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

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?

• In your title you write "unit step", whereas in your question it is "unit impulse". I assume that your question is about the computation of the Fourier transform of the unit step function. And whatever your textbook says, you don't need the sign function to compute the Fourier transform of the step function. Of course, if you already have the Fourier transform of the sign function, you can use it for computing the Fourier transform of the step function. Oct 13, 2015 at 16:22
• What confuses me is that the signum function and the unit step are the same --- modulo a scaling factor of 2 and a DC offset of -1.
– Peter K.
Oct 13, 2015 at 16:29
• The way you're trying to compute the Fourier transform of the unit step works (by taking the limit). Why you're getting a wrong result can only be answered if you show us the details of your computation and your result. I assume you get $1/j\omega$, which comes from computing the limit in the wrong way. Oct 13, 2015 at 16:34

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 $(3)$.

• could you slightly elaborate on how $\displaystyle\lim_{a \rightarrow 0} \frac{a}{a^2+w^2}= \pi\delta(\omega)$ Oct 14, 2015 at 5:33
• @Talasila: If you click on the link and scroll down you see the 'Poisson kernel', which has exactly the same form as the real part in (5). It is a nascent delta function, which means that in the limit it becomes a Dirac delta. Oct 14, 2015 at 7:16
• @MattL. Is $\pi\delta(\omega)$ singularity value? Feb 11, 2020 at 9:24
• @jomegaA: Not sure what you mean; it's not a value, it's just the Dirac delta at $\omega=0$. Feb 11, 2020 at 11:17