# Half Wave Symmetry

A function is half-wave (odd) symmetric if
$$f\left(t-\frac{T}{2}\right)=-f(t) \tag{1}$$
which means that each half-cycle is the mirror image of the next half cycle. Notice that functions $\cos n \omega_{0} t$ and $\sin n \omega_{0} t$ satisfy Eq. (1) for odd values of $n$ and therefore possess half-wave symmetry when $n$ is odd. Figure $1$ shows other examples of half-wave symmetric functions.
Fig. 1: Typical examples of half-wave odd symmetric functions.
The functions in Figs. 2(a) and 2(b) are also half-wave symmetric. Notice that for each function, one half-cycle is the inverted version of the adjacent half-cycle.
Fig. 2: Typical examples of odd periodic functions. These functions are also half-wave odd symmetric functions.
The Fourier coefficients become
$$\begin{array}{l}a_{0}=0 \\a_{n}=\left\{\begin{array}{ll}\frac{4}{T} \int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t, & \text { for } n \text { odd } \\0, & \text { for } n \text { even }\end{array}\right. \\b_{n}=\left\{\begin{array}{ll}\frac{4}{T} \int_{0}^{T / 2} f(t) \sin n \omega_{0} t d t, & \text { for } n \text { odd } \\0, & \text { for } n \text { even }\end{array}\right.\end{array} \tag{2}$$
showing that the Fourier series of a half-wave symmetric function contains only odd harmonics. To derive Eq. (2), we apply the property of half-wave symmetric functions in Eq. (1) in evaluating the Fourier coefficients in Eqs. (A), (B), and (C).
Equations from Page ( Trigonometric Fourier Series )
$$a_{0}=\frac{1}{T} \int_{0}^{T} f(t) d t \tag{A}$$ $$a_{n}=\frac{2}{T} \int_{0}^{T} f(t) \cos n \omega_{0} t d t \tag{B}$$ $$b_{n}=\frac{2}{T} \int_{0}^{T} f(t) \sin n \omega_{0} t d t \tag{C}$$
Thus,
$$a_{0}=\frac{1}{T} \int_{-T / 2}^{T / 2} f(t) d t=\frac{1}{T}\left[\int_{-T / 2}^{0} f(t) d t+\int_{0}^{T / 2} f(t) d t\right] \tag{3}$$
We change variables for the integral over the interval $-T / 2 < t < 0$ by letting $x=t+T / 2$, so that $d x=d t$; when $t=-T / 2, x=0$; and when $t=0, x=T / 2$. Also, we keep Eq. (1) in mind; that is, $f(x-T / 2)=-f(x)$. Then,
\begin{aligned}a_{0} &=\frac{1}{T}\left[\int_{0}^{T / 2} f\left(x-\frac{T}{2}\right) d x+\int_{0}^{T / 2} f(t) d t\right] \\&=\frac{1}{T}\left[-\int_{0}^{T / 2} f(x) d x+\int_{0}^{T / 2} f(t) d t\right]=0\end{aligned} \tag{4}
confirming the expression for $a_{0}$ in Eq. (2). Similarly,
$$a_{n}=\frac{2}{T}\left[\int_{-T / 2}^{0} f(t) \cos n \omega_{0} t d t+\int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t\right] \tag{5}$$
We make the same change of variables that led to Eq. (4) so that Eq. (5) becomes
$$\begin{array}{c}a_{n}=\frac{2}{T}\left[\int_{0}^{T / 2} f\left(x-\frac{T}{2}\right) \cos n \omega_{0}\left(x-\frac{T}{2}\right) d x\right. \\\left.+\int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t\right]\end{array} \tag{6}$$
Since $f(x-T / 2)=-f(x)$ and
\begin{aligned}\cos n \omega_{0}\left(x-\frac{T}{2}\right) &=\cos \left(n \omega_{0} t-n \pi\right) \\&=\cos n \omega_{0} t \cos n \pi+\sin n \omega_{0} t \sin n \pi \\&=(-1)^{n} \cos n \omega_{0} t\end{aligned} \tag{7}
substituting these in Eq. (6) leads to
\begin{aligned}a_{n} &=\frac{2}{T}\left[1-(-1)^{n}\right] \int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t \\&=\left\{\begin{array}{ll}\frac{4}{T} \int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t, & \text { for } n \text { odd } \\0, & \text { for } n \text { even }\end{array}\right.\end{aligned} \tag{8}
confirming Eq. (2). By following a similar procedure, we can derive $b_{n}$ as in Eq. (2).Table $1$ summarizes the effects of these symmetries on the Fourier coefficients.
 Symmetry $a_{0}$ $a_{n}$ $b_{n}$ Remarks Even $a_{0} \neq 0$ $a_{n} \neq 0$ $b_{n}=0$ Integrate over T / 2 and multiply by 2 to get the coefficients. Odd $a_{0}=0$ $a_{n}=0$ $b_{n} \neq 0$ Integrate over T / 2 and multiply by 2 to get the coefficients. Half-wave $a_{0}=0$ $a_{2 n}=0 , a_{2 n+1} \neq 0$ $b_{2 n}=0 , b_{2 n+1} \neq 0$ Integrate over T / 2 and multiply by 2 to get the coefficients.