Time-Reversal Property of Fourier Transform

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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

$$\mathrm{x(t) \:=\: u(-t)}$$

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,

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