Signal and systems miscellaneous Easy Questions and Answers | Page - 45

Signal and systems miscellaneous

The value of ∫^∞_–∞ e^{– 2t} s′(t)dt—

1. 0
1. 1
1. 2
1. None of these

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Given ∫^∞_{– ∞} e^{– 2t} s′(t)dt
By applying the property of impulse function
∫^∞_{– ∞} f(t) s′(t)dt = – d f(t )|_{at t = 0}
dt

= – d e^{– 2t} |_{at, t = 0}
dt

= – (– 2)e^{– 2t} | _{at, t = 0}
= 2·e^{– 2·0} = 2
Hence, alternative (C) is the correct choice.

Correct Option: C

Given ∫^∞_{– ∞} e^{– 2t} s′(t)dt
By applying the property of impulse function
∫^∞_{– ∞} f(t) s′(t)dt = – d f(t )|_{at t = 0}
dt

= – d e^{– 2t} |_{at, t = 0}
dt

= – (– 2)e^{– 2t} | _{at, t = 0}
= 2·e^{– 2·0} = 2
Hence, alternative (C) is the correct choice.

The causal systems is/are—
(i) y(n) = x²(n) u(n)
(ii) y(n) = x(|n|)
(iii) y(n) = x(n) – x(n² – n)
_N
(iv) y(n) = ∑ x(n – k)
^{k = 1}

1. (i), (ii) and (iii)
1. (i) and (iv)
1. (iii) and (iv)
1. (i) and (iv)

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Given
(i) y(n) = x²(n) u(n)
This system is causal for any value of n.
(ii) y(n) = x(|n|) If we put any negative value of n, say n = – 2, we get
y(– 2) = x(|– 2|) = x(2)
i.e., output depends on the future value of input.
Hence, this system is non-causal.
(iii) y(n) = x(n) – x(n² – n)
This system is causal for n = 1, 2 and non-causal for values other than 0, 1 and 2.
Let n = 3, we get
y(3) = x(3) – x(3² – 3)
y(3) = x(3) – x(6)
which is non-causal
_N
(iv) y(n) = ∑ x(n – k)
^{k = 1}

Since, the limit of k is positive, therefore, this system is causal for any value of k.
Hence, alternative (D) is the correct choice.

Correct Option: D

Given
(i) y(n) = x²(n) u(n)
This system is causal for any value of n.
(ii) y(n) = x(|n|) If we put any negative value of n, say n = – 2, we get
y(– 2) = x(|– 2|) = x(2)
i.e., output depends on the future value of input.
Hence, this system is non-causal.
(iii) y(n) = x(n) – x(n² – n)
This system is causal for n = 1, 2 and non-causal for values other than 0, 1 and 2.
Let n = 3, we get
y(3) = x(3) – x(3² – 3)
y(3) = x(3) – x(6)
which is non-causal
_N
(iv) y(n) = ∑ x(n – k)
^{k = 1}

Since, the limit of k is positive, therefore, this system is causal for any value of k.
Hence, alternative (D) is the correct choice.

If x₁(t) = 2 sin πt + cos 4πt
x₂(t) = sin 5πt + 3 sin 13πt, then—

1. x₁(t) and x₂(t) both are periodic
1. x₁(t) and x₂(t) both are non-periodic
1. x₁ is periodic, but x₂ is not periodic
1. x₁ is not periodic, but x₂ is periodic

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Given x₁(t) = 2 sin πt + cos 4πt
Fundamental period of
x₁′(t)=T₁′ = 2π = 2
π

Fundamental period of
x₁′′(t)=T₁′′ = 2π = 1
4π 2

Ratio = T₁′ = 2 = 4 = integer (or rational)
T₂′′ 1 1
2

Therefore, x₁(t) is periodic.
Now, for signal
x₂(t) = sin 5πt + 3 sin 13πt
Fundamental period of
x₂′(t)=T₂′ = 2π = 2
5π 5

Fundamental period of
x₂′′(t) T₂′′ = 2π = 2
13π 13

Ratio = 2 = = 2·6
T₁′ 5 13
T₂′′ 1 5
13

i.e., rational. Therefore, x₂(t) is positive.
Hence, alternative (A) is the correct answer.

Correct Option: A

Given x₁(t) = 2 sin πt + cos 4πt
Fundamental period of
x₁′(t)=T₁′ = 2π = 2
π

Fundamental period of
x₁′′(t)=T₁′′ = 2π = 1
4π 2

Ratio = T₁′ = 2 = 4 = integer (or rational)
T₂′′ 1 1
2

Therefore, x₁(t) is periodic.
Now, for signal
x₂(t) = sin 5πt + 3 sin 13πt
Fundamental period of
x₂′(t)=T₂′ = 2π = 2
5π 5

Fundamental period of
x₂′′(t) T₂′′ = 2π = 2
13π 13

Ratio = 2 = = 2·6
T₁′ 5 13
T₂′′ 1 5
13

i.e., rational. Therefore, x₂(t) is positive.
Hence, alternative (A) is the correct answer.

The Laplace transform of the function—
f(t) = sin at cos bt

1. a + b + a – b
  s² + (a + b)² s² + (a – b)²
1. s + a + b + a – b
  s² + (a + b)² s² + (a – b)²
1. 1 a + b + a – b
  2 s² + (a + b)² s² + (a – b)²
1. 2 a + b + a – b
  s² + (a + b)² s² + (a – b)²

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sin at cos bt can be written as
= 1 {sin (a + b)t + sin (a – b)t}
2

By taking Laplace transform, we get L{sin at·cos bt}
= L 1 {sin (a + b)t + sin (a – b)t}
2

= 1 a + b + a – b
2 s² + (a + b)² s² + (a – b)²

Correct Option: A

sin at cos bt can be written as
= 1 {sin (a + b)t + sin (a – b)t}
2

By taking Laplace transform, we get L{sin at·cos bt}
= L 1 {sin (a + b)t + sin (a – b)t}
2

= 1 a + b + a – b
2 s² + (a + b)² s² + (a – b)²

For the question 88, what will be inverse z-transform for
ROC, |z| < 1
2

1. 2 δ(n) – 1 ⁿ u(– n – 1) + 1 (3)ⁿ u(– n – 1)
  3 2 3
1. 2 δ(n) + 1 ⁿ u(– n – 1) – 1 (3)ⁿ u(– n – 1)
  3 2 3
1. 2 δ(n) + 1 ⁿ u(n) – 1 (3)ⁿ u(n)
  3 2 3
1. 2 δ(n) + 1 ⁿ u(n) + 1 (3)ⁿ u(n)
  3 2 3

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In the region |z| < 1 3, both the poles are exterior,
2

i.e., x(n) is anti-causal and hence
X(n) = 2 δ(n) + 1 ⁿ u(– n – 1) – 1 (3)ⁿ u(– n – 1)
3 2 3

Correct Option: B

In the region |z| < 1 3, both the poles are exterior,
2

i.e., x(n) is anti-causal and hence
X(n) = 2 δ(n) + 1 ⁿ u(– n – 1) – 1 (3)ⁿ u(– n – 1)
3 2 3