Editing Heaviside step function (section)

==Discrete form==

An alternative form of the unit step, defined instead as a function <math>H : \mathbb{Z} \rarr \mathbb{R}</math> (that is, taking in a discrete variable {{mvar|n}}), is:

<math display="block">H[n]=\begin{cases} 0, & n < 0, \\ 1, & n \ge 0, \end{cases} </math>

or using the half-maximum convention:<ref>{{cite book |last=Bracewell |first=Ronald Newbold |date=2000 |title=The Fourier transform and its applications |language=en |location=New York |publisher=McGraw-Hill |isbn=0-07-303938-1 |page=61 |edition=3rd}}</ref>

<math display="block">H[n]=\begin{cases} 0, & n < 0, \\ \tfrac12, & n = 0,\\ 1, & n > 0, \end{cases} </math>

where {{mvar|n}} is an [[integer]]. If {{mvar|n}} is an integer, then {{math|''n'' < 0}} must imply that {{math|''n'' ≤ &minus;1}}, while {{math|''n'' > 0}} must imply that the function attains unity at {{math|1=''n'' = 1}}. Therefore the "step function" exhibits ramp-like behavior over the domain of {{closed-closed|&minus;1, 1}}, and cannot authentically be a step function, using the half-maximum convention.  

Unlike the continuous case, the definition of {{math|''H''[0]}} is significant.

The discrete-time unit impulse is the first difference of the discrete-time step

<math display="block"> \delta[n] = H[n] - H[n-1].</math>

This function is the cumulative summation of the [[Kronecker delta]]:

<math display="block"> H[n] = \sum_{k=-\infty}^{n} \delta[k] </math>

where

<math display="block"> \delta[k] = \delta_{k,0} </math>

is the [[degenerate distribution|discrete unit impulse function]].