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The Heaviside step function, H, also called the unit step function, is a discontinuous function whose value is zero for negative argument and one for positive argument. It seldom matters what value is used for H(0), since $H$ is mostly used as a distribution. Some common choices can be seen below.

The function is used in the mathematics of control theory and signal processing to represent a signal that switches on at a specified time and stays switched on indefinitely. It was named in honor of the English polymath Oliver Heaviside.

It is the cumulative distribution function of a random variable which is almost surely 0. (See constant random variable.)

The Heaviside function is an antiderivative of the Dirac delta function: H′ = δ. This is sometimes written as

- $H(x)\; =\; int\_\{-infty\}^x\; \{\; delta(t)\}\; mathrm\{d\}t$

We can also define an alternative form of the unit step as a function of a discrete variable n:

- $H[n]=begin\{cases\}\; 0,\; \&\; n\; <\; 0\; 1,\; \&\; n\; ge\; 0\; end\{cases\}$

where n is an integer.

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

- $delta[n]\; =\; H[n]\; -\; H[n-1].$

This function is the cumulative summation of the Kronecker delta:

- $H[n]\; =\; sum\_\{k=-infty\}^\{n\}\; delta[k]\; ,$

where

- $delta[k]\; =\; delta\_\{k,0\}\; ,$

is the discrete unit impulse function.

For a smooth approximation to the step function, one can use the logistic function

- $H(x)\; approx\; frac\{1\}\{2\}\; +\; frac\{1\}\{2\}tanh(kx)\; =\; frac\{1\}\{1+mathrm\{e\}^\{-2kx\}\}$,

- $H(x)=lim\_\{k\; rightarrow\; infty\}frac\{1\}\{2\}(1+tanh\; kx)=lim\_\{k\; rightarrow\; infty\}frac\{1\}\{1+mathrm\{e\}^\{-2kx\}\}$

There are many other smooth, analytic approximations to the step function. They include:

- $H(x)\; =\; lim\_\{k\; rightarrow\; infty\}\; frac\{1\}\{2\}\; +\; frac\{1\}\{pi\}arctan(kx)$

- $H(x)\; =\; lim\_\{k\; rightarrow\; infty\}\; frac\{1\}\{2\}\; +\; frac\{1\}\{2\}operatorname\{erf\}(kx)$

Beware that while these approximations converge pointwise towards the step function, the implied distributions do not strictly converge towards the delta distribution. In particular, the measurable set

- $bigcup\_\{n=0\}^\{infty\}[2^\{-2n\};2^\{-2n+1\}]$

Often an integral representation of the Heaviside step function is useful:

- $H(x)=lim\_\{\; epsilon\; to\; 0^+\}\; -\{1over\; 2pi\; mathrm\{i\}\}int\_\{-infty\}^infty\; \{1\; over\; tau+mathrm\{i\}epsilon\}\; mathrm\{e\}^\{-mathrm\{i\}\; x\; tau\}\; mathrm\{d\}tau\; =lim\_\{\; epsilon\; to\; 0^+\}\; \{1over\; 2pi\; mathrm\{i\}\}int\_\{-infty\}^infty\; \{1\; over\; tau-mathrm\{i\}epsilon\}\; mathrm\{e\}^\{mathrm\{i\}\; x\; tau\}\; mathrm\{d\}tau$

The value of the function at 0 can be defined as H(0) = 0, H(0) = ½ or H(0) = 1. H(0) = ½ is the most consistent choice used, since it maximizes the symmetry of the function and becomes completely consistent with the sign function. This makes for a more general definition:

- $H(x)\; =\; frac\{1+sgn(x)\}\{2\}\; =$

1, & x > 0end{cases}

To remove the ambiguity of which value to use for H(0), a subscript specifying the value may be used:

- $H\_a(x)\; =$

a, & x = 0

1, & x > 0end{cases}

The derivative of the Heaviside step function is the Dirac delta function: $dH(x)/dx\; =\; delta(x)$

- $$

Here the $frac\{1\}\{s\}$ term must be interpreted as a distribution that takes a test function $phi$ to the Cauchy principal value of $intlimits^\{infty\}\_\{-infty\}\; phi(x)/x,\; dx$.

- Rectangular function
- Step response
- Dirac delta
- Sign function
- Negative and non-negative numbers
- Laplace transform

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Last updated on Friday July 18, 2008 at 23:07:20 PDT (GMT -0700)

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This article is licensed under the GNU Free Documentation License.

Last updated on Friday July 18, 2008 at 23:07:20 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

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