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Signals and Systems – Time-Shifting 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-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,
$$\mathrm{x\left ( t \right )\overset{FT}{\leftrightarrow}X\left ( \omega \right )}$$
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,
$$𝑡 = (𝑢 + 𝑡_{0}) and 𝑑𝑡 = 𝑑𝑢$$
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,
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,
$$\mathrm{e^{-a\left | t-2 \right |}\overset{FT}{\leftrightarrow} e^{-j2\omega}\left ( \frac{2a}{a^{2}+\omega ^{2}} \right )}$$
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