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This mock test of GATE Mock Test Electronics Engineering (ECE)- 5 for GATE helps you for every GATE entrance exam.
This contains 65 Multiple Choice Questions for GATE GATE Mock Test Electronics Engineering (ECE)- 5 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

Choose the word or phrase which is nearest in meaning to the key word:

** Pilfer**

Solution:

QUESTION: 2

The following question comprises two words that have a certain relationship between them followed by four pairs of words. Select the pair that has same relationship as the original pair of words

Fox : Cunning

Solution:

QUESTION: 3

Replace the phrase printed in bold to make it grammatically correct ?

A twenty-first century economy **" cannot be held"** hostage by power cuts nor travel on nineteenth century roads.

Solution:

QUESTION: 4

Improve the sentence with suitable options by replacing the underlined word.

He "__lay"__ on the grass enjoying the sunshine.

Solution:

QUESTION: 5

Fill in the blanks :

At times, we are all ______ to be mistaken.

Solution:

QUESTION: 6

There are two solutions of wine and water, the concentration of wine being 0.4 and 0.6 respectively. If five liters of the first solution be mixed with fifteen liters of the second, find the concentration of wine in the resultant solution.

Solution:

Resultant concentration of wine

QUESTION: 7

Rajiv bought some apples at the rate of 25 for Rs. 400 and an equal number at the rate of 30 for Rs. 270. He then sold all at the rate of 10 for Rs. 180. Find his profit/loss percentage

Solution:

Let Rajiv bought x apples of each kind.

Total cost price of Rajiv =

Total selling price of Rajiv =

SP > CP , Hence he makes profit;

Profit percentage =

QUESTION: 8

Out of 10 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is alteast one graduate among them ?

Solution:

P (at least one graduate) = 1 – P (no graduate)

QUESTION: 9

2 men and 3 boys can do a piece of work in 10 days, while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work ?

Solution:

Let work done by 1 man in 1 day = x units

Let work done by 1 boy in 1 day = y units

2 men and 1 boy can finish work

No. of days taken by 2 men and 1 boy = = 12.5 days

QUESTION: 10

In an exam, 60% of the candidates passed in Maths and 70% candidates passed in English and 10% of the candidates failed in both the subjects, 300 candidates passed in both the subjects. Find the total number of candidates appeared in the exam, if they took test in only two subjects that is Math and English.

Solution:

Let total number of students appear be X.

X = 750

QUESTION: 11

In the figure, the value of resistor R is (I 5) ohms, where I is the current in amperes. The value of current I is

Solution:

150 – IR = 0

150 – I (I +5) = 0

150 – I^{2} - 5I = 0

I^{2} +5I – 150 = 0

I^{2} +15I – 10I - 150 = 0

I (I +15) – 10 (I +15) = 0

(I +15) (I-10) = 0

I = 10A & I = -15A

QUESTION: 12

The velocity of propagation of electromagnetic wave in an underground cable with relative permittivity of 9 is

Solution:

QUESTION: 13

The function shown in the figure can be represented as

Solution:

QUESTION: 14

Consider an LTI system with impulse response h(t) = e^{-5t }u(t). Determine the output of the system for the input x(t) = 5 u(t)

Solution:

y(t) = h(t) * x(t)

= e^{-5t} u(t) * 5u(t)

y(t) = (1 - e^{-5t}) u(t)

QUESTION: 15

The value of integral is

Solution:

= e^{0} × 0

= 0

QUESTION: 16

A rectangular pulse x(t) is shown in the figure. Determine the value of

Solution:

According to Parswell’s theorem

QUESTION: 17

If and f(t) is as t → ∞ . Determine the value of K

Solution:

⇒K = 2

QUESTION: 18

In the signal flow graph shown in the given figure, the value of C/R ratio is

Solution:

There are three path

QUESTION: 19

What is the open-loop transfer function for an unity feedback system having root locus shown in the following figure ?

Solution:

QUESTION: 20

The digital circuit shown in the figure works as a

Solution:

For D flip-flop

Q_{n 1} = D

Here, D = X Q_{n}

So Q_{n 1} = X Q_{n}

So it becomes T flip-flop

QUESTION: 21

The simplified form of Boolean function F(A,B,C) implemented by the given 3 × 8 decoder is

Solution:

QUESTION: 22

The contents of the accumulator in an 8085 microprocessor is altered after the execution of the instruction

Solution:

Compare operation does not affect the accumulator and ORAA does not change the content of accumulator.

QUESTION: 23

Assuming that all the diodes are ideal in figure, the current in the diode D_{2} is

Solution:

Assume both the diodes are ON

So, I_{1} = -1A

But current can not flow from n to p side in a diode so D_{1} will be off and equivalent circuit becomes

Therefore,

QUESTION: 24

Determine the upper and lower threshold voltage of the schmitt trigger shown below :

Solution:

QUESTION: 25

The transistor in the given circuit, should always be in saturation region. Take V_{CE} (sat) = 0V and V_{BE} = 0.7V. the minimum value of R_{C} is

Solution:

QUESTION: 26

A differential amplifier has input at non inverting terminal is 1050 μV and at inverting terminal 950 μV. What is the error in the differential output if CMRR is 1000.

Solution:

V_{0} = A_{d}V_{d} A_{cm}V_{c}

QUESTION: 27

Which one of the following type of negative feedback increases the input resistance and decreases the output resistance of an amplifier ?

Solution:

QUESTION: 28

The spectral density and auto correlation function of white noise are respectively.

Solution:

Auto corelation function and PSD are fourier transform pair

Here, PSD = N_{0}

QUESTION: 29

For a random variable x having the probability density function is shown in the figure, what are the variance of random variable.

Solution:

= 4/3

QUESTION: 30

An amplitude modulated signal is given as, S(t) = 100 Cos 2pf_{c}t 50 Cos 2pf_{m}t.Cos 2pf_{c}t, where f_{c } f_{m}. and f_{c} is the carrier frequency and f_{m} is the modulating signal frequency. The power efficiency is.

Solution:

S(t) = 100 Cos 2pf_{c}t 50 Cos 2pf_{m}t.Cos 2pf_{c}t

= 100 [1 0.5 Cos 2pf_{m}t] Cos 2pf_{c}t

Here modulation index, μ = 0.5

= 11.11%

QUESTION: 31

Signal x(t) = 4Sinc^{2} 100t. Sin 500pt .At what sampling frequency should this signal be sampled to avoid aliasing ?

Solution:

So, f_{m} = f_{1} f_{2}

= 350 Hz

So, f_{s} = 2f_{m}

= 2 × 350

= 700 Hz

QUESTION: 32

The value of is

Solution:

QUESTION: 33

The solution of differential equation dy = (1-y) dx is

Solution:

dy = (1 – y) dx

y = 1 ce^{-x}

QUESTION: 34

The rank of the matrix A is 2, when K is

Solution:

= 2 {3k} - 3{2k} 4{2´15 - 3´16}

= 0 {for any value of k}

QUESTION: 35

Given If C is a counter-clockwise path in the z-plane such that |z - i| = 1, the value of is

Solution:

Poles are at z = -1, -2

There is no any pole inside the circle, so integration will be zero

QUESTION: 36

In the circuit shown in figure, determine the maximum power absorbed by the load resistance R_{L}

Solution:

R_{th} = 4 ohms

For V_{th}

V_{2} – V_{3} V_{2} – V_{1} 12 = 0

But V_{1} = 2V

So, 2 V_{2} – V_{3 }– V_{1} 12 = 0

2 V_{2} – V_{3 }– 2 12 = 0

2 V_{2} – V_{3 }= -10…………....2

From equation 1 and 2

V_{3} – V_{2 }= 12………………..1

-V_{3} 2V_{2 }= -10…………….2

V_{2} = 2 Volt

So V_{3} = 14 Volt

So V_{th} = 14V

QUESTION: 37

In the circuit shown below, the switch is closed at t = 0. What is the initial value of the current through the capacitor C.

Solution:

At t = 0^{-}, the equivalent circuit is shown below :

At t = 0^{ }, the equivalent circuit

So, i_{c} = 4 – 3.2 = 0.8A

QUESTION: 38

Determine the z-parameter for the network shown below.

Solution:

Since it is symmetrical ladder network

QUESTION: 39

At which frequency (in rad/sec) the thevenins equivalent impedance looking at terminals AB becomes purely resistive?

Solution:

From the circuit it is clear, that Z_{AB} becomes real when circuit is in resonance and resonance frequency of this circuit is given as

= 1000 rad/sec

QUESTION: 40

Consider if y(t) = Ae^{-t} for then value of A is

Solution:

QUESTION: 41

The 4-point DFT of sequence x[n] = [1,2,2,1] is

Solution:

QUESTION: 42

The output y(t) of a continuous-time system for the input x(t) is given by . Which one of the following is correct ?

Solution:

It is an integrator that generally an unstable system because integrator produces an unbounded output and for causal

Let at t = 2

It means y(2) is depends on x(4) that means present output is depends on future input, so it is non-causal

QUESTION: 43

For the NMOS and PMOS shown in the figure are matched and = 1mA/V^{2} and V_{ n} = -V_{ p} = 1 V. Determine V_{0}

Solution:

For the circuit, it is clear that since Q_{N} and Q_{P} are perfectly matched and are operating at |V_{GS}| equal values of i.e. are 2.5V, the symmetrical which dictates that V_{0} = 0V

QUESTION: 44

Determine the region of operation of BJT in the given circuit.

Solution:

Since, V_{EB} = 0.7

Here V_{B} = 0, So, V_{E} = 0.7 V

So, V_{C} = -10 4.6 × 1

= - 5.4 V

Since, V_{EC} = 0.7 5.4 = 6.1V

That is greater than 0.2V So BJT is in active region.

QUESTION: 45

In the silicon BJT circuit shown below. Assume the emitter area of transistor Q_{2} is two time of area of Q_{1}. The value of I_{o} { Let b of the transistors are very large}

Solution:

= 2mA

Since β is very large,

So, I_{ref} = I_{c1}

i.e. I_{c1 }= 2mA

Also the emitter current area of transistor Q_{2} is times of Q_{1}

So, I_{c2} = 2 I_{c1}

= 2 × 2 mA

= 4 mA

So, I_{0} = +4 mA

QUESTION: 46

Determine the current i_{o} in the given circuit.

Solution:

Circuit can be redrawn

QUESTION: 47

Consider a bar of silicon in which a hole concentration profile describe by P(x) = P_{0}.e^{-x/Lp}. Find the current at x = 0, given P_{0} = 10^{16}/cm^{3}, L_{p} = 1 mm and cross-sectional area of the bar is 100 mm^{2} and D_{p} = 12 cm^{2}/sec.

Solution:

= 192 A/cm^{2}

So the current I_{p},

I_{p} = J_{p} × A

= 192 × 10^{-8}

= 192 mA

QUESTION: 48

A system with the open loop transfer function is connected in a negative feedback gain of unity. What is the value of K for which the system is unstable.

Solution:

The characteristic equation

S (S^{3} 8S^{2} 24S 32) K = 0

Routh – Hurwitz array

S^{4} 1 24 K

S^{3} 8 32 O

S^{2} 20 K

S' (20×32-8K) 0

S^{0} K

So for stable system,

K > 0

and 20×32 - 8K > 0

K < 80

So for unstable system K > 80

i.e. K = 90

QUESTION: 49

In the system shown below x(t) = e^{-2t} u(t). Determine the output y(t) in steady state.

Solution:

Here, x(t) = e^{-2t} u(t)

QUESTION: 50

The polar diagram of a conditionally stable system for open loop gain K = 1 is shown in the figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for

Solution:

System is stable in the region 0.1 to 5 on the left side of 10 as number of enrichment there is zero

0.1K < 1 ⇒ K < 10

5K > 1 ⇒ K > 0.2

0.2 < K < 10

Also 1 > 10 K

K < 0.1

So for the stable system 0.2 < K < 10 and K < 0.1

QUESTION: 51

The sensitivity of transfer function with respect to the parameter K is -0.5, Determine the value of K

Solution:

8K^{2} 10K 3 = -4K

8K^{2} 14K 3 = 0

4K (2K 3) 1 (2K 3) = 0

(2K 3) (4K 1) = 0

K = -1.5 and K = -0.25

QUESTION: 52

Given two binary FSK signals i = 1, 2.The minimum frequency separation of these two signals such that they are orthogonal is

Solution:

For the two signals to be orthogonal

QUESTION: 53

A block code is transmitted in NRZ invent-one-one format with voltage levels

6V → ‘1’

-6V → ‘0’

Three code words are sent : 10010, 11110, 01001. Over these three transmitted code words the average voltage is

Solution:

There are total 15 digits so, In which 8 are 1’s and 7 are 0’s

So average value = = 0.4V

QUESTION: 54

A signal is given by,

γ(t) = s(t) n(t)

where s(t) = 20 Cos2π4000t 40 Cos2π2000t and the noise n(t) is white with power spectral density η_{0} = 1 W/Hz. The total received signal is put through a band pass filter with pass band between 100 to 110 Hz. The value of SNR at the output of filter is

Solution:

Signal power

= 200 800

= 1000

and noise power, P_{n} = n_{0} × (BW)

= 1 × 10

= 10W

So, SNR = 1000/10 = 100

In dB = 10log100 = 20dB

QUESTION: 55

The base band signal m(t) is recovered from the DSB-Sc signal s(t) = m(t). Cos2πf_{c}t, by multiplying s(t) by the locally generated carrier Cos (2πf_{c}t φ). The product is passed through a LPF which rejects the double frequency signal. If the base band signal is band limited to 4KHz, what is the minimum value of f_{c} for which m(t) can be recovered by filtering ?

Solution:

In order to recover m(t) from s(t) by filter method it is necessary that the lowest frequency contained in the first term of s(t) must be greater than the highest frequency contained in the second term that is

So minimum value of f_{c} = 4 KHz

QUESTION: 56

A plane wave in free space is incident normally on a large block of material . which occupies z > 0. If the incident electric field is = 10 Cos (ωt z) V/m, the reflected magnetic field is :

Solution:

QUESTION: 57

A plane wave with is incident normally on a thick plane conductor lying in the X-Y plane. Its conductivity is 6 × 10^{6} s/m and surface impedance is 5 × 10^{-4}

?. The skin depth of conductor is

Solution:

QUESTION: 58

The volume of the paralleopipe whose sides are given by is

Solution:

Volume of paralleopipe

2 {-1} 2 {-1 3} 0 = -2 4 = 2

QUESTION: 59

A number in 4 bit 2 5 complement representation is 1101. This number when stored using 8-bits will beA number in 4 bit 2’5 complement representation is 1101. This number when stored using 8-bits will be

Solution:

The number in 4 bit 1101

So original number, 0011

Now the original number in 8-bit,

00000011

So, in 2s complement, 11111101

QUESTION: 60

Four variable K-map is shown in the figure. The essential prime implicants are

Solution:

QUESTION: 61

An 8085 microprocessor is interfaced to a 8KB RAM as shown below. The address range of the RAM is

Solution:

Range of Address

The RAM will be active when is 0

when, A_{13} A_{14} A_{15} = 011 as decoder is active low.

So the range of RAM address is from COOOH to DFFFH

QUESTION: 62

If z = x – iy and = p iq, then is

Solution:

= P iq

Z = P^{3} (iq)^{3} 3P (iq) (P iq)

x – iy = P^{3} – 3Pq^{2} i (3P^{2}q – q^{3})

X = P^{3} – 3Pq^{2}

QUESTION: 63

The Newton – Raphson method is used to find the roots of the equation f(x) = x - cosπx If the initial guess for the root is 0.5, then the value of x after the first iteration is

Solution:

f(x) = X - Cosπx

f(0.5) = 0.5 – Cos 0.5π

= 0.5

= 0.38

QUESTION: 64

The eign value of the matrix are,

Solution:

QUESTION: 65

The value of is

Solution:

### ECE Branch GATE Paper 2011 Page : 1 GATE 2011 Electronics and Communication Engineering

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