GATE ECE 2021

Q11GATE 2021MCQ1MEngineering Mathematics

The vector function $F\left ( r \right )=-x\hat{i}+y\hat{j}$ is defined over a circular arc C shown in the figure. The line integral of $\int _{C} F\left ( r \right ).dr$ is

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📖 Explanation

To solve this line integral, we first express it in terms of its components. Given $\vec{F} = -x\hat{i} + y\hat{j}$ and $d\vec{r} = dx\hat{i} + dy\hat{j}$ , the integral becomes $\int_C -x\,dx + y\,dy$ .

The path $C$ is a circular arc of radius 1, which is best described using polar coordinates. We can set $x = \cos\theta$ and $y = \sin\theta$ . The angle $\theta$ ranges from $0$ to $45^\circ$ , or $0$ to $\pi/4$ radians. The differentials are then $dx = -\sin\theta\,d\theta$ and $dy = \cos\theta\,d\theta$ .

Substituting these into our integral, we get:
$\int_{0}^{\pi/4} [-\cos\theta(-\sin\theta\,d\theta) + \sin\theta(\cos\theta\,d\theta)] = \int_{0}^{\pi/4} 2\sin\theta\cos\theta\,d\theta$
Using the double-angle identity, $2\sin\theta\cos\theta = \sin(2\theta)$ , the integral simplifies to:
$\int_{0}^{\pi/4} \sin(2\theta)\,d\theta = \left[-\frac{\cos(2\theta)}{2}\right]_{0}^{\pi/4}$
Finally, evaluating this definite integral gives $-\frac{1}{2}[\cos(\pi/2) - \cos(0)] = -\frac{1}{2}[0 - 1] = \frac{1}{2}$ .

Q12GATE 2021MCQ1MEngineering Mathematics

Consider the differential equation given below. $\frac{dy}{dx}+\frac{x}{1-x^{2}}y=x\sqrt{y}$ The integrating factor of the differential equation is

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📖 Explanation

This is a Bernoulli differential equation. To solve it, we first transform it into a linear equation. Start by dividing the entire equation by $\sqrt{y}$ to isolate the terms.
$\frac{1}{\sqrt{y}}\frac{dy}{dx} + \frac{x}{1-x^2}\sqrt{y} = x$
Now, let's use the substitution $u = \sqrt{y}$ . The derivative with respect to $x$ is $\frac{du}{dx} = \frac{1}{2\sqrt{y}}\frac{dy}{dx}$ . We can see that $\frac{1}{\sqrt{y}}\frac{dy}{dx} = 2\frac{du}{dx}$ . Substituting these into our equation yields:
$2\frac{du}{dx} + \frac{x}{1-x^2}u = x$
To find the integrating factor, we need the standard linear form $\frac{du}{dx} + P(x)u = Q(x)$ . Dividing by 2 gives:
$\frac{du}{dx} + \frac{x}{2(1-x^2)}u = \frac{x}{2}$
The integrating factor is $e^{\int P(x)dx}$ , with $P(x) = \frac{x}{2(1-x^2)}$ .
$I.F. = e^{\int \frac{x}{2(1-x^2)}dx} = e^{-\frac{1}{4}\ln(1-x^2)} = e^{\ln((1-x^2)^{-1/4})} = (1-x^2)^{-1/4}$

Q13GATE 2021MCQ1MCommunication Systems

Two continuous random variables X and Y are related as $Y=2X+3$ Let $\sigma ^{2}_{X}$ and $\sigma ^{2}_{Y}$ denote the variances of X and Y, respectively. The variances are related as

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📖 Explanation

We are asked to find the relationship between the variance of $Y$ and the variance of $X$ , given $Y=2X+3$ .

The variance of $Y$ is defined as $\sigma_Y^2 = E[(Y - E[Y])^2]$ . First, let's find the mean of $Y$ using the properties of expectation: $E[Y] = E[2X+3] = 2E[X] + 3$ .

Now, substitute the expressions for $Y$ and $E[Y]$ back into the variance definition:
$\sigma_Y^2 = E[((2X+3) - (2E[X]+3))^2]$

Simplifying the term inside the brackets, the $+3$ and $-3$ cancel out, leaving:
$\sigma_Y^2 = E[(2X - 2E[X])^2]$

Factor out the 2 and apply the square:
$\sigma_Y^2 = E[4(X - E[X])^2] = 4 E[(X - E[X])^2]$

Since $E[(X - E[X])^2]$ is the definition of the variance of $X$ , $\sigma_X^2$ , we conclude that:
$\sigma_Y^2 = 4\sigma_X^2$

Q14GATE 2021MCQ1MSignals and Systems

Consider a real-valued base-band signal x(t), band limited to $\text{10 kHz}$ . The Nyquist rate for the signal $y\left ( t \right )=x\left ( t \right )x\left ( 1+\dfrac{t}{2} \right )$ is

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📖 Explanation

To find the Nyquist rate, we first need to determine the maximum frequency ( $f_{max}$ ) of the signal $y(t)$ . The signal is a product, $y(t) = x(t)x(1+t/2)$ . Multiplying signals in the time domain means their spectra are convolved, and the maximum frequency of the resulting signal is the sum of the individual maximum frequencies.

Let's find the maximum frequency of each component. For the baseband signal $x(t)$ band-limited to $10 \text{ kHz}$ , its spectrum spans from $-5 \text{ kHz}$ to $5 \text{ kHz}$ , making its maximum frequency $f_1 = 5 \text{ kHz}$ .

The second component, $x(1+t/2)$ , is time-scaled. The scaling factor of $1/2$ on $t$ expands the signal's spectrum by a factor of 2. Thus, its maximum frequency is $f_2 = 2 \times 5 \text{ kHz} = 10 \text{ kHz}$ .

The maximum frequency of $y(t)$ is the sum: $f_{max} = f_1 + f_2 = 5 \text{ kHz} + 10 \text{ kHz} = 15 \text{ kHz}$ .
The Nyquist rate is twice this maximum frequency, which is $2 \times 15 \text{ kHz} = 30 \text{ kHz}$ .

Q15GATE 2021MCQ1MSignals and Systems

Consider two 16-point sequences x[n] and h[n]. Let the linear convolution of x[n] and h[n] be denoted by y[n], while z[n] denotes the 16-point inverse discrete Fourier transform (IDFT) of the product of the 16-point DFTs of x[n] and h[n]. The value(s) of k for which z[k]=y[k] is/are

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📖 Explanation

The sequence $z[n]$ is the result of multiplying the DFTs of $x[n]$ and $h[n]$ and then taking the inverse DFT. By the convolution theorem, this process is equivalent to the 16-point circular convolution of $x[n]$ and $h[n]$ . In contrast, $y[n]$ is the standard linear convolution.

Circular convolution can be understood as a time-aliased or "wrapped-around" version of the linear convolution. The linear convolution of two 16-point sequences, $y[n]$ , results in a sequence of length $16+16-1=31$ . When this 31-point result is mapped to a 16-point circular convolution, aliasing occurs for most points. However, the final point of the circular result, at index $k=N-1$ , is uniquely determined by the corresponding linear convolution point because there are no later points to wrap around and add to it. For $N=16$ , this means $z[k]$ and $y[k]$ are guaranteed to be equal only at $k=15$ .

Q16GATE 2021MCQ1MElectronic Devices

A bar of silicon is doped with boron concentration of $10^{16} \text{cm}^{-3}$ and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of $10^{20} \text{cm}^{-3} s^{-1}$ . If the recombination lifetime is $100 \;\mu s$ , intrinsic carrier concentration of silicon is $10^{10} \text{cm}^{-3}$ and assuming $100\%$ ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is

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📖 Explanation

First, let's establish the initial state of the p-type silicon. The equilibrium hole concentration is approximately the acceptor doping level, $p_o \approx N_A = 10^{16} \text{cm}^{-3}$ . The corresponding equilibrium electron concentration is found using the mass-action law, $n_o = n_i^2 / p_o = (10^{10})^2 / 10^{16} = 10^4 \text{cm}^{-3}$ .

When light generates electron-hole pairs, it creates an excess concentration of carriers. In steady state, this excess is given by $\delta n = \delta p = g_{op} \times \tau = 10^{20} \times (100 \times 10^{-6}) = 10^{16} \text{cm}^{-3}$ .

The total concentrations under illumination are the sum of the equilibrium and excess carriers. The new hole concentration is $p = p_o + \delta p = 10^{16} + 10^{16} = 2 \times 10^{16} \text{cm}^{-3}$ . The new electron concentration is $n = n_o + \delta n = 10^4 + 10^{16}$ , which is approximately $10^{16} \text{cm}^{-3}$ because the excess concentration is much larger than the initial concentration.

Finally, the product of the steady-state concentrations is $np \approx (10^{16}) \times (2 \times 10^{16}) = 2 \times 10^{32} \text{cm}^{-6}$ .

Q17GATE 2021MCQ1MElectronic Devices

The energy band diagram of a p-type semiconductor bar of length L under equilibrium condition (i.e.. the Fermi energy level $E_{F}$ is constant) is shown in the figure. The valance band $E_{V}$ is sloped since doping is non-uniform along the bar. The difference between the energy levels of the valence band at the two edges of the bar is $\Delta$ . If the charge of an electron is q, then the magnitude of the electric field developed inside this semiconductor bar is

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📖 Explanation

The presence of a non-uniform doping profile creates an internal, or built-in, electric field ( $\mathcal{E}$ ) within the semiconductor. This field is what causes the energy bands to slope. The relationship between the electric field and the gradient of an energy band is given by $\mathcal{E} = \frac{1}{q} \frac{dE}{dx}$ .

From the diagram, the valence band energy $E_V$ changes linearly with position. The slope of this band, $\frac{dE_V}{dx}$ , can be calculated as the total change in energy ( $\Delta$ ) divided by the length of the bar ( $L$ ). This gives a constant slope of $\frac{\Delta}{L}$ .

Substituting this slope into the electric field equation yields the magnitude of the field:
$\mathcal{E} = \frac{1}{q} \left( \frac{\Delta}{L} \right) = \frac{\Delta}{qL}$ .

Q18GATE 2021MCQ1MElectronic Devices

In the circuit shown in the figure, the transistors $M_{1}$ and $M_{2}$ are operating in saturation. The channel length modulation coefficients of both the transistors are non-zero. The transconductance of the $\text{MOSFETs} M_{1}$ and $M_{2}$ are $g_{m1}$ and $g_{m2}$ , respectively, and the internal resistance of the $\text{MOSFETs} M_{1}$ and $M_{2}$ are $r_{01}$ and $r_{02}$ , respectively. Ignoring the body effect, the ac small signal voltage gain $\left ( \partial V_{out}/\partial V_{\in} \right )$ of the circuit is

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📖 Explanation

This circuit is configured as a common-source amplifier where transistor $M_2$ provides the amplification and $M_1$ acts as an active load. The small-signal voltage gain ( $A_v$ ) for this topology is the product of the amplifier's transconductance and the total output resistance ( $R_{out}$ ), given by $A_v = -g_{m2} R_{out}$ .

The output resistance $R_{out}$ is determined by looking into the output node. It is the parallel combination of the resistance looking "up" into the drain of $M_2$ and the resistance looking "down" into the load $M_1$ .

The resistance looking into the drain of $M_2$ is simply its output resistance, $r_{o2}$ . The load $M_1$ is configured as a diode-connected transistor, which has a small-signal resistance of $1/g_{m1}$ in parallel with its own output resistance, $r_{o1}$ .

Therefore, the total output resistance is the parallel combination of all three components: $R_{out} = r_{o2} \parallel r_{o1} \parallel \frac{1}{g_{m1}}$ .

Substituting this back into the gain formula yields the overall voltage gain: $A_v = -g_{m2} \left( \frac{1}{g_{m1}} \parallel r_{o1} \parallel r_{o2} \right)$ .

Q19GATE 2021MCQ1MAnalog Circuits

For the circuit with an ideal OPAMP shown in the figure. $V_{\text{REF}}$ is fixed. If $V_{\text{OUT}}=1$ volt for $V_{\text{IN}}-0.1$ volt and $V_{\text{OUT}}=6$ volt for $V_{\text{IN}}=1$ volt, where $V_{\text{OUT}}$ is measured across $R_{L}$ connected at the output of this OPAMP, the value of $R_{F}/R_{\text{IN}}$ is

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📖 Explanation

For an ideal op-amp, the voltages at the inverting ( $V^-$ ) and non-inverting ( $V^+$ ) inputs are equal due to the virtual short concept. The voltage $V^+$ is determined by a voltage divider connected to the fixed $V_{REF}$ , making $V^+$ a constant. The voltage at the inverting terminal, $V^-$ , can be expressed using superposition as $V^- = \frac{V_{IN}R_F + V_{OUT}R_{IN}}{R_{IN} + R_F}$ .

Since $V^+$ is constant, the expression for $V^-$ must yield the same value for both sets of given conditions. Therefore, we can equate them:
$\frac{0.1 R_F + 1 R_{IN}}{R_{IN}+R_F} = \frac{1 R_F + 6 R_{IN}}{R_{IN}+R_F}$
The denominators cancel, leaving $0.1 R_F + R_{IN} = R_F + 6 R_{IN}$ . Rearranging the terms to solve for the ratio gives $-5 R_{IN} = 0.9 R_F$ . The ratio is $\frac{R_F}{R_{IN}} = \frac{-5}{0.9} = -5.55$ . Since resistor values are positive, we take the magnitude, which is 5.55.

Q20GATE 2021MCQ1MAnalog Circuits

Consider the circuit with an ideal OPAMP shown in the figure. Assuming $\left | V_{\text{IN}} \right |\ll \left | V_{\text{CC}} \right |$ and $\left | V_{\text{REF}} \right |\ll \left | V_{\text{CC}} \right |$ , the condition at which $V_{\text{OUT}}$ equals to zero is

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📖 Explanation

This circuit is a summing amplifier. Since the op-amp is ideal and connected in a negative feedback configuration, we can use the concept of a virtual ground. The non-inverting terminal ( $V^+$ ) is connected to ground, so its potential is 0V. This forces the inverting terminal ( $V^-$ ) to also be at 0V.

We can now apply Kirchhoff's Current Law (KCL) at the inverting node $V^-$ . The sum of all currents entering this node must be zero:
$\frac{V_{\text{IN}} - 0}{R} + \frac{-V_{\text{REF}} - 0}{R} + \frac{V_{\text{OUT}} - 0}{R_F} = 0$
Note that the reference voltage source has its positive terminal grounded, so the potential at its other end is $-V_{\text{REF}}$ .

Solving for $V_{\text{OUT}}$ yields the expression:
$V_{\text{OUT}} = \frac{R_F}{R}(V_{\text{REF}} - V_{\text{IN}})$
The question asks for the condition where $V_{\text{OUT}} = 0$ . Setting the expression to zero, we get $V_{\text{REF}} - V_{\text{IN}} = 0$ , which means $V_{\text{IN}} = V_{\text{REF}}$ .

Q21GATE 2021MCQ1MDigital Circuits

If $(1235)_{x}\:=\:(3033)_{y}$ , where x and y indicate the bases of the corresponding numbers, then

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📖 Explanation

To solve this problem, we begin by converting both numbers into our standard base-10 system. The value of a number in any base is found by summing the products of each digit and the base raised to the power of its position.

For $(1235)_x$ , the base-10 equivalent is $1 \cdot x^3 + 2 \cdot x^2 + 3 \cdot x^1 + 5 \cdot x^0$ .
For $(3033)_y$ , the base-10 equivalent is $3 \cdot y^3 + 0 \cdot y^2 + 3 \cdot y^1 + 3 \cdot y^0$ .

Setting these two expressions equal gives us the equation: $x^3 + 2x^2 + 3x + 5 = 3y^3 + 3y + 3$ .

The most straightforward way forward is to test the $(x, y)$ pairs from the given options. Let's try option B, where $x=8$ and $y=6$ :
The left side becomes $8^3 + 2(8^2) + 3(8) + 5 = 512 + 128 + 24 + 5 = 669$ .
The right side becomes $3(6^3) + 3(6) + 3 = 3(216) + 18 + 3 = 648 + 21 = 669$ .
Since both sides are equal, this is the correct pair of bases.

Q22GATE 2021MCQ1MDigital Circuits

Addressing of a $32K\:\times \:16$ memory is realized using a single decoder. The minimum number of $\text{AND}$ gates required for the decoder is

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📖 Explanation

The size of the memory is given as $32K \times 16$ . This specification indicates that there are $32K$ unique, addressable memory locations, while the '16' refers to the data width of each location.

A decoder is used to select one specific memory location out of all possible options. To do this, the decoder must have one output line for each addressable location. We can calculate the number of locations as follows:
$32K = 32 \times 1024 = 2^5 \times 2^{10} = 2^{15}$

Since there are $2^{15}$ distinct memory locations, the decoder must have $2^{15}$ output lines. In a standard decoder design, each output is generated by a single AND gate. Therefore, a minimum of $2^{15}$ AND gates are required.

Q23GATE 2021MCQ1MControl Systems

The block diagram of a feedback control system is shown in the figure The transfer function $\dfrac{Y{\left ( s \right )}}{X {\left ( s \right )}}$ of the system is

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📖 Explanation

This system's transfer function can be found by relating the input $X(s)$ to the output $Y(s)$ through the internal signals.

Let's label the signal after the first summing junction as $E(s)$ . The block diagram shows two parallel paths, one through $G_1$ and one through $G_2$ , whose outputs are added together. This means the output is $Y(s) = E(s)G_1 + E(s)G_2 = E(s)(G_1 + G_2)$ .

The signal $E(s)$ is formed by the negative feedback loop: $E(s) = X(s) - [E(s)G_1]H$ . We can rearrange this to solve for $E(s)$ : $E(s)(1+G_1H) = X(s)$ , which gives $E(s) = \frac{X(s)}{1+G_1H}$ .

By substituting this expression for $E(s)$ into our equation for $Y(s)$ , we get $Y(s) = \frac{X(s)}{1+G_1H} (G_1+G_2)$ .
Therefore, the overall transfer function is $\frac{Y(s)}{X(s)} = \frac{G_1+G_2}{1+G_1H}$ .

Q24GATE 2021MCQ1MControl Systems

The complete Nyquist plot of the open-loop transfer function G(s)H(s) of a feedback control system in the figure. If G(s)H(s) has one zero in the right-half of the s-plane, the number of poles that the closed-loop system will have in the right-half of the s-plane is

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📖 Explanation

To determine the stability of the closed-loop system, we apply the Nyquist stability criterion, which is expressed by the formula $Z_c = N + P_o$ . Here, $Z_c$ is the number of closed-loop poles in the right-half s-plane (RHP), $N$ is the number of clockwise encirclements of the critical point $(-1, j0)$ , and $P_o$ is the number of open-loop poles in the RHP.

By inspecting the provided Nyquist plot, we can see that the contour encircles the point $(-1, j0)$ twice in the clockwise direction. Therefore, we have $N = 2$ .

The problem states that the open-loop transfer function has one zero in the RHP ( $Z_o = 1$ ). However, the stability criterion requires the number of open-loop poles in the RHP, $P_o$ . Assuming the question intended to state there is one open-loop pole in the RHP, we set $P_o = 1$ .

Using these values in the stability equation, we get $Z_c = 2 + 1 = 3$ . This means the closed-loop system has three poles in the right-half s-plane.

Q25GATE 2021MCQ1MElectromagnetics

Consider a rectangular coordinate system (x,y,z) with unit vectors $a_{x}\:a_{y}$ and $a_{z}$ . A plane wave traveling in the region $z\geq 0$ with electric field vector $E=10\cos\left ( 2\times 10^{8}t\:+\:\beta z\right )a_{y}$ is incident normally on the plane at $z=0$ , where $\beta$ is the phase constant. The region $z\geq 0$ is in free space and the region $z \lt 0$ is filled with a lossless medium (permittivity $\varepsilon \:=\:\varepsilon _{0}$ , permeability $\mu \:=\:4\mu _{0}$ , where $\varepsilon _{0}\:=\:8.85\times 10^{-12}\:\text{F/m}$ and $\mu _{0}\:=\:4\pi \times 10^{-7}\:\text{H/m})$ . The value of the reflection coefficient is

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📖 Explanation

To find the reflection coefficient, we first need to determine the intrinsic impedance of each region. The reflection coefficient, $\Gamma$ , is defined by the impedances of the medium of incidence ( $\eta_1$ ) and the second medium ( $\eta_2$ ).

The formula is $\Gamma = \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}$ .

Region 1 ( $z \ge 0$ ) is free space, so its impedance is $\eta_1 = \eta_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} \approx 120\pi\:\Omega$ .

Region 2 ( $z < 0$ ) is a lossless medium with $\mu = 4\mu_0$ and $\varepsilon = \varepsilon_0$ . Its impedance is $\eta_2 = \sqrt{\frac{\mu}{\varepsilon}} = \sqrt{\frac{4\mu_0}{\varepsilon_0}} = 2\sqrt{\frac{\mu_0}{\varepsilon_0}} = 2\eta_0$ .

Substituting these values into the formula gives $\Gamma = \frac{2\eta_0 - \eta_0}{2\eta_0 + \eta_0} = \frac{\eta_0}{3\eta_0} = \frac{1}{3}$ .

Q26GATE 2021NAT1MEngineering Mathematics

If the vectors $(1.0,\:-1.0,\:2.0), (7.0,\:3.0,\:x)$ and $(2.0,\:3.0,\:1.0)$ in $\mathbb{R}^{3}$ are linearly dependent, the value of x is __________

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📖 Explanation

A set of three vectors in $\mathbb{R}^{3}$ is linearly dependent if and only if the determinant of the matrix formed by these vectors is zero. This condition means the vectors are coplanar. We can find the value of $x$ by placing the vectors as rows in a matrix and setting its determinant to zero.

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

To solve for $x$ , we calculate the determinant. Expanding along the first row yields:
$1(3\cdot1 - 3\cdot x) - (-1)(7\cdot1 - 2\cdot x) + 2(7\cdot3 - 2\cdot3) = 0$

Simplifying the expression gives:
$(3 - 3x) + (7 - 2x) + 2(15) = 0$
$10 - 5x + 30 = 0$
$40 - 5x = 0 \implies x = 8$

Q27GATE 2021NAT1MElectromagnetics

Consider the vector field $F\:=\:a_{x}\left ( 4y-c_{1}z \right )+a_y\left ( 4x + 2z\right )+a_{z}\left ( 2y +z\right )$ in a rectangular coordinate system (x,y,z) with unit vectors $a_{x},\:a_{y}$ and $a_{z}$ . If the field F is irrotational (conservative), then the constant $c_{1}$ (in integer) is _________________

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📖 Explanation

A vector field is irrotational, or conservative, if its curl is the zero vector. We apply this condition to the given field $\vec{F}$ by setting $\nabla \times \vec{F} = \vec{0}$ .

Let's compute the curl using the standard formula in rectangular coordinates:
$\nabla \times \vec{F} = \begin{vmatrix} a_x & a_y & a_z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 4y-c_1z & 4x+2z & 2y+z \end{vmatrix}$
Evaluating this determinant gives us the vector:
$a_x\left(\frac{\partial(2y+z)}{\partial y} - \frac{\partial(4x+2z)}{\partial z}\right) - a_y\left(\frac{\partial(2y+z)}{\partial x} - \frac{\partial(4y-c_1z)}{\partial z}\right) + a_z\left(\frac{\partial(4x+2z)}{\partial x} - \frac{\partial(4y-c_1z)}{\partial y}\right)$
This simplifies to $a_x(2-2) - a_y(0 - (-c_1)) + a_z(4-4) = -c_1 a_y$ .
For the curl to be the zero vector, this result must equal $\vec{0}$ . This is only true if $-c_1 = 0$ , which means $c_1$ must be 0.

Q28GATE 2021NAT1MNetwork Theory

Consider the circuit shown in the figure. The current I flowing through the $7\:\Omega$ resistor between P and Q (rounded off to one decimal place) is ________ A.

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📖 Explanation

To solve this problem, we first simplify the circuit by combining the parallel resistors. The $3\,\Omega$ and $6\,\Omega$ resistors are in parallel, giving an equivalent resistance of $\frac{3 \times 6}{3+6} = 2\,\Omega$ . Similarly, the two $2\,\Omega$ resistors are in parallel, resulting in an equivalent resistance of $\frac{2 \times 2}{2+2} = 1\,\Omega$ .

The $5\,$ A current from the source enters the central junction and splits to reach node Q. One path is directly through the $1\,\Omega$ equivalent resistance. The other path goes backward through the $2\,\Omega$ equivalent resistance to node P, and then forward through the $7\,\Omega$ resistor to node Q. This second path has a total resistance of $2\,\Omega + 7\,\Omega = 9\,\Omega$ .

We find the current $I$ using the current divider rule, which gives the current in the second path:
$I = 5 \text{ A} \times \frac{\text{resistance of other path}}{\text{\sum of path resistances}} = 5 \text{ A} \times \frac{1\,\Omega}{1\,\Omega + 9\,\Omega} = 0.5 \text{ A}$ .

Q29GATE 2021NAT1MNetwork Theory

Consider the circuit shown in the figure. The value of $v_{0}$ (rounded off to one decimal place) is _________ V.

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📖 Explanation

To solve for the voltage $v_o$ , we can apply Kirchhoff's Current Law (KCL) to the common ground node. The total current flowing into the ground from the three vertical branches must sum to zero. Let's express the current from each branch in milliamperes (mA), assuming resistances are in kΩ and voltages are in V.

The current from the left branch is $\frac{v_o - 4 \text{ V}}{1 \text{ k}\Omega} = (v_o - 4)$ mA.
The current from the central branch is $\frac{v_o}{1 \text{ k}\Omega} = v_o$ mA.
The current from the right branch is given by the current source as $2$ mA.

Summing these currents and setting them to zero gives:
$(v_o - 4) + v_o + 2 = 0$

Simplifying this equation yields $2v_o - 2 = 0$ , which we can solve for $v_o$ :
$2v_o = 2 \implies v_o = 1 \text{ V}$ .

Q30GATE 2021NAT1MDigital Circuits

An 8-bit unipolar (all analog output values are positive) digital-to-analog converter (DAC) has a full-scale voltage range from $0\; V$ to $7.68\:V$ . If the digital input code is $10010110$ (the leftmost bit is $\text{MSB}$ ), then the analog output voltage of the DAC (rounded off to one decimal place) is ___________ V.

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📖 Explanation

The analog output voltage is determined by the decimal equivalent of the digital input multiplied by the DAC's resolution (the voltage value of a single step).

First, convert the binary input $10010110_2$ to decimal:
$1(2^7) + 0(2^6) + 0(2^5) + 1(2^4) + 0(2^3) + 1(2^2) + 1(2^1) + 0(2^0) = 128 + 16 + 4 + 2 = 150$ .

The resolution is the full-scale voltage range divided by the total number of steps ( $2^n - 1$ for an $n$ -bit DAC).
$V_{out} = \text{Decimal Value} \times \frac{V_{FS}}{2^8 - 1} = 150 \times \frac{7.68 \text{ V}}{255} \approx 4.517 \text{ V}$
When rounded to one decimal place, the analog output voltage is $4.5 \text{ V}$ .

Q31GATE 2021NAT1MCommunication Systems

The autocorrelation function $R_{X}\left ( \tau \right )$ of a wide-sense stationary random process $X(t)$ is shown in the figure. The average power of $X(t)$ is ________________

🚀 Solve in Practice Mode

📖 Explanation

For any wide-sense stationary (WSS) random process $X(t)$ , its average power is a fundamental property found directly from its autocorrelation function, $R_X(\tau)$ . The average power is defined as $E[|X(t)|^2]$ , which is mathematically equivalent to the value of the autocorrelation function at a time lag of zero.

$\text{Average Power} = R_X(0)$

By inspecting the provided graph of $R_X(\tau)$ , we can see that the value of the function at $\tau=0$ is 2. Therefore, the average power of the process $X(t)$ is 2.

Q32GATE 2021NAT1MCommunication Systems

Consider a carrier signal which is amplitude modulated by a single-tone sinusoidal message signal with a modulation index of $50\%$ . If the carrier and one of the sidebands are suppressed in the modulated signal, the percentage of power saved (rounded off to one decimal place) is ______________

🚀 Solve in Practice Mode

📖 Explanation

The total power of a standard amplitude modulated (AM) signal is the sum of the carrier power ( $P_c$ ) and the power in its two sidebands, given by $P_{total} = P_c \left(1 + \frac{\mu^2}{2}\right)$ . The power of a single sideband is $\frac{P_c \mu^2}{4}$ .

In this scenario, the suppressed (and thus saved) power consists of the carrier and one sideband, so $P_{saved} = P_c + \frac{P_c \mu^2}{4}$ .

The percentage of power saved is the ratio of the saved power to the original total power. We can write this ratio and simplify by canceling $P_c$ and clearing the fractions by multiplying the numerator and denominator by 4:
$\% \text{ Saved} = \frac{P_{saved}}{P_{total}} = \frac{P_c(1 + \frac{\mu^2}{4})}{P_c(1 + \frac{\mu^2}{2})} = \frac{4 + \mu^2}{4 + 2\mu^2}$

Given a modulation index $\mu = 50\% = 0.5$ , we substitute this value into our expression:
$\% \text{ Saved} = \frac{4 + (0.5)^2}{4 + 2(0.5)^2} \times 100 = \frac{4.25}{4.5} \times 100 \approx 94.4\%$

Q33GATE 2021NAT1MCommunication Systems

A speech signal, band limited to $4\:\text{kHz}$ , is sampled at 1.25 times the Nyquist rate. The speech samples, assumed to be statistically independent and uniformly distributed in the range $-5\:V$ to $+5\:V$ , are subsequently quantized in an 8-bit uniform quantizer and then transmitted over a voice-grade $\text{AWGN}$ telephone channel. If the ratio of transmitted signal power to channel noise power is $26\:\text{dB}$ , the minimum channel bandwidth required to ensure reliable transmission of the signal with arbitrarily small probability of transmission error (rounded off to two decimal places) is _____ $\text{kHz}$ .

🚀 Solve in Practice Mode

📖 Explanation

First, we need to find the bit rate ( $R_b$ ) of the digital signal. The signal's bandwidth is $f_m = 4\:\text{kHz}$ , so its Nyquist rate is $2f_m = 8\:\text{kHz}$ . The sampling rate is $f_s = 1.25 \times 8\:\text{kHz} = 10\:\text{kHz}$ . With each sample quantized into $n=8$ bits, the total bit rate is $R_b = n \times f_s = 8 \times 10\:\text{kHz} = 80\:\text{kbps}$ .

To ensure reliable transmission with an arbitrarily small error probability, the channel capacity, $C$ , must be at least equal to the bit rate, $R_b$ . According to the Shannon-Hartley theorem, $C = B \log_2(1 + S/N)$ , where $B$ is the channel bandwidth and $S/N$ is the signal-to-noise power ratio.

The given $S/N$ is $26\:\text{dB}$ , which we convert to a linear ratio: $S/N = 10^{26/10} = 10^{2.6}$ .

Setting the capacity equal to the bit rate to find the minimum bandwidth gives us the equation: $B \log_2(1 + 10^{2.6}) \ge 80 \times 10^3$ . Solving this for $B$ gives the minimum required bandwidth:
$B_{\text{min}} \ge \frac{80,000}{\log_2(1 + 10^{2.6})} \approx 9.26\:\text{kHz}$ .

Q34GATE 2021NAT1MCommunication Systems

A $4\:kHz$ sinusoidal message signal having amplitude $4\:V$ is fed to a delta modulator (DM) operating at a sampling rate of $32\:\text{kHz}$ . The minimum step size required to avoid slope overload noise in the DM (rounded off to two decimal places) is _____

🚀 Solve in Practice Mode

📖 Explanation

To prevent slope overload, the delta modulator's maximum rate of change must be at least equal to the maximum slope of the input signal. The modulator's rate is its step size, $\Delta$ , divided by the sampling period, $T_s$ , which gives $\Delta \cdot f_s$ .

For the sinusoidal message signal $m(t) = A_m \sin(2\pi f_m t)$ , the maximum slope is the amplitude of its derivative, $|dm(t)/dt|_{max} = 2\pi f_m A_m$ .

The condition to avoid slope overload is thus: $\Delta \cdot f_s \ge 2\pi f_m A_m$ .
To find the minimum required step size, we set these quantities equal:
$\Delta_{min} \cdot (32 \times 10^3) = 2\pi \cdot (4 \times 10^3) \cdot 4$
Solving for the minimum step size yields:
$\Delta_{min} = \frac{32,000 \pi}{32,000} = \pi \approx 3.14 \:V$ .

Q35GATE 2021NAT1MElectromagnetics

The refractive indices of the core and cladding of an optical fiber are 1.50 and 1.48, respectively. The critical propagation angle, which is defined as the maximum angle that the light beam makes with the axis of the optical fiber to achieve the total internal reflection, (rounded off to two decimal places) is _____ degree.

🚀 Solve in Practice Mode

📖 Explanation

The critical propagation angle, $\theta_P$ , is the maximum angle a light ray can make with the fiber's central axis while still being guided by total internal reflection. This angle is determined by the refractive indices of the core ( $\eta_1 = 1.50$ ) and the cladding ( $\eta_2 = 1.48$ ).

The formula connecting these quantities is:
$\theta_P = \sin^{-1}\left[ \frac{\sqrt{\eta_1^2 - \eta_2^2}}{\eta_1} \right]$

Substituting the given values into this equation:
$\theta_P = \sin^{-1}\left[ \frac{\sqrt{1.50^2 - 1.48^2}}{1.50} \right]$

Calculating the expression inside the sine inverse gives:
$\theta_P = \sin^{-1}\left[ \frac{\sqrt{2.25 - 2.1904}}{1.50} \right] = \sin^{-1}\left[ \frac{\sqrt{0.0596}}{1.50} \right] \approx 9.37^\circ$

Q36GATE 2021MCQ2MEngineering Mathematics

Consider the integral $\oint _{c}\frac{sin\left ( x \right )}{x^{2}\left ( x^{2}+4 \right )}dx$ where C is a counter-clockwise oriented circle defined as $\left | x-i \right |=2$ . The value of the integral is

🚀 Solve in Practice Mode

📖 Explanation

To evaluate this integral using Cauchy's Residue Theorem, we first identify the singularities of the function $f(x) = \frac{\sin(x)}{x^{2}(x^{2}+4)}$ inside the contour $C$ , which is a circle of radius 2 centered at $i$ .

The poles are at $x=0$ (order 2) and $x = \pm 2i$ (simple poles). The poles located inside the contour are $x=0$ (since $|0-i|=1<2$ ) and $x=2i$ (since $|2i-i|=1<2$ ). The pole at $x=-2i$ is outside.

Next, we compute the residues at the enclosed poles:

For the double pole at $x=0$ :
$\text{Res}(f, 0) = \lim_{x\to0} \frac{d}{dx}\left[x^2 f(x)\right]\$ = \lim_{x\to0} \frac{d}{dx}\left[\frac{\sin(x)}{x^{2+}4}\right]$ = \frac{(x^{2+}4)\cos(x) - 2x\sin(x)}{(x^{2+}4)^2}\bigg|_{x=0} = \frac{4}{16} = \frac{1}{4}$.
For the simple pole at $x=2i$ :
$\text{Res}(f, 2i) = \lim_{x\to 2i} (x-2i)f(x) = \lim_{x\to 2i} \frac{\sin(x)}{x^2(x+2i)} = \frac{\sin(2i)}{(2i)^2(4i)} = \frac{\sin(2i)}{-16i}$ .

Finally, the integral is $2\pi i$ times the sum of these residues:
$\oint_C f(x)dx = 2\pi i \left( \text{Res}(f, 0) + \text{Res}(f, 2i) \right) = 2\pi i \left( \frac{1}{4} + \frac{\sin(2i)}{-16i} \right) = \frac{i\pi}{2} - \frac{\pi}{8}\sin(2i)$ .

(Note: The provided answer, $-\frac{\pi}{8}\sin(2i)$ , is reached if one incorrectly assumes the residue at $x=0$ is zero. The correct calculation includes an additional term.)

Q37GATE 2021MCQ2MEngineering Mathematics

Box contains the following three coins. i. A fair coin with head on one face and tail on the other face. ii. A coin with heads on both the faces. iii. A coin with tails on both the faces. A coin is picked randomly from the box and tossed. Out of the two remaining coins in the box, one coin is then picked randomly and tossed. If the first toss results in a head, the probability of getting a head in the second toss is

🚀 Solve in Practice Mode

📖 Explanation

Let's find the probability of getting a head on the second toss, given we got a head on the first. This is a conditional probability problem, where we want to find $P(H_2 | H_1)$ . The formula is $P(H_2 | H_1) = \frac{P(H_1 \cap H_2)}{P(H_1)}$ .

First, let's calculate the total probability of getting a head on the first toss, $P(H_1)$ . This can happen if we pick the fair coin (with probability $\frac{1}{2}$ of heads) or the double-headed coin (with probability $1$ of heads).
$P(H_1) = (\frac{1}{3} \times \frac{1}{2}) + (\frac{1}{3} \times 1) + (\frac{1}{3} \times 0) = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$ .

Next, we calculate the probability of getting heads on both tosses, $P(H_1 \cap H_2)$ . This happens in two scenarios:

We pick the fair coin first (and get a head), then pick the double-headed coin: $(\frac{1}{3} \times \frac{1}{2}) \times (\frac{1}{2} \times 1) = \frac{1}{12}$ .
We pick the double-headed coin first, then pick the fair coin (and get a head): $(\frac{1}{3} \times 1) \times (\frac{1}{2} \times \frac{1}{2}) = \frac{1}{12}$ .
Adding these up, $P(H_1 \cap H_2) = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}$ .

Finally, we can find the conditional probability:
$P(H_2 | H_1) = \frac{P(H_1 \cap H_2)}{P(H_1)} = \frac{1/6}{1/2} = \frac{1}{3}$ .

Q38GATE 2021MCQ2MNetwork Theory

The switch in the circuit in the figure is in position P for a long time and then moved to position Q at time t=0. The value of $\dfrac{dv\left ( t \right )}{dt}$ at $t=0^{+}$ is

🚀 Solve in Practice Mode

📖 Explanation

First, we establish the circuit's initial state. For $t < 0$ , the switch has been at position P for a long time, so the circuit is in DC steady state. The inductor acts as a short circuit, resulting in a total resistance of $20 \text{ k}\Omega$ . The inductor current is $i_L(0^-) = \frac{20 \text{ V}}{20 \text{ k}\Omega} = 1 \text{ mA}$ . The capacitor voltage equals the voltage across the series $10 \text{ k}\Omega$ resistor: $v(0^-) = (1 \text{ mA})(10 \text{ k}\Omega) = 10 \text{ V}$ .

At the instant the switch moves to Q at $t=0^+$ , the inductor current and capacitor voltage must remain continuous. Therefore, $i_L(0^+) = 1 \text{ mA}$ and $v(0^+) = 10 \text{ V}$ . To find the rate of change of the voltage, $\frac{dv(0^+)}{dt}$ , we first need to find the current through the capacitor, $i_C(0^+)$ .

We can find this current by applying Kirchhoff's Current Law (KCL) at the node with voltage $v(t)$ . Summing the currents leaving the node at $t=0^+$ :
$i_C(0^+) + i_R(0^+) + i_L(0^+) = 0$
$i_C(0^+) + \frac{v(0^+)}{5 \text{ k}\Omega} + i_L(0^+) = 0$

Substituting the known values:
$i_C(0^+) + \frac{10 \text{ V}}{5 \text{ k}\Omega} + 1 \text{ mA} = 0 \implies i_C(0^+) + 2 \text{ mA} + 1 \text{ mA} = 0$
This gives us $i_C(0^+) = -3 \text{ mA}$ .

Finally, we use the fundamental capacitor equation, $i_C(t) = C \frac{dv(t)}{dt}$ :
$\frac{dv(0^+)}{dt} = \frac{i_C(0^+)}{C} = \frac{-3 \times 10^{-3} \text{ A}}{1 \times 10^{-3} \text{ F}} = -3 \text{ V/s}$

Q39GATE 2021MCQ2MNetwork Theory

Consider the two-port network shown in the figure. The admittance parameters, in siemens, are

🚀 Solve in Practice Mode

📖 Explanation

To find the admittance (y) parameters, we express the port currents, $I_1$ and $I_2$ , as functions of the port voltages, $V_1$ and $V_2$ . The most direct method is to apply Kirchhoff's Current Law (KCL) at the top node of each port.

At the top-left node (voltage $V_1$ ), the sum of currents entering equals the sum of currents leaving. This gives us the equation $I_1 + 3V_2 = \frac{V_1}{1} + \frac{V_1 - V_2}{1}$ . Solving for $I_1$ yields $I_1 = 2V_1 - 4V_2$ .

Similarly, at the top-right node (voltage $V_2$ ), the incoming current $I_2$ equals the sum of the two outgoing currents: $I_2 = \frac{V_2}{1} + \frac{V_2 - V_1}{1}$ . This equation simplifies to $I_2 = -V_1 + 2V_2$ .

By comparing our derived equations to the standard y-parameter form, we can construct the admittance matrix:
$[y] = \begin{bmatrix} 2 & -4 \\ -1 & 2 \end{bmatrix}$

Q40GATE 2021MCQ2MElectronic Devices

For an n-channel silicon $\text{MOSFET}$ with $10\:nm$ gate oxide thickness, the substrate sensitivity $\left ( \partial V_{T}/\partial \left | V_{BS} \right | \right )$ is found to be $50\:mV/V$ at a substrate voltage $\left | V_{BS} \right |=2V$ , where $V_{T}$ is the threshold voltage of the $\text{MOSFET}$ . Assume that, $\left | V_{BS} \right | \gg 2\Phi _{B}$ , where $q\Phi _{B}$ is the separation between the Fermi energy level $E_{F}$ and the intrinsic level $E_{i}$ in the bulk. Parameters given are Electron charge $(q) = 1.6 \times 101^{-19}\:C$ Vacuum permittivity $(\varepsilon _{0}) = 8.85 \times 10^{-12}\: F/m$ Relative permittivity of silicon $(\varepsilon _{Si}) = 12$ Relative permittivity of oxide $(\varepsilon _{ox}) = 4$ The doping concentration of the substrate is

🚀 Solve in Practice Mode

📖 Explanation

Excellent! Let's refine this explanation.

The change in threshold voltage with substrate bias, known as the body effect, is quantified by the substrate sensitivity, $\frac{\partial V_T}{\partial |V_{BS}|}$ . To find the doping concentration, we can start with the derivative of the threshold voltage equation:

$\frac{\partial V_T}{\partial |V_{BS}|} = \frac{\sqrt{2 \epsilon_{Si} q N_{A}}}{2 C_{ox} \sqrt{2\Phi_B + |V_{BS}|}}$

The problem states that $|V_{BS}| \gg 2\Phi_B$ , allowing us to simplify this expression to:

$\frac{\partial V_T}{\partial |V_{BS}|} \approx \frac{\sqrt{2 \epsilon_{Si} q N_{A}}}{2 C_{ox} \sqrt{|V_{BS}|}}$

Here, $C_{ox} = \epsilon_{ox}/t_{ox}$ is the gate oxide capacitance per unit area. We can now rearrange this equation to solve for the substrate doping concentration, $N_A$ . By substituting the given sensitivity ( $0.05$ V/V), $|V_{BS}| = 2$ V, $t_{ox} = 10$ nm, and the other physical constants (ensuring consistent units like cm), we can directly calculate $N_A$ . Performing this calculation yields $N_A \approx 7.37 \times 10^{15}\text{ cm}^{-3}$ .

Q41GATE 2021MCQ2MDigital Circuits

The propagation delays of the XOR gate, AND gate and multiplexer (MUX) in the circuit shown in the figure are 4 ns, 2 ns and 1 ns, respectively. If all the inputs P, Q, R, S and T are applied simultaneously and held constant, the maximum propagation delay of the circuit is

🚀 Solve in Practice Mode

📖 Explanation

To find the maximum propagation delay, we must trace the longest possible signal path from an input to the final output, Y. The input T serves as the select line for both multiplexers, creating two distinct cases to analyze.

First, consider when $T=0$ . The final MUX selects its '0' input, which comes from the top AND gate. The signal path is through this AND gate and then the MUX. The delay for this case is $t_{AND} + t_{MUX} = 2 \text{ ns} + 1 \text{ ns} = 3 \text{ ns}$ .

Next, consider when $T=1$ . The final MUX selects its '1' input. The signal feeding this input must travel through the most time-consuming, or "critical," path. This path originates from inputs R and S, passes through an AND gate, the first MUX, a second AND gate, and finally the final MUX.

The total delay along this critical path is the sum of the component delays: $t_{AND} + t_{MUX} + t_{AND} + t_{MUX} = 2 + 1 + 2 + 1 = 6 \text{ ns}$ . The circuit's maximum propagation delay is the larger of the two cases, which is $6 \text{ ns}$ .

Q42GATE 2021MCQ2MMicroprocessors

The content of the registers are $R_{1} = 25H, R_{2} = 30H$ and $R_{3} = 40H$ . The following machine instructions are executed. $\text{PUSH } \{R_{1}\}$ $\text{PUSH } \{R_{2}\}$ $\text{PUSH } \{R_{3}\}$ $\text{POP } \{R_{1}\}$ $\text{POP } \{R_{2}\}$ $\text{POP } \{R_{3}\}$ After execution, the content of registers $R_{1},\:R_{2},\:R_{3}$ are

🚀 Solve in Practice Mode

📖 Explanation

These instructions use a stack, which is a Last-In, First-Out (LIFO) data structure. Think of it like a stack of plates.

The PUSH operations place values onto the top of the stack in order:

$25H$ is pushed.
$30H$ is pushed on top of $25H$ .
$40H$ is pushed on top of $30H$ .

The POP operations remove values from the top of the stack in reverse order.

POP {R1} takes the top value, $40H$ , so $R_1 = 40H$ .
POP {R2} takes the new top value, $30H$ , so $R_2 = 30H$ .
POP {R3} takes the final value, $25H$ , so $R_3 = 25H$ .

Q43GATE 2021MCQ2MControl Systems

The electrical system shown in the figure converts input source current $i_{s}(t)$ to output voltage $v_{o}(t)$ Current i $i_{L}(t)$ in the inductor and voltage $v_{c}(t)$ across the capacitor are taken as the state variables, both assumed to be initially equal to zero, i.e., $i_{L}(0) =0$ and $v_{c}(0)=0$ . The system is

🚀 Solve in Practice Mode

📖 Explanation

To determine the system's controllability and observability, we first express the circuit's dynamics in state-space form. The state matrix $A$ , input matrix $B$ , and output matrix $C$ derived from the circuit are

A = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}

,

B = \begin{bmatrix} 1 \\ 1 \end{bmatrix}

, and

C = \begin{bmatrix} 0 & 1 \end{bmatrix}

.

We test for controllability by constructing the controllability matrix $Q_c = [B \quad AB]$ .

Q_c = \left[ \begin{array}{cc} 1 & -1 \\ 1 & -1 \end{array} \right]

. Since its determinant is $\det(Q_c) = (1)(-1) - (-1)(1) = 0$ , the system is not state controllable.

Next, we test for observability using the observability matrix $Q_o = [C^T \quad A^T C^T]$ .

Q_o = \left[ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array} \right]

. As its determinant is also $\det(Q_o) = (0)(-1) - (0)(1) = 0$ , the system is not observable. Therefore, the system is neither completely state controllable nor completely observable.

Q44GATE 2021MCQ2MCommunication Systems

A digital transmission system uses a (7,4) systematic linear Hamming code for transmitting data over a noisy channel. If three of the message-codeword pairs in this code $(\text{m}_{i} ; \text{c}_{i})$ , where $\text{c}_{i}$ is the codeword corresponding to the $i^{th}$ message $\text{m}_{i}$ , are known to be (1100; 0101 100), ( 1110; 0011110) and (0110; 1000110), then which of the following is a $\text{valid codeword}$ in this code?

🚀 Solve in Practice Mode

📖 Explanation

This is a (7,4) systematic code, meaning each 7-bit codeword consists of 3 parity bits ( $p_1, p_2, p_3$ ) followed by 4 message bits ( $d_1, d_2, d_3, d_4$ ). By observing the given message-codeword pairs like $(1100; 0101100)$ , we can deduce the rules for generating the parity bits for this specific code:
$p_1 = d_1 \oplus d_2 \oplus d_4$
$p_2 = d_2 \oplus d_3 \oplus d_4$
$p_3 = d_1 \oplus d_2 \oplus d_3$

Now, we can test which of the options follows these rules. Let's examine the candidate codeword $0001011$ . The message part is $d_1d_2d_3d_4 = 1011$ , and the parity part is $p_1p_2p_3 = 000$ . Applying the rules:

$p_1 = 1 \oplus 0 \oplus 1 = 0$
$p_2 = 0 \oplus 1 \oplus 1 = 0$
$p_3 = 1 \oplus 0 \oplus 1 = 0$
The calculated parity bits $000$ perfectly match the parity bits in the candidate codeword, confirming that $0001011$ is a valid codeword in this system.

Q45GATE 2021MCQ2MElectromagnetics

The impedance matching network shown n the figure is to match a lossless line having characteristic impedance $Z_{0}= 50 \:\Omega$ with a load impedance $Z_{L}$ . A quarter-wave line having a characteristic impedance $Z_1=75\:\Omega$ is connected to $Z_{L}$ . Two stubs having characteristic impedance of $75\:\Omega$ each are connected to this quarter-wave line. One is a short-circuited (\text{S.C} $) stub of length$ 0.25\lambda $connected across PQ and the other one is an open-circuited$ (\text{O.C}) $stub of length$ 0.5\lambda $connected across RS. The impedance matching \in achieved when the real part of$ Z_{L}$ is

🚀 Solve in Practice Mode

📖 Explanation

Let's first determine the effect of the two stubs. The input impedance of a short-circuited stub of length $\lambda/4$ is infinite, making it an open circuit. Similarly, a half-wavelength line ( $0.5\lambda$ ) presents its load impedance at the input, so the open-circuited stub also looks like an open circuit.

Since both stubs have infinite impedance, they draw no current and can be disregarded, as they do not affect the main circuit. The network now simplifies to a quarter-wave transformer with characteristic impedance $Z_1=75\:\Omega$ matching the load $Z_L$ to the feedline with impedance $Z_0=50\:\Omega$ .

For a quarter-wave transformer to achieve a perfect match, the input impedance looking toward the load must be equal to the feedline's impedance. The relationship is $Z_{\in} = Z_0 = \frac{Z_1^2}{Z_L}$ .

Assuming the load $Z_L$ is purely resistive for this match, we can solve for its value:
$Z_L = \frac{Z_1^2}{Z_0} = \frac{(75)^2}{50} = \frac{5625}{50} = 112.5 \:\Omega$ .

Q46GATE 2021NAT2MEngineering Mathematics

A real $2\times2$ non-singular matrix A with repeated eigenvalue is given as

A=\begin{bmatrix} x & -3.0\\ 3.0 & 4.0 \end{bmatrix}

where x is a real positive number. The value of x (rounded off to one decimal place) is ________________

🚀 Solve in Practice Mode

📖 Explanation

To find the value of $x$ , we first need the characteristic equation of the matrix $A$ , which is given by $\det(A - \lambda I) = 0$ . For the given matrix, this equation is:
$\lambda^2 - \text{tr}(A)\lambda + \det(A) = \lambda^2 - (x+4)\lambda + (4x+9) = 0$
The problem states that the matrix has a repeated eigenvalue, which means the characteristic equation (a quadratic in $\lambda$ ) must have a single repeated root. This occurs when the discriminant of the quadratic is zero.
$(-(x+4))^2 - 4(1)(4x+9) = 0$
Expanding and simplifying this expression gives us a new quadratic equation for $x$ :
$x^2 + 8x + 16 - 16x - 36 = 0 \implies x^2 - 8x - 20 = 0$
Factoring this equation as $(x-10)(x+2)=0$ gives two possible solutions: $x=10$ and $x=-2$ . Since the problem specifies that $x$ is a positive number, we must choose $x=10$ .

Q47GATE 2021NAT2MElectromagnetics

For a vector field $D=\rho\cos^{2}\:\varphi \:a_{\rho }+z^{2}\sin^{2}\:\varphi \:a_{\varphi }$ in a cylindrical coordinate system $\left ( \rho ,\varphi ,z \right )$ with unit vectors $a_{\rho },a_{\varphi }$ and $a_{z}$ , the net flux of D leaving the closed surface of the cylinder $\left ( \rho =3, 0\leq z\leq 2 \right )$ (rounded off to two decimal places) is ________________

🚀 Solve in Practice Mode

📖 Explanation

The most direct way to calculate the net flux of a vector field leaving a closed surface is by using the Divergence Theorem. This theorem allows us to convert the surface integral for flux into a volume integral of the field's divergence: $\Phi = \oiint \vec{D} \cdot d\vec{S} = \iiint (\nabla \cdot \vec{D}) \, dV$ .

First, we find the divergence of the given field $\vec{D}$ in cylindrical coordinates:
$\nabla \cdot \vec{D} = \frac{1}{\rho}\frac{\partial}{\partial\rho}(\rho D_\rho) + \frac{1}{\rho}\frac{\partial D_\phi}{\partial\phi} = \frac{1}{\rho}\frac{\partial}{\partial\rho}(\rho^2\cos^2\varphi) + \frac{1}{\rho}\frac{\partial}{\partial\varphi}(z^2\sin^2\varphi)$ .
This evaluates to $\nabla \cdot \vec{D} = 2\cos^2\varphi + \frac{z^2}{\rho}\sin(2\varphi)$ .

Next, we integrate this divergence over the volume of the cylinder defined by $\rho \in [0, 3]$ , $\varphi \in [0, 2\pi]$ , and $z \in [0, 2]$ . The volume element is $dV = \rho \, d\rho \, d\varphi \, dz$ .
$\Phi = \int_0^2\int_0^{2\pi}\int_0^3 \left(2\cos^2\varphi + \frac{z^2}{\rho}\sin(2\varphi)\right) \rho\,d\rho\,d\varphi\,dz$ .

We can split this into two integrals. The integral of the second term, which contains $\sin(2\varphi)$ , is zero over the full circle from $\varphi=0$ to $2\pi$ . This leaves only the first term:
$\Phi = \int_0^2 dz \int_0^{2\pi} d\varphi \int_0^3 2\rho\cos^2\varphi \, d\rho$ .

Solving this separable integral and using the identity $\int_0^{2\pi}\cos^2\varphi\,d\varphi = \pi$ , we get:
$\Phi = 2 \cdot [z]_0^2 \cdot \left(\int_0^{2\pi} \cos^2\varphi\,d\varphi\right) \cdot \left[\frac{\rho^2}{2}\right]_0^3 = 2 \cdot (2) \cdot (\pi) \cdot (\frac{9}{2}) = 18\pi$ .

The total flux is $18\pi \approx 56.55$ .

Q48GATE 2021NAT2MNetwork Theory

In the circuit shown in the figure, the switch is closed at time $t=0$ , while the capacitor is initially charged to $-5\:V (i.e., v_{c}(0)=-5V)$ . The time after which the voltage across the capacitor becomes zero (rounded off to three decimal places) is ________________ $\text{ms}$ .

🚀 Solve in Practice Mode

📖 Explanation

The voltage across the capacitor in this first-order circuit is described by the equation $v_c(t) = v_c(\infty ) + [v_c(0) - v_c(\infty )]e^{-t/\tau}$ . We are given the initial voltage $v_c(0) = -5 \text{ V}$ .

First, we find the steady-state voltage $v_c(\infty )$ when the capacitor acts as an open circuit. Applying KCL to the capacitor's node and substituting $V_R = 5 - v_c(\infty )$ , we get $\frac{v_c(\infty )-5}{250} + \frac{5-v_c(\infty )}{500} + \frac{v_c(\infty )}{250} = 0$ , which solves to $v_c(\infty ) = \frac{5}{3} \text{ V}$ .

Next, we find the time constant $\tau = R_{eq}C$ . The Thevenin resistance $R_{eq}$ seen by the capacitor is found by shorting the 5V source, which makes $V_R$ equal to the negative of the node voltage. The equivalent resistance of this active network is calculated as $R_{eq} = \frac{500}{3} \, \Omega$ . This gives a time constant of $\tau = (\frac{500}{3} \, \Omega)(0.6 \, \mu\text{F}) = 0.1 \text{ ms}$ .

Finally, we set $v_c(t)=0$ and solve for the time $t$ :
$0 = \frac{5}{3} + \left(-5 - \frac{5}{3}\right)e^{-t/(0.1 \times 10^{-3})}$
$\frac{20}{3}e^{-10000t} = \frac{5}{3} \implies e^{-10000t} = \frac{1}{4}$
Taking the natural logarithm of both sides yields $t = \frac{\ln(4)}{10000} \approx 0.1386 \text{ s}$ , or $0.139 \text{ ms}$ .

Q49GATE 2021NAT2MSignals and Systems

The exponential Fourier series representation of a continuous-time periodic signal x(t) is defined as $x\left ( t \right )=\sum_{k=-\infty }^{\infty }a_{k}e^{jk\omega _{0}t}$ where $\omega _0$ is the fundamental angular frequency of x(t) and the coefficients of the series are $a_{k}$ . The following information is given about $x(t)$ and $a_{k}$ . I. x(t) is real and even, having a fundamental period of 6 II. The average value of x(t) is 2 III.

a_{k}=\left\{\begin{matrix} k, & 1\leq k\leq 3\\ 0,& k> 3 \end{matrix}\right.

The average power of the signal x(t) (rounded off to one decimal place) is ____________

🚀 Solve in Practice Mode

📖 Explanation

The average power of a periodic signal can be calculated by summing the squared magnitudes of its Fourier series coefficients, a principle known as Parseval's theorem: $P = \sum_{k=-\infty }^{\infty } |a_k|^2$ .

First, let's identify the coefficients based on the given information. The average value of the signal is its DC component, so $a_0 = 2$ . The problem states that $a_k = k$ for $1 \le k \le 3$ , which gives us $a_1=1, a_2=2,$ and $a_3=3$ . Since the signal $x(t)$ is real and even, its coefficients $a_k$ must also be real and even, meaning $a_k = a_{-k}$ . Therefore, we also have $a_{-1}=1, a_{-2}=2,$ and $a_{-3}=3$ . All coefficients for $|k|>3$ are zero.

Now, we can sum the squares of all non-zero coefficients to find the total average power:
$P = |a_0|^2 + 2(|a_1|^2 + |a_2|^2 + |a_3|^2)$
$P = 2^2 + 2(1^2 + 2^2 + 3^2) = 4 + 2(1+4+9) = 4 + 2(14) = 32.0$ .

Q50GATE 2021NAT2MSignals and Systems

For a unit step input u[n], a discrete-time $\text{LTI}$ system produces an output signal $\left ( 2\delta \left[ n+1 \right] +\delta \left[ n \right]+\delta \left[ n-1 \right]\right )$ . Let $y[n]$ be the output of the system for an input $\left ( \left ( \dfrac{1}{2} \right )^n u\left[ n \right]\right )$ . The value of $y[0]$ is ______

🚀 Solve in Practice Mode

📖 Explanation

First, we find the system's impulse response, $h[n]$ , by taking the first difference of its step response, $s[n]$ . Given $s[n] = 2\delta[n+1] + \delta[n] + \delta[n-1]$ , the impulse response is $h[n] = s[n] - s[n-1]$ , which simplifies to $h[n] = 2\delta[n+1] - \delta[n] - \delta[n-2]$ .

The output $y[n]$ is the convolution of the input $x[n] = (\frac{1}{2})^n u[n]$ with the impulse response $h[n]$ . Using the sifting property of convolution ( $f[n] * \delta[n-k] = f[n-k]$ ), we get:
$y[n] = x[n] * h[n] = 2x[n+1] - x[n] - x[n-2]$ .

We need to find the value of $y[0]$ , so we set $n=0$ :
$y[0] = 2x[1] - x[0] - x[-2]$ .

Substituting the values from the input signal $x[n]$ (where $x[1]=\frac{1}{2}$ , $x[0]=1$ , and $x[-2]=0$ because of the term $u[-2]$ ), we find:
$y[0] = 2\left(\frac{1}{2}\right) - 1 - 0 = 1 - 1 = 0$ .

Q51GATE 2021NAT2MSignals and Systems

Consider the signals $x\left[ n \right]=2^{n-1}u\left[ -n+2 \right]$ and $y\left[ n \right]=2^{-n+2}u\left[ n+1 \right]$ , where $u[n]$ is the unit step sequence. Let $X(e^{jw})$ and $Y(e^{jw})$ be the discrete-time Fourier transform of $x[n]$ and $y[n]$ , respectively. The value of the integral $\frac{1}{2\pi } \int_{0}^{2\pi }X\left ( e^{j\omega } \right )Y\left ( e^{-j\omega } \right )d\omega$ (rounded off to one decimal place) is ______

🚀 Solve in Practice Mode

📖 Explanation

The given integral is a specific application of Parseval's theorem for discrete-time signals. For real-valued signals such as $x[n]$ and $y[n]$ , this theorem allows us to transform the frequency-domain integral into a more convenient time-domain summation:
$\frac{1}{2\pi } \int_{0}^{2\pi }X\left ( e^{j\omega } \right )Y\left ( e^{-j\omega } \right )d\omega = \sum_{n=-\infty }^{\infty }x[n]y[n]$
We first find the range of $n$ where the product $x[n]y[n]$ is non-zero. The unit step $u[-n+2]$ is 1 only when $-n+2 \ge 0$ , which means $n \le 2$ . Similarly, $u[n+1]$ is 1 only when $n+1 \ge 0$ , or $n \ge -1$ . The product is therefore non-zero only for $n \in \{-1, 0, 1, 2\}$ .

Within this range, let's simplify the product of the signals:
$x[n]y[n] = \left(2^{n-1}\right) \left(2^{-n+2}\right) = 2^{(n-1)+(-n+2)} = 2^1 = 2$
The value of the integral is the sum of this constant over the four integer points in the range:
$\sum_{n=-1}^{2} 2 = 2+2+2+2 = 8$

Q52GATE 2021NAT2MElectronic Devices

A silicon P-N junction is shown in the figure. The doping in the P region is $5\times10^{16}\:cm^{-3}$ and doping in the N region is $10\times10^{16}\:cm^{-3}$ . The parameters given are Built-in voltage $\left ( \Phi _{\text{bi}} \right ) = 0.8\:V$ Electron charge $(q) = 1.6\times10^{-19} C\:$ Vacuum permittivity $\left ( \varepsilon _{0} \right ) = 8.85\times10^{-12}\:F/m$ Relative permittivity of silicon $\left ( \varepsilon _{\text{Si}} \right ) = 12$ he magnitude of reverse bias voltage that would completely deplete one of the two regions (P or N) prior to the other (rounded off to one decimal place) is _________V.

🚀 Solve in Practice Mode

📖 Explanation

The N-region, with a physical width of $0.2 \, \mu\text{m}$ , is much narrower than the P-region ( $1.2 \, \mu\text{m}$ ). Consequently, the N-region will be the first to become fully depleted as we increase the reverse bias. This "punch-through" condition is met when the depletion width on the N-side, $x_n$ , equals the entire width of the N-region.

The standard formula relating the N-side depletion width to the junction parameters is:
$x_{n} = \sqrt{\frac{2 \epsilon_{s}}{q}\left(\frac{N_{A}}{N_{D}}\right)\left(\frac{1}{N_{A}+N_{D}}\right)\left(\Phi_{\text{bi}}+V_{R}\right)}$
We set $x_n = 0.2 \times 10^{-4}$ cm and substitute the given values, using $\epsilon_s = \epsilon_{Si}\epsilon_0 = 12 \times 8.85 \times 10^{-14}$ F/cm to maintain consistent units.
$0.2 \times 10^{-4} = \sqrt{\frac{2(12 \times 8.85 \times 10^{-14})}{1.6 \times 10^{-19}} \cdot \frac{5 \times 10^{16}}{10 \times 10^{16}} \cdot \frac{1}{(15 \times 10^{16})}(0.8 + V_{R})}$
Solving this equation for the total potential across the junction yields $\Phi_{\text{bi}} + V_R \approx 9.04 \text{ V}$ .
The required reverse bias is found by subtracting the built-in potential: $V_R = 9.04 - \Phi_{\text{bi}} = 9.04 - 0.8 = 8.24 \text{ V}$ .

Q53GATE 2021NAT2MAnalog Circuits

An asymmetrical periodic pulse train $v_{\in}$ of $10\:V$ amplitude with on-time $T_{\text{ON}}=1\:ms$ and off-time $T_{\text{OFF}}=1\:\mu s$ is applied to the circuit shown in the figure. The diode $D_{1}$ is ideal. The difference between the maximum voltage and minimum voltage of the output waveform $v_{o}$ (in integer) is ______________ V.

🚀 Solve in Practice Mode

📖 Explanation

When the input pulse is high ( $v_{\in} = 10 \text{ V}$ ), the ideal diode is forward-biased, acting as a short circuit and clamping the output $v_o$ to $0 \text{ V}$ . During this interval, the capacitor charges until the voltage across it is $V_C = v_{\in} - v_o = 10 - 0 = 10 \text{ V}$ .

When the input pulse is low ( $v_{\in} = 0 \text{ V}$ ), the diode is reverse-biased and acts as an open circuit. The capacitor starts to discharge through the resistor with a time constant $\tau = RC = (500 \text{ k}\Omega)(20 \text{ nF}) = 10 \text{ ms}$ . Since the off-time of the pulse ( $T_{\text{OFF}} = 1 \:\mu\text{s}$ ) is much smaller than the discharging time constant, the capacitor voltage remains practically constant at $10 \text{ V}$ .

The circuit therefore functions as a clamper, with the output voltage given by $v_o = v_{\in} - V_C$ .
The maximum output voltage is $v_{o,max} = 10 \text{ V} - 10 \text{ V} = 0 \text{ V}$ .
The minimum output voltage is $v_{o,min} = 0 \text{ V} - 10 \text{ V} = -10 \text{ V}$ .
The difference is $v_{o,max} - v_{o,min} = 0 - (-10) = 10 \text{ V}$ .

Q54GATE 2021NAT2MElectronic Devices

For the transistor $M_{1}$ in the circuit shown in the figure, $\mu_{n} C_{\text{ox}} = 100\:\mu A/V^{2}$ and $(W/L)=10$ , where $\mu_{n}$ is the mobility of electron, $C_{\text{ox}}$ is the oxide capacitance per unit area , W is the width and L is the length. The channel length modulation coefficient is ignored. If the gate-to-source voltage $V_{\text{GS}}\text{ is } 1\:V$ to keep the transistor at the edge of saturation, then the threshold voltage of the transistor (rounded off to one decimal place) is _______ V.

🚀 Solve in Practice Mode

📖 Explanation

The problem states the transistor is at the edge of saturation, which is the key condition where $V_{DS} = V_{GS} - V_T$ . Given $V_{GS}=1 \text{ V}$ , this simplifies to $V_{DS} = 1 - V_T$ .

Let's find expressions for the drain current $I_D$ and the drain-source voltage $V_{DS}$ from the circuit. The saturation current is $I_D = \frac{1}{2}\mu_n C_{ox}(\frac{W}{L})(V_{GS}-V_T)^2 = \frac{1}{2}(100\mu)(10)(1-V_T)^2 = 0.5(1-V_T)^2$ mA. Applying KVL at the drain, $V_{DS} = V_{DD} - I_D R_D = 3 - (20 \text{ k}\Omega)I_D$ . If $I_D$ is in mA, this becomes $V_{DS} = 3 - 20 I_D$ .

Substitute the expression for $I_D$ into the KVL equation: $V_{DS} = 3 - 20 \times \$ [0.5(1-V_T)^2]$ = 3 - 10(1-V_T)^2 $. Now, equate our two expressions for$ V_{DS} $:$ 1 - V_T = 3 - 10(1-V_T)^2 $. Let's set$ x=1-V_T $to solve this quadratic equation$ 10x^{2+}x-3=0 $, which gives$ x=0.5 $or$ x=-0.6$.

These two solutions for $x$ correspond to $V_T = 0.5$ V and $V_T=1.6$ V. For an NMOS to be on, we must have $V_{GS} > V_T$ . Since $V_{GS}=1$ V, the only valid solution is $V_T=0.5$ V.

Q55GATE 2021NAT2MAnalog Circuits

A circuit with an ideal $\text{OPAMP}$ is shown in the figure. A pulse $V_{\text{IN}}$ of $20\:ms$ duration is applied to the input. The capacitors are initially uncharged. The output voltage $V_{\text{OUT}}$ of this circuit at $t=0^{+}$ (in integer) is _______ V.

🚀 Solve in Practice Mode

📖 Explanation

We need to determine the output voltage at the instant the input pulse begins, which is $t=0^+$ . The capacitors start uncharged, and the voltage across a capacitor cannot change instantaneously. Therefore, at $t=0^+$ , the capacitors act as short circuits.

This behavior directly applies the input voltage to the op-amp's inverting terminal, so $V^- = V_{IN}(0^+) = 5 \text{ V}$ . The non-inverting terminal is connected to ground, meaning $V^+ = 0 \text{ V}$ .

The op-amp functions as a comparator. Since the voltage at the inverting input is greater than at the non-inverting input ( $V^- > V^+$ ), the output is driven to its negative saturation voltage.

Given the supply rails, the negative saturation voltage is $-12 \text{ V}$ . Thus, $V_{OUT} = -12 \text{ V}$ .

Q56GATE 2021NAT2MDigital Circuits

The propagation delay of the exclusive $-\text{OR} (\text{XOR})$ gate in the circuit in the figure is $3\:ns$ . The propagation delay of all the flip-flops is assumed to be zero. The clock $(\text{Clk})$ frequency provided to the circuit is $500\: \text{MHz}$ . Starting from the initial value of the flip-flop outputs $Q_{2}Q_{1}Q_{0} = 1\;1\;1$ with $D_{2}=1$ , the minimum number of triggering clock edges after which the flip-flop outputs $Q_{2}Q_{1}Q_{0}$ becomes $1\; 0\; 0$ (in integer) is

🚀 Solve in Practice Mode

📖 Explanation

The clock frequency of $500 \text{ MHz}$ gives a clock period of $T_{clk} = 1/500 \text{ MHz} = 2 \text{ ns}$ . The XOR gate's propagation delay is $T_{XOR} = 3 \text{ ns}$ . Since the gate delay is longer than the clock period, the output of the XOR gate, $D_2$ , will not be ready for the next clock edge. The value sampled by the flip-flop for $D_2$ at clock edge 'n' is the one calculated from the state existing at clock edge 'n-2'.

Let's trace the state of the flip-flops, $Q_2Q_1Q_0$ , after each triggering clock edge.
Initial state: $111$ . We are given that $D_2=1$ for the first edge.

After edge 1: Inputs $(D_2, D_1, D_0)$ are $(1, Q_2, Q_1) = (1,1,1)$ . The new state is $111$ .
After edge 2: The XOR result from the first cycle ( $1 \oplus 1 = 0$ ) isn't ready. $D_2$ is still the value from before the first edge, which is 1. The inputs are $(1,1,1)$ , so the state remains $111$ .
After edge 3: The value of $D_2$ from two cycles ago (based on state $111$ ) is now ready: $D_2 = 1 \oplus 1 = 0$ . The inputs are $(0,1,1)$ , so the state becomes $011$ .
After edge 4: The value of $D_2$ from two cycles ago (based on state $111$ ) is $D_2 = 1 \oplus 1 = 0$ . The inputs are $(0,0,1)$ , so the state becomes $001$ .
After edge 5: The value of $D_2$ from two cycles ago (based on state $011$ ) is $D_2 = 0 \oplus 1 = 1$ . The inputs are $(1,0,0)$ , so the state becomes $100$ .

The target state of $100$ is reached after 5 clock edges.

Q57GATE 2021NAT2MNetwork Theory

The circuit in the figure contains a current source driving a load having an inductor and a resistor in series, with a shunt capacitor across the load. The ammeter is assumed to have zero resistance. The switch is closed at time t = 0. Initially, when the switch is open, the capacitor is discharged and the ammeter reads zero ampere. After the switch is closed, the ammeter reading keeps fluctuating for some time till it settles to a final steady value. The maximum ammeter reading that one will observe after the switch is closed (rounded off to two decimal places) is _______________ A.

🚀 Solve in Practice Mode

📖 Explanation

This circuit forms a parallel RLC network driven by a 1 A step current. The ammeter measures the current in the RL branch, which exhibits a second-order system response. The key parameter for this response is the damping ratio, which we can calculate as $\xi = \frac{R}{2}\sqrt{\frac{C}{L}}$ . Plugging in the component values, we get a damping ratio of $\xi = 0.25$ .

Since $\xi < 1$ , the system is underdamped, meaning the current will oscillate and overshoot its final value before settling. The final steady-state current is 1 A, as the inductor will act as a short circuit to DC. The maximum fractional overshoot is given by the formula $M_p = e^{-\pi \xi / \sqrt{1-\xi^2}}$ , which yields approximately 0.44.

Therefore, the maximum current observed is the sum of the steady-state value and the overshoot:
$I_{max} = I_{final} + (M_p \times I_{final}) = 1 \text{ A} + (0.44 \times 1 \text{ A}) = 1.44 \text{ A}$ .

Q58GATE 2021NAT2MControl Systems

A unity feedback system that uses proportional-integral (\text{PI} $) control is shown \in the figure. The stability of the overall system is controlled by tuning the$ \text{PI} $control parameters$ K_{P} $and$ K_{I} $. The maximum value of$ K_{I}$ that can be chosen so as to keep the overall system stable or, in the worst case, marginally stable (rounded off to three decimal places) is ______

🚀 Solve in Practice Mode

📖 Explanation

To determine the stability limits, we first find the characteristic equation of the closed-loop system, which is $1 + G(s)H(s) = 0$ . This yields the polynomial $s^4 + 4s^3 + 5s^2 + (2 + 2K_P)s + 2K_I = 0$ . We then apply the Routh-Hurwitz criterion to find the conditions on $K_P$ and $K_I$ for stability.

For all elements in the first column of the Routh array to be positive, we derive the following constraints: $K_P > -1$ , $K_P < 9$ , and $K_I > 0$ . The key stability condition from the array is $K_I < \frac{-K_P^2 + 8K_P + 9}{8}$ . To find the maximum possible value for $K_I$ , we need to maximize this expression with respect to $K_P$ .

By taking the derivative of the expression with respect to $K_P$ and setting it to zero, we find the optimal value is $K_P = 4$ . This value lies within the stable range for $K_P$ . Substituting $K_P=4$ back into the expression gives the maximum value for $K_I$ at the edge of stability:
$K_{I,\text{max}} = \frac{-(4)^2 + 8(4) + 9}{8} = \frac{-16 + 32 + 9}{8} = \frac{25}{8} = 3.125$ .

Q59GATE 2021NAT2MCommunication Systems

A sinusoidal message signal having root mean square value of $4\:V$ and frequency of $\text{1 kHz}$ is fed to a phase modulator with phase deviation constant 2 rad/ volt. If the carrier signal is $c\left ( t \right )=2\cos\left ( 2\pi 10^{6} t\right )$ , the maximum instantaneous frequency of the phase modulated signal (rounded off to one decimal place) is ______ $\text{Hz}$ .

🚀 Solve in Practice Mode

📖 Explanation

The instantaneous frequency of a phase-modulated signal, $f_i(t)$ , is the carrier frequency plus a deviation term: $f_i(t) = f_c + \frac{K_p}{2\pi}\frac{d}{dt}m(t)$ . The maximum instantaneous frequency occurs when the derivative of the message signal, $\frac{d}{dt}m(t)$ , reaches its peak.

First, let's define the message signal, $m(t)$ . Given an RMS voltage of $4$ V and frequency of $1$ kHz, the peak voltage is $V_{peak} = 4\sqrt{2}$ V. So, $m(t) = 4\sqrt{2} \sin(2\pi \cdot 10^3 t)$ .

Next, we find the derivative: $\frac{d}{dt}m(t) = 4\sqrt{2} (2\pi \cdot 10^3) \cos(2\pi \cdot 10^3 t)$ . The maximum value of this derivative is its amplitude, which is $\left[\frac{d}{dt}m(t)\right]_{\max} = 4\sqrt{2}(2\pi \cdot 10^3)$ .

Finally, we substitute the known values into the equation for maximum frequency:
$(f_i)_{\max} = f_c + \frac{K_p}{2\pi}\left[\frac{d}{dt}m(t)\right]_{\max} = 10^6 + \frac{2}{2\pi} \left( 4\sqrt{2} \cdot 2\pi \cdot 10^3 \right)$
$(f_i)_{\max} = 10^6 + 8\sqrt{2} \cdot 10^3 \approx 1,000,000 + 11313.7 = 1,011,313.7 \text{ Hz}$

Q60GATE 2021NAT2MCommunication Systems

Consider a superheterodyne receiver tuned to $600 \text{ kHz}$ . If the local oscillator feeds a $1000 \text{ kHz}$ signal to the mixer, the image frequency (in integer) is ____________________ $\text{kHz}$ .

🚀 Solve in Practice Mode

📖 Explanation

In a superheterodyne receiver, the first step is to calculate the intermediate frequency (IF). This is found by taking the difference between the local oscillator frequency ( $f_{lo}$ ) and the desired signal frequency ( $f_s$ ).
$\text{IF} = f_{lo} - f_s = 1000 \text{ kHz} - 600 \text{ kHz} = 400 \text{ kHz}$
The image frequency ( $f_{si}$ ) is an unwanted signal that, when mixed with the local oscillator, produces the same IF. When the local oscillator is set higher than the signal ( $f_{lo} > f_s$ ), the image frequency is located at a distance of $2 \times \text{IF}$ above the desired signal.
$f_{si} = f_s + 2 \times \text{IF} = 600 + 2(400) = 1400 \text{ kHz}$

Q61GATE 2021NAT2MEngineering Mathematics

In a high school having equal number of boy students and girl students, $75\%$ of the students study Science and the remaining $25\%$ students study Commerce. Commerce students are two times more likely to be a boy than are Science students. The amount of information gained in knowing that a randomly selected girl student studies Commerce (rounded off to three decimal places) is ______ bits.

🚀 Solve in Practice Mode

📖 Explanation

Let's first find the probability that a randomly selected girl studies Commerce, denoted as $P(C|G)$ . We are given that $P(B) = P(G) = 1/2$ , $P(S) = 3/4$ , and $P(C) = 1/4$ . Let's define the probability that a Science student is a boy as $P(B|S) = x$ . The problem states a Commerce student is twice as likely to be a boy, so $P(B|C) = 2x$ .

Using the law of total probability, the overall probability of selecting a boy is $P(B) = P(B|S)P(S) + P(B|C)P(C)$ . Plugging in the values gives $\frac{1}{2} = x(\frac{3}{4}) + 2x(\frac{1}{4})$ , which simplifies to $\frac{1}{2} = \frac{5x}{4}$ , yielding $x = 2/5$ .

This means the probability of a Commerce student being a girl is $P(G|C) = 1 - P(B|C) = 1 - 2x = 1 - 4/5 = 1/5$ . Now we apply Bayes' theorem to find our target probability: $P(C|G) = \frac{P(G|C)P(C)}{P(G)} = \frac{(1/5)(1/4)}{1/2} = \frac{1}{10}$ .

The information gained is given by $I = \log_2\left(\frac{1}{p}\right)$ . Therefore, the information gained is $I = \log_2\left(\frac{1}{1/10}\right) = \log_2(10) \approx 3.322$ bits.

Q62GATE 2021NAT2MCommunication Systems

A message signal having peak-to-peak value of $2\:V$ , root mean square value of $0.1\:V$ and bandwidth of $\text{5 kHz}$ is sampled and fed to a pulse code modulation $(\text{PCM})$ system that uses a uniform quantizer. The $\text{PCM}$ output is transmitted over a channel that can support a maximum transmission rate of $\text{50 kbps}$ . Assuming that the quantization error is uniformly distributed, the maximum signal to quantization noise ratio that can be obtained by the $\text{PCM}$ system (rounded off to two decimal places) is ______

🚀 Solve in Practice Mode

📖 Explanation

To find the maximum signal-to-quantization-noise ratio (SQNR), we must calculate the ratio of the signal power to the minimum possible quantization noise power. The signal power, $S$ , is the square of the RMS value: $S = (0.1)^2 = 0.01$ .

Minimizing the noise power requires using the maximum number of bits per sample, $n$ . The bit rate, $R_b = n \times f_s$ , is limited by the channel capacity of $50 \text{ kbps}$ . Given the signal bandwidth of $5 \text{ kHz}$ , the minimum (Nyquist) sampling rate is $f_s = 2 \times 5 \text{ kHz} = 10 \text{ kHz}$ . The channel constraint thus becomes $n \times 10 \text{ kbps} \le 50 \text{ kbps}$ , which means the maximum number of bits per sample is $n_{max} = 5$ .

The quantization noise power is $N_Q = \frac{\Delta^2}{12}$ . We find the minimum step size, $\Delta_{min}$ , using $n_{max}$ and the peak-to-peak voltage: $\Delta_{min} = \frac{V_{p-p}}{2^{n_{max}}} = \frac{2}{2^5} = \frac{1}{16}$ . This gives a minimum noise power of $(N_Q)_{min} = \frac{(1/16)^2}{12} = \frac{1}{3072}$ .

Finally, the maximum SQNR is $\left(\frac{S}{N_Q}\right)_{max} = \frac{S}{(N_Q)_{min}} = \frac{0.01}{1/3072} = 30.72$ .

Q63GATE 2021NAT2MCommunication Systems

Consider a polar non-return to zero $(\text{NRZ})$ waveform, using $+2\:V$ and $-2\:V$ for representing binary '1' and '0' respectively, is transmitted in the presence of additive zero-mean white Gaussian noise with variance $0.4\:V^{2}$ . If the a priori probability of transmission of a binary '1' is 0.4, the optimum threshold voltage for a maximum a posteriori $(\text{MAP})$ receiver (rounded off to two decimal places) is ______ V.

🚀 Solve in Practice Mode

📖 Explanation

The optimal decision threshold for a MAP receiver is the voltage where the posterior probabilities of sending a '0' or a '1' are equal. For binary signals in Gaussian noise, this threshold, $V_{Th}$ , is calculated as:
$V_{Th} = \frac{s_1+s_0}{2} + \frac{\sigma^2}{s_1-s_0}\ln\left(\frac{P(0)}{P(1)}\right)$
From the problem, the signal levels are $s_1 = +2$ V and $s_0 = -2$ V. The prior probabilities are $P(1)=0.4$ and $P(0)=0.6$ , and the noise variance is $\sigma^2=0.4$ .

The first term, $\frac{s_1+s_0}{2} = \frac{2-2}{2} = 0$ , represents the midpoint voltage. The threshold is shifted from this midpoint because the probabilities are unequal.

Plugging the values into the formula:
$V_{Th} = 0 + \frac{0.4}{2 - (-2)} \ln\left(\frac{0.6}{0.4}\right)$
$V_{Th} = \frac{0.4}{4} \ln(1.5) = 0.1 \times 0.4054 \approx 0.0405 \text{ V}$
The threshold is slightly positive, biased towards the less probable symbol ('1') to minimize the overall error probability.

Q64GATE 2021NAT2MElectromagnetics

A standard air-filled rectangular waveguide with dimensions $\text{a=8 cm, b=4 cm}$ , operates at $\text{3.4 GHz}$ . For the dominant mode of wave propagation, the phase velocity of the signal in $v_{p}$ . The value (rounded off to two decimal places) of $v_{p}/c$ , where c denotes the velocity of light, is _____

🚀 Solve in Practice Mode

📖 Explanation

First, we must find the cutoff frequency ( $f_c$ ) for the dominant TE $_{10}$ mode, which is the lowest frequency that can propagate through the waveguide. This frequency is determined by the wider dimension, $a$ , using the formula $f_c = c/(2a)$ . Given $a=8$ cm, we calculate $f_c = (3 \times 10^8) / (2 \times 0.08) = 1.875$ GHz.

Since the operating frequency ( $f = 3.4$ GHz) is above the cutoff frequency, the wave will propagate. The phase velocity ( $v_p$ ) is related to the speed of light ( $c$ ) by how far the operating frequency is from the cutoff. The ratio $v_p/c$ is given by the equation:
$\frac{v_p}{c} = \frac{1}{\sqrt{1 - (f_c/f)^2}}$
Substituting our calculated and given frequencies:
$\frac{v_p}{c} = \frac{1}{\sqrt{1 - (1.875/3.4)^2}} \approx 1.198$

Q65GATE 2021NAT2MElectromagnetics

An antenna with a directive gain of $6\text{ dB}$ is radiating a total power of $\text{16 kW}$ . The amplitude of the electric field in free space at a distance of $\text{8 km}$ from the antenna in the direction of $\text{6 dB}$ gain (rounded off to three decimal places) is ______ $\text{V/m}$ .

🚀 Solve in Practice Mode

📖 Explanation

First, we must convert the antenna's directive gain from decibels (dB) to a linear ratio for our calculation. Using the relationship $G_{dB} = 10 \log_{10}(G_d)$ , a $6 \text{ dB}$ gain corresponds to a linear gain of $G_d = 10^{6/10} = 10^{0.6} \approx 3.981$ .

Next, we can find the electric field amplitude, $E_m$ , using the formula that relates it to gain, radiated power ( $P_{\text{rad}}$ ), and distance ( $r$ ):
$E_m = \frac{\sqrt{60 G_d P_{\text{rad}}}}{r}$

Substituting the given values, $P_{\text{rad}} = 16 \times 10^3 \text{ W}$ and $r = 8 \times 10^3 \text{ m}$ , along with our calculated linear gain, we get:
$E_m = \frac{\sqrt{60 \times 3.981 \times (16 \times 10^3)}}{8 \times 10^3} \approx 0.244 \text{ V/m}$

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