GATE ECE 2017 Set 2

Q1GATE 2017MCQ1MDigital Circuits

Consider the circuit shown in the figure. The Boolean expression F implemented by the circuit is

📖 Explanation

To determine the Boolean expression for the output $F$ , let's analyze the circuit in two stages.

First, examine the left multiplexer. With $X$ as the select line, the output (let's call it $F_1$ ) is $Y$ when $X=0$ and $0$ when $X=1$ . This can be written as the expression $F_1 = \bar{X}Y$ .

Next, this signal $F_1$ is an input to the second multiplexer. This MUX has $Z$ as its select line. Its output $F$ is $F_1$ when $Z=0$ , and the inverse of $F_1$ (due to the NOT gate) when $Z=1$ . This gives us $F = \bar{Z}F_1 + Z\bar{F_1}$ .

Now, substitute the expression for $F_1$ into this equation: $F = \bar{Z}(\bar{X}Y) + Z(\overline{\bar{X}Y})$ .
Applying De Morgan's theorem to the inverted term gives $\overline{\bar{X}Y} = X+\bar{Y}$ .
Plugging this back in, we get $F = \bar{X}Y\bar{Z} + (X+\bar{Y})Z$ . Distributing the $Z$ term yields the final expression: $F = \bar{X}Y\bar{Z} + XZ + \bar{Y}Z$ .

Q2GATE 2017MCQ1MSignals and Systems

An LTI system with unit sample response h[n]= $5\delta [n]-7\delta [n-1]+7\delta [n-3]-5\delta [n-4]$ is a

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📖 Explanation

To understand the filter's behavior, we analyze its frequency response, $H(e^{j\omega})$ , which is the DTFT of the impulse response $h[n]$ .
The frequency response is $H(e^{j\omega}) = 5 - 7e^{-j\omega} + 7e^{-j3\omega} - 5e^{-j4\omega}$ .

Let's test the response at the extreme ends of the frequency spectrum.
At the lowest frequency, $\omega = 0$ (DC):
$H(e^{j0}) = 5 - 7(1) + 7(1) - 5(1) = 0$ .

At the highest frequency, $\omega = \pi$ :
$H(e^{j\pi}) = 5 - 7e^{-j\pi} + 7e^{-j3\pi} - 5e^{-j4\pi} = 5 - 7(-1) + 7(-1) - 5(1) = 5+7-7-5 = 0$ .

Since the filter's gain is zero at both the lowest and highest frequencies, it must be attenuating these extremes while passing a band of frequencies in the middle. This is the characteristic behavior of a band-pass filter.

Q3GATE 2017NAT1MNetwork Theory

In the circuit shown, V is a sinusoidal voltage source. The current I is in phase with voltage V. The ratio $\frac{amplitude \; of \; voltage \; across \; the \; capacitor} {amplitude \; of \; voltage \; across \; the \; resistor}$ is

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📖 Explanation

The problem states that the source voltage $V$ and the circuit current $I$ are in phase. This is the defining condition for resonance in a series RLC circuit, meaning the net reactance is zero.

The ratio we need to find, the voltage amplitude across the capacitor ( $V_C$ ) to the voltage amplitude across the resistor ( $V_R$ ), is equal to the circuit's quality factor, $Q$ .

The quality factor is calculated using the formula $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$ .

Substituting the given component values, $R = 5\,\Omega$ , $L = 5\,\text{H}$ , and $C = 5\,\text{F}$ :

$\frac{V_C}{V_R} = Q = \frac{1}{5} \sqrt{\frac{5}{5}} = \frac{1}{5} \times \sqrt{1} = 0.2$

Q4GATE 2017MCQ1MDigital Circuits

In a DRAM,

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Q5GATE 2017NAT1MElectronic Devices

Consider an n-channel MOSFET having width W, length L, electron mobility in the channel $\mu_{n}$ and oxide capacitance per unit area $C_{ox}$ . If gate-to-source voltage $V_{GS}$ =0.7V, drain-tosource voltage $V_{DS}=0.1V,(\mu_{n}C_{ox})=100\mu A/V^{2}$ ,threshold voltage $V_{TH}=0.3 V$ and (W/L)=50,then the transconductance $g_{m}$ (in mA/V) is ___________.

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📖 Explanation

First, we need to identify the MOSFET's region of operation. The overdrive voltage is $V_{GS} - V_{TH} = 0.7V - 0.3V = 0.4V$ . Since this overdrive voltage is greater than the drain-to-source voltage, $V_{DS} = 0.1V$ , the MOSFET is operating in the linear (triode) region.

In the linear region, the drain current $I_D$ is given by the equation:
$I_D = \mu_n C_{ox} \frac{W}{L} \left[ (V_{GS} - V_{TH})V_{DS} - \frac{V_{DS}^2}{2} \right]$

Transconductance, $g_m$ , is found by taking the partial derivative of $I_D$ with respect to $V_{GS}$ :
$g_m = \frac{\partial I_D}{\partial V_{GS}} = \mu_n C_{ox} \frac{W}{L} V_{DS}$

Now, we can substitute the given values into this expression for $g_m$ :
$g_m = (100 \mu A/V^2) \times (50) \times (0.1 V)$
$g_m = 500 \mu A/V = 0.5 \text{ mA/V}$

Q6GATE 2017MCQ1MElectromagnetics

Two conducting spheres S1 and S2 of radii a and b (b>a) respectively, are placed far apart and connected by a long, thin conducting wire, as shown in the figure. For some charge placed on this structure, the potential and surface electric field on S1 are $V_{a}$ and $E_{a}$ , and that on S2 are $V_{b}$ and $E_{b}$ , respectively, which of the following is CORRECT?

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📖 Explanation

Because the two spheres are connected by a conducting wire, they form a single system that must be at a uniform potential once electrostatic equilibrium is reached. Therefore, the potential on sphere S1 must equal the potential on sphere S2, so $V_a = V_b$ .

Using the formula for potential on a sphere, $V = \frac{Q}{4\pi\epsilon_0 R}$ , the equality $V_a=V_b$ implies $\frac{Q_a}{a} = \frac{Q_b}{b}$ . Now, let's compare the electric fields using the formula $E = \frac{Q}{4\pi\epsilon_0 R^2}$ . The ratio of the fields is $\frac{E_a}{E_b} = \frac{Q_a/a^2}{Q_b/b^2} = \frac{Q_a}{Q_b} \cdot \frac{b^2}{a^2}$ . Substituting $\frac{Q_a}{Q_b} = \frac{a}{b}$ from our potential analysis, we get $\frac{E_a}{E_b} = \frac{a}{b} \cdot \frac{b^2}{a^2} = \frac{b}{a}$ . Since we are given that $b > a$ , the ratio $\frac{b}{a}$ is greater than 1, which means $E_a > E_b$ .

Q7GATE 2017MCQ1MDigital Circuits

For the circuit shown in the figure, P and Q are the inputs and Y is the output. The logic implemented by the circuit is

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📖 Explanation

Let's analyze the circuit's operation based on the input P. When $P=0$ , the top PMOS transistor turns on, connecting the output Y to the input Q. When $P=1$ , the bottom NMOS transistor turns on, connecting Y to the inverted input $\bar{Q}$ .

Combining these conditions yields the Boolean expression $Y = \bar{P}Q + P\bar{Q}$ . This is the definition of the Exclusive OR (XOR) function.

However, this pass-transistor logic design has a known flaw. An NMOS transistor passes a "weak" logic 1 (a degraded voltage), and a PMOS passes a "weak" logic 0. Because the circuit produces degraded output levels for some input combinations, it is a poor implementation. This ambiguity regarding its practical performance is the likely reason the question was invalidated in its original context.

Q8GATE 2017MCQ1MAnalog Circuits

An n-channel enhancement mode MOSFET is biased at $V_{GS}>V_{TH} \; and \; V_{DS}>(V_{GS}-V_{TH})$ , where $V_{GS}$ is the gate-to-source voltage, $V_{DS}$ is the drain-to-source voltage and $V_{TH}$ is the threshold voltage. Considering channel length modulation effect to be significant, the MOSFET behaves as a

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📖 Explanation

The bias conditions $V_{GS} > V_{TH}$ and $V_{DS} > (V_{GS} - V_{TH})$ place the MOSFET in the saturation region of operation. In this mode, the device primarily acts as a current source, where the drain current is controlled by the gate-to-source voltage, $V_{GS}$ .

If channel length modulation were ignored, the output impedance would be infinite. However, the problem states this effect is significant. Channel length modulation causes the drain current to have a slight dependence on the drain-to-source voltage, $V_{DS}$ . This small change in current for a change in voltage is precisely what defines a finite output impedance ( $r_o$ ). Therefore, the MOSFET behaves as a current source with a finite output impedance.

Q9GATE 2017NAT1MNetwork Theory

A connection is made consisting of resistance A in series with a parallel combination of resistances B and C. Three resistors of value 10 $\Omega$ , 5 $\Omega$ , 2 $\Omega$ are provided. Consider all possible permutations of the given resistors into the positions A, B, C, and identify the configurations with maximum possible overall resistance, and also the ones with minimum possible overall resistance. The ratio of maximum to minimum values of the resistances (up to second decimal place) is ____________.

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📖 Explanation

The total resistance of the circuit is given by the formula $R_{total} = R_A + \frac{R_B R_C}{R_B + R_C}$ .

To maximize the total resistance, we should select the largest resistor for the series component, $R_A$ , as it adds its full value. Thus, for the maximum configuration, $R_A = 10 \Omega$ , and $R_{max} = 10 + \frac{5 \times 2}{5+2} = \frac{80}{7} \Omega$ .

Conversely, to minimize the total resistance, we should choose the smallest resistor for the series part. Setting $R_A = 2 \Omega$ gives the minimum configuration: $R_{min} = 2 + \frac{10 \times 5}{10+5} = \frac{16}{3} \Omega$ .

Finally, the ratio of the maximum to minimum resistance is $\frac{R_{max}}{R_{min}} = \frac{80/7}{16/3} = \frac{15}{7} \approx 2.143$ .

Q10GATE 2017MCQ1MElectronic Devices

An npn bipolar junction transistor (BJT) is operating in the active region. If the reverse bias across the base - collector junction is increased, then

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📖 Explanation

In the active region, the collector-base junction is reverse-biased. Increasing this reverse bias voltage widens the depletion region, which extends into the base and reduces its effective width. This phenomenon is called the Early effect or base-width modulation.

A narrower base means that electrons injected from the emitter travel a shorter distance to reach the collector. This reduces the probability that they will recombine with holes in the base. Since base current ( $I_B$ ) is largely due to this recombination, a smaller recombination rate leads to a smaller $I_B$ for a given collector current ( $I_C$ ). As a result, the common-emitter current gain, $\beta = I_C/I_B$ , increases.

Q11GATE 2017NAT1MControl Systems

Consider the state space realization

\begin{bmatrix} x_{1}(t)\\ x_{2}(t) \end{bmatrix}=\begin{bmatrix} 0 & 0\\ 0&-9 \end{bmatrix} \begin{bmatrix} x_{1}(t)\\ x_{2}(t) \end{bmatrix}+ \begin{bmatrix} 0\\ 45 \end{bmatrix}u(t)

, with the initial condition

\begin{bmatrix} x_{1}(0)\\ x_{2}(0) \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}

where u(t) denotes the unit step function. The value of $\lim_{t\rightarrow \infty }|\sqrt{x_{1}^{2}(t)+x_{2}^{2}(t)}|$ is _______.

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📖 Explanation

Let's analyze the system by examining each state equation individually from the given matrix form.

The first state equation is $\dot{x}_1(t) = 0$ . Since the initial condition is $x_1(0)=0$ , the state $x_1(t)$ will remain at 0 for all time.

This simplifies the expression we need to evaluate: $\lim_{t\rightarrow \infty }|\sqrt{x_{1}^{2}(t)+x_{2}^{2}(t)}| = \lim_{t\rightarrow \infty }|\sqrt{0+x_{2}^{2}(t)}| = |\lim_{t\rightarrow \infty }x_{2}(t)|$ . We only need to find the steady-state value of $x_2(t)$ .

The second state equation is $\dot{x}_2(t) = -9x_2(t) + 45u(t)$ . Taking the Laplace transform with $x_2(0)=0$ yields $sX_2(s) = -9X_2(s) + \frac{45}{s}$ , which gives $X_2(s) = \frac{45}{s(s+9)}$ .

Using the Final Value Theorem to find the steady-state value:
$\lim_{t \to \infty } x_2(t) = \lim_{s \to 0} sX_2(s) = \lim_{s \to 0} s\left(\frac{45}{s(s+9)}\right) = \frac{45}{9} = 5$ .

Therefore, the final value of the expression is $|5| = 5$ .

Q12GATE 2017NAT1MEngineering Mathematics

The rank of the matrix

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

is ___________.

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📖 Explanation

The rank of a matrix is its number of linearly independent rows. We can find this by using row operations to convert the matrix into row echelon form and then counting the non-zero rows.

A key observation here is that the sum of the first four rows ( $R_1+R_2+R_3+R_4$ ) equals the negative of the fifth row ( $-R_5$ ). This means $R_1+R_2+R_3+R_4+R_5 = \mathbf{0}$ , proving the rows are linearly dependent and the rank must be less than 5.

Let's formally demonstrate this. We can combine the first four rows into the fourth row through a series of operations ( $R_4 \to R_4+R_1$ , then $R_4 \to R_4+R_3$ , then $R_4 \to R_4+R_2$ ). This yields:
$\begin{bmatrix} 1& -1 & 0& 0&0 \\ 0& 0& 1 &-1 &0 \\ 0 & 1 &-1 & 0&0 \\ 0 & 0& 0& -1 &1 \\ 0 & 0 & 0 &1 &-1 \end{bmatrix}$
Now, adding the new fourth row to the fifth ( $R_5 \to R_5+R_4$ ) creates a zero row. To finalize the echelon form, we swap the second and third rows ( $R_2 \leftrightarrow R_3$ ):
$\begin{bmatrix} 1& -1 & 0& 0&0 \\ 0& 1 & -1 & 0&0 \\ 0& 0& 1 &-1 &0 \\ 0 & 0& 0& -1 &1 \\ 0 & 0 & 0 &0 &0 \end{bmatrix}$
The resulting matrix has four non-zero rows, so the rank is 4.

Q13GATE 2017NAT1MElectromagnetics

A two - wire transmission line terminates in a television set. The VSWR measured on the line is 5.8. The percentage of power that is reflected from the television set is ______________

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📖 Explanation

The Voltage Standing Wave Ratio (VSWR) quantifies the mismatch between the transmission line and the load. We can use it to find the magnitude of the reflection coefficient, $|\Gamma|$ , which represents the fraction of the incident wave that is reflected. The governing formula is:

$|\Gamma| = \frac{\text{VSWR} - 1}{\text{VSWR} + 1}$

Substituting the given VSWR of 5.8, we get:

$|\Gamma| = \frac{5.8 - 1}{5.8 + 1} = \frac{4.8}{6.8} \approx 0.7059$

The fraction of power reflected is the square of the reflection coefficient's magnitude, $|\Gamma|^2$ . To express this as a percentage:

$\text{Percentage of Reflected Power} = |\Gamma|^2 \times 100 = (0.7059)^2 \times 100 \approx 49.83\%$

Q14GATE 2017MCQ1MSignals and Systems

The input $x(t)$ and the output $y (t)$ of a continuous-time system are related as $y(t)=\int_{t-T}^{t}x(u)du$ . The system is

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📖 Explanation

This system describes a moving average, calculating the integral of the input signal over a sliding window of fixed duration $T$ . The system is linear because the integral operator itself is linear, satisfying the principles of superposition and homogeneity.

To test for time-invariance, we compare two scenarios. First, a time-shifted output is $y(t-t_0) = \int_{t-t_0-T}^{t-t_0}x(u)du$ . Second, the output for a time-shifted input $x(t-t_0)$ is $\int_{t-T}^{t}x(u-t_0)du$ . By substituting a new variable $\tau=u-t_0$ , this second integral becomes $\int_{t-t_0-T}^{t-t_0}x(\tau)d\tau$ . This result is identical to the shifted output $y(t-t_0)$ . Since a shift in the input produces an identical shift in the output, the system is time-invariant.

Q15GATE 2017MCQ1MControl Systems

Which of the following statements is incorrect?

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Q16GATE 2017MCQ1MEngineering Mathematics

The residues of a function $f(z)=\frac{1}{(z-4)(z+1)}$ are

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📖 Explanation

Residue at, z=4 is $=\lim _{x \rightarrow 1}(z-4) \frac{1}{(z-4)(z+1)^{3}}=\frac{1}{(4+1)^{3}}=\frac{1}{125}$ Residue at z=-1 is $=\lim _{z \rightarrow-1} \frac{1}{2 !} \frac{d^{2}}{d z^{2}}\left((z-1)^{3} \frac{1}{(z-4)(z+1)^{3}}\right)$ $=\lim _{z \rightarrow-1} \frac{1}{2 !}\left(\frac{2}{(z-4)^{3}}\right)=\frac{1}{(-1-4)^{3}}=\frac{-1}{125}$

Q17GATE 2017NAT1MCommunication Systems

A sinusoidal message signal is converted to a PCM signal using a uniform quantizer. The required signal-to-quantization noise ratio (SQNR) at the output of the quantizer is 40dB. The minimum number of bits per sample needed to achieve the desired SQNR is _______

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📖 Explanation

For sinusoidal input is applied to a PCM system, $\mathrm{SQNR}=6 n+1.8$ n= number of bits per sample Given that, the required $\mathrm{SQNR}=40 \mathrm{dB}$

\begin{aligned} \mathrm{so}, 6 n+1.8 &\geq 40 \\ n &\geq \frac{38.2}{6}\\ n_{\min }&=7 \\ \because \; &n \text{ is always an integer value} \end{aligned}

Q18GATE 2017MCQ1MEngineering Mathematics

The general solution of the differential equation $\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}-5y=0$ in terms of arbitrary constants $K_{1} \; and \; K_{2}$ is

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📖 Explanation

The general solution of the differential equation

\begin{aligned} \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-5 y&=0 \\ \left(D^{2}+2 D-5\right) y&=0 \end{aligned}

Auxillary equation is

\begin{aligned} m^{2}+2 m-5 &=0 \\ m &=\frac{-2 \pm \sqrt{4+20}}{2}=\frac{-2 \pm \sqrt{24}}{2} \\ &=\frac{-2 \pm 2 \sqrt{6}}{2}=-1 \pm \sqrt{6} \\ \text { Solution is } \quad y &=K_{1} e^{(-1+\sqrt{6}) x}+K_{2} e^{(-1-\sqrt{6}) x} \end{aligned}

Q19GATE 2017MCQ1MCommunication Systems

Which one of the following graphs shows the Shannon capacity (channel capacity) in bits of a memory less binary symmetric channel with crossover probability P?

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📖 Explanation

The Shannon capacity $C$ of a memoryless binary symmetric channel is defined by the crossover probability $p$ . The formula is $C(p) = 1 - H(p)$ , where $H(p)$ is the binary entropy function, which can be written out as $C(p) = 1 + p\log_2(p) + (1-p)\log_2(1-p)$ .

Let's test some key values of $p$ . When the channel is perfect ( $p=0$ ) or perfectly predictable ( $p=1$ ), the capacity is maximal, $C=1$ . When the channel is most noisy and random ( $p=0.5$ ), the capacity is minimal, $C=0$ . The logarithmic terms mean the graph is a smooth curve, not linear. This describes a convex, U-shaped curve that is symmetric about $p=0.5$ .

Q20GATE 2017NAT1MAnalog Circuits

The output $V_{0}$ of the diode circuit shown in the figure is connected to an averaging DC voltmeter. The reading on the DC voltmeter in Volts, neglecting the voltage drop across the diode, is ____________.

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📖 Explanation

This circuit is a classic half-wave rectifier. Since we're told to neglect the diode's voltage drop, it acts as a perfect switch, passing the positive half-cycles of the input sine wave to the output and blocking the negative ones.

An averaging DC voltmeter measures the average DC value of the signal it's connected to. For a half-wave rectified sine wave, the average value is given by the formula $V_{avg} = \frac{V_{m}}{\pi}$ , where $V_{m}$ is the peak amplitude of the wave.

The input voltage is specified as $10 \sin \omega t$ , so its peak value $V_m$ is $10$ V. Therefore, the reading on the DC voltmeter will be:
$V_{reading} = \frac{10}{\pi} \approx 3.183 \text{ V}$

Q21GATE 2017NAT1MCommunication Systems

Consider the random process X(t)=U+Vt, where U is a zero-mean Gaussian random variable and V is a random variable uniformly distributed between 0 and 2. Assume that U and V are statistically independent. The mean value of the random process at t = 2 is ____________

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📖 Explanation

\begin{aligned} X(t) &=U+V t \\ \text { At } t=2, \quad X(t) &=X(2)=U+2 V \\ E[X(t)] &=E[U+2 V] \\ &=E[U]+2 E[V]\\ \text{Given that,}\\ E[U]&=0\\ \end{aligned}

Also given that, V is uniformly distributed between 0 and 2

\begin{aligned} \text{So, } \quad E[V]&=\int_{-\infty }^{\infty } f_{v}(V) d v=\int_{0}^{2}\left(\frac{1}{2}\right) d v=1 \\ \text{So, } \quad E[X(f)]&=0+2(1)=2 \end{aligned}

Q22GATE 2017NAT1MControl Systems

For the system shown in the figure, Y(s)/X(s)= __________.

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📖 Explanation

To determine the system's transfer function, we start by writing an algebraic expression for the output signal, $Y(s)$ . By tracing the signals through the block diagram, we can see how $Y(s)$ is constructed.

The output $Y(s)$ is the sum of two signals at the rightmost summing junction:

The signal from the gain block, which is $G(s)[X(s) - Y(s)]$ .
A direct feedforward signal from the input, which is $X(s)$ .

Combining these gives the equation: $Y(s) = G(s)[X(s) - Y(s)] + X(s)$ .
Our goal is to find the ratio $Y(s)/X(s)$ , so we rearrange the terms by grouping $Y(s)$ on the left and $X(s)$ on the right:
$Y(s) + G(s)Y(s) = G(s)X(s) + X(s)$
Factoring out $Y(s)$ and $X(s)$ gives: $Y(s)[1 + G(s)] = X(s)[G(s) + 1]$ .
Finally, solving for the transfer function $\frac{Y(s)}{X(s)}$ shows that the terms cancel, resulting in a value of 1.

Q23GATE 2017NAT1MEngineering Mathematics

The smaller angle (in degrees) between the planes x + y + z =1 and 2x - y + 2z = 0 is ________.

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📖 Explanation

If $\theta$ is the angle between

\begin{array}{l} a_{1} x+b_{1} y+c_{1} z+d_{1} =0 \\ a_{2} x+b_{2} y+c_{2} z+d_{2} =0 \text { then }\\ \cos \theta=\frac{a_{1} a_{2}+b_{4} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ a_{1}=1 \quad b_{1}=1 \quad c_{1}=1 \quad d_{1}=-1 \\ a_{2}=2 \quad b_{2}=-1 \quad c_{2}=2 \quad d_{2}=0\\ \cos \theta=\frac{(1)(2)+(1)(-1)+(1)(2)}{\sqrt{(1)^{2}+(1)^{2}+(1)^{2}} \sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}}\\ =\frac{2-1+2}{\sqrt{3} \sqrt{9}}=\frac{3}{3 \sqrt{3}}=\frac{1}{\sqrt{3}}\\ \theta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)=54.73^{\circ} \end{array}

Q24GATE 2017NAT1MAnalog Circuits

Consider the circuit shown in the figure. Assume base-to- emitter voltage $V_{BE}$ =0.8 V and common base current gain $(\alpha)$ of the transistor is unity. The value of the collector- to - emitter voltage $V_{CE}$ (in volt) is _______.

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📖 Explanation

To analyze this circuit, we first simplify the base voltage divider network by finding its Thevenin equivalent voltage, $V_{Th}$ . This voltage is $V_{Th} = \frac{16 \text{ k}\Omega}{16 \text{ k}\Omega + 44 \text{ k}\Omega} \times 18 \text{ V} = 4.8 \text{ V}$ .

Since the common base current gain $\alpha=1$ , it implies an infinite common emitter gain ( $\beta$ ), making the base current $I_B$ effectively zero. This also means the collector current $I_C$ is equal to the emitter current $I_E$ .

Applying KVL around the base-emitter loop, we get $V_{Th} = V_{BE} + I_E R_E$ . We can now find the emitter current: $I_E = \frac{V_{Th} - V_{BE}}{R_E} = \frac{4.8 \text{ V} - 0.8 \text{ V}}{2 \text{ k}\Omega} = 2 \text{ mA}$ .

Since $I_C = I_E = 2 \text{ mA}$ , we find the collector-to-emitter voltage, $V_{CE}$ , by applying KVL on the output loop:
$V_{CE} = V_{CC} - I_C R_C - I_E R_E = 18 \text{ V} - (2 \text{ mA} \times 4 \text{ k}\Omega) - (2 \text{ mA} \times 2 \text{ k}\Omega) = 18 - 8 - 4 = 6 \text{ V}$ .

Q25GATE 2017MCQ1MAnalog Circuits

In the figure, D1 is a real silicon pn junction diode with a drop of 0.7V under forward bias condition and D2 is a zener diode with breakdown voltage of -6.8 V. The input [latex ]V_{in}(t)[/latex] is a periodic square wave of period T, whose one period is shown in the figure. Assuming 10 $\tau \lt \lt T$ , where $\tau$ is the time constant of the circuit, the maximum and minimum values of the output waveform are respectively,

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📖 Explanation

This circuit is a biased clamper. The capacitor charges to its steady-state voltage during the positive input swing when both diodes conduct. This occurs when the silicon diode D1 is forward-biased ( $0.7$ V) and the Zener diode D2 is in breakdown ( $6.8$ V), clamping the output at their combined voltage. The maximum output voltage is therefore $V_{out,max} = 0.7\text{V} + 6.8\text{V} = 7.5\text{V}$ .

At the moment of clamping, when the input is at its peak ( $V_{\in} = +14\text{V}$ ), we can determine the capacitor's steady-state voltage, $V_C$ , using KVL: $V_{\in,max} = V_C + V_{out,max}$ . This gives $V_C = 14\text{V} - 7.5\text{V} = 6.5\text{V}$ .

During the negative input swing ( $V_{\in} = -14\text{V}$ ), the diodes are reverse-biased and act like an open circuit. The output voltage is then given by $V_{out}(t) = V_{\in}(t) - V_C$ . The minimum output voltage is thus $V_{out,min} = V_{\in,min} - V_C = -14\text{V} - 6.5\text{V} = -20.5\text{V}$ .

Q26GATE 2017MCQ2MEngineering Mathematics

If the vector function $\bar{F}=\hat{a_{x}}(3y-k_{1}z)+\hat{a_{y}}(k_{2}x-2z)-\hat{a_{z}}(k_{3}x+z)$ is irrotational, then the values of the constants $k_{1}, k_{2}$ and $k_{3}$ respectively, are

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \vec{F}=& \hat{a}_{x}\left(3 y-k_{1} z\right)+\hat{a}_{y}\left(k_{2} x-2 z\right)-\hat{a}_{2}\left(k_{3} y+z\right) \\ & \nabla \times \vec{F}=0 \quad \text { (irrotational) } \\ & \quad r \times \vec{F}=\left| \begin{array}{ccc} \dot{a}_{x} & \hat{a}_{y} & \hat{a}_{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 3 y-k_{1} z & k_{2} x-2 z & -\left(k_{3} y+z\right) \end{array}\right| \\ \end{aligned}

\begin{aligned} =\hat{a}_{x}\left[\frac{\partial}{\partial y}\left[-\left(k_{3} y+z\right)\right]-\frac{\partial}{\partial z}\left(k_{2} x-2 z\right)\right] \\ -\hat{a}_{y}\left[\frac{\partial}{\partial x}\left[-\left(k_{3} y+z\right)\right]-\frac{\partial}{\partial z}\left(3 y-k_{1} z\right)\right] \\ +\hat{a}_{z}\left[\frac{\partial}{\partial x}\left(k_{2} x-2 z\right)-\frac{\partial}{\partial y}\left(3 y-k_{1} z\right)\right] \\ \hat{a}_{x}\left[-k_{3}+2\right]-\hat{a}_{y}\left[k_{1}\right]+\hat{a}_{z}\left[k_{2}-3\right]=0 \\ \Rightarrow \quad k_{3}=2, k_{1}=0, k_{2}=3 \\ \text{or}\quad k_{1}=0, k_{2}=3, k_{3}=2 \end{aligned}

Q27GATE 2017NAT2MCommunication Systems

The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \text { Given that, } P_{C}&=5 \mathrm{kW} \text { for } \mu_{(\mathrm{max})}=0.5\\ \text { So, } \quad P_{t(\max )}&=P_{c}\left[1+\frac{(0.5)^{2}}{2}\right]\$\\ &=5\left[1+\frac{0.25}{2}\right] \mathrm{kW}=5.625 \mathrm{kW} \\ \text { For } \mu & =0.4,\\ &\begin{aligned} P_{ \text {Qrex } )}\left(1+\frac{\mu^{2}}{2}\right)=&P_{t(\max )} \\ P_{C(\text{rax})}=& \frac{5.625}{1+\frac{(0.4)^{2}}{2}} \mathrm{kW} \\ & =5.208 \mathrm{kW} \end{aligned} \end{aligned}

Q28GATE 2017NAT2MSignals and Systems

Consider an LTI system with magnitude response

|H(f)|=\left\{\begin{matrix} 1-\frac{|f|}{20} & |f|\leq 20\\ 0 & |f|> 20 \end{matrix}\right.

And phase response Arg {H(f)}=-2f If the input to the system is $x(t)=8cos(20\pi t+\frac{\pi }{4})+16sin(40\pi t+\frac{\pi }{8})+24cos(80\pi t+\frac{\pi }{16})$ Then the average power of the output signal y(t) is _________.

🚀 Solve in Practice Mode

📖 Explanation

The input signal $x(t)$ is a sum of three sinusoids with frequencies $f_1=10$ Hz, $f_2=20$ Hz, and $f_3=40$ Hz. We must examine how the LTI system's filter affects each component.

The filter's magnitude response $|H(f)|$ is zero for all frequencies $|f| \geq 20$ Hz. Therefore, the input components at $20$ Hz and $40$ Hz are completely attenuated, and only the $10$ Hz component passes through.

At $f=10$ Hz, the filter's gain is $|H(10)| = 1 - \frac{|10|}{20} = 0.5$ .
The original amplitude of the $10$ Hz component is $8$ . The output amplitude is this original amplitude scaled by the filter's gain: $A_{out} = 8 \times |H(10)| = 8 \times 0.5 = 4$ .

The output signal $y(t)$ is thus a single sinusoid with an amplitude of $4$ . The average power of this signal is given by $P_y = \frac{A_{out}^2}{2} = \frac{4^2}{2} = 8$ W.

Q29GATE 2017NAT2MElectronic Devices

A MOS capacitor is fabricated on p-type Si (silicon) where the metal work function is 4.1 eV and electron affinity of Si is 4.0 eV. $E_{C}-F_{F}$ =0.9 eV, where $E_{C}$ and $E_{F}$ are the conduction band minimum and the Fermi energy levels of Si, respectively. Oxide $\in_{r}=3.9,\in_{o}=8.85 \times 10^{-14}$ F/cm, oxide thickness $t_{ox} = 0.1 \mu m$ and electronic charge $q =1.6 \times 10^{-19} C$ . If the measured flat band voltage of the capacitor is -1V, then the magnitude of the fixed charge at the oxide-semiconductor interface, in $nC/cm^{2}$ , is __________.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \theta_{0} &=4.1 \mathrm{eV} \\ e_{K} &=4 \mathrm{eV} \\ E_{C}-E_{F P} &=0.9 \mathrm{eV} \\ e \phi_{s} &=e x+0.9=4.9 \mathrm{eV} \\ \phi_{m} &=4.1 \mathrm{V}, \quad \phi_{s}=4.9 \mathrm{V} \\ \phi_{m s} &=\phi_{m}-\phi_{s}=4.1-4.9=-0.8 \mathrm{v} \\ C_{o x} &=\frac{\epsilon_{O x}}{t_{o x}}=\frac{3.9 \times 8.85 \times 10^{-14}}{0.1 \times 10^{-4}}\\ &=34.515 \times 10^{-9} \mathrm{F} / \mathrm{cm}^{2} \\ V_{F B} &=\phi_{\text {rns }}-\frac{\phi_{o x}^{\prime}}{C_{O x}} \\ \frac{\phi_{O x}^{\prime}}{C_{O x}} &=-0.8+1=0.2 \mathrm{V} \\ \phi_{o x}^{\prime} &=0.2 \times C_{0 x} \\ &=0.2 \times 34.515 \times 10^{-9} \mathrm{C} / \mathrm{cm}^{2} \\ &=6.903 \mathrm{nC} / \mathrm{cm}^{2}\\ \end{aligned}

Q30GATE 2017NAT2MElectromagnetics

An electron ( $q_1$ ) is moving in free space with velocity $10^{5}$ m/s towards a stationary electron ( $q_2$ ) far away. The closest distance that this moving electron gets to the stationary electron before the repulsive force diverts its path is ___________ $\times 10^{-8}$ m. [Given, mass of electron $m=9.11 \times 10^{-31}$ kg, charge of electron $e=-1.6 \times 10^{-19}$ C, and permittivity $\varepsilon_{0}=(1/36\pi)\times 10^{-9}F/m]$

🚀 Solve in Practice Mode

📖 Explanation

r is the distance at which kinetic energy of $q_{1}$ becomes zero [(because kinetic energy (KE) is converted into potential energy (PE)]. When $q_{1}$ reaches 'r', it starts diverting. Kinetic energy, $K E=\frac{1}{2} m v^{2}$ and work done in moving $q_{1}$ charge to distance 'r' is

\begin{aligned} q_{1} v_{2}&=q_{1} \frac{q_{2}}{4 \pi \varepsilon_{0} r} \\ &\qquad\left(q_{1}=q_{2}=-1.6 \times 10^{-19} \mathrm{C}\right)\\ Now, \quad \frac{1}{2} m v^{2}&=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r} \\ \end{aligned}

$\Rightarrow r=\frac{\left(2 \times -1.6 \times 10^{-19}\right) \times \left(-1.6 \times 10^{-19}\right)}{4 \pi \times \frac{10^{-9}}{36 \pi} \times 9.11 \times 10^{-31} \times \left(10^{5}\right)^{2}}$ $\simeq 5.06 \times 10^{-8} \mathrm{m}$

Q31GATE 2017MCQ2MEngineering Mathematics

The values of the integrals $\int_{0}^{1}(\int_{0}^{1}\frac{x-y}{(x+y)^{3}}dy)dx$ and $\int_{0}^{1}(\int_{0}^{1}\frac{x-y}{(x+y)^{3}}dx)dy$ are

🚀 Solve in Practice Mode

📖 Explanation

Let's evaluate the two integrals step-by-step.

For the first integral, $I_1 = \int_{0}^{1} \left( \int_{0}^{1} \frac{x-y}{(x+y)^{3}} dy \right) dx$ , we begin with the inner integral over $y$ . By rewriting the integrand as $\frac{2x - (x+y)}{(x+y)^3} = \frac{2x}{(x+y)^3} - \frac{1}{(x+y)^2}$ , we can integrate with respect to $y$ to get $\frac{-x}{(x+y)^2} + \frac{1}{x+y}$ . Evaluating this from $y=0$ to $y=1$ yields $\frac{1}{(x+1)^2}$ . The outer integral is then $\int_{0}^{1} \frac{dx}{(x+1)^2} = \left[-\frac{1}{x+1}\right]_{0}^{1} = -\frac{1}{2} - (-1) = 0.5$ .

For the second integral, $I_2 = \int_{0}^{1} \left( \int_{0}^{1} \frac{x-y}{(x+y)^{3}} dx \right) dy$ , we integrate over $x$ first. This time, we rewrite the integrand as $\frac{(x+y) - 2y}{(x+y)^3} = \frac{1}{(x+y)^2} - \frac{2y}{(x+y)^3}$ . Integrating with respect to $x$ gives $-\frac{1}{x+y} + \frac{y}{(x+y)^2}$ . Evaluating this from $x=0$ to $x=1$ results in $\frac{-1}{(1+y)^2}$ . The final outer integral is $\int_{0}^{1} \frac{-1}{(1+y)^2} dy = \left[\frac{1}{1+y}\right]_{0}^{1} = \frac{1}{2} - 1 = -0.5$ .

Thus, the values of the two integrals are 0.5 and -0.5, respectively.

Q32GATE 2017NAT2MEngineering Mathematics

Passengers try repeatedly to get a seat reservation in any train running between two stations until they are successful. If there is 40% chance of getting reservation in any attempt by a passenger, then the average number of attempts that passengers need to make to get a seat reserved is _________.

🚀 Solve in Practice Mode

📖 Explanation

This problem describes a sequence of repeated, independent trials until a success is achieved, which is a classic example of a geometric distribution. Let the probability of success (getting a reservation) be $p = 0.4$ , and the probability of failure be $q = 1 - p = 0.6$ .

We want to find the average, or expected number of attempts, $E[X]$ . This is the sum of each possible number of attempts multiplied by its probability:
$E[X] = 1(p) + 2(qp) + 3(q^2p) + \dots$

To find the value of this infinite series, we can use a helpful algebraic trick. Let's subtract $qE[X]$ from $E[X]$ :
$E[X] - qE[X] = (p + 2qp + 3q^2p + \dots) - (qp + 2q^2p + \dots) = p + qp + q^2p + \dots$

The left side simplifies to $(1-q)E[X] = pE[X]$ . The right side is a geometric series that sums to $\frac{p}{1-q} = \frac{p}{p} = 1$ .
Therefore, we get the simple relation $pE[X] = 1$ , which means $E[X] = \frac{1}{p}$ .
Plugging in the given probability, the average number of attempts is $\frac{1}{0.4} = 2.5$ .

Q33GATE 2017NAT2MDigital Circuits

Figure I shows a 4-bits ripple carry adder realized using full adders and Figure II shows the circuit of a full-adder (FA). The propagation delay of the XOR, AND and OR gates in Figure II are 20 ns, 15 ns and 10 ns respectively. Assume all the inputs to the 4-bit adder are initially reset to At t=0, the inputs to the 4-bit adder are changed to $X_{3}X_{2}X_{1}X_{0}=1100,\; Y_{3}Y_{2}Y_{1}Y_{0}=0100 \; and \; Z_{0}=1$ . The output of the ripple carry adder will be stable at t (in ns) = ___________

🚀 Solve in Practice Mode

📖 Explanation

To determine when the ripple-carry adder's output becomes stable, we must find the longest delay path for the given inputs. The critical path is typically the carry propagation chain.

First, let's analyze the delay for the carry-out, $Z_{n+1}$ , of a single full adder. A carry is generated ( $Z_{n+1}=1$ regardless of $Z_n$ ) when $X_n=1$ and $Y_n=1$ . This path involves an AND gate and an OR gate, taking $15 \text{ ns} + 10 \text{ ns} = 25 \text{ ns}$ . A carry is propagated ( $Z_{n+1}=Z_n$ ) when $X_n \oplus Y_n=1$ . This path involves an XOR, AND, and OR gate.

Now, we trace the carry signal stage by stage:

Stages 0 & 1: For $n=0$ and $n=1$ , the inputs are $X_nY_n=00$ . These stages don't generate or propagate a carry, so $Z_1$ and $Z_2$ remain 0. Their values are stable instantly at $t=0$ .
Stage 2: With inputs $X_2=1, Y_2=1$ , this stage generates a carry. The carry-out $Z_3$ becomes stable after the 25 ns delay for carry generation. Thus, $Z_3$ is ready at $t = 25 \text{ ns}$ .
Stage 3: This stage has inputs $X_3=1, Y_3=0$ , which means it propagates its incoming carry. The final carry $Z_4$ will be stable after $Z_3$ propagates through this stage. The arrival time of $Z_3$ is 25 ns. The path through this stage involves an XOR, an AND, and an OR. The total time for $Z_4$ is $\max(t_{XOR}, t_{Z_3\text{ arrival}}) + t_{AND} + t_{OR} = \max(20 \text{ ns}, 25 \text{ ns}) + 15 \text{ ns} + 10 \text{ ns} = 25 + 15 + 10 = 50 \text{ ns}$ .

The last sum bit, $S_3$ , stabilizes at 45 ns. Since $Z_4$ is the last output to settle, the entire adder is stable at 50 ns.

Q34GATE 2017NAT2MElectromagnetics

The permittivity of water at optical frequencies is 1.75 $\varepsilon_{0}$ . It is found that an isotropic light source at a distance d under water forms an illuminated circular area of radius 5m, as shown in the figure. The critical angle is $\theta_{c}$ . The value of d (in meter) is _____________

🚀 Solve in Practice Mode

📖 Explanation

The illuminated circular area on the surface is limited by total internal reflection. Its radius is determined by light rays from the source that strike the water-air interface at the critical angle, $\theta_c$ . We can find this angle using the given permittivities of water ( $\epsilon_1=1.75\varepsilon_0$ ) and air ( $\epsilon_2=\varepsilon_0$ ).

The relationship is $\sin \theta_c = \sqrt{\epsilon_2/\epsilon_1} = \sqrt{\varepsilon_0/(1.75\varepsilon_0)}$ , which gives $\theta_c = \sin^{-1}(1/\sqrt{1.75}) \approx 49.1^\circ$ .

From the right-angled triangle shown in the figure, we can see that the depth $d$ and the radius (5 m) are related by the tangent of the critical angle.
$\tan \theta_c = \frac{\text{radius}}{d} = \frac{5}{d}$
Solving for the depth $d$ , we find:
$d = \frac{5}{\tan(49.1^\circ)} \approx \frac{5}{1.1547} \approx 4.33 \text{ m}$

Q35GATE 2017MCQ2MControl Systems

A unity feedback control system is characterized by the open-loop transfer function $G(s)=\frac{10K(s+2)}{s^{3}+3s^{2}+10}$ The Nyquist path and the corresponding Nyquist plot of G(s) are shown in the figures below. If $0 \lt K \lt 1$ , then the number of poles of the closed-loop transfer function that lie in the right half of the s-plane is

🚀 Solve in Practice Mode

📖 Explanation

To determine the number of unstable closed-loop poles, we use the Nyquist stability criterion, $N = P - Z$ . In this formula, $Z$ is the number of closed-loop poles in the right-half s-plane (RHP), $P$ is the number of open-loop poles in the RHP, and $N$ is the number of times the Nyquist plot encircles the critical point $(-1, 0)$ .

First, we find $P$ by checking the stability of the open-loop transfer function, $G(s)$ . The poles of $G(s)$ are the roots of the denominator, $s^3 + 3s^2 + 10 = 0$ . Applying the Routh-Hurwitz criterion, the first column of the Routh array has the values [1, 3, -10/3, 10]. There are two sign changes, which means there are two open-loop poles in the RHP. Therefore, $P=2$ .

Next, we determine $N$ from the provided Nyquist plot. The plot shows an intersection with the negative real axis at $-K$ . Since the problem states $0 < K < 1$ , the critical point $(-1, 0)$ is to the left of this intersection and is not enclosed by the plot. Thus, the number of encirclements is $N=0$ .

Finally, we substitute our values into the Nyquist criterion: $N = P - Z$ becomes $0 = 2 - Z$ . Solving for $Z$ , we find $Z=2$ . This means there are two unstable poles in the closed-loop system.

Q36GATE 2017MCQ2MSignals and Systems

The signal $x(t)=sin(14000\pi t)$ , where t is in seconds is sampled at a rate of 9000 samples per second. The sampled signal is the input to an ideal low pass filter with frequency response H(f) as follows :

H(f)=\left\{\begin{matrix} 1, & |f|\leq 12 kHz\\ 0, & |f|\geq 12 kHz \end{matrix}\right.

What is the number of sinusoids in the output and their frequencies in kHz?

🚀 Solve in Practice Mode

📖 Explanation

The input signal is $x(t) = \sin(14000\pi t)$ . We can find its frequency, $f_m$ , from the standard form $\sin(2\pi f_m t)$ , which gives $2\pi f_m = 14000\pi$ , so $f_m = 7$ kHz.

Sampling this signal at $f_s = 9$ kHz creates spectral replicas centered at integer multiples of the sampling frequency. The frequencies of these components in the sampled signal's spectrum are given by $|f_m \pm n f_s|$ for integers $n=0, 1, 2, ...$ .

Let's calculate the first few of these frequencies:
For $n=0$ , we get the original frequency: $7$ kHz.
For $n=1$ , we get aliased frequencies: $|7 - 9| = 2$ kHz and $7 + 9 = 16$ kHz.
For $n=2$ , we get another aliased frequency: $|7 - 2 \times 9| = |7 - 18| = 11$ kHz.

The ideal low-pass filter passes all frequencies $|f| \leq 12$ kHz. From the components we calculated {2, 7, 11, 16, ...} kHz, only 2 kHz, 7 kHz, and 11 kHz fall within the filter's passband. Therefore, the output consists of three sinusoids at these frequencies.

Q37GATE 2017NAT2MControl Systems

A unity feedback control system is characterized by the open-loop transfer function $G(s)=\frac{2(s+1)}{s^{3}+ks^{2}+2s+1}$ The value of k for which the system oscillates at 2 rad/s is ________.

🚀 Solve in Practice Mode

📖 Explanation

First, we need to find the characteristic equation of the closed-loop system. For a unity feedback system, the closed-loop transfer function is $T(s) = \frac{G(s)}{1+G(s)}$ . The denominator of $T(s)$ set to zero gives us the characteristic equation.
$1 + G(s) = 1 + \frac{2(s+1)}{s^{3}+ks^{2}+2s+1} = \frac{s^{3}+ks^{2}+4s+3}{s^{3}+ks^{2}+2s+1} = 0$
So, the characteristic equation is $s^{3}+ks^{2}+4s+3 = 0$ .

A system oscillates when it is marginally stable, which means it has poles on the imaginary axis. We can use the Routh-Hurwitz stability criterion to find the value of $k$ that causes this. For a third-order system of the form $as^{3+}bs^{2+}cs+d=0$ , the system is marginally stable when the product of the inner coefficients equals the product of the outer coefficients, i.e., $bc=ad$ .

Applying this condition to our equation ( $1s^{3+}ks^{2+}4s+3=0$ ), we get:
$k \cdot 4 = 1 \cdot 3$
$4k = 3 \implies k = 0.75$

Q38GATE 2017NAT2MNetwork Theory

Consider the circuit shown in the figure. The Thevenin equivalent resistance (in $\Omega$ ) across P - Q is _________.

🚀 Solve in Practice Mode

📖 Explanation

To find the Thevenin resistance ( $R_{Th}$ ), we first deactivate the independent 10V source by replacing it with a short circuit. We then apply a test source across terminals P-Q to determine the equivalent resistance. The circuit analysis yields two key relationships involving the controlling current $i_0$ and a voltage $V_x$ .

From Kirchhoff's Voltage Law (KVL), we find $V_x = 2i_0 + 1$ . From Ohm's law, we also have $V_x = i_0$ .

Equating these two expressions gives $2i_0 + 1 = i_0$ , which solves to $i_0 = -1$ A. Substituting this value back, we find the voltage to be $V_x = -1$ V. Assuming a 1 A test current was used to generate this voltage, the Thevenin resistance is calculated as:
$R_{Th} = \frac{V_x}{I_{test}} = \frac{-1 \text{ V}}{1 \text{ A}} = -1 \, \Omega$ .

Q39GATE 2017NAT2MSignals and Systems

The transfer function of a causal LTI system is H(s) = 1/s. If the input to the system is $x(t) =[sin(t) / \pi t] u(t)$ , where u(t) is a unit step function, the system output y(t) as $t \rightarrow \infty$ is______.

🚀 Solve in Practice Mode

📖 Explanation

The system's transfer function, $H(s) = 1/s$ , corresponds to a pure integrator in the time domain. This means the output $y(t)$ is simply the integral of the input signal $x(t)$ . We are asked to find the steady-state value of the output, which is the value of this integral as $t \rightarrow \infty$ .

$\lim_{t \to \infty } y(t) = \int_{0}^{\infty } x(\tau) d\tau = \int_{0}^{\infty } \frac{\sin(\tau)}{\pi \tau} d\tau$

We can factor out the constant $1/\pi$ . The remaining integral is the famous Dirichlet integral, whose value is a standard result.

$\lim_{t \to \infty } y(t) = \frac{1}{\pi} \int_{0}^{\infty } \frac{\sin(\tau)}{\tau} d\tau = \frac{1}{\pi} \left( \frac{\pi}{2} \right) = \frac{1}{2} = 0.5$

Q40GATE 2017MCQ2MEngineering Mathematics

An integral I over a counter clock wise circle C is given by $I=\oint_{0}^{c}\frac{z^{2}-1}{z^{2}+1}e^{z}dz$ . If C is defined as |z| = 3, then the value of $I$ is

🚀 Solve in Practice Mode

📖 Explanation

This integral can be evaluated using Cauchy's Residue Theorem. The function $f(z) = \frac{z^{2}-1}{z^{2}+1}e^{z}$ has simple poles where the denominator $z^{2+}1=0$ , which occurs at $z=i$ and $z=-i$ . Since the contour is a circle of radius 3 centered at the origin, both poles lie inside it.

We first compute the residue at each pole.
Residue at $z=i$ : $\text{Res}(f, i) = \lim_{z \to i} (z-i)f(z) = \lim_{z \to i} \frac{z^{2-}1}{z+i}e^z = \frac{i^{2-}1}{2i}e^i = \frac{-2}{2i}e^i = i e^i$ .
Residue at $z=-i$ : $\text{Res}(f, -i) = \lim_{z \to -i} (z+i)f(z) = \lim_{z \to -i} \frac{z^{2-}1}{z-i}e^z = \frac{(-i)^{2-}1}{-2i}e^{-i} = \frac{-2}{-2i}e^{-i} = -i e^{-i}$ .

The integral is $2\pi i$ times the sum of the enclosed residues:
$I = 2\pi i \left( i e^i - i e^{-i} \right) = 2\pi i^2 (e^i - e^{-i}) = -2\pi (e^i - e^{-i})$ .
Using the identity $e^{i\theta} - e^{-i\theta} = 2i\sin(\theta)$ , this simplifies to $I = -2\pi (2i\sin(1)) = -4\pi i \sin(1)$ .

Q41GATE 2017MCQ2MCommunication Systems

Consider a binary memory less channel characterized by the transition probability diagram shown in the figure. The channel is

🚀 Solve in Practice Mode

📖 Explanation

The channel's behavior is described by the transition probability matrix

P(Y|X) = \begin{bmatrix} 0.25 & 0.75 \\ 0.25 & 0.75 \end{bmatrix}

. A crucial observation is that both rows of this matrix are identical. This means the probability of receiving a 0 or 1 is completely independent of the input bit that was sent. For any arbitrary input distribution, the output distribution is fixed at $[0.25, 0.75]$ .

Because the output is statistically independent of the input, observing the output provides no information whatsoever about the input. This means the mutual information between input and output, $I(X;Y)$ , is always zero. A channel with zero capacity that cannot transfer any information is defined as useless.

Q42GATE 2017MCQ2MElectronic Devices

An abrupt pn junction (located at x = 0) is uniformly doped on both p and n sides. The width of the depletion region is W and the electric field variation in the x-direction is E(x). Which of the following figures represents the electric field profile near the pn junction?

🚀 Solve in Practice Mode

Q43GATE 2017MCQ2MControl Systems

A second - order LTI system is described by the following state equations, $\frac{d}{dt}x_{1}(t)- x_{2}(t)=0$ $\frac{d}{dt}x_{2}(t)+2x_{1}(t)+3x_{2}(t)=r(t)$ Where $x_{1}(t)$ and $x_{2}(t)$ are the two state variables and r(t) denotes the input. The output $c(t) =x_{1}(t)$ . The system is.

🚀 Solve in Practice Mode

📖 Explanation

To determine the system's behavior, we'll derive its transfer function from the state-space model. First, apply the Laplace transform to the given equations, which gives us $sX_1(s) = X_2(s)$ and $sX_2(s) + 2X_1(s) + 3X_2(s) = R(s)$ . We can substitute the first equation into the second to eliminate $X_2(s)$ and create a single equation relating the input $R(s)$ to the output $X_1(s)$ . This substitution yields $s(sX_1(s)) + 2X_1(s) + 3(sX_1(s)) = R(s)$ .

Rearranging this expression gives us the system's transfer function: $\frac{X_1(s)}{R(s)} = \frac{1}{s^{2+}3s+2}$ . The denominator, $s^{2+}3s+2$ , is the characteristic equation. By comparing it to the standard second-order form $s^2 + 2\zeta\omega_n s + \omega_n^2$ , we can identify that $\omega_n^2 = 2$ and $2\zeta\omega_n = 3$ . Solving for the damping ratio $\zeta$ gives $\zeta = \frac{3}{2\omega_n} = \frac{3}{2\sqrt{2}} \approx 1.06$ . Since $\zeta > 1$ , the system is overdamped.

Q44GATE 2017NAT2MSignals and Systems

Consider the parallel combination of two LTI systems shown in the figure. The impulse responses of the systems are $h_{1}(t)=2\delta (t+2)-3\delta (t+1)$ $h_{2}(t)=\delta(t-2)$ If the input x(t) is a unit step signal, then the energy of y(t) is ____________.

🚀 Solve in Practice Mode

📖 Explanation

For systems in parallel, the overall impulse response is the sum of the individual responses.
Therefore, $h(t) = h_1(t) + h_2(t) = 2\delta(t+2) - 3\delta(t+1) + \delta(t-2)$ .

The output $y(t)$ is the convolution of the input $x(t)=u(t)$ with the overall impulse response $h(t)$ . Using the property that convolution with a delta function is a shifting operation, we get:
$y(t) = u(t) * h(t) = 2u(t+2) - 3u(t+1) + u(t-2)$ .

By sketching this signal, we see that $y(t)$ is a piecewise-constant function. It equals $2$ for $t \in [-2, -1)$ , and $-1$ for $t \in [-1, 2)$ , and is zero elsewhere.

The energy of the signal, $E_y = \int_{-\infty }^{\infty } y^2(t)dt$ , is the area under the curve of $y^2(t)$ . This area can be calculated by summing the energies over the intervals:
$E_y = \int_{-2}^{-1} (2)^2 dt + \int_{-1}^{2} (-1)^2 dt = (4 \times 1) + (1 \times 3) = 7$ .

Q45GATE 2017MCQ2MAnalog Circuits

Assuming that transistors $M_{1}$ and $M_{2}$ are identical and have a threshold voltage of 1V, the state of transistors $M_{1}$ and $M_{2}$ are respectively.

🚀 Solve in Practice Mode

📖 Explanation

Let's determine the operating region for each transistor. We'll start by assuming both $M_1$ and $M_2$ are in saturation. Since they are in series, their drain currents must be equal, $I_{D1} = I_{D2}$ . Let $V_x$ be the voltage at the drain of $M_1$ .

Using the saturation current equation $I_D = \frac{k_n}{2}(V_{GS} - V_{Th})^2$ , we can write:
$\frac{k_n}{2}(V_{GS1} - V_{Th})^2 = \frac{k_n}{2}(V_{GS2} - V_{Th})^2$
Substituting the circuit values ( $V_{GS1}=2V$ , $V_{GS2}=2.5V - V_x$ , $V_{Th}=1V$ ):
$(2 - 1)^2 = (2.5 - V_x - 1)^2 \implies 1 = (1.5 - V_x)^2$
Solving for $V_x$ gives two possibilities: $V_x = 0.5V$ or $V_x = 2.5V$ . The value $V_x=2.5V$ is invalid because it would make $V_{GS2}=0V$ , turning $M_2$ off.

Now, let's check our saturation assumption with $V_x = 0.5V$ . For $M_1$ to be in saturation, we need $V_{DS1} \geq V_{GS1} - V_{Th}$ . Here, $V_{DS1} = V_x = 0.5V$ , so the condition is $0.5V \geq 2V - 1V$ , which is false. This contradiction means our initial assumption was wrong, and $M_1$ must be in the linear region.

Since $M_1$ is being pulled into the linear region by its low drain voltage, it allows a higher current than its saturation formula would suggest. This higher current requires $M_2$ to have a larger $V_{GS2}$ (and thus a smaller $V_x$ ), which keeps $M_2$ in saturation. Therefore, $M_1$ is in the linear region and $M_2$ is in the saturation region.

Q46GATE 2017MCQ2MDigital Circuits

A programmable logic array (PLA) is shown in the figure. The Boolean function F implemented is

🚀 Solve in Practice Mode

📖 Explanation

This circuit diagram shows a Programmable Logic Array (PLA), which implements logic functions in a standard two-level, sum-of-products form.

First, we analyze the AND plane to find the product terms. The asterisks indicate which input signals are connected to each AND gate.

The top AND gate connects to $\bar{P}$ , $\bar{Q}$ , and $R$ , producing the term $\bar{P}\bar{Q}R$ .
The middle AND gate connects to $\bar{P}$ , $Q$ , and $R$ , producing $\bar{P}QR$ .
The bottom AND gate connects to $P$ , $\bar{Q}$ , and $R$ , producing $P\bar{Q}R$ .

Finally, the OR plane sums these product terms. The output $F$ is the logical OR of the three AND gate outputs, resulting in the expression $F = \bar{P}\bar{Q}R + \bar{P}QR + P\bar{Q}R$ .

Q47GATE 2017NAT2MCommunication Systems

A modulating signal given By $x(t) =5sin(4 \pi 10^{3}t-10\pi cos2\pi 10^{3}t$ )V is fed to a phase modulator with phase deviation constant $k_{p}$ =0.5rad/V. If the carrier frequency is 20 kHz, the instantaneous frequency (in kHz) at t = 0.5 ms is __________

🚀 Solve in Practice Mode

📖 Explanation

The instantaneous frequency, $f_i(t)$ , of a phase-modulated signal is the sum of the carrier frequency $f_c$ and a deviation term proportional to the derivative of the modulating signal $x(t)$ . The governing equation is $f_i(t) = f_c + \frac{k_p}{2\pi}\frac{dx(t)}{dt}$ .

First, we find the derivative of the modulating signal using the chain rule:
$\frac{dx(t)}{dt} = 5\cos(4000\pi t - 10\pi\cos(2000\pi t)) \cdot (4000\pi + 20000\pi^2\sin(2000\pi t))$

Next, we evaluate this derivative at the specified time, $t = 0.5 \text{ ms} = 0.5 \times 10^{-3}$ s. At this instant, the terms simplify: $2000\pi t = \pi$ and $4000\pi t = 2\pi$ . Substituting these gives:
$\frac{dx}{dt} = 5\cos(2\pi - 10\pi\cos(\pi)) \cdot (4000\pi + 20000\pi^2\sin(\pi))$
$\frac{dx}{dt} = 5\cos(2\pi + 10\pi) \cdot (4000\pi + 0) = 5\cos(12\pi) \cdot 4000\pi = 20000\pi$ V/s.

Now, calculate the frequency deviation:
$\Delta f = \frac{k_p}{2\pi}\frac{dx}{dt} = \frac{0.5}{2\pi}(20000\pi) = 50,000 \text{ Hz} = 50 \text{ kHz}$ .

Finally, add this deviation to the carrier frequency to find the instantaneous frequency:
$f_i(0.5\text{ms}) = f_c + \Delta f = 20 \text{ kHz} + 50 \text{ kHz} = 70 \text{ kHz}$ .

Q48GATE 2017NAT2MEngineering Mathematics

The minimum value of the function $f(x)= \frac{1}{3}x(x^{2}-3)$ in the interval $-100 \leq x \leq 100$ occurs at x=________.

🚀 Solve in Practice Mode

📖 Explanation

To find the absolute minimum value of a continuous function on a closed interval, we must test the function's values at its critical points and at the endpoints of the interval.

First, we find the critical points by setting the derivative of $f(x) = \frac{1}{3}x(x^2 - 3) = \frac{1}{3}x^3 - x$ equal to zero. The derivative is $f'(x) = x^2 - 1$ . Setting $f'(x)=0$ gives $x^2=1$ , so our critical points are $x=1$ and $x=-1$ , both of which are in the interval $[-100, 100]$ .

Now, we evaluate the function at these critical points and at the endpoints, $x=-100$ and $x=100$ .

At the left endpoint: $f(-100) = \frac{1}{3}(-100)^3 - (-100) \approx -333,233.3$
At the critical point $x=-1$ : $f(-1) = \frac{1}{3}(-1)^3 - (-1) = 2/3$
At the critical point $x=1$ : $f(1) = \frac{1}{3}(1)^3 - 1 = -2/3$
At the right endpoint: $f(100) = \frac{1}{3}(100)^3 - 100 \approx 333,233.3$

By comparing these four values, we see the smallest value is approximately $-333,233.3$ , which occurs at the endpoint $x = -100$ .

Q49GATE 2017NAT2MNetwork Theory

The switch in the circuit, shown in the figure, was open for a long time and is closed at t = 0. The current i(t) (in ampere) at t = 0.5 seconds is ________

🚀 Solve in Practice Mode

📖 Explanation

For time $t > 0$ , after the switch is closed, we analyze the circuit in the Laplace domain. The expression for the current $I(s)$ through the inductor branch can be established.

The s-domain current is given by the equation:
$I(s) = \frac{10}{s} - \frac{12.5}{5 + 2.5s}$

By factoring $2.5$ from the denominator of the second term, we can simplify this expression to:
$I(s) = \frac{10}{s} - \frac{5}{s+2}$

Taking the inverse Laplace transform yields the time-domain current, $i(t)$ :
$i(t) = (10 - 5e^{-2t})u(t)$ A

To find the current at $t = 0.5$ seconds, we substitute this value into the equation:
$i(0.5) = 10 - 5e^{-2(0.5)} = 10 - 5e^{-1} = 10 - \frac{5}{e} \approx 8.16$ A

Q50GATE 2017NAT2MAnalog Circuits

In the voltage reference circuit shown in the figure, the op-amp is ideal and the transistors $Q_{1},Q_{2}...,Q_{32}$ are identical in all respects and have infinitely large values of common-emitter current the relation $I_{C}=I_{S}\; exp \;(V_{BE}/V_{T})$ , where $I_{s}$ is the saturation current. Assume that the voltage $V_{P}$ shown in the figure is 0.7 V and the thermal voltage $V_{T}$ =26mV. The output voltage $V_{out}$ (in volts) is__________.

🚀 Solve in Practice Mode

📖 Explanation

This circuit is a type of bandgap voltage reference. Let's break down the analysis into two parts.

First, we determine the voltage $V_x$ at the op-amp's inputs. The ideal op-amp maintains equal voltages at its non-inverting (+) and inverting (-) terminals. The non-inverting input is connected to the base of $Q_1$ , so its voltage is $V_{BE1} = V_x$ . The base voltage for the other 31 transistors is given as $V_P = 0.7$ V. The difference in base-emitter voltages is related to the ratio of the currents, which are scaled by the number of transistors. This gives the core relationship: $V_x - V_P = V_T \ln(31)$ . With $V_P = 0.7$ V and $V_T = 26$ mV, we find $V_x = 0.7 + (0.026)\ln(31) \approx 0.789$ V.

Second, we analyze the feedback network to find $V_{out}$ . Applying KCL at the inverting input node (which is also at voltage $V_x$ ) gives $\frac{V_{out} - V_x}{20 \text{ k}\Omega} = \frac{V_x - V_P}{5 \text{ k}\Omega}$ . Substituting $V_P = 0.7$ V and simplifying, we get the expression $V_{out} = 5V_x - 2.8$ . Finally, we plug in our calculated value for $V_x$ to find the output voltage: $V_{out} = 5(0.789) - 2.8 = 1.145$ V.

Q51GATE 2017NAT2MDigital Circuits

The state diagram of a finite state machine (FSM) designed to detect an overlapping sequence of three bits is shown in the figure. The FSM has an input 'In' and an output 'Out'. The initial state of the FSM is $S_{0}$ . If the input sequence is 10101101001101, starting with the left-most bit, then the number times 'Out' will be 1 is __________.

🚀 Solve in Practice Mode

📖 Explanation

By inspecting the state diagram, we see that the output Out is 1 only on the transition from state $S_{2}$ with an input of 1. To reach state $S_{2}$ , the FSM must receive a '1' (to go from $S_{0}$ to $S_{1}$ ) followed by a '0' (to go from $S_{1}$ to $S_{2}$ ). Therefore, the FSM is designed to detect the sequence '101'.

The problem asks us to count the overlapping occurrences of '101' in the input stream 10101101001101. Let's identify them:

**101**01101001101
10**101**101001101
10101**101**001101
10101101001**101**

Each time this pattern is detected, the output becomes 1. Since there are four such instances, the output will be 1 a total of 4 times.

Q52GATE 2017MCQ2MElectromagnetics

Standard air-filled rectangular waveguides of dimensions a = 2.29 cm and b= 1.02 cm are designed for radar applications. It is desired that these waveguides operate only in the dominatnt $TE_{10}$ mode but not higher than 95% of the next higher cutoff frequency. The range of the allowable operating frequency f is.

🚀 Solve in Practice Mode

📖 Explanation

To determine the allowed operating frequency range, we must establish both a lower and an upper limit for single-mode operation.

The lower limit ensures the signal propagates efficiently in the dominant $TE_{10}$ mode. The cutoff frequency for this mode is $f_{c,10} = \frac{c}{2a} = \frac{3 \times 10^{10} \text{ cm/s}}{2 \times 2.29 \text{ cm}} = 6.55$ GHz. The solution's logic sets the minimum operating frequency 25% above this cutoff, at $1.25 \times 6.55 \text{ GHz} \approx 8.19$ GHz.

The upper limit prevents the excitation of higher-order modes. Since $a > b$ (specifically, $2.29 \text{ cm} > 1.02 \text{ cm}$ ), the next mode to appear after $TE_{10}$ is $TE_{20}$ . Its cutoff frequency is $f_{c,20} = \frac{c}{a} = \frac{3 \times 10^{10} \text{ cm/s}}{2.29 \text{ cm}} = 13.1$ GHz. To ensure only the $TE_{10}$ mode propagates, the operating frequency must stay below this, and the problem specifies a ceiling of 95% of this value: $0.95 \times 13.1 \text{ GHz} \approx 12.45$ GHz.

Combining these limits gives the allowable frequency range: $8.19 \text{ GHz} \leq f \leq 12.45 \text{ GHz}$ .

Q53GATE 2017NAT2MElectronic Devices

For a particular intensity of incident light on a silicon pn junction solar cell, the photocurrent density ( $J_{L}$ ) is 2.5 mA/ $cm^{2}$ and the open-circuit voltage ( $V_{oc}$ ) is 0.451 V. consider thermal voltage ( $V_{T}$ ) to be 25mV. If the intensity of the incident light is increased by 20 times,assuming that the temperature remains unchanged. $V_{oc}$ (in volts) will be ______.

🚀 Solve in Practice Mode

📖 Explanation

The open-circuit voltage ( $V_{OC}$ ) of a solar cell is related to the photocurrent density ( $J_L$ ) by the equation $V_{OC} = V_T \ln(J_L / J_S)$ , where $V_T$ is the thermal voltage and $J_S$ is the reverse saturation current density.

To find the change in $V_{OC}$ when the light intensity changes, we can use the relationship $V_{OC2} - V_{OC1} = V_T \ln(J_{L2} / J_{L1})$ .

Since the photocurrent density is directly proportional to the incident light intensity, increasing the intensity by 20 times gives a ratio of $J_{L2}/J_{L1} = 20$ .

The resulting increase in voltage is $\Delta V_{OC} = (25 \text{ mV}) \times \ln(20) \approx 75 \text{ mV}$ , or $0.075 \text{ V}$ .

The new open-circuit voltage is the sum of the initial voltage and this increase:
$V_{OC2} = V_{OC1} + \Delta V_{OC} = 0.451 \text{ V} + 0.075 \text{ V} = 0.526 \text{ V}$ .

Q54GATE 2017NAT2MAnalog Circuits

In the circuit shown, transistors $Q_{1}$ and $Q_{2}$ are biased at a collector current of 2.6mA.Assuming that transistor current gains are sufficiently large to assume collector current equal to emitter current and thermal voltage of 26 mV, the magnitude of voltage gain $V_{0}/V_{s}$ in the mid-band frequency range is _____________ (up to second decimal place).

🚀 Solve in Practice Mode

📖 Explanation

This circuit is a common-emitter amplifier ( $Q_1$ ) with emitter degeneration provided by transistor $Q_2$ . For mid-band AC analysis, the coupling capacitor effectively grounds the base of $Q_2$ , configuring it as a diode-connected transistor. This presents an effective emitter resistance to $Q_1$ of $R_E = 1/g_{m2}$ .

Since both transistors have the same collector current, their transconductances are equal: $g_m = I_C/V_T = 2.6 \text{ mA} / 26 \text{ mV} = 0.1 \text{ S}$ .

The voltage gain for this degenerated CE amplifier is $A_v = \frac{-g_{m1} R_L}{1 + g_{m1} R_E}$ . Because $g_{m1} = g_{m2} = g_m$ , the degeneration resistance is $R_E = 1/g_m$ .

Substituting this into the gain formula gives $A_v = \frac{-g_m R_L}{1 + g_m(1/g_m)} = \frac{-g_m R_L}{2}$ .

Plugging in the values, we find $A_v = \frac{-(0.1 \text{ S})(1 \text{ k}\Omega)}{2} = \frac{-100}{2} = -50$ . The magnitude is therefore 50.

Q55GATE 2017MCQ2MElectronic Devices

Two n-channel MOSFETs, T1 and T2, are identical in all respects except that the width of T2 is double that of T1. Both the transistors are biased in the saturation region of operation, but the gate overdrive voltage ( $V_{GS}-V_{TH}$ ) of T2 is double that of T1, where $V_{GS} \; and \; V_{TH}$ are the gate-to-source voltage and threshold voltage of the transistors, respectively. If the drain current and transconductance of T1 are $I_{D1} \; and \; g_{m1}$ respectively, the corresponding values of these two parameters for T2 are

🚀 Solve in Practice Mode

📖 Explanation

For a MOSFET operating in the saturation region, the drain current $I_D$ and transconductance $g_m$ are defined by the following key relationships. The drain current is proportional to the channel width ( $W$ ) and the square of the overdrive voltage ( $V_{OV} = V_{GS} - V_{TH}$ ), so $I_D \propto W \cdot V_{OV}^2$ . The transconductance is proportional to the width and the overdrive voltage itself, so $g_m \propto W \cdot V_{OV}$ .

We are given that for transistor T2, the width is doubled ( $W_2 = 2W_1$ ) and the overdrive voltage is also doubled ( $V_{OV2} = 2V_{OV1}$ ).

Using these relationships, we can find the new drain current $I_{D2}$ by setting up a ratio:
$\frac{I_{D2}}{I_{D1}} = \left(\frac{W_2}{W_1}\right) \left(\frac{V_{OV2}}{V_{OV1}}\right)^2 = (2)(2)^2 = 8$
This tells us that $I_{D2} = 8I_{D1}$ .

Similarly, we can find the new transconductance $g_{m2}$ by calculating its ratio:
$\frac{g_{m2}}{g_{m1}} = \left(\frac{W_2}{W_1}\right) \left(\frac{V_{OV2}}{V_{OV1}}\right) = (2)(2) = 4$
Therefore, the new transconductance is $g_{m2} = 4g_{m1}$ .

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