# Help with convolution bounds

1. Feb 16, 2015

### izelkay

1. The problem statement, all variables and given/known data

Mainly concerned with part (a). Here's the answer:

I understand where the answers inside the bracket came from, but I don't understand how they got their bounds (-infinity to 3, 3 to 5, and 5 to infinity)

2. Relevant equations
x is the impulse function here so y(t) = ∫h(τ)x(t-τ)dτ [-∞,∞]

3. The attempt at a solution
Since x was chosen to be the impulse function, I start by reflecting it across the y-axis and then shifting it by t to get the graph x(t-τ):

(minor mistake in my graph here: that τ-5 should be t-5 and the τ-3 should be t-3)

I then start "sliding" this graph into h(τ):

And this is where my confusion starts. If I perform the integration it'll be:

∫e^(-3τ)dτ [0, t-3]

and I get the answer they have for when 3 < t ≤ 5

So basically my question is why is it 3 < t ≤ 5 here?

2. Feb 16, 2015

### haruspex

I don't understand your method of 'sliding'.
I wrote out the integrand as $(u(t-3-\tau)-u(t-5-\tau))u(\tau)e^{-3\tau}.d\tau$. It is clear that this vanishes outside $t-3>\tau>t-5$. Write the resulting integral.

3. Feb 16, 2015

### izelkay

I'm using the "flip and shift method" like here: http://www-rohan.sdsu.edu/~jiracek/DAGSAW/4.2.html

I think I may understand the bounds now though.
In the picture I have illustrated where the integrand is from 0 to t-3, it's only valid when t-5 < 0 and t - 3> 0. In other words, 3 < t ≤ 5.

When I slide it in some more so that t-5 is now included under the graph h(τ), the integrand would be from t-5 to t-3, and it's then valid from t = 5 to infinity since h(τ) goes on forever