Time-Shifting 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,

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

Time-Shifting Property of Fourier Transform

Statement – The time shifting property of Fourier transform states that if a signal x(t) is shifted by t₀ in time domain, then the frequency spectrum is modified by a linear phase shift of slope (−ωt₀). Therefore, if,

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

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

$$\mathrm{x\left ( t \:-\: t_{0}\right )\overset{FT}{\leftrightarrow}e^{-j\omega t_{0}}X\left ( \omega \right )}$$

Proof

From the definition of Fourier transform, we have

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

$$\mathrm{\therefore F\left [ x\left ( t\:-\:t_{0} \right ) \right ]\:=\:\int_{-\infty }^{\infty }x\left (t \:-\:t_{0} \right )e^{-j\omega t}\: dt}$$

Substituting (t − t₀) = u, then,

$$\mathrm{t \:=\: (u \:+\: t_{0}) \:and\: dt \:=\: du}$$

Therefore

$$\mathrm{F\left [ x\left ( t \:-\:t_{0}\right ) \right ]\:=\:\int_{-\infty }^{\infty }x\left ( u \right )e^{-j\omega \left ( u\:+\:t_{0} \right )}\: du}$$

$$\mathrm{\Rightarrow F\left [ x\left ( t \:-\:t_{0}\right ) \right ]\:=\:e^{-j\omega t_{0}}\int_{-\infty }^{\infty }x\left ( u \right )e^{-j\omega u}\: du\:=\:e^{-j\omega t_{0}}X\left ( \omega \right )}$$

$$\mathrm{\therefore F\left [ x\left ( t \:-\:t_{0}\right ) \right ]\:=\:e^{-j\omega t_{0}}X\left ( \omega \right )}$$

Or, it can also be represented as,

$$\mathrm{x\left ( t\:-\:t_{0} \right )\overset{FT}{\leftrightarrow}e^{-j\omega t_{0}}X\left ( \omega \right )}$$

Similarly,

$$\mathrm{x\left ( t\:+\:t_{0} \right )\overset{FT}{\leftrightarrow}e^{j\omega t_{0}}X\left ( \omega \right )}$$

The time shifting property of Fourier transform has a very important implication. That is,

$$\mathrm{Magnitude,\:\left | e^{-j\omega t_{0}} X\left ( \omega \right )\right |\:=\:\left | X\left ( \omega \right ) \right |}$$

And,

$$\mathrm{Phase,\:\angle e^{-j\omega t_{0}}X\left ( \omega \right )\:=\:e^{-j\omega t_{0}}\:+\:\angle X\left ( \omega \right )\:=\:\angle \left ( -\omega t_{0} \right )\:+\:\angle X\left ( \omega \right )}$$

From this, it is clear that the shifting of a function by t₀ in time domain results in multiplying its Fourier transform by e^−jωt₀ . Hence, there is no change in the magnitude spectrum but the phase spectrum is linearly shifted.

Numerical Example

Using time-shifting property of Fourier transform, find the Fourier transform of signal [e^−a|t−2|].

Solution

Given,

$$\mathrm{x(t) \:=\: e^{-a|t \:-\:2|}}$$

Since the Fourier transform of two-sided exponential signal is defined as,

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

Now, by using time-shifting property $\mathrm{ \left [i.e.\: x\left ( t\:-\:t_{0} \right )\overset{FT}{\leftrightarrow}e^{-j\omega t_{0}}X\left ( \omega \right ) \right ]}$ of the Fourier transform, we have,

$$\mathrm{F\left [ e^{-a\left | t\:-\:2 \right |} \right ]\:=\:e^{-j2\omega}\left ( \frac{2a}{a^{2}\:+\:\omega ^{2}} \right )}$$

Or, it can also be written as,

$$\mathrm{e^{-a\left | t\:-\:2 \right |}\overset{FT}{\leftrightarrow} e^{-j2\omega}\left ( \frac{2a}{a^{2}\:+\:\omega ^{2}} \right )}$$