# Fourier Series to Fourier Transform

Discussion in 'Math' started by jag1972, Sep 12, 2012.

1. ### jag1972 Thread Starter Active Member

Feb 25, 2010
60
0
All,
At the moment I am trying to follow the process of converting the Fourier series into the fourier transform. I am stuck on the middle bit atm, the complex fourier series. I have been trying my self but have been hitting a brick wall. I would appreciate your help.

I am taking this from the book: Fourier series by W.Bolton,
ISBN 0-582-23934-6, page 108.

A periodic rectangular pulse is considered with a period of $\tau$, the amplitude is V. The complex Fourier series for this function is:

$Cn=\frac{1}{2}\int f(t) e^{-jnwot} dt$ (the limits are T/2 and -T/2)

The next step includes the rectangular function which does not span the entire range of T, therefore limits have changed.

$Cn=\frac{1}{2}\int V e^{-jnwot} dt$ (the limits are $\tau$/2 and -$\tau$/2)

$=\frac{V}{T}[\frac{e^{-jnwot}}{-jnwo}]$ (the limits are $\tau$/2 and -$\tau$/2)

frequency of the fundamental is wo =2π/T

$Cn = -\frac{V}{j2n\pi}(e ^{-jn\pi/T}-e ^{jn\pi/T})$

$Cn = \frac{V}{n\pi} sin$$\frac{n\pi\tau}{T}$

I understand the steps so far but do not understand the next step. which states that Cn can be re-written as:

$Cn = \frac{V\tau}{T}$$\frac{sin \frac{n\pi\tau}{T}}{\frac{n\pi\tau}{T}}$

I dont understand how this sinc x function was derived especially when the book states that when n=0, the amplitude is $A = \frac{V\tau}{T}$. Surely its 0.

I entered $Cn = \frac{V}{n\pi} sin$$\frac{n\pi\tau}{T}$ into MATLAB and it shows no value for when n=0 as expected but the book reckons it exists. Does anyone know why?

2. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
Going between these two lines is straightforward algebra. No tricks.

$Cn = \frac{V}{n\pi} \sin \left( \frac{n\pi\tau}{T} \right) \cdot \frac{\frac{n\pi\tau}{T}}{\frac{n\pi\tau}{T}}$

$Cn = \left( \frac{V}{n\pi} \right) \left( \frac{n\pi\tau}{T} \right) \frac{\sin \left( \frac{n\pi\tau}{T} \right)}{\frac{n\pi\tau}{T}}$

$Cn = \left( \frac{V\tau}{T} \right) \frac{\sin \left( \frac{n\pi\tau}{T} \right)}{\frac{n\pi\tau}{T}}$

Surely what is zero? The sinc function? The amplitude?

Look at the original expression for Ch. When n=o, you have 0 divided by 0. What is that? Is it zero? Is it infinity? Is it something in between?

If you ask what this function IS at EXACTLY n=0, then the result is undefined. But you can ask what the function evaluates to as n approaches zero (which you've got to grant a bit of leeway here since n is an integer, but imagine that it's not for now).

It's like the sinc(x) function, which I'll define as sin(x)/x (there are other common definitions that differ by a scaling constant). At EXACTLY x=0, we have 0/0 (any point where you divide by zero is known as a 'singularity') and the value is undefined. But as we get arbitrarily close to zero, we find that the value of the function is well defined and get arbitrarily close to 1. So we adjust the definition of the sinc(x) function and specify that the value of sinc(0) is 1. This is known as a 'removable singulatiry. because the function, (without the special definition of the value at the point of the singularity) approaches the same value from either the left or the right and so we can define the value of the function at that point to be that value.

I entered $Cn = \frac{V}{n\pi} sin$$\frac{n\pi\tau}{T}$ into MATLAB and it shows no value for when n=0 as expected but the book reckons it exists. Does anyone know why?