L
Lily Bepant
- Jan 1, 1970
- 0
Hello,
I'm stuck in a convolution question.
x(t) = u(t) - 2u(t-2) + u(t-5)
h(t) = e^(2t) u(1-t)
y(t) = x(t) * h(t)
(I'm using capital T as tao and Int as integral from minus infinity to
plus infinity)
y(t) = Int ( x(t-T) h(T) ) dT
= Int ( u(t-T)-2u(t-T-2)+u(t-T-5)[e^(2T)u(1-T)] )dT
= Int ( [e^(2T)u(1-T)] u(t-T) ) - Int ....
------------------------------------------
I'm going to set integral limits due to the interval that the step
functions are nonzero. So:
1-T > 0 -> T < 1
t-T > 0 -> T < t
But the T is smaller than both 1 and t...
so.. What do I have to take as integral limits? I'm stuck here.
I know that it's basic calculus but, because it's about signal
processing, I posted it here.
Thank you for your assistance.
Lily
I'm stuck in a convolution question.
x(t) = u(t) - 2u(t-2) + u(t-5)
h(t) = e^(2t) u(1-t)
y(t) = x(t) * h(t)
(I'm using capital T as tao and Int as integral from minus infinity to
plus infinity)
y(t) = Int ( x(t-T) h(T) ) dT
= Int ( u(t-T)-2u(t-T-2)+u(t-T-5)[e^(2T)u(1-T)] )dT
= Int ( [e^(2T)u(1-T)] u(t-T) ) - Int ....
------------------------------------------
I'm going to set integral limits due to the interval that the step
functions are nonzero. So:
1-T > 0 -> T < 1
t-T > 0 -> T < t
But the T is smaller than both 1 and t...
so.. What do I have to take as integral limits? I'm stuck here.
I know that it's basic calculus but, because it's about signal
processing, I posted it here.
Thank you for your assistance.
Lily