Fourier Series complex coefficients

[jellicorse]

I have been trying to follow how the complex Fourier coefficients are obtained; the reference I am using is at www.thefouriertransform.com. However I am unable to follow the author's working exactly and wondered if anyone could help me see where I am going wrong.

First, I understand that the coefficient c_n=\frac{1}{T}\int^T_0 f(t) e^{\frac{-i2\pi nt}{T}}dt

Given a square wave, of period 2T, and amplitude :

http://s90.photobucket.com/user/jonnburton/media/SqareWave_zpsbdf193c1.jpg.html

It can be seen directly from the graph that c_0=\frac{1}{2} (I also managed to calculate this, too.)

But I can't follow how the author obtained expressions for c_n.

This is what I have done in attempt to obtain c_n:

c_n=\frac{1}{T}\int^{\frac{T}{2}}_0 1 \cdot e^{\frac{-i2\pi nt}{T}}dt

\frac{1}{T}\left[\frac{T e^{\frac{-i2\pi nt}{T}}}{i2\pi n}\right]^{\frac{T}{2}}_0

\frac{1}{T}\left[\left(\frac{T e^{-in\pi}}{in2\pi}\right)-\left(\frac{T}{in2\pi}\right)\right]

\left(\frac{e^{-in\pi}}{in2\pi}-\frac{1}{in2\pi}\right)

However, the expression the author finds is:

\frac{1}{i\pi n} for odd n, and zero for even n.

 

[Silversonic]

You seem to be missing a minus sign in your integral.

Also, simply for consistency, the limits you originally give (between 0 and T) aren't the limits you say you end up using (between 0 and T/2).

Lastly, what is e^{-in\pi} for odd/even n? Recall e^{in\pi} = cos(n\pi) + isin(n\pi).

 

[jellicorse]

Thanks for the hints Silversonic. I will go back through it again but I think I can see how this works out now!

(I just changed the limits because I though it wasn't necessary to integrate beyond \frac{T}{2} because the function there is zero.)

 

[Silversonic]

(I just changed the limits because I though it wasn't necessary to integrate beyond \frac{T}{2} because the function there is zero.)

Okay, the picture won't load for me. That could be an issue on my end.

 

[jellicorse]

Not sure why it wasn't working; I've just typed the URL in again and I think it's OK now.