Time Scaling 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 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

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

• 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,

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