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

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  • $\begingroup$ what u(t)? xxxx $\endgroup$ – Olli Niemitalo Mar 23 '16 at 6:16
  • $\begingroup$ 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. $\endgroup$ – robert bristow-johnson Mar 23 '16 at 6:27
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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.

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  • $\begingroup$ why is the step function 1/2 in 0? I've never heard about the half-maximum convention $\endgroup$ – Behind The Sciences Mar 23 '16 at 7:22
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    $\begingroup$ @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. $\endgroup$ – Matt L. Mar 23 '16 at 8:22

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