Signal and systems miscellaneous Easy Questions and Answers | Page - 54

Signal and systems miscellaneous

If [f(t)] = F(s), then [f(t – T)] is equal to—

1. e^sTF(s)
1. e^{– sT}F(s)
1. F(s)
  1 + e^sT
1. F(s)
  1 – e^sT

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Given that,
if f(t) = F(s)
Then Fourier transform of f(t – T) can be obtained by using time shifting property.
F(t – T) = ∫^∞_–∞f(t – T)e^{– jωt} ·dt
Putting, x = t – T
⇒ t = x + T
dx = dt
= ∫^∞_–∞ f(x)e^{– jω(x + T)}dx
= e^{– jωT} ∫^∞_–∞ f(x).e^{– jωx}dx
= e^{– jωT} F.(s)
= e^{– sT} F.(s)
Hence, alternative (B) is the correct choice.

Correct Option: B

Given that,
if f(t) = F(s)
Then Fourier transform of f(t – T) can be obtained by using time shifting property.
F(t – T) = ∫^∞_–∞f(t – T)e^{– jωt} ·dt
Putting, x = t – T
⇒ t = x + T
dx = dt
= ∫^∞_–∞ f(x)e^{– jω(x + T)}dx
= e^{– jωT} ∫^∞_–∞ f(x).e^{– jωx}dx
= e^{– jωT} F.(s)
= e^{– sT} F.(s)
Hence, alternative (B) is the correct choice.

A signal f(t) has energy E, the energy of the signal f(2t) will be energy—

1. E
1. E
  2
1. 2E
1. None of these

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We know that energy of a signal is given as
E = ∫^∞_–∞ |f(t)|²dt
Consider energy for signal f(2t) is E′
then E′ = ∫^∞_–∞ |f(p)|²dp
Let 2t = p
dt = dp
2

E′ = ∫^∞_–∞ |f(t)|² dt 2

E′ = E
2

(∵ by using change of variable property)
Hence, alternative (B) is the correct choice.

Correct Option: B

We know that energy of a signal is given as
E = ∫^∞_–∞ |f(t)|²dt
Consider energy for signal f(2t) is E′
then E′ = ∫^∞_–∞ |f(p)|²dp
Let 2t = p
dt = dp
2

E′ = ∫^∞_–∞ |f(t)|² dt 2

E′ = E
2

(∵ by using change of variable property)
Hence, alternative (B) is the correct choice.