Signal and systems miscellaneous
- The given system equation
y(t) = x(t + 2) is—
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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 true statements for the system given below
y′(t + 4) + 2y(t) = x(t + 2)
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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 true statement for the system equation given below
y(t)=2x(at)
(i) If a = 1, y(t) is static causal.
(ii) If a < 1, y(t) is dynamic causal.
(iii) If a > 1, y(t) is dynamic and non-causal.
(iv) If a > 1, y(t) is dynamic and causal.
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Given equation
y(t)=2x(at) …(I)
Case I: When a = 1, equation (I) becomes
y(t)=2x(t) …(A)
Equation (A) is static and causal.
Case II: When a < 1 say 1/2 Equation (I) becomesy(t) = 2x 
1 t 
...........(B) 2
Equation (B) is causal and dynamic.
Case III: If a > 1, say a = 2
Equation (I) becomes
y(t)=2x(2t) …(C)
Equation (C) becomes non-causal and dynamic.
Hence, alternative (C) is the correct choice.Correct Option: C
Given equation
y(t)=2x(at) …(I)
Case I: When a = 1, equation (I) becomes
y(t)=2x(t) …(A)
Equation (A) is static and causal.
Case II: When a < 1 say 1/2 Equation (I) becomesy(t) = 2x 1 t ...........(B) 2
Equation (B) is causal and dynamic.
Case III: If a > 1, say a = 2
Equation (I) becomes
y(t)=2x(2t) …(C)
Equation (C) becomes non-causal and dynamic.
Hence, alternative (C) is the correct choice.
- The given system equation
y′(t) – 4y(t) + y(2t) = x(t) is—
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The given system equation
y′(t) – 4y(t) + y(2t) = x(t) is
● Causal: Since, output depends upon the present and past value of the input signal.
● The coefficient of y(t) having time factor y(2t) is also changing, which makes the system equation non-linear.
● Since, y(2t) is a time compressed circuit, which make the system equation time variant. It may be remember that any compressed or expanded circuit element makes the system equation timevarying.
Hence, alternative (B) is the correct choice.Correct Option: C
The given system equation
y′(t) – 4y(t) + y(2t) = x(t) is
● Causal: Since, output depends upon the present and past value of the input signal.
● The coefficient of y(t) having time factor y(2t) is also changing, which makes the system equation non-linear.
● Since, y(2t) is a time compressed circuit, which make the system equation time variant. It may be remember that any compressed or expanded circuit element makes the system equation timevarying.
Hence, alternative (B) is the correct choice.
- The spectral density of random process, where
Px(τ) = e– 2a|τ| is—
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We know that spectral density is nothing but a Fourier transform of auto correlation function. i.e.
Sxx(ω) = 
∞ Rxx(τ) e–jωτ.dτ -∞ = ∞ e–2a(τ) e–jωτ.dτ -∞ = 0 e2a(τ) e–jωτ.dτ + 0 e–2a(τ) e–jωτ.dτ -∞ -∞ = 
eτ(2a – τω) 
0 + eτ(2a – τω) -∞ 2a - jω -∞ - (2a + jω) 0 = 1 + 1 2a – jω 2a + jω = 4a 4a2 + ω2
Hence, alternative (B) is the correct choice.Correct Option: B
We know that spectral density is nothing but a Fourier transform of auto correlation function. i.e.
Sxx(ω) = ∞ Rxx(τ) e–jωτ.dτ -∞ = ∞ e–2a(τ) e–jωτ.dτ -∞ = 0 e2a(τ) e–jωτ.dτ + 0 e–2a(τ) e–jωτ.dτ -∞ -∞ = eτ(2a – τω) 0 + eτ(2a – τω) -∞ 2a - jω -∞ - (2a + jω) 0 = 1 + 1 2a – jω 2a + jω = 4a 4a2 + ω2
Hence, alternative (B) is the correct choice.
