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 ${\displaystyle X+Y/2.}$ 

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:

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

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.

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