# Fourrier Series

1. Apr 17, 2005

### 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.

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• ###### Pulses.gif
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2. Apr 17, 2005

### 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.

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

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

$$A_0 = \frac{Wh}{D}$$

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

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)

Last edited: Apr 17, 2005
3. Apr 17, 2005

### saltydog

I get:

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

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

#### Attached Files:

• ###### pulse.JPG
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4. Apr 17, 2005

### 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:

Last edited: Apr 17, 2005
5. Apr 17, 2005

### quasar987

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

Now our expressions are equivalent.