Duality Property of Fourier Transform

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Fourier Transform

For a continuous-time function x(t), the Fourier transform of x(t) can be defined as

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

Duality Property of Continuous-Time Fourier Transform

Statement – If a function x(t) has a Fourier transform X(ω) and we form a new function in time domain with the functional form of the Fourier transform as X(t), then it will have a Fourier transform X(ω) with the functional form of the original time function, but it is a function of frequency.

Mathematically, the duality property of CTFT states that, if

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

Then, according to duality property,

$$\mathrm{X(t)\overset{FT}{\leftrightarrow}2\pi x(-\omega)}$$

Proof

From the definition of inverse Fourier transform, we have

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

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

By replacing t = (−t) in the above equation, we get,

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

Now, on interchanging t and ω, we get,

$$\mathrm{\Rightarrow \:2\pi.x(-\omega)\:=\:\int_{-\infty}^{\infty}X(t)e^{-j\omega t}dt\:=\:F[X(t)]}$$

Therefore,

$$\mathrm{F[X(t)]\:=\:2\pi.x(-\omega)}$$

Or, it can also be represented as

$$\mathrm{X(t)\overset{FT}{\leftrightarrow}2\pi.x(-\omega)}$$

Also, for even functions,

$$\mathrm{x(-\omega)\:=\:x(\omega)}$$

Therefore, the duality property of Fourier transform for even functions states that

$$\mathrm{X(t)\overset{FT}{\leftrightarrow}2\pi x(\omega)}$$

Numerical Example

Using duality property of Fourier transform, find the Fourier transform of the following signal −

$$\mathrm{x(t)\:=\:\frac{1}{a^2\:+\:t^2}}$$

Solution

Given

$$\mathrm{x(t)\:=\:\frac{1}{a^2\:+\:t^2}}$$

The Fourier transform of a double-sided exponential function is defined as

$$\mathrm{F[e^{-a|t|}]\:=\:\frac{2a}{a^2\:+\:\omega^2}}$$

$$\mathrm{\text{Now, by using duality property }\:[i.e.,X(t)\overset{FT}{\leftrightarrow}2\pi.x(-\omega)].we\:have,}$$

$$\mathrm{F\left[\frac{2a}{a^2\:+\:t^2}\right]\:=\:2\pi e^{-a|\:-\:\omega|}}$$

$$\mathrm{\Rightarrow\: F\left[\frac{1}{a^2\:+\:t^2}\right]\:=\:\frac{1}{2a}\:\cdot\:2\pi e^{-a|\omega|}}$$

Therefore, the Fourier transform of given signal is,

$$\mathrm{F[x(t)]\:=\:F\left[\frac{1}{a^2\:+\:t^2}\right]\:=\:\frac{\pi}{a}\:\cdot\:e^{-a|\omega|}}$$