Signals and Systems – Time-Reversal Property of Fourier Transform

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

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

Time Reversal Property of Fourier Transform

Statement – The time reversal property of Fourier transform states that if a function 𝑥(𝑡) is reversed in time domain, then its spectrum in frequency domain is also reversed, i.e., if

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

Then, according to the time-reversal property of Fourier transform,

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

Proof

Form the definition of Fourier transform, we have,

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

$$\mathrm{\therefore F\left [ x\left ( -t \right ) \right ]=\int_{-\infty }^{\infty }x\left ( -t \right )e^{-j\omega t}\: dt}$$

Replacing 𝑡 by (−𝑡) by (-t) in RHS of the above equation, we get,

$$\mathrm{F\left [ x\left ( -t \right ) \right ]=\int_{-\infty }^{\infty }x\left ( t \right )e^{-j\omega t}\: dt}$$

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

$$\mathrm{\therefore F\left [ x\left ( -t \right ) \right ]=X\left ( -\omega \right )}$$

Or, it can be represented as,

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

Numerical Example

Using the time reversal property of Fourier transform, find the Fourier transform of function [𝑢(−𝑡)].

Solution

𝑥(𝑡) = 𝑢(−𝑡)

The Fourier transform of a unit step function is defined as,

$$\mathrm{F\left [ u\left (t \right ) \right ]=\pi \delta \left ( \omega \right )+\frac{1}{j\omega }}$$

Now, by using time-reversal property $\mathrm{\left [ i.e.\: \: \: x\left ( -t \right )\overset{FT}{\leftrightarrow}X\left ( -\omega \right ) \right ]}$ of Fourier transform, we get,

$$\mathrm{F\left [ u\left (-t \right ) \right ]=\left \{ F\left [ u\left ( t \right ) \right ] \right \}_{\omega =\left ( -\omega \right )}}$$

$$\mathrm{\Rightarrow F\left [ u\left (-t \right ) \right ]= \left ( \pi \delta \left ( \omega \right )+\frac{1}{j\omega } \right )_{\omega =\left ( -\omega \right )}=\pi \delta \left ( -\omega \right )+\frac{1}{j\left ( -\omega \right )}}$$

$$\mathrm{\therefore F\left [ u\left (-t \right ) \right ]= \pi \delta \left ( \omega \right )+\frac{1}{j\omega }}$$

Or,

$$\mathrm{u\left ( -t \right )\overset{FT}{\leftrightarrow}\left [ \pi \delta \left ( \omega \right )-\frac{1}{j\omega } \right ]}$$