GATE ECE 2022

Q11GATE 2022MCQ1MEngineering Mathematics

Consider the two-dimensional vector field $\vec{F}(x,y)=x\vec{i}+y\vec{j}$ , where $\vec{i}$ and $\vec{j}$ denote the unit vectors along the x-axis and the y-axis, respectively. A contour C in the x-y plane, as shown in the figure, is composed of two horizontal lines connected at the two ends by two semicircular arcs of unit radius. The contour is traversed in the counter-clockwise sense. The value of the closed path integral $\oint _c \vec{F}(x,y)\cdot (dx\vec{i}+dy\vec{j})$

🚀 Solve in Practice Mode

📖 Explanation

The integral we need to evaluate is the closed line integral $\oint_C \vec{F} \cdot d\vec{r}$ . Substituting the given vector field, this becomes $\oint_C (x dx + y dy)$ .

The vector field $\vec{F}(x,y) = x\vec{i} + y\vec{j}$ is a conservative field. This means it can be written as the gradient of a scalar potential function, $\phi(x,y)$ . For this field, the potential is $\phi(x,y) = \frac{1}{2}(x^2 + y^2)$ , since $\nabla\phi = \frac{\partial\phi}{\partial x}\vec{i} + \frac{\partial\phi}{\partial y}\vec{j} = x\vec{i} + y\vec{j}$ .

A fundamental theorem of vector calculus states that the line integral of a conservative field over any closed path is always zero. This is because the integral's value is simply the change in the potential function between the endpoint and the starting point. For a closed path, these points are identical, so the net change is zero. The specific geometry of the path is therefore irrelevant.

Q12GATE 2022MCQ1MEngineering Mathematics

Consider a system of linear equations $Ax=b$ , where

A=\begin{bmatrix} 1 & -\sqrt{2} & 3\\ -1& \sqrt{2}& -3 \end{bmatrix},b=\begin{bmatrix} 1\\ 3 \end{bmatrix}

This system of equations admits ______.

🚀 Solve in Practice Mode

📖 Explanation

Let's translate the matrix equation $Ax=b$ into its corresponding system of linear equations. Letting

x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}

, we have:
$x_1 - \sqrt{2}x_2 + 3x_3 = 1$
$-x_1 + \sqrt{2}x_2 - 3x_3 = 3$
Observe that the left-hand side of the second equation is exactly $-1$ times the left-hand side of the first. If we multiply the entire second equation by $-1$ , we get $x_1 - \sqrt{2}x_2 + 3x_3 = -3$ .

This creates a contradiction: the first equation states that the expression $x_1 - \sqrt{2}x_2 + 3x_3$ equals $1$ , while the manipulated second equation states it equals $-3$ . Since the same quantity cannot have two different values, the system is inconsistent and admits no solutions.

Q13GATE 2022MCQ1MNetwork Theory

The current $I$ in the circuit shown is ________

🚀 Solve in Practice Mode

📖 Explanation

To determine the current $I$ , we first need to find the voltage at the central node, let's call it $V_A$ . We can achieve this using Nodal Analysis, with the bottom wire serving as our 0V ground reference.

Applying Kirchhoff's Current Law (KCL) at node $A$ , we set the sum of currents leaving the node equal to the sum of currents entering.
$\frac{V_A}{2 \text{ k}\Omega} + \frac{V_A - 5 \text{ V}}{2 \text{ k}\Omega} = 10^{-3} \text{ A}$
Solving this equation for $V_A$ , we get $\frac{2V_A - 5}{2000} = 10^{-3}$ , which simplifies to $2V_A - 5 = 2$ , giving us $V_A = 3.5 \text{ V}$ .

The current $I$ is the flow from the 5V source towards node $A$ . We can now find its value using Ohm's Law on the leftmost resistor:
$I = \frac{5 \text{ V} - V_A}{2 \text{ k}\Omega} = \frac{5 - 3.5}{2000} = \frac{1.5}{2000} = 0.75 \times 10^{-3} \text{ A}$

Q14GATE 2022MCQ1MNetwork Theory

Consider the circuit shown in the figure. The current $I$ flowing through the $10\Omega$ resistor is _________.

🚀 Solve in Practice Mode

📖 Explanation

A fundamental principle of circuit theory is that current can only flow in a complete, closed loop. Let's examine the path for the current labeled $I$ . The diagram shows this current flowing from left to right along a wire. However, the left end of this wire is not connected to any component or source; it's an open circuit. Since there is no continuous path for charge to travel from, through, and back to a source, no current can flow in this segment. Therefore, the current $I$ must be zero.

Q15GATE 2022MCQ1MSignals and Systems

The Fourier transform $X(j\omega )$ of the signal $x(t)=\frac{t}{(1+t^2)^2}$ is _________.

🚀 Solve in Practice Mode

📖 Explanation

This problem is a clever application of the Fourier Transform's duality property. We begin by establishing a related transform pair using the frequency differentiation property, which states that $\mathcal{F}\{t \cdot g(t)\} = j \frac{d}{d\omega}G(j\omega)$ .

Let's start with the standard pair $\mathcal{F}\{e^{-|t|}\} = \frac{2}{1+\omega^2}$ . Applying frequency differentiation, we find the transform of $t e^{-|t|}$ :
$\mathcal{F}\{t e^{-|t|}\} = j \frac{d}{d\omega} \left(\frac{2}{1+\omega^2}\right) = \frac{-j4\omega}{(1+\omega^2)^2}$
Now we use the duality property: if $g(t) \leftrightarrow G(j\omega)$ , then $G(jt) \leftrightarrow 2\pi g(-\omega)$ . Applying this to the pair we just derived gives:
$\mathcal{F}\left\{\frac{-j4t}{(1+t^2)^2}\right\} = 2\pi(-\omega)e^{-|-\omega|} = -2\pi\omega e^{-|\omega|}$
The signal in the question, $x(t) = \frac{t}{(1+t^2)^2}$ , is a scaled version of the time function above. Using linearity, we can find its transform by dividing by the constant $-j4$ :
$X(j\omega) = \frac{1}{-j4} \left( -2\pi\omega e^{-|\omega|} \right) = \frac{2\pi}{4j}\omega e^{-|\omega|} = \frac{\pi}{2j}\omega e^{-|\omega|}$

Q16GATE 2022MCQ1MElectronic Devices

Consider a long rectangular bar of direct bandgap p-type semiconductor. The equilibrium hole density is $10^{17}cm^{-3}$ and the intrinsic carrier concentration is $10^{10}cm^{-3}$ . Electron and hole diffusion lengths are $2\mu m$ and $1\mu m$ , respectively. The left side of the bar ( $x=0$ ) is uniformly illuminated with a laser having photon energy greater than the bandgap of the semiconductor. Excess electron-hole pairs are generated ONLY at $x=0$ because of the laser. The steady state electron density at $x=0$ is $10^{14}cm^{-3}$ due to laser illumination. Under these conditions and ignoring electric field, the closest approximation (among the given options) of the steady state electron density at $x=2 \mu m$ , is _____

🚀 Solve in Practice Mode

📖 Explanation

In this p-type semiconductor, the problem describes the behavior of excess minority carriers (electrons) generated at one end. Away from the generation point ( $x>0$ ) and in steady state with no electric field, the excess electron concentration is governed by a balance between diffusion and recombination.

The solution to the carrier diffusion equation under these conditions is an exponential decay:
$\delta n(x) = \delta n(0) e^{-x/L_n}$
Here, $\delta n(x)$ is the excess electron concentration at a distance $x$ , and $L_n$ is the electron diffusion length, given as $2 \mu m$ .

The excess concentration at the source is the total concentration minus the equilibrium concentration: $\delta n(0) = n(0) - n_0$ . With $n(0) = 10^{14} cm^{-3}$ and $n_0 = n_i^2/p_0 = (10^{10})^2/10^{17} = 10^3 cm^{-3}$ , the excess concentration $\delta n(0)$ is approximately $10^{14} cm^{-3}$ .

Now, we can find the excess electron density at $x=2 \mu m$ :
$\delta n(2 \mu m) = (10^{14}) \times e^{-2\mu m / 2\mu m} = 10^{14} \times e^{-1} \approx 0.37 \times 10^{14} cm^{-3}$
Since the equilibrium density ( $10^3 cm^{-3}$ ) is negligible compared to this value, the total electron density at $x=2 \mu m$ is approximately $0.37 \times 10^{14} cm^{-3}$ .

Q17GATE 2022MCQ1MElectronic Devices

In a non-degenerate bulk semiconductor with electron density $n=10^{16}cm^{-3}$ , the value of $E_C-E_{Fn}=200meV$ , where $E_C$ and $E_{Fn}$ denote the bottom of the conduction band energy and electron Fermi level energy, respectively. Assume thermal voltage as 26 meV and the intrinsic carrier concentration is $10^{10}cm^{-3}$ . For $n=0.5 \times 10^{16}cm^{-3}$ , the closest approximation of the value of ( $E_C-E_{Fn}$ ), among the given options, is ______.

🚀 Solve in Practice Mode

📖 Explanation

The relationship between electron density ( $n$ ) and the energy separation between the conduction band edge ( $E_C$ ) and the electron Fermi level ( $E_{Fn}$ ) is given by:
$E_C - E_{Fn} = kT \ln\left(\frac{N_C}{n}\right)$
where $N_C$ is the effective density of states in the conduction band and $kT$ is the thermal energy.

We can analyze the change between the two given conditions by subtracting the equations for each state. This strategy conveniently cancels out the unknown $N_C$ . Let the initial state be 1 and the new state be 2:
$(E_C - E_{Fn})_2 - (E_C - E_{Fn})_1 = kT \ln\left(\frac{N_C}{n_2}\right) - kT \ln\left(\frac{N_C}{n_1}\right) = kT \ln\left(\frac{n_1}{n_2}\right)$

Now, we can solve for the new energy separation, $(E_C - E_{Fn})_2$ . We are given $(E_C - E_{Fn})_1 = 200 \text{ meV}$ , $n_1 = 10^{16} \text{ cm}^{-3}$ , $n_2 = 0.5 \times 10^{16} \text{ cm}^{-3}$ , and $kT = 26 \text{ meV}$ .
$(E_C - E_{Fn})_2 = 200 \text{ meV} + (26 \text{ meV}) \ln\left(\frac{1 \times 10^{16}}{0.5 \times 10^{16}}\right)$
$(E_C - E_{Fn})_2 = 200 \text{ meV} + (26 \text{ meV}) \ln(2) \approx 200 + (26 \times 0.693) \approx 200 + 18 \approx 218 \text{ meV}$

Q18GATE 2022MCQ1MAnalog Circuits

Consider the CMOS circuit shown in the figure (substrates are connected to their respective sources). The gate width ( $W$ ) to gate length ( $L$ ) ratios $\frac{W}{L}$ of the transistors are as shown. Both the transistors have the same gate oxide capacitance per unit area. For the pMOSFET, the threshold voltage is -1 V and the mobility of holes is $40\frac{cm^2}{V.s}$ . For the nMOSFET, the threshold voltage is 1 V and the mobility of electrons is $300\frac{cm^2}{V.s}$ . The steady state output voltage $V_o$ is ________.

🚀 Solve in Practice Mode

📖 Explanation

In this circuit's steady state, both the pMOS and nMOS transistors operate in the saturation region because their gates are connected to the output. This means their drain currents must be equal.

We can set the saturation current equations equal, $I_{D,p} = I_{D,n}$ , where the current is proportional to $\mu (\frac{W}{L}) (V_{GS} - V_t)^2$ . For the pMOS transistor, $V_{SG} = 4 - V_o$ and $|V_t| = 1$ V. For the nMOS, $V_{GS} = V_o$ and $V_t = 1$ V.

Plugging in the given values: $40 \cdot 5 \cdot (4 - V_o - 1)^2 = 300 \cdot 1 \cdot (V_o - 1)^2$ .

This equation simplifies to $200(3-V_o)^2 = 300(V_o-1)^2$ , which reduces to $(3-V_o)^2 = 1.5(V_o-1)^2$ .

Taking the square root and solving for $V_o$ gives $V_o = \frac{3+\sqrt{1.5}}{1+\sqrt{1.5}}$ . To determine if this value is less than 2 V, we can check the inequality $\frac{3+\sqrt{1.5}}{1+\sqrt{1.5}} < 2$ . This is equivalent to $3+\sqrt{1.5} < 2+2\sqrt{1.5}$ , which simplifies to $1 < \sqrt{1.5}$ . As this is true, $V_o$ is indeed less than 2 V.

Q19GATE 2022MCQ1MDigital Circuits

Consider the 2-bit multiplexer (MUX) shown in the figure. For OUTPUT to be the XOR of C and D, the values for $A_0,A_1,A_2 \text{ and }A_3$ are _______

🚀 Solve in Practice Mode

📖 Explanation

The output of the multiplexer is determined by the select lines, $C$ and $D$ . We want the output to implement the XOR function, $C \oplus D$ . To achieve this, we can construct the truth table for the XOR gate and map its output values directly to the multiplexer's data inputs.

The select lines $(S_1, S_0)$ correspond to $(C, D)$ .
* For $(C,D) = (0,0)$ , the MUX selects input $A_0$ . The desired output is $0 \oplus 0 = 0$ , so $A_0=0$ .
* For $(C,D) = (0,1)$ , the MUX selects input $A_1$ . The desired output is $0 \oplus 1 = 1$ , so $A_1=1$ .
* For $(C,D) = (1,0)$ , the MUX selects input $A_2$ . The desired output is $1 \oplus 0 = 1$ , so $A_2=1$ .
* For $(C,D) = (1,1)$ , the MUX selects input $A_3$ . The desired output is $1 \oplus 1 = 0$ , so $A_3=0$ .

Therefore, the required input values are $A_0=0, A_1=1, A_2=1, \text{ and } A_3=0$ .

Q20GATE 2022MCQ1MAnalog Circuits

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

🚀 Solve in Practice Mode

Q21GATE 2022MCQ1MControl Systems

Consider a closed-loop control system with unity negative feedback and $KG(s)$ in the forward path, where the gain $K=2$ . The complete Nyquist plot of the transfer function $G(s)$ is shown in the figure. Note that the Nyquist contour has been chosen to have the clockwise sense. Assume $G(s)$ has no poles on the closed right-half of the complex plane. The number of poles of the closed-loop transfer function in the closed right-half of the complex plane is _______

🚀 Solve in Practice Mode

📖 Explanation

To find the number of unstable closed-loop poles ( $Z$ ), we apply the Nyquist stability criterion. For a clockwise Nyquist contour, the formula is $Z = P + N_{CW}$ , where $P$ is the number of open-loop poles in the right-half plane (RHP) and $N_{CW}$ is the number of clockwise encirclements of the critical point $(-1, 0)$ .

The problem states $G(s)$ has no RHP poles, so $P=0$ . The open-loop transfer function is $KG(s)$ with $K=2$ . We analyze the plot of $2G(s)$ by scaling the given plot of $G(s)$ by 2. The real-axis crossings at $-0.4$ and $-0.8$ are scaled to $-0.8$ and $-1.6$ . The critical point $(-1, 0)$ lies between these new crossings. The two small loops encircle this point, and their arrows indicate a clockwise direction. Thus, we have two clockwise encirclements, making $N_{CW} = 2$ .

Substituting these values into the formula gives $Z = P + N_{CW} = 0 + 2 = 2$ .

Q22GATE 2022MCQ1MControl Systems

The root-locus plot of a closed-loop system with unity negative feedback and transfer function $KG(s)$ in the forward path is shown in the figure. Note that $K$ is varied from 0 to $\infty$ . Select the transfer function $G(s)$ that results in the root-locus plot of the closed-loop system as shown in the figure.

🚀 Solve in Practice Mode

📖 Explanation

The starting points of a root-locus plot correspond to the locations of the open-loop poles. This plot shows five branches, all beginning at the same point on the real axis. This means the system must have five repeated open-loop poles at that single location. From the figure, we can identify this point as $s = -1$ . A transfer function $G(s)$ with five poles at $s=-1$ and no finite zeros would have a denominator of $(s+1)^5$ , which directly matches the form of the given transfer function.

Q23GATE 2022MCQ1MCommunication Systems

The frequency response $H(f)$ of a linear time-invariant system has magnitude as shown in the figure. Statement I: The system is necessarily a pure delay system for inputs which are bandlimited to $-\alpha \leq f\leq \alpha$ . Statement II: For any wide-sense stationary input process with power spectral density $S_X(f)$ , the output power spectral density $S_Y(f)$ obeys $S_Y(f)=S_X(f)$ for $-\alpha \leq f\leq \alpha$ . Which one of the following combinations is true?

🚀 Solve in Practice Mode

📖 Explanation

Let's break down each statement.

For a system to be a pure delay, its frequency response must be $H(f) = e^{-j2\pi f t_d}$ . This requires two conditions: a constant magnitude of 1, and a phase that is a linear function of frequency. The problem only gives us that the magnitude $|H(f)|=1$ in the specified band. Since the phase is unknown, we cannot be certain the system is a pure delay. Therefore, Statement I is incorrect.

For Statement II, the output power spectral density (PSD), $S_Y(f)$ , is related to the input PSD, $S_X(f)$ , by the equation $S_Y(f) = |H(f)|^2 S_X(f)$ . From the figure, we see that $|H(f)|=1$ for frequencies in the range $-\alpha \leq f\leq \alpha$ . Substituting this into the equation yields $S_Y(f) = (1)^2 S_X(f) = S_X(f)$ for that band. Thus, Statement II is correct.

Q24GATE 2022MCQ1MElectromagnetics

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

🚀 Solve in Practice Mode

📖 Explanation

In a series circuit, the fundamental rule is that the current is the same through every component at any given instant. The current flowing through the resistor is a conduction current, given as $I_c = 2\sin(t+\pi/2)$ .

For an ideal capacitor, the "current" that passes between its plates is purely displacement current, denoted as $I_d$ . Because the resistor and capacitor are in series, the current passing through each must be identical. Therefore, the displacement current in the capacitor must equal the conduction current in the resistor.

$I_d = I_c = 2\sin(t+\pi/2)$

Q25GATE 2022MSQ1MEngineering Mathematics

Consider the following partial differential equation (PDE) $a\frac{\partial^2 f(x,y)}{\partial x^2} +b\frac{\partial^2 f(x,y)}{\partial y^2}=f(x,y)$ , where $a$ and $b$ are distinct positive real numbers. Select the combination(s) of values of the real parameters $\xi$ and $\eta$ such that $f(x,y)=e^{(\xi x+\eta y)}$ is a solution of the given PDE.

🚀 Solve in Practice Mode

📖 Explanation

To see if $f(x,y)=e^{(\xi x+\eta y)}$ solves the PDE, we must substitute it into the equation. First, we find the required second partial derivatives. Differentiating with respect to $x$ twice gives $\frac{\partial^2 f}{\partial x^2} = \xi^2 e^{(\xi x+\eta y)}$ , and differentiating with respect to $y$ twice gives $\frac{\partial^2 f}{\partial y^2} = \eta^2 e^{(\xi x+\eta y)}$ .

Plugging these into the PDE results in:
$a(\xi^2 e^{(\xi x+\eta y)}) + b(\eta^2 e^{(\xi x+\eta y)}) = e^{(\xi x+\eta y)}$

Since $e^{(\xi x+\eta y)}$ is never zero, we can divide the entire equation by it. This simplifies to the following condition on the parameters $\xi$ and $\eta$ :
$a\xi^2 + b\eta^2 = 1$

Our task now is to check which of the provided $(\xi, \eta)$ pairs satisfy this algebraic relationship. By substituting the values from the options, we find the pairs that make the equation true.

Q26GATE 2022MSQ1MAnalog Circuits

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

🚀 Solve in Practice Mode

📖 Explanation

This circuit is an inverting amplifier, so its voltage gain is determined by the ratio of feedback impedance to input impedance, $A_v = -Z_f / Z_{\in}$ . The transfer function is therefore:
$H(j\omega) = \frac{V_{out}}{V_{\in}} = \frac{-R_f}{R_{\in} + 1/(j\omega C)} = \frac{-2000}{1000 + 1/(j\omega \cdot 10^{-6})}$ .

To understand the filter type, we check the gain at frequency extremes. As $\omega \to 0$ , the gain approaches zero. As $\omega \to \infty$ , the gain approaches a constant value of $-2$ . Since the circuit passes high frequencies and attenuates low frequencies, it is a high-pass filter.

By simplifying the transfer function, we can find the 3 dB frequency:
$H(j\omega) = \frac{-2}{1 + 1000/(j\omega)} = \frac{-2(j\omega/1000)}{1 + j\omega/1000}$ .
This is the standard form for a first-order high-pass filter, where the 3 dB cutoff frequency, $\omega_c$ , is $1000 \text{ rad/s}$ .

Q27GATE 2022MSQ1MDigital Circuits

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

🚀 Solve in Practice Mode

📖 Explanation

To find the expressions equivalent to $x+yz$ , we simplify each option using the properties of Boolean algebra.

For option B, we apply the distributive law, $(A+B)(A+C) = A+BC$ . Letting $A=x$ , $B=y$ , and $C=z$ , the expression $(x+y)(x+z)$ simplifies directly to $x+yz$ .

For option C, we use the absorption law, $A+AB=A$ . In the expression $x+xy+yz$ , the term $x$ absorbs the term $xy$ because $x+xy = x(1+y) = x$ . This leaves us with $x+yz$ .

Options A and D are not equivalent. Option A, $x+z+xy$ , simplifies to $(x+xy)+z = x+z$ . Option D, $x+xz+xy$ , simplifies to $x(1+z+y)$ , which reduces to just $x$ .

Q28GATE 2022MCQ1MDigital Circuits

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

🚀 Solve in Practice Mode

📖 Explanation

Here is a revised explanation:

For a CMOS inverter, a steady high input ( $V_{\in} = V_{DD}$ ) places the PMOS in cutoff and turns the NMOS on. The NMOS pulls the output to ground, making $V_{out} \approx 0$ . With a high gate-source voltage ( $V_{GS} = V_{DD}$ ) but a very low drain-source voltage ( $V_{DS} \approx 0$ ), the NMOS operates in its linear (triode) region.

Regarding the other options:
(A) Noise margins are equal only for a symmetric inverter, which requires specific transistor sizing and is not always true.
(B) Dynamic power is consumed to charge and discharge the output capacitance during switching, so it is never zero in practice.
(D) Switching speed depends on the current supplied by the transistors, which is directly proportional to carrier mobility ( $\mu_n, \mu_p$ ).

Q29GATE 2022MSQ1MCommunication Systems

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

🚀 Solve in Practice Mode

📖 Explanation

While properties like scaling or exponentiation (options B and D) are one-to-one functions that preserve the entropy of a random variable, squaring it is not. The function $f(x) = x^2$ can map distinct values of $X$ to the same value, which can reduce information and therefore entropy.

To demonstrate this, consider a random variable $X$ that takes values $\{-1, 0, 1\}$ with probabilities $P(X=-1)=\frac{1}{4}$ , $P(X=0)=\frac{1}{2}$ , and $P(X=1)=\frac{1}{4}$ . The entropy is $H(X) = 1.5$ bits.

Now, let $Y = X^2$ . The possible outcomes are $0$ and $1$ . The outcome $Y=1$ occurs whether $X=-1$ or $X=1$ , so $P(Y=1) = P(X=-1)+P(X=1) = \frac{1}{4}+\frac{1}{4}=\frac{1}{2}$ . The other probability is $P(Y=0)=P(X=0)=\frac{1}{2}$ .
The entropy of $Y=X^2$ is $H(X^2) = -(\frac{1}{2}\log_2\frac{1}{2} + \frac{1}{2}\log_2\frac{1}{2}) = 1$ bit. Since $1.5 > 1$ , we have found a case where $H(X) > H(X^2)$ , so statement (C) is not necessarily true.

Q30GATE 2022MSQ1MElectromagnetics

Consider the following wave equation, $\frac{\partial^2 f(x,t)}{\partial t^2}=10000\frac{\partial^2 f(x,t)}{\partial x^2}$ Which of the given options is/are solution(s) to the given wave equation?

🚀 Solve in Practice Mode

📖 Explanation

The provided equation is a classic 1D wave equation, which has the general form $\frac{\partial^2 f}{\partial t^2}=v^2\frac{\partial^2 f}{\partial x^2}$ . By comparing this to the problem's equation, we can identify the wave speed squared as $v^2 = 10000$ , which means the wave speed is $v=100$ .

The general solution to this equation, known as d'Alembert's solution, is any function that can be expressed as a superposition of a right-traveling wave and a left-traveling wave: $f(x,t) = g(x-vt) + h(x+vt)$ .

Therefore, we are looking for options that are a sum of functions with arguments $(x-100t)$ and $(x+100t)$ . Both option A and option C perfectly match this structure. The other options are incorrect because they contain terms with arguments that correspond to a different wave speed.

Q31GATE 2022NAT1MEngineering Mathematics

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

🚀 Solve in Practice Mode

📖 Explanation

To find the median, we first determine the total number of matches by summing the frequencies from the graph. The total is $5+7+8+25+20+17+8+4+3+2+1 = 100$ matches.

The median is the middle value in a sorted dataset. For an even number of data points like 100, the median is the average of the two middle terms, which are the 50th and 51st values.

Let's find these values using cumulative frequency. The number of matches with 3 or fewer wickets is $5+7+8+25 = 45$ . This means the first 45 matches in our ordered list correspond to 0, 1, 2, or 3 wickets. The next 20 matches are all for 4 wickets, covering the 46th through 65th positions.

Since both the 50th and 51st matches fall into this group, their value is 4 wickets. The median is the average of these two values, which is $\frac{4+4}{2} = 4$ .

Q32GATE 2022NAT1MEngineering Mathematics

A simple closed path $C$ in the complex plane is shown in the figure. If $\oint C\frac{2^z}{z^2-1}dz=-i \pi A$ where $i=\sqrt{-1}$ , then the value of $A$ is ______ (rounded off to two decimal places)

🚀 Solve in Practice Mode

📖 Explanation

We are asked to evaluate a complex contour integral. The integrand, $\frac{2^z}{z^2-1}$ , has singularities where the denominator is zero, i.e., at $z=1$ and $z=-1$ .

Looking at the provided path $C$ , we see it encloses the point $z=-1$ but not $z=1$ . We can therefore use Cauchy's Integral Formula to solve this. First, we rewrite the integral to isolate the enclosed pole:
$\oint_C \frac{2^z}{(z-1)(z+1)} dz = \oint_C \frac{\left(\frac{2^z}{z-1}\right)}{z-(-1)} dz$
This integral has the form $\oint_C \frac{f(z)}{z-z_0} dz = 2\pi i f(z_0)$ , with $f(z) = \frac{2^z}{z-1}$ and $z_0 = -1$ .

Evaluating $f(z)$ at the pole $z_0 = -1$ , we get:
$f(-1) = \frac{2^{-1}}{-1-1} = \frac{1/2}{-2} = -\frac{1}{4}$
The value of the integral is thus $2\pi i \cdot f(-1) = 2\pi i \left(-\frac{1}{4}\right) = -i\frac{\pi}{2}$ .

We are given that the integral equals $-i\pi A$ . By comparison, $-i\pi A = -i\frac{\pi}{2}$ , which means $A = \frac{1}{2} = 0.5$ .

Q33GATE 2022NAT1MSignals and Systems

Let $x_1(t)=e^{-t}u(t)$ and $x_2(t)=u(t)-u(t-2)$ , where $u(\cdot)$ denotes the unit step function. If $y(t)$ denotes the convolution of $x_1(t)$ and $x_2(t)$ , then $\lim_{t\rightarrow \infty }y(t)=$ _________ (rounded off to one decimal place).

🚀 Solve in Practice Mode

📖 Explanation

To find the long-term value of the convolution $y(t)$ , we can efficiently use the Final Value Theorem of the Laplace transform. First, let's find the Laplace transforms of the signals:
$X_1(s) = \mathcal{L}\{e^{-t}u(t)\} = \frac{1}{s+1}$
$X_2(s) = \mathcal{L}\{u(t)-u(t-2)\} = \frac{1}{s} - \frac{e^{-2s}}{s} = \frac{1-e^{-2s}}{s}$

Since convolution in the time domain corresponds to multiplication in the s-domain, the transform of the output is $Y(s) = X_1(s)X_2(s) = \frac{1-e^{-2s}}{s(s+1)}$ .

Now, we apply the Final Value Theorem, which states $\lim_{t\rightarrow\infty } y(t) = \lim_{s\rightarrow 0} sY(s)$ :
$\lim_{s\rightarrow 0} s \left( \frac{1-e^{-2s}}{s(s+1)} \right) = \lim_{s\rightarrow 0} \frac{1-e^{-2s}}{s+1}$

Evaluating this limit by substituting $s=0$ yields $\frac{1-e^0}{0+1} = \frac{1-1}{1} = 0$ .

Q34GATE 2022NAT1MElectronic Devices

An ideal MOS capacitor (p-type semiconductor) is shown in the figure. The MOS capacitor is under strong inversion with $V_G=2V$ . The corresponding inversion charge density $Q_{IN}$ is $2.2\mu C/cm^2$ . Assume oxide capacitance per unit area as $C_{OX}=1.7\mu F/cm^2$ . For $V_G=4V$ , the value of $Q_{IN}$ is ______ $\mu C/cm^2$ (rounded off to one decimal place).

🚀 Solve in Practice Mode

📖 Explanation

In the strong inversion region of a MOS capacitor, the inversion charge density ( $Q_{IN}$ ) is linearly proportional to the gate voltage above the threshold voltage ( $V_T$ ). This relationship is described by the equation $Q_{IN} = -C_{OX}(V_G - V_T)$ .

Since $C_{OX}$ and $V_T$ are constant for the device, we can find the change in charge density ( $\Delta Q_{IN}$ ) resulting from a change in gate voltage ( $\Delta V_G$ ) without needing to know $V_T$ . The relationship is simply $\Delta Q_{IN} = -C_{OX} \Delta V_G$ .

Here, the gate voltage changes from $2V$ to $4V$ , so $\Delta V_G = 4V - 2V = 2V$ . The resulting change in charge density is:
$\Delta Q_{IN} = -(1.7 \mu F/cm^2) \times (2V) = -3.4 \mu C/cm^2$ .

The initial inversion charge density is given as $2.2 \mu C/cm^2$ . Since it's an inversion layer in a p-type substrate, the charge is due to electrons, so we write it as $Q_{IN_1} = -2.2 \mu C/cm^2$ . The new charge density is the sum of the initial charge and the change:
$Q_{IN_2} = Q_{IN_1} + \Delta Q_{IN} = -2.2 \mu C/cm^2 + (-3.4 \mu C/cm^2) = -5.6 \mu C/cm^2$ .

The magnitude of this new charge density is $5.6 \mu C/cm^2$ .

Q35GATE 2022NAT1MCommunication Systems

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

🚀 Solve in Practice Mode

📖 Explanation

The overall data rate is determined by the symbol rate multiplied by the average number of bits conveyed by each symbol. Let's first find the number of bits for each modulation scheme.

A QPSK symbol has $M=4$ possible states, so it carries $\log_2(4) = 2$ bits.
A 16-QAM symbol has $M=16$ states, carrying $\log_2(16) = 4$ bits.

Since the symbols alternate, the average number of bits per symbol in the stream is the simple average of the two:
$\text{Average bits per symbol} = \frac{2 \text{ bits} + 4 \text{ bits}}{2} = 3 \text{ bits}$
Finally, the raw data rate is this average multiplied by the symbol rate:
$\text{Data Rate} = 1 \text{ Mega-symbol/s} \times 3 \text{ bits/symbol} = 3.0 \text{ Mbps}$

Q36GATE 2022MCQ2MEngineering Mathematics

The function $f(x)=8 \log _e x-x^2+3$ attains its minimum over the interval $[1,e]$ at $x=$ _________. (Here $\log _e x$ is the natural logarithm of $x=$ .)

🚀 Solve in Practice Mode

📖 Explanation

To find the minimum value of $f(x)=8 \ln x-x^2+3$ on the interval $[1,e]$ , we'll use the closed interval method. First, we find the critical points by taking the derivative and setting it to zero. The derivative is $f'(x) = \frac{8}{x} - 2x$ .

Setting $f'(x)=0$ gives $\frac{8}{x} - 2x = 0$ , which solves to $x^2=4$ , so $x=2$ (we only consider the positive root). This critical point lies within our interval $[1, e]$ , since $e \approx 2.718$ .

To determine if $x=2$ is a maximum or minimum, we can use the second derivative test. The second derivative is $f''(x) = -\frac{8}{x^2} - 2$ . Since $f''(2) = -\frac{8}{4} - 2 = -4$ , which is negative, $x=2$ is a local maximum.

Because the only critical point in the interval is a local maximum, the absolute minimum must occur at one of the endpoints. We now compare the function values at $x=1$ and $x=e$ :
$f(1) = 8 \ln(1) - 1^2 + 3 = 0 - 1 + 3 = 2$
$f(e) = 8 \ln(e) - e^2 + 3 = 8 - e^2 + 3 = 11 - e^2 \approx 3.61$

Comparing these values, we see the minimum value is $2$ , which occurs at $x=1$ .

Q37GATE 2022MCQ2MEngineering Mathematics

Let $\alpha ,\beta$ be two non-zero real numbers and $v_1,v_2$ be two non-zero real vectors of size 3 x 1. Suppose that $v_1$ and $v_2$ satisfy $v_1^Tv_2=0,v_1^Tv_1=1$ , and $v_2^Tv_2=1$ . Let $A$ be the 3x3 matrix given by: $A=\alpha v_1v_1^T+ \beta v_2v_2^T$ The eigenvalues of $A$ are __________.

🚀 Solve in Practice Mode

📖 Explanation

An eigenvalue-eigenvector pair $(\lambda, v)$ satisfies the equation $Av = \lambda v$ . We can test if the vectors given in the problem are eigenvectors of $A$ .

Let's compute $Av_1$ :
$Av_1 = (\alpha v_1v_1^T + \beta v_2v_2^T)v_1 = \alpha v_1(v_1^Tv_1) + \beta v_2(v_2^Tv_1)$ .
Using the given conditions $v_1^Tv_1=1$ and $v_2^Tv_1 = (v_1^Tv_2)^T = 0$ , we find $Av_1 = \alpha v_1$ . This means $\alpha$ is an eigenvalue with eigenvector $v_1$ .

Next, let's compute $Av_2$ :
$Av_2 = (\alpha v_1v_1^T + \beta v_2v_2^T)v_2 = \alpha v_1(v_1^Tv_2) + \beta v_2(v_2^Tv_2)$ .
Using $v_1^Tv_2=0$ and $v_2^Tv_2=1$ , we get $Av_2 = \beta v_2$ . So, $\beta$ is an eigenvalue with eigenvector $v_2$ .

Since $v_1$ and $v_2$ are orthogonal vectors in $\mathbb{R}^3$ , there must be a third non-zero vector $v_3$ that is orthogonal to both. This implies $v_1^Tv_3 = 0$ and $v_2^Tv_3 = 0$ . Applying $A$ to this vector gives:
$Av_3 = \alpha v_1(v_1^Tv_3) + \beta v_2(v_2^Tv_3) = \alpha v_1(0) + \beta v_2(0) = 0$ .
Since $Av_3 = 0 = 0v_3$ , the third eigenvalue is $0$ .

Q38GATE 2022MCQ2MNetwork Theory

For the circuit shown, the locus of the impedance $Z(j\omega )$ is plotted as $\omega$ increases from zero to infinity. The values of $R_1$ and $R_2$ are:

🚀 Solve in Practice Mode

📖 Explanation

To determine the resistor values, we analyze the circuit's impedance at the frequency extremes, which correspond to the endpoints of the impedance locus plot.

At DC, or $\omega=0$ , the capacitor acts as an open circuit. The total impedance is therefore the series combination $Z(0) = R_1 + R_2$ . From the plot, the impedance at $\omega=0$ is the rightmost point on the real axis, so $R_1 + R_2 = 5 \text{ k}\Omega$ .

At infinite frequency, $\omega \to \infty$ , the capacitor's impedance $1/(j\omega C)$ approaches zero, making it a short circuit. This short bypasses resistor $R_2$ . The total impedance becomes $Z(\infty ) = R_1$ . The plot shows this is the leftmost point on the real axis, so $R_1 = 2 \text{ k}\Omega$ .

Finally, substituting the value of $R_1$ into our first equation gives $R_2 = 5 \text{ k}\Omega - 2 \text{ k}\Omega = 3 \text{ k}\Omega$ .

Q39GATE 2022MCQ2MNetwork Theory

Consider the circuit shown in the figure with input $V(t)$ in volts. The sinusoidal steady state current $I(t)$ flowing through the circuit is shown graphically (where $t$ is in seconds). The circuit element $Z$ can be ________.

🚀 Solve in Practice Mode

Q40GATE 2022MCQ2MAnalog Circuits

Consider an ideal long channel nMOSFET (enhancement-mode) with gate length $10\mu m$ and width $100\mu m$ . The product of electron mobility ( $\mu _n$ ) and oxide capacitance per unit area ( $C_{OX}$ ) is $\mu _n C_{OX}=1mA/V^2$ . The threshold voltage of the transistor is 1 V. For a gate-to-source voltage $V_{GS}=[2-\sin (2t)]V$ and drain-tosource voltage $V_{DS}=1V$ (substrate connected to the source), the maximum value of the drain-to-source current is ________.

🚀 Solve in Practice Mode

📖 Explanation

To find the maximum drain current, we first need to determine the maximum gate-to-source voltage, $V_{GS}$ . Given $V_{GS} = [2-\sin(2t)]V$ , its peak value occurs when $\sin(2t)=-1$ , which gives $V_{GS,max} = 2 - (-1) = 3V$ .

Next, we identify the MOSFET's region of operation. We compare the drain-to-source voltage, $V_{DS}=1V$ , with the overdrive voltage at peak current, $V_{GS,max} - V_T = 3V - 1V = 2V$ . Since $V_{DS} < (V_{GS,max} - V_T)$ , the transistor is in the triode (linear) region.

The maximum current is calculated using the triode region current formula:
$I_{D,max} = \mu_n C_{ox} \frac{W}{L} \left[ (V_{GS,max} - V_T)V_{DS} - \frac{1}{2}V_{DS}^2 \right]$

Substituting the given values:
$I_{D,max} = (1 \text{ mA/V}^2) \left(\frac{100\mu m}{10\mu m}\right) \left[ (3 - 1)(1) - \frac{1}{2}(1)^2 \right]$ = 10 × (2 - 0.5) = 15 \text{ mA}$.

Q41GATE 2022MCQ2MAnalog Circuits

For the following circuit with an ideal OPAMP, the difference between the maximum and the minimum values of the capacitor voltage ( $V_c$ ) is __________.

🚀 Solve in Practice Mode

📖 Explanation

This circuit is an astable multivibrator where the capacitor voltage, $V_c$ , oscillates. The voltage swings between the upper and lower trigger points ( $V_{UTP}$ and $V_{LTP}$ ) set by the positive feedback network to the non-inverting input ( $V^+$ ).

When the op-amp output is at positive saturation ( $V_0 = +15V$ ), the upper trigger point is $V_{UTP} = V^+ = \frac{R}{R+2R} \times (+15V) = 5V$ .

When the output is at negative saturation ( $V_0 = -12V$ ), the lower trigger point is $V_{LTP} = V^+ = \frac{2R}{R+2R} \times (-12V) = -8V$ .

The difference between the maximum and minimum capacitor voltage is the difference between these two thresholds: $\Delta V_c = V_{UTP} - V_{LTP} = 5V - (-8V) = 13V$ .

Q42GATE 2022MCQ2MAnalog Circuits

A circuit with an ideal OPAMP is shown. The Bode plot for the magnitude (in dB) of the gain transfer function $(A_V(j\omega )=V_{out}(j\omega )/V_{\in}(j\omega ))$ of the circuit is also provided (here, $\omega$ is the angular frequency in rad/s). The values of $R$ and $C$ are

🚀 Solve in Practice Mode

📖 Explanation

This circuit is a non-inverting amplifier configured as a first-order low-pass filter. From the Bode plot, we can extract two key pieces of information.

First, the maximum gain at low frequencies is $12$ dB. We find the linear gain, $A_{max}$ , using $20 \log_{10}(A_{max}) = 12$ , which gives $A_{max} \approx 4$ . For this amplifier topology, the gain is set by the resistors: $A_{max} = 1 + \frac{R}{1 \text{ k}\Omega}$ . Setting this equal to 4 yields $R = 3 \text{ k}\Omega$ .

Second, the corner frequency $\omega_c$ , where the gain starts to roll off, is found from the plot where $\log_{10}(\omega_c) = 3$ . This means $\omega_c = 10^3 = 1000$ rad/s. This frequency is determined by the RC network at the non-inverting input, where $\omega_c = \frac{1}{(1 \text{ k}\Omega) \cdot C}$ . Solving for C gives $C = \frac{1}{1000 \times 1000} = 10^{-6} \text{ F}$ , or $1 \mu\text{F}$ .

Q43GATE 2022MCQ2MDigital Circuits

For the circuit shown, the clock frequency is $f_o$ and the duty cycle is 25%. For the signal at the Q output of the Flip-Flop, _______.

🚀 Solve in Practice Mode

📖 Explanation

The 2-bit counter cycles through four states (00, 01, 10, 11) for every four pulses of the input clock. Since the JK flip-flop's inputs are driven by the counter's outputs, the output Q must also follow a pattern that repeats every four clock cycles. This establishes the output frequency as $f_{out} = f_o/4$ .

The entire circuit acts as a synchronous, 4-state counter. In a balanced frequency divider like this, the output waveform is symmetrical, spending an equal number of clock cycles in the high and low states. Therefore, the output Q will have a 50% duty cycle. The 25% duty cycle of the input clock is not relevant because the flip-flops are edge-triggered.

Q44GATE 2022MCQ2MControl Systems

Consider an even polynomial $p(s)$ given by $p(s)=s^4+5s^2+4+K$ , where $K$ is an unknown real parameter. The complete range of $K$ for which $p(s)$ has all its roots on the imaginary axis is ________.

🚀 Solve in Practice Mode

📖 Explanation

For all roots of an even polynomial to lie on the imaginary axis, the system must be marginally stable. We can find the conditions for this using the Routh-Hurwitz criterion.

First, we form an auxiliary polynomial from the coefficients of $p(s)$ : $A(s) = s^4 + 5s^2 + (4+K)$ . We use its derivative, $A'(s) = 4s^3 + 10s$ , to build the Routh array, as the $s^3$ row would otherwise be all zeros.

The first column of the resulting array is $[1, 4, \frac{5}{2}, c_1, 4+K]$ . For marginal stability, all entries in this column must be non-negative. This requirement imposes two constraints on $K$ .

From the $s^0$ row: $4+K \geq 0 \implies K \geq -4$ .
From the $s^1$ row, the element $c_1 = \frac{(\frac{5}{2})(10) - (4)(4+K)}{5/2}$ must be non-negative. This holds if the numerator is non-negative: $25 - 4(4+K) \geq 0$ , which simplifies to $9 \geq 4K$ , or $K \leq \frac{9}{4}$ .

Combining these two conditions gives the complete range: $-4 \leq K \leq \frac{9}{4}$ .

Q45GATE 2022MSQ2MEngineering Mathematics

Consider the following series: $\sum_{n=1}^{\infty }\frac{n^d}{c^n}$ For which of the following combinations of $c,d$ values does this series converge?

🚀 Solve in Practice Mode

📖 Explanation

To determine when the series $\sum_{n=1}^{\infty }\frac{n^d}{c^n}$ converges, we can analyze the behavior based on the value of $c$ .

When $c=1$ , the series becomes $\sum \frac{1}{n^{-d}}$ , which is a p-series. This type of series converges if the exponent is greater than 1, meaning $-d > 1$ , or $d < -1$ . Following this rule, Option D ( $d=-2$ ) converges, while Option A ( $d=-1$ ) diverges.

When $c \ne 1$ , the Ratio Test is a powerful tool. The limit of the ratio of consecutive terms is $L = \lim_{n \to \infty } \left| \frac{(n+1)^d/c^{n+1}}{n^d/c^n} \right| = \left| \frac{1}{c} \right|$ . The series converges if $L<1$ (i.e., $|c|>1$ ) and diverges if $L>1$ (i.e., $|c|<1$ ). Based on this, Option B ( $c=2$ ) converges, and Option C ( $c=0.5$ ) diverges.

Q46GATE 2022MSQ2MSignals and Systems

The outputs of four systems $(S_1,S_2,S_3,S_4)$ corresponding to the input signal $\sin (t)$ , for all time $t$ , are shown in the figure. Based on the given information, which of the four systems is/are definitely NOT LTI (linear and time-invariant)?

🚀 Solve in Practice Mode

📖 Explanation

A fundamental property of any Linear Time-Invariant (LTI) system is that the output signal must have the same frequency as the input signal when the input is a sinusoid. The system can only change the sinusoid's amplitude and phase.

Let's examine the systems with an input of $\sin(t)$ , which has a frequency of $\omega=1$ .

The output of $S_3$ is $\sin(2t)$ , which has a frequency of $\omega=2$ . Since the frequency has changed, $S_3$ is not LTI.

The output of $S_4$ is $\sin^2(t)$ . Using the identity $\sin^2(t) = \frac{1}{2}(1 - \cos(2t))$ , we see the output contains a DC component (frequency $\omega=0$ ) and a component at frequency $\omega=2$ . Since new frequencies are generated, $S_4$ is also not LTI.

Q47GATE 2022MSQ2MElectronic Devices

Select the CORRECT statement(s) regarding semiconductor devices.

🚀 Solve in Practice Mode

📖 Explanation

Here is a more insightful explanation:

In an intrinsic (pure) semiconductor, charge carriers are created when thermal energy promotes a valence electron to the conduction band, leaving a hole behind. Thus, at equilibrium, the concentration of electrons ( $n$ ) must equal the concentration of holes ( $p$ ). In a Bipolar Junction Transistor (BJT), the collector is designed to be lightly doped compared to the more heavily doped base region, which is the opposite of statement B. Statement C is a fundamental principle of current continuity: in any two-terminal device under steady-state conditions, the total current must be the same at every point along the device. Finally, for silicon above room temperature (~300 K), increased lattice vibrations impede carrier flow, causing mobility to decrease, not increase.

Q48GATE 2022MSQ2MDigital Circuits

A state transition diagram with states A, B, and C, and transition probabilities $p_1,p_2,...,p_7$ is shown in the figure (e.g., $p_1$ denotes the probability of transition from state A to B). For this state diagram, select the statement(s) which is/are universally true

🚀 Solve in Practice Mode

📖 Explanation

In any state transition diagram, the sum of probabilities for all transitions *leaving* a particular state must equal 1. This is because the system has to move somewhere, even if it's back to the same state.

Let's apply this principle to each state:
From state A, the system can go to B ( $p_1$ ), C ( $p_4$ ), or stay at A ( $p_7$ ). Therefore, $p_1 + p_4 + p_7 = 1$ .

From state B, the system can go to A ( $p_3$ ) or stay at B ( $p_2$ ). This means $p_2 + p_3 = 1$ .

From state C, the system can go to A ( $p_6$ ) or stay at C ( $p_5$ ). This gives us $p_5 + p_6 = 1$ .

Since both the sums $(p_2 + p_3)$ and $(p_5 + p_6)$ are equal to 1, they must be equal to each other: $p_2 + p_3 = p_5 + p_6$ .

Q49GATE 2022MSQ2MDigital Circuits

Consider a Boolean gate (D) where the output $Y$ is related to the inputs $A$ and $B$ as, $Y=A+\bar{B}$ , where + denotes logical OR operation. The Boolean inputs '0' and '1' are also available separately. Using instances of only D gates and inputs '0' and '1', __________ (select the correct option(s)).

🚀 Solve in Practice Mode

📖 Explanation

Let the gate's operation be defined as $D(A, B) = A + \bar{B}$ . Our first step is to create a simple NOT gate. By setting the input $A$ to '0', the output becomes $D(0, B) = 0 + \bar{B} = \bar{B}$ . This configuration allows us to invert any signal $B$ .

With a NOT gate at our disposal, we can now form more complex logic. To create a NAND gate, we can use our NOT gate to generate $\bar{A}$ and then feed it into the D gate as $D(\bar{A}, B) = \bar{A} + \bar{B}$ . By De Morgan's theorem, this is equivalent to $\overline{A \cdot B}$ , which is the NAND function.

To implement a NOR gate, we first construct an OR gate by inverting the second input: $D(A, \bar{B}) = A + \overline{(\bar{B})} = A+B$ . We then take this OR output and pass it through our NOT gate configuration, producing $\overline{A+B}$ , which is the NOR function. Since both NAND and NOR logic can be implemented, the corresponding options are correct.

Q50GATE 2022MCQ2MControl Systems

Two linear time-invariant systems with transfer functions $G_1(s)=\frac{10}{s^2+s+1}, G_2(s)=\frac{10}{s^2+s\sqrt{10}+10}$ have unit step responses $y_1(t)$ and $y_2(t)$ , respectively. Which of the following statements is/are true?

🚀 Solve in Practice Mode

📖 Explanation

To analyze the dynamic behavior of these systems, we'll compare their transfer functions to the standard second-order form.

For $G_1(s)=\frac{10}{s^2+s+1}$ , we can see that the natural frequency is $\omega_{n1}=1 \text{ rad/s}$ and the damping ratio is $\xi_1=0.5$ .

For $G_2(s)=\frac{10}{s^2+s\sqrt{10}+10}$ , the natural frequency is $\omega_{n2}=\sqrt{10} \text{ rad/s}$ and the damping ratio is also $\xi_2=0.5$ .

The percentage peak overshoot, $M_p = e^{-\pi\xi/\sqrt{1-\xi^2}}$ , is a function of the damping ratio $\xi$ alone. Since both systems have an identical damping ratio of $0.5$ , their percentage peak overshoots must be the same.

Other response characteristics, such as steady-state value ( $y_{ss}=G(0)$ ), damped frequency ( $\omega_d = \omega_n\sqrt{1-\xi^2}$ ), and settling time ( $T_s \approx 4/(\xi\omega_n)$ ), all depend on the natural frequency $\omega_n$ or DC gain, which differ between the two systems.

Q51GATE 2022MCQ2MCommunication Systems

Consider an FM broadcast that employs the pre-emphasis filter with frequency response $H_{pe}(\omega )=1+\frac{j\omega }{\omega _0}$ where $\omega _0=10^4$ rad/sec. For the network shown in the figure to act as a corresponding de-emphasis filter, the appropriate pair(s) of ( $R,C$ ) values is/are ________.

🚀 Solve in Practice Mode

📖 Explanation

A de-emphasis filter's transfer function, $H_{de}(\omega)$ , must be the reciprocal of the pre-emphasis filter's function, $H_{pe}(\omega)$ . Therefore, $H_{de}(\omega) = 1/H_{pe}(\omega)$ . Given the pre-emphasis response $H_{pe}(\omega) = 1 + j\frac{\omega}{\omega_0}$ , the required de-emphasis response is $H_{de}(\omega) = \frac{1}{1 + j\omega/\omega_0}$ .

The circuit shown is a simple RC low-pass filter. Its transfer function is found by voltage division: $H_{circuit}(\omega) = \frac{1/(j\omega C)}{R + 1/(j\omega C)} = \frac{1}{1+j\omega RC}$ .

For the circuit to act as the de-emphasis filter, we must set $H_{circuit}(\omega) = H_{de}(\omega)$ . Comparing the two expressions, we can see that we need the time constant $RC$ to be equal to $1/\omega_0$ .

With the given cutoff frequency $\omega_0 = 10^4$ rad/sec, the required product is $RC = \frac{1}{10^4} = 10^{-4}$ s. We now check the options to find the pair that satisfies this condition. Only the values $R=1k\Omega$ and $C=0.1\mu F$ yield the correct product: $(1 \times 10^3 \, \Omega)(0.1 \times 10^{-6} \, F) = 10^{-4}$ s.

Q52GATE 2022MSQ2MElectromagnetics

A waveguide consists of two infinite parallel plates (perfect conductors) at a separation of $10^{-4}$ cm, with air as the dielectric. Assume the speed of light in air to be $3 \times 10^{8}$ m/s. The frequency/frequencies of TM waves which can propagate in this waveguide is/are _______.

🚀 Solve in Practice Mode

📖 Explanation

A waveguide only allows signals with frequencies above a certain threshold, the cut-off frequency ( $f_c$ ), to propagate. For the fundamental TM mode ( $m=1$ ) in a parallel-plate waveguide, this is given by $f_c = \frac{c}{2a}$ .

First, we must express the plate separation, $a$ , in meters: $a = 10^{-4} \text{ cm} = 10^{-6} \text{ m}$ . With the speed of light $c = 3 \times 10^8$ m/s, we can now find the cut-off frequency:
$f_c = \frac{3 \times 10^8 \text{ m/s}}{2 \times 10^{-6} \text{ m}} = 1.5 \times 10^{14} \text{ Hz}$
Only waves with a frequency $f$ greater than this $f_c$ can travel through the waveguide. Of the given options, both $6 \times 10^{15}$ Hz and $8 \times 10^{14}$ Hz are greater than $1.5 \times 10^{14}$ Hz.

Q53GATE 2022NAT2MEngineering Mathematics

The value of the integral $\int \int _D 3(x^2+y^2)dxdy$ , where $D$ is the shaded triangular region shown in the diagram, is _____ (rounded off to the nearest integer).

🚀 Solve in Practice Mode

📖 Explanation

To solve this, we'll set up a double integral over the given triangular region, $D$ . We can describe this region by letting $x$ range from $0$ to $4$ . For any given $x$ within this range, the corresponding $y$ values are bounded by the lines $y=-x$ and $y=x$ . This leads to the iterated integral:
$I = \int_{0}^{4} \int_{-x}^{x} 3(x^2+y^2) \,dy\,dx$
First, we compute the inner integral with respect to $y$ , treating $x$ as a constant:
$\int_{-x}^{x} (3x^2+3y^2) \,dy = $[3x^2y + y^3]_{y=-x}^{y=x}$
Evaluating this at the limits gives $(3x^3+x^3) - (-3x^3-x^3) = 8x^3$ .
Now, we substitute this result into the outer integral and solve for $x$ :
$I = \int_{0}^{4} 8x^3 \,dx = $[2x^4]_{0}^{4} = 2(4^4) - 2(0^4) = 2(256) = 512$

Q54GATE 2022NAT2MNetwork Theory

A linear 2-port network is shown in Fig. (a). An ideal DC voltage source of 10 V is connected across Port 1. A variable resistance R is connected across Port 2. As R is varied, the measured voltage and current at Port 2 is shown in Fig. (b) as a $V_2$ versus $-I_2$ plot. Note that for $V_2=5V,I_2=0mA$ , and for $V_2=4V,I_2=-4mA$ . When the variable resistance R at Port 2 is replaced by the load shown in Fig. (c), the current $I_2$ is _______ mA (rounded off to one decimal place).

🚀 Solve in Practice Mode

📖 Explanation

To determine the current $I_2$ with the new load, we first simplify the problem by finding the Thevenin equivalent of the network at Port 2. The provided $V_2$ versus $-I_2$ plot gives us the necessary information.

The Thevenin voltage ( $V_{th}$ ) is the open-circuit voltage, which occurs when $I_2=0$ . From the graph, this is $V_{th} = 5V$ . The Thevenin resistance ( $R_{th}$ ) is the negative of the slope of the $V_2$ versus $-I_2$ plot. Using the given points, the slope is $\frac{\Delta V_2}{\Delta(-I_2)} = \frac{4V - 5V}{4mA - 0mA} = -0.25 k\Omega$ . Thus, $R_{th} = 0.25 k\Omega = 250\Omega$ .

Now, we replace the network with its Thevenin equivalent: a $5V$ source in series with a $250\Omega$ resistor. This is connected to the load from Fig. (c), which is a $1k\Omega$ resistor in series with a $10V$ source. The complete circuit becomes a single loop.

In this loop, the $5V$ Thevenin source and the $10V$ load source oppose each other. The total resistance is the sum of the two series resistors. Applying KVL, the current $I_2$ is the net voltage divided by the total resistance:
$I_2 = \frac{10V - 5V}{1k\Omega + 250\Omega} = \frac{5V}{1250\Omega} = 0.004A$
Therefore, the current $I_2$ is $4.0$ mA.

Q55GATE 2022NAT2MSignals and Systems

For a vector $\bar{x}=\left[x[0],x[1],...,x[7] \right]$ , the 8-point discrete Fourier transform (DFT) is denoted by $\bar{X}=DFT(\bar{x})=\left[X[0],X[1],...,X[7] \right]$ , where $X[k]=\sum_{n=0}^{7}x[n]exp\left ( -j\frac{2 \pi}{8}nk \right )$ Here, $j=\sqrt{-1}$ . If $\bar{x}==[1,0,0,0,2,0,0,0]$ and $\bar{y}=DFT(DFT(\bar{x}))$ , then the value of $y[0]$ is _________ (rounded off to one decimal place).

🚀 Solve in Practice Mode

📖 Explanation

To solve this problem efficiently, we use a fundamental property of the discrete Fourier transform. Applying an $N$ -point DFT twice to a sequence $x[n]$ is equivalent to scaling the time-reversed version of the original sequence. This property is expressed mathematically as:
$\bar{y} = \text{DFT}(\text{DFT}(\bar{x})) \implies y[n] = N \cdot x[(-n) \pm od{N}]$ .

For this 8-point DFT problem, the length of the sequence is $N=8$ . We are asked to find the specific value of $y[0]$ . We can find this by substituting $n=0$ into the property equation:
$y[0] = 8 \cdot x[(-0) \pm od{8}] = 8 \cdot x[0]$ .

Looking at the input signal $\bar{x}=[1,0,0,0,2,0,0,0]$ , the first element is $x[0]=1$ .
Therefore, the value of $y[0]$ is $8 \cdot 1 = 8.0$ .

Q56GATE 2022NAT2MElectronic Devices

A p-type semiconductor with zero electric field is under illumination (low level injection) in steady state condition. Excess minority carrier density is zero at $x=\pm 2l_n$ , where $l_n=10^{-4} cm$ is the diffusion length of electrons. Assume electronic charge, $q=-1.6 \times 10^{-19}C$ . The profiles of photo-generation rate of carriers and the recombination rate of excess minority carriers (R) are shown. Under these conditions, the magnitude of the current density due to the photo-generated electrons at $x=+2l_n$ is _________ $mA/cm^2$ (rounded off to two decimal places).

🚀 Solve in Practice Mode

📖 Explanation

With no electric field present, the electron current is purely a diffusion current, described by the equation $J_n = qD_n\frac{d\delta_n}{dx}$ . Our goal is to find this current at $x=+2l_n$ .

Let's analyze the region $l_n \le x \le 2l_n$ . The provided graphs show that in this specific range, both the photo-generation rate ( $G$ ) and the recombination rate ( $R$ ) are zero. Therefore, the steady-state continuity equation, $D_n\frac{d^2\delta_n}{dx^2} + G - R = 0$ , simplifies to $\frac{d^2\delta_n}{dx^2} = 0$ . This implies that the excess electron concentration, $\delta_n(x)$ , has a linear profile in this region.

To find the slope of this line, we use the boundary conditions. The problem states $\delta_n(2l_n) = 0$ . At the other boundary, $x=l_n$ , the concentration is determined by the recombination profile: $\delta_n(l_n) = R(l_n)\tau_n = (10^{20}e^{-l_n/l_n})\tau_n = 10^{20}e^{-1}\tau_n$ . The constant gradient is then $\frac{d\delta_n}{dx} = \frac{\delta_n(2l_n) - \delta_n(l_n)}{2l_n - l_n} = \frac{0 - 10^{20}e^{-1}\tau_n}{l_n} = -\frac{10^{20}e^{-1}\tau_n}{l_n}$ .

Now, we calculate the magnitude of the current density using this gradient:
$|J_n| = \left|qD_n\frac{d\delta_n}{dx}\right| = qD_n \left(\frac{10^{20}e^{-1}\tau_n}{l_n}\right)$

Using the relationship for diffusion length, $l_n^2 = D_n\tau_n$ , we can simplify the expression:
$|J_n| = q \left(\frac{l_n^2}{\tau_n}\right) \frac{10^{20}e^{-1}\tau_n}{l_n} = q l_n 10^{20} e^{-1}$

Finally, substituting the given values:
$|J_n| = (1.6 \times 10^{-19} C) \times (10^{-4} cm) \times (10^{20} cm^{-3}s^{-1}) \times e^{-1} \approx 0.589 \times 10^{-3} A/cm^2 \approx 0.59$ mA/cm $^2$ .

Q57GATE 2022NAT2MAnalog Circuits

A circuit and the characteristics of the diode (D) in it are shown. The ratio of the minimum to the maximum small signal voltage gain $\frac{\partial V_{out}}{\partial V_{\in}}$ is ________ (rounded off to two decimal places)

🚀 Solve in Practice Mode

📖 Explanation

The small-signal voltage gain depends on whether the diode is conducting (ON) or not (OFF). We analyze the circuit's equivalent resistance in these two states.

When the diode is ON, its small-signal resistance is zero, creating a short circuit across the middle $2 \text{ k}\Omega$ resistor. The gain is then from a simple voltage divider: $A_{v,min} = \frac{2}{2+2} = \frac{1}{2}$ .

When the diode is OFF, its resistance is infinite, creating an open circuit. The output is across the series combination of the middle and lower resistors. The gain becomes $A_{v,max} = \frac{2+2}{2+2+2} = \frac{4}{6} = \frac{2}{3}$ .

Finally, the ratio of the minimum to the maximum gain is $\frac{A_{v,min}}{A_{v,max}} = \frac{1/2}{2/3} = \frac{3}{4} = 0.75$ .

Q58GATE 2022NAT2MDigital Circuits

Consider the circuit shown with an ideal OPAMP. The output voltage $V_o$ is __________V (rounded off to two decimal places).

🚀 Solve in Practice Mode

📖 Explanation

To determine the output voltage $V_o$ , we can model the resistive ladder network as a Thevenin equivalent source connected to the inverting amplifier.

First, let's find the Thevenin equivalent of the ladder network at its output terminals. The Thevenin resistance, $R_{th}$ , is found by looking into the output with all voltage sources grounded. The impedance of the ladder converges such that $R_{th}$ is approximately $R$ . The Thevenin voltage, $V_{th}$ , is the open-circuit voltage at the ladder's output, which can be found using nodal analysis or superposition to be $V_{th} = 0.5$ V.

This Thevenin source, with $V_{th} = 0.5$ V and series resistance $R_{th} = R$ , drives the inverting amplifier. The total resistance in the input path to the virtual ground is the sum of the ladder's Thevenin resistance and the amplifier's input resistor, $R_{\in,total} = R_{th} + 2R = R + 2R = 3R$ .

The input current to the amplifier is $I_{\in} = \frac{V_{th}}{R_{\in,total}} = \frac{0.5 \text{ V}}{3R}$ . The output voltage is then this current multiplied by the negative feedback resistance:
$V_o = -I_{\in} \times R_f = -\left(\frac{0.5 \text{ V}}{3R}\right) \times 3R = -0.5$ V.

Q59GATE 2022NAT2MAnalog Circuits

Consider the circuit shown with an ideal long channel nMOSFET (enhancementmode, substrate is connected to the source). The transistor is appropriately biased in the saturation region with $V_{GG}$ and $V_{DD}$ such that it acts as a linear amplifier. $v_i$ is the small-signal ac input voltage. $v_A$ and $v_B$ represent the small-signal voltages at the nodes A and B, respectively. The value of $\frac{v_A}{v_B}$ is ________ (rounded off to one decimal place).

🚀 Solve in Practice Mode

📖 Explanation

To find the ratio of the small-signal voltages, we analyze the AC behavior of the circuit. A single small-signal drain current, which we'll call $i_d$ , flows from the drain (node A), through the transistor, to the source (node B).

This current generates the voltage at node B as it flows through the 2 kΩ resistor, so $v_B = i_d \cdot (2 \text{ kΩ})$ .

The same current $i_d$ flows through the 4 kΩ drain resistor. Since this current flows from the AC ground (the $V_{DD}$ line) into node A, the voltage is given by $v_A = -i_d \cdot (4 \text{ kΩ})$ .

By taking the ratio of these two voltages, the current $i_d$ cancels out:
$\frac{v_A}{v_B} = \frac{-i_d \cdot 4 \text{ kΩ}}{i_d \cdot 2 \text{ kΩ}} = -\frac{4}{2} = -2$

Q60GATE 2022NAT2MControl Systems

The block diagram of a closed-loop control system is shown in the figure. R(s),Y(s) and D(s) are the Laplace transforms of the time-domain signals $r(t),y(t),$ and $d(t)$ , respectively. Let the error signal be defined as $e(t)=r(t)-y(t)$ . Assuming the reference input $r(t)=0$ for all $t$ , the steady-state error $e(\infty )$ , due to a unit step disturbance $d(t)$ , is _________ (rounded off to two decimal places).

🚀 Solve in Practice Mode

📖 Explanation

With the reference input $r(t)=0$ , the error signal is simply the negative of the output, $e(t) = -y(t)$ , which means $E(s)=-Y(s)$ . We need to find the system's response to the disturbance. The transfer function from the disturbance $D(s)$ to the output $Y(s)$ is found from the diagram to be $\frac{Y(s)}{D(s)} = \frac{1}{s^2+10s+10}$ .

Combining these, the error in response to the disturbance is $E(s) = \frac{-1}{s^2+10s+10}D(s)$ . The problem specifies a unit step disturbance, so $D(s) = 1/s$ . To find the steady-state error $e(\infty )$ , we apply the Final Value Theorem:
$e(\infty ) = \lim_{s \to 0} sE(s) = \lim_{s \to 0} s \left( \frac{-1}{s^2+10s+10} \cdot \frac{1}{s} \right)$
$e(\infty ) = \lim_{s \to 0} \frac{-1}{s^2+10s+10} = \frac{-1}{0+0+10} = -0.10$ .

Q61GATE 2022NAT2MCommunication Systems

The transition diagram of a discrete memoryless channel with three input symbols and three output symbols is shown in the figure. The transition probabilities are as marked. The parameter $\alpha$ lies in the interval [0.25, 1]. The value of $\alpha$ for which the capacity of this channel is maximized, is ________ (rounded off to two decimal places).

🚀 Solve in Practice Mode

📖 Explanation

The capacity of this channel is given by $C = \max_{P(X)}[H(Y) - H(Y|X)]$ . Because the channel is symmetric, its capacity is achieved with a uniform input distribution, which makes the output entropy $H(Y)$ maximum at $\log_2(3)$ . The conditional entropy $H(Y|X)$ is calculated from any row of the transition matrix, giving $H(Y|X) = -[(1-\alpha)\log_2(1-\alpha) + \alpha\log_2\alpha]$ .

Therefore, the capacity as a function of $\alpha$ is:
$C(\alpha) = \log_2(3) - \left(-[(1-\alpha)\log_2(1-\alpha) + \alpha\log_2\alpha]\right) = \log_2(3) + (1-\alpha)\log_2(1-\alpha) + \alpha\log_2(\alpha)$ .

To maximize $C(\alpha)$ , we need to maximize the term $f(\alpha) = (1-\alpha)\log_2(1-\alpha) + \alpha\log_2(\alpha)$ . This function is the negative of the binary entropy function, $-H_b(\alpha)$ , which has a minimum at $\alpha=0.5$ and increases towards the boundaries of its domain, $[0, 1]$ . For the given interval $\alpha \in [0.25, 1]$ , the maximum value of $f(\alpha)$ occurs at the endpoint furthest from the minimum, which is $\alpha=1$ .

Q62GATE 2022NAT2MCommunication Systems

Consider communication over a memoryless binary symmetric channel using a (7, 4) Hamming code. Each transmitted bit is received correctly with probability $(1-\epsilon )$ , and flipped with probability $\epsilon$ . For each codeword transmission, the receiver performs minimum Hamming distance decoding, and correctly decodes the message bits if and only if the channel introduces at most one bit error. For $\epsilon =0.1$ , the probability that a transmitted codeword is decoded correctly is _________ (rounded off to two decimal places).

🚀 Solve in Practice Mode

📖 Explanation

A codeword is decoded correctly if it arrives with at most one error. For a 7-bit codeword, this means it can have either zero errors or exactly one error. We can calculate the total probability by summing the probabilities of these two mutually exclusive outcomes. Since each bit flip is an independent event with probability $\epsilon=0.1$ , this scenario follows a binomial distribution.

The probability of zero errors is $\binom{7}{0}(0.1)^0(0.9)^7 = (0.9)^7$ .
The probability of exactly one error is $\binom{7}{1}(0.1)^1(0.9)^6 = 7(0.1)(0.9)^6$ .

The total probability for correct decoding is the sum of these two cases:
$P(\text{correct}) = (0.9)^7 + 7(0.1)(0.9)^6 \approx 0.4783 + 0.3720 = 0.8503$ .

Rounding to two decimal places, the probability is 0.85.

Q63GATE 2022NAT2MCommunication Systems

Consider a channel over which either symbol $x_A$ or symbol $x_B$ is transmitted. Let the output of the channel Y be the input to a maximum likelihood (ML) detector at the receiver. The conditional probability density functions for Y given $x_A$ and $x_B$ are: $f_{Y|x_A}(y)=e^{-(y+1)}u(y+1),$ $f_{Y|x_B}(y)=e^{(y-1)}(1-u(y-1)),$ where $u(\cdot )$ is the standard unit step function. The probability of symbol error for this system is _________ (rounded off to two decimal places).

🚀 Solve in Practice Mode

📖 Explanation

A Maximum Likelihood (ML) detector chooses the symbol whose conditional PDF is larger for the received value, $y$ . We first find the decision threshold by setting the two PDFs equal, $f_{Y|x_A}(y) = f_{Y|x_B}(y)$ , which simplifies from $e^{-(y+1)} = e^{y-1}$ to the threshold $y=0$ . Considering this threshold and the ranges where each PDF is non-zero, the receiver decides $x_A$ if $y \in (-1, 0) \cup (1, \infty )$ and decides $x_B$ if $y \in (-\infty , -1) \cup (0, 1)$ .

The probability of error is found by integrating the PDF of the transmitted symbol over the incorrect decision region. If $x_A$ is sent, an error occurs if we decide $x_B$ (i.e., $y \in (0,1)$ for the relevant range):
$P(e|x_A) = \int_{0}^{1} f_{Y|x_A}(y) dy = \int_{0}^{1} e^{-(y+1)} dy = e^{-1} - e^{-2}$ .

By symmetry, if $x_B$ is sent, an error occurs if we decide $x_A$ (i.e., $y \in (-1,0)$ ):
$P(e|x_B) = \int_{-1}^{0} f_{Y|x_B}(y) dy = \int_{-1}^{0} e^{y-1} dy = e^{-1} - e^{-2}$ .

Since both conditional error probabilities are equal, the total system probability of error is simply this value, $P_e = e^{-1} - e^{-2} \approx 0.2325$ , which rounds to 0.23.

Q64GATE 2022NAT2MCommunication Systems

Consider a real valued source whose samples are independent and identically distributed random variables with the probability density function, $f(x)$ , as shown in the figure. Consider a 1 bit quantizer that maps positive samples to value $\alpha$ and others to value $\beta$ . If $\alpha ^*$ and $\beta ^*$ are the respective choices for $\alpha$ and $\beta$ that minimize the mean square quantization error, then $(\alpha ^*- \beta ^*)=$ _________ (rounded off to two decimal places).

🚀 Solve in Practice Mode

📖 Explanation

First, we must define the probability density function, $f(x)$ . The total area under the PDF must be 1. The shape consists of a triangle and a rectangle, and by solving for the height $h$ at $x=0$ with the equation $\frac{1}{2}(2)(h) + (1)(h) = 1$ , we find $h=0.5$ . This gives us the piecewise function: $f(x) = 0.5$ for $0 \le x \le 1$ and $f(x) = \frac{1}{4}x + \frac{1}{2}$ for $-2 \le x \le 0$ .

To minimize the mean square quantization error, the optimal choice for each quantization level is the centroid (conditional mean) of the PDF over its corresponding region.

The optimal value $\alpha^*$ is the centroid for the positive region $[0, 1]$ :
$\alpha^* = E[X | 0 \le X \le 1] = \frac{\int_{0}^{1} x f(x) dx}{\int_{0}^{1} f(x) dx} = \frac{\int_{0}^{1} x(0.5) dx}{0.5} = \int_{0}^{1} x dx = 0.5$ .

The optimal value $\beta^*$ is the centroid for the non-positive region $[-2, 0]$ :
$\beta^* = E[X | -2 \le X \le 0] = \frac{\int_{-2}^{0} x (\frac{1}{4}x+\frac{1}{2}) dx}{\int_{-2}^{0} (\frac{1}{4}x+\frac{1}{2}) dx} = \frac{-1/3}{1/2} = -\frac{2}{3}$ .

Finally, we compute the required difference:
$\alpha^* - \beta^* = 0.5 - (-\frac{2}{3}) = \frac{1}{2} + \frac{2}{3} = \frac{7}{6} \approx 1.17$ .

Q65GATE 2022NAT2MElectromagnetics

In an electrostatic field, the electric displacement density vector, $\vec{D}$ , is given by $\vec{D}(x,y,z)=(x^3\vec{i}+y^3\vec{j}+xy^2\vec{k})C/m^2$ , where $\vec{i},\vec{j},\vec{k}$ are the unit vectors along x-axis, y-axis, and z-axis, respectively. Consider a cubical region R centered at the origin with each side of length 1 m, and vertices at ( $\pm 0.5 m, \pm 0.5 m, \pm 0.5 m$ ). The electric charge enclosed within R is _________ C (rounded off to two decimal places).

🚀 Solve in Practice Mode

📖 Explanation

According to Gauss's Law, the total electric charge enclosed, $Q_{enc}$ , is the volume integral of the divergence of the electric displacement vector $\vec{D}$ . First, we must calculate the divergence:
$\nabla \cdot \vec{D} = \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial y}(y^3) + \frac{\partial}{\partial z}(xy^2) = 3x^2 + 3y^2$ .

Next, we integrate this expression over the cubical volume R, where the limits for $x, y,$ and $z$ all range from -0.5 to 0.5:
$Q_{enc} = \iiint_R (3x^2 + 3y^2) \,dV = 3\iiint_R x^2 \,dV + 3\iiint_R y^2 \,dV$ .

Let's evaluate the first integral by separating the variables:
$3 \left(\int_{-0.5}^{0.5} x^2 \,dx\right) \left(\int_{-0.5}^{0.5} \,dy\right) \left(\int_{-0.5}^{0.5} \,dz\right) = 3 \left[\frac{x^3}{3}\right]_{-0.5}^{0.5} \times (1) \times (1) = 0.25$ C.

Due to the symmetry of the cubical region, the second integral involving $y^2$ yields the same result, 0.25 C. The total enclosed charge is the sum of these two values:
$Q_{enc} = 0.25 + 0.25 = 0.5$ C.

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