GATE EC 2021 Question Paper with Answer Key & Detailed Solutions

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GATE - EC 2021 Answers with Explanations 

Q.1

Current population 11,02,500

increasing 5% per annum

Let assume 2 years age, the population were, x

then, x*(1+5/100)^2 = 11,02,500

Or, x = 1000000

Option C

Q.2

p/q + q/p = 3

Or, (p/q)^2 + (q/p)^2 = (p/q + q/p)^2 - 2

Or, (p/q)^2 + (q/p)^2 = 9-2=7

Option B

Q.3

Option C

Q.4

Option C

Q.5

Option B

Q.6

Option C

Q.7

Option D

Q.8

Total Volume = 1/3*pi*r^2*l = 1/3*pi*1*1 = pi/3

Option A

Q.9

Option C

Q.10

The area of the triangle is  1/2*3x*3x = 9/2*x^2

The area of the regular convex hexagon is  9/2*x^2 - 3*1/2*x*x = 6/2*x^2

Or, area of hexagon: triangle = 2:3

Option D

Electronics and Communication (EC)

Q.1

$ \mathbf{F}(x,y)=-x\hat{i}+y\hat{j} $

$$ \int_C \mathbf{F}\cdot d\mathbf{r} $$

$$ =\int_C(-x\,dx+y\,dy) $$

Parameterization: $$ x=\cos t,\qquad y=\sin t $$ $$ dx=-\sin t\,dt,\qquad dy=\cos t\,dt $$

$$ \int_{0}^{\pi/4} \left[ (-\cos t)(-\sin t) + (\sin t)(\cos t) \right]dt $$

$$ = \int_{0}^{\pi/4} 2\sin t\cos t\,dt $$

where $$ 0\le t\le\frac{\pi}{4} $$

$$ 2\sin t\cos t=\sin2t $$ $$ =\int_{0}^{\pi/4}\sin2t\,dt $$

$$ =\left[-\frac12\cos2t\right]_{0}^{\pi/4} $$ $$ =-\frac12\cos\frac{\pi}{2} +\frac12\cos0 $$ $$ =0+\frac12 =\frac12 $$

$$ \boxed{\int_C \mathbf{F}\cdot d\mathbf{r}=\frac12} $$

Option A

Q.2

Find the integrating factor of the differential equation $$ \frac{dy}{dx} + \frac{x}{1-x^2}y = x\sqrt{y} $$

Solution

The given differential equation is $$ \frac{dy}{dx} + \frac{x}{1-x^2}y = x\sqrt{y} $$ Since the equation contains the term $\sqrt{y}$, it is not linear.

Let $$ v=\sqrt{y} $$ Then, $$ y=v^2 $$ Differentiating, $$ \frac{dy}{dx} = 2v\frac{dv}{dx} $$

Substitute these values into the given equation: $$ 2v\frac{dv}{dx} + \frac{x}{1-x^2}v^2 = xv $$

Divide both sides by $2v$: $$ \frac{dv}{dx} + \frac{x}{2(1-x^2)}v = \frac{x}{2} $$ This is now a linear differential equation of the form $$ \frac{dv}{dx}+P(x)v=Q(x) $$ where $$ P(x)= \frac{x}{2(1-x^2)}. $$

The integrating factor is $$ \text{I.F.} = e^{\int P(x)\,dx} = e^{\int \frac{x}{2(1-x^2)}dx} $$

Let $$ u=1-x^2 $$ Then, $$ du=-2x\,dx $$ Therefore, $$ \int \frac{x}{2(1-x^2)}dx = -\frac14 \int \frac{du}{u} $$

Hence, $$ = -\frac14\ln|1-x^2| $$

Therefore, $$ \text{I.F.} = e^{-\frac14\ln(1-x^2)} $$ Using the identity $$ e^{\ln a}=a, $$ we obtain $$ \boxed{ \text{I.F.} = (1-x^2)^{-1/4} } $$

Hence, the correct answer is $$ \boxed{(1-x^2)^{-1/4}} $$ which corresponds to Option (B).

Q.3

Var(Y) = E(Y^2) - (E(Y))^2

= E((2X+3)^2) - (E(2X+3))^2

=4* {E(X^2) - (E(X)^2)}

or, Var(Y) = 4*Var(X)

or, σY^2 = 4*σx^2

Option B

Q.4

y(t) = x(t) x(1+t/2)

      = x(t) x((t+2)/2)

Here, t is scaled by 2

Frequency of x(t) x((t+2)/2) is (10+5) = 15 KHz

So, Nyquist rate for the signal = 15*2 = 30 KHz 

Option B

Q.5

Let the linear convolution of two discrete-time finite-length sequences $x[n]$ and $h[n]$, each of length 16, be denoted by $y[n]$. Let $z[n]$ denote the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point discrete Fourier transforms (DFTs) of $x[n]$ and $h[n]$.

The value of $k$ for which $z[k]=y[k]$ is

• (A) 0
• (B) 1
• (C) 15
• (D) 16

Solution

Step 1: Length of Linear Convolution

Both sequences have length 16.

Length of linear convolution:

$ 16+16-1=31 $

Hence the linear convolution contains the samples

$ y[0],y[1],\ldots,y[30] $

Notice that there is no sample $y[31]$.

Step 2: What does the IDFT of the Product of DFTs Give?

A standard DFT property states:

$ \text{IDFT}\{X[k]H[k]\} = \text{Circular Convolution} $

Therefore,

$ z[n] = \text{16-point circular convolution of }x[n]\text{ and }h[n]. $

So,

• $y[n]$ = Linear Convolution
• $z[n]$ = 16-point Circular Convolution

Step 3: Folding (Wrap-Around)

Since circular convolution has only 16 output samples, the extra samples of the 31-point linear convolution wrap around.

Linear Convolution Sample	Wraps To
$y[0]$	$z[0]$
$y[1]$	$z[1]$
$\vdots$	$\vdots$
$y[15]$	$z[15]$
$y[16]$	$z[0]$
$y[17]$	$z[1]$
$\vdots$	$\vdots$
$y[30]$	$z[14]$

Step 4: Write the Circular Convolution Samples

\[ \begin{aligned} z[0]&=y[0]+y[16] \\ z[1]&=y[1]+y[17] \\ &\vdots \\ z[14]&=y[14]+y[30] \\ z[15]&=y[15]+y[31] \end{aligned} \]

But,

\[ y[31]=0 \]

because the linear convolution ends at $y[30]$. Hence,

\[ z[15]=y[15]. \]

Step 5: Compare $z[k]$ and $y[k]$

k	Relation	Equal?
0	$z[0]=y[0]+y[16]$	❌
1	$z[1]=y[1]+y[17]$	❌
$\vdots$	$\vdots$
14	$z[14]=y[14]+y[30]$	❌
15	$z[15]=y[15]$	✅

Final Answer

$\boxed{k=15}$

Therefore, the correct option is (C) 15.

Q.6

A bar of silicon is doped with boron concentration of $10^{16}\,\text{cm}^{-3}$ and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of $10^{20}\,\text{cm}^{-3}\text{s}^{-1}$. If the recombination lifetime is $100\,\mu s$, intrinsic carrier concentration of silicon is $10^{10}\,\text{cm}^{-3}$, and assuming 100% ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is

• (A) $10^{20}\,\text{cm}^{-6}$
• (B) $2\times10^{20}\,\text{cm}^{-6}$
• (C) $10^{32}\,\text{cm}^{-6}$
• (D) $2\times10^{32}\,\text{cm}^{-6}$

Step-by-Step Solution

Step 1 : Identify the Type of Semiconductor

The silicon is doped with boron. Boron is an acceptor impurity, therefore the semiconductor is p-type.

Since boron is fully ionized,

\[ p_0=N_A=10^{16}\,\text{cm}^{-3} \]

This is the equilibrium hole concentration.

Step 2 : Find the Minority Electron Concentration

Use the mass-action law:

\[ n_0p_0=n_i^2 \]

Substitute the given values:

\[ n_0= \frac{(10^{10})^2}{10^{16}} = \frac{10^{20}}{10^{16}} = 10^4\;\text{cm}^{-3} \]

Step 3 : Calculate the Excess Carrier Concentration

Under steady-state illumination,

\[ \Delta n=\Delta p=G\tau \]

Given

\[ G=10^{20}\,\text{cm}^{-3}\text{s}^{-1} \] \[ \tau=100\mu s=10^{-4}s \]

Therefore,

\[ \Delta n = 10^{20}\times10^{-4} = 10^{16} \]

Hence,

\[ \boxed{\Delta n=\Delta p=10^{16}\,\text{cm}^{-3}} \]

Step 4 : Calculate the Total Electron Concentration

The total electron concentration is

\[ n=n_0+\Delta n \]

\[ n=10^4+10^{16} \]

Since

\[ 10^4\ll10^{16}, \]

we can neglect $10^4$.

\[ \boxed{n\approx10^{16}\,\text{cm}^{-3}} \]

Step 5 : Calculate the Total Hole Concentration

The total hole concentration is

\[ p=p_0+\Delta p \]

\[ p = 10^{16}+10^{16} = 2\times10^{16} \]

\[ \boxed{p=2\times10^{16}\,\text{cm}^{-3}} \]

Step 6 : Calculate the Product \(np\)

\[ np = (10^{16}) (2\times10^{16}) \]

\[ np = 2\times10^{32}\,\text{cm}^{-6} \]

Final Answer

\[ \boxed{ np=2\times10^{32}\,\text{cm}^{-6} } \]

Therefore, the correct option is (D).

Q.7

A non-uniformly doped p-type semiconductor has length L. The energy band diagram is shown below.

Given:

• Energy difference = Δ
• Length = L

Find:

Electric field inside the semiconductor.

Solution

Step 1:

The semiconductor is at equilibrium, so the Fermi level remains constant throughout the material. \[ E_F=\text{constant} \] A non-uniform acceptor concentration causes the valence band to slope, producing an internal electric field.

Step 2:

For a uniform electric field, \[ E=V/L \] \[ \Delta V=EL \] where

• \(E\) = electric field
• \(L\) = semiconductor length

Step 3:

The band diagram shows an energy difference (eV) \[ \Delta \] Energy and voltage are related by \[ \Delta=q\Delta V \] Substitute \[ \Delta V=EL \] to obtain \[ \Delta=qEL \]

Step 4:

Rearrange to obtain the electric field: \[ \boxed{E=\frac{\Delta}{qL}} \]

Final Answer \[ \boxed{E=\frac{\Delta}{qL}} \]

Q.8

The given circuit contains two NMOS transistors \(M_1\) and \(M_2\). Find the voltage gain:

\[ A_v=\frac{v_{out}}{v_{in}} \]

Solution

Step 1: Identify the amplifier

The input signal is applied to the gate of \(M_2\). Therefore \(M_2\) is the amplifying transistor.

A MOS transistor converts voltage into current:

\[ i_d=g_m v_{gs} \] For \(M_2\): \[ i_2=g_{m2}v_{in} \]

Step 2: Output resistance of the circuit

For small signal analysis, the power supply \(V_{DD}\) becomes AC ground.

AC Ground

      |
      \(r_{o2}\)
      |
Vout
      |
      \(r_{o1}\)
      |
Ground

Both \(r_{o1}\) and \(r_{o2}\) are connected between the same two nodes. Therefore they are parallel.

\[ R_{out}=r_{o1}\parallel r_{o2} \]

Step 3: Why \(g_{m1}\) is not present

For \(M_1\):

\[ v_{gs1}=0 \] Therefore the transconductance current is: \[ i_{gm1}=g_{m1}v_{gs1} \] \[ i_{gm1}=g_{m1}(0)=0 \]

So \(M_1\) does not contribute a \(g_m\) current. It only contributes its output resistance \(r_{o1}\).

Step 4: Calculate output voltage

The current generated by \(M_2\) is:

\[ i_o=g_{m2}v_{in} \] The output voltage is: \[ v_{out}=-i_oR_{out} \] Substitute: \[ v_{out} = -(g_{m2}v_{in}) (r_{o1}\parallel r_{o2}) \]

Step 5: Voltage gain

Divide by \(v_{in}\): \[ \frac{v_{out}}{v_{in}} = -g_{m2}(r_{o1}\parallel r_{o2}) \]

Important Concepts

• The transistor receiving the input contributes \(g_m\).
• The transistors connected at the output contribute \(r_o\).
• Parallel resistances appear because both resistors connect from output to AC ground.

Final Answer:

\[ \boxed{ A_v= -g_{m2}(r_{o1}\parallel r_{o2}) } \]

Q.9

For the circuit shown below, the OPAMP is ideal and \(V_{REF}\) is fixed. Find the value of:

\[ \frac{R_F}{R_{IN}} \]

Given:

• \(V_{OUT}=1V\) when \(V_{IN}=0.1V\)
• \(V_{OUT}=6V\) when \(V_{IN}=1V\)

Solution

Ideal OPAMP Rules

For an ideal op-amp:
1. Input current is zero: \[ i_+=i_-=0 \] 2. With negative feedback: \[ V_+=V_- \] Since \(V_{REF}\) is fixed, the positive input voltage is constant.

Step 1: Write the output equation

The circuit is an inverting amplifier with a reference voltage. The output equation is: \[ V_{OUT} = V_+ - \frac{R_F}{R_{IN}} (V_{IN}-V_+) \] Let: \[ K=\frac{R_F}{R_{IN}} \] Therefore: \[ V_{OUT}=V_+-K(V_{IN}-V_+) \]

Step 2: Apply first condition

Given: \[ V_{IN}=0.1V \] \[ V_{OUT}=1V \] Therefore: \[ 1=V_+-K(0.1-V_+) \]

Step 3: Apply second condition

Given: \[ V_{IN}=1V \] \[ V_{OUT}=6V \] Therefore: \[ 6=V_+-K(1-V_+) \]

Step 4: Subtract the two equations

Subtract equation 1 from equation 2: \[ 6-1 = -K(1-V_+) + K(0.1-V_+) \] \[ 5 = K(0.1-1) \] \[ 5=-0.9K \] The magnitude of the gain is: \[ K=\frac{5}{0.9} \] \[ K=5.56 \]

Step 5: Find resistor ratio

Since: \[ K=\frac{R_F}{R_{IN}} \] Therefore: \[ \boxed{ \frac{R_F}{R_{IN}}=5.56 } \]

Important Understanding

\(V_{REF}\) only shifts the DC output level. It does not change the slope:

\[ \frac{\Delta V_{OUT}}{\Delta V_{IN}} \]

The resistor ratio is determined by the change in input and output voltage.

\[ \frac{R_F}{R_{IN}} = \frac{\Delta V_{OUT}} {\Delta V_{IN}} \]

Final Answer:

\[ \boxed{ \frac{R_F}{R_{IN}}=5.56 } \]

Q.10

Consider the circuit with an ideal OPAMP shown in the figure.

Assuming |V_IN| << |V_CC| and |V_REF| << |V_CC|, determine the condition at which V_OUT = 0.

Options:

1. V_IN = V_REF
2. V_IN = 0.5 V_REF
3. V_IN = 2 V_REF
4. V_IN = 2 + V_REF

Solution

Step 1: Identify the Op-Amp Configuration

The non-inverting terminal (+) is connected to ground. Therefore,

V₊ = 0 V

Since the op-amp is ideal and operates with negative feedback, the inverting terminal is also at virtual ground.

V_- = V₊ = 0 V

---

Step 2: Apply KCL at the Inverting Node

Three currents meet at the inverting input:

• Current from V_IN
• Current from V_REF
• Feedback current through R_F

Current due to V_IN

I₁ = (V_IN - 0)/R = V_IN/R

Current due to V_REF

Observe the polarity of the V_REF source. Its positive terminal is connected to ground. Hence the resistor is connected to a voltage of -V_REF.

I₂ = (-V_REF - 0)/R = - V_REF/R

Feedback current

I_F = (0 - V_OUT)/R_F = - V_OUT/R_F

---

Step 3: Apply Kirchhoff's Current Law

I₁ + I₂ + I_F = 0

Substituting the current expressions,

V_IN/R − V_REF/R − V_OUT/R_F = 0

Rearranging,

V_OUT = − (R_F/R) (V_IN − V_REF)

---

Step 4: Condition for Zero Output

For

V_OUT = 0

we obtain

V_IN − V_REF = 0

Therefore,

V_IN = V_REF

Correct Option: (A) V_IN = V_REF

Q.11

(1235)_x = (3033)_y

Or, x³+2x²+3x+5 = 3y³+3y+3

Or, x³+2x²+3x = 3y³+3y-2

Now, check points. Answer is (8, 6)

Q.12

Number of memory locations = $32\mathrm{K\; =\; 2^15}$

$$

• Therefore, the address bus needs 15 address lines.

• A single decoder must be a $\mathrm{15\; to\; 2^15}$ decoder.

• Each output of the decoder is generated using an AND gate with 15 inputs.

n address/input lines → 2ⁿ outputs → 2ⁿ AND gates (for this basic decoder implementation).

So, the minimum number of AND gates required is:

Option C - 2^15

Q.13

the feedback is negative:

$$E(s)=X(s)-H(s)G_1(s)E(s)$$

Hence,

$$E(s)\left[1+G_1H\right]=X(s)$$

$$E(s)=\frac{X(s)}{1+G_1H}$$

At the output summing junction, the two signals are added:

$$Y(s)=G_1E(s)+G_2E(s)$$

$$Y(s)=(G_1+G_2)E(s)$$

Therefore,

$$\boxed{\frac{Y(s)}{X(s)} =\frac{G_1+G_2}{1+G_1H}}$$

$$$$Answer (C)

Q.14

For Open Loop

N = P - Z

Or, 2 = P - 1  (because open loop zero = closed loop pole)

Or, P = 3

Now for Closed Loop,

N = P-Z

or, 0 = 3-Z

or, Z = 3

Answer (d) 3

Q.15

(2η0 - η0) / (2η0 + η0) = 1/3

Answer   (a) 1/3

Q.16

If det(A) = 0, the vectors are linearly dependent.

Given, 

1 -1 2

7 3 x

2 3 1

= 0

Or, 40 - 5x = 0

Or, x = 8.0

Answer  8.0

Q.17

A vector field is called rotational if

$$\mathrm{\nabla }\times \overrightarrow{F} = 0$$

Given, F = ax(4y -c1z) + ay(4x+2z)+az(2y+z)

∇ × F =

i               j             k

∂/∂x     ∂/dy        ∂/∂z

4y -c1z  4x+2z  2y+z

Or, i(2-2) - j(0+c1) + k(4-4)

Or, c1 = 0

Answer  0

Q.18

Here the branch 2 Ω || 2 Ω = 1 Ω.

Also, 6 Ω || 3 Ω = 2 Ω, so in series with 7 Ω: 2 Ω + 7 Ω = 9 Ω.

Therefore, using the current division rule:

I = 5 × 1 / (1 + 9) = 0.5 A

Hence, the current in the 7 Ω branch is 0.5 A.

Q.19

by KVL

4 − V₀ − (V₀ + 2) = 0

V₀ = 1 Volt

by KVL

4 − V₀ − (V₀ + 2) = 0

V₀ = 1 Volt

Q.20

levels = 2^8 = 256
7.68 / (2^8 - 1) × 150 = 4.517 Volt

Q.21

Average Power = Rx(0) = 2

Q.22

The power saved = Pc + Pc.(m^2/4) 

where, m = modulation index 

So, percentage of power saved

=  {Pc + Pc.(m^2/4)} / {Pc + Pc.(m^2/2)}

= 94.4%

Q.23

Data Rate, NFs = 8*1.25* 2 * 4  KHz = 80 KHz

R <= C

R <= Blog2(1+SNR)

Or, B >= R/log2(1+SNR)

Or, B >= (80 * 10^3) /log2(1+400)

Or, B>= 9.259 KHz

Answer 9.26

Q.24

m(t) = 4*sin(2*pi*4*10^3*t)

∇. Fs >= d/dt(m(t))

Or, 

∇*32*10^3 >= 4*(2*pi*4*10^3*t)

Or, 

∇>= 3.14 

Q.25

Critical angle = arcsin(η2/η1) = arcsin(1.48/1.50) = 80.63

Angle along axis, θ = 90-80.63 = 9.36 degree

Q.26

Question 26

Q.27

For your first pick

it is H

First H → remove that coin → depending on whether it was HT or HH, the remaining coins are different

Possibility 1: If you picked HT
Remove it → remaining coins are: HH, TT

Probability of H = 1/2

Possibility 2: If you picked HH
Remove it → remaining coins are: HT, TT

Probability of H = 1/3

Now, because you already got H on the first pick, HH was twice as likely as HT:

• HT chance = 1/3

• HH chance = 2/3

Therefore:

$$\frac{\mathrm{1/}}{3}\times \frac{\mathrm{1/}}{\mathrm{2 }}+ \frac{\mathrm{2/}}{3}\times \frac{\mathrm{1/}}{\mathrm{4\; =\; 1/3}}$$

Q.28

$$\frac{\mathrm{}}{}$$

Current at t(0-) = 20/(5+5+10) = 1 mA

Voltage across capacitor,

v(t) = 10 kilo-ohm * 1 mA = 10 V

at t = 0

ic(t=0+) = - (2 + 1) mA = - 3mA

ic = C*d(vc(t)) / dt

Or, d(vc(t)) / dt = -3 mA /  1mF = -3 V / sec

Q.29

Node A

-I1 -3V2 + (V1 - V2)/1 + V2 = 0

Or, I1 = 2V1 - 4V3

So, y11 = 2, y12 = -4

Node B

-I2+V2/1+(V2-V1) = 0

Or, I2 = -V1 + 2V2

So, y21 = -1, y22 = 2

Answer c

Q.30

Answer (a)

Q.31

For T= 0, 

For T = 1,

Answer (c) 6 ns

Q.32

Answer (a). See the solution

Q.33

Answer (d).  See the solution

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