Why is the signum function $=2u(t) - 1$?
The signum function is defined by

$$\text{sgn}(t)=\begin{cases}-1,&amp;t&lt;0\\0,&amp;t=0\\1,&amp;t&gt;0\end{cases}$$

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

$$u(t)=\begin{cases}0,&amp;t&lt;0\\\frac12,&amp;t=0\\1,&amp;t&gt;0\end{cases}$$

From these two definitions it should be obvious that

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

must hold.