Signal and systems miscellaneous Easy Questions and Answers | Page - 49

Signal and systems miscellaneous

The Laplace transform of a given wave—

1. 1 (1 – e^{– sT})
  s
1. 1 (1 – e^{– sT})²
  s
1. 1 (1 + e^{– sT})²
  s
1. 1 (1 – e^sT)²
  s

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For the given waveform,

L{f(t)} = ∫^∞₀ f(t)e^{– st} dt
= ∫^T₀ 1·e– st dt + ∫ 2T T (– 1) e– stdt
= e^{– st} – e^{– s}t ^2T
– s – s _T

= 1 [e^{– st} – 1 – e^{– 2st} + e^{– st}]
s

= 1 [1 – 2e^st + e^2st]
s

= 1 (1 – e^{– sT})²
s

Correct Option: B

For the given waveform,

L{f(t)} = ∫^∞₀ f(t)e^{– st} dt
= ∫^T₀ 1·e– st dt + ∫ 2T T (– 1) e– stdt
= e^{– st} – e^{– s}t ^2T
– s – s _T

= 1 [e^{– st} – 1 – e^{– 2st} + e^{– st}]
s

= 1 [1 – 2e^st + e^2st]
s

= 1 (1 – e^{– sT})²
s

The Laplace transform of the given wave as shown below is—

1. A· e^{– πs} – 1
  (s² + 1)
1. A· e^{– πs} + 1
  (s² + 1)
1. A· e^πs – 1
  (s² + 1)
1. A· e^πs + 1
  (s² + 1)

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The function for the given waveform is
f(t) = A sin t for 0 < t < π
= 0, for t > π

By definition, we have
L{f(t)} = ∫^∞₀ f(t)e^{– st}dt
= ∫^π₀ A sin t e^{– st}dt
= A ∫^π₀ sin t e^{– st}dt
= A [e^{– st}(– s sin t – cos t)]^π₀
(s² + 1)

= A e^{– sπ} + 1
(s²+ 1)

Hence, alternative (B) is the correct choice.

Correct Option: B

The function for the given waveform is
f(t) = A sin t for 0 < t < π
= 0, for t > π

By definition, we have
L{f(t)} = ∫^∞₀ f(t)e^{– st}dt
= ∫^π₀ A sin t e^{– st}dt
= A ∫^π₀ sin t e^{– st}dt
= A [e^{– st}(– s sin t – cos t)]^π₀
(s² + 1)

= A e^{– sπ} + 1
(s²+ 1)

Hence, alternative (B) is the correct choice.

The convolution integral when—
f₁(t) = e^{– 2t}
and f₂(t)= 2t is—

1. t – 1 + e^{– 2t} u(t)
  2 2
1. t + 1 - e^{– 2t} u(t)
  2 2
1. t – 1 - e^{– 2t} u(t)
  2 2
1. t + 1 + e^2t u(t)
  2 2

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We know that
f₁(t)*f₂(t) = ∫^t₀ f₁(τ ).f₂(t – τ )dτ
= ∫^t₀ 2τ·e^-2(t – τ) dτ
= e^{- 2t} ∫^t₀ 2τ·e^2τdτ
= 2e^{- 2t} τ· e^2τ ∫^t₀ τ· e^2τ · dτ ^t
2 2 ₀

= 2e^{- 2t} τe^2τ – e^2τ ^t
2 4 ₀

= 2e^{- 2t} te² – e^2t + 1
2 4 4

= t 1 + e^{- 2t} u(t)
2 2

Hence, alternative (A) is the correct answer.

Correct Option: A

We know that
f₁(t)*f₂(t) = ∫^t₀ f₁(τ ).f₂(t – τ )dτ
= ∫^t₀ 2τ·e^-2(t – τ) dτ
= e^{- 2t} ∫^t₀ 2τ·e^2τdτ
= 2e^{- 2t} τ· e^2τ ∫^t₀ τ· e^2τ · dτ ^t
2 2 ₀

= 2e^{- 2t} τe^2τ – e^2τ ^t
2 4 ₀

= 2e^{- 2t} te² – e^2t + 1
2 4 4

= t 1 + e^{- 2t} u(t)
2 2

Hence, alternative (A) is the correct answer.

The output response y(t) of the RC network shown in fig. is—

1. 1^{– et/RC}
1. 1 + e^{– t/RC}
1. 1 – e^{– t/RC}
1. 1 + e^t/RC

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The transfer function of the network

H(s) = 1
Cs
R + 1
Cs

= 1 · 1
RC s + 1
RC

Therefore, the impulse response of the network is
h(t) = 1 e^{– t/RC} for t ≥ 0
RC

Given, f(t) = 0, for t < 0
1, for t ≥ 0

= ^T x(τ) h(t – τ)·dτ
₀

= ^t 1· 1 e^{– (t – τ )/RC}dτ
₀ RC

= e^{– t/RC} ^t e^τ/RCdτ
RC ₀

= e^{-(t – τ)/RC} [RC·e^τ/RCdt]^t₀
RC

= e^{- t/RC} [e^t/RC – 1]
RC

= 1 – e^{– t/RC}
Hence, alternative (C) is the correct choice.

Correct Option: C

The transfer function of the network

H(s) = 1
Cs
R + 1
Cs

= 1 · 1
RC s + 1
RC

Therefore, the impulse response of the network is
h(t) = 1 e^{– t/RC} for t ≥ 0
RC

Given, f(t) = 0, for t < 0
1, for t ≥ 0

= ^T x(τ) h(t – τ)·dτ
₀

= ^t 1· 1 e^{– (t – τ )/RC}dτ
₀ RC

= e^{– t/RC} ^t e^τ/RCdτ
RC ₀

= e^{-(t – τ)/RC} [RC·e^τ/RCdt]^t₀
RC

= e^{- t/RC} [e^t/RC – 1]
RC

= 1 – e^{– t/RC}
Hence, alternative (C) is the correct choice.

What is the inverse Laplace transform of e^–as ?
s

1. e^{– at}
1. u(t – a)
1. δ(t – a)
1. (t – a) u(t – a)

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We know that
L{u(t)} = 1
3

L{u(t – a)} = e^{– as}
s

Therefore, (B) is the correct choice.

Correct Option: B

We know that
L{u(t)} = 1
3

L{u(t – a)} = e^{– as}
s

Therefore, (B) is the correct choice.