Signal and systems miscellaneous Easy Questions and Answers | Page - 5

Signal and systems miscellaneous

The function x(t) is shown in the given figure. If X(jω) is the Fourier transform of the x(t), then |X(jω)| at ω = 0 will be—

1. 0.5
1. 1.5
1. 2
1. 3

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Given waveform of x(t) is a combination of four shifted step function i.e.,

i.e., x(t) = u t + 1 + u t + 1 - u t - 1 -u (t - 1).......A
2 2 2

F[x(t)] = e^jω + e^jω/2 − e^−jω/2 − e^−jω
jω jω jω jω

X(jω) = e^{jω − jω/2} + e^jω/2 − e ^jω/2
jω jω

X(jω) = 2 sin ω + 2 sin ω
ω ω 2

|X(jω)|_{at ω = 0} = lim_{ω → 0} 2 sin ω + 2 sin ω
ω ω 2

= lim_{ω → 0} 2.cos ω + 2 cos ω . 1
1 2 2

= 2 + 1 = 3.
Hence, alternative (D) is the correct choice.

Correct Option: D

Given waveform of x(t) is a combination of four shifted step function i.e.,

i.e., x(t) = u t + 1 + u t + 1 - u t - 1 -u (t - 1).......A
2 2 2

F[x(t)] = e^jω + e^jω/2 − e^−jω/2 − e^−jω
jω jω jω jω

X(jω) = e^{jω − jω/2} + e^jω/2 − e ^jω/2
jω jω

X(jω) = 2 sin ω + 2 sin ω
ω ω 2

|X(jω)|_{at ω = 0} = lim_{ω → 0} 2 sin ω + 2 sin ω
ω ω 2

= lim_{ω → 0} 2.cos ω + 2 cos ω . 1
1 2 2

= 2 + 1 = 3.
Hence, alternative (D) is the correct choice.

System represented by equation
y(t + 4) + 2y(t) = x(t + 2) is:
(i) causal
(ii) linear
(iii) time invariant
The correct statements are—

1. (i) and (ii) only
1. (ii) and (iii) only
1. (i) and (iii) only
1. All of these

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Given equation y(t + 4) + 2y(t) = x(t + 2)
● is causal since output at any instant of time depends only on present and past values of there input signal.
● is linear since there is no constant term in the given equation.
● is time invariant since there is no any time factor in the given equation.
Hence, alternative (D) is the correct choice.

Correct Option: D

Given equation y(t + 4) + 2y(t) = x(t + 2)
● is causal since output at any instant of time depends only on present and past values of there input signal.
● is linear since there is no constant term in the given equation.
● is time invariant since there is no any time factor in the given equation.
Hence, alternative (D) is the correct choice.

The given system equation
y(t) = x(t + 2) is—

1. causal
1. time-variant
1. dynamic
1. None of these

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The given system equation y(t) = x(t + 2) is—
● non-causal
● time invariant
● dynamic because input and output argument are not same i.e.,
different. Hence, alternative (C) is the correct choice.

Correct Option: C

The given system equation y(t) = x(t + 2) is—
● non-causal
● time invariant
● dynamic because input and output argument are not same i.e.,
different. Hence, alternative (C) is the correct choice.

The true statements for the system given below
y′(t + 4) + 2y(t) = x(t + 2)

1. dynamic and causal
1. time-invariant and non-causal
1. dynamic and non-causal
1. time-invariant and causal

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Given system y′(t + 4) + 2y(t) = x(t + 2) is causal and dynamic.
Causal: Since, output dependes upon the present and past input only.
Dynamic: Because input and output arguments are different.

Correct Option: A

Given system y′(t + 4) + 2y(t) = x(t + 2) is causal and dynamic.
Causal: Since, output dependes upon the present and past input only.
Dynamic: Because input and output arguments are different.

The true statement for the system equation given below
y(t)=2x(at)
(i) If a = 1, y(t) is static causal.
(ii) If a < 1, y(t) is dynamic causal.
(iii) If a > 1, y(t) is dynamic and non-causal.
(iv) If a > 1, y(t) is dynamic and causal.

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

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Given equation
y(t)=2x(at) …(I)
Case I: When a = 1, equation (I) becomes
y(t)=2x(t) …(A)
Equation (A) is static and causal.
Case II: When a < 1 say 1/2 Equation (I) becomes
y(t) = 2x 1 t ...........(B)
2

Equation (B) is causal and dynamic.
Case III: If a > 1, say a = 2
Equation (I) becomes
y(t)=2x(2t) …(C)
Equation (C) becomes non-causal and dynamic.
Hence, alternative (C) is the correct choice.

Correct Option: C

Given equation
y(t)=2x(at) …(I)
Case I: When a = 1, equation (I) becomes
y(t)=2x(t) …(A)
Equation (A) is static and causal.
Case II: When a < 1 say 1/2 Equation (I) becomes
y(t) = 2x 1 t ...........(B)
2

Equation (B) is causal and dynamic.
Case III: If a > 1, say a = 2
Equation (I) becomes
y(t)=2x(2t) …(C)
Equation (C) becomes non-causal and dynamic.
Hence, alternative (C) is the correct choice.