# Heaviside step function

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The Heaviside step function

Fungsi Heaviside step, ngaran nu dipake keur ngahargaan ka Oliver Heaviside, nyaeta a fungsi diskontinyu numana nileyna nyaeta nol keur asupan negatip sarta hiji keur nu sejenna:

$H(x)=\left\{\begin{matrix} 0 : x < 0 \\ 1 : x \ge 0 \end{matrix}\right.$

The function is used in the mathematics of signal processing to represent a signal that switches on at a specified time and stays switched on indefinitely. Its derivative is formally the Dirac delta function.

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

The Heaviside function is the integral of the Dirac delta function. The value of H(0) is of very little importance, since the function is often used within an integral. Some writers give H(0) = 0, some H(0) = 1. H(0) = 0.5 is often used, since it maximizes the symmetry of the function. This makes the definition:

$H(x)=\left\{\begin{matrix} 0 : x < 0 \\ \frac{1}{2} : x = 0 \\ 1 : x > 0 \end{matrix}\right.$

This is sometimes notated by a subscript, as in H0.5(x), which means that H(0) = 0.5. This notation is also used to represent a completely different concept, however; an x offset:

$H_c(x) = H(x - c)\,\!$

where c is a positive offset in the x-dimension of the transition from 0 to 1. In other words, H3(x) = H(x − 3) would be zero until x = 3, and would transition to 1 for x > 3. The meaning of the subscript should be given in context.

The question of the Fourier transform of H is an interesting example for the theory of distributions. It is often stated that it is 1/x, up to a normalizing constant. But near x=0 that cannot be justified: the definition must be given in terms of principal value limit, and the transform isn't to be treated simply as a function. The corresponding convolution operator is the Hilbert transform.

Often an integral representation of the step function is useful,

$H(x)=-{1\over 2i\pi}\int_{-\infty}^\infty {d\tau \over \tau+i\epsilon} e^{-i\tau x},$

in the limit $\epsilon\to 0$.