# What is the Fourier Transform of $\operatorname{sgn}(t) \cdot \operatorname{sgn}(t)$?

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!

• What is the Fourier Transform of $x(t)=1 \quad \forall t$? Settle that first. Oct 16, 2023 at 20:36
• 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! Oct 16, 2023 at 21:22
• @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? Oct 17, 2023 at 4:30
• @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. Oct 17, 2023 at 4:32
• @TimWescott it is supposed to represent the Cauchy principal value of the integral considering there are singularities on the interval. Oct 19, 2023 at 8:47

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 $$(2)$$.

• //"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. Oct 16, 2023 at 21:30
• 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 ? Oct 16, 2023 at 21:31
• @Hilmar it's infinitely thin and not a dirac delta. So what does that change in an integral? Oct 16, 2023 at 21:32
• @robertbristow-johnson: I have no idea, that's why I'm asking :-) :-) :-) Oct 16, 2023 at 21:34
• @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. Oct 16, 2023 at 23:12