Why is signum function used to calculate Fourier transform of unit step function

Question

I read in a standard textbook that the Fourier transform of unit impulse function is calculated with the help of approximations and signum function as the integration of unit impulse does not converge. What's so special about signum function that it is used to calculate Fourier transform? I tried to find out an approximation as:

$$ \lim_{ a \rightarrow 0 } \int_{-\infty}^{+\infty} e^{-at} u(t) e^{-j\omega t} dt $$

But I am getting wrong result. Why is this so?

Answer 1

If somebody you trust told you that the Fourier transform of the sign function is given by

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

you could of course use this information to compute the Fourier transform of the unit step $u(t)$. Using

$$u(t)=\frac12(1+\text{sgn}(t))\tag{2}$$

(as pointed out by Peter K. in a comment), you get

$$\mathcal{F}\{u(t)\}=\frac12\left(\mathcal{F}\{1\}+\mathcal{F}\{\text{sgn}(t)\}\right)=\pi\delta(\omega)+\frac{1}{j\omega}\tag{3}$$

However, you don't need the sign function to compute the Fourier transform of the step function. As suggested in your question, using the function $e^{-at}u(t)$ and taking the limit $a\rightarrow 0^+$ will also result in the expression given in $(3)$.

You can see this as follows. The Fourier transform of $e^{-at}u(t)$, $a>0$, is given by

$$\int_0^{\infty}e^{-at}e^{-j\omega t}dt=\frac{1}{a+j\omega}\tag{4}$$

Taking the limit $a\rightarrow 0^+$ appears to give $1/j\omega$, but this is only valid for $\omega\neq 0$. Splitting the result $(4)$ in its real and imaginary part gives

$$\frac{1}{a+j\omega}=\frac{a}{a^2+\omega^2}+\frac{\omega}{j(a^2+\omega^2)}\tag{5}$$

The real part of $(5)$ is known as ($\pi$ times) a nascent delta function. It has the same form as the Poisson kernel, which in the limit becomes a Dirac delta impulse. So for $a\rightarrow 0^+$ the limit of $(5)$, and hence of $(4)$, is actually given by

$$\lim_{a\rightarrow 0^+}\frac{1}{a+j\omega}=\pi\delta(\omega)+\frac{1}{j\omega}\tag{6}$$

which equals the expression in $(3)$.