# 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

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

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

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

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

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

and

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

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

and is compared with equation (15), note that

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

and

 ${\displaystyle {X_{i}}(\omega )=A(\omega )\;\sin \phi (\omega ).}$ (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) ${\displaystyle {{\left|a\right|}^{-1}}X\left({\omega }/{a}\;\right)}$ (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) ${\displaystyle {{\left|X\left(\omega \right)\right|}^{2}}}$ (8) Parseval’s theorem ${\displaystyle \int {{\left|x\left(t\right)\right|}^{2}}\ dt}$ ${\displaystyle \int {{\left|X\left(\omega \right)\right|}^{2}}\ d\omega }$
 * denotes convolution.

Proofs of these theorems can be found in the classic reference on Fourier transforms by Bracewell (1965)[1]. 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,

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

which is explicitly given by the integral

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

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

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

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

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

and interchange the two integrals

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

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

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

Use this relation in equation (25) to get

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

then rearrange the terms to obtain

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

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

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

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

## A.2 The z-transform

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

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

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

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

 ${\displaystyle X(z)={x_{0}}+{x_{1}}z+{x_{2}}{z^{2}}+\ldots .}$ (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 ${\displaystyle {c_{k}}=\sum \limits _{j=0}^{n}{{a_{k-j}}\;{b_{j}},}\quad k=0,\;1,\;2,\;\cdots \;,m\;+\;n\;-\;1.}$

The discrete output series y(t) : (y0, y1, y2, y3) is given by

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

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

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

and

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

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

 ${\displaystyle X(z)F(z)=f_{0}x_{0}+(f_{1}x_{0}+f_{0}x_{1})z+(f_{1}x_{1}+f_{0}x_{2})z^{2}+(f_{1}x_{2})z^{3}.}$ (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

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

then,

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

and, since z = exp(-iωΔt),

 ${\displaystyle Y\left(\omega \right)=F\left(\omega \right)\ \in X\left(\omega \right).}$ (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

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

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

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

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

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

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

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

## References

1. Bracewell, R. N., 1965, The Fourier transform and its applications: McGraw-Hill Book Co.
2. Yilmaz, O. and Cumro, D., 1983, Worldwide assortment of field seismic records: Tech. Rep., Western Geophysical Company.