# Why is the signum function $=2u(t) - 1$?

Basically I couldnt find proof for this anywhere though it's a very simple and basic equation.

• what u(t)? xxxx – Olli Niemitalo Mar 23 '16 at 6:16
• heaviside unit step function. $$u(t) = \begin{cases} 1 \quad t\ge 0 \\ 0 \quad t<0 \end{cases}$$ it would have to be defined as 1/2 for t=0 if it was tied to the sgn() function. – robert bristow-johnson Mar 23 '16 at 6:27

The signum function is defined by

$$\text{sgn}(t)=\begin{cases}-1,&t<0\\0,&t=0\\1,&t>0\end{cases}$$

Using the half-maximum convention, the unit step function is defined by

$$u(t)=\begin{cases}0,&t<0\\\frac12,&t=0\\1,&t>0\end{cases}$$

From these two definitions it should be obvious that

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

must hold.

• why is the step function 1/2 in 0? I've never heard about the half-maximum convention – Behind The Sciences Mar 23 '16 at 7:22
• @BehindTheSciences: That's one possible convention for defining a value at $t=0$. Often it's not necessary to bother about $t=0$, but if it is, that convention usually makes most sense. It is mentioned here. – Matt L. Mar 23 '16 at 8:22