Expression for Exponential Fourier Series Coefficients

Signals and SystemsElectronics & ElectricalDigital Electronics

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‘₯(𝑑), the exponential form of Fourier series is given by,

$$\mathrm{X(t)=\sum_{n=-\infty}^{\infty}C_n e^{jn\omega_0t}\:\:\:...(1)}$$

Where, πœ”0 = 2πœ‹⁄𝑇 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 𝑒−π‘—π‘šπœ”0𝑑 and integrate over one period, so we have,

$$\mathrm{\int_{t_0}^{t_0+T}x(t)e^{-jm\omega_0t}dt=\int_{t_0}^{t_0+T}(\sum_{n=-\infty}^{\infty}C_ne^{jn\omega_0t})e^{-jm\omega_{0}t}dt}$$

$$\mathrm{\Rightarrow\int_{t_0}^{t_0+T}x(t)e^{-jm\omega_0t}dt=\sum_{n=-\infty}^{\infty}C_n\int_{t_0}^{t_0+T}e^{jn\omega_0t}e^{-jm\omega_0t}dt}$$

$$\mathrm{\because \int_{t_0}^{t_0+T}e^{jn\omega_0t}e^{-jm\omega_0t}dt=\begin{cases}0 & for\: m \neq n\T & for\: m = n\end{cases}}$$

$$\mathrm{\therefore \int_{t_0}^{t_0+T} x(t)e^{-jm\omega_0t}dt=TC_m}$$

$$\mathrm{\Rightarrow C_m=\frac{1}{T}\int_{t_0}^{(t_0+T)}x(t)e^{-jm\omega_0t}dt}$$

Therefore, the Fourier coefficient of the exponential Fourier series 𝐢𝑛 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 𝐢0 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 𝐢𝑛 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

$$\mathrm{𝑑_0 = 0 \:and \:(𝑑_0 + 𝑇) = 2\pi}$$

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 𝐢0 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 𝐢𝑛 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,

$$\mathrm{x(t)=\sum_{n=-\infty}^{\infty}C_ne^{jn\omega_0t}=\sum_{n=-\infty}^{\infty}-j\frac{2A}{n\pi}e^{jnt};\:for\:odd\: n}$$

raja
Updated on 06-Dec-2021 13:39:07

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