GATE Electronics and Communication Online Test Series, GATE MCQ
GATE Electronics and Communication Online Test Series 1
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GATE Electronics and Communication Online Test Series 1. GATE 2021 EC Online Test, GATE Mock Test Exam 2021.GATE Electronics and Communication Full online mock test paper is free for all students. Take GATE 2021 Exams Online Test, GATE 2021 Quiz. GATE 2021 EC Series 1 online Test 2021 in English. Take GATE 2021 Mock Test in English from below, GATE Quiz 2021 Series 1. You may also find other Subjects GATE Online Test in this page. Here we provide Graduate Aptitude Test in Engineering Question and Answers 2021. This GATE EC Online test Series 1 is very helpful for exam preparation. The GATE Electronics and Communication Mock Test 2021 is now available for all candidates, who will be appearing in the national level engineering exams 2021. Now take GATE Electronics and Communication Online Test Series 1 by Click on Below “Start Quiz Button”
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Question 1 of 25
1. Question
is equal to
Correct
Incorrect

Question 2 of 25
2. Question
Consider a ufb system with forwardpath transfer function G (s) = . The system is stable for the range of K. Which of the following is correct?
Correct
G(s) = k(s+3)/s^4(s+2)
We can write as
G(s) = k(s+3)/(s^5+2s^4+ks+1)
The system is always unstable because s^{3} and s^{2} are missing here.Incorrect
G(s) = k(s+3)/s^4(s+2)
We can write as
G(s) = k(s+3)/(s^5+2s^4+ks+1)
The system is always unstable because s^{3} and s^{2} are missing here. 
Question 3 of 25
3. Question
The time signal corresponding to is
Correct
Option (1) is correct.
X (s) = 2 – 1 / (s+2) (s+3)
= 2 – 1 / (s+2) + 1 / (s+3)
X (t) = 2(t) + (e3t – e2t) u (t)Incorrect
Option (1) is correct.
X (s) = 2 – 1 / (s+2) (s+3)
= 2 – 1 / (s+2) + 1 / (s+3)
X (t) = 2(t) + (e3t – e2t) u (t) 
Question 4 of 25
4. Question
The characteristic equation of a closedloop system is s(s + 1)(s + 2) + k = 0. The centroid of the asymptotes in rootlocus will be
Correct
Option (2) is correct.
Sum of poles = 012 = 3
Sum of zeros = 0
Therefore, no. of poles – no.of zeros = N. of values for which response is infinite = 3 poles and none zeros. Hence,Incorrect
Option (2) is correct.
Sum of poles = 012 = 3
Sum of zeros = 0
Therefore, no. of poles – no.of zeros = N. of values for which response is infinite = 3 poles and none zeros. Hence, 
Question 5 of 25
5. Question
Given the z – transform X (z) = , the limit of x [∞] is
Correct
Option (1) is correct.
The function has poles at z = 1, 3/4 . Thus, the final value theorem applies.
= (z – 1)
Incorrect
Option (1) is correct.
The function has poles at z = 1, 3/4 . Thus, the final value theorem applies.
= (z – 1)

Question 6 of 25
6. Question
A MUX network is shown below.
The function Z1 is equal to 1 whenCorrect
According to the circuit of MUXs: Z_{1} = 1 when Z_{0} = 0 otherwise Z_{1} is equal to 0. Z_{0} = 0 when b = 1, at this time a = 0. So, Z_{1} = a’ . b. c .
This is the correct answer. Also one cannot represent the operation of MUX in simple binary expression unless the conditions are met.Incorrect
According to the circuit of MUXs: Z_{1} = 1 when Z_{0} = 0 otherwise Z_{1} is equal to 0. Z_{0} = 0 when b = 1, at this time a = 0. So, Z_{1} = a’ . b. c .
This is the correct answer. Also one cannot represent the operation of MUX in simple binary expression unless the conditions are met. 
Question 7 of 25
7. Question
If (211)_{x} = (152)_{8}, the value of base x is
Correct
Option (3) is correct.
2x^{2} + x + 1 = 64 + 5 x 8 + 2
x = 7Incorrect
Option (3) is correct.
2x^{2} + x + 1 = 64 + 5 x 8 + 2
x = 7 
Question 8 of 25
8. Question
An ideal pn junction diode is operating in the forward bias region. The change in diode voltage, that will cause a factor of 9 increase in current, is
Correct
I_{d} = I_{s }
V_{1} – V_{2} = V_{t} In = 0.0259 In 10 = 59.6 mV ≈ 59mVIncorrect
I_{d} = I_{s }
V_{1} – V_{2} = V_{t} In = 0.0259 In 10 = 59.6 mV ≈ 59mV 
Question 9 of 25
9. Question
If the base width of a bipolar transistor is increased by a factor of 3, what is the collector current change?
Correct
We know
I_{c} =
I_{C }
So, if W_{B} increases by a factor of 3, then I_{C} is decreased by a factor of 3.Incorrect
We know
I_{c} =
I_{C }
So, if W_{B} increases by a factor of 3, then I_{C} is decreased by a factor of 3. 
Question 10 of 25
10. Question
For a nchannel enhancementmode MOSFET, the parameters are V_{TN} = 0.8 V, k’_{n} = 80 mA/V^{2} and W/L = 5. If the transistor is biased in saturation region with I_{D} = 0.5 mA, then required V_{GB} is
Correct
0.5 =
V_{GS} = + 0.8 = 2.38 VIncorrect
0.5 =
V_{GS} = + 0.8 = 2.38 V 
Question 11 of 25
11. Question
The value of the current measured by the ammeter in figure is
Correct
4i_{2} + 6i_{3} – 2i_{1} = 0
i_{1} + i_{2} = 2
i_{2} = 5 + i_{3}
i_{1} = 5/6 A
Incorrect
4i_{2} + 6i_{3} – 2i_{1} = 0
i_{1} + i_{2} = 2
i_{2} = 5 + i_{3}
i_{1} = 5/6 A

Question 12 of 25
12. Question
In the following lattice network, the value of R_{L} for the maximum power transfer to it is
Correct
The circuit is as shown below.
T_{TH} = 7  5 + 6  9 = 6.52 Ω
For maximum power transfer
R_{L} = T_{TH} = 6.52 ΩIncorrect
The circuit is as shown below.
T_{TH} = 7  5 + 6  9 = 6.52 Ω
For maximum power transfer
R_{L} = T_{TH} = 6.52 Ω 
Question 13 of 25
13. Question
Let J = u_{z} A/m^{2}. The value of is
Correct
= = 0Incorrect
= = 0 
Question 14 of 25
14. Question
A vector field is given by E = 4zy^{2}u_{x} + 2y sin 2xu_{y} + y^{2} sin 2xu_{z}. The surface on which E_{y} = 0 is
Correct
For E_{y} = 0, 2y sin 2x = 0
y = 0
sin 2x = 0
2x = 0, π,3π
x = 0, 3π/2Incorrect
For E_{y} = 0, 2y sin 2x = 0
y = 0
sin 2x = 0
2x = 0, π,3π
x = 0, 3π/2 
Question 15 of 25
15. Question
In a nonmagnetic medium, E = 5 cos (10^{9} t – 8x) u_{x} + 4 sin (10^{9} t – 8x) u_{z} V/m. The dielectric constant of the medium is
Correct
Incorrect

Question 16 of 25
16. Question
The aerial current of an AM transmitter is 18 A when unmodulated but increases to 20 A when modulated.
The modulation index isCorrect
I_{t} = I_{c} = 0.68
Incorrect
I_{t} = I_{c} = 0.68

Question 17 of 25
17. Question
A carrier is phase modulated (PM) with frequency deviation of 10 kHz by a signal tone frequency of 1 kHz. If the signal tone frequency is increased to 2 kHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is
Correct
Initial 4 x 10 = 40 kHz
Now if deviation is d, then Δ BW = 2d = 2 x 2 = 4 kHz
New Bandwidth BW = 40 + 4 = 44 kHzIncorrect
Initial 4 x 10 = 40 kHz
Now if deviation is d, then Δ BW = 2d = 2 x 2 = 4 kHz
New Bandwidth BW = 40 + 4 = 44 kHz 
Question 18 of 25
18. Question
A dice is thrown thrice. Getting 1 or 6 is taken as a success. The mean of the number of successes is
Correct
p = , q = and n = 3
Thus, mean (np) = 1/3×3 = 1Incorrect
p = , q = and n = 3
Thus, mean (np) = 1/3×3 = 1 
Question 19 of 25
19. Question
Let f (x) = e^{x} in [0, 1], the value of c of the mean – value theorem is
Correct
f` (c) =
c = log (e – 1)Incorrect
f` (c) =
c = log (e – 1) 
Question 20 of 25
20. Question
For dy/dx = xy, we have y = 1 at x = 0. By using Euler method and taking the step size 0.1, find the value of y at x = 0.4.
Correct
x : 0 0.1 0.2 0.3 0.4
Euler method gives:
y_{n+1} = y_{n} + h (x_{n}, y_{n}) …………. (1)
n = 0 in (1) gives:
y_{1} = y_{0} + h f (x_{0}, y_{0})
Here x_{0} = 0,
y_{0} = 1,
h = 0.1
y_{1} = 1 + 0.1 f (0, 1) = 1 + 0 = 1
n = 0 in (1) gives y_{2} = y_{1} + hf (x_{1}, y_{1})
= 1 + 0.1 f (0.1, 1) = 1 + 0.1 (0.1) = 1 + 0.01
Thus y_{2} = y_{(0, 2)} = 1.01
n = 2 in (1) gives:
y_{3} = y_{2} + hf (x_{2}, y_{2}) = 1.01 + 0.1 f (0.2, 1.01)
y_{3} = y(0, 3) = 1.01 + 0.0202 = 1.0302
n = 2 in (1) gives:
y_{3} = y_{2} + hf (x_{2}, y_{2}) = 1.01 + 0.1 f (0.2, 1.01)
y_{3} = y_{(0, 3)} = 1.01 + 0.0202 = 1.0302
n = 3 in (1) gives:
y_{4} = y_{3} + hf (x_{3}, y_{3}) = 1.0302 + 0.1 f (0.3, 1.0302)
= 1.0302 + 0.03090
Hence y_{4} = y_{(0, 4)} = 1.0611Incorrect
x : 0 0.1 0.2 0.3 0.4
Euler method gives:
y_{n+1} = y_{n} + h (x_{n}, y_{n}) …………. (1)
n = 0 in (1) gives:
y_{1} = y_{0} + h f (x_{0}, y_{0})
Here x_{0} = 0,
y_{0} = 1,
h = 0.1
y_{1} = 1 + 0.1 f (0, 1) = 1 + 0 = 1
n = 0 in (1) gives y_{2} = y_{1} + hf (x_{1}, y_{1})
= 1 + 0.1 f (0.1, 1) = 1 + 0.1 (0.1) = 1 + 0.01
Thus y_{2} = y_{(0, 2)} = 1.01
n = 2 in (1) gives:
y_{3} = y_{2} + hf (x_{2}, y_{2}) = 1.01 + 0.1 f (0.2, 1.01)
y_{3} = y(0, 3) = 1.01 + 0.0202 = 1.0302
n = 2 in (1) gives:
y_{3} = y_{2} + hf (x_{2}, y_{2}) = 1.01 + 0.1 f (0.2, 1.01)
y_{3} = y_{(0, 3)} = 1.01 + 0.0202 = 1.0302
n = 3 in (1) gives:
y_{4} = y_{3} + hf (x_{3}, y_{3}) = 1.0302 + 0.1 f (0.3, 1.0302)
= 1.0302 + 0.03090
Hence y_{4} = y_{(0, 4)} = 1.0611 
Question 21 of 25
21. Question
The solution of dy/dx y tan x – y^{2} sec x = 0 is given by
Correct
dy/dx – y tan x = y^{2} sec x
Or y^{2 }dy/dx – y^{1} tan x = sec x
Put y^{1} = v to get – y^{2 }dy/dx = dv/dx
Substituting this in the given equation, we get
dv/dx v tan x = sec x or dv/dx + (tan x). v = – sec x
I. F =
v. sec x = – + c = – tan x + c
1/y = = – sin x + cos x
Or y1 = – sin x + c2 cos xIncorrect
dy/dx – y tan x = y^{2} sec x
Or y^{2 }dy/dx – y^{1} tan x = sec x
Put y^{1} = v to get – y^{2 }dy/dx = dv/dx
Substituting this in the given equation, we get
dv/dx v tan x = sec x or dv/dx + (tan x). v = – sec x
I. F =
v. sec x = – + c = – tan x + c
1/y = = – sin x + cos x
Or y1 = – sin x + c2 cos x 
Question 22 of 25
22. Question
dxdydz is equal to
Correct
Incorrect

Question 23 of 25
23. Question
A lossless transmission line with a characteristic impedance of 80 Ω is terminated by a load of 125 Ω . The length of line is 1.25 . The input impedance is
Correct
Incorrect

Question 24 of 25
24. Question
In a nonmagnetic medium (ε_{r }= 6.25), the magnetic field of an EM wave is
H = 6 cos β x cos (10^{8} t) u_{s} A/m. The corresponding electric field isCorrect
Incorrect

Question 25 of 25
25. Question
The plane 2x + 3y – 4z = 1 separates two regions. Let µr1 = 2 in region 1 defined by 2x + 3y – 4z > 1, while µr1 = 5 in region 2 where 2x + 3y – 4z < 1. The region H_{1} = 50 u_{x} – 30 u_{y} + 20 u_{z} A/m. In region 2, H_{2} will be
Correct
From the given solution Ht2 = 54.8Ux – 22.76 Uy + 10.34 Uz Hn2 = 1.93 Ux – 2.9Uy + 3.86 Uz H2 = Ht2 + Hn2 H2 = 52.87Ux 25.66 Uy + 14.2 Uz
Incorrect
From the given solution Ht2 = 54.8Ux – 22.76 Uy + 10.34 Uz Hn2 = 1.93 Ux – 2.9Uy + 3.86 Uz H2 = Ht2 + Hn2 H2 = 52.87Ux 25.66 Uy + 14.2 Uz