2-D Stationary Phase Method in ECE 6341 with Prof. David R. Jackson's Notes
Explore the 2-D Stationary Phase Method concept explained in ECE 6341 class notes from Prof. David R. Jackson's Spring 2016 session. Dive into the theoretical framework, assumptions, and mathematical derivations involved in this advanced topic in Electrical and Computer Engineering.
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ECE 6341 Spring 2016 Prof. David R. Jackson ECE Dept. Notes 30 1
2-D Stationary Phase Method ( ) ( ) ( ) = , j g x y , I f x y e dxdy S ( ( ) ) = = , 0 g x y 0 0 x 2D stationary phase point: , 0 g x y 0 0 y Assume ( ) ( ) , , x y x y 0 0 ( ) ( ) ( ) ( ) + + , , g x y g x y g x x g y y 0 0 0 0 x y 1 2 1 2 x ( ) ( ( ) 2 2 + + g x x g y y 0 0 xx yy )( ) + g x y y 0 0 xy 2
2-D Stationary Phase (cont.) Denote 1 2 1 2 g ( ) = , g x y 0 0 xx ( ) = , g x y 0 0 yy ( ) = , x y 0 0 xy Then + + ( ) ( ) ( )( ) 2 2 x x + y y + x x y y ( ) ( ) j ( ) , j g x y 0 0 0 0 ~ , I f x y e e dxdy 0 0 0 0 3
2-D Stationary Phase (cont.) = = x y x y x y Let 0 0 + 1, 1, 0 0 g g ( ) xx = ( ) ( ) where = and x x xx = + 1, 1, 0 0 g g y ( ) yy = y yy We then have + + ( ) ( ) + 2 2 + ( ) ( ) ( ) j x y xy , j g x y ~ , I f x y e e dx dy x y 0 0 0 0 4
2-D Stationary Phase (cont.) + + ( ) ( ) + 2 2 + ( ) ( ) ( ) j x y xy , j g x y ~ , I f x y e e dx dy x y 0 0 0 0 Let = = s x t y We then have ( ) st ( ) ( ) + 2 2 + + + j s t 1 x y ( ) ( ) , j g x y ~ , I f x y e e dsdt 0 0 0 0 5
2-D Stationary Phase (cont.) ( ) st ( ) ( ) + 2 2 + + + j s t 1 x y ( ) ( ) , j g x y ~ , I f x y e e dsdt 0 0 0 0 2I Complete the square: 2 ( ) 2 ( ) st t 2 2 t ( ) ( ) ( ) ( ) ( ) + + = + + 2 2 2 x s t s t x y x x y 4 2 ( ) 2 t 2 2 t ( ) ( ) ( ) ( ) x = + + 2 x s t x y y 4 6
2-D Stationary Phase (cont.) The integral I2 is then 2 2 2 t t ( ) ( ) + ( ) ( ) ( ) x + + 2 j s j t x x y y 4 2 = I e e ds dt 2 Now use t ( ) = + s s = ds ds x 2 so 2 2 t ( ) ( ) ( ) x + + 2 j t ( ) 2 y y 4 j s = I e e ds dt x 2 7
2-D Stationary Phase (cont.) The integral I2 is then in the form of the product of two 1-D integrals: 2 2 t ( ) ( ) ( ) x + + 2 j t ( ) 2 y y 4 j s = I e dt e ds x 2 8
2-D Stationary Phase (cont.) = I I I This has the form 2 t s s Integral in : + ( ) 2 j s = I e ds Use x s = u s + 1 ( ) 2 j u = e du x Recall that + ( ) ( ) 2 = j j x = /4 j e x e dx e 4 9
2-D Stationary Phase (cont.) Integral in t: 2 ( ) ( ) ( ) x + 2 1 j t y y = 4 tI e dt Define: 2 ( ) ( ) x 1 y 4 And then let ( ) = 10
2-D Stationary Phase (cont.) + ( ) ( ) y 2 j t = tI e dt Then we have Use = u t 1 ( ) ( ) y j ( ) = tI e 4 Hence 1 1 ( ) ( ) ( ) y j j ( ) ( ) ( ) , j g x y ~ , I f x y e e e x 4 4 0 0 0 0 tI sI 11
2-D Stationary Phase (cont.) We then have ( ) ( ) ( ) , j g x y ~ , I f x y e 0 0 0 0 ( ) ( ) ( ) y + j e x 4 1 1 2 ( ) ( ) x 1 4 y 12
2-D Stationary Phase (cont.) or ( ) ( ) ( ) y + j ( ) ( ) ( ) j g x y , ~ , I f x y e e x 4 0 0 0 0 1 2 ( ) ( ) x 4 y 13
2-D Stationary Phase (cont.) Important special case (often met in practice): both , 0 0 or 2 0 and 4 2 2 ( ) ( ) x = 1 1 0 In this case: y 4 4 ( ) = 1 so 14
2-D Stationary Phase (cont.) We then have: 1 )( ) ( ) ( ) ( , j g x y /2 j ~ , I f x y e e 0 0 0 0 2 4 where + , , 0 0 and and 15
2-D Stationary Phase (cont.) Hence, the final result is 1 ( ) ( ) )( ) ( , j g x y ~ , I f x y e j 0 0 0 0 2 4 where + , , 0 0 and and 16
