# 2. Full Range Fourier Series

The Fourier Series is an infinite series expansion involving trigonometric functions.

A periodic waveform `f(t)` of period `p = 2L` has a Fourier Series given by:

`f(t)` `=(a_0)/2 + sum_(n=1)^ooa_ncos((npit)/L)+sum_(n=1)^oo b_n\ sin((npit)/L)`

`=(a_0)/2+a_1cos((pit)/L)+a_2cos((2pit)/L)+a_3cos((3pit)/L)+...` `+b_1sin((pit)/L)+b_2sin((2pit)/L)+b_3sin((3pit)/L)+...`

### Helpful Revision

where

a_{n}andb_{n}are theFourier coefficients,

and

`(a_0)/2` is the

mean value, sometimes referred to as thedc level.

**Fourier Coefficients For Full Range Series Over Any Range -****L** **TO** **L**

**L**

**L**

If `f(t)` is expanded in the range `-L` to `L` (period `= 2L`) so that the range of integration is `2L`, i.e. half the range of integration is `L`, then the Fourier coefficients are given by

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

`a_n=1/Lint_(-L)^Lf(t)"cos"(npit)/Ldt`

`b_n=1/Lint_(-L)^Lf(t)\ "sin"(npit)/Ldt`

where `n = 1, 2, 3 ...`

**NOTE:** Some textbooks use

`a_0=1/(2L)int_(-L)^Lf(t)dt`

and then modify the series appropriately. It gives us the same final result.

## Dirichlet Conditions

Any periodic waveform of period `p = 2L`, can be expressed in a Fourier series provided that

(a) it has a finite number of discontinuities within the period `2L`;

(b) it has a finite average value in the period `2L`;

(c) it has a finite number of positive and negative maxima and minima.

When these conditions, called the Dirichlet conditions, are satisfied, the Fourier series for the function `f(t)` exists.

Each of the examples in this chapter obey the Dirichlet Conditions and so the Fourier Series exists.

### Example 1: Fourier Series - Square Wave

Sketch the function for 3 cycles:

`f(t)={(0, if -4<=t<0),(5, if 0<=t<4):}`

`f(t) = f(t + 8)`

Find the Fourier series for the function.

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

#### Solution:

Here's one possible way to solve it:

Alternative approach:

[**NOTE:** Whichever method we choose, *n* must take values `1, 2, 3, ...` when we are writing out the series using sigma notation.]

#### What have we done?

Let's think about what the answer in the above example means.

We are adding a series of sine terms (with decreasing amplitudes and decreasing periods) together. The combined signal, as we take more and more terms, starts to look like our original square wave:

We start with this sine curve:

`2.5+10/pi sin((pi t)/4)`

We then add the *y*-values of the second sine curve (with lower amplitude and higher frequency) to the above curve, and get the following:

Original graph

+

`10/(3pi)sin((3 pi t) /4)`

We continue as follows:

Previous result

plus

`10/(5pi)sin((5 pi t) /4)`

Previous result

plus

`10/(7pi)sin((7 pi t)/4)`

Previous result

plus

`10/(9pi)sin((9 pi t)/4)`

Previous result

plus

`10/(11pi)sin((11 pi t)/4)`

If we graph the result of adding more and more terms, we can see that our series will closely approximate the original periodic function. We graph the first 20 terms and see we are getting "very close" to the original function.:

`2.5+10/pi sum_(n=1)^20 1/((2n-1))"sin"((2n-1)pit)/4`

Apart from helping us understand what is going on, a graph can help us check our calculations.

The following video illustrates what we are doing. The equation is not exactly the same, but the concept is. The tone heard at the end is (close to) a "pure" square wave.

## Common Case: Period = 2*L*= 2*π*

If a function is defined in the range -*π* to *π* (i.e. period 2*L* = 2*π* radians), the range of integration is 2*π* and half the range is *L* = *π*.

The Fourier coefficients of the Fourier series *f*(*t*) in this case become:

`a_0=1/piint_(-pi)^pif(t)dt`

`a_n=1/piint_(-pi)^pif(t)\ cos nt\ dt`

`b_n=1/piint_(-pi)^pif(t)\ sin nt\ dt`

and the formula for the Fourier Series becomes:

`f(t)=a_0/2+sum_(n=1)^oo\ a_n\ cos nt+sum_(n=1)^oo\ b_n\ sin nt`

where `n = 1, 2, 3, ...`

### Example 2

a) Sketch the waveform of the periodic function defined as:

`f(t) = t\ \ if\ \ -π < t < π`

`f(t) = f(t + 2π)` for all `t`.

b) Obtain the Fourier series of `f(t)` and write the first 4 terms of the series.

#### What have we found?

The graph of the first 40 terms is:

`f(t) ~~ sum_(n=1)^40(2/n(-1)^(n+1)\ sin nt)`

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

## Expressing Fourier Series in other Forms

We can express the Fourier Series in different ways for convenience, depending on the situation.

### Fourier Series Expanded In Time *t* with period *T*

Let the function *f*(*t*) be periodic with period `T = 2L` where

`omega=(2pi)/T=(2pi)/(2L)=pi/L`

In this case, our lower limit of integration is `0`.

Hence the Fourier series is

`f(t)=a_0/2+sum_(n=1)^oo\ a_n\ cos n omega t+sum_(n=1)^oo\ b_n\ sin n omega t`

where

`a_0=omega/piint_0^(2pi//omega)f(t)dt`

`a_n=omega/piint_0^(2pi//omega)f(t)\ cos n omega t\ dt`

`b_n=omega/piint_0^(2pi//omega)f(t)\ sin n omega t\ dt`

(Note: half the range of integration `pi/omega`.)

### Fourier Series Expanded in Angular Displacement ω

(**Note:** ω is measured in radians here)

Let the function `f(ω)` be periodic with period `2L`.

We let ` θ = ωt`. This function can be represented as

`f(theta)=a_0/2+sum_(n=1)^oo(a_n\ cos (n pi theta)/L+b_n\ "sin" (n pi theta)/L)`

where

`a_0=1/Lint_0^(2L)f(theta)d theta`

`a_n=1/Lint_0^(2L)f(theta)\ cos"(n pi theta)/Ld theta`

`b_n=1/Lint_0^(2L)f(theta)\ sin"(n pi theta)/Ld theta`

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