Okay, in searching around, I found that int(0 to t) [ f(r) f(t-r) dr ] easily equals t*f(t), which isn't wholly clear to me, but I'll accept that's the magic of convolution. That's very different from the formulas I was given to work with, and it is neglectful of how to apply the infinity. Help?

Convolution eh? Are you a EE?
I think you have your convolution integral wrong. I thought it looked funny so I checked wikipedia and a couple of my textbooks, none of them have a 1/sqrt(2*pi). Or are you actually taking the fourier transform of something? In which case you still have the formula wrong.

One of my books (Continuous and Discrete Time Signals and Systems by Mrinal Mandal and Amir Asif) has a very similar example, but with one notable difference: one of the signals is exp(-x)u(x) where u(x) is the unit step function. I'm just going to use * to use convolution, since that's the way I've always seen it written.

More coming...

The example I have is x(t)=exp(-t)u(t) and h(t)=exp(-2t)u(t).
y(t)=x(t)*h(t)= integral(-inf,inf)(exp(-T)u(T)exp(-2(t-T)))dT.
y(t) = exp(-2T) integral(-inf,inf)( exp(T)u(t-T)dT )
now notice that u(t-T) = 1 for T<= t and 0 for T>t. so depending on T there are 2 cases: one case is 0 for all t<0, the other is 1 for t>0 this changes the limits of integration.

y(t)=exp(-2t)integral(0,t)(exp(T)dT) = exp(-2t)-1 = exp(-t) - exp(-2t) for t>0

relating this to your problem:

assuming there is a unit step in there...
exp(-x) * exp(-x)u(x) = y
y=integral(-inf,inf)[ exp(-p) exp(-x+p) u(x-p) )dp
y=integral(-inf,inf)[ exp(0)exp(-x)u(x-p) ]dp
y=exp(-x) integral(-inf,inf)[ u(x-p) ]dp
the integral from -inf to inf of a step is a line with the slope of the time offset, aka x.
y=x exp(-x)... which is what you wanted.

Does that help? I could elaborate. I love this stuff.

By the way, this seems like a very strange place to ask for help on something like this. Don't you have classmates? ;-)

Not an EE.
Yes, trying to do some Fourier transform crap along with this, where the equality I need to show is the key step.

I followed the example mostly.
I don't understand the u(x) bit, that isn't in the notes given to me at all.
I'm unsure where the u(t-T) came from in line 3 when it wasn't in line 2 (typo?)
I'm unsure of the final step, final line, "exp(-2t)-1 = exp(-t) - exp(-2t)." That doesn't compute for me.

+1's for the help so far. Can you shed a little more light on this for me?

Strange place, maybe, but someone is helping me, yeah? =)

I do have classmates, but they're as lost as I am. No, strike that. /More/ lost.

Ah, yes. Sorry. In the example I have both signals have a u(t) tacked on to them. I wrote one of them and ignored the other each time. That's why it changes both ends of the limits on the integration. Each step can only change one limit. Have you discussed anything else with unit step functions? Or is the function defined to start at x=0? If so that's the same thing as having a step.

I missed the "relating this to your problem" post. I follow everything in that one and it meshes with how I tried to reword your example with my problem. I'd like to understand your example, all the same.

I left out a couple of paragraphs of boring crap my book has that probably explains it better than I did... Also I messed up the last line pretty majorly.

It should read
y(t) = exp(-2t)integral(0,t)[exp(T)dT] = exp(-2t)[exp(t)-1]=exp(-t)-exp(-2t) for t>0

Okay, fair enough. I think I'll take this to my prof with the u(t) element and see what he says. Thanks for your help! I'll probably be posting here again tomorrow night with more confusion. -_-

I'll +1 the remainder of your posts as thanks, again.

Are you getting it now? I've helped several people with convolution in the past couple of years and most people kind of get stuck/gloss over the importance of how the step functions change the limits of integration, which is what makes a solution possible. Are you getting that or is it something else?

By the way I'm just curious, what class is this for and what's your major/school?

hahaha coldsoup, am I getting it? Lol you've got jokes! ;)

I understand how you mean for it to work, although the finer points are, naturally, obscured for me. I'll ask my prof about it tomorrow.

It's for a PDE course, actually, and all he is doing is applications, no theory. Really sucks, because I have no idea why I'm doing what I'm doing, and that's a foreign feeling for me in a math course. Also, no textbook. All his notes. Gods....

Which reminds me, he is Korean, believes in the Judeo-Christian god, and thinks his math lecture is the time and place to convert people. Whenever we plug infinity into an omega, we have a religious experience....

by the time in understand what's going on herre...my soup is gunna get cold while im busy on my...DELL...get it...HAHAHAHA!

This is pretty easy if you use Laplace transform.
http://imagepaste.nullnetwork.net/viewimage.php?id=3707

and the use of the Convolution theorem:
http://en.wikipedia.org/wiki/Convolution_theorem

I have to say, as an EE, convolutions are probably my least favorite thing to do. So much for going into signal processing : )

You can't reach that result dubmdell, it's wrong. The convolution doesn't exist because e^x only rises as fast as e^-x falls, making the product a constant and the integration ∞. So it makes sense to assume that the function isn't e^-x, but instead -

f(x) = { 0 if x < 0
{ e^-x for x >= 0

Once you assume that, what coldsoup said works. If you're taking the one-sided Laplace transform, the result should reflect that - that you've assumed e^-x is only defined on the right half of the x-axis. The double-sided Laplace transform for e^-x doesn't exist! Remember, existence criteria are everything.

Thanks for the clarification, santosh. I actually was given a piecewise function, but I did not want to simply write the whole problem out since most of the math threads I saw a couple months ago inevitably had a "don't post your homework here" comment. I figured just posting my problem spot would be better. *shrug* I'll post the whole problem next time.

About to go talk to my prof. Let's see what he says about step functions!

A step function is the same thing as what santosh said. It's defined to be 1 for x >= 0 and 0 for x<0. It's just a convenient way to tag along initial conditions and such in signal processing.

Good comment santosh, I was thinking that it had to be infinity if there wasn't some condition on it last night just to myself because the convolution of 2 infinite signals must be infinity.

He said, "I'm too busy now." Ah, academia.

Okay, so the explanation I was given is, the step function is piece wise or (as the problem was stated originally) the function itself is piece wise. My confusion came in at the use of the function definition, where I should have seen

g(x-p)={ e^-(x-p) if x-p>0
{ 0 ow

And the similar "if p>0" in the g(p) portion, meaning x>p>0, which changes our bounds like the step function would. Whew. Thanks again for the help guys. Thanks to you, I only looked like half an idiot. ;)