GATE EC 2022 Question Paper with Answer Key & Detailed Solutions

 

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GATE - EC 2022 Answers with Explanations

Q.11

Consider the two-dimensional vector field

F(x, y) = x i + y j,

where i and j denote the unit vectors along the x-axis and y-axis respectively. The contour C in the x-y plane is composed of two horizontal line segments connected at the two ends by semicircular arcs of unit radius. The contour is traversed in the counter-clockwise direction. Evaluate

∮_C F(x,y) · (dx i + dy j)

Contour (Approximate)

Solution

The given vector field is

F(x,y) = x i + y j.

The required line integral is

∮_C (x dx + y dy).

Observe that

x dx + y dy = d((x² + y²)/2).

Hence the vector field is conservative with potential function

φ(x,y) = (x² + y²)/2.

For any conservative vector field, the line integral over a closed curve is equal to the change in the potential between the initial and final points. Since a closed curve starts and ends at the same point,

∮_C F · dr = φ(final) − φ(initial) = 0.

Alternative Verification (Green's Theorem)

Let

P = x,    Q = y.

Green's theorem gives

∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dA.

Since

∂Q/∂x = 0,    ∂P/∂y = 0,

we obtain

∮_C (x dx + y dy) = ∬_R 0 dA = 0.

Final Answer: 0

Q.12

Consider a system of linear equations

Ax = b,

where

A = [ 1   -√2   3
-1   √2   -3 ]

b = [ 1
3 ]

This system of equations admits

Option	Description
(A)	a unique solution for x
(B)	infinitely many solutions for x
(C)	no solutions for x
(D)	exactly two solutions for x

Solution

Step 1: Write the System

The given matrix equation is

A = [ 1   -√2   3
-1   √2   -3 ]

b = [ 1
3 ]

Therefore, the system of equations is

x₁ − √2 x₂ + 3x₃ = 1

−x₁ + √2 x₂ − 3x₃ = 3

Step 2: Compare the Two Rows of A

Observe that

(-1, √2, -3) = -(1, -√2, 3).

Thus, the second row of the coefficient matrix is exactly the negative of the first row. Therefore,

rank(A) = 1.

Step 3: Check Whether the System is Consistent

If the second row is the negative of the first row, then the second entry of the vector b should also be the negative of the first entry.

However,

b = [ 1
3 ]

Since

3 ≠ -1,

the right-hand side is not consistent with the coefficient matrix.

Step 4: Verify by Multiplying the First Equation by −1

The first equation is

x₁ − √2x₂ + 3x₃ = 1.

Multiplying by −1 gives

−x₁ + √2x₂ − 3x₃ = −1.

But the second equation is

−x₁ + √2x₂ − 3x₃ = 3.

Since

−1 ≠ 3,

the two equations contradict each other. Therefore, the system is inconsistent.

Final Answer:

(C) No solutions for x.

Q.13 

Solution for Current I

Given:

• Voltage source = 5 V
• Three resistors = 2 kΩ each
• Current source = 1 mA (upward)

Step 1: Label the Nodes

Bottom wire is ground (0 V).
Left node: VA = 5 V
Middle node: VB
Right node: VC

Step 2: Apply KCL at Node C

(VC − VB) / 2000 = 1 × 10⁻³

Therefore,

VC = VB + 2

Step 3: Apply KCL at Node B

(VB − 5)/2000 + VB/2000 + (VB − VC)/2000 = 0

Substitute VC = VB + 2:

(VB − 5)/2000 + VB/2000 − 2/2000 = 0

Multiply by 2000:

VB − 5 + VB − 2 = 0

2VB = 7

VB = 3.5 V

Step 4: Calculate Current I

I = (5 − 3.5)/2000

I = 0.00075 A

I = 0.75 mA

Final Answer

I = 0.75 mA (flowing from left to right).

Q.14

Answer: There is no closed return path. So, I = 0

Q. 15

Fourier Transform of \(x(t)=\dfrac{t}{(1+t^2)^2}\)

We want to determine the Fourier transform of

$$ x(t)=\frac{t}{(1+t^2)^2} $$

using the Fourier transform definition

$$ X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}\,dt $$

---

Step 1 :

Let

$$ x_1(t)=\frac{1}{1+t^2} $$

Then

$$ X_1(\omega)=\pi e^{-|\omega|}. $$

Proof:

$$ \int_{-\infty}^{\infty}\frac{1}{1+t^2}e^{-j\omega t}\,dt $$

Use the Residue Theorem.

$$ \int_{-\infty}^{\infty}\frac{1}{1+z^2}e^{-j\omega z}\,dz $$

Or,

$$ \int_{-\infty}^{\infty}\frac{1}{(z+j)(z-j)}e^{-j\omega z}\,dz $$

Poles are at \(z=-j\) and \(z=j\).

For the pole at \(z=-j\):

$$ \operatorname*{Res}_{z=-j}\left(\frac{e^{-j\omega z}}{(z+j)(z-j)}\right) = \lim_{z\to -j}(z+j)\frac{e^{-j\omega z}}{(z+j)(z-j)} = \frac{e^{-\omega}}{-2j}. $$

Therefore,

$$ -2\pi j\left(\frac{e^{-\omega}}{-2j}\right) = \pi e^{-\omega}. $$

For the pole at \(z=j\):

$$ \operatorname*{Res}_{z=j}\left(\frac{e^{-j\omega z}}{(z+j)(z-j)}\right) = \lim_{z\to j}(z-j)\frac{e^{-j\omega z}}{(z+j)(z-j)} = \frac{e^{\omega}}{2j}. $$

Therefore,

$$ 2\pi j\left(\frac{e^{\omega}}{2j}\right) = \pi e^{\omega}. $$

Step 2 : Differentiate the Time-Domain Function

Differentiate $$ \frac{1}{1+t^2} $$ with respect to \(t\): $$ \frac{d}{dt} \left( \frac{1}{1+t^2} \right) = -\frac{2t}{(1+t^2)^2} $$ Rearranging, $$ \boxed{ \frac{t}{(1+t^2)^2} = -\frac12 \frac{d}{dt} \left( \frac{1}{1+t^2} \right) } $$ Thus, the required function is simply one-half of the derivative of the known function.

Step 3 : Apply the Differentiation Property

The differentiation property of the Fourier transform is $$ \boxed{ \mathcal F \left\{ \frac{dx(t)}{dt} \right\} = j\omega X(\omega) } $$ Applying it, $$ \mathcal F \left\{ \frac{d}{dt} \left( \frac{1}{1+t^2} \right) \right\} = j\omega \left( \pi e^{-|\omega|} \right) $$

Step 4 : Multiply by \(-\frac12\)

Since $$ \frac{t}{(1+t^2)^2} = -\frac12 \frac{d}{dt} \left( \frac1{1+t^2} \right), $$ its Fourier transform becomes $$ X(\omega) = -\frac12 \left[ j\omega \pi e^{-|\omega|} \right] $$

Step 5 : Simplify

Collect the constants: $$ X(\omega) = -\frac{j\pi\omega}{2} e^{-|\omega|} $$

Final Answer

$$ \boxed{ \mathcal F \left\{ \frac{t}{(1+t^2)^2} \right\} = -\frac{j\pi\omega}{2} e^{-|\omega|} } $$

Or,

$$ \boxed{ \mathcal F \left\{ \frac{t}{(1+t^2)^2} \right\} = \frac{\pi\omega}{2j} e^{-|\omega|} } $$

Option (A)

Q. 16

 

For electron diffusion in a semiconductor, the excess electron concentration decays exponentially with distance:

$$\Delta n(x)=\Delta n(0)e^{-x/L_n}$$

where:

• $\Delta n(x)$ = excess electron concentration at distance $x$
• $\Delta n(0)$ = excess electron concentration at $x=0$
• $L_n$ = electron diffusion length

At the diffusion length $x=L_n$

​10^14 * e^(2/2) = .367 * 10^14 cm^-3

Option (1)

Q. 17

Question

In a non-degenerate bulk semiconductor with electron density n = 10¹⁶ cm⁻³, the value of (E_C − E_Fn) = 200 meV, where E_C and E_Fn denote the bottom of the conduction band and electron Fermi level energy, respectively. Assume the thermal voltage as 26 meV and the intrinsic carrier concentration as 10¹⁰ cm⁻³.

For n = 0.5 × 10¹⁶ cm⁻³, the closest approximation of the value of (E_C − E_Fn) is:

1. 226 meV
2. 174 meV
3. 218 meV
4. 182 meV

---

Step-by-Step Solution

Step 1: Write the Carrier Concentration Equation

For a non-degenerate semiconductor, the electron concentration is given by:

n = ni exp[(EFn − Ei) / kT]

Alternatively,

EC − EFn = kT ln(NC / n)

Since the effective density of states N_C remains constant at a fixed temperature, we can compare two operating conditions directly.

Step 2: Use the Difference Formula

The change in the Fermi level position is:

(EC − EFn)2 − (EC − EFn)1 = kT ln(n1 / n2)

Step 3: Substitute the Given Values

Electron concentration (n₁)	1016 cm−3
Electron concentration (n₂)	0.5 × 1016 cm−3
Initial value of (EC − EFn)	200 meV
Thermal voltage (kT)	26 meV

Step 4: Calculate the Energy Shift

ΔE = 26 × ln(1016 / (0.5 × 1016))

Since,

1016 / (0.5 × 1016) = 2

Therefore,

ΔE = 26 × ln(2)

Using,

ln(2) = 0.693

ΔE = 26 × 0.693 = 18.02 meV

Step 5: Find the New Value of (E_C − E_Fn)

Add the energy shift to the original value:

(EC − EFn) = 200 + 18

(EC − EFn) = 218 meV

Step 6: Physical Explanation

The electron concentration decreases from 10¹⁶ to 0.5 × 10¹⁶ cm⁻³. As the electron density decreases, the Fermi level moves farther away from the conduction band. Therefore, (E_C − E_Fn) increases.

Final Answer

(E_C − E_Fn) = 218 meV

Correct Option: (C) 218 meV

Q. 18

 

Consider the CMOS circuit shown in the figure (substrates are connected to their respective sources). The gate width (𝑊) to gate length (𝐿) ratios W/L of the transistors are as shown. Both the transistors have the same gate oxide capacitance per unit area. For the pMOSFET, the threshold voltage is -1 V and the mobility of holes is 40 cm2 /V.s . For the nMOSFET, the threshold voltage is 1 V and the mobility of electrons is 300 cm2 /V.s . The steady state output voltage 𝑉0 is ________. 

Answer: Mosfet current at saturation 

μCox/2 * (W/L)*(Vgs - Vth)^2

Given:

300*1*(V0 - 1)^2 = 40*5*(4-V0-1)^2

Or, V0 = 1.898V

So, Correct answer is Option C (less than 2V)

Q. 19

Consider the 2-bit multiplexer (MUX) shown in the figure. For OUTPUT to be the XOR of C and D, the values for 𝐴0, 𝐴1, 𝐴2, and 𝐴3 are ____________.

Answer: The Output will be 

CD + CD 

To select A0, C=0 & D=0 ; Output = 0 (=A0)

To select A1, C=0 & D=1 ; Output = 1 (=A1)

To select A2, C=1 & D=0 ; Output = 1 (=A2)

To select A3, C=1 & D=1 ; Output = 0 (=A3)

So, Correct Answer is Option (3)

Q. 20

 

The ideal long channel nMOSFET and pMOSFET devices shown in the circuits have threshold voltages of 1 V and −1 V, respectively. The MOSFET substrates are connected to their respective sources. Ignore leakage currents and assume that the capacitors are initially discharged. For the applied voltages as shown, the steady state voltages are _________________.

Answer: for nMOSFET, Vgs ≥ Vth  Or, (Vg - Vs) ≥ Vth

So, the maximum output voltage is limited to Vg - Vth or 5-1 = 4V

For nMOSFET, Vgs  less than or equal to Vth  Or, (Vg - Vs) <= Vth

If you were using a pMOS to pull an output down to ground (passing a "0"), it would be limited. It would turn off once the output reached: Vout,min =Vg + |Vth| or, 5V

Q. 21

The point of interest is -1/K = -0.5 on the real axis. Counting the encirclements:

• Outer Loop: Clockwise, enclosing -0.5. (Count = +1)
• Left Inner Loop: Also clockwise, enclosing -0.5. (Count = +1)
• Right Inner Loop: Counter-clockwise, but -0.5 lies outside. (Count = 0)

Total Encirclements (N) = 2

Using the stability relation:

Z = N + P = 2 + 0 = 2

Hence, the closed-loop transfer function has 2 poles in the closed right-half plane.

Correct Option: (C) 2

Q. 22

Since five Root Locus branches originate from the same point, the system must have five coincident open-loop poles at that location. This is because Root Locus branches always start from open-loop poles, and the number of Root Locus branches equals the number of open-loop poles (or zeros, whichever is greater).

Hence, the system has five repeated real poles, which corresponds to Option (A): 1/(s+1)^5

Q. 23

The frequency response 𝐻(𝑓) of a linear time-invariant system has magnitude as shown in the figure.

 Statement I: The system is necessarily a pure delay system for inputs which are bandlimited to −𝛼≤𝑓≤𝛼.

Statement II: For any wide-sense stationary input process with power spectral density 𝑆H(𝑓), the output power spectral density 𝑆x(𝑓) obeys 𝑆y(𝑓)=𝑆x(𝑓) for −𝛼≤𝑓≤𝛼. Which one of the following combinations is true? 

Answer: Option 1. Both Statements are correct

Solution:

System has magnitude 1 and bandlimited, so statement 1 is true.

𝑆y(𝑓)=𝑆x(𝑓) is true as System has magnitude 1

Q. 24

 

In a circuit, there is a series connection of an ideal resistor and an ideal capacitor. The conduction current (in Amperes) through the resistor is 2sin(𝑡 + 𝜋/2). The displacement current (in Amperes) through the capacitor is _________.

Answer: Option C (2sin(𝑡 + 𝜋/2)

Solution:

The displacement current through the capacitor
In a series circuit containing an ideal resistor and an ideal capacitor, the displacement current through the capacitor is equal to the conduction current through the resistor at every instant. No actual charge carriers move through the dielectric between the capacitor plates. Instead, the changing electric field in the dielectric gives rise to the displacement current. For an ideal capacitor, the displacement current through the dielectric is exactly equal to the conduction current in the connecting wires and through the resistor. Consequently, if the conduction current is sinusoidal, the displacement current is also sinusoidal with the same amplitude, frequency, and phase.

Q. 25

Given PDE:

\(a\frac{\partial^2 f}{\partial x^2} + b\frac{\partial^2 f}{\partial y^2} = f(x,y)\)

Assume:

\(f=e^{\xi x+\eta y}\)

Step 1: Calculate derivatives

\[ \frac{\partial^2 f}{\partial x^2}=\xi^2 f \]

\[ \frac{\partial^2 f}{\partial y^2}=\eta^2 f \]

Step 2: Substitute into PDE

\[ a(\xi^2 f)+b(\eta^2 f)=f \]

Step 3: Take \(f\) common

\[ (a\xi^2+b\eta^2)f=f \]

Step 4: Cancel \(f\)

\[ \boxed{a\xi^2+b\eta^2=1} \]

Answer: Option A & B

Q. 26

An ideal OPAMP circuit with a sinusoidal input is shown in the figure. The 3 dB frequency is the frequency at which the magnitude of the voltage gain decreases by 3 dB from the maximum value. Which of the options is/are correct?

 Answer: Option B and C

Solution:

Vout/Vin = -Rf/Rin = -2/(1+1/jwC)

at, w = infinite, Vout/Vin = -2

at, w = 0, Vout/Vin = 0

So, it is highpassfilter

and 3 dB cutoff frequency of the OPAMP is 

1/2*pi*RC = 1000 rad/sec

Q. 27

Select the Boolean function(s) equivalent to 𝑥+𝑦𝑧, where 𝑥,𝑦, and 𝑧 are Boolean variables, and + denotes logical OR operation. 

 (A) 𝑥+𝑧+𝑥𝑦 

 (B) (𝑥+𝑦)(𝑥+𝑧) 

 (C) 𝑥+𝑥𝑦+𝑦𝑧 

 (D) 𝑥+𝑥𝑧+𝑥𝑦 

Answers: Options B and C

Solution:

Option B: (𝑥+𝑦)(𝑥+𝑧) = x.x + x.z + x.y + yz = x(1+y+z) + yz =  x+yz

Option C: 𝑥+𝑥𝑦+𝑦𝑧 = x(1+y)+yz = x+yz

Q. 28

Q. Select the correct statement(s) regarding CMOS implementation of NOT gates. 

 (A) Noise Margin High (𝑁𝑀H) is always equal to the Noise Margin Low (𝑁𝑀H), irrespective of the sizing of transistors. 

 (B) Dynamic power consumption during switching is zero. 

 (C) For a logical high input under steady state, the nMOSFET is in the linear regime of operation. 

 (D) Mobility of electrons never influences the switching speed of the NOT gate.  

Answer: Option C

Solution

Option A : NMH and NML are generally not equal because electron mobility is about 2–3 times hole mobility.

Designers often make pMOS wider (roughly 2–3×) to equalize them.

Option B: Whenever the output changes, the load capacitor charges/discharges.

Therefore dynamic power is non-zero during switching.

Option C: (Correct) Hence the nMOS is indeed in the linear (triode) region at steady state. 

Option D: Higher electron mobility ⇒ higher current ⇒ faster discharge ⇒ lower delay.
Electron mobility directly affects switching speed.

Q. 29

 

Q. Let 𝐻(𝑋) denote the entropy of a discrete random variable 𝑋 taking 𝐾 possible distinct real values. Which of the following statements is/are necessarily true? 

 (A) 𝐻(𝑋)≤logH𝐾 bits 

 (B) 𝐻(𝑋)≤𝐻(2𝑋) 

 (C) 𝐻(𝑋)≤𝐻(𝑋^2) 

 (D) 𝐻(𝑋)≤𝐻(2^X) 

Answer: Option 1,2, and 4

Solution: 

Option A: (Correct) if K symbols having equal probability the maximum entropy = log2(K); other wise 𝐻(𝑋)≤logH𝐾 bits 

Option B: (Correct) -plog2p ≤ -2plog2p

Option C: (false) Think about X^2 = 4 then. X = 1,-1. Whenever different outcomes merge into one outcome, entropy decreases.

Option D: (Correct) If every value of 2^X produces a unique output, the probabilities are just relabeled. The uncertainty is unchanged, so the entropy stays the same.

If exactly one value of XXX maps to yyy, the probability stays the same.
If two or more values of XXX map to yyy, you add their probabilities.

Q. 30

Consider the following wave equation,

\[ \frac{\partial^2 f(x,t)}{\partial t^2} = 10000 \frac{\partial^2 f(x,t)}{\partial x^2} \]

Which of the given options is/are solutions to the above wave equation?

• (A) \[ f(x,t)=e^{-(x-100t)^2}+e^{-(x+100t)^2} \]
• (B) \[ f(x,t)=e^{-(x-100t)} +\frac12e^{-(x+1000t)} \]
• (C) \[ f(x,t)=e^{-(x-100t)} +\sin(x+100t) \]
• (D) \[ f(x,t)=e^{j100\pi(-100x+t)} + e^{j100\pi(100x+t)} \]

---

Step 1: Identify the Wave Speed

The given wave equation is \[ \frac{\partial^2 f}{\partial t^2} = 10000 \frac{\partial^2 f}{\partial x^2}. \] Since \[ 10000=100^2, \] the wave speed is \[ c=100. \] Hence, the general solution of the one-dimensional wave equation is \[ f(x,t)=F(x-100t)+G(x+100t), \] where \(F\) and \(G\) are any twice differentiable functions. Therefore, every valid solution must be expressible as a function of

• \(x-100t\)
• \(x+100t\)

---

Step 2: Verify Each Option

Option (A)

\[ f(x,t) = e^{-(x-100t)^2} + e^{-(x+100t)^2} \] Observe that \[ F(u)=e^{-u^2}, \qquad G(v)=e^{-v^2}. \] Therefore, \[ f(x,t)=F(x-100t)+G(x+100t), \] which exactly matches the general solution.

✔ Option (A) is a solution.

Option (B)

\[ f(x,t) = e^{-(x-100t)} + \frac12e^{-(x+1000t)} \] The second term contains \[ x+1000t \] instead of \[ x+100t. \] Let \[ g=e^{-(x+1000t)}. \] Then \[ g_{tt} = 1000^2g = 10^6g. \] Also, \[ g_{xx}=g. \] Hence \[ 10000g_{xx} = 10^4g. \] Since \[ 10^6g \neq 10^4g, \] the wave equation is not satisfied.

✘ Option (B) is NOT a solution.

Option (C)

\[ f(x,t) = e^{-(x-100t)} + \sin(x+100t) \] The first term depends on \[ x-100t. \] The second term depends on \[ x+100t. \] Thus, \[ f(x,t) = F(x-100t) + G(x+100t), \] which matches the general solution.

✔ Option (C) is a solution.

Option (D)

\[ f(x,t) = e^{j100\pi(-100x+t)} + e^{j100\pi(100x+t)} \] Consider the first exponential. Its argument is \[ t-100x, \] which is **not** of the form \[ x-100t \quad\text{or}\quad x+100t. \] Differentiate: \[ f_{tt}=k^2e^{ku}, \] where \[ u=t-100x, \qquad k=j100\pi. \] Also, \[ f_{xx} = 10000k^2e^{ku}. \] Hence, \[ 10000f_{xx} = 10^8k^2e^{ku} \neq f_{tt}. \] Therefore the wave equation is not satisfied.

✘ Option (D) is NOT a solution.

Final Answer

✔ (A) is correct.
✔ (C) is correct.

Correct Options: \[ \boxed{A \text{ and } C} \]

Q. 31

The bar graph shows the frequency of the number of wickets taken in a match by a bowler in her career. For example, in 17 of her matches, the bowler has taken 5 wickets each. The median number of wickets taken by the bowler in a match is __________ (rounded off to one decimal place).  

Solving for the Median

Fundamental Concepts

• Median: The "middle" value in a sorted data set.
• Frequency: How many times a specific value occurs.
• Cumulative Frequency: The running total of frequencies. It helps us locate the middle position in a large data set without listing every single number.

Step 1: Create a Frequency Table

We extract the values from the bar graph and calculate the cumulative frequency (CF).

Number of Wickets (x)	Frequency (f)	Cumulative Frequency (CF)
0	5	5
1	7	12
2	8	20
3	25	45
4	20	65
5	17	82
6	8	90
7	4	94
8	3	97
9	2	99
10	1	100

Step 2: Determine Total Number of Data Points

Summing all frequencies gives us N = 100. Since N is even, the median is the average of the two middle terms.

Step 3: Find the Median Positions

For an even N, we look for the values at positions:

• Position 1: N / 2 = 50^th match
• Position 2: (N / 2) + 1 = 51^st match

Step 4: Locate the Values

Looking at our table:

• Up to the 45^th match, the bowler has taken 3 wickets or fewer.
• From the 46^th to the 65^th match, the bowler has taken 4 wickets.
• Both the 50^th and 51^st values fall within this range.

Step 5: Final Calculation

Median = (Value at 50^th + Value at 51^st) / 2
Median = (4 + 4) / 2 = 4

Rounding to one decimal place as requested:

Median = 4.0

Q. 32

A simple closed path 𝐶 in the complex plane is shown in the figure if

\[ \oint_C \frac{2^z}{z^2-1}\,dz=-i\pi A \]

where 𝑖=√−1, then the value of 𝐴 is ______ (rounded off to two decimal places).

\[ \oint_C \frac{2^z}{z^2-1}\,dz=-i\pi A \]

\[ \oint_C \frac{2^z}{z^2-1}\,dz = \oint_C \frac{2^z}{(z+1)(z-1)}\,dz \]

Residue at \(z=1\):

\[ \operatorname{Res}_{z=1} \frac{2^z}{(z+1)(z-1)} = \lim_{z\to1} (z-1) \frac{2^z}{(z+1)(z-1)} = \frac{2}{2} =1 \]

\[ \oint_C \frac{2^z}{z^2-1}\,dz = 2\pi i \]

Residue at \(z=-1\):

\[ \operatorname{Res}_{z=-1} \frac{2^z}{(z+1)(z-1)} = \lim_{z\to-1} (z+1) \frac{2^z}{(z+1)(z-1)} = \frac{2^{-1}}{-2} = -\frac14 \]

\[ \oint_C \frac{2^z}{z^2-1}\,dz = -\frac14(2\pi i) = -\frac12\pi i \]

Given:

\[ \oint_C \frac{2^z}{z^2-1}\,dz = -i\pi A \]

Therefore,

\[ -i\pi A = -\frac12\pi i \]

\[ A=\frac12=0.50 \]

Note: Here only \(z=-1\) lies inside the contour.

Q. 33

Let 𝑥1(𝑡)=𝑒^(-t)𝑢(𝑡) and 𝑥2(𝑡)=𝑢(𝑡)−𝑢(𝑡−2), where 𝑢(⋅) denotes the unit step function. If 𝑦(𝑡) denotes the convolution of 𝑥1(𝑡) and 𝑥2(𝑡), then lim(x→0) 𝑦(𝑡) = _________ (rounded off to one decimal place).

Solution

Let

\[ x_1(t)=e^{-t}u(t) \]

and

\[ x_2(t)=u(t)-u(t-2) \]

where \(u(t)\) is the unit step function.

\[ y(t)=x_1(t)*x_2(t) \]

Find the final value of \(y(t)\).

Laplace Transforms

\[ \mathcal{L}\{e^{-t}u(t)\} =\frac{1}{s+1} \]

\[ \mathcal{L}\{u(t)-u(t-2)\} =\frac{1}{s}-\frac{e^{-2s}}{s} =\frac{1-e^{-2s}}{s} \]

Using Convolution Property

\[ Y(s)=X_1(s)X_2(s) \]

\[ Y(s) = \frac{1}{s+1} \cdot \frac{1-e^{-2s}}{s} \]

Applying Final Value Theorem

\[ \lim_{t\to\infty}y(t) = \lim_{s\to0}sY(s) \]

\[ = \lim_{s\to0} s \left( \frac{1}{s+1} \cdot \frac{1-e^{-2s}}{s} \right) \]

\[ = \lim_{s\to0} \frac{1-e^{-2s}}{s+1} \]

Using the limit

we obtain

\[ \boxed{\lim_{t\to\infty}y(t)=0} \]

Q. 34

Question: An ideal MOS capacitor (p-type semiconductor) is shown in the figure. The MOS capacitor is under strong inversion with V_G1 = 2 V. The corresponding inversion charge density (Q_inv1) is 2.2 µC/cm². Assume oxide capacitance per unit area (C_ox) as 1.7 µF/cm². For V_G2 = 4 V, the value of Q_inv2 is ______ µC/cm² (rounded off to one decimal place).

The Catch: Once a MOS capacitor enters Strong Inversion, the surface potential (φ_s) becomes pinned. Any further increase in gate voltage (V_G) results in a linear increase in inversion charge (Q_inv), while the depletion charge remains constant.

Step 1: Identify Given Parameters

• Initial Gate Voltage (V_G1): 2 V
• Initial Inversion Charge (Q_inv1): 2.2 µC/cm²
• Oxide Capacitance (C_ox): 1.7 µF/cm²
• Target Gate Voltage (V_G2): 4 V

Step 2: Use the Strong Inversion Relationship In strong inversion, the relationship between the change in voltage and the change in charge is defined by:

ΔQ_inv = C_ox × ΔV_G

Which can be rewritten as:

Q_inv2 - Q_inv1 = C_ox(V_G2 - V_G1)

Step 3: Substitute the Values Substitute the known values into the equation to find Q_inv2:

Q_inv2 = 2.2 + [1.7 × (4 - 2)]

Step 4: Final Calculation

Q_inv2 = 2.2 + [1.7 × 2]
Q_inv2 = 2.2 + 3.4
Q_inv2 = 5.6

Final Answer: 5.6 µC/cm²

Q. 35

A symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).  

Answer: 

Since alternative QPSK and 16-QAM used, data rate of uncoded data is (2+4)/2 = 3 Mbps

Number of bits in one QPSK symbol is 2 and

number of bits in one 16-QAM symbol is 4

Q. 36

The function f(x) = 8 log_e x − x² + 3 attains its minimum over the interval [1, e] at x = ______.
(Here log_e x is the natural logarithm of x.)

Solution:

Given:
f(x) = 8 ln(x) − x² + 3

Find the minimum of f(x) on the interval [1, e].

Step 1: Find the derivative

f'(x) = \(\frac{d}{dx}\) [8 ln(x) − x² + 3] = \(\frac{8}{x}\) − 2x

Step 2: Set derivative to zero to find critical points

f'(x) = 0 ⇒ \(\frac{8}{x}\) − 2x = 0
⇒ \(\frac{8}{x}\) = 2x
⇒ 8 = 2x²
⇒ x² = 4
⇒ x = ±2

Since the domain is [1, e], and e ≈ 2.718, we consider only x = 2.

Step 3: Evaluate f(x) at critical point and interval endpoints

• At x = 1:
  f(1) = 8 ln(1) − 1² + 3 = 0 − 1 + 3 = 2
• At x = 2:
  f(2) = 8 ln(2) − 4 + 3 = 8 × 0.693 − 1 = 5.544 − 1 = 4.544 (approx)
• At x = e:
  f(e) = 8 ln(e) − e² + 3 = 8 × 1 − 7.389 + 3 = 11 − 7.389 = 3.611 (approx)

Step 4: Compare values

f(1) = 2
f(2) ≈ 4.544
f(e) ≈ 3.611

Minimum value of f(x) on [1, e] is at x = 1.

Final answer:

x = 1

Q.37

Let α, β be two non-zero real numbers and v₁, v₂ be two non-zero real vectors of size 3 × 1. Suppose that v₁ and v₂ satisfy v₁^Tv₂ = 0, v₁^Tv₁ = 1, and v₂^Tv₂ = 1. Let A be the 3 × 3 matrix given by:

A = αv₁v₁^T + βv₂v₂^T

The eigenvalues of A are ________.

(A) 0, α, β

(B) 0, α + β, α - β

(C) 0, (α + β)/2, √(αβ)

(D) 0, 0, √(α² + β²)

Solution

To solve for the eigenvalues of matrix A, we test the given vectors as potential eigenvectors:

1. Testing v₁: Multiplying A by v₁:
   Av1 = (αv1v1T + βv2v2T)v1 = αv1(v1Tv1) + βv2(v2Tv1)
   Using the given conditions (v₁^Tv₁ = 1 and v₂^Tv₁ = 0), we get:
   Av1 = αv1(1) + βv2(0) = αv1
   Thus, α is an eigenvalue.


2. Testing v₂: Similarly, multiplying A by v₂:
   Av2 = αv1(v1Tv2) + βv2(v2Tv2) = αv1(0) + βv2(1) = βv2
   Thus, β is an eigenvalue.


3. Finding the Third Eigenvalue: Since v₁ and v₂ are in ℝ³, there exists a vector v₃ orthogonal to both.
   Av3 = αv1(0) + βv2(0) = 0
   Thus, 0 is the third eigenvalue.

Correct Answer: (A) 0, α, β

Q.52

Option (a)

The cutoff frequency is:

\[ f_{\text{cutoff}} = \frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}} \]

For the cutoff frequency of a TM wave,

\[ \frac{c}{2a} = 1.5 \times 10^{14} \ \text{Hz} \]

Therefore, frequencies above:

\[ 1.5 \times 10^{14} \ \text{Hz} \]

will propagate.

Q.53

Solving the Integral Using Polar Coordinates

We want to evaluate the integral:

$$\iint_D 3(x^2 + y^2)\, dx\, dy$$

In polar coordinates, we use:

• \(x^2 + y^2 = r^2\)
• \(dx\,dy = r\,dr\,d\theta\)

So the integrand becomes:

$$3r^2 \cdot r = 3r^3$$

1. Describe the Region \(D\) in Polar Coordinates

The triangular region lies between the lines:

• \(y = x \Rightarrow \theta = \pi/4\)
• \(y = -x \Rightarrow \theta = -\pi/4\)

The right boundary is \(x = 4\):

$$r \cos\theta = 4 \quad \Rightarrow \quad r = \frac{4}{\cos\theta}$$

Thus the region in polar form is:

$$-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}, \quad 0 \le r \le \frac{4}{\cos\theta}$$

2. Set Up the Integral

$$\int_{\theta=-\pi/4}^{\pi/4} \int_{0}^{4/\cos\theta} 3r^3 \, dr \, d\theta$$

3. Integrate with Respect to \(r\)

$$\int_{0}^{4/\cos\theta} 3r^3 \, dr = \frac{3}{4}\left(\frac{4}{\cos\theta}\right)^4 = \frac{192}{\cos^4\theta}$$

4. Integrate with Respect to \(\theta\)

$$192 \int_{-\pi/4}^{\pi/4} \sec^4\theta \, d\theta$$

Use the substitution \(u = \tan\theta\), \(du = \sec^2\theta \, d\theta\):

$$192 \int_{-1}^{1} (u^2 + 1)\, du$$

$$\int_{-1}^{1}(u^2 + 1)\, du = \left[\frac{u^3}{3} + u\right]_{-1}^{1} = \frac{8}{3}$$

$$192 \cdot \frac{8}{3} = 512$$

Final Answer:

512

Q.54

Calculation:

First, find V_th and R_th from the graph.

From the graph

V₂ = −(1 / 4m)·I₂ + 5

When I₂ = 0

V₂ = V_osc = V_th = 5

Replace the network with its Thevenin's equivalent and replace the variable resistance with the load:

I₂ = (10 − 5) / 1250 = 4 mA

Q.55

DFT Problem Solution

Given the sequence:

x̄ = [1, 0, 0, 0, 2, 0, 0, 0]

The problem asks to compute:

y = DFT(DFT(x̄))

and to find y[0].

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Step 1 — First DFT

The DFT is defined as:

X(k) = Σ x[n] · exp(-j·2πnk/8), n = 0..7

The only non-zero samples are x[0] = 1 and x[4] = 2, so:

X(k) = 1 + 2·exp(-jπk) = 1 + 2(-1)^k

Thus:

X = [3, -1, 3, -1, 3, -1, 3, -1]

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Step 2 — Second DFT

We need:

y[0] = Σ X[k], k = 0..7

Summing the values:

• Four values are 3 → sum = 12
• Four values are -1 → sum = -4

y[0] = 12 - 4 = 8

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Final Answer: 8.0

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