Signal and systems miscellaneous Easy Questions and Answers | Page - 6

Signal and systems miscellaneous

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 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 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.

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.

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.