4
$\begingroup$

Last I had an exam and there was the following question: Find $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}(1+i)\mathrm{F}(\omega)e^{iwt}\mathrm{d}\omega = e^{-2t}H(t)$, where $\mathrm{F}(\omega)$ is the Fourier transform of $f$, and $H(t)$ is the Heaviside function.

The left side of the equation clearly has the form of an inverse transform, so I thought I'd just take the $(1+i)$ out from the integral, since it's a constant, so that $\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\mathrm{F}(\omega)e^{iwt}\mathrm{d}\omega = \frac{e^{-2t}H(t)}{1+i}$. Then $\mathrm{F}(\omega)(t)$ would be the transform of $\frac{e^{-2t}H(t)}{1+i}$ and therefore $f(t)$ would be $\frac{e^{-2t}H(t)}{1+i}$. But this function is not Real! (And we have not worked with Fourier transforms of functions of a complex variable)

Do you think there is a mistake in the question (it's not unusual at all, believe it or not!), or am I missing something or doing something wrong? Because if instead of $(1+i)$ there was something else, even a function of $\omega$, say $g(\omega)$, then I'd just transform $e^{-2t}H(t)$, divide it by $g(\omega)$, antitransform and get my $f(t)$, as long as that I don't get a complex function.

$\endgroup$
  • 1
    $\begingroup$ Why do you believe that $f$ should be restricted to real-valued functions? You just found one that is not. $\endgroup$ – Mark Viola Jul 26 '15 at 19:57
  • $\begingroup$ Because that's what the question asks for! $\endgroup$ – Chrischpo Jul 26 '15 at 20:35
  • 2
    $\begingroup$ Your question(s) regarded whether the exam question had a mistake or whether you missed something or did anything wrong. Your analysi is correct. There is no reason to restrict $f$ to real-valued functions. Your analysis corroborated that. Well done. $\endgroup$ – Mark Viola Jul 26 '15 at 20:42
  • 1
    $\begingroup$ There is a difference between complex valued function and function of a complex variable. In your case $f$ is a complex valued function of a real variable, that is , $f\colon\mathbb{R}\to\mathbb{C}$. $\endgroup$ – Julián Aguirre Jul 27 '15 at 9:52
0
$\begingroup$

First of all, I think the title is a misleading. The Fourier transform (and its inverse) is defined as an operator on complex functions of real variable (i.e. $\mathscr{F}:\mathbb{C^R\rightarrow C^R}$, $f\mapsto F$, where $f, F: \mathbb{R\rightarrow C}$ are integrable functions). For this reason I think (contrary to what the title suggests) that the inverse Fourier transform should yield a complex function, it just so happens that this result can sometimes lie in the subset of real functions. This point has been made clear in the comments.

Now, you've calculated yourself the function $f$ that satisfies the expression which is in can be recast as the inverse Fourier transform of $F$. Note $f$ has a transform:$$F(\omega)=\frac{1}{2+i\omega}\cdot\frac{1}{1+i}=\frac{1}{(2-\omega)+i(2+\omega)}$$As you can check.

You can also see in this other MSE post that for most "reasonable" functions the direct and inverse Fourier transforms are unique.Both $f$ and $F$ are piecewise continuous functions which do satisfy such uniqueness property. This means that the function $f$ you got (from the inverse transform expression) is the only possible answer.

So, I do think there is a mistake in the question as the result is a complex function rather than a real one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.