Electronics And Communication - ECE 2022 GATE Paper (Practice Test) - Electronics and Communication Engineering (ECE) MCQ

Where x is a common multiplication factor.

As R gets Rs.1000 more than S.

Then, 4x - 3x = 1000

x = 1000

Now, share of Q is 2x = 2 ´1000 = Rs.2000

Share of Q is Rs.2000.

Hence, the correct option is (C).

There is a Trapezium PQRS,

Length of PQ = 11 cm

Length of QR = 4 cm

Length of PS = 3 cm

Length of RS = 6 cm

Let us consider, a line segment RA which is parallel to PS.

Now, length of side of AQ = 5 cm

PA || SR , PS || AR

So, length of side AR = 3 cm

Length of side RQ = 4 cm

Length of side AQ = 5 cm (Sides of ΔARQ follows the property of Pythagoras theorem)

Let us consider a perpendicular RT which is shortest path of PQ and RS

AQ x RT = 12 sq. cm

5 x RT = 12 sq. cm

RT = 2.4 cm

Hence, the correct option is (D).

The maximum number of square without a “hole in the interior” that can be formed within the 4 x 4 grid can be counted as.

(ABIJ, BCHI, CDGH, DEFG, FONG, HMLI, ILKJ, KLST, LMRS, MNQR, NOPQ, PUVQ, QVWR,

RWSX, SXYT, ACMK, JHRT, KMWY, LNVX, MOUW)

Total number of squares are 20.

Hence, the correct option is (D).

The art gallery engages a security guard to ensure that the items displayed are protected.

The location where the security guard posted is identified such that all the inner space of the gallery is within the line of sight of the security guard.

The condition is if the security guard does not move around the posted location and has 360⁰ view,

The set of all possible location among the locations P, Q, R and S are Q and S from where the security guard can be posted to watch over the entire inner space of the gallery

Hence, the current option is (B).

Option (B), in this option, it is clearly mentioned in given passage that “using chemical to kill mosquitoes may have undesired consequences but it is not clear if using genetically modified mosquitoes has any negative consequence” so this option is correct.

Option (C), in this option, it is mentioned “genetic engineering is dangerous” which is not mentioned so this option is also eliminated.

Option (D), in this option, it is mentioned that “both using genetically modified mosquitoes and chemicals have undesired consequence and can be dangerous” which is not related to given passage so this option is also eliminated. Hence, the correct option is (B).

2x - 1 > 7

2x > 8

x > 4

And second Inequality, 2x - 9 < 1

2x - 9 < 1

2x < 10

x < 5

By these two Inequalities we have value of x as 4 < x < 5.

Hence, the correct option is (D).

On plotting points on graph,

According to the given data a Quadrilateral PSQR will form as shown in below figure,

The distance formula between two points =

We can see clearly,

Length of diagonal

So, Quadrilateral PSQR follows property of square.

Area of Quadrilateral PSQR = Side²

Area of Quadrilateral PSQR = (√8)² = 8 Sq. unit

and also by,

Area of Rhombus = 1/2(product of both diagonals)

= 1/2(4 x 4) Sq. unit = 8 Sq. unit

Hence, the correct option is (C).

The recorded statement from students

Statement of P : R has copied in the exam.

Statement of Q : S has copied in the exam.

Statement of R : P did not copy in the exam.

Statement of S : only one of us is telling the truth.

Statement of T : R is telling the truth.

The investigating team have authentic information that S never lies.

S says only one of us is tilling the truth.

Hence, statement of all other students will be wrong.

Statement of P: R copied in the exam.

Statement of Q : S copied in the exam.

Statement of R : P copied in the exam.

Statement of T : P copied in the exam.

So, P copied in the exam.

Hence, the correct option is (B).

X : Rotation of the square by 180 degree with respect to the S-Q axis

Y : Rotation of the square by 180 degree with respect to the P-R axis

Z : Rotation of the square by 90 degree clockwise with respect to the axis perpendicular going into the screen and passing through the point T.

Following three distinct sequences of operation are as follows.

As per the information and operations sequence of operation 1 and 3 are equivalent.

Hence, the correct option is (C).

Method 1

We have a closed contour so we will use stokes theorem i.e.,

Here, F_x = x, F_y = y, F_z = 0.

Hence, the correct option is (A)

Method 2

Applying green’s theorem,

M = x, N = y

From equation (i),

R₂ → R₂ + R₁ ,

We can see that, of the above augmented matrix.

Matrix A has only one linearly independent row, ρ( A) = 1

and matrix [ A :b] has two linearly independent rows

ρ( A:b) = 2

Here, ρ( A) ≠ ρ( A: b)

So, system has no solution for x.

Hence, the correct option is (B).

Method 1

Applying KVL in loop-1 shown in above figure,

-5 + 2000I + 2000(I + 10^-3) = 0

5 = 2000I + 2000(I + 10^-3)

5 = 2000I + 2000I + 2

4000I = 5 - 2

4000I = 3

Hence, the correct option is (B).

Method 2

Assuming voltage V at node x as shown below,

Applying KCL at node x,

2V - 5 = 2

2V = 5 + 2 = 7

V = 3.5V

From figure,

I = 0.75 x 10^-3 A

Hence, the correct option is (B).

Apply KVL in loop-1,

2I₁ + 3 + I₁ = 0

I₁ = -1A

I₁ can flow in loop-1 only, so current I need to be zero. Otherwise it violets the KCL.

So, the value of I = 0 .

Hence, the correct option is (C).

We know that,

Put a = 1 on both side,

From multiplication by ' t' property of Fourier transform,

From duality property,

Hence, the correct option is (B).

(i) p-type semiconductor

(ii) Equilibrium hole density, p₀ = 10¹⁷ cm^-3

(iii) Intrinsic carrier concentration, n_i = 10¹⁰ cm^-3

(iv) Electron diffusion length, L_n = 2μm

(v) Hole diffusion length, L_p = 1μm

(vi) Excess electron density at x = 0 , n_p (0) = 10¹⁴ cm^-3

Consider a bar of direct bandgap p-type semiconductor with light is illuminated at left side of the bar at x = 0 .

From Mass action law, np = n_i² Electron or minority carrier concentration is given by,

Electron concentration at any distance ' x' is given by,

Electron concentration (density) at x = 2μm is,

n_p (x = 2) = 10³ +10¹⁴ e^-2/2

n_p (x = 2) = 10³ +10¹⁴ e^-1

n_p (x = 2) = 10³ + 10¹⁴ x 0.37

n_p (x = 2) ≅ 0.37 x 10¹⁴ cm^-3

Hence, the correct option is (C).

(i) Electron density, n₁ = 10¹⁶ cm^-3

(ii) (E_C - E_Fn)₁ = 200 meV

(iii) Intrinsic carrier concentration, n_i = 10¹⁰ cm^-3

(iv) We have to find ( E_c -E_Fn)₂ for electron density, n₂ = 0.5 x 10¹⁶ cm^-3

Since, electron density for n - type semiconductor is given by,

Where, n = Electron density, N_c = Effective density of states at the edge of conduction band.

Dividing equations (i) and (ii),

(E_C - E_Fn)₂ - 200 x 10^-3 = 26 In 2

(E_C - E_Fn)₂ = 26 In 2 + 200

(E_C - E_Fn)₂ = 218 meV

Hence, the correct option is (D).

(i) Both transistors have same gate oxide capacitance

(ii) V_Thp = -1 V

(iii) V_ThN = 1 V

For n MOSFET :

Condition for saturation is given by,

If this inequality is satisfied then n MOSFET will be operate in saturation region.

From given circuit,

Therefore n MOSFET is operated in saturation region.

For p MOSFET :

Condition for saturation is given by

If this inequality is satisfied then p MOSFET will be operate in saturation region.

From given circuit,

Therefore p MOSFET is operated in saturation region.

We can say that, whenever gate and drain terminal of n MOSFET and p MOSFET connected together, both MOSFET will be operated in saturation region.

Since, current flow through gate terminal of MOSFETs is zero, hence both MOSFET will have same drain current.

300(V₀ - 1)² = 40 x 5(4 - V₀ - 1)²

3(V₀ - 1)² = 2(3 - V₀)²

3V²₀ + 3 - 6V₀ = 18 + 2V²₀ - 12V₀

V²₀ + 6V₀ - 15 = 0

V₀ = 1.89V,- 7.89 V

Taking negative value of V₀ .

V₀ = -7.89 V

For n MOSFET :

Hence, n MOSFET will be OFF and drain current I_DN = 0

Since, drain current through n MOSFET is zero so we do not need to check whether p MOSFET is ON or OFF.

Taking positive value of V₀ .

V₀ = 1.89 V

For n MOSFET :

Hence, n MOSFET will be ON for V₀ = 1.89 V

Now we will check n MOSFET is operate either in saturation or linear region

The condition for saturation is given by,

1.89V ≥ 1.89 - 1V

1.89 ≥ 0.89

Therefore, n MOSFET is operates in saturation region for V₀ = 1.89 V

For p MOSFET :

Hence, p MOSFET will be ON for V₀ = 1.89 V

Now we will check p MOSFET is operate either in saturation or in linear region. V₀ = 1.89 V

The conditions for saturation is given by,

2.11V ≥ 2.11V - 1V

2.11 ≥ 1.11

Therefore, p MOSFET is operates in saturation region for V₀ = 1.89 V .

Hence, the correct option is (D).

Output ( F) of given 4 x 1 MUX is,

Here, S₁ = C and S₀ = D

To make we have to take

A₀ = 0 , A₁ = 1, A₂ = 1 and A₃ = 0

Hence, the correct option is (C).

(i) V_Tn = 1 V

(ii) V_Tp = -1 V

In NMOS transistor, gate voltage (V_G) work as a controlled input. When NMOS work as a switch then,

If V_G = 0 V, then NMOS will be OFF and if V_G = +5 V , then NMOS will be ON.

(i) For V_DS < />_GS - V_Tn

V_D - V_S < />_G - V_S - V_Tn

V_D - V_G - V_Tn

When NMOS Will satisfy above equation, then NMOS will be in linear region

and hence it will be ON.

Therefore, V_D = V_S

(ii) For V_DS ≥ V_GS- V_Tn

V_D - V_S ≥ v_GS - V_Tn

V_D ≥ V_G - V_Tn

When NMOS will satisfy above equation, then NMOS will be in saturation region and hence it will be ON.

Therefore, V_D = V_S = V_G - V_Tn

V_S = V_G - V_Tn

For NMOS pass transistor logic

If V_D - V_G ≥ -V_Tn

Then, V_G - V_S = V_Tn

If V_D - V_G < -="" />_Tn

Then, V_S = V_D

V_D = 5 V

V_G = 5V

V_S = V₁

V_D - V_G = 5 - 5 = 0

V_D - V_G ≥ -V_Tn

0 ≥ -1 [True]

Hence, V_G - V_S = V_Tn

5 - V₁ = 1

5 - 1 = V₁

V₁ = 4 V

For PMOS pass transistor logic,

If V_G - V_D ≥ - |V_Tp|

Then, V_S - V_G = -|V_Tp|

If V_G - V_D < />_Tp|

Then, V_S = V_D

V_G = -5 V

V_D = 5 V

V_S = V₂

V_G - V_D = -5 -5 = -10

V_G - V_D < />_Tp|

-10 < -14="" />

Hence, V_S = V_D

V₂ = 5 V

Hence, the correct option is (B).

Nyquist plot of G(s) is shown below,

Given, G(s) has no poles in the right half of s – plane i.e. P = 0 .

Modified closed loop control system is shown below,

Modified open loop transfer function is KG(s)

K = 2 (given)

The Nyquist plot for K = 2 is,

Number of anticlockwise encirclement about critical point (-1+ j0)

N = -2

Number of right hand poles of open loop transfer function P = 0 .

Nyquist stability criteria is given by,

N = P- Z (A.C.W.)

-2 = 0 - Z

Z = 2

Hence, the number of closed loop poles in the right half of complex plane is 2.

Hence, the correct option is (C).

Form above figure, at s = -1 , five root locus branches are originating following the asymptotic angles 36^o,108^o,180^o, 252^o , 324^o .

Hence, there will be five poles at s = -1

Since, all the five root locus branches are terminating at infinity, hence there will be five virtual zeros i.e. no physical zero.

Therefore OLTF is given by,

Hence, the correct option is (D).

Let x(t) be the input and y(t) be the output of a purely delay system.

So, y(t ) = x(t- t₀ ) where t₀ is the time delay.

Taking Fourier transform on both sides we get.

|H(f)| = 1 for all frequencies ……(i)

From the given frequency response it can be concluded that

|H(f)| = 1 for - α ≤ f ≤ α …………(ii)

= 0 otherwise

On comparing (i) and (ii) we can conclude that the given system will not act as a purely delay system.

Thus, statement 1 is incorrect.

The output power spectral density of an L.T.I. system is related to input power spectral density as

S_Y(f) = |H(f)|² S_X(f)

But |H(f )| = 1 for - α ≤ f ≤ α

S_Y(f) = (1)² x S_X(f) for - α ≤ f ≤ α

S_Y(f) = S_X(f) for - α ≤ f ≤ α

So, statement 2 is correct.

Hence, the correct option is (B).

Total current in capacitor, I = I_C + I_D

Where, outside capacitor plates we have only conduction current I_C and no displacement current. On the other hand inside the capacitor there is no conduction current i.e. I_C = 0 and there is only displacement current.

So, I_conduction = I_displacement

∵ E = σ/ε₀ [For the parallel plate capacitor]

I_D = I_resistor

So, they will be in-phase, resistor and capacitor are in series so both current will be equal. Hence, the correct option is (D).

Differentiating two-times equation (ii) with respect to x,

Multiply a on both sides,

Differentiating two-times equation (ii) with respect to y,

Multiply b on both sides,

Adding equation (iii) and (iv),

Compare equation (i) and (v),

aε² +bη² = 1 …..(vi)

Equation (vi) must be satisfied if

Hence, the correct options are (A) & (C).

The transfer function of above circuit,

Put s = jω ,

At ω = 0, T (0) = 0

At ω = ∞, T(∞) = R₂/R₁

So, circuit behave as HPF, because it passes only high frequency and block low frequency. The 3 dB frequency of practical differentiator circuit is,

Hence, the correct options are (C) & (D).

Given : (x + yz)

Method 1

Choosing from options :

From option (A) :

x + xz+ xy

x(1+ z+ y) (∵ 1+ Anything = 1)

So, option (A) is incorrect.

From option (B) :

x + z+ xy

x(1+ y)+ z (∵ 1+ Anything=1)

x +z

So, option (B) is incorrect.

From option (C) :

(x + y) (x+ z)

x + xz + yx + yz

x(1+ z+ y) (∵ 1+ Anything = 1)

x + yz

So, option (C) is correct.

From option (D) :

x + xy+ yz

x(1+ y)+ yz

x + yz ( ∵ 1+ Anything = 1)

So, option (D) is correct.

Hence, the correct options are (C) & (D).

Method 2

Given : (x + yz) …..(i)

From Boolean algebra, A + BC = ( A + B)( A + C)

So, equation (i) can write as,

x + yz = (x + y)(x + z)

Again from Boolean algebra, A + BC = A + AB + BC

So, equation (i) can write as,

x + yz = x + xy + yz

Hence, the correct options are (C) & (D).

For option (A) :

Noise margin high ( NM_H) and Noise margin low ( NM_L) is depends on the following parameters of MOSFET.

(i) Size of n MOSFET and p MOSFET

(ii) Threshold voltage of n MOSFET and p MOSFET.

Noise Margin High is equal to Noise Margin Low if

If

So, option (A) is incorrect.

For option (B) :

Mobility is the ability of charge carrier to move freely or be easily moved.

The mobility of electrons influences the switching speed because propagation delay (τ) depends on mobility.

Propagation delay,

If mobility of electrons (μ_n) increases, τ_PHL decreases. Therefore propagation delay (τ) decrease. So, option (B) is incorrect.

For option (C) :

The transfer characteristic of practical CMOS inverter is given by,

For a logical high input under steady state, the n MOSFET is in the linear regime of operation. So, option (C) is correct.

For option (D) :

Power dissipation in CMOS inverter :

(i) Static power :

(a) Static power exists due to leakage current in stable state or steady state of CMOS inverter.

(b) P_D(static) = V_DD x I_leakage

where , I_leakage = Leakage_(NMOS) + I_{leakage(PMOS)}

(c) Static power is the power consumed by the MOSFET during stable state i.e. it is the power consumed by MOSFET during continuously ON at logic 1 and continuously OFF at logic 0.

(ii) Dynamic capacitive power :

(a) It is the power consumed by the MOSFET during its transition state i.e. power consumed by the MOSFET during logic 1 to logic 0 or logic 0 to logic 1.

(b) It depends on charging and discharging of capacitive load.

(c) It is also referred as switching power dissipation.

(d) Energy stored in PMOS,

Energy stored in NMOS,

Energy stored in CMOS,

Dynamic power dissipation of CMOS inverter,

f_SW = Switching frequency

f_SW = αf

Where, α = Activity factor ( 0 < α="" /><1 )="" and="" f="operating">

If α is not mentioned then we assume α = 1.

Therefore, dynamic power consumption during switching is non-zero.

So, option (D) is incorrect.

Hence, the correct option is (C).

H (X) is entropy of a discrete random variable X taking K possible distinct real values.

Let variable X is taking values as x_i so set of possible values is {x₁, x₂ , x₃.......x_k } .

The entropy will be,

Case-1 : When all values are equiprobable i.e. P(x_i) = 1/k for each distinct ' K' values, then entropy will be given as,

H ( X ) = log K

When we choose base as 2,

H ( X ) = log₂ K bits

We know that, in case of equal probability the entropy will be highest,

So, H(X) ≤ log₂ K is always true for any base value.

Hence, option (A) is correct.

Case-2 : Assuming a new random variable Y = 2X which is mapped by random variable X as shown below,

So, mapping will be one to one as we have linear relation between X and Y.

The random variable X = {x₁, x₂ , x₃.......x_k } in the same way random variable Y

Y = {y₁ = 2x₁, y₂ = 2x₂ , .......... y_K = 2x_K }

Along with values the probability of occurring each value will also be mapped and hence x_i and corresponding y_i will have identical probability, i. e. P(x_i) = P(y_i)

We can say that P( x₁ ) = P(y₁), P(x₂) = P(y₂)..........P(x_K) = P(y_K)

Probability of random variable X and Y are same, because one to one mapping therefore, entropy of random variable X and Y are same. i. e. H (Y = 2X ) = H ( X )

So we will never have situation of H ( X ) < h="" (2x="" />

Hence, option (B) is correct

Case-3 : Assume new random variable Y = 2^X , This Y = 2^X will give one to one mapping because different values of X provide different values of Y so that probability of random variable Y is same as probabilities of random variable X so that entropy

H(Y = 2^X) = H(X)

So, H ( X ) < h="" />^X ) never occur

So, option (C) is also correct.

Case-4 : Assuming new random variable Y =X² .

Possibility-01 : Suppose we have 3 positive values of random variable X is x₁ = 1 , x₂ = 2, x₃ = 3 and assuming corresponding probabilities of x₁, x₂, x₃ are respectively

i.e. X = {1, 2, 3}

Probability,

Thus, random variable, Y =X²= {1, 4, 9}

Here we can say different values of X gives different values of Y, it shows one to one mapping is possible between X and Y.

So, probabilities of random variable Y is same as probability of random variable X.

It means,

Thus we can say that entropy of H(X) = H(Y= X² )

Possibility-02 : Suppose we have 3 negative values of random variable X is x₁ = -1 , x₂ = -2 , x₃ = -3 and assuming corresponding probabilities of x₁, x₂, x₃ are respectively

i.e. X = {-1, - 2, - 3}

Probability,

Thus, random variable, Y =X²= {1, 4, 9}

Here we can say different values of X gives different values of Y, it shows one to one mapping is possible between X and Y.

So, probabilities of random variable Y is same as probability of random variable X,

Thus we can say that entropy of H(X) = H(Y= X²) .

Possibility-03 : Suppose we have both positive and negative values of random varible X is x₁ = -1 , x₂ = 1, x₃ = 2 and assuming corresponding probabilities of x₁ , x₂ and x₃ are 1/5, 1/2 and 3/10.

i.e. X = {-1, 1, 2}

Thus random variable, Y =X²= {1, 1, 4}

Here we see different values of X gives same value of Y, it show one to one mapping is not possible here.

So, random variable Y have only 2-values i.e. Y = [1, 4]

So, probability of P(y₁ = 1) is sum of probability of P(x₁ =-1) and P(x₂ = 1) so, P(y₁ = 1) = 1/5 + 1/2 = 7/10

Probability of P(y₂ = 4) remains same as probability of P(x₃ = 2) .

i.e. P(y₂ = 4) = P(x₃ = 2) = 3/10

So entropy of H (Y) is

Entropy of H (X) is,

= 0.4643 + 0.5 + 0.521089

=1.4854

From Case-4, it is clear that H (X) = H (X)² is possible only when random variable X have all positive or negative values and X can take combination of positive and negative values then H (X) > H (X²) that why option (D) is incorrect.

Hence, the correct options are (A), (B) & (C).

where f (x, t) is associated electric or magnetic field associated with wave E(x, t) or H (x, t) and wave is traveling in x direction.

Given options are the wave equation in general solution form, would satisfy its partial differential equation.

Choosing from options :

Hence, the correct options are (A) & (B).