Scalar Wave Transformation Analysis and Solutions
Scalar wave transformation involving incident waves and orthogonal functions are explored. Derivations, integrations, and coefficient determinations are discussed in detail. Solutions for when x=0 and properties of the scalar functions are also addressed.
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ECE 6341 Spring 2016 Prof. David R. Jackson ECE Dept. Notes 25 1
Scalar Wave Transformation z Incident wave: ( ) = jkz , , x y z e y This is a scalar function, e.g. the pressure of a sound wave. x = = cos jkz jkr e e Notes: No variation m = 0 Must be finite at the origin (only jn) Must be finite on the z axis (only Pn) = n ( ) ( ) cos n n a j kr P n = 0 Note: Spherical Bessel functions are used here, not Schelkunoff Bessel functions. 2
Scalar Wave Transformation (cont.) (cos )sin P Multiply both sides by and integrate. m Orthogonality: 2 = , n m (cos )sin = + (cos ) P P d 2 0 1 n n m n m 0 ( ) ( ) Harrington Eq. 6.41 Hence 2 ( ) = (cos ) sin cos jkr a j kr e P d m m m + 2 1 m 0 We can now relabel m n. 3
Scalar Wave Transformation (cont.) Let = = = x u kr cos d sin du We then have 1 2 = jxu ( ) ( ) a j x e P u du n n n + 2 1 n + 1 + 1 = jxu ( ) e P u du n 1 4
Scalar Wave Transformation (cont.) The coefficients are therefore determined from + 1 2 = jxu ( ) ( ) a j x e P u du n n n + 2 1 n 1 To find the coefficients, take the limit as x 0. 5
Scalar Wave Transformation (cont.) Recall that x = ( ) ( ) x ( ) x 1 j x J J x ( ) 1/2 + n n + 2 x 2 1 Note: ) 1 n + ( = Therefore, as x 0, we have ! n 1/2 + n x ( ) ~ j x n 1 2 2 x 1/2 + + + n 2 1 n 6
Scalar Wave Transformation (cont.) or n x ( ) ~ j x n 1 2 + + + 1 n 2 1 n As x 0 we therefore have + 1 2 = jxu ( ) ( ) a j x e P u du n n n + 2 1 n 1 + 1 n 2 x = jxu ( ) a e P u du n n + 1 2 2 1 n + + + 1 n 2 1 n 1 7
Scalar Wave Transformation (cont.) Note: If we now let x 0 we get zero on both sides (unless n=0). Solution: Take the derivative with respect to x (n times) before setting x=0. n x ( ) ~ j x n 1 2 + + + 1 n 2 1 n n ! d dx n ( ) ~ j x n 1 2 n + + + 1 n 2 1 n 8
Scalar Wave Transformation (cont.) Hence + 1 2 ( ) 0 ( ) ( ) n = ( ) n n j a ju P u du n n + 2 1 n 1 = + 1 2 ! n ( ) ( ) n n a j u P u du n n + 1 2 2 1 n + + + 1 n 2 1 n 1 1 ( ) ( ) ( ) n = n 2 j u P u du n 0 Define 1 ( ) n I u P u du n n 0 9
Scalar Wave Transformation (cont.) Hence + 2 1 1 2 1 n ( ) n + = + + 1 n 2 2 1 a j I n n n 2 ! n Next, we try to evaluate In: 1 ( ) = n I u P u du n n 0 1 n 1 n d ( ) n = 2 n 1 u u du n 2 ! n du 0 (Rodriguez s formula) 10
Scalar Wave Transformation (cont.) 1 n 1 n d du ( ) n = 2 n 1 I u u du Therefore n n 2 ! n 0 1 1 dg du df du 1 = f du fg g du Integrate by parts n times: 0 0 0 1 1 n ( ) n ( ) n = 2 1 ! 1 I n u du n 2 ! n 0 Notes: m d du n m d du ( ) d du n = 2 = 1 0, = u m n n n ! u n 0, u m n m n m = 1 u = 0 u 11
Scalar Wave Transformation (cont.) or 1 1 2 ( ) n = 2 1 I u du n n 0 Schaum s outline Mathematical Handbook Eq. (15.24): 1 2 ( ) + 1 n 1 ( ) n = 2 1 x dx 1 2 + + 2 1 n 0 12
Scalar Wave Transformation (cont.) Hence 1 2 ( ) + 1 n = I n 1 2 + + + 1 n 2 1 n 13
Scalar Wave Transformation (cont.) We then have 1 2 ( ) + 1 n + 2 1 1 2 1 n ( ) n + = + + 1 n 2 2 1 a j n n 1 2 2 ! n + + + 1 n 2 1 n 1 2 = Note: Hence, )1 1 ( ) ( )( n = + + 1 2 n a j n n ! n 14
Scalar Wave Transformation (cont.) ( ) 1 + = ! n n Now use ( ) ( ) n = + 2 1 n a j n so, Hence n ( ) ( ) ( ) ( ) n = + cos jkz 2 1 e j n j kr P n n = 0 15
Acoustic Scattering z Rigid sphere = 0r a = n a y x Acoustic PW ( ) = pressure of sound wave , , x y z so, 16
Acoustic Scattering (cont.) = i jkz e 2 s = k s = of sound wave = + i s s i = n n r a = 17
Acoustic Scattering (cont.) We have = i ( ) (cos ) a j kr P n n n = 0 n = + n ( ) (2 1) n a j n where Choose = (2) n s ( ) (cos ) b h kr P n n = 0 n Hence, n ( ) j ka = b a (2) n h n n ( ) ka 18
Acoustic Scattering (cont.) Incident wave Real part of pressure scattered by a sphere for ka = 1.0 http://www.paraffinalia.co.uk/Software/examples.shtml 19
