Fourier transform of signal x(t) is given as
x(jω) = ∫^∞_{– ∞} x(t) e^{– jωt} dt
or x(jω) = ∫^∞_{– ∞} e^{– b|t|} .e^{– jωt} dt
x(jω) = ∫⁰_{– ∞} e^bt.e^{– jωt} dt + ∫^∞₀ e^bt.e^{– jωt} dt

or x(jω) =		e(b – jω)t		0	+		e– (b + jω)t		∞
		b – jω		– ∞			– (b + jω)		0

Correct Option: A

Fourier transform of signal x(t) is given as
x(jω) = ∫^∞_{– ∞} x(t) e^{– jωt} dt
or x(jω) = ∫^∞_{– ∞} e^{– b|t|} .e^{– jωt} dt
x(jω) = ∫⁰_{– ∞} e^bt.e^{– jωt} dt + ∫^∞₀ e^bt.e^{– jωt} dt

or x(jω) =		e(b – jω)t		0	+		e– (b + jω)t		∞
		b – jω		– ∞			– (b + jω)		0

Stationary processes are those for which ensemble average is independent to time average i.e.,
E[x(t₁)] = E[x(t₂)] = …… = E[x(t)]
Hence, alternative (D) is the correct choice.

Correct Option: D

Stationary processes are those for which ensemble average is independent to time average i.e.,
E[x(t₁)] = E[x(t₂)] = …… = E[x(t)]
Hence, alternative (D) is the correct choice.

Given discrete time system
y(n) = x(n²)
The order to solve such type of problems, check causality, linearity and time invariant one by one The given system
● y(n) = x(n²) is non-causal
Because for any negative value of n the output goes to future. For example:
y(– 2) = x(– 2)² = x(4)
i.e., non-causal system
● y(n) = x(n²)
To check linearity let a1 and a2 are the weights added to the input and H[x(n²)] = y (n) is the response of a discrete-time system.
Since, H[a₁x₁(n²) + a₂x₂(n²)]
= a₁H[x₁(n²)] + a₂H[x₂(n²)]
Hence, the given system is linear.
● For a system to be time invariant time delay of the input signal leads to an identical time shifts in the input signal.
Let y₁(n) = x₁(n²)
Consider input obtained by shifting x1(n) by time, n₀
i.e., x₁(n) = x₁(n – n₀)²
or y₂(n) = x₂(n²) = x₁(n – n₀)² …(A)
Similarly, when output changes by same amount n₀.
y₁(n – n₀) = x₁(n – n₀)² …(B)
From equations (A) and (B)
y₁(n – n₀) = y₂(n)
Therefore, this system is time invariant.
Hence, alternative (C) is the correct answer.

Correct Option: C

Given discrete time system
y(n) = x(n²)
The order to solve such type of problems, check causality, linearity and time invariant one by one The given system
● y(n) = x(n²) is non-causal
Because for any negative value of n the output goes to future. For example:
y(– 2) = x(– 2)² = x(4)
i.e., non-causal system
● y(n) = x(n²)
To check linearity let a1 and a2 are the weights added to the input and H[x(n²)] = y (n) is the response of a discrete-time system.
Since, H[a₁x₁(n²) + a₂x₂(n²)]
= a₁H[x₁(n²)] + a₂H[x₂(n²)]
Hence, the given system is linear.
● For a system to be time invariant time delay of the input signal leads to an identical time shifts in the input signal.
Let y₁(n) = x₁(n²)
Consider input obtained by shifting x1(n) by time, n₀
i.e., x₁(n) = x₁(n – n₀)²
or y₂(n) = x₂(n²) = x₁(n – n₀)² …(A)
Similarly, when output changes by same amount n₀.
y₁(n – n₀) = x₁(n – n₀)² …(B)
From equations (A) and (B)
y₁(n – n₀) = y₂(n)
Therefore, this system is time invariant.
Hence, alternative (C) is the correct answer.

Correct Option: B

Given that, x(n) = 6 cos		2πn
		4

⇒ x(0) = 6 cos		2π	· 0		= 6
		4

⇒ x(0) = 6 cos		2π	· 1		= 6 x 0 = 0
		4

⇒ x(0) = 6 cos		2π	· 2		= - 6
		4

⇒ x(0) = 6 cos		2π	· 3		= 6 x 0 = 0
		4

Since, the given signal
x(t) = A cos (ω₀t + φ) is periodic signal.
Hence, the given signal is power signal.

Correct Option: B

Since, the given signal
x(t) = A cos (ω₀t + φ) is periodic signal.
Hence, the given signal is power signal.