2.2 Half-wave symmetry

Half-wave symmetry

Periodic signals having half-wave symmetry have the property

It turns out that signals with this type of symmetry only have odd-numbered harmonics, the even harmonics are zero. To see this, lets look at the formula for the coefficients $a_n$ :

Making the substitution $\tau =t-T/2$ in $I_2$ gives

The quantity $cos\left(n{\Omega }_0(\tau +T/2)\right)=cos(n\Omega \tau +n\pi )$ can be simplified using the trigonometric identity

We have

Therefore

and we can write:

From this expression we find that $a_n=0$ whenever $n$ is even. In fact, we have

A similar derivation leads to

A good choice of $t_0$ can lead to a considerable savings in time when calculating the Fourier Series of half-wave symmetric signals. Note that half-wave symmetric signals need not have odd or even symmetry for the above formulae to apply. If a signal has half-wave symmetry and in addition has odd or even symmetry, then some additional simplification is possible. Consider the case when a half-wave symmetric signal also has even symmetry. Then clearly $b_n=0$ , and [link] applies. However since the integrand in [link] is the product of two even signals, $x\left(t\right)$ and $cos\left(n{\Omega }_{0}t\right)$ , it too has even symmetry. Therefore, instead of integrating from, say, $-T/4$ to $T/4$ , we need only integrate from 0 to $T/4$ and multiply the result by 2. Therefore the formula for $a_n$ for an even, half-wave symmetric signal becomes:

For an odd half-wave symmetric signals, a similar argument leads to