Dated: 30-10-2024

`Functions`¹ With `Periods`² other than \(2 \pi\)

There are sometimes functions¹ which have \(T\) period² which is other than \(2 \pi\).
We know that in physics, the oscillations are basically function¹ of time and they repeat after \(T\) interval.

\[f(t) = f(t + T)\]

If the frequency measured in hertz (Hz) is defined as

\[f = \frac 1 T\]

and angular velocity as

\[\omega = 2 \pi f\]

then

\[\omega = \frac {2 \pi}{T} \implies 2 \pi = \omega \cdot T\]

therefore, the angle \(x\) at any given time \(t\) will be

\[x = \omega \cdot t\]

The fourier series³ takes the form

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

Fourier Coefficients

\[a_{0}=\frac{2}{T}\int_{0}^{T}f(t)dt=\frac{\omega}{\pi}\int_{0}^{ \frac {2\pi}\omega}f(t)dt\]

\[a_{n}=\frac{2}{T}\int_{0}^{T}f(t)\cos(n\omega t)dt=\frac{\omega}{\pi}\int_{0}^{\frac{2\pi}\omega}f(t)\cos(n\omega t)dt\]

\[b_{n}=\frac{2}{T}\int_{0}^{T}f(t)\sin(n\omega t)dt=\frac{\omega}{\pi}\int_{0}^{\frac{2\pi}\omega}f(t)\sin(n\omega t)dt\]

Half Wave Rectifier

A sinusoidal voltage \(E \sin (\omega t)\) is passed through a half wave rectifier which clips the negative voltage onto the \(x\) axis.
Find the fourier series³ for the resulting periodic function.

\[u(t) = 0 \quad - \frac T 2 < t < 0\]

\[u(t) = E \sin (\omega t) \quad 0 < t < \frac T 2\]

Solution

\(a_0\)

\[a_{0}=\frac{2}{T}\int_{- \frac {T}2}^{\frac T 2}u(t)dt\]

\[=\frac{2}{T}\left(\int_{- \frac T 2}^{0}0~dt+\frac{E}{T}\int_{0}^{\frac T 2}\sin(\omega t)dt\right)\]

\[=\frac{2E}{T^{2}}\left[-\frac{1}{\omega}\cos(\omega t)\right]_{0}^{\frac T 2}\]

\[=\frac{2E}{\pi\omega}\left(1-\cos~\frac{\pi\omega T}{2}\right)\]

\[=\frac{\omega}{\pi}E\left|\frac{-\cos~\omega~t}{\omega}\right|_{0}^{\frac \pi \omega}\]

\[=\frac{2E}{\pi}\]

\(a_1\)

\[a_{n}=\frac{2}{T}\int_{\frac T 2}^{\frac T 2}u(t)\cos(n\omega t)dt\]

\[=\frac{2E}{T}\int_{0}^{\frac T 2}\sin(\omega t)\cos(n\omega t)dt\]

\[=\frac{\omega E}{2\pi}\int_{0}^{\frac {2\pi}\omega}2\sin(\omega t)\cos(n\omega t)dt\]

\[=\frac{\omega E}{2\pi}\int_{0}^{\frac \pi \omega}(\sin(1+n)\omega t+\sin(1-n)\omega t)dt\]

\[ a_n = \frac{\omega E}{2\pi} \left[ -\frac{\cos (1+n) \, \omega t}{(1+n) \omega} - \frac{\cos (1-n) \, \omega t}{(1-n) \omega} \right]_{0}^{\frac{\pi}{\omega}} \]

\[ = \frac{\omega E}{2\pi} \left[ -\frac{\cos (1+n) \, \pi + 1}{(1+n)\omega} + \frac{-\cos (1-n) \, \pi + 1}{(1-n)\omega} \right] \]

\[ = \frac{\omega E}{2\pi \omega} \left[ \frac{-\cos (1+n) \, \pi + 1}{(1+n)} + \frac{-\cos (1-n) \, \pi + 1}{(1-n)} \right] \]

If \(n\) is Odd

\[a_n = 0\]

If \(n\) is even

\[a_n = \frac{E}{2\pi} \left( \frac{2}{1+n} + \frac{2}{1-n} \right) \]

\[= \frac{E}{2\pi} \left[ \frac{2 - 2n + 2 + 2n}{(1+n)(1-n)} \right] \]

\[= \frac{2E}{(1-n)(1+n)\pi}\]

\[ = \frac{2E}{(1-n^2)\pi} \]

\(b_n\)

\[b_{n}=\frac{2}{T}\int_{-\frac T 2}^{\frac T 2}u(t)\sin(n\omega t)dt\]

\[=\frac{2E}{T}\int_{0}^{\frac T 2}\sin(\omega t)\sin(n\omega t)dt\]

\[=\frac{\omega E}{2\pi}\int_{0}^{\frac {2\pi}\omega}2\sin(\omega t)\sin(n\omega t)dt\]

\[=\frac{\omega E}{2\pi}\int_{0}^{\frac \pi \omega}\left(\cos(1+n)\omega t-\cos(1-n)\omega t\right)dt\]

If \(n = 1\)

\[b_n = - \frac{\omega E}{2 \pi} \int_0^{\frac \pi \omega} (\cos (2 \omega t) - 1)dx\]

\[= - \frac{\omega E}{2 \pi} \left|\frac{\sin (2 \omega t)}{2 \omega} - 1\right|_0 ^ {\frac \pi \omega}\]

\[= - \frac{\omega E}{2 \pi} \left(- \frac \pi \omega\right)\]

\[= \frac E 2\]

If \(n \ne 1\)

\[=-\frac{\omega E}{2\pi}\left[\frac{\sin(1+n)\omega t}{(1+n)\omega}-\frac{\sin(1-n)\omega t}{(1-n)\omega}\right]_{0}^{\frac \pi \omega}\]

\[=-\frac{\omega E}{2\pi}\left[\frac{\sin(1+n)\pi}{(1+n)\omega}-\frac{\sin(1-n)\pi}{(1-n)\omega}\right]\]

\[= 0\]

\[u(t)=\frac{1}{2}a_{0}+\sum_{n=2}^{\infty}a_{n}\cos(n\omega t)\]

\[u(t)=\frac{E}{\pi}+\frac{E}{2}\sin(\omega t)-\frac{2E}{\pi}\left(\frac 1 {1.3} \cos (2 \omega t) + \frac 1 {3.5} \cos (4 \omega t)+ \ldots\right)\]

Functions1 With Periods2 other than \(2 \pi\)