# Derivation of Fourier Transform from Fourier Series

## Fourier Series

Consider a periodic signal π(π‘) be periodic with period T, then the Fourier series of the function π(π‘) is defined as,

$$\mathrm{g(t)=\sum_{n=-\infty}^{\infty}C_{n}e^{jn\omega_{0}t}\:\:\:\:....(1)}$$

Where, πΆπ is the Fourier series coefficient and is given by,

$$\mathrm{C_{n}=\frac{1}{T}\int_{\frac{-T}{2}}^{\frac{T}{2}}g(t)e^{-jn\omega_{0}t}dt\:\:\:\:....(2)}$$

## Derivation of Fourier Transform from Fourier Series

Let π₯(π‘) be a non-periodic signal and let the relation between π₯(π‘) and π(π‘) is given by,

$$\mathrm{X(t)=\lim_{T\rightarrow \infty}g(t)\:\:\:\:.....(3)}$$

Where, T is the time period of the periodic signal π(π‘).

By rearranging eq. (2), we get,

$$\mathrm{TC_n=\int_{\frac{-T}{2}}^{\frac{T}{2}}g(t)e^{-jn\omega_{0}t}dt}$$

The term πΆπ represents the magnitude of the component of frequency nω0.

Let nω0 = ω at π → ∞. Then, we have,

$$\mathrm{\omega_0=\frac{2\pi}{t}|_{T\rightarrow \infty}\rightarrow 0}$$

Thus, the discrete Fourier spectrum becomes continuous and hence the summation becomes integral and [π(π‘) → π₯(π‘)]. Therefore, at π → ∞,

$$\mathrm{TC_n=\lim_{T\rightarrow \infty}\int_{\frac{-T}{2}}^{\frac{T}{2}}g(t)e^{-j\omega t}dt}$$

$$\mathrm{\Rightarrow TC_n=\int_{-\infty}^{\infty}[\lim_{T\rightarrow \infty}g(t)]e^{-j\omega t}dt\:\:\:\:.....(4)}$$

From equations (3) & (4), we have,

$$\mathrm{\Rightarrow TC_n=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt=X(\omega)\:\:\:\:....(5)}$$

Therefore, the Fourier transform of the non-periodic signal is

$$\mathrm{X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\:\:\:\:....(6)}$$

The function X(ω) represents the frequency spectrum of function π₯(π‘) and is called the spectral density function.

## Inverse Fourier Transform

The Fourier series of a periodic function π(π‘) is defined as,

$$\mathrm{g(t)=\sum_{n=-\infty}^{\infty}C_n e^{jn\omega_0 t}}$$

$$\mathrm{\because C_n=\frac{TC_n}{T}=\frac{X(\omega)}{T}\:\:\:[from\:eq.(5)]}$$

$$\mathrm{\therefore g(t)=\sum_{n=-\infty}^{\infty}\frac{X(\omega)}{T}e^{jn\omega_0 t}}$$

$$\mathrm{\because n\omega_0=\omega\:and\:T=\frac{2\pi}{\omega_0}}$$

$$\mathrm{\therefore g(t)=\sum_{n=-\infty}^{\infty}\frac{X(\omega)}{(2\pi/\omega_0)}e^{jn\omega_0t}=\sum_{n=-\infty}^{\infty}\frac{X(\omega)}{2\pi}e^{jn\omega_0t}\omega_0\:\:\:....(7)}$$

Hence, from equations (3) & (7), we have

$$\mathrm{x(t)=\lim_{T\rightarrow \infty}g(t)=\lim_{T\rightarrow \infty}\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}X(\omega)e^{jn\omega_0t}\omega_0\:\:\:....(8)}$$

As π → ∞, we have,

$$\mathrm{\omega_0=\frac{2\pi}{T}|_{T\rightarrow \infty}\rightarrow 0}$$

Thus, π0 can be represented by ππ and the summation becomes integration. Therefore, eq. (8) can be written as,

$$\mathrm{x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega\:\:\:....(9)}$$

Here, π₯(π‘) is known as the inverse Fourier transform of X(ω).

The expressions in equation (6) for X(ω) and in eq. (9) for π₯(π‘) are known as Fourier transform pair and can be represented as,

$$\mathrm{X(\omega)=F[x(t)]}$$

$$\mathrm{}$$

And

$$\mathrm{x(t)=F^{-1}[X(\omega)]}$$

The Fourier transform pair can also be represented as

$$\mathrm{x(t)\overset{FT}{\leftrightarrow}X(\omega)}$$

Updated on: 06-Dec-2021

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