Fourrier Series

[AntonVrba]

I am 52, not yet senile and would be greatfulif someone can give me the Fourier series of two isolated pulses, pulswidth "w" spaced "2D" apart or at +- D.

Thanks in advance

[quasar987]

I will try it, but don't have a program to check my calculations.

This function is symetrical. So the coeficients B_n are 0. The period is 2D.

A_n = \frac{2}{D}\int_{0}^{D} cos\left( \frac{n\pi x}{D}\right)y(x)dx = \frac{2h}{D}\int_{D-W/2}^{D} cos\left( \frac{n\pi x}{D}\right)dx = \frac{2h}{D}\left[\frac{D}{n\pi}sin\left( \frac{n\pi x}{D}\right)\right]_{D-W/2}^{D}

A_n = \frac{2h}{n\pi}sin\left( \frac{n\pi W}{2D} - n\pi \right) & n=1,2,...

A_0 = \frac{Wh}{D}

\Rightarrow y(x) = \frac{Wh}{2D} + \frac{2h}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}sin\le ft( \frac{n\pi W}{2D} - n\pi \right) cos\left( \frac{n\pi x}{D}\right)

This makes sense, because the bigest W can be is 2D wide. In this case, the function is y(x) = h, which is correct. (h is the height of the pulse)

[saltydog]

I get:

y(x)=\frac{Wh}{2D}+\frac{2h}{\pi}\sum_{n=1}^{\inft y}\frac{1}{n}Sin[\frac{n\pi W}{2D}]Cos[\frac{n\pi}{D}(x-c-\frac{W}{2})]

with:

c=\frac{2D-W}{2}

I've included a plot for:

W=5
D=10
h=1

for the first 25 terms of the series

[quasar987]

Our expressions are equivalent, in case you haven't noticed.

-c-\frac{W}{2} = \frac{-2D+W-W}{2} = D

and

\cos\left(\frac{n\pi}{D}(x+D)\right) = \cos\left(\frac{n\pi x}{D}+n\pi\right)=-\cos\left(\frac{n\pi x}{D} \right)

So it's cool.


Edit: Wait, that's not true :grumpy:

[quasar987]

Ok I corrected an error in my original post due to this false identity that I had used.

Now our expressions are equivalent.