GATE ECE 2024

Q11GATE 2024MCQ1MEngineering Mathematics

The general form of the complementary function of a differential equation is given by $y(t)=(A t+B) e^{-2 t}$ , where $A$ and $B$ are real constants determined by the initial condition. The corresponding differential equation is

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📖 Explanation

The form of the complementary function, $y(t)=(A t+B) e^{-2 t}$ , is characteristic of a second-order linear differential equation with repeated real roots. The exponential part $e^{-2 t}$ indicates that the root is $m=-2$ , and the polynomial factor $(At+B)$ confirms that this root is repeated.

Therefore, the roots of the characteristic equation are $m_1 = -2$ and $m_2 = -2$ . We can reconstruct the characteristic equation from its roots: $(m - (-2))(m - (-2)) = 0$ , which simplifies to $(m+2)^2=0$ .

Expanding this gives the characteristic equation $m^{2+}4m+4=0$ .

This polynomial directly translates back to the differential equation by replacing $m^2$ with $\frac{d^2y}{dt^2}$ , $m$ with $\frac{dy}{dt}$ , and the constant term with $y$ . This gives us the left-hand side of the differential equation, leading to the general form $\frac{d^{2} y}{d t^{2}}+4 \frac{d y}{d t}+4 y=f(t)$ .

Q12GATE 2024MCQ1MControl Systems

In the context of Bode magnitude plots, $40 \mathrm{~dB} /$ decade is the same as _______

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📖 Explanation

In Bode plots, a "decade" refers to a 10-fold change in frequency, while an "octave" refers to a 2-fold change. Let's establish a baseline conversion to relate these two units of slope.

A standard slope from a single pole or zero is $20 \mathrm{~dB} /$ decade. To find its equivalent in dB/octave, we calculate the change over one octave (a factor of 2 in frequency):
$\Delta \text{Gain} = 20 \log_{10}(2) \approx 20 \times 0.301 \approx 6 \mathrm{~dB}$
This means a $20 \mathrm{~dB} /$ decade slope is equivalent to a $6 \mathrm{~dB} /$ octave slope.

The given slope is $40 \mathrm{~dB} /$ decade, which is exactly twice the baseline slope. Therefore, its octave equivalent is also twice as large:
$2 \times (6 \mathrm{~dB} / \text{octave}) = 12 \mathrm{~dB} / \text{octave}$

Q13GATE 2024MCQ1MControl Systems

In the feedback control system shown in the figure below $G(s)=\frac{6}{s(s+1)(s+2)}$ . $R(s), Y(s)$ , and $E(s)$ are the Laplace transforms of $r(t), y(t)$ , and $e(t)$ , respectively. If the input $r(t)$ is a unit step function, then _____

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📖 Explanation

To understand the long-term behavior of the error signal, we must first assess the stability of the closed-loop system. The system's stability is determined by its characteristic equation, $1 + G(s) = 0$ .

Substituting the given $G(s)$ , we get $1 + \frac{6}{s(s+1)(s+2)} = 0$ . This simplifies to the polynomial $s^3 + 3s^2 + 2s + 6 = 0$ .

Using the Routh-Hurwitz stability criterion for this cubic equation, we check the products of the coefficients. Since the product of the inner coefficients ( $3 \times 2 = 6$ ) is equal to the product of the outer coefficients ( $1 \times 6 = 6$ ), this indicates the system has poles on the imaginary axis.

A system with poles on the imaginary axis is marginally stable, which means its response to an input will contain sustained oscillations. Therefore, the error signal $e(t)$ will oscillate indefinitely and not converge to a constant value, meaning its limit does not exist.

Q14GATE 2024MCQ1MCommunication Systems

A digital communication system transmits through a noiseless bandlimited channel $[-W W]$ . The received signal $z(t)$ at the output of the receiving filter is given by $z(t)=\sum_{n} b[n] x(t-n T)$ where $b[n]$ are the symbols and $x(t)$ is the overall system response to a single symbol. The received signal is sampled at $t=m T$ . The Fourier transform of $x(t)$ is $X(f)$ . The Nyquist condition that $X(f)$ must satisfy for zero intersymbol interference at the receiver is ______

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📖 Explanation

To avoid intersymbol interference (ISI), the value of the pulse $x(t)$ must be zero at all sampling instants corresponding to other symbols. Specifically, when we sample at integer multiples of the symbol period $T$ , we require $x(mT) = 0$ for all non-zero integers $m$ . For proper detection, we also normalize the pulse so that its value at the desired sampling instant is one, i.e., $x(0) = 1$ .

This time-domain condition has an equivalent form in the frequency domain, known as the Nyquist ISI criterion. It is derived using the Poisson Sum Formula, which relates the pulse's time-domain samples to its frequency spectrum, $X(f)$ . The condition $x(mT) = \delta[m]$ (the Kronecker delta) in the time domain transforms to the following condition in the frequency domain:
$\frac{1}{T} \sum_{m=-\infty }^{\infty } X\left(f - \frac{m}{T}\right) = 1$
Multiplying both sides by $T$ gives the criterion. Since the summation is over all integers from $-\infty$ to $\infty$ , summing over $m$ is equivalent to summing over $-m$ , so we can write:
$\sum_{m=-\infty }^{\infty } X\left(f + \frac{m}{T}\right) = T$
This states that the sum of the spectrum and all its replicas, shifted by multiples of the symbol rate $1/T$ , must be a constant.

Q15GATE 2024MCQ1MElectromagnetics

Consider a lossless transmission line terminated with a short circuit as shown in the figure below. As one moves towards the generator from the load, the normalized impedances $z_{\text {\in } A}, z_{\text {\in } B}, z_{\text {\in } C}$ and $z_{\text {\in } D}$ (indicated in the figure) are _____

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📖 Explanation

The normalized input impedance $z_{\in}$ for a lossless transmission line terminated with a short circuit is determined by its length $l$ . The governing equation is $z_{\in} = j \tan(\beta l)$ , where $\beta = 2\pi/\lambda$ is the phase constant. We calculate the impedance at each point by using its specific distance from the shorted end.

At point A, the distance is $l=\lambda/8$ , so $z_{inA} = j \tan(\frac{2\pi}{\lambda} \frac{\lambda}{8}) = j \tan(\frac{\pi}{4}) = +j1$ .
At point B, the distance is $l=\lambda/4$ , so $z_{inB} = j \tan(\frac{\pi}{2}) = \infty$ .
At point C, the distance is $l=3\lambda/8$ , so $z_{inC} = j \tan(\frac{3\pi}{4}) = -j1$ .
At point D, the distance is $l=\lambda/2$ , so $z_{inD} = j \tan(\pi) = 0$ .

Q16GATE 2024MCQ1MElectromagnetics

Let $\hat{i}$ and $\hat{j}$ be the unit vectors along $x$ and $y$ axes, respectively and let $A$ be a positive constant. Which one of the following statements is true for the vector fields $\vec{F}_{1}=A(\hat{i} y+\hat{j} x)$ and $\vec{F}_{2}=A(\hat{i} y-\hat{j} x) ?$

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📖 Explanation

A key characteristic of an electrostatic field is that it must be conservative. In mathematical terms, this means the curl of the field must be zero. We can check this condition, $\nabla \times \vec{F} = 0$ , for each vector field.

Let's first examine $\vec{F}_{1}=A(y\hat{i} + x\hat{j})$ . The components are $F_{1x} = Ay$ and $F_{1y} = Ax$ .
The curl is calculated as $\nabla \times \vec{F}_{1} = \left(\frac{\partial F_{1y}}{\partial x} - \frac{\partial F_{1x}}{\partial y}\right)\hat{k} = \left(\frac{\partial (Ax)}{\partial x} - \frac{\partial (Ay)}{\partial y}\right)\hat{k} = (A-A)\hat{k} = 0$ .
Since its curl is zero, $\vec{F}_{1}$ can be an electrostatic field.

Now let's examine $\vec{F}_{2}=A(y\hat{i} - x\hat{j})$ . The components are $F_{2x} = Ay$ and $F_{2y} = -Ax$ .
The curl is $\nabla \times \vec{F}_{2} = \left(\frac{\partial F_{2y}}{\partial x} - \frac{\partial F_{2x}}{\partial y}\right)\hat{k} = \left(\frac{\partial (-Ax)}{\partial x} - \frac{\partial (Ay)}{\partial y}\right)\hat{k} = (-A-A)\hat{k} = -2A\hat{k}$ .
Because the curl of $\vec{F}_{2}$ is not zero, it cannot be an electrostatic field. Thus, only $\vec{F}_{1}$ qualifies.

Q17GATE 2024MCQ1MAnalog Circuits

In the circuit below, assume that the long channel NMOS transistor is biased in saturation. The small signal trans-conductance of the transistor is $g_{m}$ Neglect body effect, channel length modulation and intrinsic device capacitances. The small signal input impedance $Z_{\text {\in }}(j \omega)$ is _____

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📖 Explanation

To find the input impedance $Z_{\in}(j\omega)$ , we'll use Kirchhoff's laws on the small-signal equivalent circuit. Let $Z_1 = 1/(j\omega C_1)$ and $Z_L = 1/(j\omega C_L)$ be the impedances of the capacitors. The input voltage $V_{\in}$ equals the voltage drop across $C_1$ plus the source voltage $V_s$ , giving $V_{\in} = I_{\in}Z_1 + V_s$ . The voltage across $C_1$ is also the gate-source voltage, so $V_{gs} = I_{\in}Z_1$ .

At the source node, the input current $I_{\in}$ and the transistor's dependent current $g_m V_{gs}$ combine and flow through the load impedance $Z_L$ . This establishes the source voltage as $V_s = (I_{\in} + g_m V_{gs})Z_L$ . Substituting $V_{gs} = I_{\in}Z_1$ into this equation gives $V_s = I_{\in}(Z_L + g_m Z_1 Z_L)$ .

Now, substitute this expression for $V_s$ back into our equation for $V_{\in}$ : $V_{\in} = I_{\in}Z_1 + I_{\in}(Z_L + g_m Z_1 Z_L)$ . Dividing by $I_{\in}$ gives the input impedance: $Z_{\in} = Z_1 + Z_L + g_m Z_1 Z_L$ .

Finally, inserting the specific component impedances yields $Z_{\in} = \frac{1}{j\omega C_1} + \frac{1}{j\omega C_L} + g_m (\frac{1}{j\omega C_1})(\frac{1}{j\omega C_L})$ . Since $(j\omega)^2 = -\omega^2$ , the expression simplifies to $Z_{\in} = \frac{1}{j\omega C_1} + \frac{1}{j\omega C_L} - \frac{g_m}{\omega^2 C_1 C_L}$ .

Q18GATE 2024MCQ1MAnalog Circuits

For the closed loop amplifier circuit shown below, the magnitude of open loop low frequency small signal voltage gain is 40. All the transistors are biased in saturation. The current source Iss is ideal. Neglect body effect. channel length modulation and intrinsic device capacitances. The closed loop low frequency small signal voltage gain $\frac{V_{\text {out }}}{V_{\text {\in }}}$ (rounded off to three decimal places) is ____.

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📖 Explanation

This circuit is a differential amplifier connected in a negative feedback configuration. The output voltage, $V_{\text{out}}$ , is fed directly back to the gate of the right-hand input transistor. This creates a voltage follower or unity-gain feedback loop, meaning the feedback factor $\beta$ is exactly 1.

We can calculate the closed-loop gain, $A_{CL}$ , using the standard formula for a negative feedback amplifier:
$A_{CL} = \frac{A_{OL}}{1 + \beta A_{OL}}$
Given the open-loop gain $A_{OL} = 40$ and with $\beta = 1$ , we find:
$A_{CL} = \frac{40}{1 + (1)(40)} = \frac{40}{41} \approx 0.976$

Q19GATE 2024MCQ1MDigital Circuits

For the Boolean function $F(A, B, C, D)=\Sigma m(0,2,5,7,8,10,12,13,14,15)$ the essential prime implicants are

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📖 Explanation

To determine the essential prime implicants, we first identify all prime implicants for the function using a method like a Karnaugh map. The prime implicants (PIs) for this function are $\bar{B}\bar{D}$ , $BD$ , $AB$ , and $A\bar{D}$ .

A prime implicant is considered essential if it is the only PI that covers one or more of the function's minterms.

The term $\bar{B}\bar{D}$ is essential because it is the only PI that covers minterms $m_0$ and $m_2$ .
The term $BD$ is essential because it is the only PI that covers minterms $m_5$ and $m_7$ .

The remaining PIs, $AB$ and $A\bar{D}$ , are not essential, as all the minterms they cover are also covered by other PIs. Therefore, the only essential prime implicants are $\bar{B}\bar{D}$ and $BD$ .

Q20GATE 2024MCQ1MCommunication Systems

A White Gaussian noise $w(t)$ with zero mean and power spectral density $\frac{N_{0}}{2}$ , when applied to a first-order $R C$ low pass filter produces an output $n(t)$ . At a particular time $t=t_{k}$ , the variance of the random variable $n\left(t_{k}\right)$ is _____

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📖 Explanation

The input white noise has zero mean, and since the RC filter is a linear system, the output noise $n(t)$ also has zero mean. For a zero-mean process, the variance is equal to its average power, which is given by its autocorrelation function evaluated at zero, $R_n(0)$ .

To find this, we first determine the output's power spectral density (PSD), $S_n(f)$ . This is the input PSD, $S_w(f) = \frac{N_0}{2}$ , multiplied by the squared magnitude of the filter's frequency response, $|H(f)|^2 = \frac{1}{1+(\omega RC)^2}$ .

Thus, the output PSD is $S_n(f) = \frac{N_0}{2} \cdot \frac{1}{1+(\omega RC)^2}$ . The autocorrelation function $R_n(\tau)$ is the inverse Fourier transform of this PSD, which for this standard form is $R_n(\tau) = \frac{N_0}{4RC}e^{-|\tau|/RC}$ . Finally, the variance is found by setting $\tau=0$ , which gives $\text{Var}[n(t_k)] = R_n(0) = \frac{N_0}{4RC}$ .

Q21GATE 2024MCQ1MSignals and Systems

A causal and stable LTI system with impulse response $h(t)$ produces an output $y(t)$ for an input signal $x(t)$ . A signal $x(0.5 t)$ is applied to another causal and stable LTI system with impulse response $h(0.5 t)$ . The resulting output is ______

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📖 Explanation

Let's solve this problem using the frequency domain, where convolution becomes multiplication. The Fourier transform of the original output is $Y(\omega) = X(\omega)H(\omega)$ .

Now, consider the new system. The input is $x(0.5t)$ and the impulse response is $h(0.5t)$ . We'll use the time-scaling property of the Fourier transform, which states that if $f(t) \leftrightarrow F(\omega)$ , then $f(at) \leftrightarrow \frac{1}{|a|}F(\frac{\omega}{a})$ .

Applying this property, the transform of the new input is $2X(2\omega)$ , and the transform of the new impulse response is $2H(2\omega)$ .

The new output's transform, let's call it $Y_{new}(\omega)$ , is the product of these two:
$Y_{new}(\omega) = (2X(2\omega)) \cdot (2H(2\omega)) = 4X(2\omega)H(2\omega)$ .

Since $Y(\omega)=X(\omega)H(\omega)$ , we can express $Y_{new}(\omega)$ in terms of $Y(\omega)$ as $Y_{new}(\omega) = 4Y(2\omega)$ .

To find the time-domain signal, we recognize that the transform of $2y(0.5t)$ is $2 \cdot [2Y(2\omega)] = 4Y(2\omega)$ . This matches our result for $Y_{new}(\omega)$ , so the new output is $2y(0.5t)$ .

Q22GATE 2024MCQ1MElectronic Devices

For non-degenerately doped n-type silicon, which one of the following plots represents the temperature $(T)$ dependence of free electron concentration $(n)$ ?

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Q23GATE 2024MCQ1MAnalog Circuits

In the circuit shown, the $n: 1$ step-down transformer and the diodes are ideal. The diodes have no voltage drop in forward biased condition. If the input voltage (in Volts) is $V_{s}(t)=10 \sin \omega t$ and the average value of load voltage $V_{L}(t)$ (in Volts) is $2.5 / \pi$ , the value of $n$ is ______

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📖 Explanation

This circuit is a center-tapped full-wave rectifier. The average DC voltage across the load, $V_{L,avg}$ , is related to the peak voltage across half of the secondary winding, which we'll call $V_m$ . The standard formula is $V_{L,avg} = \frac{2V_m}{\pi}$ .

We are given that the average load voltage is $2.5 / \pi$ . We can use this to find $V_m$ :
$\frac{2.5}{\pi} = \frac{2V_m}{\pi} \implies V_m = \frac{2.5}{2} = 1.25$ V.

The total peak voltage across the entire secondary winding, $V_{sec,peak}$ , is twice $V_m$ , so $V_{sec,peak} = 2 \times 1.25 = 2.5$ V. The peak primary voltage from the source $V_s(t) = 10 \sin \omega t$ is $V_{pri,peak} = 10$ V. The turns ratio $n$ is simply the ratio of the peak primary to peak secondary voltages.

$n = \frac{V_{pri,peak}}{V_{sec,peak}} = \frac{10}{2.5} = 4$ .

Q24GATE 2024MSQ1MSignals and Systems

For a causal discrete-time LTI system with transfer function $H(z)=\frac{2 z^{2}+3}{\left(z+\frac{1}{3}\right)\left(z-\frac{1}{3}\right)}$ which of the following statements is/are true?

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📖 Explanation

Let's analyze the properties of this system based on its transfer function, $H(z)=\frac{2 z^{2}+3}{(z+\frac{1}{3})(z-\frac{1}{3})}$ .

First, for stability (A), we check the location of the poles, which are at $z = \pm \frac{1}{3}$ . Since both poles have a magnitude less than 1, they lie inside the unit circle, meaning the causal system is stable.

For the minimum phase property (B), we check the zeros. The zeros are the roots of $2z^{2+}3=0$ , which are $z = \pm j\sqrt{3/2}$ . The magnitude of these zeros is $|z| = \sqrt{1.5} > 1$ . Since at least one zero is outside the unit circle, the system is non-minimum phase.

To find the initial value (C), we use the initial value theorem: $h(0) = \lim_{z \to \infty } H(z)$ . This limit equals the ratio of the leading coefficients of the numerator and denominator, which is $\frac{2}{1} = 2$ .

Finally, for the final value (D), we can use the final value theorem because the system is stable: $h(\infty ) = \lim_{z \to 1}(z-1)H(z) = \lim_{z \to 1} (z-1)\frac{2 z^{2}+3}{z^2 - 1/9} = 0$ .

Q25GATE 2024MSQ1MEngineering Mathematics

Let $\rho(x, y, z, t)$ and $u(x, y, z, t)$ represent density and velocity, respectively, at a point $(x, y, z)$ and time $t$ . Assume $\frac{\partial \rho}{\partial t}$ is continuous. Let $V$ be an arbitrary volume in space enclosed by the closed surface $S$ and $\hat{n}$ be the outward unit normal of $S$ . Which of the following equations is/are equivalent $\frac{\partial \rho}{\partial t}+\nabla \cdot(\rho u)=0$ ?

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Q26GATE 2024MSQ1MElectronic Devices

The free electron concentration profile $n(x)$ in a doped semiconductor at equilibrium is shown in the figure, where the points $A, B$ , and $C$ mark three different positions. Which of the following statements is/are true?

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Q27GATE 2024NAT1MMicroprocessors

A machine has a 32-bit architecture with 1-word long instructions. It has 24 registers and supports an instruction set of size 40. Each instruction has five distinct fields, namely opcode, two source register identifiers, one destination register identifier, and an immediate value. Assuming that the immediate operand is an unsigned integer, its maximum value is ______.

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📖 Explanation

Let's determine how many bits are available for the immediate value within the 32-bit instruction. First, we calculate the bits needed for the other fields. To encode 40 unique instructions, the opcode requires $\lceil \log_2(40) \rceil = 6$ bits. To identify one of 24 registers, each of the three register fields (two source, one destination) needs $\lceil \log_2(24) \rceil = 5$ bits.

The total bits used by the opcode and register fields are $6 + (3 \times 5) = 21$ bits. The number of bits remaining for the immediate value is the total instruction size minus these used bits: $32 - 21 = 11$ bits. For an unsigned integer, the maximum value that can be represented with $n$ bits is $2^n - 1$ . Therefore, with 11 bits, the maximum value is $2^{11} - 1 = 2048 - 1 = 2047$ .

Q28GATE 2024NAT1MCommunication Systems

An amplitude modulator has output (in Volts) $s(t)=A \cos (400 \pi t)+B \cos (360 \pi t)+B \cos (440 \pi t)$ The carrier power normalized to $1 \Omega$ resistance is 50 Watts. The ratio of the total sideband power to the total power is $1 / 9$ . The value of $B$ (in Volts. rounded off to two decimal places) is ______

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📖 Explanation

The output signal is a standard AM wave. First, we find the carrier amplitude, $A$ , from the given carrier power, $P_C = 50$ W. The power of a sinusoid with amplitude $A$ across a $1\,\Omega$ resistor is $P_C = A^2/2$ . Solving for $A$ , we get $A = \sqrt{2 \times 50} = 10$ V.

Next, we determine the modulation index, $\mu$ . The ratio of sideband power to total power is given by the formula $\frac{P_{SB}}{P_t} = \frac{\mu^2}{2+\mu^2}$ . We are told this ratio is $1/9$ . Solving the equation $\frac{\mu^2}{2+\mu^2} = \frac{1}{9}$ gives $9\mu^2 = 2+\mu^2$ , which simplifies to $\mu=1/2$ .

Finally, the amplitude $B$ of each sideband is related to the carrier amplitude $A$ and modulation index $\mu$ by $B = \frac{A \mu}{2}$ . Substituting the values we found, we calculate $B = \frac{10 \times (1/2)}{2} = 2.5$ V.

Q29GATE 2024NAT1MDigital Circuits

In a number system of base $r$ , the equation $x^{2}-12 x+37=0$ has $x=8$ as one of its solutions. The value of $r$ is ______

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Q30GATE 2024NAT1MEngineering Mathematics

Let $R$ and $R^{3}$ denote the set of real numbers and the three dimensional vector space over it, respectively. The value of $\alpha$ for which the set of vectors

\left\{\left[\begin{array}{lll} 2 & -3 & \alpha \end{array}\right],\left[\begin{array}{lll} 3 & -1 & 3 \end{array}\right],\left[\begin{array}{lll} 1 & -5 & 7 \end{array}\right]\right\}

does not form a basis of $R^{3}$ is ______

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📖 Explanation

For a set of three vectors in $R^3$ to form a basis, they must be linearly independent. The set does not form a basis if the vectors are linearly dependent, which occurs when the determinant of the matrix formed by these vectors is zero. Let's construct the matrix and set its determinant to zero to find the value of $\alpha$ .

$\begin{vmatrix} 2 & -3 & \alpha \\ 3 & -1 & 3 \\ 1 & -5 & 7 \end{vmatrix} = 0$

Expanding the determinant along the first row gives us the equation:
$2((-1)(7) - (3)(-5)) - (-3)((3)(7) - (3)(1)) + \alpha((3)(-5) - (-1)(1)) = 0$
$2(8) + 3(18) - 14\alpha = 0$
$16 + 54 - 14\alpha = 0$
$70 = 14\alpha \implies \alpha = 5$

Q31GATE 2024NAT1MNetwork Theory

In the given circuit, the current $I_{x}$ (in $\mathrm{mA}$ ) is _____

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📖 Explanation

To solve this circuit, we'll use nodal analysis. Let the node voltage to the left of the dependent source be $V_1$ and to the right be $V_2$ , with the bottom wire as our 0V reference. The dependent voltage source creates a supernode enclosing $V_1$ and $V_2$ .

Applying Kirchhoff's Current Law (KCL) to this supernode gives:
$\frac{V_1}{1 \text{k}\Omega} + \frac{V_2}{1 \text{k}\Omega} = 5 \text{ mA} - 2 \text{ mA} \implies V_1 + V_2 = 3 \text{ V}$

Next, we write the constraint equation from the dependent source itself: $V_2 - V_1 = 1000 I_0$ . The controlling current $I_0$ is defined by Ohm's law as $I_0 = V_1 / (1 \text{k}\Omega)$ . Substituting this into the constraint gives $V_2 - V_1 = 1000 \left(\frac{V_1}{1000}\right) = V_1$ , which simplifies to $V_2 = 2V_1$ .

Substituting $V_2 = 2V_1$ into our KCL equation yields $V_1 + 2V_1 = 3$ , giving $V_1=1 \text{ V}$ and thus $V_2=2 \text{ V}$ . Finally, the current $I_x$ is found using Ohm's law on the rightmost resistor: $I_x = \frac{V_2}{1 \text{k}\Omega} = \frac{2 \text{ V}}{1000 \Omega} = 2 \text{ mA}$ .

Q32GATE 2024NAT1MNetwork Theory

In the circuit given below, the switch S was kept open for a sufficiently long time and is closed at time $t=0$ . The time constant (in seconds) of the circuit for $t > 0$ is ______

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📖 Explanation

The time constant of an RL circuit is given by $\tau = L/R_{eq}$ , where $R_{eq}$ is the equivalent Thévenin resistance seen by the inductor. We analyze the circuit for $t>0$ , which means the switch is closed.

To find $R_{eq}$ , we must deactivate all independent sources. This involves replacing the 5A current source with an open circuit.

With the source opened, the circuit seen from the inductor's terminals consists of the $2 \, \Omega$ resistor in series with the parallel combination of the two $4 \, \Omega$ resistors.

The equivalent resistance is calculated as $R_{eq} = 2 \, \Omega + (4 \, \Omega \parallel 4 \, \Omega) = 2 + \frac{4 \times 4}{4+4} = 4 \, \Omega$ .

Therefore, the time constant is $\tau = \frac{L}{R_{eq}} = \frac{3 \text{ H}}{4 \, \Omega} = 0.75$ seconds.

Q33GATE 2024NAT1MCommunication Systems

Suppose $X$ and $Y$ are independent and identically distributed random variables that are distributed uniformly in the interval $[0,1]$ . The probability that $X \geq Y$ is ______

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📖 Explanation

The question asks for the probability that $X \geq Y$ . This is equivalent to finding the probability that their difference is non-negative, or $P(X - Y \geq 0)$ .

Let's define a new random variable $Z = X - Y$ . Since $X$ and $Y$ are independent and have the same distribution, the probability distribution of their difference, $Z$ , is symmetric around its mean. The mean of this difference is $E[Z] = E[X] - E[Y]$ . For a uniform distribution on $[0,1]$ , the mean is $0.5$ , so $E[Z] = 0.5 - 0.5 = 0$ .

Because the distribution of $Z$ is symmetric about 0, exactly half of the probability lies on the positive side and half on the negative side. Therefore, the probability that $Z$ is greater than or equal to 0 must be $1/2$ .

Q34GATE 2024NAT1MCommunication Systems

A source transmits symbols from an alphabet of size 16. The value of maximum achievable entropy (in bits) is ______

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📖 Explanation

The entropy of a source measures its average uncertainty. This uncertainty is greatest when every symbol is equally likely to occur.

For a source with an alphabet of size $M$ , the maximum possible entropy ( $H_{\text{max}}$ ) is calculated using the formula:
$H_{\text{max}} = \log_2(M)$

In this case, the alphabet size is $M=16$ . Plugging this value into the formula gives us:
$H_{\text{max}} = \log_2(16) = 4$ bits.

Q35GATE 2024NAT1MAnalog Circuits

As shown in the circuit, the initial voltage across the capacitor is $10 \mathrm{~V}$ , with the switch being open. The switch is then closed at $t=0$ . The total energy dissipated in the ideal Zener diode ( $V_{z}=5 \mathrm{~V}$ ) after the switch is closed (in $\mathrm{mJ}$ , rounded off to three decimal places) is ________

🚀 Solve in Practice Mode

📖 Explanation

When the switch closes at $t=0$ , the capacitor discharges, pushing current through the resistor and Zener diode. The ideal Zener diode acts as a constant 5 V voltage drop as long as the capacitor voltage, $V_C$ , is above 5 V. The diode will stop conducting and dissipating energy once $V_C$ falls to 5 V.

A straightforward way to find the total energy dissipated in the Zener is to use the total charge, $\Delta Q$ , that passes through it. The charge that flows out of the capacitor is determined by its voltage change:
$\Delta Q = C \times \Delta V_C = (10 \mu\text{F}) \times (10 \text{ V} - 5 \text{ V}) = 50 \mu\text{C}$ .

Since the Zener voltage is constant at $V_Z = 5 \text{ V}$ during this entire process, the energy it dissipates is simply the product of its voltage and the total charge that flowed through it:
$E_Z = V_Z \times \Delta Q = 5 \text{ V} \times 50 \mu\text{C} = 250 \mu\text{J}$ .

Converting this to millijoules gives the final answer of $0.25 \text{ mJ}$ .

Q36GATE 2024MCQ2MEngineering Mathematics

Consider the Earth to be a perfect sphere of radius $R$ . Then the surface area of the region, enclosed by the $60^{\circ} \mathrm{N}$ latitude circle, that contains the north pole in its interior is _______

🚀 Solve in Practice Mode

📖 Explanation

To find the surface area of this polar cap, we'll use integration in spherical coordinates. The area element on a sphere of radius $R$ is given by $dS = R^2 \sin\theta \, d\theta \, d\phi$ . The polar angle $\theta$ is measured from the north pole, so $\theta=0$ at the pole itself. The boundary of our region is the $60^\circ \mathrm{N}$ latitude circle. This latitude corresponds to a polar angle of $\theta = 90^\circ - 60^\circ = 30^\circ$ .

The area is found by integrating the area element over the appropriate ranges. The azimuthal angle $\phi$ sweeps from $0$ to $2\pi$ to cover the full circle, while the polar angle $\theta$ ranges from $0$ to $30^\circ$ .

$S = \int_{0}^{2\pi} \int_{0}^{30^\circ} R^2 \sin\theta \, d\theta \, d\phi$
$S = 2\pi R^2 \int_{0}^{30^\circ} \sin\theta \, d\theta = 2\pi R^2 [-\cos\theta]_{0}^{30^\circ}$
$S = 2\pi R^2 (-\cos 30^\circ - (-\cos 0^\circ)) = 2\pi R^2 \left(1 - \frac{\sqrt{3}}{2}\right)$
Simplifying this expression gives the final area: $S = (2 - \sqrt{3})\pi R^2$ .

Q37GATE 2024MCQ2MControl Systems

Consider a unity negative feedback control system with forward path gain $G(s)=\frac{K}{(s+1)(s+2)(s+3)}$ as shown. The impulse response of the closed-loop system decays faster than $e^{-t}$ if ______

🚀 Solve in Practice Mode

📖 Explanation

The rate at which the system's impulse response decays is governed by the real part of its closed-loop poles. For the response to decay faster than $e^{-t}$ , all closed-loop poles must be located to the left of the line $\text{Re}(s) = -1$ in the s-plane.

The system's characteristic equation is $1 + G(s) = 0$ , which is $(s+1)(s+2)(s+3) + K = 0$ .

To analyze stability relative to the $s=-1$ line, we shift the axis of interest to the imaginary axis by substituting $s = z-1$ . The condition for all poles to be left of $s=-1$ now becomes the condition for all poles of the new system to be in the left-half of the $z$ -plane (\text{Re}(z)<0$).

The transformed characteristic equation is $(z-1+1)(z-1+2)(z-1+3)+K = 0$ , which simplifies to $z^3 + 3z^2 + 2z + K = 0$ .

Applying the Routh-Hurwitz criterion to this new equation, the first column of the Routh array must be positive for stability. This yields the conditions $K>0$ and $(6-K)/3 > 0$ , which together require $0 < K < 6$ .

Q38GATE 2024MCQ2MControl Systems

A satellite attitude control system, as shown below, has a plant with transfer function $G(s)=\frac{1}{s^{2}}$ cascaded with a compensator $C(s)=\frac{K(s+\alpha)}{s+4}$ , where $K$ and $\alpha$ are positive real constants. In order for the closed-loop system to have poles at $-1 \pm j \sqrt{3}$ , the value of $\alpha$ must be ____

🚀 Solve in Practice Mode

📖 Explanation

First, we determine the characteristic equation of the closed-loop system, which is given by $1 + C(s)G(s) = 0$ . Substituting the given functions, we get $1 + \frac{K(s+\alpha)}{s^2(s+4)} = 0$ . This simplifies to the polynomial $s^3 + 4s^2 + Ks + K\alpha = 0$ .

Next, we construct the desired characteristic equation from the specified poles. The complex conjugate poles at $-1 \pm j\sqrt{3}$ correspond to the quadratic factor $(s+1)^2 + (\sqrt{3})^2 = s^2 + 2s + 4$ . Since the system is third-order, there must be a third real pole, which we can call $-p$ .

The full desired characteristic equation is then $(s+p)(s^2 + 2s + 4) = 0$ , which expands to $s^3 + (p+2)s^2 + (2p+4)s + 4p = 0$ .

By comparing the coefficients of the two polynomial forms of the characteristic equation, we can find the unknown values. Matching the $s^2$ terms, we have $4 = p+2$ , which means the third pole is at $p=2$ . From the $s^1$ terms, we get $K = 2p+4 = 2(2)+4 = 8$ . Finally, comparing the constant terms, $K\alpha = 4p$ , which becomes $8\alpha = 4(2) = 8$ . Thus, the value of $\alpha$ must be 1.

Q39GATE 2024MCQ2MElectromagnetics

A uniform plane wave with electric field $\vec{E}(x)=A_{y} \hat{a}_{y} e^{-j \frac{2 \pi x}{3}} \mathrm{~V} / \mathrm{m}$ is travelling in the air (relative permittivity, $\epsilon_{r}=1$ and relative permeability, $\mu_{r}=1$ ) in the $+x$ direction $\left(A_{y}\right.$ is a positive constant, $\hat{a}_{y}$ is the unit vector along the $y$ axis). It is incident normally on an ideal electric conductor (conductivity, $\sigma=\infty$ ) at $x=0$ . The position of the first null of the total magnetic field in the air (measured from $x=0$ , in metres) is ________

🚀 Solve in Practice Mode

📖 Explanation

When a wave reflects from a perfect electric conductor at $x=0$ , the incident and reflected waves interfere to form a standing wave in the air for $x<0$ . The reflection coefficient for the electric field is $\Gamma=-1$ , which means the total electric field is a null (zero) at the conductor's surface.

In a standing wave, the nulls of the total magnetic field occur at the locations of the antinodes (maxima) of the total electric field. The first electric field antinode is located a quarter-wavelength ( $\lambda/4$ ) from the electric field null at the boundary.

From the incident wave's expression, $e^{-j \frac{2 \pi x}{3}}$ , we can identify the phase constant as $\beta = \frac{2\pi}{3}$ . We can find the wavelength using the relationship $\beta = \frac{2\pi}{\lambda}$ , which gives $\lambda=3$ meters.

Therefore, the first magnetic field null is located at a distance of $\lambda/4$ from the conductor, in the region $x<0$ . This position is $x = -\frac{\lambda}{4} = -\frac{3}{4}$ meters.

Q40GATE 2024MCQ2MDigital Circuits

A 4-bit priority encoder has inputs $D_{3}, D_{2}, D_{1}$ and $D_{0}$ in descending order of priority. The two-bit output $A B$ is generated as 00,01, 10, and 11 corresponding to inputs $D_{3}$ , $D_{2}, D_{1}$ and $D_{0}$ , respectively. The Boolean expression of the output bit $B$ is _____

🚀 Solve in Practice Mode

📖 Explanation

A priority encoder generates an output corresponding to the highest-priority active input. The inputs $D_3, D_2, D_1, D_0$ are in descending order of priority. To find the expression for output bit $B$ , let's first determine from the problem's specifications when $B$ is high.

The specified outputs $AB$ are $00, 01, 10, 11$ for inputs $D_3, D_2, D_1, D_0$ respectively. This non-standard mapping defines our truth table for $B$ :

If $D_3$ is active, $AB=00 \implies B=0$ .
If $D_2$ is active (and $D_3=0$ ), $AB=01 \implies B=1$ .
If $D_1$ is active (and $D_3=D_2=0$ ), $AB=10 \implies B=0$ .
If $D_0$ is active (and $D_3=D_2=D_1=0$ ), $AB=11 \implies B=1$ .

The expression for $B$ is the sum of the conditions where it is 1.
$B = (\overline{D_3}D_2) + (\overline{D_3}\overline{D_2}\overline{D_1}D_0)$

This expression can be simplified. A K-map with a don't-care for the all-zero input case reveals the simplified form. Alternatively, using Boolean algebra, we can simplify $B = \overline{D_3}D_2 + \overline{D_3}\overline{D_2}\overline{D_1}$ (a known simplification from the K-map). Factoring gives $B = \overline{D_3}(D_2 + \overline{D_2}\overline{D_1})$ . Using the absorption identity $X+\overline{X}Y = X+Y$ , this becomes $B = \overline{D_3}(D_2 + \overline{D_1})$ , which expands to $\overline{D_3}D_2 + \overline{D_3}\overline{D_1}$ .

Q41GATE 2024MCQ2MDigital Circuits

The propagation delay of the $2 \times 1 \mathrm{MUX}$ shown in the circuit is $10 \mathrm{~ns}$ . Consider the propagation delay of the inverter as $0 \mathrm{~ns}$ . If $S$ is set to 1 then the output $Y$ is ______

🚀 Solve in Practice Mode

📖 Explanation

With the select line $S$ set to 1, the MUX output $Y$ is fed by its '1' input, which is the inverted version of $Y$ itself. This feedback loop creates an oscillator.

For the output to complete one full cycle, it must transition twice. First, from low to high, which takes one MUX propagation delay of $10 \mathrm{~ns}$ . Second, from high back to low, which takes another $10 \mathrm{~ns}$ delay.

The total period $T$ of the resulting square wave is the sum of these two delays:
$T = 10 \mathrm{~ns} + 10 \mathrm{~ns} = 20 \mathrm{~ns}$

The frequency is the reciprocal of the period:
$f = \frac{1}{T} = \frac{1}{20 \times 10^{-9} \mathrm{~s}} = 50 \mathrm{~MHz}$

Q42GATE 2024MCQ2MDigital Circuits

The sequence of states $\left(Q_{1} Q_{0}\right)$ of the given synchronous sequential circuit is ______

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Q43GATE 2024MCQ2MEngineering Mathematics

Let $z$ be a complex variable. If $f(z)=\frac{\sin (\pi z)}{z^{2}(z-2)}$ and $C$ is the circle in the complex plane with $|z|=3$ then $\oint_{c} f(z) d z$ is ______

🚀 Solve in Practice Mode

📖 Explanation

To evaluate this integral, we apply Cauchy's Residue Theorem. The function $f(z)=\frac{\sin (\pi z)}{z^{2}(z-2)}$ has two singularities inside the circle $|z|=3$ : a simple pole at $z=2$ and a pole of order 2 at $z=0$ .

First, we calculate the residue at the simple pole $z=2$ :
$\text{Res}(f, 2) = \lim_{z \to 2} (z-2)f(z) = \lim_{z \to 2} \frac{\sin(\pi z)}{z^2} = \frac{\sin(2\pi)}{4} = 0$ .

Next, we find the residue at the pole of order 2 at $z=0$ using the derivative formula:
$\text{Res}(f, 0) = \frac{d}{dz}\left[z^2 f(z)\right]_{z=0} = \frac{d}{dz}\left[\frac{\sin(\pi z)}{z-2}\right]_{z=0}$ .
Applying the quotient rule, we get $\left[\frac{\pi(z-2)\cos(\pi z) - \sin(\pi z)}{(z-2)^2}\right]_{z=0} = \frac{-2\pi}{4} = -\frac{\pi}{2}$ .

The sum of the residues is $0 + (-\frac{\pi}{2}) = -\frac{\pi}{2}$ .
By the Residue Theorem, the integral is $2\pi j$ times the sum of the enclosed residues:
$\oint_{c} f(z) d z = 2\pi j \left(-\frac{\pi}{2}\right) = -\pi^2 j$ .

Q44GATE 2024MCQ2MSignals and Systems

Consider two continuous time signals $x(t)$ and $y(t)$ as shown below If $X(f)$ denotes the Fourier transform of $x(t)$ , then the Fourier transform of $y(t)$ is ____

🚀 Solve in Practice Mode

📖 Explanation

To solve this, we first need to express the signal $y(t)$ in terms of $x(t)$ by identifying the transformations involved: time scaling, time shifting, and amplitude scaling.

By comparing the two plots, we can see that $y(t)$ is an amplitude-inverted, time-stretched, and shifted version of $x(t)$ . Specifically, the shape of $y(t)$ is obtained by flipping $x(t)$ over the time axis (multiplying by -1), stretching it horizontally by a factor of 4, and then shifting it. This sequence of operations gives the relationship $y(t) = -x\left(\frac{t-2}{4}\right)$ , which can be rewritten as $y(t) = -x\left(\frac{t}{4} - \frac{1}{2}\right)$ .

Now, we use the properties of the Fourier transform. The time-scaling and time-shifting property states that if $x(t) \leftrightarrow X(f)$ , then $x(at - b) \leftrightarrow \frac{1}{|a|} X\left(\frac{f}{a}\right) e^{-j2\pi f (b/a)}$ .

For our expression $x\left(\frac{t}{4} - \frac{1}{2}\right)$ , we have $a = 1/4$ and $b = 1/2$ . Applying the property, the transform is:
$\frac{1}{1/4} X\left(\frac{f}{1/4}\right) e^{-j2\pi f (1/2)/(1/4)} = 4 X(4f) e^{-j4\pi f}$ .

Since $y(t)$ includes a negative sign, we multiply the resulting transform by -1. Therefore, the Fourier transform of $y(t)$ is $-4 X(4f) e^{-j4\pi f}$ .

Q45GATE 2024MCQ2MCommunication Systems

A source transmits a symbol $s$ , taken from $(-4,0,4)$ with equal probability, over an additive white Gaussian noise channel. The received noisy symbol $r$ is given by $r=s+w$ , where the noise $w$ is zero mean with variance 4 and is independent of $s$ . Using $Q(x)=\frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty } e^{\frac{-t^{2}}{2}} d t$ , the optimum symbol error probability is ______

🚀 Solve in Practice Mode

📖 Explanation

Since the symbols $s \in \{-4, 0, 4\}$ are transmitted with equal probability, the total symbol error probability $P_e$ is the average of the conditional error probabilities for each symbol. For optimal detection in AWGN, we use a minimum distance decoder, placing decision boundaries midway between symbols, at $r=-2$ and $r=2$ . The noise standard deviation is $\sigma = \sqrt{\text{variance}} = \sqrt{4} = 2$ .

If $s=-4$ is sent, an error occurs if the received signal $r = -4+w > -2$ , which means the noise $w>2$ . The probability of this is $P(e|s=-4) = Q\left(\frac{2}{\sigma}\right) = Q(1)$ .
By symmetry, if $s=4$ is sent, an error occurs if $r = 4+w < 2$ , meaning $w<-2$ . The probability is also $P(e|s=4) = Q\left(\frac{2}{\sigma}\right) = Q(1)$ .
If $s=0$ is sent, an error occurs if the noise pushes the signal past either boundary, i.e., $|w|>2$ . This probability is $P(e|s=0) = P(w>2) + P(w<-2) = 2Q\left(\frac{2}{\sigma}\right) = 2Q(1)$ .

Averaging these probabilities gives the total error rate:
$P_e = \frac{1}{3} \left[ P(e|s=-4) + P(e|s=4) + P(e|s=0) \right] = \frac{1}{3}[Q(1) + Q(1) + 2Q(1)] = \frac{4}{3}Q(1)$ .

Q46GATE 2024MCQ2MDigital Circuits

A full scale sinusoidal signal is applied to a 10-bit ADC. The fundamental signal component in the ADC output has a normalized power of $1 \mathrm{~W}$ , and the total noise and distortion normalized power is $10 \mu \mathrm{W}$ . The effective number of bits (rounded off to the nearest integer) of the ADC is ____

🚀 Solve in Practice Mode

📖 Explanation

The performance of an Analog-to-Digital Converter (ADC) is often measured by its Signal-to-Noise and Distortion Ratio (SINAD). We first calculate this ratio from the given powers:
$\text{SINAD} = \frac{\text{Signal Power}}{\text{Noise Power}} = \frac{1 \mathrm{~W}}{10 \times 10^{-6} \mathrm{~W}} = 10^5$
Next, we convert this power ratio to the decibel (dB) scale, which is standard for such metrics:
$\text{SINAD}_{dB} = 10 \log_{10}(10^5) = 50 \mathrm{~dB}$
The SINAD of an ADC is directly related to its Effective Number of Bits (ENOB), denoted as $n$ , by the well-known formula for a sinusoidal input:
$\text{SINAD}_{dB} = 6.02n + 1.76$
By substituting our calculated SINAD, we can solve for the ENOB:
$50 = 6.02n + 1.76 \implies 6.02n = 48.24 \implies n \approx 8.01$
Rounding this value to the nearest integer gives an effective number of bits of 8.

Q47GATE 2024MCQ2MCommunication Systems

The information bit sequence (11101010 1) is to be transmitted by encoding with Cyclic Redundancy Check 4 (CRC-4) code, for which the generator polynomial is $C(x)=x^{4}+x+1$ . The encoded sequence of bits is ______

🚀 Solve in Practice Mode

📖 Explanation

First, we translate the generator polynomial $C(x)=x^{4}+x+1$ into its binary representation, which is $10011$ . The degree of this polynomial is $r=4$ , which tells us that the Cyclic Redundancy Check (CRC) will be 4 bits long.

To prepare the data for encoding, we append $r=4$ zeros to the end of the information sequence $111010101$ . This gives us the new sequence $1110101010000$ .

Next, we perform modulo-2 binary division of this appended sequence by the generator $10011$ .
$\frac{1110101010000}{10011}$
The remainder from this division is $1100$ . This 4-bit remainder is our CRC checksum.

The final encoded sequence is created by replacing the four appended zeros with this 4-bit remainder. We append the checksum $1100$ to the original data $111010101$ to get the transmitted codeword: $1110101011100$ .

Q48GATE 2024MCQ2MSignals and Systems

A continuous time signal $x(t)=2 \cos (8 \pi t+\pi / 3)$ is sampled at a rate of $15 \mathrm{~Hz}$ . The sampled signal $x_{s}(t)$ when passed through an LTI system with impulse response $h(t)=\left(\frac{\sin 2 \pi t}{\pi t}\right) \cos (38 \pi t-\pi / 2)$ produces an output $x_{0}(t)$ . The expression for $x_{0}(t)$ is _____

🚀 Solve in Practice Mode

📖 Explanation

The original signal, with frequency $\omega_0 = 8\pi$ rad/s, is sampled at $\omega_s = 2\pi(15) = 30\pi$ rad/s. This sampling process creates spectral replicas (images) of the original signal centered at integer multiples of the sampling frequency, $n\omega_s$ . The frequencies present in the sampled signal are therefore $n\omega_s \pm \omega_0$ .

The LTI system is a band-pass filter, as its impulse response is a sinc function modulated by a cosine. The filter's center frequency is $38\pi$ rad/s, which corresponds to the spectral replica at $\omega_s + \omega_0 = 30\pi + 8\pi$ . The filter will therefore isolate this component.

The replica at $38\pi$ has an amplitude of $A \times f_s = 2 \times 15 = 30$ and the original phase of $\pi/3$ . So, the effective input to the filter is $30 \cos(38\pi t + \pi/3)$ . The filter's complex gain at its center frequency is determined by the modulating term, $\cos(38\pi t - \pi/2)$ , which contributes a gain of $1/2$ and a phase shift of $-\pi/2$ .

Applying this to the filter's input, the output amplitude becomes $30 \times \frac{1}{2} = 15$ , and the output phase becomes $\pi/3 - \pi/2 = -\pi/6$ . Thus, the final output signal is $x_o(t) = 15 \cos(38\pi t - \pi/6)$ .

Q49GATE 2024MCQ2MAnalog Circuits

The opamps in the circuit shown are ideal, but have saturation voltages of $\pm 10 \mathrm{~V}$ . Assume that the initial inductor current is $0 \mathrm{~A}$ . The input voltage $\left(V_{i}\right)$ is a triangular signal with peak voltages of $\pm 2 \mathrm{~V}$ and time period of $8 \mu \mathrm{s}$ . Which one of the following statements is true?

🚀 Solve in Practice Mode

📖 Explanation

The first op-amp circuit is a non-inverting Schmitt trigger. Its upper and lower trip points are determined by the resistor ratio and saturation voltage, calculated as $V_{UT/LT} = \pm\frac{10 \text{ k}\Omega}{100 \text{ k}\Omega} (\pm 10 \text{ V}) = \pm 1 \text{ V}$ .

The input triangular wave has a period of $8 \mu \text{s}$ and peaks of $\pm 2 \text{ V}$ . This means its slope is $1 \text{ V}/\mu \text{s}$ as it rises from $0 \text{ V}$ . It crosses the upper trip point of $+1 \text{ V}$ at $t=1 \mu \text{s}$ , causing the output $V_{01}$ to switch. This results in a square wave for $V_{01}$ that is delayed by $1 \mu \text{s}$ relative to the input's zero-crossing.

The second stage is an integrator, since $V_{02}(t) = -\frac{R}{L}\int V_{01}(t)dt$ . With a square wave input ( $V_{01}$ ), the output $V_{02}$ will be a triangular waveform. However, we must check for saturation. The rate of change of the output is $\frac{dV_{02}}{dt} = -\frac{1 \text{ k}\Omega}{1 \text{ mH}}(\pm 10 \text{ V}) = \mp 10 \text{ V}/\mu \text{s}$ .

To swing from $-10 \text{ V}$ to $+10 \text{ V}$ (a $20 \text{ V}$ change), it takes $\frac{20 \text{ V}}{10 \text{ V}/\mu \text{s}} = 2 \mu \text{s}$ . The half-period of the $V_{01}$ square wave is $4 \mu \text{s}$ . Since the output reaches its saturation limit in just $2 \mu \text{s}$ , it will remain saturated for the rest of the half-period, clipping the peaks of the triangular wave and creating a trapezoidal waveform.

Q50GATE 2024MCQ2MAnalog Circuits

In the circuit below, the opamp is ideal. If the circuit is to show sustained oscillations, the respective values of $R_{1}$ and the corresponding frequency of oscillation are ______

🚀 Solve in Practice Mode

📖 Explanation

This circuit is a standard Wien bridge oscillator. The frequency of oscillation for this configuration is determined by the positive feedback network and is given by $f_{o} = \frac{1}{2 \pi R C}$ .

For sustained oscillations, the total loop gain must be exactly 1. The RC feedback network has an attenuation of $1/3$ at the oscillation frequency. To compensate, the non-inverting amplifier's gain must be 3.

The gain of the op-amp stage is $A = 1 + \frac{R_{1}}{R}$ . Setting the gain to 3, we get the condition for oscillation:
$1 + \frac{R_{1}}{R} = 3$
Solving this equation gives us $R_{1} = 2R$ .

Q51GATE 2024MCQ2MAnalog Circuits

In the circuit shown below, the transistors $M_{1}$ and $M_{2}$ are biased in saturation. Their small signal transconductances are $g_{m 1}$ and $g_{m 2}$ respectively. Neglect body effect, channel length modulation and intrinsic device capacitances. Assuming that capacitor $C_{1}$ is a short circuit for $A C$ analysis, the exact magnitude of small signal voltage gain $\left|\frac{v_{\text {out }}}{v_{\text {\in }}}\right|$ is _______

🚀 Solve in Practice Mode

📖 Explanation

To determine the overall voltage gain, we can analyze the circuit in two main steps. First, we find the gain of the output stage, and then we find how the input signal is attenuated before reaching that stage.

The output voltage $v_{out}$ is generated by a common-source amplifier built around transistor $M_2$ . Its voltage gain is given by the product of its transconductance and output resistance, so $v_{out} = -g_{m2} v_{gs2} R_D$ . Here, $v_{gs2}$ is the small-signal voltage at the gate of $M_2$ .

Next, let's find $v_{gs2}$ in terms of $v_{\in}$ . For AC analysis, capacitor $C_1$ is a short circuit, and the diode-connected transistor $M_1$ acts as a resistance of $1/g_{m1}$ . The input voltage $v_{\in}$ is therefore applied to a voltage divider formed by $R_S$ in series with the combined resistance of $R_B$ and $M_1$ .

Applying the voltage divider rule, the voltage at the gate of $M_2$ is:
$v_{gs2} = v_{\in} \cdot \frac{R_B + 1/g_{m1}}{R_S + (R_B + 1/g_{m1})}$

Finally, we substitute this expression for $v_{gs2}$ into our equation for $v_{out}$ and solve for the gain $\frac{v_{out}}{v_{\in}}$ :
$\frac{v_{out}}{v_{\in}} = -g_{m2} R_D \left( \frac{R_B + 1/g_{m1}}{R_S + R_B + 1/g_{m1}} \right)$

The question asks for the magnitude, so we simply take the absolute value of this expression.

Q52GATE 2024MSQ2MElectronic Devices

Which of the following statements is/are true for a BJT with respect to its DC current gain $\beta$ ?

🚀 Solve in Practice Mode

Q53GATE 2024MSQ2MControl Systems

Consider a system $S$ represented in state space as

\frac{d x}{d t}=\left[\begin{array}{ll} 0 & -2 \\ 1 & -3 \end{array}\right] x+\left[\begin{array}{l} 1 \\ 0 \end{array}\right] r, y=\left[\begin{array}{ll} 2 & -5 \end{array}\right] x

Which of the state space representations given below has/have the same transfer function as that of $S$ ?

🚀 Solve in Practice Mode

📖 Explanation

First, we find the transfer function for the given system $S$ using the standard formula $G(s) = C(sI-A)^{-1}B$ . The calculation results in the transfer function $G(s) = \frac{2s+1}{s^{2+}3s+2}$ . A single transfer function can have infinitely many state-space representations, so we need to check which of the options realize this same function.

One common representation is the diagonal (or parallel) form. We can find this by decomposing the transfer function into partial fractions: $G(s) = \frac{2s+1}{(s+1)(s+2)} = \frac{-1}{s+1} + \frac{3}{s+2}$ . This structure directly corresponds to the state-space model in Option C, which describes two decoupled first-order systems.

Another standard representation is the controllable canonical form. The matrices for this form are built directly from the coefficients of the transfer function's numerator and denominator. The system in Option A is a realization of this form for $G(s)$ . Therefore, both systems in options A and C share the same transfer function as the original system $S$ .

Q54GATE 2024MSQ2MEngineering Mathematics

Let $F_{1}, F_{2}$ , and $F_{3}$ be functions of $(x, y, z)$ . Suppose that for every given pair of points $A$ and $B$ in space, the line integral $\int_{C}\left(F_{1} d x+F_{2} d y+F_{3} d z\right)$ evaluates to the same value along any path $C$ that starts at $A$ and ends at $B$ . Then which of the following is/are true?

🚀 Solve in Practice Mode

📖 Explanation

The problem statement tells us the line integral of the vector field $\vec{F} = \langle F_1, F_2, F_3 \rangle$ is path-independent. This is the definition of a conservative vector field. Let's examine the consequences based on this fact.

First, by the Fundamental Theorem for Line Integrals, a field is conservative if and only if it is the gradient of a scalar potential function $f$ . This means $\vec{F} = \nabla f$ , so $F_1 = \frac{\partial f}{\partial x}$ , $F_2 = \frac{\partial f}{\partial y}$ , and $F_3 = \frac{\partial f}{\partial z}$ . is true.

An equivalent definition of a conservative field is that its line integral over any closed path $\Gamma$ is zero. This means $\oint_{\Gamma} \vec{F} \cdot d\vec{r} = 0$ , making option B true.

Furthermore, a necessary condition for a field to be conservative is that it must be irrotational, meaning its curl is zero: $\nabla \times \vec{F} = \vec{0}$ . Calculating the curl and setting its components to zero yields the exact set of equations in option D.

Option C, which states that the divergence is zero ( $\nabla \cdot \vec{F} = 0$ ), is a separate property. A field can be conservative without having zero divergence, so this option is not necessarily true.

Q55GATE 2024MSQ2MEngineering Mathematics

Consider the matrix

\left[\begin{array}{ll}1 & k \\ 2 & 1\end{array}\right]

, where $k$ is a positive real number. Which of the following vectors is/are eigenvector(s) of this matrix?

🚀 Solve in Practice Mode

📖 Explanation

An eigenvector $v$ and its corresponding eigenvalue $\lambda$ for a matrix $A$ satisfy the equation $Av = \lambda v$ , or $(A - \lambda I)v = 0$ .

First, we find the eigenvalues by solving the characteristic equation $\det(A - \lambda I) = 0$ . For the given matrix, this is $(1-\lambda)^2 - 2k = 0$ , which yields the eigenvalues $\lambda = 1 \pm \sqrt{2k}$ .

Let's find the eigenvector for $\lambda = 1 + \sqrt{2k}$ by solving $(A-\lambda I)v=0$ for

v = \begin{bmatrix} x \\ y \end{bmatrix}

. This gives the system

\left[\begin{array}{cc} -\sqrt{2k} & k \\ 2 & -\sqrt{2k} \end{array}\right]\left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \end{array}\right]

. The first row implies $-\sqrt{2k}x + ky = 0$ , which simplifies to $y = \sqrt{2/k}x$ . Setting $x=1$ gives the eigenvector

\begin{bmatrix} 1 \\ \sqrt{2/k} \end{bmatrix}

.

Following the same process for the eigenvalue $\lambda = 1 - \sqrt{2k}$ leads to the relation $y = -\sqrt{2/k}x$ . Setting $x=1$ gives the eigenvector

\begin{bmatrix} 1 \\ -\sqrt{2/k} \end{bmatrix}

.

Q56GATE 2024MSQ2MSignals and Systems

The radian frequency value(s) for which the discrete time sinusoidal signal $x[n]=A$ $\cos (\Omega n+\pi / 3)$ has a period of 40 is/are _____

🚀 Solve in Practice Mode

📖 Explanation

For a discrete-time sinusoid to be periodic with period $N$ , its radian frequency $\Omega$ must satisfy the relationship $\Omega = \frac{2\pi m}{N}$ for some positive integer $m$ . This integer $m$ represents the number of complete cycles the sinusoid undergoes over one period of $N$ samples.

Given that the desired period is $N=40$ , we can establish the general form for the frequency:
$\Omega = \frac{2\pi m}{40} = m \cdot \frac{\pi}{20} = 0.05\pi m$
This equation shows that any valid frequency $\Omega$ must be an integer multiple of $0.05\pi$ . We can now check which of the options satisfy this condition for an integer $m$ :

$0.15\pi$ is valid for $m=3$ .
$0.30\pi$ is valid for $m=6$ .
$0.45\pi$ is valid for $m=9$ .
The option $0.225\pi$ is not an integer multiple of $0.05\pi$ and is therefore not a valid frequency for this period.

Q57GATE 2024NAT2MCommunication Systems

Let $X(t)=A \cos \left(2 \pi f_{0} t+\theta\right)$ be a random process, where amplitude $A$ and phase $\theta$ are independent of each other, and are uniformly distributed in the intervals $[-2,2]$ and $[0,2 \pi]$ , respectively. $X(t)$ is fed to an 8-bit uniform mid-rise type quantizer. Given that the autocorrelation of $X(t)$ is $R_{x}(\tau)=\frac{2}{3} \cos \left(2 \pi f_{0} \tau\right)$ , the signal to quantization noise ratio (in $\mathrm{dB}$ , rounded off to two decimal places) at the output of the quantizer is ______

🚀 Solve in Practice Mode

📖 Explanation

The signal's average power, $S$ , is the value of its autocorrelation function at $\tau=0$ . Thus, the signal power is $S=R_x(0)=2/3$ .

The quantization noise power is given by $N_q = \Delta^2/12$ . Here, the step size $\Delta$ is determined by the quantizer's range and number of levels. Since the signal's amplitude $A$ is in $[-2, 2]$ , the full-scale range is $V_{pp} = 2 - (-2) = 4$ . For an 8-bit quantizer, the step size is $\Delta = \frac{V_{pp}}{2^8} = \frac{4}{2^8} = \frac{1}{2^6}$ .

This gives a noise power of $N_q = \frac{(1/2^6)^2}{12} = \frac{1}{12 \times 2^{12}}$ .

Now, we can find the signal-to-quantization noise ratio (SQNR):
$\frac{S}{N_q} = \frac{2/3}{1/(12 \times 2^{12})} = \frac{2}{3} \times 12 \times 2^{12} = 8 \times 2^{12} = 2^3 \times 2^{12} = 2^{15} = 32,768$
Finally, converting this ratio to decibels yields:
$\text{SQNR}_{\text{dB}} = 10 \log_{10}(32,768) \approx 45.15 \text{ dB}$

Q58GATE 2024NAT2MElectromagnetics

A lossless transmission line with characteristic impedance $Z_{0}=50 \Omega$ is terminated with an unknown load. The magnitude of the reflection co-efficient is $|\Gamma|=0.6$ . As one moves towards the generator from the load, the maximum value of the input impedance magnitude looking towards the load (in $\Omega$ ) is ______

🚀 Solve in Practice Mode

📖 Explanation

On a mismatched transmission line, the input impedance looking towards the load varies with position. The maximum value of this impedance, $Z_{max}$ , occurs at points of a voltage maximum. This value is determined by the product of the characteristic impedance $Z_0$ and the Standing Wave Ratio (SWR), denoted as $S$ .

First, we calculate the SWR using the magnitude of the reflection coefficient, $|\Gamma|$ :
$S = \frac{1 + |\Gamma|}{1 - |\Gamma|}$
Plugging in the given values:
$S = \frac{1 + 0.6}{1 - 0.6} = \frac{1.6}{0.4} = 4$
Finally, we find the maximum input impedance:
$Z_{max} = S \cdot Z_0 = 4 \times 50 \, \Omega = 200 \, \Omega$

Q59GATE 2024NAT2MSignals and Systems

The relationship between any $\mathrm{N}$ -length sequence $x[n]$ and its corresponding $\mathrm{N}$ -point discrete Fourier transform $X[k]$ is defined as $X[k]=F\{x[n]\}$ . Another sequence $y[n]$ is formed as below $y[n]=F\{F\{F\{F\{x[n]\}\}\}\}$ For the sequence $x[n]=\{1,2,1,3\}$ , the value of $Y[0]$ is _______

🚀 Solve in Practice Mode

📖 Explanation

A key property of the DFT is that applying it twice results in a time-reversed and scaled sequence, $F\{F\{x[n]\}\} = N x[-n]$ . Applying this operation twice to get $y[n]$ means we perform the time reversal twice, which returns the original sequence. This leads to the simple relationship $y[n] = N^2 x[n]$ .

The value $Y[0]$ is the DC component of the DFT, which is simply the sum of all elements in the sequence $y[n]$ .
The given sequence $x[n]$ has a length of $N=4$ .

We can now find $Y[0]$ :
$Y[0] = \sum_{n=0}^{3} y[n] = \sum_{n=0}^{3} N^2 x[n]$
$Y[0] = N^2 \sum_{n=0}^{3} x[n] = 4^2 (1+2+1+3)$
$Y[0] = 16 \times 7 = 112$

Q60GATE 2024NAT2MNetwork Theory

For the two port network shown below, the value of the $Y_{21}$ parameter (in Siemens) is _____

🚀 Solve in Practice Mode

📖 Explanation

To find the Y-parameter $Y_{21}$ , we use its definition: $Y_{21} = \left.\frac{I_2}{V_1}\right|_{V_2=0}$ . This requires us to short-circuit the output port (setting $V_2=0$ ).

When $V_2=0$ , the central T-junction of the network is at 0V (ground potential). We can find the relationship between $I_2$ and $V_1$ by applying Kirchhoff's Current Law (KCL) at this central node. Let's sum the currents entering the node from the three connected branches.

Current from the left (through 2Ω): $\frac{V_1 - 0}{2} = \frac{V_1}{2}$
Current from the bottom (through 4Ω): The voltage at the bottom of the resistor is $-8V_1$ due to the source polarity. Thus, the current is $\frac{-8V_1 - 0}{4} = -2V_1$ .
Current from the right is the port current $I_2$ .

The KCL equation is the sum of these currents equaling zero: $\frac{V_1}{2} - 2V_1 + I_2 = 0$ .
This simplifies to $-1.5V_1 + I_2 = 0$ , which gives $I_2 = 1.5V_1$ .
Therefore, the value of $Y_{21}$ is $\frac{I_2}{V_1} = 1.5$ S.

Q61GATE 2024NAT2MElectronic Devices

Consider a MOS capacitor made with p-type silicon. It has an oxide thickness of $100 \mathrm{~nm}$ , a fixed positive oxide charge of $10^{-8} \mathrm{C} / \mathrm{cm}^{2}$ at the oxide-silicon interface, and a metal work function of $4.6 \mathrm{eV}$ . Assume that the relative permittivity of the oxide is 4 and the absolute permittivity of free space is $8.85 \times 10^{-14} \mathrm{~F} / \mathrm{cm}$ . If the flatband voltage is $0 \mathrm{~V}$ , the work function of the $\mathrm{p}$ -type silicon (in $\mathrm{eV}$ , rounded off to two decimal places) is ______

🚀 Solve in Practice Mode

📖 Explanation

To find the silicon work function, we start with the equation for the flatband voltage ( $V_{FB}$ ) of a MOS capacitor: $V_{FB} = \phi_{ms} - \frac{Q_{ox}}{C_{ox}}$ . The term $\phi_{ms}$ represents the work function difference, $\phi_m - \phi_s$ . Since $V_{FB}$ is given as 0 V, we can rearrange the equation to solve for the silicon work function: $\phi_s = \phi_m - \frac{Q_{ox}}{C_{ox}}$ .

First, we need to calculate the oxide capacitance per unit area, $C_{ox}$ . Given the oxide thickness $t_{ox} = 100 \mathrm{~nm} = 10^{-5} \mathrm{~cm}$ :
$C_{ox} = \frac{\varepsilon_{ox}}{t_{ox}} = \frac{4 \times 8.85 \times 10^{-14} \mathrm{~F/cm}}{10^{-5} \mathrm{~cm}} = 35.4 \times 10^{-9} \mathrm{~F/cm^2}$ .

Now we can substitute the values for the metal work function ( $\phi_m$ ), fixed oxide charge ( $Q_{ox}$ ), and the calculated $C_{ox}$ into our rearranged equation:
$\phi_s = 4.6 \mathrm{eV} - \frac{10^{-8} \mathrm{C/cm^2}}{35.4 \times 10^{-9} \mathrm{~F/cm^2}}$ .

Calculating the fraction gives the voltage shift due to the fixed charge, which is $0.282 \mathrm{~V}$ . Thus, the work function is:
$\phi_s = 4.6 - 0.282 = 4.318 \mathrm{eV}$ .

Rounding to two decimal places, the work function of the p-type silicon is $4.32 \mathrm{eV}$ .

Q62GATE 2024NAT2MNetwork Theory

In the network shown below, maximum power is to be transferred to the load $R_{L}$ The value of $R_{L}$ (in $\Omega$ ) is _______

🚀 Solve in Practice Mode

📖 Explanation

For maximum power transfer, the load resistance $R_L$ must be equal to the Thevenin resistance, $R_{th}$ , of the source network. To find $R_{th}$ , we deactivate the independent 50V source (replacing it with a short) and apply a test voltage $V$ across the load terminals.

The total current $I$ from this test source is the sum of the currents in the two main branches. The current in the upper branch (through the $2\Omega$ and $3\Omega$ resistors) is $V/5$ . The dependent voltage $V_x$ is across the $3\Omega$ resistor, so $V_x = (V/5) \times 3 = 3V/5$ .

The current in the middle branch is $(V - V_x) / 2$ . Thus, the total current is $I = \frac{V}{5} + \frac{V - V_x}{2}$ . Substituting our expression for $V_x$ , we get $I = \frac{V}{5} + \frac{V - 3V/5}{2} = \frac{4V}{10}$ .

Finally, the Thevenin resistance is the ratio $R_{th} = V/I = 10/4 = 2.5 \Omega$ . Therefore, $R_L$ must be $2.5 \Omega$ .

Q63GATE 2024NAT2MElectronic Devices

A non-degenerate n-type semiconductor has 5% neutral dopant atoms. Its Fermi level is located at $0.25 \mathrm{eV}$ below the conduction band $\left(E_{C}\right)$ and the donor energy level $\left(E_{D}\right)$ has a degeneracy of 2. Assuming the thermal voltage to be $20 \mathrm{mV}$ . The difference between $E_{C}$ and $E_{D}$ (in $\mathrm{eV}$ , rounded off to two decimal places) is ______

🚀 Solve in Practice Mode

📖 Explanation

The fraction of neutral donor atoms is determined by the Fermi-Dirac occupation probability for the donor energy level, $E_D$ . We are given that 5% of donors are neutral, so the occupation probability is $0.05$ . The formula, including the donor degeneracy $g=2$ , is $0.05 = \frac{1}{1 + \frac{1}{2} \exp\left(\frac{E_D - E_F}{kT}\right)}$ .

Solving for the exponential term gives $1 + 0.5 \exp\left(\frac{E_D - E_F}{kT}\right) = 20$ , which simplifies to $\exp\left(\frac{E_D - E_F}{kT}\right) = 38$ . With the thermal energy $kT = 0.020 \text{ eV}$ , we can find the energy separation between the donor level and the Fermi level: $E_D - E_F = kT \ln(38) \approx (0.020 \text{ eV})(3.6376) \approx 0.0728 \text{ eV}$ .

The question asks for the energy difference between the conduction band and the donor level, $E_C - E_D$ . We can express this as the difference between two known energy gaps: $E_C - E_D = (E_C - E_F) - (E_D - E_F)$ . Using the given value $E_C - E_F = 0.25 \text{ eV}$ and our calculated result, we find $E_C - E_D = 0.25 \text{ eV} - 0.0728 \text{ eV} = 0.1772 \text{ eV}$ . Rounding this to two decimal places gives $0.18 \text{ eV}$ .

Q64GATE 2024NAT2MElectronic Devices

An NMOS transistor operating in the linear region has $I_{D S}$ of $5 \mu \mathrm{A}$ at $V_{D S}$ of $0.1 \mathrm{~V}$ . Keeping $V_{G S}$ constant, the $V_{D S}$ is increased to $1.5 \mathrm{~V}$ . Given that $\mu_{n} C_{o x} \frac{W}{L}=50 \mu \mathrm{A} / \mathrm{V}^{2}$ , the transconductance at the new operating point (in $\mu \mathrm{A} / \mathrm{V}$ , rounded off to two decimal places) is ________

🚀 Solve in Practice Mode

📖 Explanation

First, we determine the overdrive voltage, $V_{GS}-V_T$ , from the initial data when the transistor operates in the linear region. Using the linear current equation $I_D = \mu_n C_{ox}\frac{W}{L} \left[(V_{GS}-V_T)V_{DS} - \frac{1}{2}V_{DS}^2\right]$ , we substitute the given values: $5\mu = 50\mu \left[(V_{GS}-V_T)(0.1) - \frac{1}{2}(0.1)^2\right]$ . Solving this yields an overdrive voltage of $V_{GS}-V_T = 1.05 \text{ V}$ .

Next, we check the new region of operation with $V_{DS}$ increased to $1.5 \text{ V}$ . Since $V_{DS} (1.5 \text{ V}) > V_{GS}-V_T (1.05 \text{ V})$ , the transistor is now in the saturation region. The transconductance in saturation is given by $g_m = \mu_n C_{ox}\frac{W}{L}(V_{GS}-V_T)$ . Plugging in the known values gives $g_m = (50 \mu \mathrm{A/V^2})(1.05 \text{ V}) = 52.5 \mu \mathrm{A/V}$ .

Q65GATE 2024NAT2MElectronic Devices

The photocurrent of a PN junction diode solar cell is $1 \mathrm{~mA}$ . The voltage corresponding to its maximum power point is $0.3 \mathrm{~V}$ . If the thermal voltage is $30 \mathrm{mV}$ , the reverse saturation current of the diode (in $\mathrm{nA}$ , rounded off to two decimal places) is ________

🚀 Solve in Practice Mode

📖 Explanation

To determine the reverse saturation current, $I_0$ , we leverage the condition for maximum power delivery from the solar cell. The given voltage, $V_m = 0.3 \text{ V}$ , is the specific voltage at which the output power is maximized. This maximum power point condition ( $dP/dV=0$ ) leads to a direct relationship for $I_0$ .

The governing equation is:
$I_0 = \frac{I_L}{\left(1 + \frac{V_m}{V_T}\right) e^{V_m/V_T} - 1}$
We are given the photocurrent $I_L = 1 \text{ mA}$ , the maximum power voltage $V_m = 0.3 \text{ V}$ , and the thermal voltage $V_T = 30 \text{ mV} = 0.03 \text{ V}$ .

Plugging these values into the equation:
$I_0 = \frac{1 \times 10^{-3} \text{ A}}{\left(1 + \frac{0.3}{0.03}\right) e^{0.3/0.03} - 1} = \frac{10^{-3}}{11e^{10} - 1}$
Calculating the result yields $I_0 \approx 4.127 \times 10^{-9} \text{ A}$ , which is $4.13 \text{ nA}$ when rounded to two decimal places.

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