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