GATE Mock Test Electronics Engineering (ECE)- 2

65 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | GATE Mock Test Electronics Engineering (ECE)- 2

Attempt GATE Mock Test Electronics Engineering (ECE)- 2 | 65 questions in 180 minutes | Mock test for GATE preparation | Free important questions MCQ to study GATE ECE (Electronics) 2023 Mock Test Series for GATE Exam | Download free PDF with solutions

A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in total from both as interest. The rate of interest per annum is.

Solution: Let the rate % = R

According to the question

100R + 120 R = 2200

2200 R = 2200

R = 10%

Hence required rate % = 10%


Amit rows a boat 9 kilometres in 2 hours down-stream and returns upstream in 6 hours. The speed of the boat (in kmph) is:


Down-stream rate = 9/2 = 4.5 kmph

Upstream rate = 9/6= 1.5 kmph

The speed of the boat = (4.5 – 1.5) kmph = 3 kmph


The sentences given with blanks are to be filled with an appropriate word(s). Four alternatives are suggested for each question. For each question, choose the correct alternative and click the button corresponding to it.

They abandoned their comrades ______the wolves.

Solution: When something is left in between more than 2 things or persons, ‘among’ is used. There are a number of wolves in the sentence.

Hence the correct answer is option D.


In the following question, out of the four alternatives, select the word opposite in meaning to the given word.


Solution: Turgid = swollen and distended or congested; pompous

Bloated = inflated

Humble = plain, simple

Puffy = inflated

Tumescent = swollen

Hence, humble is the correct answer.


Direction: Study the following information carefully and answer the given questions:

The following pie chart shows the percentage distribution of total number of Apples (Dry + Wet) sold in different days of a week :

The pie chart 2 shows the percentage distribution of total number of Apples (Wet) sold in different days of a week.

Find the ratio of Apples (Dry + Wet) sold on Tuesday to that of the Apples (Wet) sold on Thursday?

Solution: Required ratio = 13% of 7000: 23% of 4500 = 182:207

Directions: In each of the questions below, some statements are given followed by some conclusions. You have to consider the statements to be true even if they seem to be at variance with commonly known facts. You have to decide which of the following conclusions logically follows from the given statements. Give answer.


Some jacket is shirt.

Some shirts are trouser.

No shoes are t-shirt.

All trousers are shoes.

All pants are t-shirt.


I. All shirts being t-shirt is a possibility.

II. Some shoes are trouser.

III. Some jackets are t-shirt.

IV. Some shirts being t-shirt is a possibility.



The question below consists of a set of labelled sentences. These sentences, when properly sequenced form a coherent paragraph. Select the most logical order of sentences from among the options.

P: July 1969 was to see a transformed Indira Gandhi.

Q: Quite a few people contributed with ideas.

R: She sounded the bugle through her historic ‘Note on Economic Policy and Programme’ that was circulated among delegates at Bangalore on July 9, 1969.

S: But the pivot was P.N.Haksar who gave shape, structure and substance with the help of some of his colleagues in the Prime Minister’s Secretariat.


P is clearly the first sentence as it introduces the main person of the paragraph, i.e. Indira Gandhi. The pronoun ‘she’ in R refers to Indira Gandhi mention in P. Thus, R forms the second statement. Now, in the sentence R, ‘delegates’ is mentioned which says about the people, who are given in the sentence Q. So, sentence Q must follow R.

Thus, the sequence after rearrangement is PRQS and option A is the correct answer.


Direction: Which of the following is the MOST SIMILAR in meaning to the given word?



‘Remnant’ means ‘remainder’ which is same as ‘residue’.


Direction: Read the information carefully and give the answer of the following questions:

Total number of children = 2000

If two-ninths of the children who play football are females, then the number of male football children is approximately what percent of the total number of children who play cricket?


57.4% (If 2/9 th of children play football are female , then male children = 1 - 2/9 = 7/9th of children play football;

Let's take z% of male football children equal to cricket children , then

⇒ z% of cricket children = 7/9 th of football;

⇒ z% of (23% of 2000) = 7/9 th of (17% of 2000);

⇒ z% of 23 = 7/9 th of 17

⇒ z = 7× 17 × 100 = 57.4%

Short-cut :

Required percentage = (7/9)×17×100/ 23 = 57.4%


Direction: Study the following information carefully and answer the given questions.

Seven people, P, Q, R, S, T, W and X sitting in a straight line facing north, not necessarily in the same order. R sits at one of the extreme ends of the line. T has as many people sitting on his right, as on his left. S sits third to the left of X. Q sits on the immediate left of W. Q does not sit at any of the extreme ends of the line.

If all the people are made to sit in alphabetical order from right to left, the positions of how many people will remain unchanged?


From the given conditions, we can conclude

After arranging in alphabetical order,

Hence, only Q's position will remain unchanged.


The figure shows the circuit of a gate in the resistor transistor logic (RTL) Then the circuit represent which gate?


Hence gate is OR Gate.



*Answer can only contain numeric values

For the given p+ -n junction diode the minority carrier life time in the n -region is ΥP = 1.1 μsec. If the excess minority charges present in the n -region at any time t is given as QP(t) then QP(t) at t = 1.5 μsec is _____ μC?

Solution: For t < />

I0 = 2A

Diode in FB has diffusion capacitance, as current increases the charge inside increase. Hence voltage increase.



Y = life time of holes in N-region

= 1.1μsec

Multiplying equation (1) by C gives

And t = 1.5 \musec


*Answer can only contain numeric values

For the given network, the power given to the load is ____ kW? [Assume R =1 Ω and ideal diode]

V = 100. sinωt

Solution: (3,4)

Using mirror symmetry

The resistive network between AB can be simplified as.

Further simplifying


A current source with an internal resistance of RS, supplied power to load RL. The plot of power delivered as a function of load RL is?


Hence option B is correct.


where (x) denotes greatest integer



For given system

Angle of arrival is?

Solution: Zeros:- s= -3 ± j

Poles:- s = -1 ± j

ϕ’ = 180 - tan-11 = 135°

ϕA = 180° + ϕ

Where ϕ = SP - SZ

= 180 + 135 - 90

= 225°

∴ ϕA = 180° + ϕ = 405°

= 45°

Hence ϕA = ±45°


A system is shown in figure as follows

Where, hk (n) = δ [n – k.(1/2)k]

Then value of overall impulse response of above system h(n) at h = 2 is ______?

Solution: We have properties

Therefore h(n) can be drawn as

∴ at n = 2h(n) = 1

*Answer can only contain numeric values

A square wave with frequency = 5 kHz

This signal is passed through an ideal low-pass filter with 0 dB of passband gain and 10 kHz of cut-off frequency. The filtered signal is subsequently buried additively into a zero-mean noise processed with one-sided power spectral density (PSD) of 20 nW Hz–1. Upto a frequency of 2.5 MHz. The PSD of noise is zero beyond 2.5 MHz. The signal to noise ratio of the output is _______dB.

Solution: We have

Now, Noise power = Area

= One sided PSD × Frequency range


The output of filter is a

Solution: x(t) can be drawn like that

Hence impulse response of matched filter h(t) = x(T – t) is

∴ output y(t) = x(t) * h(t)

∴ Hence option D.


For a MOD -30 ripple counter, the maximum operating frequency is 8 MHz. The propagation delay of each flip flop must be _________nsec?

Solution: Mod-30 counter required

∴ Propagation delay pe flip flops

*Answer can only contain numeric values

For a successive approximation type of ADC shown.

If Va = 8V and resolution of D/A = 0.25 volt

Then the error voltage between Va and SAR output at 5th clock is_______volt?

Solution: Resolution = 0.25

Va = 8v

VDAC = Resolution × Decimal equivalent of binary o/p of SAR

∴at 5th clock max o/p = 7.75 volt

∴ error = 8 – 7.75 = 0.25 volt.


For the transfer function G(jω) = 6 + jω. The corresponding Nyquist plot for positive frequency has the form

Solution: As real part σ = 6

Only imaginary port increases with ω.

Hence option D is correct.


Consider a random process as z(t) = x(t) + y(t)

Where x(t) and y(t) are two random processes with zero mean and they are individually WSS. The PSD of x(t) is Sx(f) and y(t) is Sy(f). Then PSD of z(t) is?

Solution: We know

Rz(T) = E[z(t).z(t − T)]

Here z(t) = x(t) + y(t)

∴ Rz(T) = E[(x(t) + y(t))⋅(x(t − T) + y(t − T))]

= E[x(t)⋅x(t − T)] + E[y(t)⋅x(t − T)] + E[x(t)⋅y(t − T)] + E[y(t)⋅y(t − T)]

∴ Rz(T) = Rx(T) + Ry(T) + Rxy(T) + Ryx(T)

We know

Rz(T) ⟶ PSD Sz(f)

∴ S2(f) = Sx(f) + Sy(f) + S(f) + Sxy(f)

Hence option D is correct.


A digital circuit is designed in a way that it accept 3-bit number and generate it’s square. The circuit is to be designed using ROM with minimum hardware. Then the size of ROM is _______?

Solution: Truth table is

It can be seen that

D1 = 0 and D0 = z

Hence ROM size = 23 × 4 = 8 × 4

The design is


Eight voice signals, each limited to 4 kHz and sampled at Nyquist rate, are converted into binary PCM signal using 256 quantization levels. The bit transmission rate for the time-division multiplexed signal will be (in kbps)____

Solution: Nyquist rate = 4×103×2 = 8×103 samples/sec

256 levels = 8 bits.

So one sample is represented by 8 -bit. So total = 8×8×103 bits/sec

For 8 voice signal = 8×8×8×103 bits/sec

= 512 kbps


When a relatively small d.c. voltage is applied across a thin slice of Gallium Arsenide with thickness of order of few tenths of a micrometer and the voltage gradient across the slice is in excess of about 3300 V/cm, then the negative resistance will manifest itself. Which of the following devices operate based on the above phenomenon?

Solution: The above-mentioned phenomenon occurs in Gunn diode and it is called the GUNN EFFECT. Gunn Effect is exhibited by semiconductor materials like gallium arsenide, indium phosphide, cadmium telluride and indium arsenide. Thus, option B is the correct answer.

A signal of bandwidth 100 MHz and strength 10mW is transmitted through a transmitter through a cable that has 40 dB loss. If N0/2 = 0.5 × 10−20ω/Hz then the signal to noise ratio at the input of the receiver is



In a superheterodyne AM receiver, the image channel selectivity is determined by :

Solution: Image frequency rejection can be improved by placing more no of tank circuits, in between Antenna and the IF amplifier and by increasing their selectivity against image frequency.

Consider a medium such that μ=μ0,ε = 81ε0,σ = 20S/m. The frequency at which the conduction current density is 10 times the displacement current density in magnitude is ____(GHz)


on putting all these values.

= 36 GHz


A 78 Ω lossless planar line was designed but did not meet a requirement. The fraction of the widths of strip (in %)should be added or removed to get the characteristic impedance of 75Ω is ___


i.e. the width must be increased by 4%.


A 16K × 8 ROM IC is interfaced to 8085 microprocessor. What is the address space?


16K × 8 ⇒ 24 × 210 × 8 = 214 × 8

i.e. 14 address input lines and 8 data lines

With 14 address means there are 14 different address lines out of 16 address lines

Starting address should be 0000 H

Hence starting address will be

0000 0000 0000 0000 = 0000 H

And end address

0011 1111 1111 1111 = 3FFF H

Hence address space is 0000 H – 3FFF H


A 16 bit computer can address 1M word and has 128K words installed from 64K 1 chips. Calculate the no. of chips required and no. of bits of address bus that allows to select the memory chips


As it is a 64K memory chips hence 16 bits are needed to address them

i.e. 64K = 216

As we want 128K × 16 from 64K × 1 chips

Now as we have two rows of 16 chips each, we will need 1 bit to difference one row from another. So A16 will be used to select the memory chips


Maximum number of Self Dual possible for 4-variable function.

Solution: For n variable Boolean function maximum number of self-dual possible is


A minimized Boolean expression is a combination of

Solution: Points to Remember

The prime implicant which contains at least one 1 which cannot be covered by any other prime implicant is called an essential prime implicant (EPI).

The prime implicant whose each 1 is covered by at least one EPI is called Redundant prime implicant (RPI).

The prime implicant which is neither EPI nor RPI is called selective prime implicant (SPI).


If then which of the following holds


Partially differentiating with respect to y,

Partially differentiating with respect to x,


*Answer can only contain numeric values

Calculate the value of


*Answer can only contain numeric values

If dt/dx = sin(t + x) + cos(t + x) then find out the value of t at x = π/3 correct upto 2 decimal places. It is given that t = π/2 at x = 0



A poisson variate satisfies P(x = 1) = 0.5 P(x = 2). Then find the value of P(x = 4) is



Find the general solution of the equation


This is Cauchy’s Homogeneous Equation,



A discrete time signal with input as x[n] has impulse response h[n]=kδ[n] where k is constant. The output y[n] of the system is given as:


For a discrete time signal with input as x[n] has impulse response h[n] =kδ[n], the output y[n] of the system is given as





Taking Fourier transform on both sides

Y(ω)=X(ω) x H(ω)

Therefore H(ω)= Y(ω)/X(ω)

This is known as transfer function of the system.

h[n] = F-1[H(ω)]


Find the complete solution of the state equation given below

Given the initial condition as


Consider the state equation be,


We know that,

State transition matrix,


For the given system, determine the overall response Y(s) of the system corresponding to both inputs.


When multiple inputs are present in a linear system, output is computed by treating each input independently. The overall response of the system is the obtained by adding the outputs corresponding to each input.

Assuming X2(s) = 0,

The block diagram can be reduced to,

Reducing the inner feedback loop,

The block diagram is reduced to


The block diagram is reduced to,

Thus, YX2(s) = output due to X1(s) acting alone

Hence, the response of the system,


An integral controller is used for speed control with a set-point of 120 rpm within a range of 100 to 150 rpm. The controller output is 22% initially. The constant KI for controller output is 0.15%/sec for unit percentage error. If the speed goes to 105 rpm, then the controller output after two seconds for a constant error ep is (in %).


Where r = set point value and b = measured value

The controller output for constant error will be


A junction diode is fabricated in which the p and n regions are doped equally with 5 × 1016 atoms/cm3. Assume ni = 1.5 × 1010/cm3. If the cross-sectional area of the junction is 20 μm2, the magnitude of the charge stored on either side of the junction while no external bias being applied would be – (Assume ∈s = 1.04 x 10-14 F/cm, V0 (barrier potential) = 0.78V)


We have the formula to find out Charge stored on either side on a junction given as follows,

We don’t have the value of W with us, it needs to be calculated. This can be easily calculated using the following formula,

is the applied Voltage and V0 is the barrier potential.

here V = 0 as no external bias is applied.

therefore,VB = V0 = 0.78V

Then the value of W is 2.01 x 10-6 cm.

So the value of QJ is found out to be 1.61 x 10-15 C


A thin P-type silicon sample is uniformly illuminated with light which generates excess carriers. The recombination rate is directly proportional to


Recombination rate, R = B(nn0 + nn)(Pn0 + Pn)Electron and hole concentrations respectively under thermal equilibrium nn0 & Pn0 Excess elements and hole concentrations respectively. Thus, option B is the correct answer.


Two ideal pn junction have exactly the same electrical and physical parameters except for the band gap of the semiconductor materials. The first has a bandgap energy of 0.525 eV and a forward-bias current of 10 mA with Va = 0.255 V. The second pn junction diode is to be designed such that the diode current I = 10μA at a forward-bias voltage of Va = 0.32 V. The bandgap energy of second diode would be_____ (eV)



Consider the given figure and calculate the feedback factor of the wave generator if R1 = 5kΩ, R2 = 20kΩ, R = 500Ω and C = 10 µF. Also calculate the one cycle period?


The given circuit is that of a square wave generator.

Given: R1 = 5kΩ, R2 = 20kΩ, R = 500Ω and C = 10 µF.

To calculate feedback factor:

The period T is given by,

T = 0.01 x 0.405

T = 4.05 x 10-3s

T = 4.05ms


Calculate the hysteresis width (in Volts) for an inverting Schmitt trigger with feedback fraction 0.5. Assume the supply voltage to be 10V.


For a inverting Schmitt trigger.

Given: β=0.5, Vcc = 10V

Hysteresis Voltage VH = VUT - VLT

VUT = +β x Vcc

VLT = -β x Vcc

VH = 2 x β x Vcc

VH = 2 x 0.5 x 10

VH = 10V


The combination of resistors is shown in figure below. If R1 is operated at T and R2 is operated at temperature (T+t) ,then the equivalent noise temperature if the circuit is


The circuit can be redrawn as


Decide the stability of the system whose characteristic equation is given by s5 + 2s4 + 5s3 + 10s2 + 4s + 8 = 0


s3 row elements are all zero, thus the even polynomial above is given by:

f(s) = 2s4 + 10s2 + 8

f′(s) = 8s3 + 20s

There is no sign change in the first column, thus no zeros of the polynomial are in

the RHP, with 4 poles on the imaginary axis and one in LHP. Hence the system response is converging oscillations, hence stable.


Consider the electronic PID controller shown in figure below

If R1 = R2 = R and C1 = C2 = C2 = C, then find Kp, Kd and Ki


*Answer can only contain numeric values

Two point charges of 10 µC at point M and 20 µC at point N exerts a force of 4.8 N and 10 N respectively on a point charge of 40 µC at point O. Determine the electric field intensity at point O.

Solution: The electric field intensity is given as the force per unit charge.

E = F/Q

The electric field intensity due to two point charges Q1 at point M and Q2 at point N is the sum of the forces on the test charge Qt at point O, caused by the point charges Q1 and Q2 acting alone. Thus,


An op-amp is used in a feedback configuration having open loop forward gain value of 50 as shown. The feedback gain is β in the circuit, the feedback circuit is given below. For which of the combination of R1 and R2 respectively would the configuration work as an oscillator?


Here, we can see from the figure that the feedback is connected in a negative feedback configuration. Then, overall gain is given by,

Where, Af is the open loop forward gain. β is the gain provided by the feedback component. According to Barkhausen Criteria, in this situation the given arrangement would work as an oscillator only if Af∗β would be ′−1′

Thus, here β should be -0.02. Among the given op-amp configurations for feedback element, given that gain for op-amp as inverting amplifier İs given as

Only C provides the required gain β.


Find the h parameters of the following circuit


For parallel connection Y parameters gets added Y = Y1 + Y2


The expression for unit tangent ∅ = e2x casyz at the or igin at the or igin in the direction of the tangent to curve x = a sin t, y = a cos⁡ t,z = 2α sin ⁡t, at t = T/4



Root locus of 2s(s+4)+k(s+8) is a circle. The co-ordinates of the centre of the circle?


Characteristic Equation is

Now for the centre of the root locus we have to find the breakaway point of the plot For breakaway point,

On solving for s, s1 = −1.1255 and s2 = -12.74

So, the centre will be at and the coordinates are (-7, 0)

*Answer can only contain numeric values

In the following table, x is a discrete random variable and p(x) is the probability density. The difference between the variance & the standard deviation is


Difference of variance & standard deviation = 3.06 - 1.749 = 1.311


Find the state transition diagram for the logic circuit shown below-


Hence option C is correct


Consider the following circuit consisting of three D-flip flops.

If all the flip-flops were reset to 0 at power on, then the total number of different output states represented by the counter ABC is equal to-


So it requires 4 states

*Answer can only contain numeric values

Consider a silicon PN junction at T = 300K with current density 3.6 10-11 A/cm2. If the photocurrent density is 10 mA/cm2. Then find the open-circuit voltage (in V) of a solar cell when the solar intensity is increased by a factor of

Assume thermal voltage VT = 26 mV

Solution: Given photo current density of laser is JL = 10mA/cm2

Current density JS = 3.6 × 10−11A/cm2since the solar intensity is increased by 10 times

JL = 10 × 10mA/cm2 = 100mA/cm2The open-circuit voltage of solar cell is


DTFT of x(n) is X(ω). Hence x(n) = (0.5)nu(n). Then the time signal corresponding to X2(ω) is



A varying magnetic flux linking a coil is φ=1/3 λt3. If the value of λ = −1Wb/s2, then at what time the induced EMF is 9V?



Determine length of a cavity resonator(rectangular) if it has dimensions a = 3cm, b = 2cm, and is excited by TE101 mode at 30GHz frequency.


Given: f = 30GHz = 30 × 109Hz

We have TE101 mode, which means: m = 1,n = 0 and p = 1 Also, given 1 = 3cm = 0.03m and b = 2cm = 0.02m.

Taking square on both sides, we get:

Now, we have a = 0.03m. So, on putting the values and solving for 'd' we get:

(1/d)2 = 38888.89 ⇒ d = 5.07 x 10-3 m

*Answer can only contain numeric values

If directive gain of a half-wave dipole is 2.64 W/W, and ohmic losses is 8Ω, then determine its power gain(in dB).

Solution: We know that the radiation resistance for a half-wave dipole is given as

Rrad = 73Ω

Now, antenna efficiency is given as:

ƞr = (Rrad) / (Rrad + Rloss) = 73 / (73 + 8) = 0.901.

Also, we know that efficiency of antenna can also be written as:

ƞr = Gp/ Gd = Power gain / directive gain

Now, we have, ƞr = 0.901 and directive gain Gd = 2.64

So, power gain Gp can be given as:

0.901 = Gp/ (2.64)

So, Gp = 2.37

Gp(in db) can be give as : Gp(dB)= 10log10 Gp = 10log(2.37)= 3.74 dB.

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