# Fourier transforms of the unit impulse and boxcar

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 9 295 - 366 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 9.3a

Because a unit impulse ${\displaystyle \delta ({\textit {t}})}$ is zero except at ${\displaystyle t=0}$, where it equals ${\displaystyle +1}$, we can apply the Fourier transform equation (9.3c) and find that ${\displaystyle \delta ({\textit {t}})\leftrightarrow }$ ${\displaystyle +1}$. Show that

{\displaystyle {\begin{aligned}\delta (t-t_{0})\leftrightarrow e^{-{\hbox{j}}\varpi t_{0}}.\end{aligned}}}

### Background

Equations (9.1a,b,c) apply to a harmonic function which repeats after every interval ${\displaystyle T}$. If we let ${\displaystyle T}$ increase, the repetitions occur at longer intervals and in the limit when ${\displaystyle T=\infty ,}$ the function becomes aperiodic, that is, it no longer repeats. To see the effect of this on equation (9.1c), we replace ${\displaystyle \alpha _{n}}$ in equation (9.1c) with the right-hand side of equation (9.1h). The result is

 {\displaystyle {\begin{aligned}g(t)=\sum \limits _{n=-\infty }^{+\infty }\left({\frac {1}{T}}\int _{-T/2}^{+T/2}g(t)e^{-{\hbox{j}}\omega _{n}t}\mathrm {d} t\right)e^{{\hbox{j}}\omega _{n}t}.\end{aligned}}} (9.3a)

We now let ${\displaystyle T\to +\infty }$ so that ${\displaystyle \omega _{0}\to 0}$, which causes the two adjacent frequencies, ${\displaystyle \omega _{n}}$ and ${\displaystyle \omega _{n+1}}$, differing by ${\displaystyle \omega _{0}}$, to approach each other. In the limit ${\displaystyle \omega _{n}}$ becomes a continuous variable ${\displaystyle \omega }$, ${\displaystyle 1/T}$ becomes ${\displaystyle {\hbox{d}}\omega /2\pi }$, the summation becomes an integral, and equation (9.3a) becomes

 {\displaystyle {\begin{aligned}g(t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{+\infty }e^{+{\hbox{j}}\omega t}\left[\int \limits _{-\infty }^{+\infty }g(t)e^{-{\hbox{j}}\omega t}{\hbox{d}}t\right]\mathrm {d} \omega .\end{aligned}}} (9.3b)

It is convenient to represent the inner integral by a symbol and write equation (9.3b) as two equations:

 {\displaystyle {\begin{aligned}G(f)=\int \limits _{-\infty }^{+\infty }g(t)e^{-{\hbox{j}}2\pi ft}\,\mathrm {d} t\quad \mathrm {or} \quad G(\omega )=\int \limits _{-\infty }^{+\infty }g(t)e^{-{\hbox{j}}\omega t}\mathrm {d} t,\end{aligned}}} (9.3c)

 {\displaystyle {\begin{aligned}g(t)=\int \limits _{-\infty }^{+\infty }G(f)e^{{\hbox{j}}2\pi ft}\mathrm {d} f\quad \mathrm {or} \quad g(t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{+\infty }G(\omega )e^{{\hbox{j}}\omega t}\mathrm {d} \omega .\end{aligned}}} (9.3d)

The function ${\displaystyle G(\omega )}$ [or ${\displaystyle G(f)}$] is the Fourier transform of ${\displaystyle g(t)}$ while ${\displaystyle g(t)}$ is the inverse Fourier transform of ${\displaystyle G(\omega )}$ [or ${\displaystyle G(f)}$]. The relation between ${\displaystyle g(t)}$ and ${\displaystyle G(\omega )}$ can be indicated by a double arrow:

 {\displaystyle {\begin{aligned}g(t)\leftrightarrow G(\omega )\quad \mathrm {or} \quad g(t)\leftrightarrow G(f).\end{aligned}}} (9.3e)

The unit impulse ${\displaystyle \delta (t)}$, also called the Dirac delta, is by definition zero everywhere except at ${\displaystyle t=0}$ where it equals ${\displaystyle +1}$; similarly, ${\displaystyle \delta (t-t_{0})}$ is zero except when ${\displaystyle (t-t_{0})=0}$, where it equals ${\displaystyle +1}$. In digital notation (see problem 9.4) we write ${\displaystyle \delta _{t}}$, ${\displaystyle \delta _{(t-t_{0})}}$, but the meaning is the same.

The convolution ${\displaystyle g(t)*h(t)}$ is discussed in problem 9.2. The convolution theorem [see Sheriff and Geldart, 1995, equation (15.145)] states that

 {\displaystyle {\begin{aligned}g(t)*h(t)\leftrightarrow G(\omega )H(\omega ),\end{aligned}}} (9.3f)

where ${\displaystyle G(\omega )}$ and ${\displaystyle H(\omega )}$ are the Fourier transforms of ${\displaystyle g(t)}$ and ${\displaystyle h(t)}$. For digital functions we use ${\displaystyle z}$-transforms (see Sheriff and Geldart, 1995, section 15.5.3), the equivalent of equation (9.3f) being

 {\displaystyle {\begin{aligned}g_{t}*h_{t}\leftrightarrow G(z)H(z).\end{aligned}}} (9.3g)

The boxcar, written ${\displaystyle {\hbox{box}}_{2t_{0}}(t)}$, is a function whose value is everywhere zero except in the interval from ${\displaystyle -t_{0}}$ to ${\displaystyle +t_{0}}$, where its value is ${\displaystyle +1}$. A boxcar in the frequency domain is written ${\displaystyle {\hbox{box}}_{2f_{0}}(f)}$ or ${\displaystyle {\hbox{box}}_{2\omega _{0}}(\omega )}$.

### Solution

The transform of ${\displaystyle \delta (t-t_{0})}$ is given by equation (9.3c):

 {\displaystyle {\begin{aligned}\delta (t-t_{0})\leftrightarrow \int \limits _{-\infty }^{+\infty }\delta (t-t_{0})e^{-{\hbox{j}}\omega t}\mathrm {d} t=e^{-{\hbox{j}}\omega t_{0}}.\end{aligned}}} (9.3h)

## Problem 9.3b

Show that

 {\displaystyle {\begin{aligned}\delta (t)*g(t)=g(t),\quad \delta _{t}*g_{t}=g_{t},\end{aligned}}} (9.3i)
 {\displaystyle {\begin{aligned}\delta (t-\tau )*g(t)=g(t-\tau ),\quad \delta _{t-\tau }*g_{t}=g_{t-\tau }.\end{aligned}}} (9.3j)

### Solution

Convolution involves replacing each element of one function with the other function and since ${\displaystyle \delta (t)}$ involves only one nonzero element at zero time, replacing an element at ${\displaystyle t=0}$ with ${\displaystyle g(t)}$ gives ${\displaystyle g(t)}$, thus proving equation (9.3i). Likewise, replacing an element at time ${\displaystyle \tau }$ with ${\displaystyle g(t)}$ gives ${\displaystyle g(t-\tau )}$, thus proving equation (9.3j).

An alternative proof for both cases above can be obtained by using transforms. Because the transform of both ${\displaystyle \delta (t)}$ and ${\displaystyle \delta _{t}}$ is ${\displaystyle +1}$, equations (9.3f, g) show that the transforms of ${\displaystyle \delta (t)*g(t)}$ and ${\displaystyle \delta _{t}*g_{t}}$ give ${\displaystyle G(\omega )}$ and ${\displaystyle G(z)}$, respectively.

To prove equations (9.3j), we first use equation (9.3h) and get

{\displaystyle {\begin{aligned}\delta (t-\tau )*g(t)\leftrightarrow e^{-{\hbox{j}}\omega \tau }G(\omega )\leftrightarrow g(t-\tau ),\end{aligned}}}

using Sheriff and Geldart, 1995, equation (15.136). Next we use Sheriff and Geldart, 1995, problem (15.27) to write

{\displaystyle {\begin{aligned}\delta _{t-\tau }*g_{t}\leftrightarrow z^{\tau }G(z)\leftrightarrow g_{t-\tau }.\end{aligned}}}

## Problem 9.3c

Show that a boxcar of height ${\displaystyle h}$ extending from ${\displaystyle -f_{0}}$ to ${\displaystyle +f_{0}}$ in the frequency domain has the transform

{\displaystyle {\begin{aligned}h\ \mathrm {box} _{2f_{0}}(f)\leftrightarrow A{\frac {\sin(2\pi f_{0}t)}{2\pi f_{0}t}}=A\sin {\hbox{c}}(2\pi f_{0}t),\end{aligned}}}

where ${\displaystyle A=2hf_{0}=}$ area of the boxcar.

### Solution

Using equation (9.3d) a boxcar in the frequency domain becomes ${\displaystyle g(t)}$ in the time domain:

 {\displaystyle {\begin{aligned}g(t)&=\int _{-\infty }^{+\infty }h\mathrm {box} _{2f_{0}}(f)e^{{\hbox{j}}2\pi ft}\mathrm {d} f=h\int _{-f_{0}}^{+f_{0}}e^{{\hbox{j}}2\pi ft}\,\mathrm {d} f\\&=\left({\frac {h}{2\pi jt}}\right)e^{{\hbox{j}}2\pi ft}|_{+f_{0}}^{-f_{0}}={\frac {h}{2\pi jt}}(e^{{\hbox{j}}2\pi f_{0}t}-e^{-{\hbox{j}}2\pi f_{0}t}).\\&={\frac {h}{\pi t}}\sin(2\pi f_{0}t)={\frac {2f_{0}h}{(2\pi f_{0}t)}}\sin \,(2\pi f_{0}t)=A\sin c(2\pi f_{0}t),\end{aligned}}} (9.3k)

where ${\displaystyle A=2hf_{0}}$ and sinc ${\displaystyle x=(\sin x)/x.}$

## Problem 9.3d

Calculate the transform of the pair of displaced boxcars in Figure 9.3b. Discuss the relation between your result and equation (9.3k) for a single boxcar centered at the origin.

### Solution

Equation (9.3d) gives for the transform of the pair of boxcars:

 {\displaystyle {\begin{aligned}f(t)&={\frac {h}{2\pi }}\left[\int \limits _{-\omega _{2}}^{-\omega _{1}}e^{{\hbox{j}}\omega t}d\omega +\int \limits _{+\omega _{1}}^{+\omega _{2}}e^{{\hbox{j}}\omega t}d\omega \right]={\frac {h}{2\pi jt}}\left[e^{{\hbox{j}}\omega t}|_{-\omega _{2}}^{-\omega _{1}}+e^{{\hbox{j}}\omega t}|_{+\omega _{1}}^{+\omega _{2}}\right]\\&={\frac {h}{2\pi jt}}[(e^{-{\hbox{j}}\omega _{1}t}-e^{-{\hbox{j}}\omega _{2}t})+(e^{{\hbox{j}}\omega _{2}t}-e^{-{\hbox{j}}\omega _{1}t})]={\frac {h}{\pi t}}(\sin \omega _{2}t-\sin \omega _{1}t)\\&=(h/\pi )(\omega _{2}\sin \mathrm {c} \ \omega _{2}t-\omega _{1}\sin \mathrm {c} \ \omega _{1}t)\\&=A_{2}\sin \mathrm {c} \ \omega _{2}t-A_{1}\sin \mathrm {c} \ \omega _{1}t,\end{aligned}}} (9.3l)

where ${\displaystyle A_{2}=2hf_{2}}$, ${\displaystyle A_{1}=2hf_{1}}$, and we have used Euler’s formulas (see Sheriff and Geldart, 1995, problem 15.12a) to get the sines. Thus, the transform is the difference between two sinc functions corresponding to the upper and lower limiting frequencies of the boxcars.

Note that equation (9.3${\displaystyle \ell }$) can be regarded as giving the result of two boxcars, one extending from ${\displaystyle -\omega _{2}}$ to ${\displaystyle +\omega _{2}}$, the other extending from ${\displaystyle -\omega _{1}}$ to ${\displaystyle +\omega _{1}}$, the effect of the latter being subtracted from that of the former.

To compare equation (9.3l) with equation (9.3k), we set ${\displaystyle \omega _{1}=0}$. Then sinc ${\displaystyle \omega _{1}t=\sin \,0/0=1}$, so ${\displaystyle \omega _{1}}$ sinc ${\displaystyle \omega _{1}t=0}$ and ${\displaystyle f(t)}$ becomes

{\displaystyle {\begin{aligned}f(t)=2hf_{2}\sin \mathrm {c} \ \omega _{2}t.\end{aligned}}}

Setting ${\displaystyle f_{2}=f_{0}}$ gives equation (9.3k).