4
$\begingroup$

I applied the FT to a piecewise function defined as:

$$f(t) = \begin{cases} \sin(t), & \text{-π≤t≤π} \\ 0, & \text{otherwise} \end{cases} $$

and got $F[f(t)]= \frac{2i\sin(\pi\omega)}{\omega^2-1} $

I thought it'd be a interesting to attempt to applying the inverse FT to try to get back to the piecewise function. However I am an engineer and have limited knowledge of complex variables.

When applying the definition of the inverse Fourier transform I get:

$$f(t)= \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{2i\sin(\pi\omega)}{\omega^2-1} e^{i\omega t} dt. $$

Is anyone able to help me do this? or give me a hint of where to start? I have no idea how to choose the contour and which direction I should be closing it.

1 Answers 1

6

I assume there is a typo and you meant $$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty}\frac{2i \sin(\pi \omega)}{\omega^{2}-1} \, e^{i t \omega} \, d {\color{red}{\omega}}. \tag{1} $$

And I assume the definition of the Fourier transform you're using is $$\mathcal{F}[f(t)](\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} \, dt. $$

We can express $(1)$ as the difference of two Cauchy principal value integrals.

Specifically, $$ \begin{align} f(t) &= \frac{1}{2\pi} \int_{-\infty}^{\infty}\frac{2i \sin(\pi \omega)}{\omega^{2}-1} \, e^{i t \omega} \, d {\color{red}{\omega}} \\ &= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{e^{i \pi \omega} - e^{- i \pi \omega}}{w^{2}-1} \, e^{i t \omega } \, d \omega \\ &= \frac{1}{2 \pi} \left( \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega - \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t- \pi)}}{w^{2}-1} \, d \omega \right). \end{align}$$

If $a \ge 0$, the magnitude of the function $e^{iaz} $ is bounded in the upper half-plane.

And if $a \le 0$, the magnitude of $e^{iaz}$ is bounded in the lower half-plane.

So if ${\color{red}{t \ge - \pi}}$, we can evaluate the first integral by integrating the function $$h(z) = \frac{e^{i z(t+ \pi)}}{z^{2}-1}$$ around a closed semicircular contour in the upper half-plane that has small half-circle indentations around the simple poles at $z=-1$ and $z=1$.

Letting the radii of the indentations go to zero and the radius of the big arc go to infinity, we get $$ \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega - i \pi \, \text{Res} [h(z), -1] - \, i \pi \, \text{Res} [h(z), 1] =0,$$

which implies that $$ \begin{align} \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega &= i \pi \, \text{Res} [h(z), 1] + \, i \pi \, \text{Res} [h(z), -1] \\ &= i \pi \left( \frac{e^{i (t+ \pi)}}{2} - \frac{e^{-i (t+ \pi)}}{2} \right) \\ &= - \pi \sin (t+ \pi) \\ &=\pi \sin t. \end{align}$$

I used fact that if a function $f(z)$ has a simple pole at $z_{0}$, then $$\lim_{r \to 0} \int_{C_{r}} f(z) \, dz = i \alpha \, \text{Res}[f(z), z_{0}],$$ where $C_{r}$ is an arc of the circle $|z- z_{0}|=r$ of angle $\alpha$. (This is sometimes referred to as the fractional residue theorem.)

And if ${\color{red}{ t \le \pi}}$, we can evaluate the second integral by integrating the function $$ g(z) = \frac{e^{i z(t- \pi)}}{z^{2}-1}$$ around a similar contour in the lower half-plane.

This causes us to detour around the poles in the opposite direction (i.e., counterclockwise.)

We end up with $$\begin{align} \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t- \pi)}}{w^{2}-1} \, d \omega &= {\color{red}{-}}i \pi \, \text{Res} [g(z), 1] {\color{red}{-}}i \pi \, \text{Res} [g(z), -1] \\ &= -i \pi \left( \frac{e^{i (t- \pi)}}{2} - \frac{e^{-i (t- \pi)}}{2} \right) \\ &= \pi \sin (t- \pi) \\ &=-\pi \sin t. \end{align}$$

So if ${\color{red}{|t| \le \pi}}$, $$\frac{1}{2\pi} \int_{-\infty}^{\infty}\frac{2i \sin(\pi \omega)}{\omega^{2}-1} \, e^{i t \omega} \, d \omega= \frac{1}{2 \pi} \left(\pi \sin t + \pi \sin t \right) = \sin t.$$

If $t< - \pi$, both $h(z)$ and $g(z)$ have to be integrated in the lower-half plane.

And if $t > \pi$, both $h(z)$ and $g(z)$ have to be integrated in the upper-half plane.

In both cases, this results in $f(t)=0$.

  • 0
    Thanks for the thorough reply and correction of my typos. What determines whether $e^{iaz} $ is bounded in the upper half or lower half of the plane? How do you know which direction (clockwise or anti-clockwise) to close the contour in? EDIT: Additionally for the first PV integral you have found the residues at the two poles. Is it not required to show that the contour integral around the "big arc" is zero? – 2017-02-11
  • 0
    $$|e^{iaz}|=|e^{ia(x+iy)}| = |e^{iax}e^{-ay}| = |e^{iax}|| e^{-ay}|= e^{-ay}$$ So in the upper half plane, where $y >0$, $|e^{iaz}|$ never exceeds $1$ if $a \ge 0$. Similarly, in the lower half plane, where $y <0$, $|e^{iaz}|$ never exceeds $1$ one if $a \le 0$. The reason this is important is because then the [estimation lemma](https://en.wikipedia.org/wiki/Estimation_lemma) can be used to argue that integral along the big arc in both cases vanishes in the limit. – 2017-02-11
  • 1
    Alternatively, you could just use [Jordan's lemma](https://en.wikipedia.org/wiki/Jordan's_lemma). And it doesn't really matter in which direction you go around the contour. If we went in the opposite direction, we would get $$\operatorname{PV} \int_{\infty}^{-\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega + i \pi \, \text{Res} [f(z), -1] + \, i \pi \, \text{Res} [f(z), 1] =0,$$ which leads to the same result. – 2017-02-11
  • 0
    I think that I almost understand your explanation. As I understand it, a function of the form $e^{iaz} $, where $a > 0$, will tend to zero for $y > 0$. This is why we choose the upper-half plane to evaluate the integral so that the big arc essentially vanishes through integration. Is that correct? Secondly I don't quite understand what you're showing through your modulus calculations. Specifically at the last equals sign: $$|e^{iax}|| e^{-ay}|= e^{-ay}$$ How does the modulus of the two functions equal just one of the functions? Apologies if this is trivial. – 2017-02-11
  • 1
    $$|e^{iax}|= \left|\cos(ax) + i \sin(ax) \right| =\sqrt{\cos^{2}(ax) + \sin^{2}(ax)} = 1$$ And if $a \ge 0$, then along a big arc of radius $R$ in the upper half-plane , we have $$\left|\frac{e^{iaz}}{z^{2}-1}\right| \le \frac{1}{\left|{(Re^{it})^{2}-1}\right|} \le \frac{1}{\left||R^{2}e^{2it}|-|-1| \right|} = \frac{1}{R^{2}-1},$$ where $0 \le t \le \pi$. (The second inequality is just the reverse triangle inequality.) Combine this with the fact that the length of the big arc is $\pi R$, and then apply the estimation lemma to show that the integral along the big arc vanishes as $R \to \infty$. – 2017-02-11
  • 0
    That's excellent, thanks again. I feel that my complex variables knowledge is really coming along with this question! One more question - when you are calculating the residues why is it that the residues are multiplied by $i \pi $ as opposed to the standard $2\pi i$? – 2017-02-12
  • 1
    @JoeyWheeler The simple poles, which are on the real axis, are not inside the contour. We're avoiding them with small half-circle indentations, and then letting the radii of these half-circles go to zero. The general picture is something like [THIS](http://mathfaculty.fullerton.edu/mathews/c2003/integralsindentedcontour/IntegralsIndentedContourMod/Images/IntegralsIndentedContourMod_gr_517.gif), except, in our case, there are only two poles on the real axis and no poles inside the contour. – 2017-02-12
  • 1
    As I mentioned in my answer, there is a theorem about this that is sometimes called the fractional residue theorem. The reason it's $i \pi$ and not $2 \pi i$ is because they're half-circles and not entire circles. – 2017-02-12
  • 0
    I notice that in the first PV integral that when you changed the direction of the contour, you also changed the limits of integration. Is there a rule of thumb that determines when the direction of the contour affects the limits of integration? Do both the integrals have to be calculated using different direction contours (clockwise/anti-clockwise)? I've confused myself from the image you sent. If the first integral is like the image you sent and the second integral is in the lower-half plane with a different direction contour, wouldn't both integrals cancel each other out to zero? – 2017-02-12
  • 1
    @JoeyWheeler If $|t| \le \pi$, the first PV integral evaluates to $\pi \sin t$ and the second PV integral evaluates to $- \pi \sin t$. But $$f(t) = \frac{1}{2 \pi} \left( \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega \color{red}- \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t- \pi)}}{w^{2}-1} \, d \omega \right) = \frac{1}{2\pi} \left(\pi \sin t - (- \pi \sin t) \right) = \sin t.$$ – 2017-02-12
  • 1
    And if you change the direction of the first contour, you're now traversing the real axis from $\infty$ to $-\infty$ instead of from $- \infty$ to $\infty$. But $$\operatorname{PV} \int_{\infty}^{-\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega = -\operatorname{PV}\int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega , $$ and you end up with the exact same result. – 2017-02-12
  • 0
    I can't see why in the first PV integral, when you use Cauchy's Integral Theorem, you subtract two residues from it rather than adding the residues like you did in the second PV integral. I also fail to see how the two PV integrals cancel each other out in the cases where $t<- \pi$ or if $t >\pi$. – 2017-02-13
  • 0
    @JoeyWheeler For the first PV integral, the small half-circles around the simple poles on the real axis are clockwise-oriented. That's why it's negative. Look at the [picture](http://mathfaculty.fullerton.edu/mathews/c2003/integralsindentedcontour/IntegralsIndentedContourMod/Images/IntegralsIndentedContourMod_gr_517.gif) again. If you draw a similar picture for a contour in the lower-half plane, the small half-circles have to be counterclockwise-oriented so that the poles don't fall inside the contour. – 2017-02-13
  • 0
    If $t > \pi$, both $\frac{e^{i z(t+ \pi)}}{z^{2}-1}$ and $\frac{e^{i z(t- \pi)}}{z^{2}-1}$ have to be integrated in the upper-half plane. So if $t > \pi$, $$\int_{-\infty}^{\infty}\frac{e^{i \omega(t+ \pi)}}{\omega^{2}-1} \, d \omega = \int_{-\infty}^{\infty}\frac{e^{i \omega(t- \pi)}}{\omega^{2}-1} \, d \omega, $$ which leads to $$f(t) = \frac{1}{2 \pi} \left( \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t+ \pi)}}{w^{2}-1} \, d \omega \color{red}- \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i \omega(t- \pi)}}{w^{2}-1} \, d \omega \right) = 0. $$ – 2017-02-13
  • 0
    Similarly, if $t < - \pi$, both $\frac{e^{i z(t+ \pi)}}{z^{2}-1} $ and $\frac{e^{i z(t- \pi)}}{z^{2}-1} $ have to be integrated in the lower half-plane. – 2017-02-13
  • 0
    The last statement in my answer is indeed not quite stated correctly. – 2017-02-13