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File:Degf19.jpg
F-19. La función de tiempo a la izquierda es la Transformada de Fourier de las funciones en frecuencia de la derecha y viceversa. Muchos otros pares podrían ser mostrados. Las de arriba son todas las funciones pares y por lo tanto tienen fase cero. Las transformadas para funciones impares reales son imaginarias,i.e., tienen un corrimiento de +π/2. las transformadas de aquellas funciones que no son ni pares ni impares implican variaciones de fase con la frecuencia. Nota "f"=1/"t".

Formula que convierte una función en tiempo "g"("t") (onda, traza sísmica, etc.) en su representación en el dominio de la frecuencia "G"("t") y viceversa. "G"("t") y "g"("t") constituyen un Par Transformada de Fourier; Ver abajo F-19. Un ejemplo es

${\displaystyle g(t)\leftrightarrow G(f)=\int \limits _{-\infty }^{\infty }g(t)e^{-j2\pi ft}dt=\int \limits _{-\infty }^{\infty }g(t)\cos(j2\pi ft)dt-j\int \limits _{-\infty }^{\infty }g(t)\sin(j2\pi ft)dt}$
${\displaystyle g(t)=\int \limits _{-\infty }^{\infty }G(f)e^{j2\pi ft}dt=\int \limits _{-\infty }^{\infty }G(f)\cos(j2\pi ft)dt+j\int \limits _{-\infty }^{\infty }G(f)\sin(j2\pi ft)dt}$

Obtener "G"("t") de "g"("t") se llama Análisis de Fourier y obtener "g"("t") de "G"("t") se llama Síntesis de Fourier. "G"("t") es el espectro complejo, la parte real transformada coseno y la parte imaginaria es la transformada seno cuando "g"("t") es real. Otra expresión para G(t) es

${\displaystyle G(f)=\left\vert A(f)\right\vert e^{j\gamma (f)}}$

Donde las funciones "A"("f") e γ("f") son reales. Ellas son, el espectro de amplitud y el espectro de fase de "g"("t"):

${\displaystyle A(f)={\sqrt {[{\text{real part of }}G(f)]^{2}+[{\text{imaginary part of }}G(f)]^{2}}}}$
${\displaystyle \gamma (f)=\arctan {\bigg (}{\frac {{\text{imaginary part of }}G(f)}{{\text{real part of }}G(f)}}{\bigg )}}$

γ("f") está en el primer y segundo cuadrante si la parte imaginaria es positive, en el primer o cuarto cuadrante si la parte real es positiva. Puede asumirse que una traza "h"("t") que se extiende solamente de 0 a "T" puede ser repetida infinitamente y entonces expandida en una Serie de Fourier de periodo "T":

${\displaystyle h(t)=\sum _{n=0}^{\infty }a_{n}\cos(2\pi n({\frac {t}{T}}))+\sum _{n=1}^{\infty }b_{n}\sin(2\pi n({\frac {t}{T}}))}$

Donde

${\displaystyle a_{n}={\frac {2}{T}}\int \limits _{0}^{T}h(t)\cos(2\pi n({\frac {t}{T}}))dt,}$
${\displaystyle b_{n}={\frac {2}{T}}\int \limits _{0}^{T}h(t)\sin(2\pi n({\frac {t}{T}}))dt,}$

y

${\displaystyle h(t)\leftrightarrow H_{n}=\left\vert A_{n}\right\vert e^{j\gamma _{n}}}$
${\displaystyle A_{n}={\sqrt {a_{n}^{2}+b_{n}^{2}}},\gamma _{n}=\arctan {\Bigg (}{\frac {b_{n}}{a_{n}}}{\Bigg )}}$

The same rules for quadrants apply to γn as expressed for γ(f); a0 is the zero-frequency component (or dc shift). The frequency spectrum is discrete if the function is periodic. If ht is a sampled time series sampled at intervals of time t2, then we can stop summing when n>2T/t2 (see sampling theorem). In this case an and bn can be expressed as sums:

${\displaystyle a_{n}={\frac {2}{T}}\sum _{n=0}^{2{\frac {T}{t_{2}}}}h_{t}\cos(2\pi n({\frac {t}{T}}))}$

and

${\displaystyle b_{n}={\frac {2}{T}}\sum _{n=1}^{2{\frac {T}{t_{2}}}}h_{t}\sin(2\pi n({\frac {t}{T}}))}$

Also see phase response and fast Fourier transform.

F-20. Equivalence of Fourier transform operations. Doing the time operation is equivalent to doing the frequency operation on the transform of the data. Note: g(t)↔G(f) and h(t)↔H(f).

Operations in one domain have equivalent operations in the transform domain (see Figure F-20). Computations can sometimes be carried out more economically in one domain than the other and Fourier transforms provide a means of accomplishing this. The Fourier-transform relations can be generalized for more than one dimension (see Figure F-21). For example,

${\displaystyle G(k,\omega )=\int \limits _{-\infty }^{\infty }\int \limits _{-\infty }^{\infty }g(x,t)e^{-j(kx+\omega t)}dxdt}$

and

${\displaystyle g(x,t)={\frac {1}{\pi }}\int \limits _{-\infty }^{\infty }\int \limits _{-\infty }^{\infty }G(k,\omega )e^{j(kx+\omega t)}d\omega dk}$

The 1/4π factor is sometimes distributed between the two integrals; where calculations involve an arbitrary scaling factor, the 1/4π factor may be dropped entirely.

Fourier transforms are discussed in Sheriff and Geldart (1995, 277, 532–533)[1]. Theorems relating to Fourier transforms are shown in Figure F-22.

F-22. Fourier transform theorems.

## Notations and sign conventions

The notations and sign conventions used above are common in the electrical engineering world. Many exploration geophysicists may be familiar, however, with the conventions used by physicists and mathematicians. These differ subtly from those used above. Many geophysicists whose education may in those fields, or who draw from the scientific results from those fields prefer conventions different from those of electrical engineers. It is, therefore, important to include a short discussion of these conventions.

It is more common in the world of mathematics and theoretical physics for the convention of ${\displaystyle i={\sqrt {-1}}}$ rather than ${\displaystyle j}$. Electrical engineers prefer to use the letter ${\displaystyle j}$ because the letter ${\displaystyle i}$ is reserved for current. Mathematicians and mathematical physicists prefer using the angular frequency ${\displaystyle \omega =2\pi f}$ where the units are radians per time, rather than cycles per time. This means that there may be a factor of ${\displaystyle 2\pi }$ discrepancy between computations using the differing conventions. Finally, there may be different sign conventions on the exponent of the exponentials in the Fourier transform definitions. Because signals recorded in the space-time domain are causal, meaning that there are no arrivals before time ${\displaystyle t=0,}$ the forward temporal transform integration will start at ${\displaystyle 0}$ rather than ${\displaystyle -\infty }$.

## 1D transforms

Putting all of these together, we obtain a common notational convention for the forward Fourier transform in time as

${\displaystyle {\tilde {f}}(\omega )=\int _{-\infty }^{\infty }f(t)e^{i\omega t}\;dt}$

and the inverse Fourier transform

${\displaystyle f(t)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }{\tilde {f}}(\omega )e^{-i\omega t}\;d\omega .}$

The 1D spatial forward Fourier transform differs from the temporal transform in that the integration is on infinite limits and the sign of the exponent in the exponential is negative

${\displaystyle {\tilde {f}}(k)=\int _{-\infty }^{\infty }f(x)e^{-ikx}\;dx}$

and the inverse spatial Fourier transform is similarly different

${\displaystyle f(t)={\frac {1}{2\pi }}\int _{-\infty }^{\infty }{\tilde {f}}(k)e^{ikx}\;dk.}$

## Transforms in ${\displaystyle n}$ spatial dimensions

In ${\displaystyle n}$ dimensions, the spatial transforms become

${\displaystyle {\tilde {f}}(\mathbf {k} )=\int _{-\infty }^{\infty }...{\mbox{n total integrations}}...\int _{-\infty }^{\infty }f(x)e^{-i\mathbf {k} \cdot \mathbf {x} }\;d\mathbf {x} }$

and the inverse spatial Fourier transform is similarly different

${\displaystyle f(\mathbf {x} )={\frac {1}{(2\pi )^{n}}}\int _{-\infty }^{\infty }...{\mbox{n total integrations}}...\int _{-\infty }^{\infty }{\tilde {f}}(\mathbf {k} )e^{i\mathbf {k} \cdot \mathbf {x} }\;d\mathbf {k} .}$

Here we have used the conventions that ${\displaystyle \mathbf {x} =(x_{1},x_{2},...,x_{n}),}$ ${\displaystyle d\mathbf {x} =dx_{1}dx_{2}...dx_{n},}$ ${\displaystyle \mathbf {k} =(k_{1},k_{2},...,k_{n}),}$ ${\displaystyle d\mathbf {k} =dk_{1}dk_{2}...dk_{n},}$ and ${\displaystyle \mathbf {k} \cdot \mathbf {x} =k_{1}x_{1}+k_{2}x_{2}+...+k_{n}x_{n}.}$

## Transforms in 1 temporal and 3 spatial dimensions

In 3 dimensions of space and 1 dimension of time, as is encountered in problems dealing with the wave equation, we have the forward Fourier transform from ${\displaystyle (\mathbf {x} ,t)\rightarrow (\mathbf {k} ,\omega )}$

${\displaystyle {\tilde {F}}(\mathbf {k} ,\omega )=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{0}^{\infty }f(\mathbf {x} ,t)e^{-i(\mathbf {k} \cdot \mathbf {x} -\omega t)}\;dt\,d^{3}x}$

and the corresponding inverse Fourier transform

${\displaystyle f(\mathbf {x} ,t)={\frac {1}{(2\pi )^{n+1}}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{\infty }^{\infty }{\tilde {F}}(\mathbf {k} ,\omega )e^{i(\mathbf {k} \cdot \mathbf {x} -\omega t)}\;d\omega \,d^{3}k.}$

Here, ${\displaystyle d^{3}x=dx_{1}dx_{2}dx_{3}}$ and ${\displaystyle d^{3}k=dk_{1}dk_{2}dk_{3}.}$

## References

1. Sheriff, R. E. and Geldart, L. P., 1995, Exploration Seismology, 2nd Ed., Cambridge Univ. Press.