Chapter Contents

• Fourier Series
• Helpful Revision
• 1. Overview of Fourier Series
• 2. Full Range Fourier Series
• 3. Fourier Series of Even and Odd Functions
• Fourier Series Graph Interactive
• 4. Fourier Series of Half Range Functions
• 5. Harmonic Analysis
• 6. Line Spectrum
• 7. Fast Fourier Transform
• Fourier Series Problem Solver

• Comments, Questions?

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3. Fourier Series of Even and Odd Functions

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

In some of the problems that we encounter, the Fourier coefficients a_o, a_n or b_n 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`

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 b_n has zero value:

`b_n= 0`

So we only need to calculate a₀ and a_n 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

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:

Answer

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.

Answer