3. Fourier Series of Even and Odd Functions

This section can make our lives a lot easier because it reduces the work required.

Revision

Go back to Even and Odd Functions for more information.

In some of the problems that we encounter, the Fourier coefficients ao, an or bn become zero after integration.

Finding zero coefficients in such problems is time consuming and can be avoided. With knowledge of even and odd functions, a zero coefficient may be predicted without performing the integration.

Even Functions

Recall: A function `y = f(t)` is said to be even if `f(-t) = f(t)` for all values of `t`. The graph of an even function is always symmetrical about the y-axis (i.e. it is a mirror image).

Example 1 - Even Function

`f(t) = 2\ cos\ πt`

Cosine curve

Fourier Series for Even Functions

For an even function `f(t)`, defined over the range `-L` to `L` (i.e. period = `2L`), we have the following handy short cut.

Since

`b_n=1/Lint_(-L)^Lf(t)\ sin {:(npit)/L:}dt`

and

`f(t)` is even,

it means the integral will have value 0. (See Properties of Sine and Cosine Graphs.)

So for the Fourier Series for an even function, the coefficient bn has zero value:

`b_n= 0`

So we only need to calculate a0 and an when finding the Fourier Series expansion for an even function `f(t)`:

`a_0=1/Lint_(-L)^Lf(t)dt`

`a_n=1/Lint_(-L)^Lf(t)cos{:(n pi t)/L:}dt`

An even function has only cosine terms in its Fourier expansion:

`f(t)=a_0/2+sum_(n=1)^oo\ a_n\ cos\ (n pi t)/L`

Fourier Series for Odd Functions

Recall: A function `y = f(t)` is said to be odd if `f(-t) = - f(t)` for all values of t. The graph of an odd function is always symmetrical about the origin.

Example 2 - Odd Function

f(t) = sin t

sine curve

Fourier Series for Odd Functions

For an odd function `f(t)` defined over the range `-L` to `L` (i.e. period `= 2L`), we find that `a_n= 0` for all `n`.

We have:

`a_n=1/Lint_(-L)^Lf(t)\ cos{:(n pi t)/L:}dt`

So the zero coefficients in this case are: `a_0= 0` and `a_n= 0`. The coefficients `b_n` are given by:

`b_n=1/Lint_(-L)^L\ f(t)\ sin{:(n pi t)/L:}dt`

The Fourier Series for an odd function is:

`f(t)=sum_(n=1)^oo\ b_n\ sin{:(n pi t)/L:}`

An odd function has only sine terms in its Fourier expansion.

Exercises

1. Find the Fourier Series for the function for which the graph is given by:

square wave - odd

2. Sketch 3 cycles of the function represented by

`f(t)={(0,if\ -1<=t<-1/2),(cos\ 3pit,if\ -1/2<=t<1/2),(0,if\ 1/2<=t<1):}`

and `f(t) = f(t + 2)`.

Find the Fourier Series for this function.

Interactive: You can explore this example using the interactive Fourier Series graph.

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