The rectangular function (also known as the rectangle function, rect function, Pi function, Heaviside Pi function,^[1] gate function, unit pulse, or the normalized boxcar function) is defined as^[2]

$${\displaystyle \operatorname {rect} \left({\frac {t}{a))\right)=\Pi \left({\frac {t}{a))\right)=\left\((\begin{array}{rl}0,&{\text{if ))|t|>{\frac {a}{2))\\{\frac {1}{2)),&{\text{if ))|t|={\frac {a}{2))\\1,&{\text{if ))|t|<{\frac {a}{2)).\end{array))\right.}$$

Alternative definitions of the function define ${\textstyle \operatorname {rect} \left(\pm {\frac {1}{2))\right)}$ to be 0,^[3] 1,^[4][5] or undefined.

Its periodic version is called a rectangular wave.

History

The rect function has been introduced by Woodward^[6] in ^[7] as an ideal cutout operator, together with the sinc function^[8][9] as an ideal interpolation operator, and their counter operations which are sampling (comb operator) and replicating (rep operator), respectively.

Relation to the boxcar function

The rectangular function is a special case of the more general boxcar function:

$${\displaystyle \operatorname {rect} \left({\frac {t-X}{Y))\right)=H(t-(X-Y/2))-H(t-(X+Y/2))=H(t-X+Y/2)-H(t-X-Y/2)}$$

where ${\displaystyle H(x)}$ is the Heaviside step function; the function is centered at ${\displaystyle X}$ and has duration ${\displaystyle Y}$, from ${\displaystyle X-Y/2}$ to ${\displaystyle X+Y/2.}$

Fourier transform of the rectangular function

The unitary Fourier transforms of the rectangular function are^[2]

$${\displaystyle \int _{-\infty }^{\infty }\operatorname {rect} (t)\cdot e^{-i2\pi ft}\,dt={\frac {\sin(\pi f)}{\pi f))=\operatorname {sinc} (f),}$$

${\displaystyle \mathrm {sinc} }$

$${\displaystyle {\frac {1}{\sqrt {2\pi ))}\int _{-\infty }^{\infty }\operatorname {rect} (t)\cdot e^{-i\omega t}\,dt={\frac {1}{\sqrt {2\pi ))}\cdot {\frac {\sin \left(\omega /2\right)}{\omega /2))={\frac {1}{\sqrt {2\pi ))}\operatorname {sinc} \left(\omega /2\right),}$$

${\displaystyle \omega }$

${\displaystyle \mathrm {sinc} }$

For ${\displaystyle \mathrm {rect} (x/a)}$, its Fourier transform is

$${\displaystyle \int _{-\infty }^{\infty }\operatorname {rect} \left({\frac {t}{a))\right)\cdot e^{-i2\pi ft}\,dt=a{\frac {\sin(\pi af)}{\pi af))=a\ \mathrm {sinc} {(af)}.}$$

Relation to the triangular function

We can define the triangular function as the convolution of two rectangular functions:

$${\displaystyle \mathrm {tri} =\mathrm {rect} *\mathrm {rect} .\,}$$

Use in probability

Viewing the rectangular function as a probability density function, it is a special case of the continuous uniform distribution with ${\displaystyle a=-1/2,b=1/2.}$ The characteristic function is

$${\displaystyle \varphi (k)={\frac {\sin(k/2)}{k/2)),}$$

and its moment-generating function is

$${\displaystyle M(k)={\frac {\sinh(k/2)}{k/2)),}$$

where ${\displaystyle \sinh(t)}$ is the hyperbolic sine function.

Rational approximation

The pulse function may also be expressed as a limit of a rational function:

$${\displaystyle \Pi (t)=\lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1)).}$$

Demonstration of validity

First, we consider the case where ${\textstyle |t|<{\frac {1}{2)).}$ Notice that the term ${\textstyle (2t)^{2n))$ is always positive for integer ${\displaystyle n.}$ However, ${\displaystyle 2t<1}$ and hence ${\textstyle (2t)^{2n))$ approaches zero for large ${\displaystyle n.}$

It follows that:

$${\displaystyle \lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1))={\frac {1}{0+1))=1,|t|<{\tfrac {1}{2)).}$$

Second, we consider the case where ${\textstyle |t|>{\frac {1}{2)).}$ Notice that the term ${\textstyle (2t)^{2n))$ is always positive for integer ${\displaystyle n.}$ However, ${\displaystyle 2t>1}$ and hence ${\textstyle (2t)^{2n))$ grows very large for large ${\displaystyle n.}$

It follows that:

$${\displaystyle \lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1))={\frac {1}{+\infty +1))=0,|t|>{\tfrac {1}{2)).}$$

Third, we consider the case where ${\textstyle |t|={\frac {1}{2)).}$ We may simply substitute in our equation:

$${\displaystyle \lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1))=\lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{1^{2n}+1))={\frac {1}{1+1))={\tfrac {1}{2)).}$$

We see that it satisfies the definition of the pulse function. Therefore,

$${\displaystyle \mathrm {rect} (t)=\Pi (t)=\lim _{n\rightarrow \infty ,n\in \mathbb {(} Z)}{\frac {1}{(2t)^{2n}+1))={\begin{cases}0&{\mbox{if ))|t|>{\frac {1}{2))\\{\frac {1}{2))&{\mbox{if ))|t|={\frac {1}{2))\\1&{\mbox{if ))|t|<{\frac {1}{2)).\\\end{cases))}$$

Dirac delta function

The rectangle function can be used to represent the Dirac delta function ${\displaystyle \delta (x)}$.^[10] Specifically,

$${\displaystyle \delta (x)=\lim _{a\to \infty }{\frac {1}{a))\mathrm {rect} ({\frac {x}{a))).}$$

${\displaystyle g(x)}$

${\displaystyle a}$

$${\displaystyle g_{avg}(0)={\frac {1}{a))\int \limits _{-\infty }^{\infty }dx\ g(x)\mathrm {rect} ({\frac {x}{a))).}$$

${\displaystyle g(0)}$

$${\displaystyle g(0)=\lim _{a\to 0}{\frac {1}{a))\int \limits _{-\infty }^{\infty }dx\ g(x)\mathrm {rect} ({\frac {x}{a)))}$$

$${\displaystyle g(0)=\int \limits _{-\infty }^{\infty }dx\ g(x)\delta (x).}$$

${\displaystyle \delta (t)}$

$${\displaystyle \delta (f)=\int _{-\infty }^{\infty }\delta (t)\cdot e^{-i2\pi ft}\,dt=\lim _{a\to 0}{\frac {1}{a))\int _{-\infty }^{\infty }\mathrm {rect} ({\frac {t}{a)))\cdot e^{-i2\pi ft}\,dt=\lim _{a\to 0}\mathrm {sinc} {(af)}.}$$

${\displaystyle f=1/a}$

${\displaystyle a}$

${\displaystyle \delta (t)}$

$${\displaystyle \delta (f)=1,}$$