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I am trying to find Fourier Transform of:

$$f(t)=4te^{-t^2}$$.

I found in MatLab that $\mathfrak{F}\left \{ f(t) \right \}=i\sqrt{2}e^{- \frac{w^2}{4}}w$ .So is this possible to come to same result by using shifting theorem since the function is Gaussian.

What I was thing was: Suppose $g(t)=-2e^{-t^2}$ and it's first derivative is $g'(t)=4te^{-t^2}$ so by using these facts I am trying to use properties of Fourier Transforms to come to same result.

Melina
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1 Answers1

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we know that $f'(x) = -2 a x f(x) \implies \ln f(x) = -a x^2 + cste \implies f(x) = C e^{-a x^2}$.

now if $g(t) = e^{-a t^2}$ then $g'(t) = -2 a t g(t)$ and if $a > 0$, $|g(t)|$ and $|g'(t)|$ fastly decreases when $t \to \infty$ so :

$$i \omega \int_{-\infty}^\infty g(t) e^{- i \omega t }dt = \int_{-\infty}^\infty g'(t) e^{- i \omega t }dt = -\int_{-\infty}^\infty 2 a t g(t) e^{- i \omega t }dt = -2 a i \frac{d}{d\omega} \left(\int_{-\infty}^\infty g(t) e^{- i \omega t }dt\right) $$

hence $$\int_{-\infty}^\infty g(t) e^{- i \omega t }dt = C e^{-\omega^2/(4 a)}$$ and $C = 1/\sqrt{2 a}$ is obtained from the Parseval theorem

reuns
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  • I am sorry what is stopping me to make this as accepted answer is that it is not clear what you wanted to show. – Melina Feb 02 '16 at 23:30
  • @Melina : really ? only from the properties of the Fourier transform and the differential equation that $g(t)$ respects I proved what was its Fourier transform. this is what you asked (the Fourier transform of $g'(t)$ being easily deduced from that : multiply by $i\omega$..) – reuns Feb 02 '16 at 23:32
  • Yeah, because function was f(t)=4te^{-t^2} and its Fourier Transform is $\mathfrak{F}=i\sqrt{2}e^{-\frac{w^2}{4}}w$. So you can see differences? I really appreciate what you have done but ... – Melina Feb 02 '16 at 23:35
  • @Melina : understand what I wrote first (and don't come on that forum excepting people will resolve freely your exercices) – reuns Feb 02 '16 at 23:35
  • Sorry, in last step when you take the derivative d/dw of $\frac{1}{\sqrt{2}}e^{-\frac{w^2}{4}}$ in our case a=1 in the end result is $\frac{i}{\sqrt{2}}e^{-\frac{w^2}{4}}$ so sqrt{2} must be nominator not denominator. :( – Melina Feb 02 '16 at 23:53
  • you are boring me for a constant which is not even false, so say thank you and work – reuns Feb 02 '16 at 23:57