GATE 2025 EC Question Paper (Available)- Download Solution PDF with Answer Key

Understanding the Majority Function.

A majority function for three variables \( x, y, z \) is defined as: \[ f(x,y,z) = 1 when at least two inputs are 1. \]

The sum of minterms where the function evaluates to 1 is: \[ \Sigma(3,5,6,7) \]

Using Boolean algebra, the canonical sum-of-products (SOP) expression for the majority function is: \[ f(x,y,z) = xy + yz + xz \]

We can also derive this using simplifications:

1. Expanding in terms of the majority condition:
\[ xy + yz + xz \]


2. Alternative simplification using XOR:
\[ f(x,y,z) = x(y \oplus z) + yz \]

Since both derivations are equivalent, the minimal expression is:
\[ xy + yz + xz \] Quick Tip: In logic circuits, the majority function is useful in voting systems and fault-tolerant computing, where outputs depend on the majority of inputs.

Finding the Pattern in the Sequence.

The given sequence: \[ 3, 9, 19, 33, \_ \]
Finding first-order differences: \[ 9 - 3 = 6, \quad 19 - 9 = 10, \quad 33 - 19 = 14 \]

Finding second-order differences: \[ 10 - 6 = 4, \quad 14 - 10 = 4 \]

Since the second-order difference is constant (\(4\)), the sequence follows a quadratic pattern.

The next first-order difference is: \[ 18 = 14 + 4 \]

Thus, the next term in the sequence is: \[ 33 + 18 = 51 \] Quick Tip: A constant second-order difference indicates a quadratic sequence. Use this method to predict future terms efficiently.

Understanding Rank in an Upper Triangular Matrix.

For a matrix in upper triangular form, its rank is equal to the number of nonzero diagonal elements.

The diagonal elements of \( A \) are: \[ 2, 6, \alpha, \gamma \]

Since rank(A) must be at least 3, we require at least three of these elements to be nonzero.

- \( 2 \) and \( 6 \) are already nonzero.

- At least one of \( \alpha \) or \( \gamma \) must be nonzero.

- \( \beta \) does not affect rank, so it can be any value.


Thus, the possible values are: \[ \alpha \neq 0 \quad or \quad \gamma \neq 0, \quad \beta is arbitrary. \] Quick Tip: For an upper triangular matrix, rank is simply the number of nonzero diagonal elements.

Understanding the Circuit.

The given circuit consists of:

- A Full Adder (FA)

- An XOR gate that inverts \( y \) (i.e., \( y' = y \oplus 1 = \bar{y} \))

- A Carry-in value of 1


Binary Subtraction via Two’s Complement: \[ A - B = A + \bar{B} + 1 \]
Here:

- \( x \) is directly passed to the FA.

- \( y \) is complemented via XOR with 1.

- Carry-in = 1.


This is the standard way to perform subtraction using addition, confirming that the circuit acts as a subtractor. Quick Tip: Subtraction can be implemented using addition by converting the second operand into its Two’s Complement form (\(\bar{B} + 1\)).

Understanding Frequency Domain Parameters.

The given diagram represents a frequency response plot, likely from Nyquist Stability Criterion or Bode Plot Analysis.

1. Sinusoidal Component \( \sin \omega_c \):

- Indicates the system's response at critical frequency \( \omega_c \).

- This often corresponds to the gain crossover frequency in a Bode plot.


2. Phase Angle \( \theta \) at \( \omega_p \):

- \( \omega_p \) is the phase crossover frequency, where phase margin is measured.

- It is a key factor in stability determination.


Thus, the correct answer is: \[ \sin \omega_c, \quad \theta is \omega_p \] Quick Tip: In control systems, the gain crossover frequency (\( \omega_c \)) and phase crossover frequency (\( \omega_p \)) help analyze stability margins using Bode and Nyquist plots.

Applying Contour Integration Theorems.

By Cauchy's Integral Theorem (CIT): \[ \oint_C f(z) dz = 0, \quad if f(z) is analytic inside and on C. \]
Checking the options:

1. \( \cos z \) is entire (analytic everywhere), so \( \oint_C \cos z dz = 0 \).
(A) is correct.


2. \( \sec z = \frac{1}{\cos z} \) has poles where \( \cos z = 0 \) (e.g., \( z = \frac{\pi}{2}, -\frac{\pi}{2} \)).
- If a pole exists inside \( C \), the integral is nonzero.
(B) is correct.


3. \( z^n \) is analytic for all \( n \neq -1 \), so \( \oint_C z^n dz = 0 \) for \( n \neq -1 \).
(C) is correct.


4. \( e^z \) is entire (analytic everywhere), so \( \oint_C e^z dz = 0 \).
(D) is correct.

Quick Tip: Cauchy's Integral Theorem states that if a function is analytic inside and on a closed contour, its integral is zero. However, if a function has poles inside the contour, then we apply Residue Theorem.

Expectation of the Sum of Two Dice.

Let \( X_1 \) and \( X_2 \) be the outcomes of two fair six-sided dice.
The sum of outcomes is: \[ X = X_1 + X_2 \]


For a single fair die: \[ E[X_1] = E[X_2] = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5 \]


By linearity of expectation: \[ E[X] = E[X_1] + E[X_2] = 3.5 + 3.5 = 7 \]


Thus, the expected value of \( X \) is: \[ \mathbf{7} \] Quick Tip: For any two independent random variables \( X \) and \( Y \), the expectation follows the additive property: \[ E[X + Y] = E[X] + E[Y] \] This property simplifies expectation calculations in probability theory.

Step 1: Calculate the Reflection Coefficient.
\[ \Gamma = \frac{Z_L - Z_0}{Z_L + Z_0} \]


Substituting values: \[ \Gamma = \frac{(50 - j75) - 50}{(50 - j75) + 50} = \frac{-j75}{100 - j75} \]


Multiplying by the complex conjugate: \[ \Gamma = \frac{5625 - j7500}{15625} \]

\[ |\Gamma| = \sqrt{\left( \frac{5625}{15625} \right)^2 + \left( \frac{-7500}{15625} \right)^2} = 0.6 \]


Step 2: Compute Power Delivered to Load.

The formula for power delivered:

\[ P_L = P_i (1 - |\Gamma|^2) \]

\[ P_L = 10 (1 - 0.36) = 10 \times 0.64 = 6.4 mW \]


Thus, the final power delivered is:
\[ \mathbf{6.4 mW} \] Quick Tip: In transmission line problems, power delivered is calculated using reflection coefficient, which accounts for power loss due to mismatched impedances.

Effect of Time Scaling on Fourier Series.

Given \( y(t) = f(\alpha t) \), the Fourier series of \( f(t) \) is:

\[ f(t) = \sum_{n=-\infty}^{\infty} C_k e^{-jn\omega_0 t} \]


Applying time scaling:

\[ y(t) = f(\alpha t) = \sum_{n=-\infty}^{\infty} C_k e^{-jn\omega_0 (\alpha t)} \]


Rewriting:

\[ y(t) = \sum_{n=-\infty}^{\infty} C_k e^{-jn (\alpha \omega_0) t} \]


Comparing with the standard Fourier series form:

\[ y(t) = \sum_{n=-\infty}^{\infty} d_k e^{-jn\omega_0' t} \]


We get:
- \( \omega_0' = \alpha \omega_0 \) (frequency scales by \( \alpha \)).

- Fourier coefficients remain the same: \( d_k = C_k \).


Checking the given options:


1. \( C_k = d_k \) (Correct)

2. \( C_k = \alpha d_k \) (Incorrect)

3. Time period transformation:
\[ T' = \frac{T_0}{\alpha} \]
(Correct)

4. \( T' = \alpha T_0 \) (Incorrect).


Thus, the correct answers are (A) and (C). Quick Tip: In time-scaling \( y(t) = f(\alpha t) \), - The time period scales as \( T' = \frac{T_0}{\alpha} \). - The fundamental frequency scales as \( \omega_0' = \alpha \omega_0 \). - The Fourier coefficients remain unchanged.

Step 1: Understanding the Timing Parameters.

The given timing parameters from the circuit:

- Setup Time (\( t_s \)) = 2 ns

- Hold Time (\( t_h \)) = 0 ns

- Clock-to-Q Propagation Delay (\( t_{pcq} \)) = 2 ns

- Logic Gate Propagation Delay = 1 ns


Step 2: Compute the Minimum Clock Period.

The minimum clock period for a sequential circuit is given by:

\[ T_{min} = t_{pcq} + t_{pd(logic)} + t_{setup} \]


Substituting values:
\[ T_{min} = 2 ns + 1 ns + 2 ns = 5 ns \]


Step 3: Compute the Maximum Clock Frequency.
\[ f_{max} = \frac{1}{T_{min}} = \frac{1}{5 \times 10^{-9}} Hz \]

\[ f_{max} = 200 MHz \]


Thus, the maximum clock frequency is:
\[ \mathbf{200} MHz \] Quick Tip: For a synchronous sequential circuit, the maximum clock frequency is determined by the longest combinational delay path between flip-flops plus setup time constraints.