I'm new here and i hope you can help-me. I started studying the Fourier Transform now at my university and i have a lot of doubts about this subject. My teacher sent to us one list of exercises and i don't know how can i solve this question:
$$ x(t) = \delta ((t-T)/a) $$
I don't know how using the properties of scale and shifting in this case, if you can explain to me the steps to solve this question i will really enjoy. Thanks for your help :)
this is the image of the question:
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Remember that $$\mathcal F \{\sigma (t-a)\} (s) = \int_{-\infty}^{\infty} e^{-2\pi i s t} \sigma(t-a)\mathrm {d}t $$ Now use the substituion $u = t-a \Rightarrow \mathrm{d} u = \mathrm{d} t $ and $t = u +a$ to get $$\int_{-\infty}^{\infty} e^{-2\pi i s (u+a)} \sigma(u)\mathrm {d}u=\int_{-\infty}^{\infty} e^{-2\pi i s u}e^{-2\pi i s a} \sigma(u)\mathrm {d}u=e^{-2\pi i s a}\int_{-\infty}^{\infty} e^{-2\pi i s u} \sigma(u)\mathrm {d}u=e^{-2\pi i s a}\mathcal F \{\sigma (t)\} (s)$$ The final expression depends on $\sigma$.
The Fourier transform of $x$ is defined as $$\tilde x(\omega) = \int_{-\infty}^\infty x(t)e^{-i\omega t}dt $$ (maybe with a factor of $1/\sqrt{2\pi}$ in there, or a $+$ sign and/or a factor of $2\pi$ in the exponent, depending on what conventions you're using).
Plugging in $x(t) = \sigma((t-T)/a)$ gives $$ \tilde x(\omega) = \int_{-\infty}^\infty \sigma((t-T)/a)e^{-i\omega t}dt$$
Then you can do a substitution $u = (t-T)/a$ to get $$\tilde x(\omega) = a\int_{-\infty}^\infty \sigma(u)e^{-i\omega(au+T)}du = ae^{-i\omega T}\int_{-\infty}^\infty \sigma(u)e^{-ia\omega u}du = ae^{-i\omega T}\tilde\sigma(a\omega)$$ where $\tilde\sigma(\omega)$ is the Fourier transform of $\sigma.$
This is a general formula for how the Fourier transform changes when you shift and rescale.
You are indicating that the $\sigma$ is an impulse, which I'm taking to mean is a delta function centered on zero. The fourier transform is then $\tilde\sigma(\omega) = 1$ so you have $$ \tilde x(\omega) = ae^{-i\omega T}.$$