Signal and systems miscellaneous Easy Questions and Answers

Signal and systems miscellaneous

System described by differential equation—
(i) d y(t) + 2y(t) = x(t)
dt

(ii) d y(t) + y(t) + 4 = x(t)
dt

1. (i) is linear but (ii) is non-linear
1. (i) is non-linear but (ii) is linear
1. both (i) and (ii) are linear
1. both (i) and (ii) are non-linear

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(i) Let the response of the system to x₁(t) be y₁(t) and the response of the system to x₂(t) be y₂(t).
Thus, for the input x₁(t) the describing equation is
dy₁(t) + 2y₁(t) = x₁(t) = F[y₁(t)]
dt

and for the input
dy₂(t) + 2y₂(t) = x₂(t) = F[y₂(t)]
dt

Multiplying these equations by a₁ and a₂ respectively, and adding yields,
a₁ dy₁(t) + a₂ dy₂(t) + 2a₁y₁(t) + 2a₂y₂(t) = a₁x₁(t) + a₂x₂(t)
dt dt

d [a₁y₁(t) + a₂y₂(t)] + 2 [a₁y₁(t) + a₂y₂(t)] = a₁x₁(t) + a₂x₂(t)
dt

i.e., F[y₁(t)] + F[y₂(t)] = F[y₁(t) + y₂(t)]
Hence, the system is linear. (ii) Similarly, for equation
d y(t) + 4 = x(t)
dt

dy₁(t) + y₁(t) + 4 = x₁(t) = F[y₁(t)]
dt

and for input x₂(t).
Multiplying these equations by a₁ and a₂ respectively, and adding yields,
dy₂(t) + y₂(t) + 4 = x₂(t) = F[y₂(t)]
dt

i.e., d + [a₁y₁(t) + a₂y₂(t)] + 2 [a₁y₁(t) + a₂y₂(t)] + 4(a₁ + a₂) ≠ a₁.x₁(t) + a₂.x₂(t)
dt

i.e., F[y₁(t)] + F[y₂(t)] ≠ F[y₁(t) + y₂(t)]
Hence, the system is non-linear.
Therefore, alternative (A) is the correct choice.

Correct Option: A

(i) Let the response of the system to x₁(t) be y₁(t) and the response of the system to x₂(t) be y₂(t).
Thus, for the input x₁(t) the describing equation is
dy₁(t) + 2y₁(t) = x₁(t) = F[y₁(t)]
dt

and for the input
dy₂(t) + 2y₂(t) = x₂(t) = F[y₂(t)]
dt

Multiplying these equations by a₁ and a₂ respectively, and adding yields,
a₁ dy₁(t) + a₂ dy₂(t) + 2a₁y₁(t) + 2a₂y₂(t) = a₁x₁(t) + a₂x₂(t)
dt dt

d [a₁y₁(t) + a₂y₂(t)] + 2 [a₁y₁(t) + a₂y₂(t)] = a₁x₁(t) + a₂x₂(t)
dt

i.e., F[y₁(t)] + F[y₂(t)] = F[y₁(t) + y₂(t)]
Hence, the system is linear. (ii) Similarly, for equation
d y(t) + 4 = x(t)
dt

dy₁(t) + y₁(t) + 4 = x₁(t) = F[y₁(t)]
dt

and for input x₂(t).
Multiplying these equations by a₁ and a₂ respectively, and adding yields,
dy₂(t) + y₂(t) + 4 = x₂(t) = F[y₂(t)]
dt

i.e., d + [a₁y₁(t) + a₂y₂(t)] + 2 [a₁y₁(t) + a₂y₂(t)] + 4(a₁ + a₂) ≠ a₁.x₁(t) + a₂.x₂(t)
dt

i.e., F[y₁(t)] + F[y₂(t)] ≠ F[y₁(t) + y₂(t)]
Hence, the system is non-linear.
Therefore, alternative (A) is the correct choice.

The system represented by equation—
(i) y(t) = 5 sin x(t)
(ii) y(t) = 7x(t) + 5

1. (i) is linear, (ii) is non-linear
1. (i) is non-linear, (ii) is linear
1. Both (i) and (ii) are linear
1. Both (i) and (ii) are non-linear

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(i) y(t) = 5 sin x(t)
F[x_₁(t)] = y₁(t) = 5 sin x₁(t)
F[x₂(t)] = y₂(t) = 5 sin x₂(t)
F[x₁(t)] + F[x₂(t)] = 5[sin x₁(t) + sin x₂(t)]}
whereas,F[x₁(t) + x₂(t)] = 5{sin [x₁(t) + x₂(t)}]
Since here,
F[x₁(t) + F] [x₂(t)] ≠ F[x₁(t) + F[x₂(t)]
and hence the system is non-linear.
(ii) y(t)=7x(t) + 5
F[x₁(t)] = y₁(t) = 7x₁(t) + 5
F[x₂(t)] = y₂(t) = 7x₂(t) + 5
Therefore,
F[x₁(t)] + F[x₂(t)] = 7[x₁(t) + x₂(t)] + 0
where, F[x₁(t) +x₂(t)] = 7[x₁(t) + x₂(t)] + 5
Since here,
F[x₁(t) +x₂(t)] ≠ F[x₁(t) + F[x₂(t)]
Hence, the system is non-linear.
Hence, alternative (D) is correct choice.

Correct Option: D

(i) y(t) = 5 sin x(t)
F[x_₁(t)] = y₁(t) = 5 sin x₁(t)
F[x₂(t)] = y₂(t) = 5 sin x₂(t)
F[x₁(t)] + F[x₂(t)] = 5[sin x₁(t) + sin x₂(t)]}
whereas,F[x₁(t) + x₂(t)] = 5{sin [x₁(t) + x₂(t)}]
Since here,
F[x₁(t) + F] [x₂(t)] ≠ F[x₁(t) + F[x₂(t)]
and hence the system is non-linear.
(ii) y(t)=7x(t) + 5
F[x₁(t)] = y₁(t) = 7x₁(t) + 5
F[x₂(t)] = y₂(t) = 7x₂(t) + 5
Therefore,
F[x₁(t)] + F[x₂(t)] = 7[x₁(t) + x₂(t)] + 0
where, F[x₁(t) +x₂(t)] = 7[x₁(t) + x₂(t)] + 5
Since here,
F[x₁(t) +x₂(t)] ≠ F[x₁(t) + F[x₂(t)]
Hence, the system is non-linear.
Hence, alternative (D) is correct choice.

If f₁(t) and f₂(t) are duration-limited signals such that
f₁(t) ≠ 0, for 5 < t < 7
= 0, otherwise

f₂(t) ≠ 0, for 1 < t < 3
= 0, otherwise

then the convolution of f₁(t) and f₂(t) is zero every where except for—

1. 1 < t < 7
1. 3 < t < 5
1. 5 < t < 21
1. 6 < t < 10

View Hint View Answer Discuss in Forum

Given
f₁(t) ≠ 0, for 5 < t < 7
= 0, otherwise

and, f₂(t) ≠ 0, for 1 < t < 3
= 0, otherwise

Assume that

f(t) = f₁(t)*f₂(t)
= ^∞ f₁(τ ).f₂(t – τ )dτ
_{– ∞}

Except these two possible range, we get zero output so range, 6 < t < 10.
Hence, alternative (D) is the correct choice.

Correct Option: D

Given
f₁(t) ≠ 0, for 5 < t < 7
= 0, otherwise

and, f₂(t) ≠ 0, for 1 < t < 3
= 0, otherwise

Assume that

f(t) = f₁(t)*f₂(t)
= ^∞ f₁(τ ).f₂(t – τ )dτ
_{– ∞}

Except these two possible range, we get zero output so range, 6 < t < 10.
Hence, alternative (D) is the correct choice.

The impulse response of system is h(t) = δ(t – 0·5). If two such system are cascaded, the impulse response of the overall system will be—

1. 0·5 δ(t – 0·25)
1. δ(t – 0·25)
1. δ(t – 1)
1. 0·5 δ(t – 1)

View Hint View Answer Discuss in Forum

Given, the impulse response of the system is
h(t) = δ(t – 0·5)

Now, according to question if two such systems are cascaded, i.e.,

then in order to determine the impulse response of the cascaded system, the over all transfer function of cascaded system must be calculate.
H(s) = H(s).H(s) = e^{– 0·5s}.e^{– 0·5s} = e^{– s}
x(t) = δ(t)
⇒ X(s)= 1
Now, Y(s) = H(s).X(s) = e^{– s}.1 = e^{– s} and impulse response.
y(t) = δ(t – 1)
Hence, alternative (C) is the correct choice.

Correct Option: C

Given, the impulse response of the system is
h(t) = δ(t – 0·5)

Now, according to question if two such systems are cascaded, i.e.,

then in order to determine the impulse response of the cascaded system, the over all transfer function of cascaded system must be calculate.
H(s) = H(s).H(s) = e^{– 0·5s}.e^{– 0·5s} = e^{– s}
x(t) = δ(t)
⇒ X(s)= 1
Now, Y(s) = H(s).X(s) = e^{– s}.1 = e^{– s} and impulse response.
y(t) = δ(t – 1)
Hence, alternative (C) is the correct choice.

Which one of the following is the response y(t) of a causal LTI system described by
H(s) = (s + 1)
s² + 6s + 10

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

1. y(t) = e^{– 3t} sin t u(t)
1. y(t) = e^{– 3t(t – 1)} sin (t – 1) u(t – 1)
1. y(t) = sin (t – 1) u(t – 1)
1. None of these

View Hint View Answer Discuss in Forum

Given
H(s) = (s + 1)
s² + 6s + 10

x(t) = e^–t u(t)
then
X(s) = 1
s + 1

∴ Y(s) = H(s) X(s)
or
Y(s) = (s + 1) . 1
s² + 6s + 10 (s + 1)

or

Y(s) = 1
s² + 6s + 10

or
Y(s) = 1
(s + 3)² + 1

(rearranging the terms)
or
y(t) = e^–3t sin t.u(t)
Hence, alternative (A) is the correct choice.

Correct Option: A

Given
H(s) = (s + 1)
s² + 6s + 10

x(t) = e^–t u(t)
then
X(s) = 1
s + 1

∴ Y(s) = H(s) X(s)
or
Y(s) = (s + 1) . 1
s² + 6s + 10 (s + 1)

or

Y(s) = 1
s² + 6s + 10

or
Y(s) = 1
(s + 3)² + 1

(rearranging the terms)
or
y(t) = e^–3t sin t.u(t)
Hence, alternative (A) is the correct choice.

a₁	dy₁(t)	+ a₂	dy₂(t)	+ 2a₁y₁(t) + 2a₂y₂(t) = a₁x₁(t) + a₂x₂(t)
	dt		dt

	d	[a₁y₁(t) + a₂y₂(t)] + 2 [a₁y₁(t) + a₂y₂(t)] = a₁x₁(t) + a₂x₂(t)
	dt

i.e.,	d	+ [a₁y₁(t) + a₂y₂(t)] + 2 [a₁y₁(t) + a₂y₂(t)] + 4(a₁ + a₂) ≠ a₁.x₁(t) + a₂.x₂(t)
	dt

(i)	d	y(t) + 2y(t) = x(t)
	dt

(ii)	d	y(t) + y(t) + 4 = x(t)
	dt

	dy₁(t)	+ 2y₁(t) = x₁(t) = F[y₁(t)]
	dt

	dy₂(t)	+ 2y₂(t) = x₂(t) = F[y₂(t)]
	dt

Signal and systems miscellaneous

Signal and systems miscellaneous

Signals and Systems

Correct Option: A

Correct Option: D

Correct Option: D

Correct Option: C

Correct Option: A