Appendix/Mathematics/StepFunction: Difference between revisions

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=Unit Step Function and Its Derivative=
=Unit Step Function and Its Derivative=


==Standard Presentation==
The unit &#8212; or, [https://en.wikipedia.org/wiki/Heaviside_step_function Heaviside] &#8212; step function, <math>H(x)</math>, is defined such that,
The unit &#8212; or, [https://en.wikipedia.org/wiki/Heaviside_step_function Heaviside] &#8212; step function, <math>H(x)</math>, is defined such that,


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<math>\int_{-\infty}^x \delta(\xi)d\xi    \, .</math>
<math>\int_{-\infty}^x \delta(\xi)d\xi    \, .</math>
   </td>
   </td>
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==Sign of a Function==
Notice that the sign of <math>x</math>, may be written in terms of the step function as,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>\sgn[x] = \frac{|x|}{x}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>2H(x) - 1  \, .</math>
  </td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">
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  <td align="right">
<math>\frac{d}{dx} \biggl[ \sgn(x)\biggr] = \frac{d}{dx}\biggl[ \frac{|x|}{x} \biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>2\delta(x) \, .</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">{{ Hunter2003 }}, &sect;2.2, immediately following Eq. (3)</td>
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Latest revision as of 21:12, 11 August 2023


Unit Step Function and Its Derivative

Standard Presentation

The unit — or, Heaviside — step function, H(x), is defined such that,

H(x)={0;x<01;x>0


[MF53], Part I, §2.1 (p. 123), Eq. (2.1.6)

Heaviside Function

In evaluating this function at x=0, we will adopt the half-maximum convention and set H(0)=12. As has been pointed out in, for example, a relevant Wikipedia discussion, the derivative of the unit step function is,

dH(x)dx

=

δ(x),

where, δ(x) is the Dirac Delta function. Hence, the unit step function is sometimes written as,

H(x)

=

xδ(ξ)dξ.

Sign of a Function

Notice that the sign of x, may be written in terms of the step function as,

sgn[x]=|x|x

=

2H(x)1.

Hence,

ddx[sgn(x)]=ddx[|x|x]

=

2δ(x).

📚 Hunter (2003), §2.2, immediately following Eq. (3)

See Also

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