CN104950332A

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Info

Publication number: CN104950332A
Application number: CN201510342764.3A
Authority: CN
Inventors: 梁立锋; 张宏兵
Original assignee: Hohai University HHU
Current assignee: Hohai University HHU
Priority date: 2015-06-18
Filing date: 2015-06-18
Publication date: 2015-09-30
Anticipated expiration: 2035-06-18
Also published as: CN104950332B

Abstract

本发明公开了一种弹性多层介质中平面波反射系数的计算方以下法，步骤如下(1)设平面谐和波从介质n+1向目的层系入射，则在n+1介质内产生反射纵波及横波，在介质1内产生透射纵波及横波，各层介质均可写成标量位与向量位形式；(2)确定位移、应力与位移位存在的关系；(3)把步骤(1)中的标量位和向量位带入到步骤(2)中获得第n+1层内及第1层的位移和应力关系式；(4)根据(3)中获得的第n+1层内及第1层的位移和应力关系式得到包含各层弹性系数的位移、应力传递矩阵，对其进行求解，可获得反射及透射系数。本发明充分考虑了弹性层系多次波和转换波及厚度、频率对反射系数的影响，适宜于薄层AVO分析模拟，计算效率高。

The invention discloses a method for calculating the plane wave reflection coefficient in an elastic multilayer medium. The steps are as follows: (1) Assuming that the plane harmonic wave is incident from medium n+1 to the target layer system, reflection will be generated in the n+1 medium The longitudinal wave and the transverse wave generate the transmitted longitudinal wave and transverse wave in the medium 1, and each layer of medium can be written in the form of scalar bit and vector bit; (2) determine the relationship between displacement, stress and displacement; (3) put step (1) The scalar bit and vector bit in step (2) to obtain the displacement and stress relational expressions in the n+1th layer and the first layer; (4) according to the n+1th layer obtained in (3) and The displacement and stress relational expressions of the first layer obtain the displacement and stress transfer matrix including the elastic coefficients of each layer, and solve it to obtain the reflection and transmission coefficients. The invention fully considers the effects of elastic layer multiple waves and converted wave thicknesses and frequencies on reflection coefficients, is suitable for thin-layer AVO analysis and simulation, and has high calculation efficiency.

Description

Translated fromChinese

一种弹性多层介质中平面波反射系数的计算方法A Calculation Method of Plane Wave Reflection Coefficient in Elastic Multilayer Media

技术领域technical field

本发明涉及一种弹性多层介质中平面波反射系数的计算方法，属于薄层AVO模拟技术领域。The invention relates to a calculation method of plane wave reflection coefficient in an elastic multi-layer medium, belonging to the technical field of thin-layer AVO simulation.

背景技术Background technique

模拟薄层AVO分为两大类，(1)时间域褶积模型，对某一特定角度情况下，对时间域采样点逐个计算反射系数，利用子波与反射系数进行褶积，形成该角度对应地震道，该方法虽然计算快捷，公式简单，容易实现，但是难以体现薄层效应，模拟精度较低。(2)采用弹性波方程的积分解有限差分法模拟薄层波场，可模拟出薄层效应，但是计算效率较低，受到建模影响较大，工业生产难以实用化。Simulated thin-layer AVO is divided into two categories, (1) time-domain convolution model, for a certain angle, the reflection coefficient is calculated for the time-domain sampling points one by one, and the wavelet and reflection coefficient are used to convolve to form the angle Corresponding to seismic traces, although this method is fast in calculation, simple in formula and easy to implement, it is difficult to reflect the thin layer effect and the simulation accuracy is low. (2) Using the integral solution finite difference method of the elastic wave equation to simulate the thin-layer wave field can simulate the thin-layer effect, but the calculation efficiency is low, which is greatly affected by the modeling, and it is difficult to be practical in industrial production.

常规的Zoeppritz方程计算反射系数都是基于单界面的，无厚度变量，虽然也能计算AVO特征，但不能直接分析厚度对各频率分量的影响效果。The conventional Zoeppritz equation calculates the reflection coefficient based on a single interface and has no thickness variable. Although it can also calculate AVO characteristics, it cannot directly analyze the effect of thickness on each frequency component.

发明内容Contents of the invention

针对现有技术存在的不足，本发明目的是提供一种弹性多层介质中平面波反射系数的计算方法，充分考虑了弹性层系多次波和转换波及厚度、频率对反射系数的影响，适宜于薄层AVO分析模拟，计算效率高，模拟精度高。In view of the deficiencies in the prior art, the purpose of the present invention is to provide a calculation method for the reflection coefficient of plane waves in elastic multilayer media, which fully considers the influence of multiple waves and converted waves of the elastic layer system and the thickness and frequency on the reflection coefficient, and is suitable for Thin-layer AVO analysis and simulation, high calculation efficiency and high simulation accuracy.

为了实现上述目的，本发明是通过如下的技术方案来实现：In order to achieve the above object, the present invention is achieved through the following technical solutions:

本发明的一种弹性多层介质中平面波反射系数的计算方法，具体包括以下几个步骤：The calculation method of plane wave reflection coefficient in a kind of elastic multi-layer medium of the present invention specifically comprises the following steps:

(1)设平面谐和波P_e从介质n+1向目的层系以入射，则在n+1介质内产生反射纵波和反射横波，其反射系数分别为P_r和S_r，在介质1内产生透射纵波和透射横波，其透射系数分别为P_t和S_t，各层介质均可写成标量位与向量位形式；(1) Assuming that the plane harmonic wave P_e travels from the medium n+1 to the target layer by incident, the reflected longitudinal wave and reflected transverse wave are generated in n+1 medium, and their reflection coefficients are P_r and S_r respectively, and the transmitted longitudinal wave and transmitted transverse wave are generated in medium 1, and their transmission coefficients are P_t and S_t respectively. Layer media can be written in the form of scalar bits and vector bits;

(2)在二维情况下，确定位移、应力与位移位之间的关系式；(2) In the two-dimensional case, determine the relationship between displacement, stress and displacement;

(3)把步骤(1)中涉及到标量位和向量位带入到步骤(2)中，可以获得第n+1层内及第1层的位移和应力关系式；(3) Bringing the scalar bit and vector bit involved in step (1) into step (2), the displacement and stress relational expressions in the n+1th layer and the first layer can be obtained;

(4)根据(3)中获得的第n+1层内及第1层的位移和应力关系式，可以得到包含各层弹性系数的位移、应力传递矩阵，然后对其进行求解，即可获得反射系数、透射系数。(4) According to the relationship between displacement and stress in the n+1th layer and the first layer obtained in (3), the displacement and stress transfer matrix including the elastic coefficient of each layer can be obtained, and then solved to obtain Reflection coefficient, transmission coefficient.

步骤(1)具体包括以下步骤：Step (1) specifically includes the following steps:

设夹层包括n－1个水平层，层序编号为2、3、……n，下部介质为1层；各层中纵波和横波波速分别用带下标的α,β表示，下标为层序号，取x坐标轴与第n层底界面相重合；设一个波自n+1层方向入射到夹层顶面，记纵波入射波为P_e，此时，在n+1层中有一纵波反射波P_r，横波反射波S_r，在1层中有纵波透射波P_t和横波透射波S_t，设纵波和横波到各层入射角It is assumed that the interlayer includes n-1 horizontal layers, the sequence numbers are 2, 3, ... n, and the lower medium is 1 layer; the wave speeds of longitudinal waves and shear waves in each layer are represented by α and β with subscripts respectively, and the subscripts are layer numbers , take the x-coordinate axis to coincide with the bottom interface of the nth layer; suppose a wave is incident on the top surface of the interlayer from the direction of the n+1 layer, denote the incident longitudinal wave as P_e , at this time, there is a longitudinal reflected wave in the n+1 layer P_r , S-wave reflected wave S_r , there are P-wave transmitted wave P_t and S-wave transmitted wave S_t in one layer, and the incident angles of P-wave and S-wave to each layer are set

则在n+1层介质中总的位移位为：Then the total displacement in the n+1 layer medium is:

${φ φ}_{n no + + 11} = = {φ φ}_{e e} + + {φ φ}_{r r} = = (({A A}_{11}^{n no + + 11} {e e}^{- - {jd jd}^{n no + + 11} z z} + + {A A}_{22}^{n no + + 11} {e e}^{{jd jd}^{n no + + 11} z z})) {e e}^{j j ((σ σ x x - - ω ω t t))}$

${ψ ψ}_{n no + + 11} = = {ψ ψ}_{r r} = = {B B}_{22}^{n no + + 11} {e e}^{{js js}^{n no + + 11} z z} {e e}^{j j ((σ σ x x - - ω ω t t))} - - - - - - ((22 - - 11 - - 11))$

其中： $d^{n + 1} = k^{n + 1} {cosi}_{d}^{n + 1} = \sqrt{{(k^{n + 1})}^{2} - σ^{2}}, s^{n + 1} = K^{n + 1} {cosi}_{s}^{n + 1} = \sqrt{{(K^{n + 1})}^{2} - σ^{2}},$ σ＝ω/c，kⁿ⁺¹＝ω/α_n+1，Kⁿ⁺¹＝ω/β_n+1in: $d^{no + 1} = k^{no + 1} {cosi}_{d}^{no + 1} = \sqrt{{(k^{no + 1})}^{2} - σ^{2}}, {the s}^{no + 1} = K^{no + 1} {cosi}_{the s}^{no + 1} = \sqrt{{(K^{no + 1})}^{2} - σ^{2}},$ σ=ω/c, kⁿ⁺¹ = ω/α_n+1 , Kⁿ⁺¹ = ω/β_n+1

其中，ω为频率，φ_e为入射纵波位函数、φ_r反射纵波位函数、为第(n+1)层入射纵波的振幅、e为常数、j为虚数单位、z为方向、为反射纵波的振幅、x为方向、t为时间、ψ_r代表反射横波的位函数、为反射横波的振幅；Among them, ω is the frequency, φ_e is the incident longitudinal wave potential function, φ_r the reflected longitudinal wave potential function, is the amplitude of the incident longitudinal wave on the (n+1) layer, e is a constant, j is an imaginary number unit, z is a direction, is the amplitude of the reflected longitudinal wave, x is the direction, t is the time, ψ_r represents the potential function of the reflected shear wave, is the amplitude of the reflected shear wave;

C为波沿分界面方向的视速度，根据斯奈尔定律，对分界面上的各个波都是相等的；C is the apparent velocity of the wave along the direction of the interface, according to Snell's law, all waves on the interface are equal;

在n层介质中纵波和横波位函数表达式分别为：In n-layer media, the expressions of the longitudinal wave and shear wave potential functions are respectively:

${φ φ}_{n no} = = {φ φ}_{e e} + + {φ φ}_{r r} = = (({A A}_{11}^{n no} {e e}^{- - {jd jd}^{n no} z z} + + {A A}_{22}^{n no} {e e}^{{jd jd}^{n no} z z})) {e e}^{j j ((σ σ x x - - ω ω t t))}$

${ψ ψ}_{n no} = = {ψ ψ}_{e e} + + {ψ ψ}_{r r} = = {B B}_{11}^{n no} {e e}^{- - {js js}^{n no} z z} {e e}^{j j ((σ σ x x - - ω ω t t))} + + {B B}_{22}^{n no} {e e}^{{js js}^{n no} z z} {e e}^{j j ((σ σ x x - - ω ω t t))} - - - - - - ((22 - - 11 - - 22))$

${d d}^{n no} = = {k k}^{n no} {cosi cosi}_{d d}^{n no} = = \sqrt{{(({k k}^{n no}))}^{22} - - {σ σ}^{22}} - - - - - - ((22 - - 11 - - 33))$

${s the s}^{n no} = = {K K}^{n no} {cosi cosi}_{s the s}^{n no} = = \sqrt{{(({K K}^{n no}))}^{22} - - {σ σ}^{22}} - - - - - - ((22 - - 11 - - 44))$

其中，为入射纵波振幅、为反射纵波振幅、ψ_e为入射横波的位函数、为入射横波振幅、为反射横波振幅、kⁿ第n层纵波的水平波数、Kⁿ第n层横波的水平波数；in, is the incident longitudinal wave amplitude, is the amplitude of the reflected P-wave, ψ_e is the potential function of the incident S-wave, is the incident shear wave amplitude, is the amplitude of the reflected shear wave, the horizontal wavenumber of the k^nth layer of longitudinal waves, and the horizontal wavenumber of^Kn ’s nth layer of shear waves;

则在1层介质中透射波位函数表达式分别为：Then the expression of the transmitted wave potential function in a layer of medium is:

${φ φ}_{11} = = {φ φ}_{t t} = = {A A}_{11}^{11} {e e}^{- - {jd jd}^{11} z z} {e e}^{j j ((σ σ x x - - ω ω t t))}$

${ψ ψ}_{11} = = {ψ ψ}_{t t} = = {B B}_{11}^{11} {e e}^{- - {js js}^{11} z z} {e e}^{j j ((σ σ x x - - ω ω t t))} - - - - - - ((22 - - 11 - - 55))$

其中， $d^{1} = \sqrt{{(k^{1})}^{2} - σ^{2}}, s^{1} = \sqrt{{(K^{1})}^{2} - σ^{2}},$ in, $d^{1} = \sqrt{{(k^{1})}^{2} - σ^{2}}, {the s}^{1} = \sqrt{{(K^{1})}^{2} - σ^{2}},$

φ_t为透射纵波位函数、为透射纵波振幅、ψ_t为透射横波位函数、为透射横波振幅、k¹为第一层纵波水平波数、K¹为第一层横波水平波数。φ_t is the transmitted longitudinal wave potential function, is the transmitted longitudinal wave amplitude, ψ_t is the transmitted shear wave potential function, is the amplitude of the transmitted shear wave, k¹ is the horizontal wavenumber of the first-layer longitudinal wave, and K¹ is the horizontal wavenumber of the first-layer shear wave.

步骤(2)中，在二维情况，位移、应力与位移位的关系式为：In step (2), in the two-dimensional case, the relationship between displacement, stress and displacement is:

$u u = = \frac{\partial \partial φ φ}{\partial \partial x x} - - \frac{\partial \partial ψ ψ}{\partial \partial z z}$

$ω ω = = \frac{\partial \partial φ φ}{\partial \partial z z} + + \frac{\partial \partial ψ ψ}{\partial \partial x x}$

${σ σ}_{z z z z} = = λ λ ((\frac{\partial \partial u u}{\partial \partial x x} + + \frac{\partial \partial ω ω}{\partial \partial z z})) + + 22 μ μ \frac{\partial \partial ω ω}{\partial \partial z z}$

${τ τ}_{z z x x} = = μ μ ((\frac{\partial \partial u u}{\partial \partial z z} + + \frac{\partial \partial ω ω}{\partial \partial x x})) - - - - - - ((22 - - 11 - - 66))$

其中，u代表位移、σ_zz代表沿z轴的主应变、τ_zx代表剪应力、z为坐标方向、λ和μ拉梅常数。Among them, u represents the displacement, σ_zz represents the principal strain along the z axis, τ_zx represents the shear stress, z is the coordinate direction, λ and μ Lame constants.

步骤(3)具体包括以下步骤：Step (3) specifically comprises the following steps:

设n层厚度为h，计算n层中的位移分量和应力分量，取其z＝h的值，为第n层顶面上的位移和应力分量值，记为u⁽ⁿ⁾、ω⁽ⁿ⁾、将式(2-1-2)、(2-1-3)、(2-1-4)带入(2-1-6)式，可以得到矩阵：其中p＝dⁿh，Q＝sⁿh，dⁿ为第n层纵波垂直波数、h为厚度、sⁿ为第n层横波垂直波数；Let the thickness of layer n be h, calculate the displacement component and stress component in layer n, take the value of z=h as the value of displacement and stress component on the top surface of layer n, denoted as u⁽ⁿ⁾ , ω^{(n )} , Put the formulas (2-1-2), (2-1-3), (2-1-4) into the formula (2-1-6), you can get the matrix: where p=dⁿ h, Q=sⁿ h, dⁿ is the vertical wave number of the nth layer of longitudinal wave, h is the thickness, sⁿ is the vertical wave number of the nth layer of shear wave;

$(\begin{matrix} {u u}^{((n no))} \\ {ω ω}^{((n no))} \\ {σ σ}_{z z z z}^{((n no))} \\ {τ τ}_{z z x x}^{((n no))} \end{matrix}) = = (({B B}_{i i j j})) (\begin{matrix} {A A}_{22}^{n no} + + {A A}_{11}^{n no} \\ {A A}_{22}^{n no} - - {A A}_{11}^{n no} \\ {B B}_{22}^{n no} - - {B B}_{11}^{n no} \\ {B B}_{22}^{n no} + + {B B}_{11}^{n no} \end{matrix}) {e e}^{j j ((σ σ x x - - ω ω t t))} - - - - - - ((22 - - 11 - - 77))$

其中in

$(({B B}_{i i j j})) = = (\begin{matrix} j j σ σ cos cos p p & - - σ σ sin sin p p & - - j j s the s cos cos Q Q & s the s sin sin Q Q \\ - - d d sin sin p p & j j d d cos cos p p & - - σ σ sin sin Q Q & jσ jσ c c o o s the s Q Q \\ - - (({λ λ}_{n no} {k k}_{00}^{22} + + 22 {μ μ}_{n no} {d d}^{22})) cos cos p p & - - j j (({λ λ}_{n no} {k k}_{00}^{22} + + 22 {μ μ}_{n no} {d d}^{22})) sin sin p p & - - 22 {μ μ}_{n no} σ σ s the s cos cos Q Q & - - 22 {jμ jμ}_{n no} σ σ s the s sin sin Q Q \\ - - j j d d σ σ sin sin p p & - - d d σ σ cos cos p p & \frac{j j}{22} (({s the s}^{22} - - {σ σ}^{22})) sin sin Q Q & \frac{11}{22} (({s the s}^{22} - - {σ σ}^{22})) c c o o s the s Q Q \end{matrix})$

${k k}_{00}^{22} = = {σ σ}^{22} + + {d d}^{22} - - - - - - ((22 - - 11 - - 88))$

其中，s为第1层横波垂直波数和d为第1层纵波垂直波数、λ_n和μ_n为第n层的拉梅常数 Among them, s is the vertical wavenumber of the shear wave in the first layer and d is the vertical wavenumber of the longitudinal wave in the first layer, λ_n and μ_n are the Lame constants of the nth layer

取位移和应力分量在z＝0处的值，可得n层底面上的位移分量和应力分量，根据分界面位移和应力连续条件，它应与(n－1)层顶面的相应值相等，记为u^(n-1)、ω^(n-1)、当z＝0时，p＝0，Q＝0，由矩阵(2-1-8)可以得到Taking the value of the displacement and stress components at z=0, the displacement component and stress component on the bottom surface of layer n can be obtained. According to the continuous condition of displacement and stress at the interface, it should be equal to the corresponding value on the top surface of (n-1) layer , recorded as u^(n-1) , ω^(n-1) , When z=0, p=0, Q=0, it can be obtained from the matrix (2-1-8)

${(({B B}_{i i j j}))}_{p p = = 00,, Q Q = = 00} = = (\begin{matrix} j j σ σ & 00 & - - j j s the s & 00 \\ 00 & j j d d & 00 & j j σ σ \\ - - (({λ λ}_{n no} {k k}_{00}^{22} + + 22 {μ μ}_{n no} {d d}^{22})) & 00 & - - 22 {μ μ}_{n no} σ σ s the s & 00 \\ 00 & - - d d σ σ & 00 & \frac{11}{22} (({s the s}^{22} - - {σ σ}^{22})) \end{matrix}) - - - - - - ((22 - - 11 - - 99))$

(n－1)层顶面上的位移与应力分量值可表示为：The displacement and stress components on the top surface of (n-1) layer can be expressed as:

$(\begin{matrix} {u u}^{((n no - - 11))} \\ {ω ω}^{((n no - - 11))} \\ {σ σ}_{z z z z}^{((n no - - 11))} \\ {τ τ}_{z z x x}^{((n no - - 11))} \end{matrix}) = = {(({B B}_{i i j j}))}_{Q Q = = 00}^{p p = = 00} (\begin{matrix} {A A}_{22}^{n no} + + {A A}_{11}^{n no} \\ {A A}_{22}^{n no} - - {A A}_{11}^{n no} \\ {B B}_{22}^{n no} - - {B B}_{11}^{n no} \\ {B B}_{22}^{n no} + + {B B}_{11}^{n no} \end{matrix}) {e e}^{j j ((σ σ x x - - ω ω t t))} - - - - - - ((22 - - 11 - - 1010))$

由上述公式可以建立起(n)层与(n－1)层顶面的位移分量和应力分量之间的关系，为此，求解方程组(2-1-10)可有：From the above formula, the relationship between the displacement component and the stress component of the top surface of (n) layer and (n-1) layer can be established. For this reason, the equations (2-1-10) can be solved as follows:

$(\begin{matrix} {A A}_{22}^{n no} + + {A A}_{11}^{n no} \\ {A A}_{22}^{n no} - - {A A}_{11}^{n no} \\ {B B}_{22}^{n no} - - {B B}_{11}^{n no} \\ {B B}_{22}^{n no} + + {B B}_{11}^{n no} \end{matrix}) {e e}^{j j ((σ σ x x - - ω ω t t))} = = (({b b}_{i i j j})) (\begin{matrix} {u u}^{((n no - - 11))} \\ {ω ω}^{((n no - - 11))} \\ {σ σ}_{z z z z}^{((n no - - 11))} \\ \frac{11}{22 {μ μ}_{n no - - 11}} {τ τ}_{z z x x}^{((n no - - 11))} \end{matrix}) - - - - - - ((22 - - 11 - - 1111))$

其中(b_ij)表示的逆矩阵：where (b_ij ) means The inverse matrix of :

$(({b b}_{i i j j})) = = (\begin{matrix} \frac{- - 22 j j σ σ}{{K K}^{22}} & 00 & - - \frac{11}{{μ μ}_{n no} {K K}^{22}} & 00 \\ 00 & \frac{- - j j (({s the s}^{22} - - {σ σ}^{22}))}{{dK dK}^{22}} & 00 & \frac{- - 22 σ σ}{{dK dK}^{22}} \\ \frac{j j (({λ λ}_{n no} {k k}_{00}^{22} / / {μ μ}_{n no} + + 22 {d d}^{22}))}{{sK k}^{22}} & 00 & \frac{- - σ σ}{{μ μ}_{n no} {K K}^{22} s the s} & 00 \\ 00 & \frac{- - 22 j j σ σ}{{K K}^{22}} & 00 & \frac{22}{{K K}^{22}} \end{matrix})$

其中，K为横波水平波数Among them, K is the horizontal wave number of the shear wave

将式(2-1-11)代入(2-1-7)可得Substitute formula (2-1-11) into (2-1-7) to get

$(\begin{matrix} {u u}^{((n no))} \\ {ω ω}^{((n no))} \\ {σ σ}_{z z z z}^{((n no))} \\ \frac{11}{22 {μ μ}_{n no}} {τ τ}_{z z x x}^{((n no))} \end{matrix}) = = (({B B}_{i i j j})) (({b b}_{i i j j})) (\begin{matrix} {u u}^{((n no - - 11))} \\ {ω ω}^{((n no - - 11))} \\ {σ σ}_{z z z z}^{((n no - - 11))} \\ \frac{11}{22 {μ μ}_{n no - - 11}} {τ τ}_{z z x x}^{((n no - - 11))} \end{matrix}) - - - - - - ((22 - - 11 - - 1212))$

取记号(a_ij)表示矩阵乘积(B_ij)(b_ij)，同样为4×4阶方阵，各元素为：Take the notation (a_ij ) to represent the matrix product (B_ij )(b_ij ), which is also a 4×4 order square matrix, and each element is:

a₁₁＝2sin²γcos p-cos2γcos Qa₁₁ ＝2sin² γcos p-cos2γcos Q

a₁₂＝j(tanθcos2γsin p+sin2γsin Q)a₁₂ ＝j(tanθcos2γsin p+sin2γsin Q)

${a a}_{1313} = = \frac{j j s the s i i n no θ θ}{ρ ρ α α ω ω} ((cos cos Q Q - - cos cos p p))$

${a a}_{1414} = = \frac{22 β β}{ω ω} ((t t a a n no θ θ s the s i i n no γ γ sin sin p p - - c c o o s the s γ γ sin sin Q Q))$

${a a}_{21 twenty one} = = j j ((\frac{β β c c o o s the s θ θ}{α α c c o o s the s γ γ} s the s i i n no 22 γ γ sin sin p p - - t t a a n no γ γ c c o o s the s 22 γ γ sin sin Q Q))$

a₂₂＝cos2γcos p+2sin²γcos Qa₂₂ ＝cos2γcos p+2sin² γcos Q

${a a}_{23 twenty three} = = \frac{11}{ρ ρ α α ω ω} ((c c o o s the s θ θ sin sin p p + + t t a a n no γ γ s the s i i n no θ θ sin sin Q Q))$

${a a}_{24 twenty four} = = \frac{22 j j β β}{ω ω} s the s i i n no γ γ ((cos cos Q Q - - cos cos p p))$

a₃₁＝2jωρβsinγ(cosp-cosQ)cos2γa₃₁ ＝2jωρβsinγ(cosp-cosQ)cos2γ

${a a}_{3232} = = - - ρ ρ ω ω ((\frac{{αcos α cos}^{22} 22 γ γ}{c c o o s the s θ θ} sin sin p p + + 44 {βcosγsin βcosγsin}^{22} γ γ sin sin Q Q))$

a₃₃＝cos2γcos p+2sin²γcos Qa₃₃ ＝cos2γcos p+2sin² γcos Q

a₃₄＝2jρβ²(cos2γtanθsin p-sin2γsin Q)a₃₄ ＝2jρβ² (cos2γtanθsin p-sin2γsin Q)

${a a}_{4141} = = - - ω ω ((\frac{22}{α α} {sin sin}^{22} γ γ c c o o s the s θ θ s the s i i n no p p + + \frac{11}{22 β β} \frac{{cos cos}^{22} 22 γ γ}{cos cos γ γ} sin sin Q Q))$

${a a}_{4242} = = \frac{j j ω ω}{α α} s the s i i n no θ θ c c o o s the s 22 γ γ ((cos cos p p - - cos cos Q Q))$

${a a}_{4343} = = \frac{j j}{22 ρ ρ} ((\frac{s the s i i n no 22 θ θ}{{α α}^{22}} sin sin p p - - \frac{c c o o s the s 22 γ γ}{{β β}^{22}} t t a a n no γ γ sin sin Q Q))$

a₄₄＝2sin²γcos p+cos2γcos Q (2-1-13) a₄₄ ＝2sin² γcos p+cos2γcos Q (2-1-13)

其中，γ＝i_s、θ＝i_d分别表示波在层中的入射角，所以有sinγ＝σ/K，sinθ＝σ/k，α,β为纵横波速度，ρ为介质密度，在式(2-1-12)中建立了(n)层和(n－1)层顶面位移分量和应力分量之间的关系，计算(a_ij)矩阵元素，在公式(2-1-13)将使用(n)层的参数数值，为此，在式(2-1-12)中系数矩阵用上角码(n)表示，可有：Among them, γ=i_s and θ=id represent the incident angles of waves in the layer respectively, so sinγ=σ/K, sinθ=σ/k,_α , β are the longitudinal and transverse wave velocities, ρ is the medium density, in the formula In (2-1-12), the relationship between the displacement component and the stress component of the top surface of the (n) layer and (n－1) layer is established, and the (a_ij ) matrix elements are calculated in the formula (2-1-13) The parameter value of layer (n) will be used. For this reason, the coefficient matrix in the formula (2-1-12) is represented by the upper corner code (n), which can be:

$(\begin{matrix} {u u}^{((n no))} \\ {ω ω}^{((n no))} \\ {σ σ}_{z z z z}^{((n no))} \\ \frac{11}{22 {μ μ}_{n no}} {τ τ}_{z z x x}^{((n no))} \end{matrix}) = = (({a a}_{i i j j}^{n no})) (\begin{matrix} {u u}^{((n no - - 11))} \\ {ω ω}^{((n no - - 11))} \\ {σ σ}_{z z z z}^{((n no - - 11))} \\ \frac{11}{22 {μ μ}_{n no - - 11}} {τ τ}_{z z x x}^{((n no - - 11))} \end{matrix}) - - - - - - ((22 - - 11 - - 1414))$

得到了夹层各分界面上的位移分量和应力分量的递推公式，满足关系式(2-1-14)等价于满足夹层分界面上的位移与应力连续条件。The recursive formulas of displacement components and stress components on each interface of the interlayer are obtained, and satisfying the relation (2-1-14) is equivalent to satisfying the continuity condition of displacement and stress on the interface of the interlayer.

步骤(4)具体包括以下步骤：Step (4) specifically comprises the following steps:

为了确定夹层顶面反射系数和透过夹层在夹层底面以下传播的透射系数，根据公式(2-1-14)建立(n)层顶面和(1)层顶面上的位移分量和应力分量的关系：In order to determine the reflection coefficient of the top surface of the interlayer and the transmission coefficient of the transmission through the interlayer below the bottom surface of the interlayer, the displacement components and stress components on the top surface of (n) layer and (1) layer top surface are established according to the formula (2-1-14) Relationship:

$(\begin{matrix} {u u}^{((n no))} \\ {ω ω}^{((n no))} \\ {σ σ}_{z z z z}^{((n no))} \\ \frac{11}{22 {μ μ}_{n no}} {τ τ}_{z z x x}^{((n no))} \end{matrix}) = = (({a a}_{i i j j}^{n no})) (({a a}_{i i j j}^{n no - - 11})) ... ... (({a a}_{i i j j}^{22})) (\begin{matrix} {u u}^{((11))} \\ {ω ω}^{((11))} \\ {σ σ}_{z z z z}^{((11))} \\ \frac{11}{22 {μ μ}_{11}} {τ τ}_{z z x x}^{((11))} \end{matrix}) - - - - - - ((22 - - 11 - - 1515))$

对(2-1-15)变形得Pair (2-1-15) deformed

$(\begin{matrix} {u u}^{((n no))} \\ {ω ω}^{((n no))} \\ {σ σ}_{z z z z}^{((n no))} \\ {τ τ}_{z z x x}^{((n no))} \end{matrix}) = = (\begin{matrix} {A A}_{1111} & {A A}_{1212} & {A A}_{1313} & {A A}_{1414} \\ {A A}_{21 twenty one} & {A A}_{22 twenty two} & {A A}_{23 twenty three} & {A A}_{24 twenty four} \\ {A A}_{3131} & {A A}_{3232} & {A A}_{3333} & {A A}_{3434} \\ {A A}_{4141} & {A A}_{4242} & {A A}_{4343} & {A A}_{4444} \end{matrix}) (\begin{matrix} {u u}^{((11))} \\ {ω ω}^{((11))} \\ {σ σ}_{z z z z}^{((11))} \\ {τ τ}_{z z x x}^{((11))} \end{matrix}) - - - - - - ((22 - - 11 - - 1616))$

将式(2-1-1)、式(2-1-5)代入式(2-1-16)得到矩阵方程(2-1-17)，求解后可以获得夹层的反射系数和透射系数；Substituting formula (2-1-1) and formula (2-1-5) into formula (2-1-16) to obtain matrix equation (2-1-17), after solving, the reflection coefficient and transmission coefficient of the interlayer can be obtained;

考虑到A_ij是复数，令A₁₁＝R₁₁+iI₁₁，A₁₂＝R₁₂+iI₁₂，A₁₃＝R₁₃+iI₁₃，…Considering that A_ij is a complex number, let A₁₁ =R₁₁ +iI₁₁ , A₁₂ =R₁₂ +iI₁₂ , A₁₃ =R₁₃ +iI₁₃ , ...

$(\begin{matrix} i i σ σ \\ - - i i {d d}^{n no + + 11} \\ - - {ρ ρ}_{n no + + 11} {β β}_{n no + + 11} (({K K}_{n no + + 11}^{22} - - 22 {σ σ}^{22})) \\ 22 {ρ ρ}_{n no + + 11} {β β}_{n no + + 11}^{22} σ σ {d d}^{n no + + 11} \end{matrix}) = = (\begin{matrix} {B B}_{1111} & {B B}_{1212} & {B B}_{1313} & {B B}_{1414} \\ {B B}_{21 twenty one} & {B B}_{22 twenty two} & {B B}_{23 twenty three} & {B B}_{24 twenty four} \\ {B B}_{3131} & {B B}_{3232} & {B B}_{3333} & {B B}_{3434} \\ {B B}_{4141} & {B B}_{4242} & {B B}_{4343} & {B B}_{4444} \end{matrix}) (\begin{matrix} V V \\ {V V}^{/ /} \\ {W W}^{/ /} \\ W W \end{matrix}) - - - - - - ((22 - - 11 - - 1717))$

其中，ρ_n+1第n+1层的密度、第n+1层的横波水平波数Among them, ρ_n+1 the density of layer n+1, Horizontal wavenumber of shear wave at layer n+1

其中B称为解矩阵，它的各元素为：Among them, B is called the solution matrix, and its elements are:

B₁₁＝B₂₂＝-iσ，B₁₂＝is_n+1B₁₁ =B₂₂ =-iσ, B₁₂ =is_n+1

B₂₁＝id_n+1， $B_{31} = - B_{42} = ρ_{n + 1} β_{n + 1}^{2} (K_{n + 1}^{2} - 2 σ^{2})$ B₂₁ =id_n+1 , $B_{31} = - B_{42} = ρ_{no + 1} β_{no + 1}^{2} (K_{no + 1}^{2} - 2 σ^{2})$

${B B}_{3232} = = 22 {ρ ρ}_{n no + + 11} {β β}_{n no + + 11}^{22} {σs σs}_{n no + + 11},, {B B}_{4141} = = 22 {ρ ρ}_{n no + + 11} {β β}_{n no + + 11}^{22} {σd σd}_{n no + + 11}$

${B B}_{n no 44} = = - - {D D.}_{n no 11} σ σ + + {D D.}_{n no 22} {d d}_{11} - - {C C}_{n no 33} {ρ ρ}_{11} {β β}_{11}^{22} (({K K}_{11}^{22} - - 22 {σ σ}^{22})) + + {C C}_{n no 44} 22 {ρ ρ}_{11} {β β}_{11}^{22} {σd σd}_{11} ((n no = = 11,, 22,, 33,, 44))$

$\begin{matrix} {B B}_{n no 33} = = [[- - {D D.}_{n no 11} {s the s}_{11} - - {D D.}_{n no 22} σ σ + + {C C}_{n no 33} 22 {ρ ρ}_{11} {β β}_{11}^{22} {σs σs}_{11} {+ + C C}_{n no 44} {ρ ρ}_{11} {β β}_{11}^{22} (({K K}_{11}^{22} - - 22 {σ σ}^{22}))]] + + \\ + + i i [[{C C}_{n no 11} {s the s}_{11} + + {C C}_{n no 22} σ σ + + {D D.}_{n no 33} 22 {ρ ρ}_{11} {β β}_{11}^{22} {σs σs}_{11} {+ + D D.}_{n no 44} {ρ ρ}_{11} {β β}_{11}^{22} (({K K}_{11}^{22} - - 22 {σ σ}^{22}))]] ((n no = = 11,, 22,, 33,, 44)) \end{matrix}$

其中，D_n1为复数；Wherein, D_n1 is a complex number;

A_n1的虚部、D_n2为复数A_n2的虚部、d₁为厚度、C_n3为复数A_n3的虚部、ρ₁为第一层的密度、表示第一层横波的水平波数、C_n4为复数A_n4的虚部、s₁第一层横波垂直波数、C_n1为复数A_n1的虚部、C_n2为复数A_n2的虚部解出解矩阵表达式公式(2-2-17)即可得到平面谐和波在弹性层系的纵波反射系数R_pp、横波反射系数V^/、横波透射系数W^/、纵波透射系数W。The imaginary part of A_n1 , D_n2 is the imaginary part of the complex number_A_n2 , d1 is the thickness, C_n3 is the imaginary part of the complex number A_n3 , ρ1 is the density of the_first layer, Indicates the horizontal wave number of the first layer of shear wave, C_n4 is the imaginary part of the complex number A_n4 , s₁ the vertical wave number of the first layer of shear wave, C_n1 is the imaginary part of the complex number A_n1 , C_n2 is the imaginary part of the complex number A_n2 Solve the solution The matrix expression formula (2-2-17) can obtain the longitudinal wave reflection coefficient R_pp , the shear wave reflection coefficient V^/ , the shear wave transmission coefficient W^/ , and the longitudinal wave transmission coefficient W of the plane harmonic wave in the elastic layer system.

本发明基于弹性层系的反射系数谱理论，应用弹性层系的位移和应力递推公式，针对固体多层介质，推导出了纵、横波反射、透射系数公式，该方法可以计算出不同入射角度的平面波反射系数，在石油物探上，可以用弹性层系模型代表地质上的薄(互)层地层结构，因此本发明的方法可以用于计算薄(互)层AVO(Amplitude versus offset)，可以解决多次波、转换波对薄层AVO的影响，同时可以定量模拟不同储层厚度、不同频率变化引起的AVO变化，同时可以用于解释地震道集资料与常规商业化软件基于井zoeppritz方程正演道集不匹配的问题；本发明可以做为目前石油勘探工业商业化软件的一个重要补充。The present invention is based on the reflection coefficient spectrum theory of the elastic layer system, applies the displacement and stress recurrence formulas of the elastic layer system, and deduces the longitudinal and transverse wave reflection and transmission coefficient formulas for solid multilayer media, and the method can calculate different incident angles plane wave reflection coefficient, in oil geophysical prospecting, the thin (inter) layer formation structure on the geological surface can be represented by the elastic layer model, so the method of the present invention can be used to calculate thin (inter) layer AVO (Amplitude versus offset), can Solve the influence of multiple waves and converted waves on thin layer AVO, and at the same time, it can quantitatively simulate the AVO changes caused by different reservoir thicknesses and different frequency changes, and can also be used to interpret seismic gather data and normal commercial software based on well zoeppritz equation. The problem of the incompatibility of lecture collection; the present invention can be used as an important supplement to the commercialization software of the current petroleum exploration industry.

附图说明Description of drawings

图1为层状介质的反射和透射模型；Figure 1 is a reflection and transmission model of a layered medium;

图2为单个薄层的反射和透射；Figure 2 shows the reflection and transmission of a single thin layer;

图3为不同储层厚度情况下的储层AVO特征模拟； Figure 3 is the simulation of reservoir AVO characteristics under different reservoir thicknesses;

图4为不同入射波频率情况下的储层AVO特征模拟； Fig. 4 is the simulation of reservoir AVO characteristics under different incident wave frequencies;

图5为单层介质与传统商业化软件相符。Figure 5 shows single-layer media consistent with traditional commercial software.

具体实施方式Detailed ways

为使本发明实现的技术手段、创作特征、达成目的与功效易于明白了解，下面结合具体实施方式，进一步阐述本发明。In order to make the technical means, creative features, goals and effects achieved by the present invention easy to understand, the present invention will be further described below in conjunction with specific embodiments.

常规的Zoeppritz方程计算反射系数都是基于单界面的，无厚度变量，虽然也能计算AVO特征，但不能直接分析厚度对各频率分量的影响效果。弹性多层介质的平面波反射系数充分考虑了弹性层系多次波和转换波及厚度、频率对反射系数的影响，因此更适宜于研究薄层、薄互层的AVO分析及正演模拟。因此本发明详细介绍了弹性多层介质中的平面波反射系数及正演计算方法，并对单层模型进行了推导验证。The conventional Zoeppritz equation calculates the reflection coefficient based on a single interface and has no thickness variable. Although it can also calculate AVO characteristics, it cannot directly analyze the effect of thickness on each frequency component. The plane wave reflection coefficient of elastic multilayer media fully considers the effects of multiple waves and converted waves of the elastic layer system and the thickness and frequency on the reflection coefficient, so it is more suitable for AVO analysis and forward modeling of thin layers and thin interlayers. Therefore, the present invention introduces the plane wave reflection coefficient in the elastic multi-layer medium and the forward calculation method in detail, and deduces and verifies the single-layer model.

弹性多层介质中的平面波反射系数计算方法如下：The calculation method of plane wave reflection coefficient in elastic multilayer medium is as follows:

在两个半无限弹性介质中间的一个夹层。该夹层由多个水平层次构成。设有一个平面波自上部介质入射夹层顶面，则产生一个夹层反射波在上部介质中传播，同时波亦将透过夹层，产生一个在下部介质中传播的透射波。计算夹层的反射系数和透射系数。A sandwich between two semi-infinitely elastic media. The mezzanine consists of several horizontal layers. If a plane wave enters the top surface of the interlayer from the upper medium, a reflected wave of the interlayer will be generated and propagate in the upper medium, and the wave will also pass through the interlayer to generate a transmitted wave propagating in the lower medium. Calculate the reflection and transmission coefficients of the interlayer.

如图1所示，夹层由n－1个水平层构成，层序编号为2、3、……n；下部介质为1层。各层中纵波和横波波速分别用带下标的α,β表示，下标为层序号。取x坐标轴与第n层底界面相重合。有一个P－SV波系统自n+1层方向入射到夹层顶面，记纵波入射波为P_e，此时，在n+1层中有一纵波反射波P_r，横波反射波S_r；在1层中有纵波透射波P_t和横波透射波S_t。设纵波和横波到各层入射角As shown in Figure 1, the interlayer is composed of n-1 horizontal layers, and the sequence numbers are 2, 3, ... n; the lower medium is 1 layer. The velocities of longitudinal waves and shear waves in each layer are denoted by α and β with subscripts respectively, and the subscripts are layer numbers. Take the x-coordinate axis to coincide with the bottom interface of the nth layer. There is a P-SV wave system incident on the top surface of the interlayer from the n+1 layer direction, and the longitudinal wave incident wave is denoted as P_e . At this time, there is a longitudinal wave reflection wave P_r and a shear wave reflection wave S_r in the n+1 layer; Layer 1 has a longitudinal wave transmitted wave P_t and a transverse wave transmitted wave S_t . Let the incident angles of longitudinal wave and shear wave to each layer

C为波沿分界面方向的视速度，根据斯奈尔定律，对分界面上的各个波都是相等的。C is the apparent velocity of the wave along the interface, which is equal to all waves on the interface according to Snell's law.

其中 $d^{1} = \sqrt{{(k^{1})}^{2} - σ^{2}}, s^{1} = \sqrt{{(K^{1})}^{2} - σ^{2}}$ in $d^{1} = \sqrt{{(k^{1})}^{2} - σ^{2}}, {the s}^{1} = \sqrt{{(K^{1})}^{2} - σ^{2}}$

在二维情况，位移、应力与位移位的关系式为：In the two-dimensional case, the relationship between displacement, stress and displacement is:

设n层厚度为h，计算n层中的位移分量和应力分量，取其z＝h的值，为第n层顶面上的位移和应力分量值，记为u⁽ⁿ⁾、ω⁽ⁿ⁾、将式(2-1-2)、(2-1-3)、(2-1-4)带入(2-1-6)式，可以得到矩阵：其中p＝dⁿh，Q＝sⁿh，Let the thickness of layer n be h, calculate the displacement component and stress component in layer n, take the value of z=h as the value of displacement and stress component on the top surface of layer n, denoted as u⁽ⁿ⁾ , ω^{(n )} , Put the formulas (2-1-2), (2-1-3), (2-1-4) into the formula (2-1-6), you can get the matrix: where p=dⁿ h, Q=sⁿ h,

其中in

令一方面，取位移和应力分量在在z＝0处的值，可得n层底面上的位移分量和应力分量。根据分界面位移和应力连续条件，它应与(n－1)层顶面的相应值相等，记为u^(n-1)、ω^(n-1)、当z＝0时，p＝0，Q＝0，由矩阵(2-1-8)可以得到On the one hand, taking the values of the displacement and stress components at z=0, the displacement components and stress components on the bottom surface of the n-layer can be obtained. According to the interface displacement and stress continuity conditions, it should be equal to the corresponding value on the top surface of (n-1) layer, denoted as u^(n-1) , ω^(n-1) , When z=0, p=0, Q=0, it can be obtained from the matrix (2-1-8)

由上述公式可以建立起(n)层与(n－1)层顶面的位移分量和应力分量之间的关系。为此，求解方程组(2-1-10)可有：The relationship between the displacement component and the stress component of the top surface of (n) layer and (n-1) layer can be established by the above formula. For this reason, solving the equation system (2-1-10) can have:

$(\begin{matrix} {A A}_{22}^{n no} + + {A A}_{11}^{n no} \\ {A A}_{22}^{n no} - - {A A}_{11}^{n no} \\ {B B}_{22}^{n no} - - {B B}_{11}^{n no} \\ {B B}_{22}^{n no} + + {B B}_{11}^{n no} \end{matrix}) {e e}^{j j ((σ σ x x - - ω ω t t))} = = (({b b}_{i i j j})) (\begin{matrix} {u u}^{((n no - - 11))} \\ {ω ω}^{((n no - - 11))} \\ {σ σ}_{z z z z}^{((n no - - 11))} \\ \frac{11}{22 {μ μ}_{n no - - 11}} {τ τ}_{z z x x}^{((n no - - 11))} \end{matrix})$

$((22 - - 11 - - 1111))$

其中(b_ij)表示的逆矩阵：where (b_ij ) means The inverse matrix of :

将式(2-1-11)代入(2-1-7)可得Substitute (2-1-11) into (2-1-7) to get

a₁₁＝2sin²γcos p-cos2γcos Qa₁₁ ＝2sin² γcos p-cos2γcos Q

a₂₂＝cos2γcos p+2sin²γcos Qa₂₂ ＝cos2γcos p+2sin² γcos Q

a₃₃＝cos2γcos p+2sin²γcos Qa₃₃ ＝cos2γcos p+2sin² γcos Q

${a a}_{4141} = = - - ω ω ((\frac{22}{α α} {sin sin}^{22} γ γ c c o o s the s θ θ sin sin p p + + \frac{11}{22 β β} \frac{{cos cos}^{22} 22 γ γ}{cos cos γ γ} sin sin Q Q))$

其中γ＝i_s、θ＝i_d分别表示波在层中的入射角，所以有sinγ＝σ/K，sinθ＝σ/k。其它参数，如α,β为纵横波速度，ρ为介质密度，ω为频率。在式(2-1-12)中建立了(n)层和(n－1)层顶面位移分量和应力分量之间的关系，计算(a_ij)矩阵元素，在公式(2-1-13)将使用(n)层的参数数值。为此，在式(2-1-12)中系数矩阵用上角码(n)表示，可有：Among them, γ=i_s and θ=id represent the incident angles of waves in the layer respectively, so there are_sinγ =σ/K and sinθ=σ/k. Other parameters, such as α, β are the speed of longitudinal and transverse waves, ρ is the medium density, and ω is the frequency. In the formula (2-1-12), the relationship between the displacement component and the stress component of the top surface of the (n) layer and (n-1) layer is established, and the (a_ij ) matrix elements are calculated, and in the formula (2-1- 13) The parameter values of (n) layers will be used. For this reason, in the formula (2-1-12), the coefficient matrix is represented by the upper corner code (n), which can be:

得到了夹层各分界面上的位移分量和应力分量的递推公式。满足关系式(2-1-14)等价于满足夹层分界面上的位移与应力连续条件。The recursive formulas of the displacement components and stress components on each interface of the interlayer are obtained. Satisfying the relationship (2-1-14) is equivalent to satisfying the displacement and stress continuity conditions on the interlayer interface.

对(2-1-15)变形得Pair (2-1-15) deformed

将式(2-1-1)、式(2-1-5)代入式(2-1-16)得到矩阵方程(2-1-17)，求解后可以获得夹层的反射系数和透射系数。Substitute Equation (2-1-1) and Equation (2-1-5) into Equation (2-1-16) to obtain matrix equation (2-1-17). After solving, the reflection coefficient and transmission coefficient of the interlayer can be obtained.

$\begin{matrix} {B B}_{n no 33} = = [[- - {D D.}_{n no 11} {s the s}_{11} - - {D D.}_{n no 22} σ σ + + {C C}_{n no 33} 22 {ρ ρ}_{11} {β β}_{11}^{22} {σs σs}_{11} + + {C C}_{n no 44} {ρ ρ}_{11} {β β}_{11}^{22} (({K K}_{11}^{22} - - 22 {σ σ}^{22}))]] + + \\ + + i i [[- - {C C}_{n no 11} {s the s}_{11} - - {C C}_{n no 22} σ σ + + {D D.}_{n no 33} 22 {ρ ρ}_{11} {β β}_{11}^{22} {σs σs}_{11} + + {D D.}_{n no 44} {ρ ρ}_{11} {β β}_{11}^{22} (({K K}_{11}^{22} - - 22 {σ σ}^{22}))]] ((n no = = 11,, 22,, 33,, 44)) \end{matrix}$

解出解矩阵表达式公式(2-2-17)即可得到平面谐和波在弹性层系的纵波反射系数R_pp、横波反射系数V^/、横波透射系数W^/、纵波透射系数W。Solve the solution matrix expression formula (2-2-17) to get the longitudinal wave reflection coefficient R_pp , the shear wave reflection coefficient V^/ , the shear wave transmission coefficient W^/ , and the longitudinal wave transmission coefficient W of the plane harmonic wave in the elastic layer system.

图3为单个薄层模型所对应的纵波反射系数随角度变化的成果，图中，横坐标为角度，从0度递增到60度，纵坐标为纵波反射系数，薄层厚度5米到30米之间变化，从图中可以看出不同储层厚度对应的薄层AVO形态相似，但四条曲线的截距不同，四条曲线的梯度也不同，这要求我们判断薄层的AVO时要考虑薄层的厚度变化。Figure 3 shows the results of the longitudinal wave reflection coefficient changing with the angle corresponding to a single thin layer model. In the figure, the abscissa is the angle, increasing from 0 to 60 degrees, and the ordinate is the longitudinal wave reflection coefficient. The thickness of the thin layer is 5 meters to 30 meters It can be seen from the figure that the AVO shapes of the thin layers corresponding to different reservoir thicknesses are similar, but the intercepts of the four curves are different, and the gradients of the four curves are also different, which requires us to consider the thin layer AVO when judging the AVO of the thin layer change in thickness.

图4为单个薄层模型所对应的纵波反射系数随角度变化的成果，图中，横坐标为角度，从0度递增到45度，纵坐标为纵波反射系数，频率由10hz到90hz之间变化，从图中可以看出不同频率对应的薄层AVO形态相似，10条曲线都是单调递增的，但10条曲线的梯度各不相同，这要求我们判断薄层的AVO时要考虑入射波的主频变化，尤其同一套薄储层在埋深变化较大的情况下。Figure 4 shows the results of the longitudinal wave reflection coefficient changing with the angle corresponding to a single thin layer model. In the figure, the abscissa is the angle, increasing from 0 to 45 degrees, and the ordinate is the longitudinal wave reflection coefficient, and the frequency changes from 10hz to 90hz , it can be seen from the figure that the thin-layer AVO corresponding to different frequencies is similar in shape, and the 10 curves are all monotonically increasing, but the gradients of the 10 curves are different, which requires us to consider the incident wave when judging the AVO of the thin layer The main frequency changes, especially when the buried depth of the same set of thin reservoirs varies greatly.

为检验上述理论的正确性，按上面的理论推导垂直入射单个薄层情况下的反射系数。In order to test the correctness of the above theory, the reflection coefficient in the case of normal incidence of a single thin layer is deduced according to the above theory.

设两个半无限介质中间有一个厚度为h的固体夹层，计算该夹层的反射系数和透射系数。如图2所示，在1和3两个半无限介质中间夹有一个薄层，其厚度为h。有一纵波自介质3方向垂直入射到薄层顶面，将产生一个在介质3中传播的反射波p_r和在介质1中传播的透射波p_t，根据图中所示的坐标系，各波表达式为：Assuming that there is a solid interlayer with thickness h between two semi-infinite media, calculate the reflection coefficient and transmission coefficient of the interlayer. As shown in Figure 2, there is a thin layer sandwiched between the two semi-infinite media 1 and 3, the thickness of which is h. A longitudinal wave is vertically incident on the top surface of the thin layer from the direction of medium 3, which will generate a reflected wave p_r propagating in medium 3 and a transmitted wave p_t propagating in medium 1. According to the coordinate system shown in the figure, each wave The expression is:

${φ φ}_{e e} = = {Ae Ae}^{- - j j ω ω t t} {e e}^{- - j j ω ω ((z z - - h h)) / / {α α}_{33}}$

${φ φ}_{r r} = = {Be be}^{- - j j ω ω t t} {e e}^{j j ω ω ((z z - - h h)) / / {α α}_{33}}$

${φ φ}_{t t} = = {Ce Ce}^{- - j j ω ω t t} {e e}^{- - j j ω ω z z / / {α α}_{11}} - - - - - - ((22 - - 22 - - 11))$

对第3介质有位函数：There are bit functions for the 3rd medium:

φ₃＝φ_e+φ_r (2-2-2)φ₃ ＝φ_e +φ_r (2-2-2)

对第1介质有位函数：There are bit functions for the first medium:

φ₁＝φ_t (2-2-3)φ₁ ＝φ_t (2-2-3)

它们应满足的边界条件，根据(2-1-15)，为The boundary conditions they should satisfy, according to (2-1-15), are

$(\begin{matrix} {ω ω}^{((22))} \\ {σ σ}_{z z z z}^{((22))} \end{matrix}) = = (\begin{matrix} {a a}_{22 twenty two} & {a a}_{23 twenty three} \\ {a a}_{3232} & {a a}_{3333} \end{matrix}) (\begin{matrix} {ω ω}^{((11))} \\ {σ σ}_{z z z z}^{((11))} \end{matrix}) - - - - - - ((22 - - 22 - - 44))$

其中考虑了垂直入射的情况下，μ＝0，τ_zx＝0。将式(2-2-2)，式(2-2-3)带入式(2-2-4)式，在z＝0上有：Here, when the vertical incidence is considered, μ=0, τ_zx =0. Bring formula (2-2-2), formula (2-2-3) into formula (2-2-4) formula, there is on z=0:

${ω ω}^{((11))} = = - - j j \frac{ω ω}{{α α}_{11}} {ce ce}^{- - j j ω ω t t}$

${σ σ}_{z z z z}^{((11))} = = - - {ρ ρ}_{11} {ω ω}^{22} {ce ce}^{- - j j ω ω t t}$

在z＝h上可有：On z=h there are:

${ω ω}^{((33))} = = - - j j \frac{ω ω}{{α α}_{33}} ((A A - - B B)) {e e}^{- - j j ω ω t t}$

${σ σ}_{z z z z}^{((33))} = = - - {ρ ρ}_{33} {ω ω}^{22} ((A A + + B B)) {e e}^{- - j j ω ω t t}$

边界条件方程组为：The boundary condition equations are:

$(\begin{matrix} \frac{j j}{{α α}_{33}} & {a a}_{22 twenty two} \frac{j j}{{α α}_{11}} + + {a a}_{23 twenty three} {ρ ρ}_{11} ω ω \\ - - {ρ ρ}_{33} ω ω & {a a}_{3232} \frac{j j}{{α α}_{11}} + + {a a}_{3333} {ρ ρ}_{11} ω ω \end{matrix}) (\begin{matrix} \frac{B B}{A A} \\ \frac{C C}{A A} \end{matrix}) = = (\begin{matrix} \frac{j j}{{α α}_{33}} \\ {ρ ρ}_{33} ω ω \end{matrix}) - - - - - - ((22 - - 22 - - 55))$

方程组的解就是所求的反射系数和透射系数，它们是：The solution to the system of equations is the desired reflection and transmission coefficients, which are:

${R R}_{P P P P} = = \frac{B B}{A A} = = \frac{- - {a a}_{3232} + + {jωρ jωρ}_{11} {α α}_{11} {a a}_{3333} - - {jωρ jωρ}_{33} {α α}_{33} {a a}_{22 twenty two} - - {ρ ρ}_{33} {α α}_{33} {ρ ρ}_{11} {α α}_{11} {ω ω}^{22} {a a}_{23 twenty three}}{- - {a a}_{3232} + + {jωρ jωρ}_{11} {α α}_{11} {a a}_{3333} + + {jωρ jωρ}_{33} {α α}_{33} {a a}_{22 twenty two} + + {ρ ρ}_{33} {α α}_{33} {ρ ρ}_{11} {α α}_{11} {ω ω}^{22} {a a}_{23 twenty three}} - - - - - - ((22 - - 22 - - 66))$

${T T}_{P P P P} = = \frac{{α α}_{33}}{{α α}_{11}} \frac{C C}{A A} = = \frac{22 {jωρ jωρ}_{33} {α α}_{33}}{- - {a a}_{3232} + + {jωρ jωρ}_{11} {α α}_{11} {a a}_{3333} + + {jωρ jωρ}_{33} {α α}_{33} {a a}_{22 twenty two} + + {ρ ρ}_{33} {α α}_{33} {ρ ρ}_{11} {α α}_{11} {ω ω}^{22} {a a}_{23 twenty three}} - - - - - - ((22 - - 22 - - 77))$

其中T_PP透射系数取透射波与入射波的位移振幅比。将薄层介质2参数及入射角数据，θ＝0，γ＝0，带入式(2-1-13)，可有：The T_PP transmission coefficient is the ratio of the displacement amplitude of the transmitted wave to the incident wave. Put the thin-layer medium 2 parameters and incident angle data, θ=0, γ=0, into the formula (2-1-13), it can be:

${a a}_{22 twenty two} = = c c o o s the s ((\frac{ω ω}{{α α}_{22}} h h))$

${a a}_{23 twenty three} = = \frac{11}{{ρ ρ}_{22} {α α}_{22}} \frac{11}{ω ω} s the s i i n no ((\frac{ω ω}{{α α}_{22}} h h))$

${a a}_{3232} = = - - {ρ ρ}_{22} {α α}_{22} ω ω s the s i i n no ((\frac{ω ω}{{α α}_{22}} h h))$

${a a}_{3333} = = c c o o s the s ((\frac{ω ω}{{α α}_{22}} h h)) - - - - - - ((22 - - 22 - - 88))$

将式(2-2-8)代入式(2-2-6)，式(2-2-7)，可得Substituting formula (2-2-8) into formula (2-2-6) and formula (2-2-7), we can get

${R R}_{P P P P} = = \frac{{ρ ρ}_{22} {α α}_{22} sin sin \frac{ω ω}{{α α}_{22}} h h + + {jρ jρ}_{11} {α α}_{11} cos cos \frac{ω ω}{{α α}_{22}} h h - - {jρ jρ}_{33} {α α}_{33} cos cos \frac{ω ω}{{α α}_{22}} h h - - \frac{{ρ ρ}_{33} {α α}_{33} {ρ ρ}_{11} {α α}_{11}}{{ρ ρ}_{22} {α α}_{22}} sin sin \frac{ω ω}{{α α}_{22}} h h}{{ρ ρ}_{22} {α α}_{22} sin sin \frac{ω ω}{{α α}_{22}} h h + + {jρ jρ}_{11} {α α}_{11} cos cos \frac{ω ω}{{α α}_{22}} h h + + {jρ jρ}_{33} {α α}_{33} cos cos \frac{ω ω}{{α α}_{22}} h h + + \frac{{ρ ρ}_{33} {α α}_{33} {ρ ρ}_{11} {α α}_{11}}{{ρ ρ}_{22} {α α}_{22}} sin sin \frac{ω ω}{{α α}_{22}} h h} - - - - - - ((22 - - 22 - - 99))$

${T T}_{P P P P} = = \frac{22 {jρ jρ}_{33} {α α}_{33}}{{ρ ρ}_{22} {α α}_{22} s the s i i n no \frac{ω ω}{{α α}_{22}} h h + + {jρ jρ}_{11} {α α}_{11} c c o o s the s \frac{ω ω}{{α α}_{22}} h h + + {jρ jρ}_{33} {α α}_{33} c c o o s the s \frac{ω ω}{{α α}_{22}} h h + + \frac{{ρ ρ}_{33} {α α}_{33} {ρ ρ}_{11} {α α}_{11}}{{ρ ρ}_{22} {α α}_{22}} s the s i i n no \frac{ω ω}{{α α}_{22}} h h} - - - - - - ((22 - - 22 - - 1010))$

当h＝0时，可以得到一个分界面时纵波垂直入射时的反射系数和透射系数：When h=0, the reflection coefficient and transmission coefficient when the longitudinal wave is vertically incident at a boundary can be obtained:

${R R}_{P P P P} = = \frac{{ρ ρ}_{11} {α α}_{11} - - {ρ ρ}_{33} {α α}_{33}}{{ρ ρ}_{11} {α α}_{11} + + {ρ ρ}_{33} {α α}_{33}}$

${T T}_{P P P P} = = \frac{22 {ρ ρ}_{33} {α α}_{33}}{{ρ ρ}_{11} {α α}_{11} + + {ρ ρ}_{33} {α α}_{33}} - - - - - - ((22 - - 22 - - 1111))$

其结果很明显是正确的。The result is clearly correct.

图5是利用zoppritz方程计算的单界面情况下的反射系数随角度变化的情况，图中，横坐标为角度，范围从0度到60度，纵坐标为反射系数，整个递减变化。为了验证本发明正确性，我们用本发明的计算方法，把中间薄层厚度取为0米，则本发明的模型退变为单界面模型，利用本发明的程序进行计算反射系数随角度变化情况，经过比较，本发明的方法与zoeppritz方程对于单界面模型两者是相同的，这也验证了本发明的正确性。Figure 5 shows the change of reflection coefficient with angle in the case of a single interface calculated by using the zoppritz equation. In the figure, the abscissa is the angle, ranging from 0 degrees to 60 degrees, and the ordinate is the reflection coefficient, which changes gradually. In order to verify the correctness of the present invention, we use the calculation method of the present invention to take the thickness of the middle thin layer as 0 meters, then the model of the present invention degenerates into a single interface model, and the program of the present invention is used to calculate the variation of reflection coefficient with angle , after comparison, the method of the present invention is the same as the zoeppritz equation for the single interface model, which also verifies the correctness of the present invention.

以上显示和描述了本发明的基本原理和主要特征和本发明的优点。本行业的技术人员应该了解，本发明不受上述实施例的限制，上述实施例和说明书中描述的只是说明本发明的原理，在不脱离本发明精神和范围的前提下，本发明还会有各种变化和改进，这些变化和改进都落入要求保护的本发明范围内。本发明要求保护范围由所附的权利要求书及其等效物界定。The basic principles and main features of the present invention and the advantages of the present invention have been shown and described above. Those skilled in the industry should understand that the present invention is not limited by the above-mentioned embodiments, and what described in the above-mentioned embodiments and the description only illustrates the principles of the present invention, and the present invention will also have other functions without departing from the spirit and scope of the present invention. Variations and improvements are possible, which fall within the scope of the claimed invention. The protection scope of the present invention is defined by the appended claims and their equivalents.