Signal and systems miscellaneous Easy Questions and Answers | Page - 8

Signal and systems miscellaneous

Match List-I with List-II
List-I List-II
A. 1. Linear and time variant
B. 2. Non-linear and time variant
C. 3. Linear and time variant
D. 4. Non-linear and time variant

1. A B C D
  3 2 4 1
1. A B C D
  3 2 1 4
1. A B C D
  2 3 4 1
1. A B C D
  3 4 2 1

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represents a linear and time invariant system.

constant term 4 make the system non-linear, while it represents time invariant.

term 3i²(t) make non-linear and coefficient of di(t)/dt i.e., t makes the system time-variant.

The coefficient of i(t) i.e., makes the system time-variant while the system is linear.

Correct Option: B

represents a linear and time invariant system.

constant term 4 make the system non-linear, while it represents time invariant.

term 3i²(t) make non-linear and coefficient of di(t)/dt i.e., t makes the system time-variant.

The coefficient of i(t) i.e., makes the system time-variant while the system is linear.

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.

System described by the equation
(i) y′′(t) + 3y′(t) + 2y(t) = x(t)
(ii) y′′(t) + 3y′′(t) = x(t)

1. (i) is stable, (ii) is unstable
1. (i) is unstable, (ii) is stable
1. Both (i) and (ii) are stable
1. Both (i) and (ii) are unstable

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Given equation
(i) y′′(t) + 3y′(t) + 2y(t) = x(t)
or (s2 + 3s + 1) Y(s) = X(s)
Y(s) = 1
X(s) s² + 3s + 1

Roots of characteristic equation
s² + 3s + 1 = 0
(s + 1) (s + 2) = 0
s = – 1, – 2
Here, all the roots are lie on the left hand side of s-plane hence the system is stable.
(ii) y′′′(t) + 3y′′(t) = x(t)
(s³ + 3s²) Y(s) = X(s)
Y(s) = 1 = 1
X(s) s³ + 3s2 s² + (s + 3)

Since, the two poles are lie on the origin therefore, the given system is unstable.
Hence, alternative (A) is the correct choice.

Correct Option: A

Given equation
(i) y′′(t) + 3y′(t) + 2y(t) = x(t)
or (s2 + 3s + 1) Y(s) = X(s)
Y(s) = 1
X(s) s² + 3s + 1

Roots of characteristic equation
s² + 3s + 1 = 0
(s + 1) (s + 2) = 0
s = – 1, – 2
Here, all the roots are lie on the left hand side of s-plane hence the system is stable.
(ii) y′′′(t) + 3y′′(t) = x(t)
(s³ + 3s²) Y(s) = X(s)
Y(s) = 1 = 1
X(s) s³ + 3s2 s² + (s + 3)

Since, the two poles are lie on the origin therefore, the given system is unstable.
Hence, alternative (A) is the correct choice.

The given system equation
y′(t) + 2y²(t) = 2x′(t) – x(t) is—

1. linear only
1. time-invariant only
1. Linear and time-invariant
1. Neither linear nor time-invariant

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Given system equation
y′(t) + 2y²(t)=2x′(t) – x(t)
Since, there is no time factor in the given equation, therefore, given system is time invariant but term y²(t) make the system equation non-linear.
Therefore, alternative (B) is the correct choice.

Correct Option: B

Given system equation
y′(t) + 2y²(t)=2x′(t) – x(t)
Since, there is no time factor in the given equation, therefore, given system is time invariant but term y²(t) make the system equation non-linear.
Therefore, alternative (B) is the correct choice.