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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:

Fourier Transform of Impulse

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    What is the question? You want to Fourier transform $x$? If so, what is $\sigma$? Or do you want the FT of $x$ in terms of the FT of $\sigma$? Do you mean to have $(t-T)/a$ in the parentheses (otherwise it's just shifting, and not rescaling and shifting)? – 2017-02-12
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    Yes, i want the Fourier transform of x... the σ is the impulse.... Yes, at the question they are in parentheses. – 2017-02-12
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    So $\sigma$ is a delta function? – 2017-02-12
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    Yes.. i put the image of the question at the post. – 2017-02-12

2 Answers 2

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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 $$ \tilde x(\omega) = ae^{-i\omega T}.$$

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    I didn't understand very well the substitution that you used, why you substitute "t" to "au+T"? and how you get ae−iωT before the integral? and do you call au+T = u in the end? – 2017-02-12
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    @DouglasNickson So that $u$ would be inside the $\sigma$ function. We have $u = (t-T)/a$ which solves to $t=au+T$ I don't call $au+T = u.$ That's where the $e^{-i\omega T} $ comes from. $e^{-i\omega(au+T)} = e^{-i\omega T} e^{-i\omega a u}$ – 2017-02-12
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    And the "a" before the integral comes from where? How you get "σ~(aω)" in the and? – 2017-02-12
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    I didn't understand the end of the integral, the FT of impulse is not 1 on zero? but, in this case the impulse was not shifting? – 2017-02-13
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    @DouglasNickson the expression $\int \sigma(u)e^{ia\omega u} = \tilde\sigma(a\omega).$ This is the definition of the FT in my first eq with $\omega$ replaced by $\omega a$. By the same token the FT of $\delta(t)$ is $\int \delta(t)e^{-i\omega t}dt$ = 1. – 2017-02-13
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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 $\sigma$.

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    Where is the "T"? why you use only "t-a" and not "t-T/a"? – 2017-02-12
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    @DouglasNickson just set $a=T/a$. – 2017-02-12
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    I can do this? because the expression is "(t-T)/a"... and whats is the FT of the impulse in this case? – 2017-02-12