π−π0.5π-0.5πtf(t)Open image in a new page

Graph of `f(t)`.

We can see from the graph that `f(t + π) = - f(t)`.

For example, we notice that `f(2) = 0.4`, approximately. If we now move `π` units to the right (or about `2 + 3.14 = 5.14`), we see that the function value is

`f(5.14) = -0.4`.

That is, `f(t + π) = - f(t)`.

This same behaviour will occur for any value of `t` that we choose.

So the Fourier Series will have odd harmonics.

This means that in our Fourier expansion we will only see terms like the following:

`f(t)=(a_0)/2+(a_1\ cos t + b_1\ sin t)` ` +\ (a_3\ cos 3t + b_3\ sin 3t)` ` +\ (a_5\ cos 5t+ b_5\ sin 5t)+...`

[Note: Don't be confused with odd functions and odd harmonics. In this example, we have an even function (since it is symmetrical about the y-axis), but because the function has the property that `f(t + π) = - f(t)`, then we know it has odd harmonics only.

The fact that it is an even function does not affect the nature of the harmonics and can be ignored.]

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