Consider a causal LTI system with frequency response H(jω)

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Consider a causal LTI system with frequency response
H(jω) = 1
jω + 3

This system produces the output to input x(t) as follows
y(t) = e^{– 2t} u(t) – e^{– 4t} u(t)
The input x(t) is

1. (2e^{– 4t} – 3e^{– 3t})u(t)
2. e^{– 4t} u(t)
3. (3e^{– 4t} – 2e^{– 3t})u(t)
4. – e^{– 4t} u(t)

Correct Option: B

x(t) = cos (4πt + θ)
X(jπ) = e^jθ πδ(ω – 2π) + e ^-jθπδ(ω + 2π)
The non zero portion of X(jω) lie outside the range – 3π ≤ ω ≤ 3π. This implies that
Y(jω) = X(jω) H(jω) = 0
∴ Y(t) = 0.

H(jω) =	1
	jω + 3