## GATE Electronics and Communication Online Test Series 1

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GATE Electronics and Communication Online Test Series 1. GATE 2019 EC Online Test, GATE Mock Test Exam 2019.GATE Electronics and Communication Full online mock test paper is free for all students. Take GATE 2019 Exams Online Test, GATE 2019 Quiz. GATE 2019 EC Series 1 online Test 2019 in English. Take GATE 2019 Mock Test in English from below, GATE Quiz 2019 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 2019. This GATE EC Online test Series 1 is very helpful for exam preparation. The GATE Electronics and Communication Mock Test 2019 is now available for all candidates, who will be appearing in the national level engineering exams 2019. 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

CorrectIncorrect - Question 2 of 25
##### 2. Question

Consider a ufb system with forward-path transfer function G (s) = .

CorrectG(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.IncorrectG(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

CorrectOption (1) is correct.

X (s) = 2 – 1 / (s+2) (s+3)

= 2 – 1 / (s+2) + 1 / (s+3)

X (t) = 2(t) + (e-3t – e-2t) u (t)IncorrectOption (1) is correct.

X (s) = 2 – 1 / (s+2) (s+3)

= 2 – 1 / (s+2) + 1 / (s+3)

X (t) = 2(t) + (e-3t – e-2t) u (t) - Question 4 of 25
##### 4. Question

The characteristic equation of a closed-loop system is s(s + 1)(s + 2) + k = 0. The centroid of the asymptotes in root-locus will be

CorrectOption (2) is correct.

Sum of poles = 0-1-2 = -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,IncorrectOption (2) is correct.

Sum of poles = 0-1-2 = -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

CorrectOption (1) is correct.

The function has poles at z = 1, 3/4 . Thus, the final value theorem applies.

= (z – 1)

IncorrectOption (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 whenCorrectAccording 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.IncorrectAccording 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 isCorrectOption (3) is correct.

2x^{2}+ x + 1 = 64 + 5 x 8 + 2

x = 7IncorrectOption (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

CorrectI

_{d}= I_{s }

V_{1}– V_{2}= V_{t}In = 0.0259 In 10 = 59.6 mV ≈ 59*mV*IncorrectI

_{d}= I_{s }

V_{1}– V_{2}= V_{t}In = 0.0259 In 10 = 59.6 mV ≈ 59*mV* - Question 9 of 25
##### 9. Question

If the base width of a bi-polar transistor is increased by a factor of 3, what is the collector current change?

CorrectWe know

I_{c}=

I_{C }

So, if W_{B}_{C}IncorrectWe know

I_{c}=

I_{C }

So, if W_{B}_{C} - Question 10 of 25
##### 10. Question

For a n-channel enhancement-mode 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}isCorrect0.5 =

V_{GS}= + 0.8 = 2.38 VIncorrect0.5 =

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

Correct4i

_{2}+ 6i_{3}– 2i_{1}= 0

i_{1}+ i_{2}= 2

i_{2}= 5 + i_{3}

i_{1}= -5/6 A

Incorrect4i

_{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 isCorrectThe circuit is as shown below.

T_{TH}= 7 || 5 + 6 || 9 = 6.52 Ω

For maximum power transfer

R_{L}= T_{TH}= 6.52 ΩIncorrectThe 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 isCorrect

= = 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 isCorrectFor E

_{y}= 0, 2y sin 2x = 0

y = 0

sin 2x = 0

2x = 0, π,3π

x = 0, 3π/2IncorrectFor 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 isCorrectIncorrect - 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 isCorrectI

_{t}= I_{c}= 0.68IncorrectI

_{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

CorrectInitial 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 kHzIncorrectInitial 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

Correctp = , q = and n = 3

Thus, mean (np) = 1/3×3 = 1Incorrectp = , 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 isCorrectf` (c) =

c = log (e – 1)Incorrectf` (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.

Correctx : 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)}= 1.0611Incorrectx : 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 byCorrectdy/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 y-1 = – sin x + c2 cos xIncorrectdy/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 y-1 = – sin x + c2 cos x - Question 22 of 25
##### 22. Question

dxdydz is equal to

CorrectIncorrect - 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

CorrectIncorrect - Question 24 of 25
##### 24. Question

In a non-magnetic 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 isCorrectIncorrect - 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 beCorrectFrom 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

IncorrectFrom 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