PYQs / GATE EC / 2024 / Set 1 / Q48 GATE EC 2024 Set 1 — Question 48 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Hard Sampling Theorem Continuous-Time Signals Signals & Systems Filters & Bandwidth LTI Systems
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Last updated 5 September 2026
Question A continuous time signal
x ( t ) = 2 cos ( 8 π t + π / 3 ) x(t) = 2 \cos(8\pi t + \pi/3) x ( t ) = 2 cos ( 8 π t + π /3 ) is sampled at a rate of
15 15 15 Hz. The sampled signal
x s ( t ) x_s(t) x s ( t ) when passed through an LTI system with impulse response
h ( t ) = ( sin 2 π t π t ) cos ( 38 π t − π / 2 ) h(t) = \left( \frac{\sin 2\pi t}{\pi t} \right) \cos(38\pi t - \pi/2) h ( t ) = ( π t sin 2 π t ) cos ( 38 π t − π /2 ) produces an output
x o ( t ) x_o(t) x o ( t ) . The expression for
x o ( t ) x_o(t) x o ( t ) is
_____ .
Correct answer (C) 15 cos(38π t - π/6)
Solution 1. Input signal x ( t ) = 2 cos ( 8 π t + π / 3 ) x(t) = 2 \cos(8\pi t + \pi/3) x ( t ) = 2 cos ( 8 π t + π /3 ) f 0 = 4 f_0 = 4 f 0 = 4 2. Sampling frequency f s = 15 f_s = 15 f s = 15 X s ( f ) = f s ∑ k = − ∞ ∞ X ( f − k f s ) X_s(f) = f_s \sum_{k=-\infty}^{\infty} X(f - kf_s) X s ( f ) = f s ∑ k = − ∞ ∞ X ( f − k f s ) 3. X ( f ) = e j π / 3 δ ( f − 4 ) + e − j π / 3 δ ( f + 4 ) X(f) = e^{j\pi/3} \delta(f-4) + e^{-j\pi/3} \delta(f+4) X ( f ) = e j π /3 δ ( f − 4 ) + e − j π /3 δ ( f + 4 ) 4. X s ( f ) = 15 ∑ k [ e j π / 3 δ ( f − 4 − 15 k ) + e − j π / 3 δ ( f + 4 − 15 k ) ] X_s(f) = 15 \sum_{k} [e^{j\pi/3} \delta(f-4-15k) + e^{-j\pi/3} \delta(f+4-15k)] X s ( f ) = 15 ∑ k [ e j π /3 δ ( f − 4 − 15 k ) + e − j π /3 δ ( f + 4 − 15 k )] 5. Impulse response h ( t ) = 2 sinc ( 2 t ) sin ( 38 π t ) h(t) = 2 \text{sinc}(2t) \sin(38\pi t) h ( t ) = 2 sinc ( 2 t ) sin ( 38 π t ) H ( f ) = rect ( f / 2 ) ∗ 1 2 j [ δ ( f − 19 ) − δ ( f + 19 ) ] = 1 2 j [ rect ( f − 19 2 ) − rect ( f + 19 2 ) ] H(f) = \text{rect}(f/2) * \frac{1}{2j} [\delta(f-19) - \delta(f+19)] = \frac{1}{2j} [\text{rect}(\frac{f-19}{2}) - \text{rect}(\frac{f+19}{2})] H ( f ) = rect ( f /2 ) ∗ 2 j 1 [ δ ( f − 19 ) − δ ( f + 19 )] = 2 j 1 [ rect ( 2 f − 19 ) − rect ( 2 f + 19 )] 6. The passband of H ( f ) H(f) H ( f ) [ 18 , 20 ] [18, 20] [ 18 , 20 ] [ − 20 , − 18 ] [-20, -18] [ − 20 , − 18 ] 7. From X s ( f ) X_s(f) X s ( f ) k = 1 k=1 k = 1 f = 4 + 15 ( 1 ) = 19 f = 4 + 15(1) = 19 f = 4 + 15 ( 1 ) = 19 k = − 1 k=-1 k = − 1 f = − 4 + 15 ( − 1 ) = − 19 f = -4 + 15(-1) = -19 f = − 4 + 15 ( − 1 ) = − 19 8. At f = 19 f=19 f = 19 X o ( 19 ) = 15 e j π / 3 ⋅ 1 2 j = 15 2 e j ( π / 3 − π / 2 ) = 15 2 e − j π / 6 X_o(19) = 15 e^{j\pi/3} \cdot \frac{1}{2j} = \frac{15}{2} e^{j(\pi/3 - \pi/2)} = \frac{15}{2} e^{-j\pi/6} X o ( 19 ) = 15 e j π /3 ⋅ 2 j 1 = 2 15 e j ( π /3 − π /2 ) = 2 15 e − j π /6 9. At f = − 19 f=-19 f = − 19 X o ( − 19 ) = 15 e − j π / 3 ⋅ − 1 2 j = 15 2 e − j π / 3 e j π / 2 = 15 2 e j π / 6 X_o(-19) = 15 e^{-j\pi/3} \cdot \frac{-1}{2j} = \frac{15}{2} e^{-j\pi/3} e^{j\pi/2} = \frac{15}{2} e^{j\pi/6} X o ( − 19 ) = 15 e − j π /3 ⋅ 2 j − 1 = 2 15 e − j π /3 e j π /2 = 2 15 e j π /6 10. x o ( t ) = 15 cos ( 38 π t − π / 6 ) x_o(t) = 15 \cos(38\pi t - \pi/6) x o ( t ) = 15 cos ( 38 π t − π /6 ) Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker. Review answer and solution without JavaScript Interactive answer checking needs JavaScript. The published solution is available below.
Correct answer (C) 15 cos(38π t - π/6)
Solution 1. Input signal x ( t ) = 2 cos ( 8 π t + π / 3 ) x(t) = 2 \cos(8\pi t + \pi/3) x ( t ) = 2 cos ( 8 π t + π /3 ) f 0 = 4 f_0 = 4 f 0 = 4 2. Sampling frequency f s = 15 f_s = 15 f s = 15 X s ( f ) = f s ∑ k = − ∞ ∞ X ( f − k f s ) X_s(f) = f_s \sum_{k=-\infty}^{\infty} X(f - kf_s) X s ( f ) = f s ∑ k = − ∞ ∞ X ( f − k f s ) 3. X ( f ) = e j π / 3 δ ( f − 4 ) + e − j π / 3 δ ( f + 4 ) X(f) = e^{j\pi/3} \delta(f-4) + e^{-j\pi/3} \delta(f+4) X ( f ) = e j π /3 δ ( f − 4 ) + e − j π /3 δ ( f + 4 ) 4. X s ( f ) = 15 ∑ k [ e j π / 3 δ ( f − 4 − 15 k ) + e − j π / 3 δ ( f + 4 − 15 k ) ] X_s(f) = 15 \sum_{k} [e^{j\pi/3} \delta(f-4-15k) + e^{-j\pi/3} \delta(f+4-15k)] X s ( f ) = 15 ∑ k [ e j π /3 δ ( f − 4 − 15 k ) + e − j π /3 δ ( f + 4 − 15 k )] 5. Impulse response h ( t ) = 2 sinc ( 2 t ) sin ( 38 π t ) h(t) = 2 \text{sinc}(2t) \sin(38\pi t) h ( t ) = 2 sinc ( 2 t ) sin ( 38 π t ) H ( f ) = rect ( f / 2 ) ∗ 1 2 j [ δ ( f − 19 ) − δ ( f + 19 ) ] = 1 2 j [ rect ( f − 19 2 ) − rect ( f + 19 2 ) ] H(f) = \text{rect}(f/2) * \frac{1}{2j} [\delta(f-19) - \delta(f+19)] = \frac{1}{2j} [\text{rect}(\frac{f-19}{2}) - \text{rect}(\frac{f+19}{2})] H ( f ) = rect ( f /2 ) ∗ 2 j 1 [ δ ( f − 19 ) − δ ( f + 19 )] = 2 j 1 [ rect ( 2 f − 19 ) − rect ( 2 f + 19 )] 6. The passband of H ( f ) H(f) H ( f ) [ 18 , 20 ] [18, 20] [ 18 , 20 ] [ − 20 , − 18 ] [-20, -18] [ − 20 , − 18 ] 7. From X s ( f ) X_s(f) X s ( f ) k = 1 k=1 k = 1 f = 4 + 15 ( 1 ) = 19 f = 4 + 15(1) = 19 f = 4 + 15 ( 1 ) = 19 k = − 1 k=-1 k = − 1 f = − 4 + 15 ( − 1 ) = − 19 f = -4 + 15(-1) = -19 f = − 4 + 15 ( − 1 ) = − 19 8. At f = 19 f=19 f = 19 X o ( 19 ) = 15 e j π / 3 ⋅ 1 2 j = 15 2 e j ( π / 3 − π / 2 ) = 15 2 e − j π / 6 X_o(19) = 15 e^{j\pi/3} \cdot \frac{1}{2j} = \frac{15}{2} e^{j(\pi/3 - \pi/2)} = \frac{15}{2} e^{-j\pi/6} X o ( 19 ) = 15 e j π /3 ⋅ 2 j 1 = 2 15 e j ( π /3 − π /2 ) = 2 15 e − j π /6 9. At f = − 19 f=-19 f = − 19 X o ( − 19 ) = 15 e − j π / 3 ⋅ − 1 2 j = 15 2 e − j π / 3 e j π / 2 = 15 2 e j π / 6 X_o(-19) = 15 e^{-j\pi/3} \cdot \frac{-1}{2j} = \frac{15}{2} e^{-j\pi/3} e^{j\pi/2} = \frac{15}{2} e^{j\pi/6} X o ( − 19 ) = 15 e − j π /3 ⋅ 2 j − 1 = 2 15 e − j π /3 e j π /2 = 2 15 e j π /6 10. x o ( t ) = 15 cos ( 38 π t − π / 6 ) x_o(t) = 15 \cos(38\pi t - \pi/6) x o ( t ) = 15 cos ( 38 π t − π /6 ) Understand the concept, then try another question Revisit Signals & Systems with concept notes, common mistakes and an original worked example before your next attempt.
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