Consider the convolution of two signals (1),h(1) y(t)=e¹u(t)* Σ 8(t−3k) k= = Σ e¹u(t)* 8(t−3k) Use the property of sampling of the impulse signal, y(t) = e(¹-3)u(t-3k) k=-x x(t)8(t-1)=x(t-t₁).

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In the step 2 they exapand the summation and say that u(t-3), u(t-6) and u(t-9) are out of range. Why? In the exercise it says that t is [0,3]

 

 

Consider the convolution of two signals x (1), h(t)
y(t)=e¹u(1)* Σ8(1-3k)
= Σe¹u(t)* 8(1-3k)
Use the property of sampling of the impulse signal, x(t)8(t−t,)=x(t−t₁).
y(t)= Σe (²-3) u(t-3k)
k=-
Transcribed Image Text:Consider the convolution of two signals x (1), h(t) y(t)=e¹u(1)* Σ8(1-3k) = Σe¹u(t)* 8(1-3k) Use the property of sampling of the impulse signal, x(t)8(t−t,)=x(t−t₁). y(t)= Σe (²-3) u(t-3k) k=-
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