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GATE SOLVED PAPER - EC 2015-1
General Aptitude
t e n . g n i r e e n i g n e y s a .E w w w
Q. 1 - Q. 5 Carry one mark each. Q. 1
Choose the word most similar in meaning to the given word: Educe (A) Exert (B) Educate (C) Extract (D) Extend
Sol. 1
Correct option is (C).
Q. 2
If log x ^ 75 h =− 13 , then the value of x is
Sol. 2
Q. 3
Sol. 3
Q. 4
Sol. 4
(A) 343 125 (C) - 25 49 Correct option is (A). log x b 5 l =− 1 3 7 5 −1 b 7 l = ^x 3 h or x 1/3 = b 7 l 5 3 Hence, x = b 7 l = 343 5 125
(B) 125 343 (D) - 49 25
Operators 4, G and " are defined by: a4b = a − b ; aZb = a + b ; a " b = ab. a+b a−b Find the value (6646) " (66Z6). (A) - 2 (B) - 1 (C) 1 (D) 2 Correct option is (C).
^66 4 6h " ^66G6h = 66 − 6 # 66 + 6 = 1 66 + 6 66 − 6 Choose the most appropriate word form the options given below to complete the following sentence. The principal presented the chief guest with a _______, as token of appreciation. (A) momento (B) memento (C) momentum (D) moment Correct option is (B).
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GATE SOLVED PAPER - EC
2015-1
t e n . g n i r e e n i g n e y s a .E t w e w n . w g n i r e e n i g n e y s a .E w w w
Q. 5
Choose the appropriate word/phrase, out of the four options given below, to complete the following sentence: Frogs _______. (A) Croak (B) Roar (C) Hiss (D) Patter
Sol. 5
Correct option is (A).
Q. 6 - Q. 10 Carry two marks each. Q. 6
A cube of side 3 units is formed using a set of smaller cubes of side 1 unit. Find the proportion of the number of faces of the smaller cubes visible to those which are NOT visible. (A) 1 : 4 (B) 1 : 3 (C) 1 : 2 (D) 2 : 3
Sol. 6
Correct option is (C). Consider the cube of side 1 unit shown below.
Now, a set of similar cubes as shown above is joint to form a cube of side 3 units, we have to determine the ratio of faces of smaller cubes visible to those which are not visible . Complete surface area of the cube of side 3 units determined as A = 6 # ^3h2 = 54 Again, the area of one face of cube of side 1 unit is
A1 = ^1 h2 = 1 So, the total number of visible faces is N visible = A = 54 A1 Again, total number of smaller cubes required to form the big cube is Number of smaller cubes = volume of cube of side 3 unit volume of cube of side 1 unit
^3h3 = = 27 ^1 h3 So, the total number of faces of smaller cubes is given as N total = 6 # (number of smaller cubes) = 6 # 27 = 162
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GATE SOLVED PAPER - EC
2015-1
t e n . g n i r e e n i g n e y s a .E t w e w n . w g n i r e e n i g n e y s a .E w w w
Therefore number of invisible faces is
Q. 7
Sol. 7
Q. 8
N invisible = N total − N visible = 162 − 54 = 108 Hence, the desired ratio is N visible = 54 = 1 or 1:2 108 2 N invisible Fill in the missing value
Correct answer is 3. In the given problem, the numbers appearing in the centre line is average of the sum of numbers appearing on left and right to the numbers. This is shown in figure below.
Hence, the unknown number is given by 3+3 = 3 2 Humpty Dumpty sits on a wall every day while having lunch. The wall sometimes breaks. A person sitting on the wall falls if the wall beaks. Which one of the statements below is logically valid and can be inferred from the above sentences ? (A) Humpty Dumpty always falls while having lunch (B) Humpty Dumpty does not fall sometimes while having lunch (C) Humpty Dumpty never falls during dinner (D) When Humpty Dumpty does not sit on the wall, the wall does not break.
Sol. 8
Correct option is (B).
Q. 9
The following question presents a sentence, part of which is underlined. Beneath the sentence you find four ways of phrasing the underline part. Following the requirements of the standard written English, select the answer that produces the most effective sentence.
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GATE SOLVED PAPER - EC
2015-1
Tuberculosis, together with its effects, ranks one of the leading causes of death in India. (A) ranks as one of the leading causes of death (B) rank as one of the leading causes of death (C) has the rank of one of the leading causes of death (D) are one of the leading causes of death Sol. 9
Correct option is (A).
Q. 10
Read the following paragraph and choose the correct statement. Climate change has reduced human security and threatened human well being. An ignored reality of human progress is that human security largely depends upon environmental security. But on the contrary, human progress seems contradictory to environmental security. To keep up both at the required level is a challenge to be addressed by one and all. One of the ways to curb the climate change may be suitable scientific innovations, while the other may be the Gandhian perspective on small scale progress with focus on sustainability. (A) Human progress and security are positively associated with environmental security. (B) Human progress is contradictory to environmental security. (C) Human security is contradictory to environmental security. (D) Human progress depends upon environmental security.
Sol. 10
Correct option is (B). END OF THE QUESTION PAPER
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GATE SOLVED PAPER - EC
2015-1
Electronics & Communication Engineering Q. 1 - Q. 25 Carry one mark each. Q. 1
A region of negative differential resistance is observed in the current voltage characteristics of a silicon PN junction if (A) Both the P-region and N-region are heavily doped (B) The N-region is heavily doped compared to the P-region (C) The P-region is heavily doped compared to the N-region (D) An intrinsic silicon region is inserted between the P-region and the N-region
Sol. 1
Correct option is (A). In case of Tunnel diode formed by PN junction, tunnel diode gives negative resistance and works in forward bias, and in tunnel diode both N and P regions are heavily doped.
Q. 2
A silicon sample is uniformly doped with donor type impurities with a concentration of 1016/cm3. The electron and hole mobilities in the sample are 1200 cm2/V-s and 400 cm2/V-s respectively. Assume complete ionization of impurities. The charge of an electron is 1.6 # 10-19 C. The resistivity of the sample (in Ω -cm) is_______.
Sol. 2
Correct answer is 0.5208 . Since, we know Resistivity ^ρ h =
1 conductivity (σ) Given doped with donor type impurities i.e. n -type. So, conductivity will be given by
Sol. 3
= eµ n N D
(charge on an electron) = 1.6 # 10−19 C 16 3 (given) = 10 /cm 2 (given) = 1200 cm /V − s −19 16 Therefore, = 1.6 # 10 # 1200 # 10 = 1.92 Hence, ρ = 1 σN = 1 = .5208 Ω−cm 1.92 A unity negative feedback system has the open-loop transfer function G (s) = s (s + 1K)(s + 3) . The value of the gain K (>0) at which the root locus crosses the imaginary axis is _______. where
Q. 3
σN e ND µn σN
Correct answer is 12.
G ^s h =
K s ^s + 1h^s + 3h
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GATE SOLVED PAPER - EC
C.L.T.F (close loop transfer function) is G ^s h = 1 + G ^s h H ^s h G ^s h = 1 + G ^s h K = s ^s + 1h^s + 3h + K So, characteristic equation is
2015-1
(for negative feedback) [H ^s h = 1]
s ^s + 1h^s + 3h + K = 0 s3 + 4s2 + 3s + K = 0 Now, we obtain the Routh array as s3 1 3 2 s K 4 s1 b K - 12 l -4 0 s K
Row of s1 to be zero for oscillatory response or for poles to be on imaginary axis. So, we have K - 12 = 0 -4 or K = 12 This is the value of gain at which root locus crosses the imaginary axis. Q. 4
Suppose A and B are two independent events with probabilities P (A) ! 0 and P (B) ! 0 . Let A and B be their complements. Which one of the following statements is FALSE? (A) P (A + B) = P (A) P (B) (B) P ^ BA h = P (A) (C) P (A , B) = P (A) + P (B) (D) P (A + B ) = P (A) P (B )
Sol. 4
Correct option is (C). A and B are two independent events with probabilities, P ^A ! 0h and P ^B ! 0h Now, we check the given options. (A) TRUE P ^A + B h = P ^AhP ^B h [By the rule of two independent events] P ^A + B h P ^AhP ^B h TRUE (B) P ^A # B h = = P ^Ah = P ^B h P ^B h (C) P ^A , B h = P ^Ah + P ^B h − P ^A + B h False = P ^Ah + P ^B h − P ^AhP ^B h (D) TRUE P ^A + B h = P ^A h P ^B h
Note: Since A and B are independent, therefore A and B will also be independent.
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Q. 5
Sol. 5
2015-1
The waveform of a periodic signal x (t) is shown in the figure.
A signal g (t) is defined by g (t) = x ^ t −2 1 h. The average power of g (t) is_____.
Correct answer is 2. x b t - 1 l = x b 1 ^t − 1hl 2 2 1 So, we first scale x ^ t h by 2 , and then shift the result by 1 unit right. The resulting waveform is obtained as
Now, average power is given by 2 Pav . = 1 g ^ t h dt T T We have the magnitude of waveform g ^ t h as shown below.
#
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GATE SOLVED PAPER - EC
2015-1
Taking the sample period − 3 < t < 3 , we obtain 3 2 g ^ t h dt Pav. = 1 6 −3
#
Pav. = 1 6
#
3
−3
g ^ t h dt 2
2 1 = 1 < b − 3 t + 3 l dt + 6 −1 2 2
#
= 3 < ^1 − t h2 dt + 8 −1
#
1
# 1
3
# 1
3
2 b 23 t − 23 l dtF
^t − 1h2 dtF
^1 − t h ^t − 1h E4 = 3 *<− +; F 8 3 3 1 −1 = 3&8 + 80 8 3 3 3 1
3 3
= 2. Q. 6
In the network shown in the figure, all resistors are identical with R = 300Ω. The resistance Rab (in Ω) of the network is________.
Sol. 6
Correct answer is 100. The equivalent resistance across terminal a -b is obtained by solving the circuit as
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2015-1
n i . o c
. a
i d
o n
©
. w
w
w
. a
i d
Rab = R 3 = 300 Ω = 100 Ω 3 Consider a system of linear equations :
o n
Hence,
Q. 7
n i . o c
. w
w
x − 2y + 3z =− 1 x − 3y + 4z = 1 − 2x + 4y − 6z = k The value of k for which the system has infinitely many solutions is______. Sol. 7
©
w
Correct answer is 2. Given equations x − 2y + 3z =− 1 x − 3y + 4z = 1 − 2x + 4y − 6z = k Now, we represent the system as
where
Ax = b R− 1V R 1 − 2 3V S S W W A = S 1 − 3 4W, b = S 1 W S− 2 4 − 6W Sk W T X T X
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2015-1
For infinitely many solution, we must have R 1 − 2 3 − 1V S W A augmented = A aug = S 1 − 3 4 1 W < Rank of A < N S− 2 4 6 k W T X or ^RAug = RAh < ^n = 3, number of columnsh R V S 1 − 2 3 − 1W Now, A aug = S 1 − 3 4 1W R2 " 2R2 + R 3 , R 3 " 2R1 + R 3 S− 2 4 − 6 kW RT1 − 2 3 − 1VX S W = S0 − 2 2 2 + k W S0 0 0 k − 2W T X For k = 2 , we have (Rank of Aug of Rank A = 2 ) < ^n = 3h Hence, the system has many solutions for k = 2 .
n i . o c
. a
i d
o n
. w
Q. 8
The polar plot of the transfer function G (s) = 10s(+s +101) for # ω < 3 will be in the (A) first quadrant (B) second quadrant (C) third quadrant (D) fourth quadrant
Sol. 8
Correct option is (A). Given transfer function, 10 ^s + 1h G ^s h = s + 10
©
w
w
n i . o c
10 ^ jw + 1h 10 − jw 10 + jw # 10 − jw
G ^ jωh =
or
0#ω#3
(Put s = jω )
10 ^10jw + 10 + w2 − jwh 100 + w2 10 = ^9jw + ^10 + w2hh 100 + w2 For ω = 0 , 1st quadrant G ^ jωh = 61 + j0@ 2 For ω = 3, G ^ jωh = lim 10j c 9ω 2 m + 10 c 10 + ω 2 m ω"3 100 + ω 100 + ω 1st quadrant = 10 + 0j Hence, the polar plot of transfer function is in the first quadrant.
. a
=
o n
i d
Q. 9
. w
w
w
Let z = x + iy be a complex variable. Consider that contour integration is performed along the unit circle in anticlockwise direction. Which one of the following statements is NOT TRUE? (A) The residue of z -z 1 at z = 1 is 12 (B)
© #
2
2
C
z dz = 0
(C) 1 2πi
# z1 dz = 1 C
(D) z (complex conjugate of z ) is an analytical function Sol. 9
Correct option is (D). The complex variable is defined as z = x + iy , where contour in unit circle in a clockwise direction. Now, we check the given options
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GATE SOLVED PAPER - EC
(A)
n = 1, (B) (C)
2015-1
f ^z h =
z ^z + 1h^z − 1h n−1 Residue = 1 lim d n − 1 ^z − 1hn f ^z h n! z " 1 dz z = lim ^z − 1h =1 z"1 ^z + 1h^z − 1h 2
TRUE
# z dz 2
TRUE =0 (By cauchy’s integral theorem; z2 is analytical in the given contour.) 1 1 dz = 1 2pi f 0 ^ h 2πi z 2pi
#
TRUE =1 (D) This is only remaining option, hence FALSE. It can be proved as well, z = x − iy = u + iv For analyticity, ux = vy and uy =− vx 1 !- 1 0 =− 0 Hence z is NOT analytical.
u = x , v =− y
Q. 10
t (t)(2pfc t) where m t (t) denotes the Consider the signal s (t) = m (t) cos (2pfc t) + m Hilbert transform of m (t) and the bandwidth of m (t) is very small compared to fc . The signal s (t) is a (A) high-pass signal (B) low-pass signal (C) band-pass signal (D) double side-band suppressed carrier signal
Sol. 10
Correct option is (C). We have the signal,
Q. 11
Sol. 11
t (t) sin (2pfc t) s ^ t h = m (t) cos (2pfc t) + m Here, s ^ t h represents SSB - Lower side band, and thus a band pass signal.
For the circuit with ideal diodes shown in the figure, the shape of the output (Vout ) for the given sine wave input (Vin ) will be
Correct option is (C).
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2015-1
n i . o c
For positive half A1 , we have
. a
i d
o n
So, Vout =− Vin For negative half A2 , both diode will be OFF. So,
. w
Vout = 0 Hence, the output is obtained as
Q. 12
©
w
w
The result of the convolution x (- t) * δ (- t - t 0) is (A) x (t + t 0)
n i . o c
(B) x (t - t 0) (C) x (− t + t 0) (D) x (- t - t 0) Sol. 12
Correct option is (D). From the convolution property,
Q. 13
Sol. 13
. a
i d
x ^ t h * δ ^t - t 0h = x ^t − t 0h
o n
Now, we replace t by - t to obtain x ^- t h * δ ^- t - t 0h = x ^− t − t 0h R1V R4 1 2V W S W S The value of p such that the vector S2W is an eigenvector of the matrix S p 2 1 W S3W S14 - 4 10W is T X T X Correct answer is 17. For eigen vector X and matrix 6A@, we define
©
. w
w
w
AX = λX From give problem, we have R V S1W X = S2W S3W X Substituting it in equation, weTget R 4 1 2V R1V R V S W S W S1W = λ S2W S p 2 1W S2W S14 - 4 10W S3W S3W 3#3 3#1 T X T X T X R V R 12 V Sl W W S or = S2lW Sp + 7W S W S 36 W S3lW 3#1 X T T X
R 4 1 2V S W A = S p 2 1W S14 − 4 10W T X
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GATE SOLVED PAPER - EC
2015-1
If λ = 12 , 3λ = 36 , and hence we obtain p + 7 = 2λ or p = 24 − 7 or p = 17 Q. 14
In the given circuit, the values of V1 and V2 respectively are
(A) 5 V, 25 V (C) 15 V, 35 V Sol. 14
(B) 10 V, 30 V (D) 0 V, 20 V
Correct option is (A). Given circuit is
By Nodal analysis V2 - V1 = V1 + V1 + 2I 4 4 4 V2 - V1 = 5 4 V2 - V1 = 20 V2 = 20 + V1 Since, Hence,
V2 = 20 + 4I V1 = 4I = 5 V V2 = 20 + 5 = 25 V
(V1 /4 = I , or V1 = 4I )
b 5 = I + I + 2I, or I = 54 l
Q. 15
In an 8085 microprocessor, the shift registers which store the result of an addition and the overflow bit are, respectively (A) B and F (B) A and F (C) H and F (D) A and C
Sol. 15
Correct option is (B). The shift registers A and F store the result of an addition and the overflow bit.
Q. 16
Negative feedback in a closed-loop control system DOES NOT (A) reduce the overall gain (B) reduce bandwidth (C) improve disturbance rejection (D) reduce sensitivity to parameter variation
Sol. 16
Correct option is (B). Negative feedback in closed-loop control system does not reduce bandwidth.
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GATE SOLVED PAPER - EC
2015-1
Q. 17
A 16 Kb ( = 16, 384 bit) memory array is designed as a square with an aspect ratio of one (number of rows is equal to the number of columns). The minimum number of address lines needed for the row decoder is______.
Sol. 17
Correct answer is 7. Memory chip = (Number of rows) # (Number of columns) = M#N The number of address line required per row decoder is n , where M = 2n or n = log 2 M Given M = N , or or Hence,
Q. 18
Sol. 18
Q. 19
Sol. 19
M # N = M # M = M 2 = 16 K M2 = 2 4 # 210 M = 128 N =7
A function f (x) = 1 − x2 + x3 is defined in the closed interval [- 1, 1]. The value of x , in the open interval (- 1, 1) for which the mean value theorem is satisfied, is (A) - 1 (B) - 1 2 3 (C) 1 (D) 1 3 2 Correct option is (B). Lagrange’s mean value theorem states that if a function f ^x h is continuous in close interval 6a, b@ and differentiable in open interval ^a, b h, then for point c in the interval, we may define f ^b h − f ^a h f l^c h = b−a Now, we have f ^x h = 1 − x2 + x3 Since, polynomial function is always continuous and differentiable, so f ^1 h − f ^− 1h f l^x h = =1 1 − ^− 1h or − 2x + 3x2 = 1 or 3x2 - 2x - 1 = 0 or x = 1, - 1 3 Hence, x =− 1 , as - 1 d ^- 1, 1h 3 3
The electric field component of a plane wave traveling in a lossless dielectric medium is given by Ey (z, t) = aty 2 cos _108 t − z2 i V/m. The wavelength (in m) for the wave is______. Correct answer is 8.8858.
E ^z, t h = aty 2 cos b108 t − z l V/m 2 Comparing with the general form, E ^z, t h = aty A cos ^ωt − βz h V/m we get β = 1 2
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Since, So,
2015-1
β = 2π = 1 λ 2 λ = 2#π# 2 = 8.885765
Q. 20
In the circuit shown, the switch SW is thrown from position A to position B at time t = 0 . The energy (in µJ ) taken from the 3 V source to charge the 0.1 µ F capacitor from 0 V to 3 V is
Sol. 20
Correct option is (C). For the given circuit, we have
E = 1 CV 2 2 = 1 # .1 # 10−6 # 3 # 3 = 0.45 µJ 2 In the circuit shown below, the Zener diode is ideal and the Zener voltage is 6 V. The output voltage V0 (in volts) is______. So,
Q. 21
Sol. 21
Correct answer is 5. We have the circuit,
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GATE SOLVED PAPER - EC
2015-1
Firstly, open zener diode and calculate VA .
VA = 5 V So, VA < VZ Hence, diode remain open and VA = 5 V = V0 Q. 22
Consider a four bit D to A converter. The analog value corresponding to digital signals of values 0000 and 0001 are 0 V and 0.0625 V respectively. The analog value (in Volts) corresponding to the digital signal 1111 is______.
Sol. 22
Correct option is 0.9375. n = number of bit = 4 Analog output = (Decimal equivalent of input) # (Resolution) 0000 " 0 V 0001 " 0.0625 V = Resolution as decimal equivalent = 1 1111 " ? Analog output = ^Decimal equivalenth # ^Resolutionh = ^15h^.0625h = .9375
Q. 23
In the circuit shown, at resonance, the amplitude of the sinusoidal voltage (in Volts) across the capacitor is_______.
Sol. 23
Correct answer is 25. For the quality factor Q , we have the voltage across capacitor and inductor as and VC = QV VL = QV , ω 0 = 1 LC At resonance, Q = ω0L = 1 L R R C = 10 .1 mH = 10 = 2.5 4 1.0 µF 4 Hence, we obtain VC = QV = 2.5 # 10 = 25 volts
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Q. 24
Sol. 24
2015-1
A sinusoidal signal of 2 kHz frequency is applied to a delta modulator. The sampling rate and step-size T of the delta modulator are 20,000 samples per second and 0.1 V, respectively. To prevent slope overload, the maximum amplitude of the sinusoidal signal (in Volts) is (A) 1 (B) 1 2π π (C) 2 (D) π π Correct option is (A). For preventing slope overload, we should have (slope of m ^ t h) # (slope of sampling) For sinusoidal signal, if Vinput = Am sin ^ω m t h m input = b dV l = Am ω m dt max step size = TFs Am ω m # T = tr sampling interval input
Q. 25
Sol. 25
Am 2π ^2 # 103h # 2 # 10 4 # .1 Am # 1 2π Consider a straight, infinitely long, current carrying conductor lying on the z-axis. Which one of the following plots (in linear scale) qualitatively represents the dependence of H φ on r , where H φ is the magnitude of the azimuthal component of magnetic field outside the conductor and r is the radial distance from the conductor ?
Correct option is (C). We know that for a straight, infinitely, long current carrying conductor, Hφ = 1 r " distance from the current element 2πr So, Hφ \ 1 r
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2015-1
Q. 26 - Q. 55 Carry two marks each. Q. 26
The input X to the Binary Symmetric Channel (BSC) shown in the figure is ‘1’ with probability 0.8. The cross-over probability is 71 . If the received bit Y = 0 , the conditional probability that ‘1’ was transmitted is______.
Sol. 26
Correct answer is 0.4. We have the Binary symmetric channel as shown below.
P 6X = 0@ = 0.2 P 6X = 1@ = 0.8 According to Baye’s theorem, Here,
P ^ YX == 01 h P ^X = 1h X = 1 Pb = Y = 0 l P ^ YX == 01 h P ^X = 1h + P ^ YX == 01 h P ^X = 1h 1 ^ 7 h^.8h = 0.4 = 1 6 ^ 7 h^.8h + ^ 7 h^.2h
Q. 27
The transmitted signal in a GSM system is of 200 kHz bandwidth and 8 users share a common bandwidth using TDMA. If at a given time 12 users are talking in a cell, the total bandwidth of the signal received by the base station of the cell will be at least (in kHz) ______.
Sol. 27
Correct answer is 400. Given width of a channel in GSM system Wchannel = 200 kHz. Since, 8 users share the channel so, width assigned per unit is Wuser = 200 = 25 kHz. 8 Now, 12 users are talking in a cell. So, the minimum number of channels in the cell is obtained as
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GATE SOLVED PAPER - EC
2015-1
N channel > 12 # Wuser Wchannel N channel > 12 # 25 200 or N channel = 2 Hence, the bandwidth of the cell is Wcell = N channel # Wchannel = 2 # 200 = 400 kHz. Q. 28
For the discrete-time system shown in the figure, the poles of the system transfer function are located at
(A) 2, 3
Sol. 28
(B) 1 , 3 2 (D) 2, 1 3
(C) 1 , 1 2 3 Correct option is (C). We have the discrete time system as shown in figure below.
The circuit is minimized as
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GATE SOLVED PAPER - EC
So, we obtain
2015-1
H ^z h =
6 6 = −1 −1 5 −1 −1 6 b1 − z l + ^z h ^6 − 5z h + z 6 = 2 6 6z − 5z + 1
Hence, poles are at z = 1, 1 2 3 Q. 29
Sol. 29
The circuit shown in the figure has an ideal opamp. The oscillation frequency and the condition to sustain the oscillations, respectively, are
(B) 1 and R1 = 4R2 CR (D) 1 and R1 = 4R2 2CR
(A) 1 and R1 = R2 CR (C) 1 and R1 = R2 2CR Correct option is (D). Given op-amp circuit is
By virtual ground property, we write
or
or
_
V+
1 jω2C
V- = Vout b
i
+
V − Vout V+ + + =0 R 2R + 1 jω C
R2 = V+ R1 + R 2 l
jw C V jwC = out V+ c jw2C + 1 + R 2jwCR + 1 m 2jwCR + 1 jw C 2jwCR + 1 Vout = jw2C + 1 + c m V+ R 1 + 2jwCR mc jw C
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2015-1
jw2CR ^1 + 2jgwCRh + ^1 + 2jwCRh + jwCR ^2jwCR + 1h H jw C R ^1 + 2jwCR h ^− j h = 2 ^1 + 2jwCRh + +3 wCR We equate imaginary part to zero, i.e. 4wCR - 1 = 0 wCR =>
^2ωCRh2 = 1
1 (Oscillation frequency) 2CR The condition to sustain the oscillation is ω =
where Here, and
Aβ = 1 A " open loop gain β " feedback gain A = Vout = 1 + R1 Vin R2 V β = + Vout 1 = Vout V+ β
(gain of non-inverting opamp)
At oscillation, 1 , 2CR Vout = 1 = 5 = A V+ β A = 1 + R1 = 5 R2 R1 = 4 R2 ω =
So, or Hence, Q. 30
R1 = 4R2
A source emits bit 0 with probability 1/3 and bit 1 with probability 2/3. The emitted bits are communicated to the receiver. The receiver decides for either 0 or 1 based on the received value R. It is given that the conditional density functions of R are as 1, −3 # x # 1 1, − 1 # x # 5 and f R/1 (r) = * 6 fR/0 (r) = * 4 0, otherwise 0, otherwise The minimum decision error probability is (A) 0
Sol. 30
(B) 1 12 (D) 1 6
(C) 1 9 Correct option is (D). Given the conditional density function of R as 1 −3 < r < 1 f R/0 ^r h = * 4 0 otherwise
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1, − 1 < r < 5 f R/1 ^r h = * 6 0, otherwise
Q. 31
Sol. 31
Decision error probability that receiver decides 0 for a transmitted bit 1 is f R/1 ^r = 0h = 1 6 Again, the decision error probability that receiver decides 1 for a transmitted bit 0 is f R/0 ^r = 1h = 1 4 Hence, the minimum decision error probability is f R/1 ^r = 0h = 1 6 For a silicon diode with long P and N regions, the accepter and donor impurity concentrations are 1 # 1017 cm-3 and 1 # 1015 cm-3 , respectively. The lifetimes of electrons in P region and holes in N region are both 100 µs. The electron and hole diffusion coefficients are 49 cm2/s and 36 cm2/s, respectively. Assume kT/q = 26 mV, the intrinsic carrier concentration is 1 # 1010 cm-3 and q = 1.6 # 10−19 C. When a forward voltage of 208 mV is applied across the diode, the hole current density (in nA/cm2) injected from P region to N regions is ______. Correct answer is 28.6 Given Acceptor concentration, NA = 1 # 1017 cm−3 Donor concentration, ND = 1 # 1015 cm−3 Lifetime, τ n = τ p = 100µs = 10−4 sec. Electron diffusion coefficient, Dn = 49 cm2 /s Hole diffusion coefficient, D p = 36 cm2 /s Intrinsic carrier concentration, ni = 1 # 1010 cm−3 Applied forward voltage, Va = 208 mV The hole current density injected from P region to N region is given by eD p pno Jp = exp d eva n − 1G Lp = kT Now, we obtain the hole concentration in n -region as 2 pno = ni ND
(1)
20
and
= 1 # 1015 = 105 cm−3 1 # 10 Lp = Dp τ p
= 36 # 10−4 = 6 # 10−2 cm Substituting the values in equation (1), we get ^1.6 # 10−19h^36h^105h Jp = exp b 208 l − 1E ; −2 26 ^6 # 10 h = 2.86 # 10−8 A/cm2 = 28.6 nA/cm2
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Q. 32
2015-1
For the NMOSFET in the circuit shown, the threshold voltage is Vth , where Vth > 0 . The source voltage Vss is varied from 0 to VDD . Neglecting the channel length modulation, the drain current ID as a function Vss is represented by
Sol. 32
Correct option is (A). Given NMOSFET circuit is
We have the following conditions VGS - VTh > 0 VGS - VTh # 0 VDD - VSS - VTh # 0 Since it is in saturation, current ID is given by ^VGS − VTh h2 W ID = µ n cos b l 2 L
(FET to be ON) (FET is OFF) (FET is OFF)
^VDD − VSS − VTh h ID = µ n cos bW l 2 L Thus, ID -VSS graph shows Parabolic relation for VSS < VDD − VTh and zero for VSS > VDD − VTh . Only graph shown in option (A) satisfies this result. 2
or
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Q. 33
In the system shown in figure (a), m (t) is a low-pass signal with bandwidth W Hz. The frequency response of the band-pass filter H (f) is shown in figure (b). If it is desired that the output signal z (t) = 10x (t), the maximum value of W (in Hz) should be strictly less than ______.
Sol. 33
Correct answer is 350. We have the input signal,
and So,
x ^ t h = m ^ t h cos ^2400πt h = m ^ t h cos ^ωt h ω = 2400 πrad . 1200 Hz y ^ t h = 10x ^ t h + x2 ^ t h
= 10m ^ t h cos ^2400pt h + m2 ^ t h cos2 ^2400pt h cos ^2wt h + 1 = 10m ^ t h cos ^wt h + m2 ^ t h; E 2
m2 ^ t h m2 ^ t h cos ^2ωt h = + 10m ^ t h cos ^ωt h + 2 2 1 444 2 444 3 1 444 2 444 3 S 6ω − W, ω + W @ +ve frequency 62ω − 2W, 2ω + 2W @ Ranges
6− θ, 2W @
From the frequency plot, we conclude the following results. Result 1 ω - W > 700 1200 - W > 700 W < 500 Result 2 ω + W < 1700 1200 + W < 1700 Result 3
W < 500 ω - W > 2W 1200 > 3W
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W < 400 Result 4 2ω - 2W > 1700 2400 - 1700 > 2W 2W < 700 W < 350 Thus, the above conclusions result in W < 350 Q. 34
In the circuit shown, I1 = 80 mA and I2 = 4 mA. Transistors T1 and T2 are identical. Assume that the thermal voltage VT is 26 mV at 27°C. At 50°C, the value of the voltage V12 = V1 − V2 (in mV) is ______.
Sol. 34
Correct answer is 83.8. Given circuit is
Current in the circuit is given by I2 = 4 mA = Is eV /ηV I2 = 4 mA = Is eV /ηV I1 = 80 mA = Is eV /ηV = Is eV /ηV = Is eV /ηV I1 = eVη−VV I2 BE
or Again,
T
2
BE
T
BE
T
T
1
So, Since, So,
T
1
2
T
VT = 26 mV at 27cC or 300 K VT at ( 50cC ) = 26 mV b 50 + 273 l − 27.993 mV 300
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Hence, we obtain
or
V1 - V2 = ηVT ln 20 = 1 # 27.993 # ln 20 V1 - V2 = 83.859
Q. 35
The electric field intensity of a plane wave traveling in free space is give by the following expression E (x, t) = ay 24π cos (ωt − k 0 x) (V/m) In this field, consider a square area 10 # 10 cm on a plane x + y = 1. The total time-averaged power (in mW) passing through the square area is ______.
Sol. 35
Correct answer is 53 to 54.
Q. 36
The maximum area (in square units) of a rectangle whose vertices lie on the ellipse x2 + 4y2 = 1 is _______.
Sol. 36
Correct answer is 1. Let x and y be the length and breadth of the rectangle as shown in figure below.
For the given ellipse, we have or Area of rectangle is
x 2 + 4y 2 = 1 x 2 = 1 − 4y 2
(1)
A = 4xy Now, we have to determine the point of maxima, so we consider z = A2 = 16x2 y2 When z will be maximum, A will also be maximum. So we determine point of maxima for z as
For maxima,
So, Again,
z = 16x2 y2 = 16y2 ^1 − 4y2h dz = 0 dy
2y - ^4h^4h y3 = 0 2y ^1 - 8y2h = 0
y =! 1 8 2 d z = 2 − 16 3 y2 ^ h^ h y = dy2 2 - 16 # 3 # 1 < 0 8
^y ! 0h 1 8
-4 < 0
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2 y = 1 as d z2 dx 8
Therefore, z is maximum at
y= 1 8
<0
1/2 A max = 4xy = 1 b1 − 4 l = 4 = 1 8 4 8
Hence,
Note: Point of maxima can also be evaluated by directly differentiating A. But is may lead to tedious calculation. Q. 37
In the given circuit, the maximum power (in Watts) that can be transferred to the load RL is _______.
Sol. 37
Correct answer is 1.649 We have the circuit,
Thevenin equivalent of the circuit is obtained as 2 2j 2 − 2j Zth = 2 || 2j = # # 2 − 2j 2 + 2j 8j + 8 = 1+j = 8 So,
VTh^rmsh = 4 # =
^2 − 2j h 2j 8j = # 2 + 2j ^2 + 2j h ^2 − 2j h
8j 2 − 2j h = 2 + 2j 8^
= 2 2 45c
For maximum power, RL = ZTh = So,
Pmax = I 2 RL =
2Ω 2 2 45c # j + 2
2
= 1.649 W
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Q. 38
A lead compensator network includes a parallel combination of R and C in the feed-forward path. If the transfer function of the compensator is Gc (s) = ss ++ 42 , the value of RC is _______.
Sol. 38
Correct answer is 0.5 Given transfer function is 1 s+1 Gc ^s h = s + 2 = 1 = 12 s + 4 2 4 s + 1G For load compensator, RC circuit is shown below.
(1)
From the circuit, we have the transfer function G ^ s h = τs + 1 βτs + 1 where So, Hence,
τ = RC 1 = RC 2
[Comparing with equation (1)]
RC = 0.5 .
Q. 39
A MOSFET in saturation has a drain current of 1 mA for VDS = 0.5 V. If the channel length modulation coefficient is 0.05 V-1 , the output resistance (in kΩ ) of the MOSFET is _______.
Sol. 39
Correct answer is 20. Given ID = 1 mA VDS = 0.5 V (channel length modulation factor) λ = 0.05 V−1 Since, ID = IDsat ^1 + λVDS h dID = I So, Dsat ^λ h dVDS 1 Hence, Rout = 1 = λIDsat 0.05 # 10−3 = 20 kΩ
Q. 40
The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (CJ ) at a reverse bias (VR ) of 1.25 V is 5pF, the value of CJ (in pF) when VR = 7.25 V is_______.
Sol. 40
Correct answer is 2.5 Given Vbi = 0.75 = built in potential Since CJ = ε r W er or CJ = 2e r ^Vbi + VRh Na + Nd 1/2 ) b Na Nd l3 e
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CJ \
or So, we may write
5 pF = CJ2 Hence,
1 Vbi + VR .75 V + 7.25 = .75 V + 1.25 V
8 2
CJ2 = 2.5 pF .
Q. 41
In the circuit shown, assume that the opamp is ideal. The bridge output voltage V0 (in mV) for δ = 0.05 is _______.
Sol. 41
Correct answer is 250. Given op-amp circuit is redrawn as
For ideal op-amp, we have V1 = 1 V Since, no current flows towards negative terminal of op-amp. So, we have I1 + I 2 = 0 Therefore, we obtain I1 =− I2 =− V1 50 =− 1 50 This current is equally divided into two branches of the bridge. So, we have V3 = V2 − 250 ^1 + δ h I1 2 and V4 = V2 − 250 ^1 − δ h I1 2 Hence, the bridge output voltage is V0 = V3 − V4
= :V2 − 250 ^1 + d h I1 D − :V2 − 250 ^1 − d h I1 D 2 2
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=− 500δ I1 2 =− 250 ^0.05hb − 1 l 50 = 0.25 V = 250 mV Q. 42
Sol. 42
Q. 43
Sol. 43
The damping ratio of a series RLC circuit can be expressed as 2 (B) 22L (A) R C 2L RC (C) R C (D) 2 L 2 L R C Correct option is (C). Damping ratio is given by ε = 1 (for series RLC circuit) 2Q where Q = ω0L = 1 b L l = 1 L R R C LC R So, ε=R C 2 L In the circuit shown, switch SW is closed at t = 0 . Assuming zero initial conditions, the value of vc (t) (in Volts) at t = 1 sec is_______.
Correct answer is 2.5284 For the given circuit, we have
Initially, capacitor oppose any voltage change across it and thus will act as short circuit, so Vc ^ t h t = 0 = 0 +
At t = 3,
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After getting fully changed, capacitor will act as open circuit So, Vc ^ t h t = 3 = 2 # 10 = 4 V 2+3 Vc ^ t h = Vc ^3h − 6Vc ^3h − Vc ^0 h@e−t/τ
Therefore,
= 4 ^1 − e− τ h t
where Hence, At t = 1, Q. 44
τ = Req C = R thevein C = ^3112h # 65 = 65 # 56 = 1 Vc ^ t h = 4 ^1 − e−t h Vc ^ t h t = 1 = 2.5284 V
Consider a uniform plane wave with amplitude (E 0 ) of 10 V/m and 1.1 GHz frequency travelling in air, and incident normally on a dielectric medium with complex relative permittivity ( ε r ) and permeability ( µ r ) as shown in the figure.
The magnitude of the transmitted electric field component (in V/m) after it has travelled a distance of 10 cm inside the dielectric region is _______. Sol. 44
Correct answer is 1. Electromagnetic wave travels from air medium to dielectric medium. In both the medium, σ =0 So, electric field of EM wave is defined as E i = E i0 e − β z E t = E t 0 e− β z 1
and Also, we have
2
E t 0 = 2h 2 h 2 + h1 Ei 0 = For air medium, we have
2 µ 2 /ε 2 µ 2 / ε 2 + µ 1 /ε 1
µ1 = µ 0 , e1 = e 0 and for dielectric medium µ 2 = µr µ0 , e 2 = e r e 0 So, we get Et 0 = 2 µ r /ε r Ei 0 µ r /ε r + 1 = 2 1+1
^µ r = 1 − j 2, ε r = 1 − j 2h
=1 Since, we have Ei0 = 10 V/m
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and So, we obtain
2015-1
f = 1.1 GHz = 1.1 # 109 Hz.
Et 0 = E i 0 = 10 V/m In dielectric medium, phase constant is β2 = ω r ε = 2π ^1.1 # 109h
µr εr 3 # 108
= 22π ^1 − j 2h2 3 = 22π ^1 − j 2h 3 Therefore, the electric field at any distance z from the interface is obtained as Et ^z h = Et 0 e−jβ z 2
= 10e−
j 22π ^1 − j 2hz 3
= 810e− 3 zB e− 3 z Hence, the magnitude of electric field at z = 10 cm is 44p
Et ^z = 0.1 mh = 10e−
j 22p
44π 0.1 3 # 4.4π
= 10e− 3 = 1 V/m
Q. 45
A vector Pv is given by Pv = x3 y2 avz . Which one of the following statements is TRUE? (A) Pv is solenoidal, but not irrotational (B) Pv is irrotational, but not solenoidal (C) Pv is neither solenoidal nor irrotational (D) Pv is both solenoidal and irrotational
Sol. 45
Correct option is (A). Given vector, Pv = x3 yatx − x2 y2 aty − x2 yzatz If divergence d : Pv = 0 , then vector Pv is solenoidal. So, we obtain 2P d : Pv = 2Px + y + 2Pz 2x 2y 2z = 3x2 y − 2x2 y − x2 y =0 Hence, it is solenoidal. Again, if curl d # Pv = 0 , then Pv is irrotational. So, we obtain atx aty atz 2 2 2 v d # P = 2x 2y 2z x3 y − x2 y2 − x2 yz
2^− x2 yz h 2^− x2 y2h 2^− x2 yz h 2^x3 y h == − − G atx − aty ; E 2y 2z 2x 2z 2^− x2 y2h 2^x3 y h + az = − G 2x 2y
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= atx ^− x2 z − 0h − aty 6− 2xyz − 0@ + atz 6− 2xy2 − x3@ = ^− x2 z h atx + ^2xyz h aty − ^2xy2h atz ! 0 Hence, it is not irrotational. Q. 46
All the logic gates shown in the figure have a propagation delay of 20 ns. Let A = C = 0 and B = 1 until time t = 0 . At t = 0 , all the inputs flip (i.e., A = C = 1 and B = 0 ) and remain in that state. For t > 0 , output Z = 1 for a duration (in ns) of
Sol. 46
Correct answer is 40. Given logic circuit is
At t = 0− ,
A = C = 0, B = 1 Z = AB 5 C Z = 050 = 0 At t = 3, A = C = 1, B = 0 Z = AB 5 C Z = 151 = 0 So, we obtain the output pulses as
Hence, Z = 1 for the duration of 40 ns.
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Q. 47
A plant transfer function is given as G (s) = ^KP + Ks h s (s 1+ 2) . When the plant operates in a unity feedback configuration, the condition for the stability of the closed loop system is (A) KP > KI > 0 (B) 2KI > KP > 0 2 I
(C) 2KI < KP Sol. 47
2015-1
(D) 2KI > KP
Correct option is (A). Given plant transfer function,
G ^s h = :KP + KI D; 1 E s s ^s + 1h P + KI = sK s2 ^s + 1h So, closed loop transfer function is sKP + KI T.F. = G = 2 1+G s ^s + 2h + sKP + KI Therefore, characteristics equation is s2 ^s + 2h + sKP + KI = s3 + 2s2 + sKP + KI For stability, we form the Routh array. 1 KP s3 2 2 KI s 1 K I - 2K P s -2 KI s0 For stability, we must have KI > 0 and KI - 2KP > 0 -2
(Routh stability criteria)
KI > 0 and KI - 2KP < 0 KI > 0 and KP > KI 2 Hence, KP > KI > 0 2 Q. 48
A 3-input majority gate is defined by the logic function M (a, b, c) = ab + bc + ca. Which one of the following gates is represented by the function M (M (a, b, c), M (a, b, c ), c) ? (A) 3-input NAND gate (B) 3-input XOR gate (C) 3-input NOR gate (D) 3-input XNOR gate
Sol. 48
Correct option is (B). 3 input majority gate is given as
M ^a, b, c h = ab + bc + ca
We have to obtain M ^M ^a, b, c h, M ^a, b, c h, c h Let
P = M ^a, b, c h = ab + bc + ca = ab : bc : ca = ^a + b h : ^b + c h : ^c + a h
(by DeMorgan’s Law)
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= ^a b + a c # b + b c h^c + a h = ^b # a c h^c + a h = b c+b a+a c+a c = b c+b a+a c Q = M ^a, b, c h = ab + bc + ca and R =C M ^P, Q, Rh = PQ + QR + RP So, = M ^b c + b a + a c, ab + bc + ca, c h = ^b c + b a + a c h^ab + bc + ca h + ^ab + bc + ca h c + c ^b c + b a + a c h = b ca + abc + abc + b ac We obtain truth table for the function as a
b
c
M
0
0
0
0
0
0
1
1
0
1
0
1
0
1
1
0
1
0
0
1
1
0
1
0
1
1
0
0
1
1
1
1
So, the function is odd number of 1’s detector. This function represent the 3-input XOR gate. Q. 49
The longitudinal component of the magnetic field inside an air-filled rectangular waveguide made of a perfect electric conductor is given by the following expression Hz (x, y, z, t) = 0.1 cos (25πx) cos (30.3πy) cos (12π # 109 t − βz)(A/m) The cross-sectional dimensions of the waveguide are given as a = 0.08 m and b = 0.033 m. The mode of propagation inside the waveguide is (A) TM12 (B) TM21 (C) TE21 (D) TE12
Sol. 49
Correct option is (C). We have the expression,
Hz ^x, y, z, t h = 0.1 cos ^25πx h cos ^30.3πy h cos ^12π # 109 t − βz h A/m and the cross sectional dimension is a = 0.08 m , b = 0.033 m Now, we compare the equation for TEm, n mode whose Hz is given as nπ y Hz ^x, y, z, t h = H 0 cos a mπx k cos a cos ωt − βz h A/m a b k ^ mπx = 25πx So, a
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m =2 nπ y = 30.3yπ b
or and
or n =1 Hence, given mode is TE21 mode. Q. 50
n i . o c
The open-loop transfer function of a plant in a unity feedback configuration is given as G (s) = (s +K8()(s +s 4−) 9) . The value of the gain K (> 0) for which − 1 + j2 lies on the root locus is _______. 2
Sol. 50
. a
Correct answer is 25.5 Given open loop transfer function, K ^s + 4h G ^s h = " ^s + 8h^s2 − 9h and K >0
i d
o n
. w
If the point ^− 1 + j2h lies on the locus, then by the magnitude criteria G ^s h H ^s h s =− 1 + j2 = 1
w
w
K ^− 1 + j2 + 4h − 1 + j 2 + 8 ^ h^− 1 + j2 + 3h^− 1 + j2 − 3h K ^3 + 2j h or 7 + j 2 ^ h^2 + 2j h^− 4 + j2h K 13 or 53 8 20 Hence, K or
Q. 51
©
=1
n i . o c =1
= 25.5403
. a
Two sequences [a , b, c ] and [A, B , C ] are related as. RAV R1 1 1 VW RSa VW S W S 2π SBW = S1 W 3−1 W 3−2WSb W where W3 = e j 3 SC W S1 W −2 W −4W Sc W 3 3 T X T XT X If another sequence [p, q , r ] is derived as, RpV R1 1 1 V R1 0 0 V RA/3V W S W S WS WS Sq W = S1 W 31 W 32WS0 W 32 0 WSB/3W Sr W S1 W 2 W 4W S0 0 W 4W SC/3W 3 3 3 X T X T XT XT then the relationship between the sequences [p, q , r ] and [a , b, c ] is (A) [p, q , r ] = [b, a , c ] (B) [p, q , r ] = [b, c , a ] (C) [p, q , r ] = [c , a , b] (D) [p, q , r ] = [c , b, a ]
i d
o n
©
Sol. 51
=1
. w
w
w
Correct option is (C). Given relation is RAV R1 1 1VW RSa VW S W S SBW = S1 W 3−1 W 3−2W Sb W SC W S1 W −2 W −4W Sc W 3 3 TSX T1 444 2 444 3X TSX P D Q
Comparing it with the DFT concept of taking fourier transform by matrix form. We may calculate that here we are taking the 3 order DFT of 6a b c@T whose
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2015-1
transformed output is 6A B C @T . So, P = DQ or Q = D−1 P = 1 D * P & IDFT 3 R V Ra V 1 1W RSAWV S W 1 S1 or Q = Sb W = S1 ^W 3−1h* ^W 3−2h*W SBW Sc W 3 SS1 ^W −2h* ^W −4h*WW SC W 3 3 Again, we have T X T XT X RpV R VR VR V S W 1 S1 1 1 1 2WS1 0 2 0 WSAW Sq W = S1 W 3 W 3 WS0 W 3 0 WSBW Sr W 3 S1 W 2 W 4WS0 0 W 4WSC W 3 3 3 TR XV X T XT X RT V 2 4 R V SpW 1 S1 W 33 W 36WSAW Sq W = S1 W 3 W 3 WSBW Sr W 3 SS1 W 4 W 8WWSC W 3 3 T X T XT X −1 + j 3 j 2π 3 So, W3 = e = cos ^120ch + j sin ^120ch = 2
n i . o c
(DFT)
(1)
. a
i d
o n
. w
(2)
w
w
Again, we consider the equation (1), Ra V R 1 1 VWRSAVW S W 1 S1 Sb W = S1 ^W 3−1h* ^W 3−2h*WSBW Sc W 3 S1 W −2 * W −4 WSC W ^ 3 h ^ 3 h RT VR V XT X RTa VX S W 1 S1 1−2 1−4WSAW or Sb W = S1 w w WSBW Sc W 3 SS1 w−4 w −6WWSC W 3 TR XV RT VXRT VX a 1 1 1 S WSAW S W 1 −1 or Sb W = S1 w w WSBW Sc W 3 SS1 w−1 w WWSC W T X T cube roots XT Xof unity is given as Since, the relation between 1 + w + w2 = 0 ; w3 = 1 2 1 1 w = w ; w = w w = w2 ; w 2 = w So, we solve the matrix equation as R VR V Ra V S W 1 S1 1 1−1WSAW Sb W = S1 w w WSBW Sc W 3 SS1 w−1 w WWSC W T X T X T Again, we consider the equation (2),X R V RpV −1 w WRSAVW S W 1 S1 w Sq W = S1 1 1 WSBW Sr W 3 SS1 w w−1WWSC W X apply T elementary XT X row operation as In above equation,T we
©
©
Again,
. w
. a
i d
o n
n i . o c
2
w
w
R1 * R 2 R VR V RpV S W 1 S1 1 −1 1 WSAW Sq W = S1 w w WSBW Sr W 3 SS1 w w−1WWSC W T X T XT X R2 * R3
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R VR V RpV 1 1 1 S WSAW S W 1 −1 Sq W = S1 w w WSBW Sr W 3 SS1 w−1 w WWSC W T X T XT X Hence, we can conclude that
n i . o c
8p q r B = 8c a bB 8q r pB = 8a b cB
or Q. 52
Sol. 52
. a
dy d2 y The solution of the differential equation 2 + 2 + y = 0 with y (0) = yl (0) = 1 dt dt is (A) (2 - t) et (B) (1 + 2t) e−t (C) (2 + t) e−t (D) (1 - 2t) et
i d
o n
. w
Correct option is (B). We have the differential equation, d 2y dy +2 +y = 0 dt dt2
©
w
w
and y ^0 h = yl^0 h = 1 Given equation is linear constant coefficient differential equation. Let d =D dt So, D2 + 2D + 1 or ^D + 1h2 or D Therefore, y^t h For t = 0 , y ^0 h Again, yl^ t h For t = 0 , yl^0 h
= ^C1 + C2 t h e−t = 1 = C1
n i . o c
i d
. a
= C2 e−t + ^C1 + C2 t h^− e−t h = 1 = C2 − C2 C 2 = 1 + C1 = 2 y ^ t h = ^1 + 2t h e−t
or Hence, Q. 53
=0 =0 =− 1, - 1
. w
o n
w
The pole-zero diagram of a causal and stable discrete-time system is shown in the figure. The zero at the origin has multiplicity 4. The impulse response of the system is h [n]. If h [0] = 1, we can conclude
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t e n . g n i r e e n i g n e y s a .E t w e w n . w g n i r e e n i g n e y s a .E w w w
(A) (B) (C) (D) Sol. 53
2015-1
h [ n] h [ n] h [ n] h [ n]
is is is is
real for all n purely imaginary for all n real for only even n purely imaginary for only odd n
Correct option is (A). Given pole-zero diagram of causal and discrete time system,
From the pole-zero plot, we obtain the transfer function as Kz 4 H ^z h = ^z − 0.5 + j 0.5h^z − 0.5 − j 0.5h^z + 0.5 + j 0.5h^z + 0.5 − j 0.5h Kz 4 = 8^z − 0.5h2 − j 2 ^0.5h2B8^z + 0.5h2 − j 2 ^0.5h2B Kz 4 = 2 ^z − z + 0.5h^z2 + z + 0.5h 4 = 2 Kz 2 ^z + 0.5h − z2 4 = 4 Kz 2 z + ^0.5h = K −4 1+z 4 Q. 54
It’s inverse z -transform will be a real function. Thus, h 6n@ is real for all n .
Which one of the following graphs describes the function f (x) = e−x (x2 + x + 1) ?
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Sol. 54
2015-1
t e n . g n i r e e n i g n e y s a .E t w e w n . w g n i r e e n i g n e y s a .E w w w
Correct option is (B).
f ^x h = e−x ^x2 + x + 1h f l^x h =− e−x ^x2 + x + 1h + e−x ^2x + 1h = e−x 6− x2 − x − 1 + 2x + 1@ = e−x 6− x2 + x @ = ^e−x h^x h^1 − x h f l^x h = 0 , x = 0 , 1 This condition is satisfied by the graph shown in option (B).
Q. 55
The Boolean expression F (X, Y, Z) = XYZ + XY Z + XYZ + XYZ converted into the canonical product of sum (POS) form is (A) (X + Y + Z)(X + Y + Z ) (X + Y + Z ) (X + Y + Z ) (B) (X + Y + Z)(X + Y + Z ) (X + Y + Z)(X + Y + Z ) (C) (X + Y + Z)(X + Y + Z ) (X + Y + Z)(X + Y + Z ) (D) (X + Y + Z ) (X + Y + Z)(X + Y + Z)(X + Y + Z)
Sol. 55
Correct option is (A). We have the SOP Boolean form,
F ^x, y, z h = xyz + xy z + xyz + xyz
So,
F = Σm ^2, 4, 6, 7h
Hence, in POS form, we have
F = ^x + y + z h^x + y + z h^x + y + z h^x + y + z h = Πm ^0, 1, 3, 5h
END OF THE QUESTION PAPER
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t e n . g n i r e e n i g n e y s a .E t w e w n . w g n i r e e n i g n e y s a .E w w w ANSWER KEY
General Aptitude
1
2
3
4
5
6
7
8
9
10
(C)
(A)
(C)
(B)
(A)
(C)
(3)
(B)
(A)
(B)
Electronics & Communication
1
2
3
4
5
6
7
8
9
10
(A)
(0.5208)
(12)
(C)
(2)
(100)
(2)
(A)
(D)
(C)
11
12
13
14
15
16
17
18
19
20
(C)
(D)
(17)
(A)
(B)
(B)
(7)
(B)
(8.8858)
(C)
21
22
23
24
25
26
27
28
29
30
(5)
(0.9375)
(25)
(A)
(C)
(0.4)
(400)
(C)
(D)
(D)
31
32
33
34
35
36
37
38
39
40
(28.6)
(A)
(350)
(83.8)
(53-54)
(1)
(1.649)
(0.5)
(20)
(2.5)
41
42
43
44
45
46
47
48
49
50
(250)
(C)
(2.5284)
(1)
(A)
(40)
(A)
(B)
(C)
(25.5)
51
52
53
54
55
(C)
(B)
(A)
(B)
(A)
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