Rectangular function

Relation to the boxcar function

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

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

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 $${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{rect}(t)\cdot e^{-i2\pi ft}dt=\frac{\sin (\pi f)}{\pi f}=\mathrm{sinc}(\pi f)={\mathrm{sinc}}_{\pi }(f),}$$ using ordinary frequency f, where ${\displaystyle {\mathrm{sinc}}_{\pi }}$ is the normalized form of the sinc function and $${\displaystyle \frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{rect}(t)\cdot e^{-i\omega t}dt=\frac{1}{\sqrt{2\pi }}\cdot \frac{\sin \left(\omega /2\right)}{\omega /2}=\frac{1}{\sqrt{2\pi }}\cdot \mathrm{sinc}\left(\omega /2\right),}$$ using angular frequency ${\displaystyle \omega }$, where ${\displaystyle \mathrm{sinc}}$ is the unnormalized form of the sinc function.

For ${\displaystyle \mathrm{rect}(x/a)}$, its Fourier transform is$${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{rect}\left(\frac{t}{a}\right)\cdot e^{-i2\pi ft}dt=a\frac{\sin (\pi af)}{\pi af}=a{\mathrm{sinc}}_{\pi }(af).}$$

Self convolution of the Rectangular function

The self convolution of the dis-continuous rectangular function results in the triangular function, a piecewise defined spline that is continuous, but not continuously differentiable. Successive convolutions of the rectangular function result in piecewise defined pulses with lower maximums which are wider and smoother, with "smoother" meaning higher-order derivatives are continuous.

A convolution of the discontinuous rectangular function with itself results in the triangular function, which is a continuous function:

$${\displaystyle \begin{array}{c}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(2\mathrm{t}/\mathrm{T})\ast \mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(2\mathrm{t}/\mathrm{T})=\mathrm{t}\mathrm{r}\mathrm{i}(\mathrm{t}/\mathrm{T})=\{\begin{array}{ll}1+t,& -T<t<0\\ 1-t,& 0<t<T\\ 0& \text{otherwise}\end{array}\end{array}}$$

Self convolution of the rectangular function applied twice yields a continuous and differentiably continuous parabolic spline:

$${\displaystyle \begin{array}{c}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(2\mathrm{t}/\mathrm{T})\ast \mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(2\mathrm{t}/\mathrm{T})\ast \mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(2\mathrm{t}/\mathrm{T})=\mathrm{t}\mathrm{r}\mathrm{i}(\mathrm{t}/\mathrm{T})\ast \mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}(2\mathrm{t}/\mathrm{T})=\{\begin{array}{ll}\frac{9}{8}+\frac{3}{2}t+\frac{1}{2}{t}^2,& -\frac{3}{2}T<t<-\frac{1}{2}T\\ \frac{3}{4}-t^2,& -\frac{1}{2}T<t<\frac{1}{2}T\\ \frac{9}{8}-\frac{3}{2}t+\frac{1}{2}{t}^2,& \frac{1}{2}T<t<\frac{3}{2}T\\ 0& \text{otherwise}\end{array}\end{array}}$$

A self convolution of the rectangular function applied three times yields a continuous, and a second order differentiably continuous cubic spline:

$${\displaystyle \begin{array}{c}\mathrm{t}\mathrm{r}\mathrm{i}(\mathrm{t}/\mathrm{T})\ast \mathrm{t}\mathrm{r}\mathrm{i}(\mathrm{t}/\mathrm{T})=\{\begin{array}{ll}\frac{4}{3}+2t+t^2+\frac{1}{6}{t}^3,& -2T<t<-T\\ \frac{2}{3}-t^2-\frac{1}{2}{t}^3,& -T<t<0\\ \frac{2}{3}-t^2+\frac{1}{2}{t}^3,& 0<t<T\\ \frac{4}{3}-2t+t^2-\frac{1}{6}{t}^3,& T<t<2T\\ 0& \text{otherwise}\end{array}\end{array}}$$

A self convolution of the rectangular function applied four times yields a continuous, and a third order differentiably continuous 4th order spline:

$${\displaystyle \begin{array}{c}{4}^{th}\text{order spline}=\{\begin{array}{ll}\frac{625}{384}+\frac{125}{48}t+\frac{25}{16}{t}^2+\frac{5}{12}{t}^3+\frac{1}{24}{t}^4,& -\frac{5}{2}T<t<-\frac{3}{2}T\\ \frac{55}{96}-\frac{5}{24}t-\frac{5}{4}{t}^2-\frac{5}{6}{t}^3-\frac{1}{6}{t}^4,& -\frac{3}{2}T<t<-\frac{1}{2}T\\ \frac{115}{192}-\frac{5}{8}{t}^2+\frac{1}{4}{t}^4,& -\frac{1}{2}T<t<\frac{1}{2}T\\ \frac{55}{96}+\frac{5}{24}t-\frac{5}{4}{t}^2+\frac{5}{6}{t}^3-\frac{1}{6}{t}^4,& \frac{1}{2}T<t<\frac{3}{2}T\\ \frac{625}{384}-\frac{125}{48}t+\frac{25}{16}{t}^2-\frac{5}{12}{t}^3+\frac{1}{24}{t}^4,& \frac{3}{2}T<t<\frac{5}{2}T\\ 0& \text{otherwise}\end{array}\end{array}}$$

Since the Fourier transform of the rectangular function is the Sinc function, the convolution theorem implies that the Fourier transform of pulses resulting from successive convolutions of the rectangular function with itself is simply the Sinc function to the order of the number of times that the convolution function was applied + 1 (i.e., the Fourier transform of the triangular function is Sinc², the Fourier transform of the parabolic spline resulting from two successive convolutions of the rectangular function with itself is Sinc³, etc.)

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

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

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:

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

Dirac delta function

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

$${\displaystyle g_{avg}(0)=\frac{1}{a}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}dxg(x)\mathrm{rect}\left(\frac{x}{a}\right).}$$ To obtain ${\displaystyle g(0)}$, the following limit is applied,

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

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