Heaviside step function: Difference between revisions
imported>Paul Wormer No edit summary |
mNo edit summary |
||
(2 intermediate revisions by one other user not shown) | |||
Line 16: | Line 16: | ||
1 &\quad\hbox{if}\quad x > a\\ | 1 &\quad\hbox{if}\quad x > a\\ | ||
0 &\quad\hbox{if}\quad x < a\\ | 0 &\quad\hbox{if}\quad x < a\\ | ||
\end{cases} | |||
\qquad\quad\hbox{and}\quad | |||
H(-x) = | |||
\begin{cases} | |||
0 &\quad\hbox{if}\quad x > 0\\ | |||
1 &\quad\hbox{if}\quad x < 0\\ | |||
\end{cases} | \end{cases} | ||
</math> | </math> | ||
Line 44: | Line 50: | ||
</math> | </math> | ||
Here the "turnover rule" for d/d''x'' is used, which may be proved by integration by parts and which holds when ''f''(''x'') vanishes at the integration limits. | Here the "turnover rule" for d/d''x'' is used, which may be proved by integration by parts and which holds when ''f''(''x'') vanishes at the integration limits. | ||
== | ==H(x) as limit of arctan== | ||
- | [[Image:Arctan Heaviside.png|right|thumb|350px|[arctan(ε/x)]/π goes to ''H''(−''x'') for decreasing positive ε.]] | ||
In the figure it is shown that | |||
:<math> | |||
\frac{1}{\pi} \lim_{\epsilon \rightarrow 0^+} \arctan(\epsilon/x) = H(-x). | |||
</math> | |||
Note here that the function arctan returns angles on the full interval 0 to 2π. In particular, if a point ''x'' + ''iy'' in the complex plane has ''x'' < 0 and ''y'' approaches zero from above, then the function arctan returns a value approaching π. Most computer languages use a two parameter function for this form of the inverse tangent. | |||
==Fourier transform== | ==Fourier transform== | ||
:<math> | :<math> | ||
\mathcal{F}(H) \equiv \sqrt{\frac{1}{2\pi}} \int_{-\infty}^{\infty} e^{-iux}H(x) dx = | \mathcal{F}(H) \equiv \sqrt{\frac{1}{2\pi}} \int_{-\infty}^{\infty} e^{-iux}H(x) dx = | ||
\sqrt{\frac{1}{2\pi}} \left( \pi \delta(u) - i | \sqrt{\frac{1}{2\pi}} \left( \pi \delta(u) - i \mathrm{P}\Big(\frac{1}{u}\Big)\right), | ||
</math> | </math> | ||
where δ(''u'') is the [[Dirac delta function]] and | where δ(''u'') is the [[Dirac delta function]] and P stands for the [[Cauchy principal value]]. | ||
===Proof=== | ===Proof=== | ||
Write | Write | ||
Line 68: | Line 80: | ||
Now use the following relation, | Now use the following relation, | ||
:<math> | :<math> | ||
\lim_{\epsilon \rightarrow 0^+} \frac{-i}{u-i\epsilon} = -i \Big | \lim_{\epsilon \rightarrow 0^+} \frac{-i}{u-i\epsilon} = -i \mathrm{P}\Big(\frac{1}{u}\Big) + \pi \delta(u) | ||
</math> | </math> | ||
and the result is proved. | and the result is proved. | ||
'''( | In order to prove the last relation we write a complex number in polar form | ||
:<math> | |||
x+iy = (x^2+y^2)^{1/2}\; e^{i\alpha}\quad \hbox{with}\quad \alpha = \arctan(y/x). | |||
</math> | |||
Take the natural logarithm and the limit for ''y'' → 0 | |||
:<math> | |||
\lim_{y \rightarrow 0^+} \ln(x+iy) = \lim_{y \rightarrow 0^+} \frac{1}{2} \ln(x^2+y^2) + i\lim_{y \rightarrow 0^+} \arctan(y/x) = \ln|x| + i \pi H(-x). | |||
</math>. | |||
Differentiation of the last expression gives gives | |||
:<math> | |||
\lim_{y \rightarrow 0^+} \frac{1}{x+iy} = \frac{1}{x} -i\pi \delta(x). | |||
</math> | |||
In fact, the functions in this expression are distributions (generalized functions) and are to be used in an integrand multiplied by a well-behaved function ''f''(''x''). Since 1/''x'' is singular for ''x'' = 0, we must take the Cauchy principal value of the integral, i.e., exclude the singularity from the integral; we make the replacement | |||
:<math> | |||
\int_{-\infty}^{\infty} \frac{f(x)}{x} \mathrm{d}x \Rightarrow \mathrm{P}\int_{-\infty}^{\infty} \frac{f(x)}{x} \mathrm{d}x, | |||
</math> | |||
and the result follows.[[Category:Suggestion Bot Tag]] |
Latest revision as of 12:01, 26 August 2024
![](http://s9.addthis.com/button1-share.gif)
In mathematics, physics, and engineering the Heaviside step function is the following function,
The function is undetermined for x = 0, sometimes one defines .
From the definition it follows immediately that
The function is named after the English mathematician Oliver Heaviside.
Derivative
Note that a block ("boxcar") function BΔ of width Δ and height 1/Δ can be given in terms of step functions (for positive Δ), namely
Knowing this, the derivative of H follows easily
where δ(x) is the Dirac delta function, which may be defined as the block function in the limit of zero width, see the article on the Dirac delta function.
The step function is a generalized function (a distribution). When H(x) is multiplied under the integral by the derivative of an arbitrary differentiable function f(x) that vanishes for plus/minus infinity, the result of the integral is minus the function value for x = 0,
Here the "turnover rule" for d/dx is used, which may be proved by integration by parts and which holds when f(x) vanishes at the integration limits.
H(x) as limit of arctan
In the figure it is shown that
Note here that the function arctan returns angles on the full interval 0 to 2π. In particular, if a point x + iy in the complex plane has x < 0 and y approaches zero from above, then the function arctan returns a value approaching π. Most computer languages use a two parameter function for this form of the inverse tangent.
Fourier transform
where δ(u) is the Dirac delta function and P stands for the Cauchy principal value.
Proof
Write
where we used
Now use the following relation,
and the result is proved.
In order to prove the last relation we write a complex number in polar form
Take the natural logarithm and the limit for y → 0
- .
Differentiation of the last expression gives gives
In fact, the functions in this expression are distributions (generalized functions) and are to be used in an integrand multiplied by a well-behaved function f(x). Since 1/x is singular for x = 0, we must take the Cauchy principal value of the integral, i.e., exclude the singularity from the integral; we make the replacement
and the result follows.