Fourier Analysis for a Continuous Time Non-periodic Signal

Fourier Series is defined in trigonometric terms but cosine and sine can be defined with complex terms also-

cos{\theta} + isin{\theta} = e^{i\theta}
cos{\theta} - isin{\theta} = e^{-i\theta}

First of all before we go further one must know that along with defining any periodic function by the sum of various harmonics of sine and cosine, these can be defined by the following as well-

f(x) = \sum_{n=-\infty}^{\infty} C_{n}e^{\frac{in\pi\omega t}{l}} ;[Let the period of any periodic function is 2l ]
where angular velocity {\omega} = \frac{2\pi}{T} ; [T is the time period; i.e time required to pass 2\pi angle once]

{\omega} = {2\pi f} [f is the frequency]
Here {\omega} is considered as the frequency.

Here one should know that C_{n}=\frac{1}{2l}\int_{-l}^{l}f(x)e^{\frac{-in\pi x}{l}}\,\mathrm{d}x
Now defining f(x) in terms of \omega is called frequency domain expression and defining in terms of time (where laplace transformation is used to express any system) is called time domain expression.
A case can arise when f(x) is not periodic,this means f(x) has no period and time period T is tending to infinity (T \rightarrow \infty ) and so \omega\rightarrow 0 . As it’s non-periodic function, so obviously its frequency can’t be defined.
Say-

1-0

 

 

This is a non-periodic function where -a\leq\omega t\leq a ,means this function doesn’t repeat after a fixed period of time.
Now f(t) =\sum_{n=-\infty}^{\infty}C_{n}e^{jn\omega t} ; [ l term only comes when period is not 2\pi ,say here time period is 2\pi ]

\therefore C_{n}=\frac{1}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)e^{-jn\omega t}\,\mathrm{d}t
As here f(t) is non-periodic function, t \rightarrow \infty and \omega\rightarrow 0 ,so 'n' becomes meaningless in the above expression.
Let n\omega \rightarrow \omega
\omega \rightarrow \triangle \omega

\therefore T \rightarrow \frac{2\pi}{\triangle \omega}

\therefore f(t) becomes-

f(t)=\sum_{\omega =-\infty}^{\infty}C_{\omega}e^{j\omega t} [\forall \omega = 0,\pm \triangle \omega , \pm 2\triangle \omega ... ] \longrightarrow (1)

\therefore C_{\omega} = \frac{\triangle \omega}{2\pi}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)e^{-j\omega t}\,\mathrm{d}t

\therefore Substituting C_{\omega} in equation (1)

f(t) = \frac{1}{2\pi}[\sum_{\omega =-\infty}^{\infty}\int_{\frac{-T}{2}}^{\frac{T}{2}}f(t)e^{-j\omega t}\,\mathrm{d}t]e^{j\omega t}\triangle \omega

\because T \rightarrow \infty ,\triangle \omega \rightarrow \,\mathrm{d}\omega and \sum \rightarrow \int

\therefore f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}[\int_{-\infty}^{\infty}f(t)e^{-j\omega t}\,\mathrm{d}t]e^{j\omega t}\,\mathrm{d}\omega

\therefore f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{j\omega t}\,\mathrm{d}\omega \Longrightarrow [Inverse Fourier Transform]

F({\omega}) = \mathscr{F}[f(t)]= \int_{-\infty}^{\infty}f(t)e^{-j\omega t}\,\mathrm{d}t \Longrightarrow [Fourier Transform]
[\mathscr{F} denotes fourier transform ]

So it has been shown that for a non-periodic function we have to first get the fourier trasform of that function (Of course firstly by verifying Dirichlet conditions) to get the fourier series of that function.

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