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I'm trying to find the IFT of the following function: $\hat{f}(\omega) = \frac{i\omega}{1+\omega^2}$. I know about the Laplace Transform, so right away I thought about looking for an inverse of $\hat{f}$ involving $\cosh$ and then do a change of variables. I managed to arrive at $f(x) = \sqrt{2\pi} \cosh(x) \theta(x)$, where $\theta$ is the Heaviside Function. Wolfram Alpha agrees that the Fourier Transform of that $f$ is $\frac{i\omega}{1+\omega^2}$. However, when I use Wolfram Alpha to calculate $\hat{f}^{-1}$, I have in return $g(x) = -\sqrt{\frac{\pi}{2}}e^{-x}[e^{2x}\theta(-x) - \theta(x)]$. Both functions, $f$ and $g$, are obviously not the same. Did I do something wrong? Is it normal for a function to have more than one inverse? In that case, if I was asked on an exam to find the IFT of a function, would both $f$ and $g$ be correct?

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    $\begingroup$ Are you sure $f \neq g$? Keep in mind that $\cosh (x) = \frac{e^x + e^{-x}}{2}$. Try doing some algebraic manipulations with this in mind. $\endgroup$
    – jgd1729
    Commented Aug 21, 2022 at 21:13
  • $\begingroup$ @jgd1729 I know, they look a bit similar, but $f$ is zero for negative values and $g$ is not. $\endgroup$
    – user715547
    Commented Aug 21, 2022 at 21:15
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    $\begingroup$ maybe there was a data-entry error somewhere? considering $x>0$ and $x<0$, your g seems to be more or less $e^{-|x|}$; the Fourier transform of that is not your $\hat f$. $\endgroup$ Commented Aug 21, 2022 at 21:34
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    $\begingroup$ Fair enough. When you say you did a change of variable, do you mean you did a change of variable from the laplace transform to the fourier? If yes, did you use the classical Laplace transform or the bilateral Laplace transform? That could account for the difference $\endgroup$
    – jgd1729
    Commented Aug 21, 2022 at 21:44
  • $\begingroup$ @jgd1729 I knew that $\mathcal{L}\{\cosh(at)\} = \frac{s}{s^2 - a^2}$. So, by multiplying $\cosh(x)$ by $\theta(x)$, I only needed to integrate on one side of the line, which allowed me to go from the Fourier case to the Laplace case. At the end, I changed $s$ for $i\omega$. Maybe I did something I shouldn't do, but still wouldn't explain why Wolfram Alpha agrees with the answer. $\endgroup$
    – user715547
    Commented Aug 21, 2022 at 21:56

2 Answers 2

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Traditionally, the Fourier transform operator is considered over the space of finite energy signals, i.e. signals that have a well defined square integral,

$$\int_{-\infty}^{\infty} f(\tau)^2\,d\tau < \infty,$$

quotienting out those signals that differ on a set of measure zero. This ensures the convergence of the Fourier integral to produce the corresponding transform $\hat{f}.$ The (continuous) Fourier transform is one-to-one and onto on this space of signals. As far as I recall, one can relax these conditions a bit, but not to the extent of your $f.$ Your signal $f$ grows unbounded on the imaginary axis in exponential order. The Fourier integral,

$$\hat{f}(\omega) = \int_{\mathbb{R}} f(x)\,e^{i\omega x}\,dx$$

simply does not converge for real values of $\omega.$ Verify that,

$$\begin{aligned} \int_{-\infty}^{\infty} f(x)\,e^{i\omega x}\,dx &=\int_{-\infty}^{\infty} \sqrt{2\pi} \cosh(x)\, \theta(x)\, e^{i\omega x}\,dx\\ &=\sqrt{2\pi} \int_{0}^{\infty} \cosh(x)\, e^{i\omega x}\,dx\\ &=\frac{\sqrt{\pi}}{\sqrt{2}} \int_{0}^{\infty} \left[ e^{x} + e^{-x}\right] e^{i\omega x}\,dx \end{aligned}$$ The second term will converge to finite value, but the first term grows unbounded inside the integral; the integral over $\mathbb{R}$ is not going to be finite for any $\omega.$

Compare this with the Laplace transform of your signal,

$$\hat{f}(s) = \int_{\mathbb{R}} f(x)\,e^{-s x}\,dx$$

which does converge but for $\mathfrak{Re}\{s\} > 1.$ That is, the domain of convergence for the Laplace transform does not contain the imaginary axis. It follows that the Fourier transform of $f(x)$ is not well-defined and does not exist!

What Wolfram Alpha computes is the square-integrable signal that does have that corresponding transform. The function $g(x)$ decays in both the negative $x$ direction and in the positive $x$ direction exponentially. First rewrite $g$ as,

$$g(x) = \left\{ \begin{array}{ll} -\sqrt{\frac{\pi}{2}}\,e^{x} & \mathrm{if } x < 0,\\ \sqrt{\frac{\pi}{2}}\,e^{-x} & \mathrm{if } x > 0, \end{array}\right. $$

where I'm leaving out the point at $x=0$ because it simply won't matter for this discussion. Observe that $g(-x) = -g(x).$ Therefore we can reduce the Fourier integral of $g$ to,

$$\begin{aligned} \int_{-\infty}^{\infty} g(x)\,e^{i\omega x}\,dx &= \int_{-\infty}^{0} g(x)\,e^{i\omega x}\,dx + \int_{0}^{\infty} g(x)\,e^{i\omega x}\,dx\\ &= \int_{0}^{\infty} -g(x)\,e^{-i\omega x}\,dx + \int_{0}^{\infty} g(x)\,e^{i\omega x}\,dx\\ &= \int_{0}^{\infty} g(x)\left[e^{i\omega x} - e^{-i\omega x}\right]\,dx\\ &= 2 i \,\int_{0}^{\infty} g(x)\,\sin(\omega\,x)\,dx\\ \end{aligned}$$

Substituting $g,$

$$\hat{g}(\omega) = 2 i \sqrt{\frac{\pi}{2}} \,\int_{0}^{\infty} e^{-x}\,\sin(\omega\,x)\,dx $$

If you squint hard enough, you'll recognize this integral as just the Laplace transform of $\sin(\omega x)$ evaluated at a point which will result in the correct transform up to a constant factor (the constant factor is because I didn't use the standard multiplier in the Fourier Transform). This Fourier integral converges for all $\omega \in \mathbb{R}.$

In summary, if you want to use transforms for your signal $f,$ you will need to use the Laplace transform to ensure you can invert it. Of course, again, you'll have to consider the class of signals you concern yourself with. Presumably, the use of the heaviside step function suggests you are concerned with one-sided signals where the one-sided Laplace transform does have that required property that you can go forward and invert and get a unique result.

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The inverse Fourier transform of $i\omega/(1+\omega^2)$ is $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i\omega x}\frac{i\omega}{1+\omega^2}d\omega. $$ This is well-defined because $i\omega/(1+\omega^2)$ is square integrable on $\mathbb{R}$, which guarantees that the Fourier transform is square integrable as well. This may be further decomposed as $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i\omega x}\frac{1}{2}\left[\frac{i}{\omega-i}+\frac{i}{\omega+i}\right]d\omega $$ If $x > 0$, then $e^{i\omega x}$ decays in the upper half-plane. If $x < 0$, then $e^{i\omega x}$ decays in the lower half-plane. So, for $x > 0$, the above may be written as the limit of a contour integral that is closed in the upper half-plane, which allows us to evaluate the above as a residue at $\omega=i$, and that is given by $$ \frac{\pi}{\sqrt{2\pi}}e^{-x}. $$ Similarly, if $x < 0$, then the integral may be closed in the lower half-plane, leading to the negative of a positively oriented contour integral evaluation at $\omega=-i$: $$ - \frac{\pi}{\sqrt{2\pi}}e^{x}. $$ So it appears that the original integral evaluates to $$ \mbox{sgn}(x)\sqrt{\frac{\pi}{2}}e^{-|x|},\;\;\; -\infty < x < \infty,\; x\ne 0, $$ where $\mbox{sgn}(x)$ is $1$ if $x > 0$, is $-1$ if $x < 0$, and is $0$ if $x=0$. The original integral does not converge absolutely for $x=0$, but it does converge to $0$ as a Cauchy principle value integral. Regardless of the details at $x=0$, the inverse transform is in $L^2(\mathbb{R})$, and everything is just fine in that regard.

Verification: The Fourier transform of our function is $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\omega x}\mbox{sgn}(x)\sqrt{\frac{\pi}{2}}e^{-|x|}dx \\ = -\frac{1}{2}\int_{-\infty}^{0}e^{-i\omega x}e^{x}dx+\frac{1}{2}\int_{0}^{\infty}e^{-i\omega x}e^{-x}dx \\ = -\frac{1}{2}\frac{1}{1-i\omega}+\frac{1}{2}\frac{1}{1+i\omega} \\ = \frac{1}{2}\left[\frac{1}{1+i\omega}-\frac{1}{1-i\omega}\right]=\frac{-i\omega}{1+\omega^2} $$ Oops! I'm off by a sign. I'll let you sort it out.

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