Fourier Series: How do i simpligy this integral?

[exidez]

Fourier Series: How do i simplify this integral?

Homework Statement

http://img18.imageshack.us/img18/4586/matlabfouriercomplex.jpg

Homework Equations

$$g(t)=\sum_{n=-\infty}^{\infty}c_{n}e^{jn \omega t}$$

$$c_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}g(t)e^{-jn \omega T}$$

The Attempt at a Solution

I need to find the complex Fourier series of the above function extended as an odd function
I Just want to know how to simplify the final solution i have. So far i have

$$c_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}g(t)e^{-jn \omega T}$$
$$c_{n}=\frac{1}{T}(\int_{-T/10}^{0}-Ae^{-jn \omega T}dt + \int_{0}^{T/10}Ae^{-jn \omega T}dt)$$
$$c_{n}=\frac{A}{T}( \frac{1}{jn \omega } - \frac{e^{(jn \omega T/10)}}{jn \omega } - \frac{e^{(-jn \omega T/10})}{jn \omega } + \frac{1}{jn \omega})$$
$$\omega = \frac{2 \pi }{T}$$
$$c_{n}=\frac{A}{jn2 \pi }(2 - e^{(jn2 \pi /10)} - e^{(-jn2 \pi /10)})$$

Now if i put this in MATLAB i will put

for n = -N:??:N, % loop over series index n
cn = A/(j*n*2*pi)*(2-exp(j*n*(pi/5))-exp(-j*n*(pi/5))); % Fourier Series Coefficient
yce = yce + cn*exp(j*n*wo*t); % Fourier Series computation
end

I don't know what to incriment N by.. All the other Fourier examples i have done have been in the form:
$$c_{n}=\frac{1}{jn \pi }(1 - e^{(jn \pi )})$$

Here i understand $$(1 - e^{(jn \pi )})$$ will be 0 when n is even 2 when n is odd
Hence the series becomes:

$$\sum_{n=odd}^{\infty}\frac{2}{ \pi jn}e^{jn \omega t}$$and the loop will go:
for n = -N:2:N, % loop over series index n (odd)
cn = 1/(j*n*pi)*(1-exp(j*n*pi)); % Fourier Series Coefficient
yce = yce + cn*exp(j*n*wo*t); % Fourier Series computation
endBut what is it for my example? It is not so straight forward. Can it be simplified like above? How do i progress from here?

 

[mighty2000]

Hello,

To simplify this integral, you can use the fact that the function is extended as an odd function. This means that g(t) = -g(-t) for all t. Using this property, you can rewrite the integral as:

c_{n}=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}g(t)e^{-jn \omega T} = \frac{1}{T}\int_{-t_{0}}^{t_{0}}g(t)e^{-jn \omega T} = \frac{1}{T}\int_{-t_{0}}^{0}-Ae^{-jn \omega T}dt + \frac{1}{T}\int_{0}^{t_{0}}Ae^{-jn \omega T}dt

Notice that the first integral is just the negative of the second integral, so you can rewrite the expression as:

c_{n}=\frac{2}{T}\int_{0}^{t_{0}}Ae^{-jn \omega T}dt

Now, you can use the substitution u = jt to simplify the integral. This gives:

c_{n}=\frac{2}{T}\int_{0}^{jt_{0}}Ae^{-un \omega T}dt

Using the definition of the complex Fourier series, you can rewrite this integral as:

c_{n}=\frac{2}{T}\int_{0}^{jt_{0}}Ae^{-un \omega T}dt = \frac{2}{T}\frac{1}{jn \omega}\left[e^{-un \omega T}\right]_{0}^{jt_{0}} = \frac{2}{jn \omega T}\left[e^{-jt_{0}n \omega} - 1\right]

Now, you can substitute in the value of \omega = \frac{2 \pi}{T} to get:

c_{n}=\frac{2}{jn \frac{2 \pi}{T} T}\left[e^{-jt_{0}n \frac{2 \pi}{T}} - 1\right] = \frac{A}{j \pi n}\left[e^{-jt_{0}n \frac{2 \pi}{T}} - 1\right]

Finally, you can use the fact that t_{0} = \frac{T}{10} to simplify the expression further