Derivation of Fourier Transform from Fourier Series

Fourier Series

Consider a periodic signal ?(?) be periodic with period T, then the Fourier series of the function ?(?) is defined as,

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

Where, ?_? 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 ?(?) be a non-periodic signal and let the relation between ?(?) and ?(?) is given by,

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

Where, T is the time period of the periodic signal ?(?).

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 ?_? represents the magnitude of the component of frequency nω₀.

Let nω₀ = ω at ? → ∞. 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 [?(?) → ?(?)]. Therefore, at ? → ∞,

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

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

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 ?(?) and is called the spectral density function.

Inverse Fourier Transform

The Fourier series of a periodic function ?(?) 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 ? → ∞, we have,

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

Thus, ?0 can be represented by ?? 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, ?(?) is known as the inverse Fourier transform of X(ω).

The expressions in equation (6) for X(ω) and in eq. (9) for ?(?) 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

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