What is the inverse Fourier transform of $F(w) = \frac{4}{4+(j2\pi f)^2} $?
I have two suggested solutions:

  1. Assume $j2\pi f = \omega$ and use the standard transform $e^{-\alpha |\tau |} = \frac{2\alpha}{\alpha ^2 + \omega ^2}$ to obtain $\frac{1}{4}e^{-2|\tau|}$

  2. Calculate $f(\tau) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\alpha}{\alpha^2-\omega^2}e^{i\omega\tau}d\omega$

For (2), how can I solve the integral in order to get the inverse Fourier transform $f(\tau)$?


In general Fourier transform (or inverse) take products into convolutions. Let $F(x)$ and $G(x)$ have transforms $f(t)$ and $g(t)$. Then the transform of $F(x)G(x)$ is $h(t)=\int f(s)g(t-s)ds$.

For your problem $f(t)=\frac{1}{16}\int e^{-a|s|-b|t-s|}ds$ where $a=2$ and $b=8$.

(Note: I may have constant wrong, but you see the point.)


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It looks like there is some problem with the answer in "1." First note that it is indeed true that the Fourier Transform $\fG(\omega)$ of the function $\fg(\tau)\eqd e^{-\alpha\abs{\tau}}$ is $\fG(\omega)\eqd\frac{2\alpha}{\alpha^2+\omega^2}$. Assuming the definition of the Fourier Transform used is $\fG(\omega)\eqd \int_{-\infty}^{\infty}\fg(\tau)e^{-j\omega\tau}\dtau$, we can check the relation as follows: \begin{align} \fG(\omega) &\eqd \int_{-\infty}^{\infty}\fg(\tau)e^{-j\omega\tau}\dtau && \text{by definition Fourier Transform} \\&\eqd \int_{-\infty}^{\infty} e^{-\alpha\abs{\tau}} e^{-j\omega\tau}\dtau && \text{by definition of $\fg(\tau)$} \\&= \int_{-\infty}^{0} e^{-\alpha(-\tau)} e^{-j\omega\tau}\dtau + \int_{0}^{\infty} e^{-\alpha(\tau)} e^{-j\omega\tau}\dtau \\&= \int_{-\infty}^{0} e^{\tau(\alpha-j\omega)}\dtau + \int_{0}^{\infty} e^{\tau(-\alpha-j\omega)}\dtau \\&= \brlr{\frac{e^{\tau(\alpha-j\omega)}}{\alpha-j\omega}}_{-\infty}^{0} + \brlr{\frac{e^{\tau(-\alpha-j\omega)}}{-\alpha-j\omega}}_{0}^{\infty} && \text{by Fundamental Theorem of Calculus} \\&= \brs{\frac{1}{\alpha-j\omega} - 0} + \brs{0- \frac{1}{-\alpha-j\omega}} \\&= \brs{\frac{1}{\alpha-j\omega}}\brs{\frac{\alpha+j\omega}{\alpha+j\omega}} + \brs{\frac{1}{\alpha+j\omega}}\brs{\frac{\alpha-j\omega}{\alpha-j\omega}} && \text{(to rationalize the denominator)} \\&= \frac{\alpha+j\omega}{\alpha^2+\omega^2} + \frac{\alpha-j\omega}{\alpha^2+\omega^2} \\&= \boxed{\frac{2\alpha}{\alpha^2+\omega^2}} \end{align}

And so the Inverse Fourier Transform (IFT) of $\frac{2\alpha}{\alpha^2+\omega^2}$ is $e^{-\alpha\abs{\tau}}$.

However, your question does not ask for the IFT of $\frac{2\alpha}{\alpha^2+\omega^2}=\frac{2\alpha}{\alpha^2+(2\pi f)^2}$, but rather for the IFT of $$ \frac{2\alpha}{\alpha^2+(j2\pi f)^2} = \frac{2\alpha}{\alpha^2+(j\omega)^2} =\frac{2\alpha}{\alpha^2-\omega^2} $$

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    $\begingroup$ can you show me the steps to get the IFT? $\endgroup$ Feb 7 at 17:09
  • $\begingroup$ I have tried to find the IFT but so far haven't been successful. I will work more on it though. Maybe someone else already knows the answer and could post it here? $\endgroup$ Feb 8 at 0:09
  • $\begingroup$ WolframAlpha has something interesting to say. In essence ... $$\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{4}{4-\omega^2}e^{i\tau\omega}d\omega = -\frac{i}{2}e^{-2i\tau}(-1+e^{4i\tau})sgn(\tau) $$ $\endgroup$ Feb 8 at 1:00
  • $\begingroup$ \begin{align} \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{4}{4-\omega^2}e^{i\tau\omega}d\omega &= -\frac{i}{2}e^{-2i\tau}(-1+e^{4i\tau})sgn(\tau) \\&= -\frac{i}{2}(-e^{-2i\tau}+e^{2i\tau})sgn(\tau) \\&= -\frac{i}{2}[2i\sin(2\tau)]sgn(\tau) && \text{by Euler's Identity} \\&= \sin(2\tau)sgn(\tau) \end{align} $\endgroup$ Feb 8 at 1:19
  • $\begingroup$ Conjecture: IFT of $\frac{2\alpha}{\alpha^2-\omega^2}$ is $\sin(\alpha\tau)\mathrm{sgn}(\tau)$ $\ldots$ but I'm not sure at this point $\ldots$ $\endgroup$ Feb 8 at 1:25

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I'll be honest with you---it looks like there is no Inverse Fourier Transform for $2\alpha^2/(\alpha^2-\omega^2)$. I'd be happy to be proven wrong...but at least consider the following...

(1) Let $s\eqd j\omega$. Then the Fourier Transform $\int_{-\infty}^{\infty} \fg(\tau)e^{-j\omega\tau}\dtau$ of $\fg(\tau)$ becomes the (Two-Sided) Laplace Transform $\int_{-\infty}^{\infty} \fg(\tau)e^{-s\tau}\dS$ of $\fg(\tau)$.

(2) Note that what we want is a function $\fg(\tau)$ with Laplace Transform $\frac{2\alpha}{\alpha^2+s^2}$. Why? Because then when we substitute back in $s=j\omega$ we would get $$\brlr{\frac{2\alpha}{\alpha^2+s^2}}_{s=j\omega} = \frac{2\alpha}{\alpha^2-\omega^2}$$ and the world would be a happier place.

(3) If only there was such a function $\fg(\tau)$ ... hmmmmm ... But wait! There is a function like that! In fact, if you happened to guess that the function $\fg(\tau)$ is $$\fg(\tau)\eqd 2\sin(\omega\tau)\step(\tau)$$ where $\step(\tau)=1$ for $\tau\geq1$ and $0$ otherwise (the step function), then you would be right! And you can check that here (where the One-Sided Laplace is used and so $\step(\tau)$ is implied).

(4) So the Inverse Laplace Transform of $\fG(s)\eqd\frac{2\alpha}{\alpha^2+s^2}$ is $\fg(\tau)\eqd 2\sin(\omega\tau)\step(\tau)$. So we just set $s=j\omega$ and we're all done here, right? Well not quite. That Laplace Transform $\fG(s)$ has a Region of Convergence of $\Re(s)>0$, where $\Re(s)$ is the real component of the complex $s$. The problem here is that the Fourier Transform exists along the imaginary axis of $s$---that is, at $s=j\omega$, which means $\Re(s)=0$ ... which is not in the Region of Convergence of $\fG(s)$ ... and so rather, it diverges or "goes to infinity" or "blows up" at $s=j\omega$.

And so, while the Laplace Transform of $\fg(\tau)$ does exist (for $\Re(s)>0$), it does not exist (it diverges) for $\Re(s)\leq0$. And so by extension, it looks like the Fourier Transform $\fG(\omega)=\frac{2\alpha}{\alpha^2-\omega^2}$ does not exist (because it is not in the Region of Convergence of $\fG(s)$).


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