0
$\begingroup$

Cheers, I am trying to find the fourier transform of the signum function, which is

$$ \operatorname{sgn}(t) \triangleq \begin{cases} 1 \qquad & t>0 \\ 0 \qquad & t=0 \\ -1 \qquad & t<0 \\ \end{cases} $$

I rewrite this as:

$$\operatorname{sgn}(t) = 2u(t) -1$$

and find it's first derivative which is:

$$\operatorname{sgn}'(t) = 2 \delta(t)$$

and using the integration rule I know that:

$$\begin{align} \mathscr{F}\{\operatorname{sgn}(t)\} &= \mathscr{F}\left\{\int_{-\infty}^t2\delta(ρ)dρ \right\} \\ &= \frac{1}{j\omega}2 + \pi X(0)\delta(\omega) \\ \end{align}$$

I need to find to prove that $X(0)=0$, as the fourier tranform of the signum function is $\frac{2}{j\omega}$, but I think this transformation always yields 1. Is what I am thinking correct, and if yes how would I go about this? I know that there are alternatives, but I am just checking to see alternative ways to prove things I already know. Thanks =)

Edit: I tried the following thing: I split the Fourier of $\operatorname{sgn}(\cdot)$ to the Fourier of the unit step and the -1 constant. Then, I get that

$$\mathscr{F}\{2u(t)\} = \frac{2}{j\omega} + \pi \delta(\omega) $$

and

$$\mathscr{F}\{1\}= \pi \delta(\omega)$$

so by subtracting, I get the correct thing. Is that the way to do it?

$\endgroup$
2
  • $\begingroup$ I think you mean $-1, t < 0)$ ? $\endgroup$
    – Hilmar
    Jan 5, 2022 at 1:31
  • $\begingroup$ i don't do angular frequency that often, but I think $$\mathscr{F}\{1\}= 2 \pi \delta(\omega)$$ $\endgroup$ Jan 5, 2022 at 4:32

2 Answers 2

1
$\begingroup$

By taking the derivative you loose all information about the DC value of the original signal. Any signal

$$x(t) = 2\cdot u(t)- a$$

has the same derivative, regardless of what $a$ is. So you do have to calculate the DC value by hand, which is simply the mean of the signal.

$$X(0) = \lim_{\tau \to \infty} \int_{-\tau}^{+\tau} \operatorname{sgn}(t) \, \mathrm{d}t = 0 $$

$\endgroup$
0
$\begingroup$

As you've shown via differentiation, the Fourier transform of all piecewise constant signals that have a jump discontinuity of height $2$ at $t=0$ is

$$X(j\omega)=\frac{2}{j\omega}+c\delta(\omega)\tag{1}$$

where $c$ is some real-valued constant.

The easiest way to see that $c=0$ for $x(t)=\textrm{sign}(t)$ is to note that $x(t)$ is an odd function, and, consequently, its Fourier transform must be purely imaginary, i.e., $c=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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