# Signals & Systems : Fourier Series Representation of Periodic Signals

### The Response of LTI Systems to Complex Exponential

The Response of LTI Systems to Complex Exponential input is the same complex exponential with only a change in amplitude , i.e.

Continuous time : $e^{st} \rightarrow H(s). e^{st}$

Discrete time : $z^n \rightarrow H(z). z^n$

Where complex amplitude factor H(s) or H(z) will be a function of the complex variable s or z . A signal for which the system output is (possibly complex) constant times the input , is referred to as an eigen function of the system , and the amplitude factor is referred to as the system eigenvalue .

If the input to continuous time LTI system is represented as a linear combination of complex exponential , i.e.

$\displaystyle x(t) = \Sigma_k a_k e^{s_kt}$

Then , output will be ,

$\displaystyle Y(t) = \Sigma_k a_k H(s_k)e^{s_kt}$

In similar fashion , if the input to a discrete time LTI system is represented as a linear combination of complex exponential i.e.

$\displaystyle x[n] = \Sigma_k a_k Z_k^n$

Then , output will be ,

$\displaystyle Y[n] = \Sigma_k a_k H(z_k) Z_k^n$

### Linear Combination of Harmonically Related Complex Exponential

A linear combination of harmonically related complex complex exponential is expressed as ,

$\displaystyle x(t) = \Sigma_{k=-\infty}^{\infty} a_k e^{j k \omega t}$

$\displaystyle x(t) = \Sigma_{k=-\infty}^{\infty} a_k e^{j k(\frac{2\pi}{T})t}$

For real x(t) , $\displaystyle x^\ast (t) = x(t)$

and , $\displaystyle x(t) = \Sigma_{k=-\infty}^{\infty} a_k^\ast e^{-j k \omega_0 t}$

Replacing k by -k , we have

$\displaystyle x(t) = \Sigma_{k=-\infty}^{\infty} a_{-k}^\ast e^{j k \omega_0 t}$

### Fourier Series Representation of a Continuous Time Periodic Signal

Fourier Series Representation of a Continuous Time Periodic Signal x(t) is given by pair of equations ,

$\displaystyle x(t) = \Sigma_{k=-\infty}^{\infty} a_k e^{j k \omega_0 t}$

$\displaystyle x(t) = \Sigma_{k=-\infty}^{\infty} a_k e^{j k (\frac{2\pi}{T}) t}$ ….(i)

$\displaystyle a_k = \frac{1}{T} \int_T x(t) e^{-j k \omega_0 t} dt$

$\displaystyle a_k = \frac{1}{T} \int_T x(t) e^{-j k (\frac{2\pi}{T}) t} dt$ ….(ii)

The equation (i) is known as Synthesis equation and equation (ii) is known as analysis equation . The set of coefficients {ak} are called Spectral coefficients or Fourier series coefficients of x(t) .

### Dirichlet Conditions

(i) Over any period , x(t) must be absolutely integrable , i.e.

$\displaystyle \int_T |x(t)|dt < \infty$

(ii) In any finite interval of time , x(t) is of bounded variation , i.e. there are no more than a finite number of maxima and minima during any single period of signal .

(iii) In any finite interval of time , there are only finite number of discontinuities . Further , each of these discontinuities is finite .

Solved Example : Consider the Signal x(t) = Sinω0t ; whose fundamental frequency is ω0 . Determine the Fourier series coefficients for this Signal .

Solution: $\displaystyle x(t) = sin\omega_0 t = \frac{1}{2 j}(e^{j\omega_0 t} – e^{-j\omega_0 t})$

$\displaystyle x(t) = \Sigma_{k=-\infty}^{\infty} a_k e^{j k \omega_0 t}$

$\displaystyle a_k = \frac{1}{T} \int_T x(t) e^{-j k \omega_0 t} dt$

On comparing we get ,

$\displaystyle a_1 = \frac{1}{2 j}$

$\displaystyle a_{-1} = -\frac{1}{2 j}$

ak = 0 , k ≠ + 1 or -1

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