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I'm reading a text which is an introductory text on Fourier transforms. The author has two expressions:

$$ F(\omega_{o}) = \frac{1}{\sigma \sqrt {2 \pi} } e^{\Large- \frac{{\omega_{o}}^2}{2\sigma^2}} $$ and $$G(\omega_{o}) = 2\pi \cos(\omega_{o}t).$$

Then he states that the inverse transformations of the functions are

$$f(t_{o}) = \frac{1}{2 \pi} e^{\Large-\frac{1}{2}\sigma^2 t_{o}^2}$$

and

$$g(t_{o}) = 2 \pi \left( \frac{\delta(t_{o} - t)}{2} + \frac{\delta(t_{o} + t)}{2} \right) .$$

Would someone be able to derive how $f(t_{o})$ and $g(t_{o})$ are obtained. I'd appreciate it because I'd probably learn a lot from seeing the steps. Thank you very much.

Tunk-Fey
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mark leeds
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1 Answers1

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For the first one, we have $$ F(\omega)=\frac{1}{\sigma\sqrt{2\pi}} e^{\Large- \frac{\omega^2}{2\sigma^2}}. $$ It seems that you are using the textbook for physics or engineering, so the inverse Fourier transform of $F(\omega)$ using the 'convention' notation in those fields is $$ \begin{align} \mathcal{F}^{-1}[F(\omega)]&=\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)\ e^{\large it\omega}\ d\omega\\ f(t)&=\frac{1}{2\pi}\cdot\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^\infty e^{\Large- \frac{\omega^2}{2\sigma^2}}\ e^{\large it\omega}\ d\omega\\ &=\frac{1}{2\pi}\cdot\frac{2}{\sigma\sqrt{2\pi}}\int_{0}^\infty e^{\Large- \left(\frac{\omega^2}{2\sigma^2}- it\omega\right)}\ d\omega. \end{align} $$ In general $$ \begin{align} \int_{x=0}^\infty e^{-(ax^2-bx)}\,dx&=\int_{x=0}^\infty \exp\left(-a\left(\left(x-\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)\right)\,dx\\ &=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x-\frac{b}{2a}\right)^2\right)\,dx. \end{align} $$ Let $u=x-\frac{b}{2a}\;\rightarrow\;du=dx$, then $$ \begin{align} \int_{x=0}^\infty e^{-(ax^2-bx)}\,dx&=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x-\frac{b}{2a}\right)^2\right)\,dx\\ &=\exp\left(\frac{b^2}{4a}\right)\int_{u=0}^\infty e^{-au^2}\,du.\\ \end{align} $$ The last form integral is Gaussian integral that equals to $\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}$. Hence $$ \int_{x=0}^\infty e^{-(ax^2-bx)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right). $$ It yields the same result for $\displaystyle\int_{x=0}^\infty e^{-(ax^2\color{red}+bx)}\,dx$. Thus $$ \begin{align} \frac{1}{2\pi}\cdot\frac{2}{\sigma\sqrt{2\pi}}\int_{0}^\infty e^{\Large- \left(\frac{\omega^2}{2\sigma^2}- it\omega\right)}\ d\omega&=\frac{1}{2\pi}\cdot\frac{2}{\sigma\sqrt{2\pi}}\cdot\frac{1}{2}\sqrt{2\pi\sigma^2}\exp\left(\frac{( it)^2\cdot2\sigma^2}{4}\right)\\ f(t)&=\color{blue}{\frac{1}{2\pi}\ e^{\Large- \frac{\sigma^2t^2}{2}}}. \end{align} $$

Tunk-Fey
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  • Note that, notation of Fourier transforms depends on the textbook and the field of study. Here are the list of notation of Fourier transforms. – Tunk-Fey May 20 '14 at 07:36
  • tung fey. thanks for the great answer and just a heads up that the last line should not have a $2 \pi$ in the exponential inside the integral. I think someone edited my latex in my question because it looked much nicer than I wrote it but I didn't want to change yours without permission. Thanks again and if you can be bothered doing the second one also at your convenience, it's appreciated. – mark leeds May 21 '14 at 06:23
  • @markleeds I've edited my answer. The $2\pi$ term appears since you didn't give me the notation of Fourier transform that you are using. It's OK, I fix my answer. For the second one, I think it's a bit difficult since it involves Delta Dirac notation (Honestly, I almost forget about that thing because it's my undergraduate stuff which has passed a few years ago :P). Besides, I am busy right now, college stuff that has deadline tomorrow. I'll try to help you when I'm done with my assignment but I cannot assure that. I hope someone here will help & give you a good answer. Good luck! – Tunk-Fey May 21 '14 at 06:49
  • Thanks Tunk_Fey. I think I'm okay because I found a lot of material on the internet about that problem so I'll read that. Thanks again. – mark leeds May 21 '14 at 13:25