Im(0j^) = pj^ sin0, where pR is 1/2 the peak to peak pressure of the bottom reflected 

 signal and is the phase. In any event when reahstic profiles are considered, it is not 

 possible to separate the direct and bottom reflected paths at low frequencies. 



The normal mode form of the wave theory representation, required for low fre- 

 quency calculations, is shown in Figure 16 (Bucker, 1970). The source is at depth z and 

 range zero and the receiver is at depth z and range r. The potential function >// is given by 



1/2' 



where 



= (t) ^lu„(^o)Un(z)exp(ik„r) 



U = Af(z) + Bg(z) , < z < Zp 



U = exp (iCj^z) + R exp [iC|3(2z|j-z)] , Zp < z < z^ . 



Cj,- = (aj/cb)2 - k2 . 



The depth function U is a sum of linearly independent solutions of the z-separated part of 

 the wave equation and k is the horizontal wave number. The kj., are those values of k for 

 which U satisfies the boundary conditions. The plane wave coefficient R is introduced into 

 the normal mode solution in the following manner. Assume that a "pseudo" isospeed 

 layer, with sound speed c^, extends from depth Zp to z^ as shown in Figure 16. Then the 

 z-component solution for the layer can be written as a downgoing plane wave exp (iS.^ z) 

 and an upgoing plane wave exp (-ii-^ z) multiplied by the plane wave reflection coefficient 

 R. The vertical wave number is ^^. The values of the coefficients A and B can be deter- 

 mined by requiring the usual interface conditions at depth z These conditions require 

 that the pressure (p3U/3t) and the vertical component of particle velocity (3U/9z) must 

 be continuous functions. Solve these two equations and take the limit z^ -* z^ to obtain 

 the values of A and B. 



Interface Conditions: 



continuity of pressure (pU) 

 continuity of vertical (-3U/9z) 

 component of particle velocity 



Solve for A and B as z -^ z^. 



A = exp (iCj, z,^) [g',3 (1 +R) - iCb gb (1 -R)] /W . 



B =exp(iC5Zb)[iCbfb(l-R)-fyi + R)]/W. 



w = fb g'b - f b % 



fb = f(Zb^ gb-g(^b) ' fb = (df/dz) , g{, = (dg/dz) . 



Note that the isospeed layer has been removed from the problem as Zp -^ z\y. However R 

 remains in the solution in A and B. 



48 



where 



