Definition of Fourier Series and Typical Examples

Baron Jean Baptiste Joseph Fourier (1768 − 1830) introducedthe idea that any periodic function can be represented by a series of sines and cosines which are harmonically related. 

To consider this idea in more detail, we should introduce some definitions and common terms.

Basic Definitions

A function f (x) is said to have period P if f (x + P) = f (x) for all x. Let the function f (x) has period 2π. In this case, it is enough to consider behavior of the function on the interval [−π, π]. 

1. Suppose that the function f (x) with period 2π is absolutely integrable on [−π, π] so that the following so-calledDirichlet integral is finite:

2. Suppose also that the function f (x) is a single valued, piecewise continuous (must have a finite number of jump discontinuities), and piecewise monotonic (must have a finite number of maxima and minima).

If the conditions 1 and 2 are satisfied, the Fourier series for the function f (x) exists and converges to the given function (see also Convergence of Fourier Series about convergence conditions.) 

At a discontinuity x0, the Fouries Series converges to

The Fourier series of the function f (x) is given by

where the Fourier coefficients a0, an, and bn are defined by the integrals

Sometimes alternative forms of the Fourier series are used. Replacing an and bn by the new variables dn and φn or dn andθn, where

we can write:

Fourier Series of Even and Odd Functions

The Fourier series expansion of an even function f (x) with the period of 2π does not involve the terms with sines and hasthe form:

where the Fourier coefficients are given by the formulas

Accordingly, the Fourier series expansion of an odd 2π-periodic function f (x) consists of sine terms only and has the form:

where the coefficients bn are

Below we consider expansions of 2π-periodic functions into their Fourier series, supposing that these expansions exist and are convergent. 

   Example 1

Let the function f (x) be 2π-periodic and suppose that it is presented by the Fourier series:

      

Calculate the coefficients a0, an, and bn.

Solution.

To define an, we integrate the Fourier series on the interval [−π, π]:

      

For all n > 0,

      

Therefore, all the terms on the right of the summation sign are zero, so we obtain

      

In order to find the coefficients an at m > 0, we multiply both sides of the Fourier series by cos mx and integrate term by term:

      

The first term on the right side is zero. Then, using the well-known trigonometric identities, we have

      

if m ≠ n. 

In case when m = n, we can write:

      

Thus,

      

Similarly, multiplying the Fourier series by sin mx and integrating term by term, we obtain the expression for bm:

      

Rewriting the formulas for an, bn, we can write the final expressions for the Fourier coefficients:

      

   Example 2

Find the Fourier series for the square 2π-periodic wave defined on the interval [−π, π]:

      

Solution.

First we calculate the constant a0:

      

Find now the Fourier coefficients for n ≠ 0:

      

Since  , we can write:

      

Thus, the Fourier series for the square wave is

      

We can easily find the first few terms of the series. By setting, for example, n = 5, we get

      

The graph of the function and the Fourier series expansion for n = 10 is shown in Figure 1.

Fig.1, n = 10 Fig.2, n = 5, n = 10

   Example 3

Find the Fourier series for the sawtooth wave defined on the interval [−π, π] and having period 2π.

Solution.

Calculate the Fourier coefficients for the sawtooth wave. Since this function is odd (Figure 2), then a0 = an = 0. Findthe coefficients bn:

      

To calculate the latter integral we use integration by parts formula:

      

Let  . Then   so the integral becomes

      

Substituting   and   for all integer values of n, we obtain

      

Thus, the Fourier series expansion of the sawtooth wave is (Figure 2 above)

      

   Example 4

Let f (x) be a 2π-periodic function such that   for  . Find the Fourier series for the parabolic wave.

Solution.

Since this function is even, the coefficients bn = 0. Then

      

Apply integration by parts twice to find:

      

Since   and   for integer n, we have

      

Then the Fourier series expansion for the parabolic wave is (Figure 3)

      

Fig.3, n = 2, n = 5 Fig.4, n = 1, n = 2

   Example 5

Find the Fourier series for the triangle wave

      

defined on the interval [−π, π].

Solution.

The constant a0 is

      

Determine the coefficients an:

      

Integrating by parts, we can write

      

Then

      

The values of sin nx at x = 0 or x = ± π are zero. Therefore,

      

When n = 2k, then  . When n = 2k + 1, then   Since the function f (x) is even, the Fourier coefficients bn are zero. Therefore, the complete Fourier expansion for the triangle wave (see Figure 4 above) is

      

   Example 6

Find the Fourier series for the function

      

defined on the interval [−π, π].

Solution.

First we find the constant a0:

      

Now we calculate the coefficients an:

      

Notice that

      

Since cos (n − 1)π = (−1)n −1, we get the following expression for the coefficients an:

      

It's seen that an = 0 for odd n. For even n, when n = 2k (k = 1,2,3,...), we have

      

Calculate the coefficients bn. Start with b1:

      

The other coefficients bn for n > 1 are zero. Indeed,

      

Thus, the Fourier series of the given function is given by

      

Graphs of the function and its Fourier expansions for n = 2 and n = 8 are shown in Figure 5.

Fig.5, n = 2, n = 8 Fig.6, n = 10

   Example 7

Find the Fourier series for the function

      

defined on the interval [−π, π].

Solution.

      

Compute the coefficients an:

      

(These results are obvious since this function is odd.) 

Calculate the coefficients bn:

      

Thus, the Fourier series expansion of the function is given by

      

The graph of the function and the Fourier series expansion for n = 10 are shown in Figure 6 above.

An even function can be expanded using half its range from

0 to L or −L to 0 or L to 2L

That is, the range of integration is L. The Fourier series of the half range even function is given by:

f(t)=a02+∑n=1∞ an cos nπtL

for n=1,2,3,... , where

a0=2L∫L0f(t)dt

an=2L∫L0f(t) cos nπtLdt

and bn=0.

Illustration

In the figure below, f(t)=t is sketched from t=0 to t=π.

An even function means that it must be symmetrical about the f(t) axis and this is shown in the following figure by the broken line between t=−π and t=0.

It is then assumed that the "triangular wave form" produced is periodic with period 2π outside of this range as shown by the red dotted lines.

Example

We are given that

f(t)={−ttifif−π≤t<00≤t<π

and f(t) is periodic with period 2π.

a) Sketch the function for 3 cycles.

b) Find the Fourier trigonometric series for f(t), using half-range series.

a) Sketch:

b) Since the function is even, we have bn = 0.

In this example, `L=pi`.

We have:

`{:(a_0,=2/Lint_0^Lf(t)dt),(,=2/piint_0^pit\ dt),(,=2/pi[t^2/2]_0^pi),(,=2/pi(pi^2)/2),(,=pi):}`

To find an, we use a result from before (see Table of Common Integrals):

`intt\ cos\ nt\ dt=1/n^2(cos\ nt+nt\ sin\ nt)`

We have:

`{: (a_n,=2/Lint_0^Lf(t)cos{:(n pi t)/L:}dt),(,=2/piint_0^pi t\ cos\ nt\ dt),(,=2/pi[1/n^2(cos\ nt+nt\ sin\ nt)]_0^pi),(,=2/(pi n^2)[(cos\ n pi+0)-(cos\ 0+0)]),(,=2/(pi n^2)[(cos\ n pi-1)]),(,=2/(pi n^2)[(-1)^n-1]):}`

When n is odd, the last line gives us `-4/(pin^2`.

When n is even, the last line equals `0`.

For the series, we need to generate odd values for n. We need to use `(2n - 1)` for `n = 1, 2, 3,...`.

So we have:

`{: (f(t),=a_0/2+sum_(n=1)^oo a_n cos{:(n pi t)/L:}),(,=pi/2-4/pi sum_(n=1)^oo(cos(2n-1)t)/((2n-1)^2)),(,=pi/2-4/pi(cos\ t+1/9cos\ 3t+1/25cos\ 5t+...)):}`

Check: The graph for the first 40 terms:

An odd function can be expanded using half its range from 0 to L, i.e. the

range of integration has value L. The Fourier series of the odd function is:

Since ao = 0 and an = 0, we have:

f(t)=∑n=1∞ bn sin nπtL for n = 1, 2, 3, ...

bn=2L∫L0f(t) sin nπtLdt

In the figure below, f(t) = t is sketched from t = 0 to t = π, as before.

An odd function means that it is symmetrical about the origin and this is

shown by the red broken lines between t = −π and t=0.

It is then assumed that the waveform produced is periodic of

period 2π outside of this range as shown by the dotted lines.

Half-Range Expansions

In many physical problems the function is only known over a finite interval,

say . To express as a Fourier series means that we need to extend the function to be valid over all . If we choose to extend the function periodically, with period , then we retrieve the results of Section 2.4.1. However, we could also extend the function in an even manner, as shown in Figure 2.5 to get a cosine series or as an odd function to get a sineseries.

Figure 2.5: (a) A function defined on a interval , where . (b) The even extension

of the function onto the interval is shown as a solid curve and the periodic extension

of period is shown as a dot-dash curve. (c) The odd extension of the function, solid curve, and the

periodic extension of period , dot-dash curve.

The odd extension of will generate a Fourier series that only involves sine terms and is called the sine half-range expansion. It is given by

(2.15)

where

and is a positive integer.

On the other hand, the even extension generates the cosine half-range expansion 

(2.16)

where

and is a positive integer.