Fourier Expansion

Harmonic Analysis

August 23rd, 2009 Rishabh Dev No comments

Harmonics formed on waves are generated as component frequencies of a fundamental frequency of the wave.

The fundamental & the higher frequencies(harmonics) generate periodic signals from the original wave. And every periodic signal can be written as a sum of the variuos harmonics using the Fourier series.

fourier_harmonics

Hence, to find the various harmonics using the fourier series, we can use...

nth harmonic : (a_ncosx+b_nsinx)

where,

$\120dpi \large a_{n}=\frac{2}{p}\sum_{i=1}^{p}y_{i}cos(nx_{i})$

$\120dpi \large b_{n}=\frac{2}{p}\sum_{i=1}^{p}y_{i}sin(nx_{i})$

where p is the number of unique values of the function y. The following example will make things a bit more clear...

Example : y is a function of x periodic with period 2pi. Some experimental values of y are given below calculated for certain values of x. Expand y to 2 harmonics.

harmonic_example

Solution :

Clearly, in the above, p=6,

& We simply need to find:

1st harmonic + 2nd harmonic = (a₁cosx+b₁sinx) + (a₂cos2x+b₂sin2x)

So, all we need is a₁, b₁, a₂ &
b₂

for which we use the formula mentioned above:

$\120dpi \large a_{1}=\frac{2}{p}\sum_{i=1}^{p}y_{i}cos(x_{i})$

$\120dpi \large b_{1}=\frac{2}{p}\sum_{i=1}^{p}y_{i}sin(x_{i})$

$\120dpi \large a_{2}=\frac{2}{p}\sum_{i=1}^{p}y_{i}cos(2x_{i})$

$\120dpi \large b_{2}=\frac{2}{p}\sum_{i=1}^{p}y_{i}sin(2x_{i})$

where x_i=0, 60, 120... & so on.

Categories: Mathematics Tags: fourier expansion, fourier harmonic analysis, fourier series, harmonic analysis, harmonic expansion, harmonics

Fourier Series for an Arbitary Interval

August 17th, 2009 Rishabh Dev No comments

In the previous post on Fourier Series, we looked at functions periodic with period 2pi. Now, we'll take a look at Fourier series for functions having an arbitary period, lets say, some period 2L.

The general formulas we would need for finding the Fourier series are as follows...

The Series:

$\150dpi \large f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty }(a_{n}cos\frac{n\pi x}{l}+b_{n}sin\frac{n\pi x}{l})$

The Constants:

$\150dpi \large a_{0}=\frac{1}{L}\int_{-L}^{L}f(x)dx$

$\150dpi \large a_{n}=\frac{1}{L}\int_{-L}^{L}f(x)cos\frac{n\pi x}{L}dx$

$\150dpi \large b_{n}=\frac{1}{L}\int_{-L}^{L}f(x)sin\frac{n\pi x}{L}dx$

These apply when the period given in the question are [-L, L]. We would modify the limits of integration in the above depending on the given interval.

Further, for functions periodic with a period 2pi, we only need to put L=pi in the above formulas. You can find those formula here.

Categories: Mathematics Tags: arbitary interval, fourier expansion, fourier series, periodic functions

The Fourier Series - Introduction & Example

August 6th, 2009 Rishabh Dev No comments

One of the equations you would come across during higher studies would be the heat equation. Its another partial differential equation which we can solve using something called the Fourier series... Thanks to Joseph Fourier who introduced this series for the same purpose.

Using the Fourier Series, we simply decompose a particular real-valued function into terms containing the sine & cosine functions.

This function, lets say f(x) is taken to be periodic with a period 2pi. We could further extent the definition of the Fourier Series for functions with arbitrary periods.

The Fourier Series of a function f(x) could we written down as...

$\150dpi \large f(x)=\frac{a_{0}}{2}+ \sum_{n=1}^{\infty }(a_{n}cosnx+b_{n}sinnx)$

In the above decomposition, $\120dpi \inline \large a_{0},a_{n},b_{n}$ are called Fourier Coefficients or Euler's Coefficients.

The former because they appear in the Fourier Series & the latter because we obtain their values using the Euler's Formula.

Now the Euler's Formula gives the the values of the Euler's coefficients & for a function with period 2pi, we have...

$\150dpi \inline \large a_{0}=\frac{1}{\Pi }\int_{c}^{c+2\Pi }f(x)dx$

$\150dpi \inline \large a_{n}=\frac{1}{\Pi }\int_{c}^{c+2\Pi }f(x)cosnxdx$

$\150dpi \inline \large b_{n}=\frac{1}{\Pi }\int_{c}^{c+2\Pi }f(x)sinnxdx$

Now, once you know the values of the coefficients(which would come out to be functions of n) you could find out the complete sum & hence, the complete Fourier series of the function f(x).

The value of 'c' here would depend on the interval specified with the function f(x).

For example, Lets say we have a function,

$\150dpi \inline \large f(x)=1+x$