Electronics And Communication - ECE 2021 GATE Paper (Practice Test) - Electronics and Communication Engineering (ECE) MCQ

Squaring on both sides, we get

Hence, the correct option is (B).

Question_Type:

Current population of city = 1102500

Rate of increment of population = 5% per year

Method 1 :

Let the population of the city 2 years ago be x . As the population has been increasing at the rate of 5% per annum, so after 2 years x becomes as,

( x × 105%) × 105% = 1102500

x = 10,00,000

Hence, the correct option is (C).

Method 2 :

Suppose 2 years ago population of city was “100”

Hence, 2 years ago population was 100 ×10000= 1000000

Hence, the correct option is (C).

Question_Type:

∴ Sides of Hexagon formed by an equilateral triangle to the same equilateral triangle = 1:3

∴ The ratio of the area of the regular convex Hexagon (PQRSTU) to the area of original equilateral triangle is

⇒ 2 : 3

Hence, the correct option is (A).

Question_Type:

It is cut along the main diagonal to get two triangles

One of the cut triangle is revolved about its short edge to form solid cone

Hence, the correct option is (A).

Question_Type:

Anticipation : Expectation or prediction to future.

Hence, the correct option is (C).

Question_Type:

Except Thursday, there are total 6 days in which one of the students spent a minimum of 10% more than the other student.

Hence, the correct option is (C).

Hence, the correct option is (C).

Both the conclusion are wrong by can’t say condition and variable of the conclusion are same. Also one conclusion is positive and another is negative. Hence they full fill the condition of “either or” case. Hence, the correct option is (C).

The signal flow graph is shown below,

Number of forward paths and its associate path factor is, 11

P₁=G₁ Δ₁= 1

P₂ =G₂ Δ ₂= 1

Number of individual loops, 11

L₁ =−G₁H

Determinant of SFG is,

Δ=1−(L₁ ) =1−(−G₁H) =1+G₁ H

From the Mason’s gain formula the transfer function of this system can be given as

∴

Hence, the correct option is (C).

V_C (0− ) =− 5 V

When switch is closed and circuit is in steady state (i.e. t =∞, and capacitor becomes open circuit) So circuit becomes as,

Apply KCL at node V_C (∞) is,

Vc(∞) 5\3 V → Steady state voltage of capacitor 3

Now, for calculating time constant ( τ ), first we calculate Rth as shown below,

Apply KCL at node N,

Thus, time constant, て=R_th ×C

Apply transient equation for voltage across capacitor is,

If V_C(t)=0, then equations (i) becomes as,

Hence, the correct answer of t is 0.1386 msec.

Queston_Type: 4

According to question, |V_IN| < />_CC| and |V_REF| < />_CC|

Here, Op-Amp is ideal and -ve feedback is present so virtual ground concept is applicable.

So, V₊V_-=0

Thus applying KCL at node N,

According to question, V_out = 0

Hence, the correct answer is option (A).

The given circuit is shown below,

Above figure can be redrawn as

From above circuit it is clear that,

V− =V_IN= + 5 V

V₊= 0 V

Here, V₋ ≠V₊

It means, voltage at both terminals of op-amp are unequal and fixed so virtual ground concept is not valid here and op-amp work as a comparator.

Thus, V₋ >V₊

So, V_out =−V_sat

V_out =−12 V

Hence, the output voltage V_out of this circuit at t = 0⁺ is −12 V.

Question_Type: 4

Given :

(i) 1V_out =1V when V_in= 0.1 V

(ii) V_out= 6V when V_in= 1V

By voltage divider rule, voltage at non-inverting terminal can be found as

Here virtual ground concept is applicable because –ve feedback is present and op-amp is ideal so due to virtual ground concept,

Applying KCL at V− ,

Condition 1 : If V_out = 1 V when V_in = 0.1 V then equestion (i) becomes as

Condition 2 : If V_out 6 V when V_in =1 V then equation (ii) becomes as

Note : As per the given data of question we get negative answer so none of the option is

matched.

From questions it is clear that, change in energy level / difference in energy level of E_C and E_V is Δ .

Hence, equation (i) becomes as

Hence, the correct option is (A).

Boron concentration N_A = 10¹⁶ cm_-3

So, N_A = p_po =10¹⁶cm^-3

Generation Rate G_p = 10²⁰ cm^-3s^-1

Recombination life time て_p = 100 μ sec

Intrinsic carrier concentration n₁ =10¹⁰ cm^-3

Using mass action law under thermal equilibrium

So minority carrier concentration when light is not exposed is 10⁴ cm^-3 When light exposed,

So after exposing light, excess carrier concentration is

So after exposing light, under equilibrium,

1. Electrons concentration :

2. Hole concentration :

Thus, product of hole concentration and electron concentration after exposing the light is

Hence, the correct option is (D).

Given

Doping in P-region N_A =5*10¹⁶ cm^-3

Doping in N-region N_D =10¹⁷ cm^-3

Given p-n junction is shown below,

So width of depletion region, W = X_n + X_p

From charge equality concept,

Suppose when N-is depleted completely, during reverse bias so,

i.e. when N is depleted completely (i.e. 0.2 μm ) then and only then P side is depleted only by 0.4 μm only. Thus

So, total depletion width (W) under reverse bias is,

VR = 8.2395 V≈ 8.24 V

Thus, the magnitude of reverse bias voltage that would completely deplete N region is 8.24 V. Hence, the correct answer for V_R is 8.24 V .

Question_Type: 4

Assume, i_L(t) = x₁ → First state variable

V_C (t) = x₂ → Second state variable

i_s (t) = u → Input of system

and V₀ (t) = y → Output of system

Apply KCL at node N,

KVL in the loop between inductor and resistor,

From equation (i), (ii) and (iii), state variable modal can be written as,

Thus matrices A, B, and C are,

Condition of controllability,

Thus, system is uncontrollable.

Condition of observability,

Thus, system is un-observable.

So, the system is neither controllable nor observable.

Hence, the correct option is (D).

Given : Directive gain

Radiation power P_rad =16kWrad

Distance r = 8km

Directive gain G_d (dB)=6=10 log G_d

G_d = 10^0.6= 3.981

So maximum electric field amplitude in free space at distance r from antenna is,

Hence, the correct answer for

Question_Type: 4

Cylinder (ρ =3, 0≤ z ≤ 2) is shown below,

From this given cylinder,

ρ = 3m, z= 0 to 2 m, φ = 0to 2π

Shaded portion shows closed surface of cylinder and unit vector that is perpendicular this surface is thus, net flux leaving the closed surface of cylinder is

Hence, total flux ψ living closed surface of the given cylinder as 56.548 C.

Question_Type: 4

So,

Assume, x =r cosθ, y = r sin θ

r = 1

x = cosθ and y = sinθ

Here, θ , Variation from 0 to 45⁰ .

Put x = cosθ,y = sin θ

dx =− sin θ dθ

dy = cosθ dθ

Thus equation (i) becomes as

Hence, the correct option is (A).

For a vector field to be irrotational its curl must be zero,

It shows because it is unit vector.

Thus, c₁ = 0 then the above equation will be satisfied. Hence, the correct answer is zero.

Question_Type: 4

Given

Refractive index of core n₁ = 1.50

Refractive index of cladding n₂ = 1.48

According to the definition of critical propagation angle given in statement of question, θ₁, which is made with optical fiber axis and leads total internal reflection (TIR) in optical fiber cable, so we can calculate θ₁ as,

Here, θ_A Acceptance angle

θ_C= Critical angle

θ₁= Critical propagation angle

Where, n₁ = refractive index of core = 1.5 2

n₂= refractive index of cladding = 1.48

Hence, the value of critical propagation angle that leads total internal reflection in optical fiber is 9.36⁰

Question_Type: 4

(i) Medium 1 is free space.

(ii) μ₁ = μ₀ , ε₁= ε₀

(iii) E₁ = 10 cos (2×10⁸ t + βz)

(iv)

(v)

Medium 2 ( z < />

Hence, the correct option is (A).

Lossless line characteristic impedance Z₀ = 50Ω

Open circuit stub characteristic impedance ( Z₁)oc = 75Ω

Short circuit stub characteristic impedance ( Z₁)sc = 75Ω

Method 1 : The given arrangement of transmission line is shown below,

Input Impedance of Λ\2 long line-1,

Thus total impedance at terminal QR is,

Z_QR = Z_L || (Z_in)_Line-1= Z_L || ∞= Z_L

Now arrangement of Transmission Line becomes as,

Thus Input Impedance of line 2 is,

Input Impedance of Line 3 is,

Thus total impedance at terminal PS is,

Z_PS = (Z_in)_Line-2 || (Z_in)_Line-3

Now Transmission Line arrangement becomes as-

Hence Z_PS work as load for main Transmission line.

For matching of main Transmission line with load ( Z_PS) ,

Hence, the correct option is (A)

Method 2 :

From given arrangement, it is clear that,

So, input impedance of both line-1 and line-3 are ∞ (i.e. open circuit) so it does not make any effect on main transmission line, so given transmission line configuration becomes as,

Thus Input Impedance at terminal PS is,

So ( Z_in)_PS work as load for main transmission line so arrangement of transmission line becomes as,

Hence ( Z_in)_PS work as load for main Transmission line so the condition, for matching the main Transmission line with load ( Z_in)_PS is

Hence, the correct option is (A)

Given circuit is shown below,

Re-arrange the above circuit as shown below,

Now above circuit reduces as,

Above circuit can be re-arranged as,

Apply current divider rule

Hence, the correct answer is 0.5 A.

Question_Type: 4

Case 1 :

Here, G(s)H(s) has one zero in the right half of s-plane i.e. Z = 1 .

According to principle of argument

N =P− Z (Anti clockwise direction) …(i)

Where,

N = Number of encirclements of Nyquist plot of G(s)H(s) about origin in anti-clockwise direction P = Number of poles of G(s)H(s) in right half of s-plane.

Z = Number of zeros of G(s)H(s) in right half of s-plane.

Here, N =−2 (anti clockwise direction)

Z = 1

From equation (i)

−2= P − 1

P = –1 …(ii)

So, poles cannot be negative in numbers so no more further discussion on this.

Case 2 : If we assume Nyquist contour in anti clockwise direction then according to principle of argument

N =P− Z (Clockwise direction) …(iii)

Where,

N = Number of encirclements of Nyquist plot of G(s)H(s) about origin in clockwise direction

P = Number of poles of G(s)H(s) in right half of s-plane.

Z = Number of zeros of G(s)H(s) in right half of s-plane.

Here, N = 2 (Clockwise direction)

Z = 1

From equation (iii)

2 = P−1

P = 3 …(iv)

According to Nyquist stability criteria

N =P− Z (Anti clockwise direction) …(v)

Where, N = Number of encirclements about (−1+ 0 j ) in clockwise direction

Z = Number of closed loop poles in right half of s-plane.

P = Number of open loop poles or poles of G(s)H(s) in right half of s-plane.

From given Nyquist plot number of encirclements about (−1+ 0 j ) is

N = 1−1= 0

Using equation (v)

0 = P−Z

Z = P

Z = 3

So here number of poles in right half of s-plane is 3.

Hence, the correct option is (D).

Given : Number of bits, n = 8,

Full scale voltage,V_FS= 7.68

Thus, analog output from 8 bit unipolar DAC is,

V_out = (Resolution)× (Decimal equivalent of digital input )

Hence, the analog output voltage of the 8 bit unipolar DAC is 4.517 Volt

Question_Type: 4

Delay of XOR gate = 4ns

Delay of AND gate = 2 ns

Delay of MUX = 1ns

Case 1 :

Assuming T = 0 then selection line of MUX S₀ = 0 , so that MUX input ‘0’ get enable so path followed by signal in the given circuit is shown by dotted lines as,

So total propagation delay τ₂ from input to output is,

τ₂₌ = (Propagation delay of AND gate) + (Propagation delay of MUX-1) +

(Propagation delay of AND gate) + (Propagation delay of MUX-2)

τ₂= 2ns +1ns + 2ns + 1ns = 6ns

Hence MUX input '1' get enable then propagation delay of given circuit τ₂= 6ns

Hence maximum delay of circuit is MAX ( τ₁ ,τ₂) = MAX ( 3ns, 6ns)= 6 ns

Hence, the correct option is (C).