Signal and systems miscellaneous Easy Questions and Answers | Page - 12

Signal and systems miscellaneous

Which one the following is the response y(t) of a causal LTI system described by
H(s) = (s + 1)
s² + 2s + 2

for a given input x(t) = e^{– t} .u(t)?

1. y(t) = e^{– t} sin t.u(t)
1. y(t) = e^{– (t – 1)} sin (t – 1).u(t – 1)
1. y(t) = sin (t – 1) u(t – 1)
1. y(t) = e^{– t} cos tu(t)

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Given
H (s) = (s + 1)
s² + 25 + 2

if x(t) = e^–tu(t)
then
X(s) = 1
s + 1

Now,
Y(s) = H(s).X(s)
or Y(s) = (s + 1) . (s + 1)
s² + 25 + 2 (s + 1)

or Y(s) = 1
(s + 1)² + 1

y(t) = e^–t.sin t.u(t)
Hence, alternative (A) is the correct choice.

Correct Option: A

Given
H (s) = (s + 1)
s² + 25 + 2

if x(t) = e^–tu(t)
then
X(s) = 1
s + 1

Now,
Y(s) = H(s).X(s)
or Y(s) = (s + 1) . (s + 1)
s² + 25 + 2 (s + 1)

or Y(s) = 1
(s + 1)² + 1

y(t) = e^–t.sin t.u(t)
Hence, alternative (A) is the correct choice.

If y(t) + ^∞ y(τ) x(t – τ )dτ = S(t) + x(t) then y(t) is—
₀

1. u(t)
1. δ(t)
1. r(t)
1. 1

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y(t) + ^∞ y(τ) x(t – τ )dτ = δ(t) + x(t) …(A)
0

L.H.S. = y(t) + ^∞ y(τ) x(t – τ )dτ
0

= y(t) + y(t) * x(t)
If we compare the equation with R.H.S part of equation (A), we conclude that
y(t) = δ(t) {∵ x(t) * δ(t) = x(0)}

Correct Option: B

y(t) + ^∞ y(τ) x(t – τ )dτ = δ(t) + x(t) …(A)
0

L.H.S. = y(t) + ^∞ y(τ) x(t – τ )dτ
0

= y(t) + y(t) * x(t)
If we compare the equation with R.H.S part of equation (A), we conclude that
y(t) = δ(t) {∵ x(t) * δ(t) = x(0)}

The impulse response of a system consists of two delta function as shown in the given figure.

The input to the system is a unit amplitude square pulse of one unit time duration. Which one of the following depicts the correct output?

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Given that impulse response i.e.
h(t) = 1.δ(t) – 1 δ(t – 1)
2

or H(s) = 1 – 1 e^–s
2

Also, given x(t) = unit amplitude pulse of one unit time duration i.e.
or x(t) = u(t) – u(t – 1)

or X(s) = 1 - 1 e^–s
s s

Now, Y(s) = H(s).X(s)

or Y(s) = 1 - 1 e^–s 1 - 1 e^–s
s s s

or Y(s) = 1 - 1 . 1 e^–s + 1 1 e^–s
s 2 s 2 s

or Y(s) = 1 - 1 - e^–s - 1 1 e^–s + 1 1 e^–2s
s s 2 s 2 s

Taking inverse Laplace transform

or y(t) = u(t) – u(t – 1) – 1/2 · u(t – 1) + 1/2 u(t – 2)
or y(t) = u(t) – 3/2 u(t – 1) + 1/2 u(t – 2)
Hence, alternative (B) is the correct choice.

Correct Option: B

Given that impulse response i.e.
h(t) = 1.δ(t) – 1 δ(t – 1)
2

or H(s) = 1 – 1 e^–s
2

Also, given x(t) = unit amplitude pulse of one unit time duration i.e.
or x(t) = u(t) – u(t – 1)

or X(s) = 1 - 1 e^–s
s s

Now, Y(s) = H(s).X(s)

or Y(s) = 1 - 1 e^–s 1 - 1 e^–s
s s s

or Y(s) = 1 - 1 . 1 e^–s + 1 1 e^–s
s 2 s 2 s

or Y(s) = 1 - 1 - e^–s - 1 1 e^–s + 1 1 e^–2s
s s 2 s 2 s

Taking inverse Laplace transform

or y(t) = u(t) – u(t – 1) – 1/2 · u(t – 1) + 1/2 u(t – 2)
or y(t) = u(t) – 3/2 u(t – 1) + 1/2 u(t – 2)
Hence, alternative (B) is the correct choice.

The amplitude response of a discrete system with a simple pole is shown in the given figure.

The must be located—

1. On the real axis at z = 1
1. On the real axis at z = – 1
1. At the origin of the z-plane
1. At z = ∞

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From given figure Function say x(n) represented as

x(n) = (– 1)n u(n) Now, X(z)
_∞
X(z) = x(n)·z^–n
^{n = – ∞}

or
_∞
X(z) = (– 1)ⁿ·z^–n
^{n = – ∞}

or
X(z) = 1 + (– 1)z^–1 + z^–2 + (– 1)^–3 + …
or
X(z) = 1
1 + z^–1

or
X(z) = z
z + 1

Here, pole at z + 1 = 0
i.e., z = – 1
Hence, alternative (B) is the correct choice.

Correct Option: B

From given figure Function say x(n) represented as

x(n) = (– 1)n u(n) Now, X(z)
_∞
X(z) = x(n)·z^–n
^{n = – ∞}

or
_∞
X(z) = (– 1)ⁿ·z^–n
^{n = – ∞}

or
X(z) = 1 + (– 1)z^–1 + z^–2 + (– 1)^–3 + …
or
X(z) = 1
1 + z^–1

or
X(z) = z
z + 1

Here, pole at z + 1 = 0
i.e., z = – 1
Hence, alternative (B) is the correct choice.

The discrete time system described by y[n] = x(n²) is—

1. Causal, linear and time-varying
1. Causal, non-linear and time varying
1. Non-causal, linear and time-invariant
1. Non-causal, non-linear and time-variant

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The given discrete-time system is y[n] = x[n²]
● is linear since there is no constant term as well as square terms in input and output.
● non-causal since for any value of n except 0 and 1, output depends upon the future value of the input for example at n = – 1
y[– 1] = x[1²] = x[1]
y[2] = x[2²] = x[4]

i.e., output depends upon the future input.
● System is time-invariant since there is no time factor term i.e., n in the given equation.
Hence, alternative (C) is the correct choice.

Correct Option: C

The given discrete-time system is y[n] = x[n²]
● is linear since there is no constant term as well as square terms in input and output.
● non-causal since for any value of n except 0 and 1, output depends upon the future value of the input for example at n = – 1
y[– 1] = x[1²] = x[1]
y[2] = x[2²] = x[4]

i.e., output depends upon the future input.
● System is time-invariant since there is no time factor term i.e., n in the given equation.
Hence, alternative (C) is the correct choice.