What is Quarter Wave Symmetry?

---

---

Quarter Wave Symmetry

A periodic function $x(t)$ which has either odd symmetry or even symmetry along with the half wave symmetry is said to have quarter wave symmetry.

Mathematically, a periodic function $x(t)$ is said to have quarter wave symmetry, if it satisfies the following condition −

$$\mathrm{x(t)\:=\:x(-t)\:or\:x(t)\:=\:-x(-t)\:and\:x(t)\:=\:-x\left (t\: \pm\: \frac{T}{2}\right )}$$

Some examples of periodic functions having quarter wave symmetry are shown in Figure-1.

The Fourier series coefficients for the function having quarter wave symmetry are evaluated as follows −

Case I – When n is odd

$$\mathrm{x(t) \:=\: -x(-t)\:and\:x(t)\:=\:-x\left (t\:\pm\: \frac{T}{2}\right )}$$

For this case,

$$\mathrm{a_{0} \:=\: 0\:\:and\:\:a_{n} \:=\: 0}$$

And,

$$\mathrm{b_{n} \:=\: \frac{8}{T} \int_{0}^{T/4}x(t)\:sin\:n\omega_{0}\:t\:dt}$$

Case II – When n is even

$$\mathrm{x(t)\:=\:x(-t)\:and\:x(t)\:=\:-x\left (t \:\pm\: \frac{T}{2}\right )}$$

For this case,

$$\mathrm{a_{0}\:=\:0\:\:and\:\:b_{n}\:=\:0}$$

And,

$$\mathrm{a_{n}\:=\:\frac{8}{T} \int_{0}^{T/4}x(t)\:cos\:n\omega_{0}\:t\:dt}$$

Numerical Example

Find the trigonometric Fourier series for the waveform shown in Figure-2 using the quarter wave symmetry.

Solution

From the waveform shown in Figure-2, it is clear that it has odd symmetry along with half wave symmetry because it satisfies the following conditions −

$$\mathrm{x(t) \:=\: -x(-t)\:and\:x(t)\:=\: -x\left (t \:\pm\: \frac{T}{2}\right )}$$

Therefore, the given waveform has quarter wave symmetry. Hence, the trigonometric Fourier series coefficients are,

$$\mathrm{a_{0}\:=\:0\:\:and\:\:a_{n}\:=\:0}$$

And the coefficient $b_{n}$ is given by,

$$\mathrm{b_{n}\:=\:\frac{8}{T} \int_{0}^{T/4}x(t)\:sin\:n\omega_{0}\:t\:dt}$$

Also, for this function only odd harmonics exist. Therefore,

• Time period of waveform,

$$\mathrm{T \:=\: 2\pi}$$

• Fundamental frequency,

$$\mathrm{\omega_{0} \:=\: \frac{2\pi}{T} \:=\: \frac{2\pi}{2\pi} \:=\: 1}$$

$$\mathrm{\therefore\:b_{n} \:=\:\frac{8}{T} \int_{0}^{T/4}x(t)\:sin\:n\omega_{0}\:t\:dt \:=\:\frac{8}{2\pi} \int_{0}^{\pi/2} \:A\:sin\:nt\:dt}$$

$$\mathrm{\Rightarrow\:b_{n} \:=\: \frac{4A}{\pi}\left[\frac{-cos\:nt}{n} \right ]_{0}^{\pi/2}\:=\: -\frac{4A}{n\pi} \:[cos(n\pi/2) \:-\: cos\:0]}$$

$$\mathrm{\therefore\:b_{n} \:=\: \begin{cases}\frac{4A}{n\pi} \:\:for\:odd\:n\\\\0 \:\:for\:even\:n\end{cases}}$$

Therefore, the trigonometric Fourier series for the waveform is,

$$\mathrm{x(t)\:=\:\frac{4A}{\pi}sin\:t \:+\: \frac{4A}{3\pi}sin\:3t \:+\: \frac{4A}{5\pi}sin\:5t\:+\:\dotso}$$