# A mathematical review of the Fourier transform

## A.1 The 1-D Fourier transform

Given a continuous function x(t) of a single variable t, its Fourier transform is defined by the integral

 $X(\omega )=\int _{-\infty }^{+\infty }x(t)\;\exp(-i\omega t)\;dt,$ (13)

where ω is the Fourier dual of the variable t. If t signifies time, then ω is angular frequency. The temporal frequency f is related to the angular frequency ω by ω = 2πf.

The Fourier transform is reversible; that is, given X(ω), the corresponding time function is

 $x(t)=\int _{-\infty }^{+\infty }X(\omega )\;\exp(i\omega t)\;d\omega .$ (14)

Throughout this book, the following sign convention is used for the Fourier transform. For the forward transform, the sign of the argument in the exponent is negative if the variable is time and positive if the variable is space. Of course, the inverse transform has the opposite sign used in the respective forward transform. For convenience, the scale factor 2π in equations (13) and (14) are omitted.

Generally, X(ω) is a complex function. By using the properties of the complex functions, X(ω) is expressed as two other functions of frequency

 $X(\omega )=A(\omega )\;\;\exp[i\phi (\omega )],$ (15)

where A(ω) and ϕ(ω) are the amplitude and phase spectra, respectively. They are computed by the following equations:

 $A(\omega )={\sqrt {X_{r}^{2}(\omega )+X_{i}^{2}(\omega )}}$ (16)

and

 $\phi (\omega )=\tan ^{-1}{\frac {X_{i}(\omega )}{X_{r}(\omega )}},$ (17)

where Xr(ω) and Xi(ω) are the real and imaginary parts of the Fourier transform X(ω). When X(ω) is expressed in terms of its real and imaginary components

 $X(\omega )=X_{r}(\omega )+iX_{i}(\omega ),$ (18)

and is compared with equation (15), note that

 ${X_{r}}(\omega )=A(\omega )\;\cos \phi (\omega ),$ (19)

and

 ${X_{i}}(\omega )=A(\omega )\;\sin \phi (\omega ).$ (20)

We now consider two functions — x(t) and f(t). Listed in Table A-1 are basic theorems that are useful in various applications of the Fourier transform.

 Operation Time Domain Frequency Domain (1) Shifting x(t − τ) exp(−iωτ)X(ω) (2) Scaling x(at) ${{\left|a\right|}^{-1}}X\left({\omega }/{a}\;\right)$ (3) Differentiation dx(t)/dt iωX(ω) (4) Addition f(t) + x(t) F(ω) + X(ω) (5) Multiplication f(t) x(t) F(ω) * X(ω) (6) Convolution f(t) * x(t) F(ω) X(ω) (7) Autocorrelation x(t) * x(−t) ${{\left|X\left(\omega \right)\right|}^{2}}$ (8) Parseval’s theorem $\int {{\left|x\left(t\right)\right|}^{2}}\ dt$ $\int {{\left|X\left(\omega \right)\right|}^{2}}\ d\omega$ * denotes convolution.

Proofs of these theorems can be found in the classic reference on Fourier transforms by Bracewell (1965). Also, some of the proofs are left to the exercises at the end of this chapter. Here, we shall derive the convolutional relation (6) for continuous functions, and the same relation for discrete functions in Section A.2. Consider convolution of two functions x(t) and f(t) with their Fourier transforms X(ω) and F(ω), respectively,

 $y(t)=f(t)\ast x(t),$ (21)

which is explicitly given by the integral

 $y(t)=\int _{-\infty }^{+\infty }f(t-t')\;x(t')\;dt'.$ (22)

The Fourier transform of the resulting function y(t) is

 $Y(\omega )=\int _{-\infty }^{+\infty }y(t)\;\;\exp(-i\omega t)\;dt.$ (23)

Substitute the convolution integral of equation (22) into equation (23)

 $Y(\omega )=\int _{-\infty }^{+\infty }{\left[{\int _{-\infty }^{+\infty }{f\left({t-t'}\right)x\left({t'}\right)\;dt'}}\right]}\quad \exp \left({-i\omega t}\right)\;dt,$ (24)

and interchange the two integrals

 $Y(\omega )=\int _{-\infty }^{+\infty }{x\left({t'}\right)}\left[{\int _{-\infty }^{+\infty }{f\left({t-t'}\right)\;\;}\exp \left({-i\omega t}\right)\;dt}\right]\;dt'.$ (25)

From the shift theorem given by entry (1) of Table A-1, we have

 $\int _{-\infty }^{+\infty }f(t-t')\;\;\exp(-i\omega t)\;dt=F(\omega )\;\;\exp(-i\omega t').$ (26)

Use this relation in equation (25) to get

 $Y(\omega )=\int _{-\infty }^{+\infty }x(t')\;[F(\omega )\;\exp(-i\omega t')]\;dt',$ (27)

then rearrange the terms to obtain

 $Y(\omega )=F(\omega )\int _{-\infty }^{+\infty }x(t')\;\exp(-i\omega t')\;dt'.$ (28)

Note that the integral in equation (28) is the Fourier transform of x(t), and therefore,

 $Y(\omega )=F(\omega )\;X(\omega ),$ (29)

which is the desired result given by entry (6) of Table A-1.

## Alternate conventions and formulations

There are a number of variations on the the conventions and notations employed by mathematicians, electrical engineers, physicists, and geophysicists in formulating the Fourier transform. The conventions employed in Yilmaz's treatment are by no means universal. It is important to see what conventions an author is employing when studying results involving the Fourier transform. As long as consistency is maintained, all results will be equivalent, no matter what conventions are used.

When there are several choices to be made, it is a guarantee that some people will choose the conventions that you don't like. The choices that may be made are notational, and formulational. Mere notational changes will have little ramifications. However, choice of scaling, as to whether or not there is a factor of $1/2\pi$ in front of the inverse transform will yield a result that has a different constant scaling factor, which must be dealt with in a self-consistent manner.

### Is the ${\sqrt {-1}}$ represented by $\pm i$ or by $\mp j?$ Among electrical engineers, for example, it is common to see the definition $j={\sqrt {-1}}$ because the symbol $i$ is often reserved to represent electric current. Particularly in older geophysical literature, or in papers and books produced by particular electrical-engineering oriented research groups, we may see the use of $j$ in this fashion.

A second convention is the choice of sign of the exponent of the exponential in the transform. For example a common convention is to choose a plus sign for the forward temporal (time to frequency) transform, requiring that a minus sign be chosen for the inverse (frequency to time) transform. The same authors often choose to use a negative sign for the forward spatial transforms, requiring a plus sign for the inverse temporal transforms. Thus, one common alternative representation for a temporal signal $s(t)$ or a spatial signal $s(x)$ $S(\omega )=\int _{-\infty }^{\infty }s(t)e^{i\omega t}\;dt\qquad \qquad s(t)=\int _{-\infty }^{\infty }S(\omega )e^{-i\omega t}\;d\omega ,$ and

$S(k)=\int _{-\infty }^{\infty }s(x)e^{-ikx}\;dx\qquad \qquad s(x)=\int _{-\infty }^{\infty }S(k)e^{ikx}\;dk.$ ### Causal Fourier transforms

Most signals encountered in the world of physical (and geophysical) measurement are causal, meaning that there is no signal before some critical time, which could be the arrival time of the earliest possible signal, or universally, time $t=0.$ So prevalent are causal temporal functions in geophysics that we may consider the

$S(\omega )=\int _{0}^{\infty }s(t)e^{i\omega t}\;dt\qquad \qquad s(t)=\int _{C}S(\omega )e^{-i\omega t}\;d\omega ,$ where $C$ is an integration contour chosen in the complex plane to yield a causal time domain result. This causal contour choice translates into a requirement of analyticity in a half-plane of $\omega$ (either upper or lower) governed by the choice of the sign of the exponent in the exponential in the inverse transform.

Indeed, causal considerations are not confined to the temporal transform. For problems involving wave propagation, there is a relationship between the frequency $\omega$ and the wavenumber $k,$ . The choice of integration path for the frequency to time transform will result in a shift of singularities in the inverse wavenumber transform. This shift will be determined by the sign convention for the inverse wavenumber to space transform.

$S(k)=\int _{-\infty }^{\infty }s(t)e^{-ikx}\;dx\qquad \qquad s(x)=\int _{-\infty }^{\infty }S(k)e^{ikx}\;dk.$ ### Input and output variables

We may wish to make a distinction between input and output variables. If we are engaged in processing that involves the application of a Fourier transform, followed by an operation in the transform domain, which is subsequently followed by an inverse Fourier transform, the effect may result in a shift either in space or time. Thus, it is to our benefit to define the forward and inverse Fourier transforms making a distinction of input or output. For example we may write

$S(\omega )=\int _{0}^{\infty }s(\tau )e^{i\omega \tau }\;d\tau \qquad \qquad s(t)={\frac {1}{2\pi }}\int _{C}S(\omega )e^{-i\omega t}\;d\omega ,$ and

$S(k)=\int _{-\infty }^{\infty }s(x)e^{-ik\xi }\;d\xi \qquad \qquad s(x)=\int _{-\infty }^{\infty }S(k)e^{ikx}\;dk.$ Here $(\xi ,\tau )$ are the input space and time variables, and $(x,t)$ are the output variables.

### What about the $1/2\pi ?$ In many formulations of the Fourier transform we see a factor of $1/2\pi$ in front of the inverse transforms. For example, our notations above would be represented as:

$S(\omega )=\int _{0}^{\infty }s(\tau )e^{i\omega \tau }\;d\tau \qquad \qquad s(t)={\frac {1}{2\pi }}\int _{C}S(\omega )e^{-i\omega t}\;d\omega ,$ and

$S(k)=\int _{-\infty }^{\infty }s(\xi )e^{-ik\xi }\;d\xi \qquad \qquad s(x)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }S(k)e^{ikx}\;dk.$ The ramifications of this extra factor of $1/2\pi$ may be seen if we cascaded the forward and inverse temporal transform:

${\frac {1}{2\pi }}\int _{-\infty }^{\infty }\int _{0}^{\infty }s(\tau )e^{i\omega \tau }\;d\tau \;e^{-i\omega t}\;d\omega =s(t)$ The result has to be $s(t)$ and we can see by rearranging the order of integrations that we have

$\int _{0}^{\infty }s(\tau )\left[{\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{-i\omega (t-\tau )}\;d\omega \right]\;d\tau =s(t).$ This implies that the expression in square brackets $[\quad ]$ is equivalent to the Dirac delta function

${\frac {1}{2\pi }}\int _{-\infty }^{\infty }e^{-i\omega (t-\tau )}\;d\omega =\delta (t-\tau ).$ Thus, any scaling factor on the inverse Fourier transform will change the definition of the Dirac Delta, and may propagate through calculations.

### Angular frequency $\omega$ versus ordinary frequency $f$ Another ramification of the choice of scaling is to consider the definition of $2\pi f=\omega .$ We can immediately seen where the factor of $1/2\pi$ comes from, noting that $d\omega =2\pi df$ $S(f)=\int _{0}^{\infty }s(\tau )e^{i2\pi f\tau }\;d\tau \qquad \qquad s(t)=\int _{C}S(\omega )e^{-i2\pi t}\;df,$ and

$S(k)=\int _{-\infty }^{\infty }s(\xi )e^{-ik\xi }\;d\xi \qquad \qquad s(x)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }S(k)e^{ikx}\;dk.$ There is no consistent convention for angular versus regular wavenumber, so the space- wavenumber transforms stay the same.

## A.2 The z-transform

A discrete time function is called a time series. When digitized, the continuous function x(t) takes the form

 $x(t)=\sum \limits _{k}{x_{k}}\delta (t-k\Delta t),\quad k=0,\;1,\;2,\;\ldots \;,$ (30)

where Δt is the sampling interval and δ(t − kΔt) is the Dirac delta function. The discrete equivalent of the Fourier integral given by equation (13) is written as a summation

 $X(\omega )=\sum \limits _{k}{x_{k}}\;\exp(-i\omega k\Delta t),\quad k=0,\;1,\;2,\;\ldots \;.$ (31)

A new variable z = exp(-iωΔt) now is defined. By substituting into equation (31) and explicitly writing the summation, we get

 $X(z)={x_{0}}+{x_{1}}z+{x_{2}}{z^{2}}+\ldots .$ (32)

Function X(z) in equation (32) is called the z-transform of x(t). It is a polynomial of the z variable. The power of z represents the time delay of the discrete samples in the time series x(t).

We now show that convolution of two time series is equivalent to the multiplication of their z-transforms. Consider two discrete time series — x(t) : (x0, x1, x2) and f(t) : (f0, f1). Convolution of the two series is obtained by using Table 1-5. The results of the convolutional process are displayed in Table A-2.

 Fixed Array: a0, a1, a2, a3, a4, a5, a6, a7 Moving Array: b0, b1, b2 Given two arrays, ai and bj: Step 1 : Reverse moving array bj. Step 2 : Multiply in the vertical direction. Step 3 : Add the products and write as output ck. Step 4 : Shift array bj one sample to the right and repeat Steps 2 and 3. Convolution Table: a0 a1 a2 a3 a4 a5 a6 a7 Output b2 b1 b0 c0 b2 b1 b0 c1 b2 b1 b0 c2 b2 b1 b0 c3 b2 b1 b0 c4 b2 b1 b0 c5 b2 b1 b0 c6 b2 b1 b0 c7 b2 b1 b0 c8 b2 b1 b0 c9 where ${c_{k}}=\sum \limits _{j=0}^{n}{{a_{k-j}}\;{b_{j}},}\quad k=0,\;1,\;2,\;\cdots \;,m\;+\;n\;-\;1.$ The discrete output series y(t) : (y0, y1, y2, y3) is given by

 ${\begin{array}{*{35}{l}}{{y}_{0}}&=&{{f}_{0}}{{x}_{0}}&&\\{{y}_{1}}&=&{{f}_{1}}{{x}_{0}}&+&{{f}_{0}}{{x}_{1}}\\{{y}_{2}}&=&{{f}_{1}}{{x}_{1}}&+&{{f}_{0}}{{x}_{2}}\\{{y}_{3}}&=&{{f}_{1}}{{x}_{2}}.&&\\\end{array}}$ (33)

The z-transforms of the two input series are expressed by

 $X(z)=x_{0}+x_{1}z+x_{2}z^{2},$ (34)

and

 $F(z)=f_{0}+f_{1}z.$ (35)

By multiplying the two polynomials of equations (34) and (35), we obtain

 $X(z)F(z)=f_{0}x_{0}+(f_{1}x_{0}+f_{0}x_{1})z+(f_{1}x_{1}+f_{0}x_{2})z^{2}+(f_{1}x_{2})z^{3}.$ (36)
 x0 x1 x2 Output f1 f0 y0 f1 f0 y1 f1 f0 y2 f1 f0 y3

By comparing the coefficients of the polynomial given by equation (36) with the output of convolution in equations (33), we find that if

 $y\left(t\right)=f\left(t\right)*x\left(t\right),$ (37)

then,

 $Y\left(z\right)=F\left(z\right)\ X\left(z\right),$ (38)

and, since z = exp(-iωΔt),

 $Y\left(\omega \right)=F\left(\omega \right)\ \in X\left(\omega \right).$ (39)

## A.3 The 2-D Fourier transform

The 2-D Fourier transform of a 2-D function, such as a wavefield P(x, t), is given by

 $P({{k}_{x}},\ \omega )=\int {\int {P}}(x,\ t)\ \ \exp(i{{k}_{x}}x-i\omega t)\ dx\ dt.$ (40)

Function P(x, t) can be reconstructed from P(kx) by the 2-D inverse Fourier transform:

 $P(x,\ t)=\int {\int {P}}({{k}_{x}},\ \omega )\ \ \exp(-i{{k}_{x}}x+i\omega t)\ d{{k}_{x}}\ d\omega .$ (41)

The integral given by equation (40) is evaluated in two steps. First, by Fourier transforming in t,

 $P(x,\omega )=\int {P}(x,t)\ \ \exp(-i\omega t)\ dt,$ (42)

then by Fourier transforming in x, we get the 2-D transform:

 $P({{k}_{x}},\omega )=\int {P}(x,\omega )\ \ \exp(i{{k}_{x}})\ dx.$ (43)