Time Scaling property of the Fourier Transform

If $ F(\omega)=\mathcal{F}[f(t) $ ], then
$$\mathcal{F}[f(a t)]=\frac{1}{|a|} F\left(\frac{\omega}{a}\right) \tag{1}$$
where $ a $ is a constant. Equation (1) shows that time expansion $ (|a|>1) $ corresponds to frequency compression, or conversely, time compression $ (|a|<1) $ implies frequency expansion. The proof of the time-scaling property proceeds as follows.
$$\mathcal{F}[f(a t)]=\int_{-\infty}^{\infty} f(a t) e^{-j \omega t} d t \tag{2}$$
If we let $ x=a t $, so that $ d x=a d t $, then
$$\mathcal{F}[f(a t)]=\int_{-\infty}^{\infty} f(x) e^{-j \omega x / a} \frac{d x}{a}=\frac{1}{a} F\left(\frac{\omega}{a}\right) \tag{3}$$
For example, for the rectangular pulse $ p(t) $, $$\mathcal{F}[p(t)]=A \tau \operatorname{sinc} \frac{\omega \tau}{2} \tag{4.1}$$ Using Eq. (1),
$$\mathcal{F}[p(2 t)]=\frac{A \tau}{2} \operatorname{sinc} \frac{\omega \tau}{4} \tag{4.2}$$
It may be helpful to plot $ p(t) $ and $ p(2 t) $ and their Fourier transforms. Since
$$p(t)=\left\{\begin{array}{ll}A, & -\frac{\tau}{2}
then replacing every $ t $ with $ 2 t $ gives
$$p(2 t)=\left\{\begin{array}{ll}A, & -\frac{\tau}{2}<2 t<\frac{\tau}{2} \\0, & \text { otherwise }\end{array}=\left\{\begin{array}{ll}A, & -\frac{\tau}{4}
showing that $ p(2 t) $ is time compressed, as shown in Fig. 1(b). To plot both Fourier transforms in Eq. (4), we recall that the sinc function has zeros when its argument is $ n \pi $, where $ n $ is an integer.
Fig. 1: The effect of time scaling: (a) transform of the pulse, (b) time compression of the pulse causes frequency expansion.
Hence, for the transform of $ p(t) $ in Eq. (4.1),
$$ \omega \tau / 2=2 \pi f \tau / 2=n \pi \rightarrow f=n / \tau $$
and for the transform of $ p(2 t) $ in Eq. (4.2),
$$ \omega \tau / 4=2 \pi f \tau / 4= $ $ n \pi \rightarrow f=2 n / \tau $$
The plots of the Fourier transforms are shown in Fig. 1, which shows that time compression corresponds with frequency expansion. We should expect this intuitively, because when the signal is squashed in time, we expect it to change more rapidly, thereby causing higher-frequency components to exist.

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