Given that yt-te-2t- x- -2d- is an LTI system- Its impulse response is ht- Then the value of h-2 is

Y(t)=∫_ ∞^te^ 2(t τ ) x(τ 2)dτ h(t)=∫_ ∞^te^ 2(t τ ) δ (τ 2)dτ h(t)=∫_ ∞^te^ 2(t 2) δ (τ 2)dτ h(t)=e^ 2(t 2)∫_ ∞^tδ (τ 2)dτ h(t)=e^ 2(t 2)u(t 2) h( 2)= ...

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Standard XII

Physics

Angular Impulse

Given that yt...

Question

Given that $y (t) = t \int - \infty e^{- 2 (t - τ)} x (τ - 2) d τ$ is an LTI system. Its impulse response is h(t). Then the value of h(-2) is_______
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Solution

The correct option is A 0
$y (t) = t \int - \infty e^{- 2 (t - τ)} x (τ - 2) d τ$

$h (t) = t \int - \infty e^{- 2 (t - τ)} δ (τ - 2) d τ$

$h (t) = t \int - \infty e^{- 2 (t - 2)} δ (τ - 2) d τ$

$h (t) = e^{- 2 (t - 2)} t \int - \infty δ (τ - 2) d τ$

$h (t) = e^{- 2 (t - 2)} u (t - 2)$

$h (- 2) = 0$

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Given that y(t)=t∫−∞e−2(t−τ)x(τ−2)dτ is an LTI system. Its impulse response is h(t). Then the value of h(-2) is_______0

The correct option is A 0y(t)=t∫−∞e−2(t−τ)x(τ−2)dτ h(t)=t∫−∞e−2(t−τ)δ(τ−2)dτ h(t)=t∫−∞e−2(t−2)δ(τ−2)dτ h(t)=e−2(t−2)t∫−∞δ(τ−2)dτ h(t)=e−2(t−2)u(t−2) h(−2)=0

Given that $y (t) = t \int - \infty e^{- 2 (t - τ)} x (τ - 2) d τ$ is an LTI system. Its impulse response is h(t). Then the value of h(-2) is_______
0