Fourier transform of time-scaled signum function

Question

I have the signals $x_1(t)=\mathrm{sgn}(9t)$ and $x_2(t)=\mathrm{sgn}(t/5)$, where $\mathrm{sgn}$ denotes the signum function: $-1$ or $+1$ depending on the sign of the argument. I computed their Fourier transforms as $$X_1(f)=\frac{1}{j\pi(f/9)}$$ and $$X_2(f)=\frac{5}{j\pi(5f)}$$ Are these transforms correct?

Answer 1

The Fourier transform of the signum function $s$ is:

$$ \hat{s}(f)=S(f)=\frac{1}{j\pi f}\,.$$

If you apply the scaling property of the Fourier transform:

$$ x(at) \leftrightarrow \frac{1}{|a|}X\left(\frac{f}{a}\right)$$ you get:

$$ X_1(f) = \frac{1}{9}S\left(\frac{f}{9}\right) = \frac{1}{9}\left(\frac{1}{j\pi f/9}\right) = \frac{1}{j\pi f}\,.$$

A factor was missing in your $X_1$, and you can further simplify $X_2$. The signum function is invariant (in time and amplitude) by the dilation of its argument $t$ by a positive factor $a$, therefore those avatars $x_1$ and $x_2$ are essentially identical, and have the same Fourier transform.

Note: the signum function is not integrable, hence its Fourier transform is defined only "in a certain sense".