Re: Complex Fourier Series

Quote:

Originally Posted by

**halfnormalled** Hello, I did a forum search and I think this is the correct place to post this. There seem to be a lot of other posts in here on this topic. The question is this:

Find the complex Fourier series of the sawtooth wave

and where

So, I'm trying to use the equation:

where

But I'm not getting the answer I'm expecting. I'm not sure what the limits of the integral should be. As the function is not defined for t<0, can I just put:

Or do I need to do something to balance out the fact that the equation doesn't run across the origin...? I'm confused....

Thanks very much in advance.

1. The function certainly is defined for t<0 (by the condition f(t) = f(t+1)).

2. However, it is fine to take the integral from 0 to 1, since that covers one complete period of this periodic function.

3. it would be helpful if you reserved the summation symbol \sum for sums, and used the integral symbol \int for integrals.

4. The function has period 1, so you should put

5. For the same reason, you should put (because always).

If you do all that, then you get the formula and you can evaluate that integral using integration by parts.

Re: Complex Fourier Series

Thanks for your reply Opalg. Much appreciated.

Apologies, I was rather rushing to post this on my lunch break at work... I've corrected the signs. Thanks for pointing it out.

Yes, I see now that the function is defined for t<0. I was confusing myself slightly. I did get to the equation you provided at the end, but I think there must have been something suspect in my working as I'm not getting the answer provided in the book. I'll keep trying today and post back here with my findings.

What I'm still not completely clear on is the limits of the integral. It's given in the definition with the following limits

Is there any reason why it goes from a negative to a positive? It made me think initially that the limits should then be

Re: Complex Fourier Series

The integral can be taken over any interval of length 1. The function has period 1, so that (for example) its values in the interval [–1/2,0] are the same as its values in the interval [1/2,1]. Therefore the integrals over those two intervals are the same, and it follows that the integral over [–1/2,1/2] is the same as the integral over [0,1].