If a function has symmetry about the vertical axis or the origin, then the computation of the Fourier coefficients may be greatly facilitated. A function f (t) which is symmetrical about the vertical axis is to be an even function and has the property

\[f(t)=f(-t)\]

For all t. that is, we may replace t by –t without changing the function. Examples of even function are t^{2} and cost, and a typical even function is shown in figure 1 (a). Evidently, we could fold the figure along the vertical axis, and the two properties of the graph would coincide.

Fig.1 (a): Even Function

A function f (t) which is symmetrical about the origin is said to be an odd function and has the property

$f(t)=-f(-t)$

In other words, replacing t by –t changes only the sign of the function. A typical odd function is shown in figure 1 (b), and other examples are t and sin t. evidently, we can fold the right half of the figure of an odd function and then rotate it about the t-axis (x-axis normally) so that it will coincide with the left half.

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Fig.1 (b): Odd Function

Now let us see how symmetry properties can help us in determining the Fourier coefficients. Evidently, in figure 1 (a) we may see by inspection that for f (t) an even function we have

$\int\limits_{-a}^{a}{f(t)dt=2}\int\limits_{0}^{a}{f(t)dt}\text{ }\cdots \text{ (3)}$

This is true because the area from –a to 0 is identical to that from 0 to a. this result may also be established analytically by writing

$\begin{align} & \int\limits_{-a}^{a}{f(t)dt=}\int\limits_{-a}^{0}{f(t)dt+\int\limits_{0}^{a}{f(t)dt}} \\ & =-\int\limits_{a}^{0}{f(-\tau )d\tau +\int\limits_{0}^{a}{f(t)dt}} \\ & =\int\limits_{0}^{a}{f(\tau )d\tau +\int\limits_{0}^{a}{f(t)dt}} \\ & =2\int\limits_{0}^{a}{f(t)dt} \\\end{align}$

In the case of f (t) an odd function, it is clear from figure 1(b) that

$\int\limits_{-a}^{a}{f(t)dt=0}\text{ }\cdots \text{ (4)}$

This is true because the area from –a to 0 is precisely the negative of that from0 to a. this result may also be obtained analytically, as was done for even functions.

In the case of the fourier coefficients we need to integrate the functions

$g(t)=f(t)\cos (n{{\omega }_{o}}t)\text{ }\cdots \text{ (5)}$

And

$h(t)=f(t)\sin (n{{\omega }_{o}}t)\text{ }\cdots \text{ }(6)$

If f (t) is even, then

$\begin{align} & g(-t)=f(-t)\cos (-n{{\omega }_{o}}t) \\ & =f(t)\cos (n{{\omega }_{o}}t) \\ & =g(t) \\\end{align}$

And

$\begin{align} & h(-t)=f(-t)sin(-n{{\omega }_{o}}t) \\ & =-f(t)sin(n{{\omega }_{o}}t) \\ & =-h(t) \\\end{align}$

Thus g (t) is even and h (t) is odd. Therefore, taking t_{o}=-T/2 in Fourier coefficients which are given below

$\begin{matrix} {{a}_{n}}=\frac{2}{T}\int\limits_{{{t}_{o}}}^{{{t}_{o}}+T}{f(t)cos(n{{\omega }_{o}}t)}dt, & \text{n=0,1,2,}\cdots & {} \\ {} & {} & {} \\ {{b}_{n}}=\frac{2}{T}\int\limits_{{{t}_{o}}}^{{{t}_{o}}+T}{f(t)\sin (n{{\omega }_{o}}t)}dt, & \text{n=1,2,3,}\cdots & {} \\\end{matrix}\text{ }$

We have, for f (t) even,

$\begin{matrix} {{a}_{n}}=\frac{4}{T}\int\limits_{0}^{T/2}{f(t)cos(n{{\omega }_{o}}t)}dt, & \text{n=0,1,2,}\cdots & {} \\ {} & {} & (7) \\ {{b}_{n}}=0, & \text{n=1,2,3,}\cdots & {} \\\end{matrix}\text{ }$

By a similar procedure we may show that if f (t) is odd, then

$\begin{matrix} {{a}_{n}}=0, & \text{n=0,1,2,}\cdots & {} \\ {} & {} & (8) \\ {{b}_{n}}=\frac{4}{T}\int\limits_{0}^{T/2}{f(t)\sin (n{{\omega }_{o}}t)}dt, & \text{n=1,2,3,}\cdots & {} \\\end{matrix}\text{ }$

In either case one set of coefficients is zero, and the other set is obtained by taking twice the integral over half the period, as described by (7) and (8).

In summary, an even function has no sine terms, and an odd function has no constant or cosine terms in its Fourier series. To illustrate further, let us find the Fourier coefficients of

$\begin{matrix} \begin{matrix} f(t)=4, \\ =-4, \\ f(t+2)=f(t) \\\end{matrix} & \begin{matrix} 0<t<1 \\ 1<t<2 \\ {} \\\end{matrix} \\\end{matrix}$

Evidently T=2, and thus ω_{o}=π. The function is odd, and therefore by (8) we have

$\begin{align} & \begin{matrix} {{a}_{n}}=0, & \text{n=0,1,2,}\cdots & {} \\ {} & {} & {} \\ {{b}_{n}}=\frac{4}{2}\int\limits_{0}^{1}{4\sin (n\pi t)}dt, & {} & {} \\\end{matrix} \\ & =\frac{8}{n\pi }[1-{{(-1)}^{n}}] \\ & \text{ } \\\end{align}$

Therefore b_{n}=0 for n even and b_{n}=16/nπ for n odd.

It often happens that the function f (t) is not periodic, so we shall consider an alternate method for this case. However, as is often the case, we may be interested only in f (t) on some finite interval (0, T), in which case we can consider it as periodic of period T, and find its Fourier series. Of course, what we have is not the Fourier series of f (t) but of its periodic extension. We may, however, use the result on the interval of interest and not be concerned either its periodic behavior elsewhere. Alternatively, we may consider (0, T) to be half the period and expand f (t) as either an even or an odd function using (7) or (8). The result called a half-range sine or cosine series, is easier to obtain and is equally valid in the interval of interest.

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