# Time Scaling Property of Fourier Transform

Signals and SystemsElectronics & ElectricalDigital Electronics

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 Scaling Property of Fourier Transform

Statement – The time-scaling property of Fourier transform states that if a signal is expended in time by a quantity (a), then its Fourier transform is compressed in frequency by the same amount. Therefore, if

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

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

$$\mathrm{x\left ( at \right )\overset{FT}{\leftrightarrow}\frac{1}{\left | a \right |} X\left ( \frac{\omega }{a} \right )}$$

• When 𝑎 > 1, then 𝑥(𝑎𝑡) is the compressed version of 𝑥(𝑡), and

• When 𝑎 < 1, the function 𝑥(𝑎𝑡) is the expanded version of 𝑥(𝑡).

Proof – By the definition of Fourier transform, we have,

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

Then, we have,

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

Putting 𝑎𝑡 = 𝑢, then

$$\mathrm{t=\frac{u}{a};\; \; and\; \; dt=\frac{du}{a}}$$

$$\mathrm{\therefore F\left [ x\left ( at \right ) \right ]=\int_{-\infty }^{\infty}x\left ( u \right )e^{-j\omega \left ( \frac{u}{a} \right )}\frac{du}{a}}$$

$$\mathrm{\Rightarrow F\left [ x\left ( at \right ) \right ]=\frac{1}{a}\int_{-\infty }^{\infty}x\left ( u \right )e^{-j\left ( \frac{\omega}{a} \right ) u }du}$$

Case I – When 𝑎 > 0, then,

$$\mathrm{F\left [ x\left ( at \right ) \right ]=\frac{1}{a}\int_{-\infty }^{\infty}x\left ( u \right )e^{-j\left ( \frac{\omega}{a} \right ) u }du=\frac{1}{a}X\left ( \frac{\omega }{a} \right )}$$

Case II – When 𝑎 < 0, then,

$$\mathrm{F\left [ x\left ( at \right ) \right ]=-\frac{1}{a}\int_{-\infty }^{\infty}x\left ( u \right )e^{j\left ( \frac{\omega}{a} \right ) u }du}$$

$$\mathrm{\Rightarrow F\left [ x\left ( at \right ) \right ]=-\frac{1}{a}\int_{-\infty }^{\infty}x\left ( u \right )e^{-j\left ( \frac{\omega}{a} \right ) u }du=-\frac{1}{a}X\left ( -\frac{\omega }{a} \right )}$$

Therefore,

$$\mathrm{F\left [ x\left ( at \right ) \right ]=\frac{1}{\left | a \right |}X\left ( \frac{\omega }{a} \right )}$$

Or, it can also be represented as,

$$\mathrm{x\left ( at \right )\overset{FT}{\leftrightarrow}\frac{1}{\left | a \right |}X\left ( \frac{\omega }{a} \right )}$$

Updated on 15-Dec-2021 11:45:52