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).
$$a_{0}=\frac{1}{T} \int_{0}^{T} f(t) d t \tag{A}$$
$$a_{n}=\frac{2}{T} \int_{0}^{T} f(t) \cos n \omega_{0} t d t \tag{B}$$
$$b_{n}=\frac{2}{T} \int_{0}^{T} f(t) \sin n \omega_{0} t d t \tag{C}$$
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.
TABLE 1: Effects of symmetry on Fourier coefficients.
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. |
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