Half Wave Symmetry

Electrical Circuit Analysis > The Fourier Series > Symmetry Considerations of The Fourier Series

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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.

TABLE 1: Effects of symmetry on 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.

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