Z jZ R j30 j50Z 50
Z jZ 50 jR 30
=++
50(R j80)
(20 jR)
= +
+50(R j80)
(20 jR)
For Zinto be real
Zin = + +
=+ +2
50(R j80) (20 jR) 50(R j80) (20 jR)
(20 jR) (20 jR) (R 400)
= +
+50(R j80)
(20 jR)
For Zinto be real
Zi = + +
=
+ +2
50(R j80) (20 jR) 50(R j80) (20 jR)
(20 jR) (20 jR) (R 400)
jR2+ j1600 = 0
or R = = 1600 40
Q.9 Which of the following equation is correct?
(a) [ ]>22E[x ] E(x) (b) [ ]
22E[x ] E(x)
(c) [ ]