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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?

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  • $\begingroup$ 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

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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.

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

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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., $c=0$.

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