# Signals & Systems – Duality Property of Fourier Transform

Signals and SystemsElectronics & ElectricalDigital Electronics

## 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

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

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

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

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

Therefore, the Fourier transform of given signal is,

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