Signal and systems miscellaneous Easy Questions and Answers | Page - 4

Signal and systems miscellaneous

The Laplace transform of i(t) is given by
I(s) = 2
s(1 + s)

As t → ∞ the value of i(t) tends to—

1. 0
1. 1
1. 2
1. ∞

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lim L{l(s)} = lim sl(s)
^{t → 0} ^{s → 0}

= lim
_{s → 0} s. 2
s(1 + s)

= lim
_{s → 0} 2 = 2
1 + s

Alternative (C) is the correct choice.

Correct Option: C

lim L{l(s)} = lim sl(s)
^{t → 0} ^{s → 0}

= lim
_{s → 0} s. 2
s(1 + s)

= lim
_{s → 0} 2 = 2
1 + s

Alternative (C) is the correct choice.

A sequence x(n) with the z-transform X(z) = z⁴ + z² – 2z + 2 – 3z^{– 4} is applied as an input to a linear, time-invariant system with the impulse response h(n) = 2δ(n – 3) where

The output at n = 2 is—

1. – 6
1. Zero
1. 2
1. – 4

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Given that,
X(z) = z⁴ + z² – 2z + 2 – 3z^{– 4}
z{h(n)} = 2z^{– 3} = H(z)
we know that
Y(z) = H(z) X(z) = 2z– 3[z⁴ + z² – 2z + 2 – 3z⁴]
= 2z + 2z^{– 1} – 2z^{– 2} + 2z^{– 3} – 2z^{– 7}
or y(n)= 2δ(n + 1) + 2δ(n – 1) – 4δ(n – 2) + 4δ(n – 3) – 6δ(n – 7)
The output at n = 2
y(2) = 2δ(3) + 2δ(1) – 4δ(0) + 4δ(– 1) – 6δ(– 5)
or y(2) = 2.0 + 2.0 – 4.1 + 4.0 + 6.0
or y(2) = – 4
Hence alternative (D) is the correct choice.

Correct Option: D

Given that,
X(z) = z⁴ + z² – 2z + 2 – 3z^{– 4}
z{h(n)} = 2z^{– 3} = H(z)
we know that
Y(z) = H(z) X(z) = 2z– 3[z⁴ + z² – 2z + 2 – 3z⁴]
= 2z + 2z^{– 1} – 2z^{– 2} + 2z^{– 3} – 2z^{– 7}
or y(n)= 2δ(n + 1) + 2δ(n – 1) – 4δ(n – 2) + 4δ(n – 3) – 6δ(n – 7)
The output at n = 2
y(2) = 2δ(3) + 2δ(1) – 4δ(0) + 4δ(– 1) – 6δ(– 5)
or y(2) = 2.0 + 2.0 – 4.1 + 4.0 + 6.0
or y(2) = – 4
Hence alternative (D) is the correct choice.

Let P be linearity, Q be time-in variance, R be causality and S be stability. A discrete-time system has the input output relationship—

where s(n) is the input and y(n) is the output. The above system has the properties—

1. P, S but not Q, R
1. P, Q S but not R
1. P, Q, R, S
1. Q, R, S but not P

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The given system is linear since there is no constant term with input x(t) and no square of input x(t).
The system is not causal since for any value of n its output depends upon the future value of the input
The system is stable, since this system BIBO stable.
The system is time variant since for different instant of time we get different output.
Hence alternative (A) is the correct choice.

Correct Option: A

The given system is linear since there is no constant term with input x(t) and no square of input x(t).
The system is not causal since for any value of n its output depends upon the future value of the input
The system is stable, since this system BIBO stable.
The system is time variant since for different instant of time we get different output.
Hence alternative (A) is the correct choice.

The Fourier series coefficient for the given periodic signal given below is—

1. 1
1. cos πK
  2
1. sin πK
  2
1. 2

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Fourier series coefficient
a_k = 1 ^T/2 x(t).e^–jkω0tdt
T _-T/2

a_k = 1 ^T/2 10.e^–jkω0tdt
T _-T/2

10 1 = 10 = 2
T 5

Correct Option: D

Fourier series coefficient
a_k = 1 ^T/2 x(t).e^–jkω0tdt
T _-T/2

a_k = 1 ^T/2 10.e^–jkω0tdt
T _-T/2

10 1 = 10 = 2
T 5

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.