GATE EC 2024 Set 1

The sequence [charm $\rightarrow$ enamor $\rightarrow$ bewitch] represents an increasing order of intensity of attraction or fascination. Similarly, the sequence [bored $\rightarrow$ _______ $\rightarrow$ weary] should represent an increasing order of intensity of fatigue or lack of interest. Options like 'baffled' (confused), 'dead' (lifeless), and 'worsted' (defeated) do not fit the intensity scale of boredom/tiredness. Therefore, jaded is the most appropriate word.

Let Aman's age and his father's age five years ago be $x$ and $4x$ respectively.Present ages:
Aman $= x + 5$
Father $= 4x + 5$Ages five years from now:
Aman $= (x + 5) + 5 = x + 10$
Father $= (4x + 5) + 5 = 4x + 10$According to the problem, the ratio five years from now is $2:5$:$$\frac{x + 10}{4x + 10} = \frac{2}{5}$$$$5(x + 10) = 2(4x + 10)$$$$5x + 50 = 8x + 20$$$$3x = 30$$$$x = 10$$Present ages calculation:
Aman $= 10 + 5 = 15$ years
Father $= 4(10) + 5 = 45$ yearsFather's age when Aman was born:
This is simply the constant age difference between them:
Age difference $= 45 - 15 = 30$ years.Therefore, when Aman was $0$ years old, his father was $30$ years old.

Using the property of logarithms $\frac{1}{\log_a b} = \log_b a$, the given equation can be rewritten as:$$\log_x 2 + \log_x 3 + \log_x 4 = 1$$Using the product rule for logarithms $\log_b m + \log_b n = \log_b (mn)$:$$\log_x (2 \times 3 \times 4) = 1$$$$\log_x 24 = 1$$By the definition of a logarithm, $x^1 = 24$, which gives $x = 24$.Therefore, the correct option is (C).

We can factorize the given expression as follows:$$3^{199} - 3^{196} = 3^{196}(3^3 - 1)$$$$= 3^{196}(27 - 1)$$$$= 3^{196} \times 26$$$$= 3^{196} \times 2 \times 13$$The prime factors of the expression are 2, 3, and 13. Among these, the greatest prime factor is 13.Therefore, the correct option is (A).

To form a coherent passage, we look for logical flow:The logical sequence is R-S-P-Q.Therefore, the correct option is (C).

The correct answer is Option D. A detailed step-by-step solution for this question is being prepared and will be available shortly after review by our verification team.

The correct answer is Option C. A detailed step-by-step solution for this question is being prepared and will be available shortly after review by our verification team.

The correct answer is Option A. A detailed step-by-step solution for this question is being prepared and will be available shortly after review by our verification team.

Let the initial dimensions of the sheets be length $L = 24$ cm and width $W = 16$ cm.Sheet A: Folded twice using FO1 (axis parallel to the initial long edge)
FO1 means the folding axis is parallel to the long edge. So, the long edge remains constant, and the short edge is halved with each fold. Length = $L = 24$ cm
Width = $W/2 = 16/2 = 8$ cm
Dimensions: 24 cm $\times$ 8 cm Length = $L = 24$ cm
Width = $(W/2)/2 = 8/2 = 4$ cm
Dimensions: 24 cm $\times$ 4 cm The perimeter of the final shape of A is $P_A = 2 \times (Length + Width) = 2 \times (24 + 4) = 2 \times 28 = 56$ cm.Sheet B: Folded twice using FO2 (axis parallel to the initial short edge)
FO2 means the folding axis is parallel to the short edge. So, the short edge remains constant, and the long edge is halved with each fold. Length = $L/2 = 24/2 = 12$ cm
Width = $W = 16$ cm
Dimensions: 12 cm $\times$ 16 cm Length = $(L/2)/2 = 12/2 = 6$ cm
Width = $W = 16$ cm
Dimensions: 6 cm $\times$ 16 cm The perimeter of the final shape of B is $P_B = 2 \times (Length + Width) = 2 \times (6 + 16) = 2 \times 22 = 44$ cm.Ratio of the perimeters of A and B:
Ratio = $P_A : P_B = 56 : 44$
Divide both sides by 4:
Ratio = $14 : 11$The final answer is $\boxed{\text{14:11}}$

The complementary function of a linear homogeneous differential equation with constant coefficients is given by $y_c(t)$.
Given $y_c(t) = (At + B)e^{-2t}$.This form of the complementary function indicates that the characteristic equation of the differential equation has a repeated root at $s = -2$.If $s = -2$ is a repeated root, the characteristic equation must be of the form $(s - (-2))^2 = 0$.
$(s + 2)^2 = 0$
$s^2 + 4s + 4 = 0$The corresponding homogeneous differential equation is obtained by replacing $s^2$ with $\frac{d^2}{dt^2}$, $s$ with $\frac{d}{dt}$, and the constant term with the coefficient of $y(t)$.
So, the homogeneous part of the differential equation is $\frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 4y = 0$.For a non-homogeneous differential equation, the right-hand side is a function of $t$, typically denoted as $f(t)$.
Thus, the corresponding differential equation is $\frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 4y = f(t)$.Comparing this with the given options, option (A) matches this form.Options (B), (C), and (D) would have different characteristic equations and thus different complementary functions:
(B) $s^2 + 4 = 0 \implies s = \pm 2j \implies y_c(t) = C_1 \cos(2t) + C_2 \sin(2t)$
(C) $s^2 + 3s + 2 = 0 \implies (s+1)(s+2) = 0 \implies s = -1, -2 \implies y_c(t) = C_1 e^{-t} + C_2 e^{-2t}$
(D) $s^2 + 5s + 6 = 0 \implies (s+2)(s+3) = 0 \implies s = -2, -3 \implies y_c(t) = C_1 e^{-2t} + C_2 e^{-3t}$Therefore, option (A) is the correct differential equation.The final answer is $\boxed{\text{A}}$

In Bode plots, a 'decade' refers to a frequency ratio of 10, and an 'octave' refers to a frequency ratio of 2.We are given a slope of 40 dB/decade. This means that for every factor of 10 increase in frequency, the magnitude changes by 40 dB.Let $G$ be the gain in dB. The slope $S$ is given by:
$S = \frac{\Delta G}{\log_{10}(\omega_2/\omega_1)}$ dB/decade
$S = \frac{\Delta G}{\log_{2}(\omega_2/\omega_1)}$ dB/octaveWe know that $\log_{10}(x) = \frac{\log_2(x)}{\log_2(10)}$.So, $S_{\text{dB/decade}} = \frac{\Delta G}{\log_{10}(\omega_2/\omega_1)}$
And $S_{\text{dB/octave}} = \frac{\Delta G}{\log_{2}(\omega_2/\omega_1)}$From these, we can relate the two slopes:
$S_{\text{dB/decade}} = S_{\text{dB/octave}} \times \log_2(10)$We are given $S_{\text{dB/decade}} = 40$ dB/decade.
We need to find $S_{\text{dB/octave}}$.$40 = S_{\text{dB/octave}} \times \log_2(10)$Calculate $\log_2(10)$:
$\log_2(10) = \frac{\ln(10)}{\ln(2)} \approx \frac{2.3025}{0.6931} \approx 3.3219$Now, substitute this value:
$40 = S_{\text{dB/octave}} \times 3.3219$
$S_{\text{dB/octave}} = \frac{40}{3.3219} \approx 12.041$ dB/octaveRounding to the nearest integer, this is 12 dB/octave.Alternatively, a slope of 20 dB/decade corresponds to a first-order system. A first-order system has a slope of $20 \log_{10}(2) \approx 20 \times 0.301 = 6.02$ dB/octave.
Since 40 dB/decade is twice 20 dB/decade, it corresponds to a second-order system. Therefore, the slope in dB/octave would be $2 \times 6.02 \approx 12.04$ dB/octave.The final answer is $\boxed{\text{A}}$

The open-loop transfer function is given as $G(s) = \frac{6}{s(s+1)(s+2)}$. For a unity feedback system, the closed-loop characteristic equation is $1 + G(s) = 0$.$$1 + \frac{6}{s(s+1)(s+2)} = 0$$$$s(s+1)(s+2) + 6 = 0$$$$s(s^2 + 3s + 2) + 6 = 0$$$$s^3 + 3s^2 + 2s + 6 = 0$$To check the stability, we apply the Routh-Hurwitz criterion:The presence of a row of zeros in the $s^1$ row indicates that the system is marginally stable. The auxiliary equation from the $s^2$ row is:$$3s^2 + 6 = 0 \implies s^2 + 2 = 0 \implies s = \pm j\sqrt{2}$$The closed-loop poles are at $s = -3$ and $s = \pm j\sqrt{2}$.The error signal $E(s)$ for a unit step input $R(s) = \frac{1}{s}$ is:$$E(s) = \frac{R(s)}{1 + G(s)} = \frac{1/s}{1 + \frac{6}{s(s+1)(s+2)}} = \frac{(s+1)(s+2)}{s^3 + 3s^2 + 2s + 6}$$The Final Value Theorem (FVT) states that $\lim_{t \to \infty} e(t) = \lim_{s \to 0} sE(s)$, provided all poles of $sE(s)$ lie in the left half of the $s$-plane. Here, $sE(s) = \frac{s(s+1)(s+2)}{s^3 + 3s^2 + 2s + 6}$ has poles at $s = -3$ and $s = \pm j\sqrt{2}$. Since there are poles on the imaginary axis, the steady-state error will be oscillatory and the limit $\lim_{t \to \infty} e(t)$ does not exist. Thus, $e(t)$ is oscillatory.

The Nyquist criterion for zero intersymbol interference (ISI) requires that the sampled impulse response $x(mT)$ satisfies:
$x(mT) = \begin{cases} 1, & m=0 \\ 0, & m \neq 0 \end{cases}$
In the frequency domain, this condition is equivalent to the periodic summation of the Fourier transform $X(f)$ of the overall system response $x(t)$ being a constant:$$\sum_{m=-\infty}^{\infty} X\left(f + \frac{m}{T}\right) = T$$This ensures that the aliased spectrum of the pulse is flat, which corresponds to a delta function in the time domain after sampling. Therefore, option (A) is the correct condition.

For a lossless transmission line terminated with a short circuit, the load impedance is $z_L = 0$. The normalized input impedance at a distance $d$ from the load is given by:$$z_{in}(d) = j \tan(\beta d) = j \tan\left(\frac{2\pi}{\lambda} d\right)$$From the figure, the distances of the points from the load (at $z=0$) are:

• Point A: $d = \lambda/8$
• Point B: $d = \lambda/8 + \lambda/8 = \lambda/4$
• Point C: $d = \lambda/4 + \lambda/8 = 3\lambda/8$
• Point D: $d = 3\lambda/8 + \lambda/8 = \lambda/2$

Calculating the normalized impedances:Thus, the correct sequence is $z_{inA} = +1j \Omega, z_{inB} = \infty, z_{inC} = -1j \Omega, z_{inD} = 0$, which corresponds to option (A).

A vector field $\vec{F}$ is an electrostatic field if it is conservative, which means its curl is zero ($\nabla \times \vec{F} = 0$).For $\vec{F}_1 = A(y\hat{i} + x\hat{j}) = Ay\hat{i} + Ax\hat{j}$:$$\nabla \times \vec{F}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ Ay & Ax & 0 \end{vmatrix} = \hat{k} \left( \frac{\partial (Ax)}{\partial x} - \frac{\partial (Ay)}{\partial y} \right) = \hat{k} (A - A) = 0$$Thus, $\vec{F}_1$ is an electrostatic field.For $\vec{F}_2 = A(y\hat{i} - x\hat{j}) = Ay\hat{i} - Ax\hat{j}$:$$\nabla \times \vec{F}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ Ay & -Ax & 0 \end{vmatrix} = \hat{k} \left( \frac{\partial (-Ax)}{\partial x} - \frac{\partial (Ay)}{\partial y} \right) = \hat{k} (-A - A) = -2A\hat{k} \neq 0$$Thus, $\vec{F}_2$ is not an electrostatic field.Therefore, only $\vec{F}_1$ is an electrostatic field.

To find the input impedance $Z_{in}(j\omega)$, we use the small-signal model of the NMOS transistor in saturation. $I_{in} = (V_g - V_s) \cdot j\omega C_1$ $I_{in} + g_m(V_g - V_s) = V_s \cdot j\omega C_L$ $I_{in} + g_m \left( \frac{I_{in}}{j\omega C_1} \right) = V_s \cdot j\omega C_L$
$I_{in} \left( 1 + \frac{g_m}{j\omega C_1} \right) = V_s \cdot j\omega C_L \implies V_s = I_{in} \frac{1 + \frac{g_m}{j\omega C_1}}{j\omega C_L}$ $V_g = (V_g - V_s) + V_s = \frac{I_{in}}{j\omega C_1} + I_{in} \frac{1 + \frac{g_m}{j\omega C_1}}{j\omega C_L}$ $Z_{in}(j\omega) = \frac{V_g}{I_{in}} = \frac{1}{j\omega C_1} + \frac{1 + \frac{g_m}{j\omega C_1}}{j\omega C_L}$
$Z_{in}(j\omega) = \frac{1}{j\omega C_1} + \frac{1}{j\omega C_L} + \frac{g_m}{(j\omega C_1)(j\omega C_L)}$
$Z_{in}(j\omega) = \frac{1}{j\omega C_1} + \frac{1}{j\omega C_L} - \frac{g_m}{\omega^2 C_1 C_L}$This matches option (A).

The given circuit is a differential amplifier configured with unity negative feedback. $$A_{CL} = \frac{A}{1 + A\beta}$$

• Open-loop gain magnitude $A = 40$
• Feedback factor $\beta = 1$ (unity feedback)

$$A_{CL} = \frac{40}{1 + 40(1)} = \frac{40}{41}$$ $$A_{CL} = \frac{40}{41} \approx 0.975609...$$ $$A_{CL} \approx 0.976$$Thus, option (A) is correct.

To find the essential prime implicants (EPIs), we first identify all prime implicants (PIs) using a K-map for the given minterms: $\{0, 2, 5, 7, 8, 10, 12, 13, 14, 15\}$.1. Identify Prime Implicants (PIs):

• Group 1: $\{m_5, m_7, m_{13}, m_{15}\} \rightarrow BD$
• Group 2: $\{m_0, m_2, m_8, m_{10}\} \rightarrow \bar{B}\bar{D}$
• Group 3: $\{m_{12}, m_{13}, m_{14}, m_{15}\} \rightarrow AB$
• Group 4: $\{m_8, m_{10}, m_{12}, m_{14}\} \rightarrow A\bar{D}$

2. Identify Essential Prime Implicants (EPIs):
A prime implicant is essential if it covers at least one minterm that is not covered by any other prime implicant.

• Minterms $m_0$ and $m_2$ are covered only by $\bar{B}\bar{D}$. Thus, $\bar{B}\bar{D}$ is an EPI.
• Minterms $m_5$ and $m_7$ are covered only by $BD$. Thus, $BD$ is an EPI.
• Minterm $m_{12}$ is covered by $AB$ and $A\bar{D}$.
• Minterm $m_{13}$ is covered by $BD$ and $AB$.
• Minterm $m_{14}$ is covered by $AB$ and $A\bar{D}$.
• Minterm $m_{15}$ is covered by $BD$ and $AB$.
• Minterm $m_8$ is covered by $\bar{B}\bar{D}$ and $A\bar{D}$.
• Minterm $m_{10}$ is covered by $\bar{B}\bar{D}$ and $A\bar{D}$.

Since every minterm covered by $AB$ and $A\bar{D}$ is also covered by at least one other prime implicant, neither $AB$ nor $A\bar{D}$ is essential. The only essential prime implicants are $BD$ and $\bar{B}\bar{D}$.Therefore, the correct option is (A).

The power spectral density (PSD) of the white Gaussian noise $w(t)$ is given as $S_w(f) = \frac{N_0}{2}$.The transfer function of a first-order RC low pass filter is given by:
$H(f) = \frac{1}{1 + j2\pi f RC}$The magnitude squared of the transfer function is:
$|H(f)|^2 = \left|\frac{1}{1 + j2\pi f RC}\right|^2 = \frac{1}{1 + (2\pi f RC)^2}$The PSD of the output noise $n(t)$ is $S_n(f) = S_w(f) |H(f)|^2$:
$S_n(f) = \frac{N_0}{2} \cdot \frac{1}{1 + (2\pi f RC)^2}$The variance of the random variable $n(t_k)$ at any time $t_k$ is equal to the total average power of the output noise, which is the integral of its PSD over all frequencies:
$\sigma_n^2 = \int_{-\infty}^{\infty} S_n(f) df = \int_{-\infty}^{\infty} \frac{N_0}{2} \cdot \frac{1}{1 + (2\pi f RC)^2} df$Let $x = 2\pi f RC$. Then, $df = \frac{dx}{2\pi RC}$.
When $f \to -\infty$, $x \to -\infty$.
When $f \to \infty$, $x \to \infty$.Substitute these into the integral:
$\sigma_n^2 = \frac{N_0}{2} \int_{-\infty}^{\infty} \frac{1}{1 + x^2} \frac{dx}{2\pi RC}$
$\sigma_n^2 = \frac{N_0}{4\pi RC} \int_{-\infty}^{\infty} \frac{1}{1 + x^2} dx$The integral $\int_{-\infty}^{\infty} \frac{1}{1 + x^2} dx$ is a standard integral:
$\int_{-\infty}^{\infty} \frac{1}{1 + x^2} dx = [\arctan(x)]_{-\infty}^{\infty} = \arctan(\infty) - \arctan(-\infty) = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$Substitute the value of the integral back into the expression for $\sigma_n^2$:
$\sigma_n^2 = \frac{N_0}{4\pi RC} \cdot \pi = \frac{N_0}{4RC}$Thus, the variance of the random variable $n(t_k)$ is $\frac{N_0}{4RC}$.The final answer is $\boxed{\text{A}}$