A Linear Time Invariant system with an impulse response h(t) produces output y(t) when input x(t) is applied. When the input x(t- $τ$ ) is applied to a system with impulse response h(t- $τ$ ) the output will be

Open in App

Solution

$Case 1: Y(s) = H(s) . X(s)$

$Case 2: input = x(t- τ) \Rightarrow X(s) . e^{- s τ}$

$impulse response = h(t - τ) \Rightarrow H(s) . e^{- s τ}$

$Output: Y'(s) = X(s) . e^{- s τ} . H(s) e^{- s τ}$

$Y'(s) = X(s) . H(s) . e^{- s 2 t} = Y(s) e^{- s 2 T}$

$\Rightarrow y'(t) = y(t - 2 τ)$

Suggest Corrections

Similar questions

y (t) |_{t = 5}

x (t) = e x p (- 2 t) u (t) + δ (t - 6)

X (t)

h (t) = t e^{- α (t)}

α

X (t)

Y (t)

R_{X Y} (τ) = t e^{- α τ} u (τ) .

τ \leq 0

h (t) = e^{- 5 t} u (t)

y (t) = e^{- 3 t} u (t) - e^{- 5 t} u (t)

y (t) = 5 e^{- 10 t} u (t)

δ (t)

Join BYJU'S Learning Program