Rectangular function

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 ${\displaystyle X+Y/2.}$

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

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

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

using ordinary frequency f, where ${\displaystyle \mathrm {sinc} }$ is the normalized form of the sinc function and

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

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

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

Note that as long as the definition of the pulse function is only motivated by its behavior in the time-domain experience, there is no reason to believe that the oscillatory interpretation (i.e. the Fourier transform function) should be intuitive, or directly understood by humans. However, some aspects of the theoretical result may be understood intuitively, as finiteness in time domain corresponds to an infinite frequency response. (Vice versa, a finite Fourier transform will correspond to infinite time domain response.)

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

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

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,

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

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

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

For a function ${\displaystyle g(x)}$, its average over the width ${\displaystyle a}$ around 0 in the function domain is calculated as,

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

To obtain ${\displaystyle g(0)}$, the following limit is applied,

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

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

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

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.

• Fourier transform
• Square wave
• Step function
• Top-hat filter
• Boxcar function