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I am wondering what the Fourier Transform of $\operatorname{sgn}(t) \cdot \operatorname{sgn}(t)$ will be, where $\operatorname{sgn}(t)$ indicates the signum function. It would seem obvious that this is equal to the FT of $1$, but I would like to see if this would be possible to find using the multiplicative property of the FT.

When I try this myself, I get stuck at trying to find the following integral:

$$\text{P.V.} \int_{-\infty}^{\infty} \frac{1}{2\pi q(q - \Omega)}\text{d}q$$

While this function is obviously equal to $0$ if $\Omega \neq 0$, the integral will diverge if $\Omega = 0$. Does this mean that we approach $0$, or is there a possibility that we approach the delta function? The former would make most sense to me but I have no idea how we could show this. Perhaps this is a better question for the math SE?

Thanks in advance for any help!

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  • $\begingroup$ What is the Fourier Transform of $x(t)=1 \quad \forall t$? Settle that first. $\endgroup$ Oct 16, 2023 at 20:36
  • $\begingroup$ Find the Fourier Transform of the unit step function. Write signum in terms of the unit step function. Use linearity to find its Fourier Transform. Convolve it with itself! $\endgroup$ Oct 16, 2023 at 21:22
  • $\begingroup$ @robertbristow-johnson that should be equal to $2\pi \delta(\Omega)$, which means we expect the same result in the frequency domain if we only change a single point in time to a non-finite value and immediately take the transform, right? $\endgroup$ Oct 17, 2023 at 4:30
  • $\begingroup$ @AhsanYousaf thanks for the reply, if I try this I still end up with the same integral besides some scaling and additional terms which can be found more easily. $\endgroup$ Oct 17, 2023 at 4:32
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    $\begingroup$ @TimWescott it is supposed to represent the Cauchy principal value of the integral considering there are singularities on the interval. $\endgroup$ Oct 19, 2023 at 8:47

1 Answer 1

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The functions $f(t)=[\textrm{sgn}(t)]^2$ and $g(t)=1$ differ only at a single point $t=0$ ($f(0)=0$ whereas $g(0)=1$). The difference $f(t)-g(t)$ is sometimes referred to as null function. The integral over a null function vanishes, and, consequently, $f(t)$ and $g(t)$ have the same Fourier transform (see Lerch's theorem).

Clearly, the Fourier transform of $g(t)$ (and hence of $f(t)$) equals $2\pi\delta(\omega)$. Of course, it must be possible to show this using the multiplication/convolution theorem. We have

$$\mathscr{F}\{\textrm{sgn}(t)\}=\frac{2}{j\omega}\tag{1}$$

(whereby it's understood that all integrals involving this correspondence are to be interpreted as principal values).

Consequently, we should expect

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{4}{\Omega (\Omega-\omega)}d\Omega=2\pi\delta(\omega)\tag{2}$$

One way to show this is via the definition of the Hilbert transform. We know that

$$\mathscr{H}\big\{\mathscr{H}\{f(x)\}\big\}=-f(x)\tag{3}$$

We also know that

$$\mathscr{H}\big\{\delta(x)\big\}=\frac{1}{\pi x}\tag{4}$$

Combining $(3)$ and $(4)$ results in

$$\delta(x) = -\mathscr{H}\left\{\frac{1}{\pi x}\right\}=\frac{1}{\pi^2}\int_{-\infty}^{\infty}\frac{1}{y(y-x)}dy\tag{5}$$

which is equivalent to $(2)$.

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  • $\begingroup$ //"Since $\textrm{sgn}(t)\cdot\textrm{sgn}(t)=1$,"// uhm, not quite. //" its Fourier transform must equal $2\pi\delta(\omega)$."// yah, but you need to dot your t's and cross your i's, Matt. $\endgroup$ Oct 16, 2023 at 21:30
  • $\begingroup$ Not sure, but isn't $\text{sgn}(0) = 0$ and hence $\text{sgn}(t) \cdot \text{sgn}(t) = 1, t \ne 0$. So there is a discontinuity at $t = 0$. Does that change anything ? $\endgroup$
    – Hilmar
    Oct 16, 2023 at 21:31
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    $\begingroup$ @Hilmar it's infinitely thin and not a dirac delta. So what does that change in an integral? $\endgroup$ Oct 16, 2023 at 21:32
  • $\begingroup$ @robertbristow-johnson: I have no idea, that's why I'm asking :-) :-) :-) $\endgroup$
    – Hilmar
    Oct 16, 2023 at 21:34
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    $\begingroup$ @Hilmar RBJ, in the end we're free as hell to define $\operatorname{sgn}$ as we want; including unambigously setting $\operatorname{sgn}(0) = 1$. It doesn't really change the integral, anyways. $\endgroup$ Oct 16, 2023 at 23:12

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