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GATE Electronics & Communication Engineering Mastery Hub: Th

Timed mock exams, detailed analytics, and practice drills for GATE Electronics & Communication Engineering Mastery Hub: The Industry Foundation.

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Q1Domain Verified
In the context of signal sampling, what is the minimum sampling frequency required to perfectly reconstruct a signal with a maximum frequency component of $f_{max}$ according to the Nyquist-Shannon sampling theorem, and what is the primary consequence of undersampling?
$f_s = 2f_{max}$, leading to aliasing.
$f_s = f_{max}/2$, leading to signal attenuation.
$f_s = f_{max}$, leading to signal distortion.
$f_s = 2f_{max}$, leading to increased signal bandwidth.
Q2Domain Verified
Consider a causal Linear Time-Invariant (LTI) system with impulse response $h(t) = e^{-at}u(t)$, where $a > 0$ and $u(t)$ is the unit step function. If an input signal $x(t) = e^{-bt}u(t)$, where $b > 0$, is applied to this system, what is the condition on $a$ and $b$ for the output signal $y(t)$ to be bounded?
$a \ge 0$ and $b \ge 0$
$a < b$
$a = b$
$a > b$
Q3Domain Verified
asks about the output being bounded for a specific input. The output $y(t)$ is the convolution of $x(t)$ and $h(t)$. When $x(t) = e^{-bt}u(t)$ and $h(t) = e^{-at}u(t)$, the convolution integral is $\int_{0}^{t} e^{-b\tau} e^{-a(t-\tau)} d\tau$. For $y(t)$ to be bounded as $t \to \infty$, the dominant exponential term in the convolution must decay. If $a > b$, the term $e^{-at}$ dominates the decay. If $b > a$, the term $e^{-bt}$ dominates. If $a=b$, the integral leads to a term proportional to $t e^{-at}$, which grows unboundedly. Thus, for $y(t)$ to be bounded, the decay rate of the impulse response must be faster than or equal to the decay rate of the input. More rigorously, the convolution integral involves terms like $e^{-(a-b)\tau}e^{-at}$. For the integral to converge and the output to be bounded, we need $a > b$. Option B is incorrect because if $a < b$, the term $e^{-bt}$ will dominate the decay, and the integral will diverge. Option C is incorrect because if $a=b$, the output will contain a term proportional to $t e^{-at}$, which grows unboundedly as $t \to \infty$. Option D is too general and doesn't consider the relative decay rates. Question: In a digital communication system, a signal is transmitted through a channel with additive white Gaussian noise (AWGN). If the signal power is $P_s$ and the noise power spectral density is $N_0/2$, what is the signal-to-noise ratio (SNR) at the receiver's output after passing through a matched filter for a signal with bandwidth $B$?
$SNR = \frac{P_s B}{N_0}$
$SNR = \frac{P_s}{N_0}$
$SNR = \frac{2 P_s B}{N_0}$
$SNR = \frac{P_s}{N_0 B}$

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This domain protocol is rigorously covered in our 2026 Elite Framework. Every mock reflects direct alignment with the official assessment criteria to eliminate performance gaps.

This domain protocol is rigorously covered in our 2026 Elite Framework. Every mock reflects direct alignment with the official assessment criteria to eliminate performance gaps.

This domain protocol is rigorously covered in our 2026 Elite Framework. Every mock reflects direct alignment with the official assessment criteria to eliminate performance gaps.

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