Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Expression for Exponential Fourier Series Coefficients
Exponential Fourier Series
A periodic signal can be represented over a certain interval of time in terms of the linear combination of orthogonal functions. If these orthogonal functions are exponential functions, then it is called the exponential Fourier series
For any periodic signal π₯(π‘), the exponential form of Fourier series is given by,
$$\mathrm{X(t)=\sum_{n=-\infty}^{\infty}C_n e^{jn\omega_0t}\:\:\:...(1)}$$
Where, π0 = 2π⁄π is the angular frequency of the periodic function.
Coefficients of Exponential Fourier Series
In order to evaluate the coefficients of the exponential series, we multiply both sides of the equation (1) by π−πππ0π‘ and integrate over one period, so we have,
$$\mathrm{\int_{t_0}^{t_0+T}x(t)e^{-jm\omega_0t}dt=\int_{t_0}^{t_0+T}(\sum_{n=-\infty}^{\infty}C_ne^{jn\omega_0t})e^{-jm\omega_{0}t}dt}$$
$$\mathrm{\Rightarrow\int_{t_0}^{t_0+T}x(t)e^{-jm\omega_0t}dt=\sum_{n=-\infty}^{\infty}C_n\int_{t_0}^{t_0+T}e^{jn\omega_0t}e^{-jm\omega_0t}dt}$$
$$\mathrm{\because \int_{t_0}^{t_0+T}e^{jn\omega_0t}e^{-jm\omega_0t}dt=\begin{cases}0 & for\: m
eq n\T & for\: m = n\end{cases}}$$
$$\mathrm{\therefore \int_{t_0}^{t_0+T} x(t)e^{-jm\omega_0t}dt=TC_m}$$
$$\mathrm{\Rightarrow C_m=\frac{1}{T}\int_{t_0}^{(t_0+T)}x(t)e^{-jm\omega_0t}dt}$$
Therefore, the Fourier coefficient of the exponential Fourier series πΆπ is given by,
$$\mathrm{ C_n=\frac{1}{T}\int_{t_0}^{t_0+T}x(t)e^{-jn\omega_0t}dt\:\:\:....(2)}$$
Equation (2) is also called as the analysis equation.
Also, the DC component πΆ0 of the exponential Fourier series is given by,
$$\mathrm{ C_0=\frac{1}{T}\int_{t_0}^{(t_0+T)}x(t)dt\:\:\:....(3)}$$
The exponential Fourier series coefficients of a periodic function x(t) have only a discrete spectrum because the values of the coefficient πΆπ exists only for discrete values of n. As the exponential Fourier series represents a complex spectrum, thus, it has both magnitude and phase spectra.
About the magnitude and phase spectra, the following points may be noted −
- The magnitude line spectrum is always an even function of n.
- The phase line spectrum is always an odd function of n.
Numerical Example
Obtain the exponential Fourier series for the waveform shown in the figure.
Solution
The waveform shown in the figure represents a periodic function with a period T = 2π and can be mathematically expressed as,
$$\mathrm{x(t)=\begin{cases}A & 0 \leq t\leq \pi\-A & \pi \leq t\leq 2\pi \end {cases}}$$
Here, let
$$\mathrm{π‘_0 = 0 \:and \:(π‘_0 + π) = 2\pi}$$
Therefore, the fundamental frequency of the function is,
$$\mathrm{\omega_0=\frac{2\pi}{T}=\frac{2\pi}{2\pi}=1}$$
Now, the exponential Fourier series coefficient πΆ0 is given by,
$$\mathrm{C_0=\frac{1}{T}\int_{0}^{T}x(t)dt}$$
$$\mathrm{\Rightarrow C_0=\frac{1}{2\pi}\int_{0}^{\pi}A\:dt+\frac{1}{2\pi}\int_{\pi}^{2\pi}-A\:dt=\frac{A}{2\pi}[(t)^{\pi}_{0}-(t)^{2\pi}_{\pi}]=0}$$
Again, the coefficient πΆπ is given by,
$$\mathrm{C_n=\frac{1}{T} \int_{0}^{T}x(t)e^{-jn\omega_0t}dt}$$
$$\mathrm{\Rightarrow C_n=\frac{1}{2\pi} \int_{0}^{\pi}A\:e^{-jnt}dt+\frac{1}{2\pi} \int_{\pi}^{2\pi}-A\:e^{-jnt}dt}$$
$$\mathrm{\Rightarrow C_n=\frac{A}{2\pi}[(\frac{e^{-jnt}}{-jn})^{\pi}_{0}-(\frac{e^{-jnt}}{-jn})^{2\pi}_{\pi}]}$$
$$\mathrm{\Rightarrow C_n=\frac{-A}{j2n\pi}[(e^{-jnt}-e^{0})-(e^{-j2n\pi}-e^{-jn\pi})]}$$
$$\mathrm{\Rightarrow C_n=\frac{-A}{j2n\pi}[\left \{ (-1)^n-1 \right \}-\left \{ 1-(-1)^n \right \}]=-j\frac{2A}{n\pi}}$$
$$\mathrm{\therefore C_n=\begin{cases}0 & ;\:for\:even\:n\-j\frac{2A}{n\pi} & ;\:for\:odd\: n \end {cases}}$$
Hence, the exponential Fourier series for the given function is,
$$\mathrm{x(t)=\sum_{n=-\infty}^{\infty}C_ne^{jn\omega_0t}=\sum_{n=-\infty}^{\infty}-j\frac{2A}{n\pi}e^{jnt};\:for\:odd\: n}$$
10K+ Views
