GATE - EC 2024 Question Paper with Answer Key and Full Explanation

 

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GATE - EC 2024 Answers with Explanations

 Q.1

 Answer: Option A

 

 Q.2

 Answer: Option A

 

Q.3

To solve this problem, we will use the given age ratios to set up equations. Let's define the current ages of Aman and his father:

• A = Aman's current age
• F = Father's current age

Step 1: Analyze the first condition (5 years ago)

Five years ago, the ratio of Aman’s age to his father’s age was 1:4. This means:

 (A - 5) / (F - 5) = 1 / 4 

Rewriting the equation:

 4(A - 5) = F - 5 

Expanding both sides:

 4A - 20 = F - 5 

Simplifying:

 4A - F = 15 

This gives us Equation 1: 4A - F = 15

Step 2: Analyze the second condition (5 years from now)

Five years from now, the ratio of Aman’s age to his father’s age will be 2:5. This gives us:

 (A + 5) / (F + 5) = 2 / 5 

Rewriting the equation:

 5(A + 5) = 2(F + 5) 

Expanding both sides:

 5A + 25 = 2F + 10 

Simplifying:

 5A - 2F = -15 

This gives us Equation 2: 5A - 2F = -15

Step 3: Solve the system of equations

We now have two equations:

• 4A - F = 15 (Equation 1)
• 5A - 2F = -15 (Equation 2)

Start by solving Equation 1 for F:

 F = 4A - 15 

Substitute this into Equation 2:

 5A - 2(4A - 15) = -15 

Simplifying:

 5A - 8A + 30 = -15 

Solving for A:

 -3A + 30 = -15 

 -3A = -45 

 A = 15 

Aman’s current age is 15 years.

Step 4: Find the father's current age

Now, substitute A = 15 into the equation F = 4A - 15:

 F = 4(15) - 15 = 60 - 15 = 45 

Aman’s father is 45 years old currently.

Step 5: Find the father's age when Aman was born

When Aman was born, his father’s age would have been:

 F - A = 45 - 15 = 30 

Aman's father was 30 years old when Aman was born.

Final Answer:

The correct option is (B) 30 years.

Q.4

We are given the equation:

            (1 / log₂ x) + (1 / log₃ x) + (1 / log₄ x) = 1
        

Our goal is to solve for x, where x > 1.

Step 1: Use the change of base formula

The change of base formula for logarithms is:

            logᵦ a = log a / log b
        

We will apply this to each term in the equation:

            (1 / log₂ x) = log 2 / log x
            (1 / log₃ x) = log 3 / log x
            (1 / log₄ x) = log 4 / log x
        

Substitute these expressions into the original equation:

            (log 2 / log x) + (log 3 / log x) + (log 4 / log x) = 1
        

Step 2: Simplify the equation

Since all terms have a common denominator of log x, we can factor it out:

            (log 2 + log 3 + log 4) / log x = 1
        

We know that log 4 = log (2²) = 2 log 2, so:

            log 2 + log 3 + log 4 = log 2 + log 3 + 2 log 2 = 3 log 2 + log 3
        

Thus, the equation becomes:

            (3 log 2 + log 3) / log x = 1
        

Step 3: Solve for log x

Multiply both sides of the equation by log x to eliminate the denominator:

            3 log 2 + log 3 = log x
        

Step 4: Solve for x

We can rewrite this expression as:

            log x = 3 log 2 + log 3
        

Which simplifies to:

            log x = log (2³) + log 3 = log 8 + log 3
        

Using the property log a + log b = log (a * b), we get:

            log x = log (8 * 3) = log 24
        

Therefore, x = 24.

Final Answer:

The value of x is 24, so the correct option is (C).

Q.5

The expression is:

        3199 - 3196
    

Step 1: Factorize the expression

We can factor out the common term 3196 from both terms:

        3199 - 3196 = 3196 (33 - 1)
    

Step 2: Simplify the expression inside the parentheses

Simplifying 33 - 1 gives:

        33 - 1 = 27 - 1 = 26
    

Thus, the expression becomes:

        3199 - 3196 = 3196 × 26
    

Step 3: Prime factorization of 26

The prime factorization of 26 is:

        26 = 2 × 13
    

Thus, we have:

        3199 - 3196 = 3196 × 2 × 13
    

Step 4: Identify the greatest prime factor

The prime factors are 3, 2, and 13. The greatest prime factor is 13.

Final Answer:

The greatest prime factor of 3199 - 3196 is 13.

Thus, the correct option is A) 13.

 

 Q.6

 Answer: Option C

 

 

 

Q.7

Step 1: Arrangement of the Chalk-Sticks

The four chalk-sticks are arranged in a tight square formation:

• Two chalk-sticks are placed side by side in one row.
• Two chalk-sticks are placed side by side in another row.

The total dimensions of the bundle are determined by the radii and length of the chalk-sticks.

Step 2: Perimeter Calculation for Duct Tape

The duct tape will cover the outer boundary of the chalk-stick bundle, which consists of the following components:

• The width of the bundle is the sum of the diameters of two chalk-sticks placed side by side. Since each chalk-stick has radius r, the total width of the bundle is 2r + 2r = 4r.
• The length of the bundle is the same as the length of one chalk-stick, which is l = 10 cm.
• The corners of the bundle have quarter-circle arcs, with a total length of 2πr (accounting for two quarter-circles). This adds up to half of the circumference of one chalk-stick.

Step 3: Total Length of the Duct Tape

The total length of duct tape required to wrap the bundle of chalk-sticks can be expressed as:

        Length of duct tape = 2r + 2r + 2r + 2r + 2πr
    

Here's why the formula is structured this way:

• 2r + 2r represents the length of the two vertical sides (each is 2r in width).
• 2r + 2r represents the length of the two horizontal sides of the bundle.
• 2πr accounts for the curved portions around the corners of the bundle.
• The four curved portions contribute to the perimeter of the circle.

Step 4: Area of the Duct Tape

The width of the duct tape is equal to the length of the chalk-stick, which is l = 10 cm. So, the area of the duct tape required is:

        Area of tape = Length × Width
                     = 10 × (4 + π)
    

The total area of duct tape required is 10(4 + π) cm².

Final Answer:

The correct area of duct tape required to wrap the bundle of chalk-sticks once is 10(4 + π) cm².

Q.8

Correct Answer: (C)

Using the approximate values from the bar chart:

Category	2003	2023
Underweight	~10%	~5%
Normal	~50%	~30%
Overweight	~30%	~40%
Obese	~20%	~25%

Checking Each Statement

(A) Ratio (Overweight/Normal):
2003: 30/50 = 0.6
2023: 40/30 ≈ 1.33 → increased ✔

(B) Ratio (Underweight/Normal):
2003: 10/50 = 0.2
2023: 5/30 ≈ 0.17 → decreased ✔

(C) Ratio (Obese/Normal):
2003: 20/50 = 0.4
2023: 25/30 ≈ 0.83 → increased, but the statement says it decreased ✘
→ Incorrect statement

(D) Percentage in Normal category:
50% → 30% → decreased ✔

Therefore, the incorrect option is (C).

Q.9

Correct Answer: (A)

Q.10

Given: Two sheets A and B of size 24 cm × 16 cm.

Operations:

• Sheet A is folded twice using FO1 (fold axis parallel to long edge).
• Sheet B is folded twice using FO2 (fold axis parallel to short edge).

---

Step 1: Calculate dimensions after folding

FO1 (fold axis ∥ long edge = 24 cm):

• Dimension along fold axis (24 cm) remains unchanged.
• Dimension perpendicular (16 cm) halves each fold.
• After 1 fold: 24 cm × 8 cm
• After 2 folds: 24 cm × 4 cm

FO2 (fold axis ∥ short edge = 16 cm):

• Dimension along fold axis (16 cm) remains unchanged.
• Dimension perpendicular (24 cm) halves each fold.
• After 1 fold: 12 cm × 16 cm
• After 2 folds: 6 cm × 16 cm

---

Step 2: Calculate perimeters of final shapes

Perimeter formula: P = 2(L + W)

P_A = 2 × (24 + 4) = 2 × 28 = 56 cm

P_B = 2 × (6 + 16) = 2 × 22 = 44 cm

---

Step 3: Calculate ratio of perimeters

56/44 = 14:11

Final Answer:Option A

Q.11

The general form of the complementary function of a differential equation is given by:

y(t) = (A t + B) e^-2t, where A and B are constants.

Step 1: First Derivative of y(t)

To find the first derivative of y(t), we apply the product rule:

    dy/dt = d/dt[(A t + B) e-2t]
           = A e-2t + (A t + B) (-2 e-2t)
           = (A - 2A t - 2B) e-2t
    

Step 2: Second Derivative of y(t)

Now, differentiate again to get the second derivative:

    d2y/dt2 = d/dt[(A - 2A t - 2B) e-2t]
                                   = (-2A) e-2t - 2 (A - 2A t - 2B) e-2t
                                   = (-4A + 4A t + 4B) e-2t
    

Step 3: Formulating the Differential Equation

The corresponding second-order linear differential equation is of the form:

    d2y/dt2 + p (dy/dt) + q y = f(t)
    

Substituting the coefficients, we get:

    d2y/dt2 + 4 (dy/dt) + 4 y = f(t)
    

Thus, the correct differential equation is:

    d2y/dt2 + 4 (dy/dt) + 4 y = f(t)
    

Answer

The correct answer is (A).

Q.12

Understanding the Relationship Between Decibels per Decade (dB/decade) and Decibels per Octave (dB/octave)

To understand the relationship between decibels per decade (dB/decade) and decibels per octave (dB/octave), we need to recall that a decade represents a tenfold increase in frequency, whereas an octave represents a twofold increase in frequency.

Relationship Between Decade and Octave

First, let's define the relationship between these units. The ratio of a decade to an octave is given by:

Number of octaves in a decade = log2(10) ≈ 3.32

This means that a change by 1 decade is equivalent to approximately 3.32 octaves.

Converting 40 dB/decade to dB/octave

If we have a slope of 40 dB/decade, we can determine its equivalent in dB/octave by dividing by the number of octaves in a decade:

dB/octave = 40 dB/decade ÷ 3.32 octaves/decade ≈ 12.05 dB/octave

Final Answer:

Therefore, 40 dB/decade is approximately equal to 12 dB/octave. Thus, the correct answer is:

Option A: 12 dB/octave

Q.13

Given System

The open-loop transfer function is given by:

\[ G(s) = \frac{6}{s(s + 1)(s + 2)} \]

For Negative Unity Feedback System

The relationship between the input \( r(t) \), error \( e(t) \), and output \( y(t) \) is:

\[ E(s) = R(s) - Y(s) \]

\[ Y(s) = G(s)E(s) \]

\[ E(s) = R(s) - G(s)E(s) \]

To find the steady-state error, we use the final value theorem:

\[ \lim_{t \to \infty} e(t) = \lim_{s \to 0} s \cdot E(s) \] = 0

We need to check whether the system is stable or not.

Stability Check Using Routh-Hurwitz Criterion

The characteristic equation for the system is:

\[ s^3 + 3s^2 + 2s + 6 = 0 \]

Routh Array

From the Routh-Hurwitz criterion, the Routh array is formed as:

\[ \begin{array}{c|cc} s^3 & 1 & 2 \\ s^2 & 3 & 6 \\ s^1 & 0 & 0 \\ \end{array} \]

We observe that there is a zero in the first column, which indicates that the system has a pole on the imaginary axis (marginally stable or oscillatory behavior).

Conclusion

Since the system has unstable poles (specifically, a pole on the imaginary axis), the steady-state error does not exist, and the system exhibits oscillatory behavior. Hence, the correct answer is:

Option D: \( \lim_{t \to \infty} e(t) \) does not exist, \( e(t) \) is oscillatory.

Q.14

Options:

• (A) \( \sum_{m=-\infty}^{\infty} X\left(f + \frac{m}{T}\right) = T \)
• (B) \( \sum_{m=-\infty}^{\infty} X\left(f + \frac{m}{T}\right) = \frac{1}{T} \)
• (C) \( \sum_{m=-\infty}^{\infty} X(f + mT) = T \)
• (D) \( \sum_{m=-\infty}^{\infty} X(f + mT) = \frac{1}{T} \)

Solution:

The problem deals with the Nyquist criterion for zero intersymbol interference (ISI) in a digital communication system. The Nyquist condition for zero ISI is mathematically expressed as:

\( \sum_{m=-\infty}^{\infty} X\left(f + \frac{m}{T}\right) = \frac{1}{T} \)

This condition ensures that there is no overlap between the frequency components of successive symbols, thus preventing ISI at the receiver.

Correct Answer:

(B) \( \sum_{m=-\infty}^{\infty} X\left(f + \frac{m}{T}\right) = \frac{1}{T} \)

Q.15

For a lossless transmission line terminated with short circuit, input impedance is given by:

Z_in = jZ₀ tan βl

Normalized impedance is given by:

Z_in = Z_in / Z₀ = jZ₀ tan βl / Z₀

Z_in = j tan βl

For normalized impedance Z_inA:

Electrical length is given by:

βl = (2π / λ) × (λ / 8) = π / 4

tan βl = tan (π / 4)

Normalized input impedance at A is given by:

Z_inA = j tan (π / 4)

Z_inA = j1Ω

For normalized impedance Z_inB:

Electrical length is given by:

βl = (2π / λ) × (λ / 4) = π / 2

tan βl = tan (π / 2) = ∞

Normalized input impedance at B is given by:

Z_inB = j tan (π / 2)

Z_inB = ∞

For normalized impedance Z_inC:

Electrical length is given by:

βl = (2λ / λ) × (3λ / 8) = 3π / 4

tan βl = tan (3π / 4)

Normalized input impedance at C is given by:

Z_inC = j tan (3π / 4)

Z_inC = -j1Ω

For normalized impedance Z_inD:

Electrical length is given by:

βl = (2π / λ) × (4λ / 8) = π

tan βl = tan π

Normalized input impedance at D is given by:

Z_inD = j tan π

Z_inD = 0

Hence, the correct option is (A).

Q.16

Given:

Let i and j be the unit vectors along the x and y axes, respectively, and let A be a positive constant. The following vector fields are given:

F₁ = A(iy + jx)

F₂ = A(iy - jx)

Analysis:

To determine which vector field represents an electrostatic field, we need to check if the curl of each field is zero. An electrostatic field is conservative, meaning its curl must be zero.

Step 1: Analyzing F₁

For F₁ = A(iy + jx), we have:

F_x = A y

F_y = A x

Now, calculating the curl of F₁:

∇ × F₁ = (∂(A x)/∂x - ∂(A y)/∂y) k

∇ × F₁ = A - A = 0

This means that F₁ is a conservative field, and hence it is an electrostatic field.

Step 2: Analyzing F₂

For F₂ = A(iy - jx), we have:

F_x = A y

F_y = -A x

Now, calculating the curl of F₂:

∇ × F₂ = (∂(-A x)/∂x - ∂(A y)/∂y) k

∇ × F₂ = -A - A = -2A

This means that the curl of F₂ is non-zero, so F₂ is not a conservative field and hence cannot represent an electrostatic field.

Conclusion:

Only F₁ is an electrostatic field because its curl is zero.

Answer:

(B) Only F₁ is an electrostatic field.

Q.17

We are tasked with calculating the small-signal input impedance, \( Z_{in}(j\omega) \), for the given circuit. Below are the steps:

Key Parameters:

• g_m: Small-signal transconductance of the MOSFET.
• C₁: Input capacitance.
• C_L: Load capacitance.

Step-by-Step Derivation:

1. Small-Signal Model:
The MOSFET is represented as a voltage-controlled current source with a transconductance of \( g_m \), and its gate is driven by the input signal. The small-signal drain current is \( i_d = g_m v_{gs} \), where \( v_{gs} \) is the small-signal gate-source voltage.

2. Impedance Calculation:
The total small-signal input impedance \( Z_{in}(j\omega) \) is determined by considering the following components:

• The capacitive impedance due to the input capacitance \( C_1 \): \( Z_C = \frac{1}{j\omega C_1} \).
• The small-signal transconductance \( g_m \), which creates feedback between the drain and gate.
• The impedance due to the load capacitance \( C_L \), which contributes \( \frac{1}{j\omega C_L} \) to the overall impedance.

3. Final Expression for Input Impedance:
By applying Kirchhoff’s current law (KCL) and considering the voltage division, the small-signal input impedance is given by the expression:

\[ Z_{in}(j\omega) = \frac{-g_m}{C_1 C_L \omega^2} + \frac{1}{j\omega C_1} + \frac{1}{j\omega C_L} \]

Conclusion:

The correct expression for the small-signal input impedance is:

\[ Z_{in}(j\omega) = \frac{-g_m}{C_1 C_L \omega^2} + \frac{1}{j\omega C_1} + \frac{1}{j\omega C_L} \]

Answer:

The correct choice is:

A:

\[ Z_{in}(j\omega) = \frac{-g_m}{C_1 C_L \omega^2} + \frac{1}{j\omega C_1} + \frac{1}{j\omega C_L} \]

Q.18

Given:

• The open-loop small-signal voltage gain \( A_{OL} \) is 40.
• All transistors are biased in saturation.
• The current source \( I_{SS} \) is ideal.
• The circuit is a differential amplifier with feedback.
• The goal is to find the closed-loop low-frequency small-signal voltage gain \( \frac{V_{out}}{V_{in}} \).

Approach:

The closed-loop gain \( A_{CL} \) of a circuit with negative feedback can be approximated by:

        A_{CL} = \frac{A_{OL}}{1 + A_{OL} \beta}
    

where \( \beta \) is the feedback factor.

Feedback Factor \( \beta \):

The feedback factor \( \beta \) for this configuration is typically the ratio of the feedback resistor to the total resistance seen at the output node. However, in this case, based on the typical configuration of a differential pair with an ideal current source and assuming no additional resistive components, the feedback factor \( \beta \) is considered approximately 1.

Closed-Loop Gain Calculation:

Using the equation for closed-loop gain:

        A_{CL} = \frac{A_{OL}}{1 + A_{OL} \beta}
    

Substituting \( A_{OL} = 40 \) and \( \beta \approx 1 \):

        A_{CL} = \frac{40}{1 + 40 \times 1} = \frac{40}{41}
    

So,

        A_{CL} \approx 0.9756
    

Final Answer:

Rounding to three decimal places, the closed-loop gain is approximately 0.976.

Answer:

The correct option is: (A) 0.976

Q.19

AB / CD	00	01	11	10
00	1	0	0	1
01	0	1	1	0
11	1	1	1	1
10	1	0	0	1

Step 1: Identify All '1' Cells

The 1's are located at minterms:
0, 2, 5, 7, 8, 10, 12, 13, 14, 15

Step 2: Create Largest Possible Groups

Group 1: Four-cell group (Row AB=11)

Covers minterms: 12, 13, 14, 15
These all share: A = 1 and B = 1
Term → AB

Group 2: Four-cell group (Column CD=00 for rows B=0)

Covers minterms: 0, 2, 8, 10
These all share: B = 0 and D = 0
Term → B'D'

Group 3: Four-cell group (Block containing 5, 7, 13, 15)

Covers minterms: 5, 7, 13, 15
These all share: B = 1 and D = 1
Term → BD

Step 3: Write final simplified function

F = AB + B'D' + BD

Q.20

Given:

A white Gaussian noise w(t) with zero mean and power spectral density S_w(f) = N₀/2

Step 1: Properties of the white Gaussian noise

• E[w(t)] = 0
• S_w(f) = N₀/2

Step 2: Transfer function of the RC Low-Pass Filter

The transfer function of the first-order RC low-pass filter is:

H(f) = 1 / (1 + j2πfRC)  ... (i)

Step 3: Mean of Output Noise

The mean of the output noise n(t) is:

E[n(t)] = E[w(t)] * H(f) = 0

Hence, the mean of output noise is zero.

Step 4: Variance of Output Noise

The variance of output noise at time t = t_k is:

Var[n(tk)] = E[n2(tk)]

Since variance equals square of output noise, the calculation follows:

Var[n(tk)] = E[n2(tk)] = Power of output noise

Step 5: Power Spectral Density of Output Noise

The power spectral density of the output noise is:

Sn(f) = Sw(f) * |H(f)|2

Sn(f) = (N0/2) * |H(f)|2

Sn(f) = (N0/2) * (1 / (1 + (2πfRC)2))

Step 6: Calculate Power of Output Noise

The power of output noise P_n is:

Pn = ∫-∞∞ Sn(f) df

Substituting the expression for S_n(f):

Pn = ∫-∞∞ (N0/2) * (1 / (1 + (2πfRC)2)) df

Step 7: Final Simplification

After solving the integral, the power of the output noise is:

Pn = (N0 / 4RC)

Step 8: Conclusion

Therefore, the variance of the output noise is:

Var[n(tk)] = E[n2(tk)] = (N0 / 4RC)

Answer:

The correct answer is (A) N₀ / 4RC.

Q.21

A causal, stable LTI system with impulse response h(t) produces output:

y(t) = x(t) * h(t)

Now the input applied is a time-scaled version x(0.5t), and the impulse response of the second system is also time-scaled: h(0.5t).

Key LTI Time-Scaling Property

x(at) * h(at) = (1 / |a|) · y(at)

Here, a = 0.5, so:

x(0.5t) * h(0.5t) = (1 / 0.5) · y(0.5t)

= 2y(0.5t)

Final Answer: Option A) 2y(0.5t)

Q.22

Non-Degenerately Doped N-Type Semiconductor

(i) At low temperature: At low temperature, negligible intrinsic electron-hole pairs exist, and the donor electrons remain bound to donor atoms.

(ii) As temperature rises: These donor electrons gain enough thermal energy to be donated to the conduction band. Around 100 K, all donor atoms are ionized.

This temperature range is called the ionization region.

Based on this behavior, the graph of free electron concentration (n) versus temperature (T) is as shown in option (A).

Hence, the correct option is (A).

Q.23

Step 1: Analyze the circuit functionality

The circuit is a full-wave rectifier utilizing a center-tapped transformer. The rectified voltage across the load resistor \(R_L\) is derived from the secondary winding of the transformer.

Step 2: Determine the relationship between primary and secondary voltages

The primary voltage is given as:

\[ V_s(t) = 10 \sin(\omega t) \]

The secondary voltage \(V_{\text{sec}}(t)\) is scaled by the transformer turns ratio \(n:1\), such that:

\[ V_{\text{sec}}(t) = \frac{10}{n} \sin(\omega t) \]

Step 3: Rectified output voltage

In a full-wave rectifier, the output is the absolute value of the secondary voltage:

\[ V_L(t) = \left|\frac{10}{n} \sin(\omega t)\right| \]

Step 4: Compute the average output voltage

For a full-wave rectified sine wave, the average voltage is:

\[ V_{\text{avg}} = \frac{2}{\pi} V_{\text{peak}} \]

Substituting \(V_{\text{peak}} = \frac{10}{n}\), we get:

\[ V_{\text{avg}} = \frac{2}{\pi} \cdot \frac{10}{n} \]

Step 5: Solve for \(n\)

Given that \(V_{\text{avg}} = \frac{2.5}{\pi}\), equate and solve for \(n\):

\[ \frac{2}{\pi} \cdot \frac{10}{n} = \frac{2.5}{\pi} \]

Simplify:

\[ \frac{10}{n} = 2.5 \]

\[ n = \frac{10}{2.5} = 4 \]

Final Answer:

The transformer turns ratio \(n\) is therefore 4, which matches option (A).

Q.24

Transfer Function:

The given transfer function is:

        H(z) = (2z^2 + 3) / ((z + 1/3)(z - 1/3))
    

1. Stability of the System

To determine if the system is stable, we need to check the poles of the transfer function.

The poles are the values of z that make the denominator zero. Set the denominator equal to zero:

        (z + 1/3)(z - 1/3) = 0
    

So the poles are at:

        z = -1/3 and z = 1/3
    

The magnitudes of the poles are:

        |z_1| = | -1/3 | = 1/3 and |z_2| = | 1/3 | = 1/3
    

Since both poles lie inside the unit circle (i.e., their magnitudes are less than 1), the system is stable.

2. Minimum Phase System

A system is minimum phase if all its poles and zeros lie inside the unit circle.

First, we find the zeros by setting the numerator equal to zero:

        2z^2 + 3 = 0 => z^2 = -3/2 => z = ±j√(3/2)
    

These zeros are imaginary, and their magnitudes are greater than 1, meaning the system is not a minimum phase system.

3. Initial Value of the Impulse Response

The initial value of the impulse response \( h[0] \) can be calculated using the final value theorem:

        h[0] = lim_{z -> ∞} (z - 1) * H(z)
    

Multiplying the numerator and denominator by \( z - 1 \) and simplifying, we find that as \( z \to ∞ \), the dominant term is \( 2z^3 / z^2 \), so:

        h[0] = 2
    

4. Final Value of the Impulse Response

The final value of the impulse response can be calculated using the final value theorem in the z-domain:

        lim_{z -> 1} (z - 1) * H(z)
    

Substitute \( z = 1 \) into the transfer function and simplify:

        H(1) = (2(1)^2 + 3) / ((1 + 1/3)(1 - 1/3)) = 5 / (4/3 * 2/3) = 5 / (8/9) = 45/8
    

The final value of the impulse response is 0, since this is the theoretical outcome based on the properties of the system.

Conclusion:

• (A) The system is stable. True
• (B) The system is a minimum phase system. False
• (C) The initial value of the impulse response is 2. True
• (D) The final value of the impulse response is 0. True

Correct Answers:

The correct answers are: (A), (C), and (D).

Q.25

We start with the continuity equation:

∂ρ/∂t + ∇·(ρu) = 0

Integrate over an arbitrary control volume V:

∫V (∂ρ/∂t) dV + ∫V ∇·(ρu) dV = 0

Rearranging:

∫V (∂ρ/∂t) dV = - ∫V ∇·(ρu) dV

Apply the divergence theorem:

∫V ∇·(ρu) dV = ∮S ρu · n̂ dS

So the equivalent integral form becomes:

∫V (∂ρ/∂t) dV = - ∮S ρu · n̂ dS

Correct Options

• (A) ✓ Correct
• (C) ✓ Correct
• (B) ✗ Incorrect sign
• (D) ✗ Incorrect sign

Final Answer

(A) and (C)

Q.26

Answer: Option A, B, an C

Q.27

Step 1: Bits needed for opcode

Instruction set size = 40
Bits needed = ⌈log₂(40)⌉ = 6 bits

Step 2: Bits needed for register identifiers

Number of registers = 24
Bits needed to identify one register = ⌈log₂(24)⌉ = 5 bits
Number of register fields = 3 (2 source + 1 destination)
Total bits for registers = 3 × 5 = 15 bits

Step 3: Bits remaining for immediate

Total instruction length = 32 bits
Bits used (opcode + registers) = 6 + 15 = 21 bits
Remaining bits for immediate = 32 − 21 = 11 bits

Step 4: Maximum unsigned value for immediate field

For an 11-bit unsigned integer, maximum value = 2¹¹ − 1 = 2047

Final Answer

The maximum immediate value is 2047.

Q.28

Given the signal

s(t) = A cos(400πt) + B cos(360πt) + B cos(440πt)

The term at 400π is the carrier, so its amplitude is A.

Step 1: Carrier Power

Carrier power (with R = 1 Ω) is:

Pc = A² / 2 = 50

Therefore:

A² = 100 → A = 10

Step 2: Sideband Power

Each sideband has amplitude B, so each contributes:

PSB₁ = B² / 2

There are two sidebands, so:

PSB = B²

Step 3: Using the Given Power Ratio

The ratio of total sideband power to total power is:

PSB / Ptotal = 1/9

But:

Ptotal = Pc + PSB = 50 + B²

Substituting:

B² / (50 + B²) = 1/9

Cross-multiplying:

9B² = 50 + B²

8B² = 50

B² = 6.25 → B = 2.50

Final Answer: B = 2.50 V

Q.29

The given equation in base r is:

x² − 12x + 37 = 0

It is given that one solution is x = 8 (in base r).

Step 1: Convert coefficients from base r to decimal

Digits 1, 2, 3, and 7 must be valid in base r, so:

r > 7

12 in base r:

12_r = 1·r + 2

37 in base r:

37_r = 3r + 7

Since x = 8 (base r), the decimal value is 8, requiring r > 8.

Therefore, r ≥ 9.

Step 2: Substitute x = 8 into the equation

x² − (12_r)x + (37_r) = 0

Substitute values:

64 − (r + 2)·8 + (3r + 7) = 0

Simplify:

64 − 8r − 16 + 3r + 7 = 0

55 − 5r = 0

Solve for r:

5r = 55
r = 11

Final Answer: 11

Q.30

We are given the vectors:

v₁ = (2, −3, α),
v₂ = (3, −1, 3),
v₃ = (1, −5, 7)

These vectors do not form a basis of ℝ³ when they are linearly dependent. This happens when the determinant of the matrix formed by them is zero.

Step 1: Form the matrix

A = | 2  -3   α |
    | 3  -1   3 |
    | 1  -5   7 |

Step 2: Compute the determinant

det(A) =

2 · | -1   3 |
    | -5   7 |

+ 3 · | 3   3 |
      | 1   7 |

+ α · | 3  -1 |
      | 1  -5 |

Step 3: Evaluate each minor

| -1 3 | = (-1)(7) - (3)(-5) = 8
| -5 7 |

| 3 3 | = 3(7) - 3(1) = 18
| 1 7 |

| 3 -1 | = 3(-5) - (-1)(1) = -14
| 1 -5 |

Step 4: Substitute into determinant

det(A) = 16 + 54 - 14α = 70 - 14α

Set det(A) = 0 for linear dependence:

70 − 14α = 0 → 14α = 70 → α = 5

Final Answer: 5

Q.31

Applying KCL:

V1 / (1 × 10³) + V2 / (1 × 10³) + 2 mA − 5 mA = 0
(V1 + V2) / 1000 = 3 mA
V1 + V2 = 3 volt    ...(i)

Applying KVL (super node):

V2 − V1 = 1000 I₀    ...(ii)
V2 − V1 = 1000 ( V1 / 1000 )
V2 = 2V1    ...(iii)

From equation (i):

V2 + V2/2 = 3
3V2 / 2 = 3
3V2 = 6
V2 = 2 volt

Finding I_x:

I_x = V2 / (1 × 10³) = 2 / 1000 = 2 mA

Hence, the correct answer is 2 mA.

Q.32

For an R–L network, the time constant (τ) is given by:

τ = L / R_TH    ...(i)

where R_TH is the Thevenin resistance across the inductor L when all independent sources are deactivated.

Circuit Diagram

Calculation

R_TH = 2Ω + (4Ω ∥ 4Ω)

R_TH = 2Ω + (4 × 4 / 8)

R_TH = 2 + 2 = 4Ω

From equation (i):

τ = 3/4 sec

τ = 0.75 sec

Hence, the correct answer is 0.75.

Q.33

1. Think of the Pair (X, Y) as a Point in a Square

Since both X and Y range from 0 to 1, every possible outcome (X, Y) lies inside the unit square:

(0,1) ─────────────── (1,1)
   │                    │
   │                    │
   │                    │
(0,0) ─────────────── (1,0)

Because the distribution is uniform and the variables are independent, every point inside this square is equally likely.

2. The Condition X ≥ Y

The condition X ≥ Y describes all points where the horizontal value is greater than or equal to the vertical value. This corresponds to the region below the line X = Y.

Y
│
│        Above the line → X < Y
│       /
│      /
│     / ← diagonal line X = Y
│    /
│   /
│  /  ← below line (X ≥ Y)
└─────────────────────── X

3. Why the Region Is a Triangle

The line X = Y connects the points (0, 0) and (1, 1). This line splits the unit square into two equal right triangles.

The region where X ≥ Y is simply the lower-right triangle.

4. Computing the Area

The square has total area:

1 × 1 = 1

Each triangle created by the diagonal has base 1 and height 1, so its area is:

(1/2) × 1 × 1 = 1/2

Since probability corresponds to area (because every point is equally likely),

P(X ≥ Y) = 1/2.

Final Intuition

The diagonal divides the square into two equal halves. One half satisfies X ≥ Y. Therefore:

Probability = 1/2.

Q.34

log₂(16) = 4

Final Answer: 4

Q.35

Given:

• Capacitance: C = 10 μF
• Initial capacitor voltage: V(0⁻) = 10 V
• Zener diode voltage: V_Z = 5 V
• Resistor: R = 10 kΩ

1. Initial Energy in the Capacitor

E_initial = (1/2) C V² = (1/2)(10 × 10⁻⁶)(10²) = 5 mJ

2. Final Energy in the Capacitor

When the capacitor discharges to 5 V, the Zener stops conducting.

E_final = (1/2) C V² = (1/2)(10 × 10⁻⁶)(5²) = 0.125 mJ

3. Total Energy Lost

E_lost = E_initial − E_final = 5 − 0.125 = 4.875 mJ

4. Energy Dissipated in the Zener Diode

The capacitor discharges from 10 V to 5 V while the Zener clamps at 5 V. The Zener receives energy:

E_Z = ∫ 5C dV from 10 to 5 = 5C(10 − 5) = 25C

Substituting C = 10 μF:

E_Z = 25 × 10 × 10⁻⁶ = 0.00025 J = 0.25 mJ

Final Answer:

0.250 mJ

Q.36

To find the surface area of the region north of latitude 60°N on a sphere of radius R, we treat it as a spherical cap.

Step 1: Convert Latitude to Colatitude

Latitude \(60^\circ \text{N}\) corresponds to a colatitude:

\(\theta = 90^\circ - 60^\circ = 30^\circ\)

Step 2: Formula for Spherical Cap Area

The area of a spherical cap cut at colatitude \(\theta\) is:

\(A = 2\pi R^2 (1 - \cos\theta)\)

Step 3: Substitute Values

Since \(\cos 30^\circ = \frac{\sqrt{3}}{2}\):

\(A = 2\pi R^2 \left(1 - \frac{\sqrt{3}}{2}\right)\)

\(A = (2 - \sqrt{3})\pi R^2\)

Final Answer: \((2 - \sqrt{3})\pi R^2\)

Q.37

Given a unity negative feedback system with forward gain

\( G(s) = \frac{K}{(s+1)(s+2)(s+3)} \)

The closed-loop impulse response decays faster than \( e^{-1} \) if all closed-loop poles satisfy:

\( \Re(s) < -1 \)

Step 1: Characteristic Equation

The characteristic equation comes from:

\( 1 + G(s) = 0 \)

Thus,

\( (s+1)(s+2)(s+3) + K = 0 \)

Expanding:

\( (s+1)(s+2) = s^2 + 3s + 2 \)
\( (s^2 + 3s + 2)(s+3) = s^3 + 6s^2 + 11s + 6 \)

So the full characteristic equation becomes:

\( s^3 + 6s^2 + 11s + (6 + K) = 0 \)

Step 2: Shift to Boundary at \( s = -1 \)

Let:

\( s = z - 1 \)

Substituting and simplifying gives the shifted polynomial:

\( z^3 + 3z^2 + 2z + K = 0 \)

We now check stability using the Routh–Hurwitz criterion.

Step 3: Routh–Hurwitz Stability

Routh table for:

\( z^3 + 3z^2 + 2z + K \)

Row	Elements
\( z^3 \)	1    2
\( z^2 \)	3    K
\( z^1 \)	\( \frac{6-K}{3} \)
\( z^0 \)	K

For stability, all first-column elements must be positive:

\( K > 0 \)
\( K < 6 \)

Thus:

\( 0 < K < 6 \)

Final Answer: Option (A) — \( 1 \le K \le 5 \)

Q.38

Solution: Closed-Loop Poles at \( -1 \pm j\sqrt{3} \)

The plant transfer function is:

\( G(s) = \frac{1}{s^2} \)

The compensator transfer function is:

\( C(s) = \frac{K(s + \alpha)}{s + 4} \)

Thus the open-loop transfer function is:

\( L(s) = \frac{K(s + \alpha)}{s^2(s + 4)} \)

Step 1: Form the Closed-Loop Characteristic Equation

The characteristic equation comes from:

\( 1 + L(s) = 0 \)

Which becomes:

\( s^2(s + 4) + K(s + \alpha) = 0 \)

Expanding:

\( s^3 + 4s^2 + Ks + K\alpha = 0 \)

Step 2: Use the Given Desired Poles

The closed-loop system must have poles at:

\( -1 \pm j\sqrt{3} \)

These correspond to the quadratic factor:

\( s^2 + 2s + 4 \)

So the full characteristic polynomial must be:

\( (s + a)(s^2 + 2s + 4) \)

Expanding this:

\( s^3 + (2 + a)s^2 + (4 + 2a)s + 4a \)

This must match:

\( s^3 + 4s^2 + Ks + K\alpha \)

Step 3: Compare Coefficients

2 + a = 4 \quad \Rightarrow \quad a = 2

4 + 2a = K \quad \Rightarrow \quad K = 8

4a = K\alpha \quad \Rightarrow \quad 8 = 8\alpha \Rightarrow \alpha = 1

Final Answer: \( \alpha = 1 \) (Option B)