PYQs / GATE EC / 2026 / Set 1 / Q46 GATE EC 2026 Set 1 — Question 46 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 Medium Energy & Power Signals Continuous-Time Signals Signals & Systems Parseval's Theorem
Signals & Systems → Continuous-Time Signals → Energy & Power Signals
Last updated 5 September 2026
Question Consider a real baseband signal
x ( t ) = e − 2 t x(t) = e^{-2t} x ( t ) = e − 2 t , for
t t t (in seconds)
≥ 0 \geq 0 ≥ 0 .
If 99% of energy of
x ( t ) x(t) x ( t ) lies within
B B B Hz, then which of the following options is TRUE for the value of
B B B ?
Correct answer (B) 63/π Hz < B < 64/π Hz
Solution The given signal is
x ( t ) = e − 2 t u ( t ) x(t) = e^{-2t} u(t) x ( t ) = e − 2 t u ( t ) .
1. Total Energy (E t o t a l E_{total} E t o t a l E t o t a l = ∫ − ∞ ∞ ∣ x ( t ) ∣ 2 d t = ∫ 0 ∞ ( e − 2 t ) 2 d t = ∫ 0 ∞ e − 4 t d t = [ e − 4 t − 4 ] 0 ∞ = 1 4 = 0.25 J E_{total} = \int_{-\infty}^{\infty} |x(t)|^2 dt = \int_{0}^{\infty} (e^{-2t})^2 dt = \int_{0}^{\infty} e^{-4t} dt = \left[ \frac{e^{-4t}}{-4} \right]_0^\infty = \frac{1}{4} = 0.25 \text{ J} E t o t a l = ∫ − ∞ ∞ ∣ x ( t ) ∣ 2 d t = ∫ 0 ∞ ( e − 2 t ) 2 d t = ∫ 0 ∞ e − 4 t d t = [ − 4 e − 4 t ] 0 ∞ = 4 1 = 0.25 J 2. Energy Spectral Density (S x ( f ) S_x(f) S x ( f ) The Fourier transform of
x ( t ) x(t) x ( t ) is
X ( f ) = 1 2 + j 2 π f X(f) = \frac{1}{2 + j2\pi f} X ( f ) = 2 + j 2 π f 1 .
S x ( f ) = ∣ X ( f ) ∣ 2 = 1 4 + 4 π 2 f 2 S_x(f) = |X(f)|^2 = \frac{1}{4 + 4\pi^2 f^2} S x ( f ) = ∣ X ( f ) ∣ 2 = 4 + 4 π 2 f 2 1 3. Energy within Bandwidth B B B E B E_B E B For a real baseband signal, the energy within
B B B Hz is typically calculated over the frequency range
[ − B , B ] [-B, B] [ − B , B ] .
E B = ∫ − B B S x ( f ) d f = ∫ − B B 1 4 + 4 π 2 f 2 d f = 2 ∫ 0 B 1 4 ( 1 + π 2 f 2 ) d f E_B = \int_{-B}^{B} S_x(f) df = \int_{-B}^{B} \frac{1}{4 + 4\pi^2 f^2} df = 2 \int_{0}^{B} \frac{1}{4(1 + \pi^2 f^2)} df E B = ∫ − B B S x ( f ) df = ∫ − B B 4 + 4 π 2 f 2 1 df = 2 ∫ 0 B 4 ( 1 + π 2 f 2 ) 1 df E B = 1 2 ∫ 0 B 1 1 + ( π f ) 2 d f = 1 2 [ 1 π tan − 1 ( π f ) ] 0 B = 1 2 π tan − 1 ( π B ) E_B = \frac{1}{2} \int_{0}^{B} \frac{1}{1 + (\pi f)^2} df = \frac{1}{2} \left[ \frac{1}{\pi} \tan^{-1}(\pi f) \right]_0^B = \frac{1}{2\pi} \tan^{-1}(\pi B) E B = 2 1 ∫ 0 B 1 + ( π f ) 2 1 df = 2 1 [ π 1 tan − 1 ( π f ) ] 0 B = 2 π 1 tan − 1 ( π B ) Given
E B = 0.99 E t o t a l E_B = 0.99 E_{total} E B = 0.99 E t o t a l :
1 2 π tan − 1 ( π B ) = 0.99 × 1 4 \frac{1}{2\pi} \tan^{-1}(\pi B) = 0.99 \times \frac{1}{4} 2 π 1 tan − 1 ( π B ) = 0.99 × 4 1 tan − 1 ( π B ) = 0.99 × 2 π 4 = 0.495 π \tan^{-1}(\pi B) = \frac{0.99 \times 2\pi}{4} = 0.495\pi tan − 1 ( π B ) = 4 0.99 × 2 π = 0.495 π π B = tan ( 0.495 π ) ≈ 63.66 \pi B = \tan(0.495\pi) \approx 63.66 π B = tan ( 0.495 π ) ≈ 63.66 B = tan ( 0.495 π ) π ≈ 63.66 π Hz B = \frac{\tan(0.495\pi)}{\pi} \approx \frac{63.66}{\pi} \text{ Hz} B = π tan ( 0.495 π ) ≈ π 63.66 Hz 5. Conclusion:
Since
63 < 63.66 < 64 63 < 63.66 < 64 63 < 63.66 < 64 , we have:
63 π < B < 64 π \frac{63}{\pi} < B < \frac{64}{\pi} π 63 < B < π 64 Thus, option (B) is correct.
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 (B) 63/π Hz < B < 64/π Hz
Solution The given signal is
x ( t ) = e − 2 t u ( t ) x(t) = e^{-2t} u(t) x ( t ) = e − 2 t u ( t ) .
1. Total Energy (E t o t a l E_{total} E t o t a l E t o t a l = ∫ − ∞ ∞ ∣ x ( t ) ∣ 2 d t = ∫ 0 ∞ ( e − 2 t ) 2 d t = ∫ 0 ∞ e − 4 t d t = [ e − 4 t − 4 ] 0 ∞ = 1 4 = 0.25 J E_{total} = \int_{-\infty}^{\infty} |x(t)|^2 dt = \int_{0}^{\infty} (e^{-2t})^2 dt = \int_{0}^{\infty} e^{-4t} dt = \left[ \frac{e^{-4t}}{-4} \right]_0^\infty = \frac{1}{4} = 0.25 \text{ J} E t o t a l = ∫ − ∞ ∞ ∣ x ( t ) ∣ 2 d t = ∫ 0 ∞ ( e − 2 t ) 2 d t = ∫ 0 ∞ e − 4 t d t = [ − 4 e − 4 t ] 0 ∞ = 4 1 = 0.25 J 2. Energy Spectral Density (S x ( f ) S_x(f) S x ( f ) The Fourier transform of
x ( t ) x(t) x ( t ) is
X ( f ) = 1 2 + j 2 π f X(f) = \frac{1}{2 + j2\pi f} X ( f ) = 2 + j 2 π f 1 .
S x ( f ) = ∣ X ( f ) ∣ 2 = 1 4 + 4 π 2 f 2 S_x(f) = |X(f)|^2 = \frac{1}{4 + 4\pi^2 f^2} S x ( f ) = ∣ X ( f ) ∣ 2 = 4 + 4 π 2 f 2 1 3. Energy within Bandwidth B B B E B E_B E B For a real baseband signal, the energy within
B B B Hz is typically calculated over the frequency range
[ − B , B ] [-B, B] [ − B , B ] .
E B = ∫ − B B S x ( f ) d f = ∫ − B B 1 4 + 4 π 2 f 2 d f = 2 ∫ 0 B 1 4 ( 1 + π 2 f 2 ) d f E_B = \int_{-B}^{B} S_x(f) df = \int_{-B}^{B} \frac{1}{4 + 4\pi^2 f^2} df = 2 \int_{0}^{B} \frac{1}{4(1 + \pi^2 f^2)} df E B = ∫ − B B S x ( f ) df = ∫ − B B 4 + 4 π 2 f 2 1 df = 2 ∫ 0 B 4 ( 1 + π 2 f 2 ) 1 df E B = 1 2 ∫ 0 B 1 1 + ( π f ) 2 d f = 1 2 [ 1 π tan − 1 ( π f ) ] 0 B = 1 2 π tan − 1 ( π B ) E_B = \frac{1}{2} \int_{0}^{B} \frac{1}{1 + (\pi f)^2} df = \frac{1}{2} \left[ \frac{1}{\pi} \tan^{-1}(\pi f) \right]_0^B = \frac{1}{2\pi} \tan^{-1}(\pi B) E B = 2 1 ∫ 0 B 1 + ( π f ) 2 1 df = 2 1 [ π 1 tan − 1 ( π f ) ] 0 B = 2 π 1 tan − 1 ( π B ) Given
E B = 0.99 E t o t a l E_B = 0.99 E_{total} E B = 0.99 E t o t a l :
1 2 π tan − 1 ( π B ) = 0.99 × 1 4 \frac{1}{2\pi} \tan^{-1}(\pi B) = 0.99 \times \frac{1}{4} 2 π 1 tan − 1 ( π B ) = 0.99 × 4 1 tan − 1 ( π B ) = 0.99 × 2 π 4 = 0.495 π \tan^{-1}(\pi B) = \frac{0.99 \times 2\pi}{4} = 0.495\pi tan − 1 ( π B ) = 4 0.99 × 2 π = 0.495 π π B = tan ( 0.495 π ) ≈ 63.66 \pi B = \tan(0.495\pi) \approx 63.66 π B = tan ( 0.495 π ) ≈ 63.66 B = tan ( 0.495 π ) π ≈ 63.66 π Hz B = \frac{\tan(0.495\pi)}{\pi} \approx \frac{63.66}{\pi} \text{ Hz} B = π tan ( 0.495 π ) ≈ π 63.66 Hz 5. Conclusion:
Since
63 < 63.66 < 64 63 < 63.66 < 64 63 < 63.66 < 64 , we have:
63 π < B < 64 π \frac{63}{\pi} < B < \frac{64}{\pi} π 63 < B < π 64 Thus, option (B) is correct.
Understand the concept, then try another question Revisit Signals & Systems with concept notes, common mistakes and an original worked example before your next attempt.
More questions on Continuous-Time Signals ← Q45 Full paper Q47 → 