Half Wave Symmetry

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

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Introduction

In signal analysis and electrical engineering, the Fourier series is a powerful mathematical tool used to represent any periodic signal as a sum of sine and cosine functions. Recognizing symmetry properties in a waveform helps simplify the computation of Fourier coefficients. One important type of symmetry is half-wave symmetry — frequently encountered in power electronics, waveform analysis, and communication systems.

What Is Half-Wave Symmetry?

A periodic function ( f(t) ) with period (T) is said to exhibit half-wave symmetry if shifting it by half its period results in the negative of the original signal: $$ \boxed{f\left(t - \frac{T}{2}\right) = -f(t)} $$

This means each half-cycle of the waveform is the inverted mirror image of the next half-cycle.

For example, certain square and pulse waveforms satisfy this condition, which makes their Fourier analysis much easier than for arbitrary periodic signals.

Figure $ 1 $ shows other examples of half-wave symmetric functions.

Fig. 1: Typical examples of half-wave odd symmetric functions.

Effect of Half-Wave Symmetry on Fourier Coefficients

In general, the Fourier series of a periodic waveform ( f(t) ) is given by:

$$ f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\left[a_n\cos(n\omega_0 t)+b_n\sin(n\omega_0 t)\right] \tag{1} $$

Where:

($a_0$,$a_n$,$b_n$) are Fourier coefficients
($\omega_0 = \frac{2\pi}{T}$) is the fundamental angular frequency

For functions with half-wave symmetry:

✔ The average value (DC component) is zero $$ a_0 = 0 $$ ✔ Even harmonics (cosine and sine terms for even (n)) vanish $$ a_{2n}=0,\quad b_{2n}=0 $$ ✔ Only odd harmonics remain $$ a_{2n+1} \neq 0,\quad b_{2n+1} \neq 0 $$ The Fourier coefficients for odd (n) become:

$$ \begin{aligned} a_{n} &= \frac{4}{T}\int_{0}^{T/2} f(t)\cos(n\omega_0 t),dt,\quad (n\ \text{odd})\\ b_{n} &= \frac{4}{T}\int_{0}^{T/2} f(t)\sin(n\omega_0 t),dt,\quad (n\ \text{odd})\ \end{aligned} $$

For even (n), both coefficients are zero.

$$\begin{array}{l}a_{0}=0 \\a_{n}=\left\{\begin{array}{ll}\frac{4}{T} \int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t, & \text { for } n \text { odd } \\0, & \text { for } n \text { even }\end{array}\right. \\b_{n}=\left\{\begin{array}{ll}\frac{4}{T} \int_{0}^{T / 2} f(t) \sin n \omega_{0} t d t, & \text { for } n \text { odd } \\0, & \text { for } n \text { even }\end{array}\right.\end{array} \tag{2}$$

These results show that the Fourier series of a half-wave symmetric function contains only odd harmonics, simplifying both the analysis and synthesis of periodic signals. 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.

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.

Derivation of half-wave symmetric functions

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.

Why Half-Wave Symmetry Simplifies Analysis

The alternating sign property of half-wave symmetry $f(t-T/2)=-f(t)$causes many mathematical cancellations when computing Fourier integrals:

The average value over one period becomes zero because positive and negative halves cancel out.
Integration of symmetric intervals leads to zero for even harmonic terms.
Only the contributions from odd harmonics survive.

This symmetry reduces computational effort and highlights patterns in the frequency content of real waveforms used in power converters and signal processing.

Typical Examples of Half-Wave Symmetric Functions

Waveforms with half-wave symmetry include:

Symmetrical square waves
Certain pulse trains with alternating polarities
Waveforms where each half-cycle is the inverted replica of its neighbor

These waveforms feature prominently in AC power systems, digital signal generation, and communications.

Benefits of Half-Wave Symmetry in Engineering

Recognizing half-wave symmetry in signals offers multiple advantages:

✅ Simplifies Fourier coefficient calculations
✅ Reduces computational complexity
✅ Reveals which harmonics are present
✅ Helps in filter and harmonic analysis design

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