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Rectangular function

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]

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Rectangular function with a = 1

{\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.

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.

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.}

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Plot of normalized {\displaystyle \operatorname {sinc} (x)} function (i.e. {\displaystyle \operatorname {sinc} (\pi x)}) with its spectral frequency components.

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} _{\pi }(f),} using ordinary frequency f, where {\displaystyle \operatorname {sinc} _{\pi }} is the normalized form[10] of the sinc function and {\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 }}}\cdot \operatorname {sinc} \left(\omega /2\right),} using angular frequency {\displaystyle \omega }, where {\displaystyle \operatorname {sinc} } is the unnormalized form of the sinc function.

For {\displaystyle \operatorname {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\ \operatorname {sinc} _{\pi }{(af)}.}

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

{\displaystyle \operatorname {tri(t/T)} =\operatorname {rect(2t/T)} *\operatorname {rect(2t/T)} .\,}

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.

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 \operatorname {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}}}

The rectangle function can be used to represent the Dirac delta function {\displaystyle \delta (x)}.[11] Specifically,{\displaystyle \delta (x)=\lim _{a\to 0}{\frac {1}{a}}\operatorname {rect} \left({\frac {x}{a}}\right).}For a function {\displaystyle g(x)}, its average over the width {\displaystyle a} around 0 in the function domain is calculated as,

{\displaystyle g_{avg}(0)={\frac {1}{a}}\int \limits _{-\infty }^{\infty }dx\ g(x)\operatorname {rect} \left({\frac {x}{a}}\right).} To obtain {\displaystyle g(0)}, the following limit is applied,

{\displaystyle g(0)=\lim _{a\to 0}{\frac {1}{a}}\int \limits _{-\infty }^{\infty }dx\ g(x)\operatorname {rect} \left({\frac {x}{a}}\right)} and this can be written in terms of the Dirac delta function as, {\displaystyle g(0)=\int \limits _{-\infty }^{\infty }dx\ g(x)\delta (x).}The Fourier transform of the Dirac delta function {\displaystyle \delta (t)} is

{\displaystyle \delta (f)=\int _{-\infty }^{\infty }\delta (t)\cdot e^{-i2\pi ft}\,dt=\lim _{a\to 0}{\frac {1}{a}}\int _{-\infty }^{\infty }\operatorname {rect} \left({\frac {t}{a}}\right)\cdot e^{-i2\pi ft}\,dt=\lim _{a\to 0}\operatorname {sinc} {(af)}.} where the sinc function here is the normalized sinc function. Because the first zero of the sinc function is at {\displaystyle f=1/a} and {\displaystyle a} goes to infinity, the Fourier transform of {\displaystyle \delta (t)} is

{\displaystyle \delta (f)=1,} means that the frequency spectrum of the Dirac delta function is infinitely broad. As a pulse is shorten in time, it is larger in spectrum.

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