Fourier Series help, again!

cunnins4

Hi lads, need a little help with some fourier series stuff. Just can't quite figure these three out.

Any help's appreciated as always!



that's the three of them. They're all from my lecturer's notes, but he just jumps from one line to the next and I can't get a reply from him. I just need to figure out the little bits in the first two, I can almost get them out, but the third one just doesn't come out for me at all. On the second one, I don't know what happened to the j? I keep getting it in the denominator.

Michael Collins

Hey cunnins4, we've met before!

The first one is perfect. The Dirac delta impulse just has a spike where its argument is 0. So in this case the argument is zero when w = - w0. Any integral with a delta impluse is trivial since the only output can be where the impuse happens (since its zero everywhere else).

In the 2nd I reckon you've just forgotten that the exponential version of sine actually has a j beside the two on the bottom: sin(x) = (e^jx - e^-jx)/2j

In the third you are just taking a train of impulses (seperated by T in the time domain) and integrating them in order to get the Fourier Series (which you can get because it's periodic with period T). So

w0 = 2.pi /T

ak = 1/T * int (0 -> T) { p(t) e^-jk(2.pi/T)t dt }

As per the first part, the only output will be when the delta argument is 0:

=> ak = 1/T * 1 for all k

since, over the range 0 -> T the delta argument is only 0 at zero (we don't count the one at T). So you evaluate the integrand at t = 0 and the exponential goes to 1.

=> ak = 1/T for all k

So the answer follows fairly obviously from that, since you just add all the exponentials at each k together to get the signal back in terms of its Fourier Series components.

cunnins4

cheers mick, i guessed you'd be first one in with help. Much appreciated again!

Yup, forgot the j on the second one. Doh!

Think I've the third one sorted in my head now, i'll give it a crack in the morning just to make sure.

Was studying this stuff for my exam next week and this question came up. Hey presto didn't need any help, really understand it now. You're a legend man!

cunnins4

Thanks again Mick, the third one only just clicked with me now. The only other question I have is why don't we include the impulse at T?

Cheers!

Michael Collins

Its just because you take a full period. So I chose 0 <= t < T, note the strict inequality at T there.

If it helps, you could have taken -T/2 <= t < T/2, then there would be no ambiguity.

The Dirac delta impulse is quite odd in that integration over a infinitesimal range can alter the integral - this is not the case for normal functions. (You may hear people calling it the Dirac delta fuction, but this is as misnomer, as it isn't a function for this very reason).

cunnins4

Michael Collins wrote: »

If it helps, you could have taken -T/2 <= t < T/2, then there would be no ambiguity.

Ahhh, that makes it much clearer.

Michael Collins wrote: »

The Dirac delta impulse is quite odd in that integration over a infinitesimal range can alter the integral - this is not the case for normal functions. (You may hear people calling it the Dirac delta fuction, but this is as misnomer, as it isn't a function for this very reason).

I don't have the notes to hand, but I'm certain that's what our lecturer called it last year!

Thanks again-I was also looking at yet another thread you helped me on earlier in the year about the ROC of the Laplace Transform-You've really helped me this year, I'll dedicate any success in this exam to your good self!;)