How Do You Determine Integration Limits for Convolution Integrals?

[Abdulwahab Hajar]

Homework Statement

Hi all, I hope you all can help me
so I'm studying for my signals course and I encounter this example in the book, and the answer is there but the solution isn't... The convolution integral exists for 3 intervals and I could evaluate the first two just fine... however I can't find the limits of integration of the third.
The third one corresponds to the fourth in the picture which is 2T < t < 3T
The question and answer is shown in the picture attached

Thank you!

Homework Equations

No relevant equations

The Attempt at a Solution

I believe that the lower limit should be -2T + t, and the upper limit should be to T.

 

[BvU]

I believe that the lower limit should be -2T + t, and the upper limit should be to T

Limit for what ? This will become clear when you post your relevant equation: the definition of a convolution. Per that definition, the integration limits for $\tau$ are $-\infty$ and $+\infty$. Of course for most $\tau$ the contribution is zero. Except for $\tau\in [0,T]$

 

[Abdulwahab Hajar]

Limit for what ? This will become clear when you post your relevant equation: the definition of a convolution. Per that definition, the integration limits for $\tau$ are $-\infty$ and $+\infty$. Of course for most $\tau$ the contribution is zero. Except for $\tau\in [0,T]$

Yes actually the relevant equation would be
x(t) * h(t) = (from -∞ to +∞) ∫x(k)h(t - k)dk where K was used in this case instead of tau to avoid confusion...
However as shown in the file attached there is more than one interval of integration...
I need the limits of the fourth one where 2T < t < 3T

 

[BvU]

You mean: 2T < t - k < 3T

Remember: k is your integration variable, NOT t ! And its limits are $-\infty$ and $+\infty$ .

 

[Abdulwahab Hajar]

You mean: 2T < t - k < 3T

Remember: k is your integration variable, NOT t ! And its limits are $-\infty$ and $+\infty$ .

true, it's limits are from -∞ to ∞
however, what is the range on which evaluated integral is nonzero... when 2T < t (is constant) < 3T
if you check the picture I attached there are many intervals on which the integral is evaluated
on the fourth interval 2T < t < 3T, when is the range on which the integral is nonzero?

 

[BvU]

The point is that k is the integration variable and t - k is the argument of the function h(). You want to intgrate the section where the argument is in $[2T, 3T]$