In Figure E-2 for both x and y positive (protected region) the signs 

 of the upper limit of the integrals in equations (E-ll) and (E-12) are 

 negative. For x positive and y negative both limits are positive. For x 

 negative and y positive the upper limit of the integral in (E-ll) is 

 positive and that in E-12 is negative. As a sirrplified solution, Putnam 

 and Arthur give 



(E-15) 



e-^'^ f (u) 



(u = u^) 



F(x,y) 



Using the simplified solution and comparing equations (E-9)and (E-l5) it 

 can be seen that the modulus of f(u) (written |f (u)| ) determines the 

 relative height of waves with a barrier present to those without a barrier. 

 That is, the ratio diffracted wave height - vt b If Cu)| (E-16) 



incident wave height I ^ '^1 



Also from equation (E-9) and E-l$) diffracted waves differ in phase from 

 undiffracted waves by the argument of f(u) (written arg f (u) ). i.e. 

 equation (E-l5) may be written 



.ik(y- i£i/M ) 



F(x,y) 



f(u) 



(E-17) 



Both the modulus and argument of f (u) may be determined from tabulated 

 values of the Fresnel integrals C(u) and S(u) since 



C(u) - iS(u) = / e"^"^ /^ dv for u>0 (E-18) 



and it can be shown that for u < 0, f (-u) = 1-f (u) (E-19) 



The modulus and argument of f (u) are plotted on Figure E-3 as a function 

 of u. Equation (E-13) may be solved for X/L. 



1 ' 



z: 



-ir 



-3Tr 



F 













- 









Ar\ 



/: 



1.2 

 1.0 

 0.8 _ 



0.6 r 

 0.4 

 0.2 

 0.0 



u). 



Argument^ f (u ) 





/ 



x: 



■ 



X 





/■ 





■ 



/ 



>1 



y 



/ 





■ 



- X 







— -^Modulus f(u) 



- 1 



f fll) 1 



■ 



, , , 











- 



-2 -1 

 GURE E-3 FUNCTIONS OF 



u^ 

 u: ARGUMENT f(u 



1 i 



AND MODULUS 



> 



f( 



E-U 



