Understanding Cylindrical Wave Functions in Electromagnetics

ECE 6341 Spring 2016 Prof. David R. Jackson ECE Dept. Notes 8 1

Cylindrical Wave Functions 2 + 2 = 0 k Helmholtz equation: ( ) = , , or z A F z z 2 2 2 1 1 + + + + 2 = 0 k 2 2 2 2 z Separation of variables: ( ) ( ) = ( ) ( ) Z z , , z R let Substitute into previous equation and divide by 2

Cylindrical Wave Functions (cont.) 2 2 2 1 1 + + + + 2 = 0 k 2 2 2 2 z ( ) ( ) = ( ) ( ) Z z , , z R 2 2 2 1 1 R R Z + + + + = 2 0 Z Z RZ R k R Z 2 2 2 2 z Divide by let 1 1 R R R R Z Z + + + + = 2 0 k 2 3

Cylindrical Wave Functions (cont.) 1 1 R R R R Z Z + + + + = 2 0 k (1) 2 or 1 1 Z Z R R R R = 2 k 2 ( ) f z ( , ) g Hence, f(z) =constant = - kz2 4

Cylindrical Wave Functions (cont.) Hence = Z Z 2 k z = = jk z ( ) ( ) ,sin( ),cos( ) Z z h k z e k z k z z z z z Next, to isolate the -dependent term, multiply Eq. (1) by 2: 1 1 R R R R + + + + = 2 2 z 2 2 0 k k 2 5

Cylindrical Wave Functions (cont.) Hence 1 R R R R = + 2 2 2 k k (2) z ( ) g ( ) f Hence, = = 2 constant so j = = ( ) ,sin( ),cos( ) h e 6

Cylindrical Wave Functions (cont.) From Eq. (2) we now have 1 R R R R = + 2 2 2 2 k k z The next goal is to solve this equation for R( ). First, multiply by R and collect terms: ( ) + + = 2 2 2 2 z 2 0 R R k k R R 7

Cylindrical Wave Functions (cont.) 2 2 2 k k k Define z ( ) 2 + + = 2 2 0 R R k R Then, = = x k Next, define ( ) ( ) y x R dR d dy dx dx d ( ) ( ) = = = R y x k Note that ( ) p ( ) = 2 R y x k and 8

Cylindrical Wave Functions (cont.) Then we have + + = 2 2 2 0 x y xy x y Bessel equation of order ( ), x Y x ( ) J Two independent solutions: Hence = + ( ) y x ( ) x ( ) AJ BY x Therefore ( ) = ( ), ( ) R J k Y k 9

Cylindrical Wave Functions (cont.) Summary ( ) ( ) = ( ) ( ) Z z , , z R = jk z ( ) ,sin( ),cos( ) Z z e k z k z z z z j = ,sin( ),cos( ) e ( ) = ( ), ( ) R J k Y k = 2 2 2 k k k z 10

References for Bessel Functions M. R. Spiegel, Schaum s Outline Mathematical Handbook, McGraw-Hill, 1968. M. Abramowitz and I. E. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards, Government Printing Office, Tenth Printing, 1972. N. N. Lebedev, Special Functions & Their Applications, Dover Publications, New York, 1972. 11

Properties of Bessel Functions 1 1 (0) J is finite n =0 0.8 n =1 0.6 0 n =2 0.4 J0 x ( ) Jn (x) J1 x ( ) 0.2 Jn 2 x ( ) 0 0.2 0.4 0.403 0.6 0 1 2 3 4 5 6 7 8 9 10 0 xx 10 12

Bessel Functions (cont.) 1 0.521 n =0 0 n =1 1 n =2 2 Yn (x) Y0 x ( ) (0) Y is infinite Y1 x ( ) 3 Yn 2 x ( ) 4 5 6 6.206 7 0 1 2 3 4 5 6 7 8 9 10 0 xx 10 13

Bessel Functions (cont.) Small-Argument Properties (x 0): Ax 1, 2,... ( ) x , J 1, 2,... = ( ) x , J Ax ( ) , 0 Y x Bx ( ) x = 0( ) Y x ln , 0 C For order zero, the Bessel function of the second kind Y0 behaves as ln(x) rather than algebraically. 14

Bessel Functions (cont.) Non-Integer Order: n = ( ) y x ( ), x J ( ) x J Two linearly independent solutions Bessel equation is unchanged by Note: ( ) x J is a always a valid solution These are linearly independent when is not an integer. ( ) x , ( ) x 0 J Ax J A x x as 1 2 15

Bessel Functions (cont.) Symmetry property = n ( 1) = n ( ) x ( ) J J x n n The functions Jn and J-n are no longer linearly independent. ( 1) = n ( ) x ( ) Y Y x n n 16

Bessel Functions (cont.) Frobenius solution : ( ( ) k + 2 k 1 + ( x = ( ) x J ) ! = ! 2 ) 1 + k k = 0 k ! z z This is valid for any (including = n). Ferdinand Georg Frobenius (October 26, 1849 August 3, 1917) was a German mathematician, best known for his contributions to the theory of differential equations and to group theory (Wikipedia). 17

Bessel Functions (cont.) Definition of Y ( )cos( x ) ( ) x J J ( ) Y x sin( ) - 2, -1, 0, 1, 2 (This definition gives a nice asymptotic behavior as x .) = ( ) lim ( ) Y x Y x For integer order: n n 18

Bessel Functions (cont.) From the limiting definition, we have, as n: ( ) k n 2 n k 1 ! 2 1 1 n x x ( ) x = + ( ) ln Y x J n n 2 ! 2 k = 0 k k n + 2 1 1 x ( ) ( ) k ( ) k + + 1 n k ( ) + ! ! 2 k n k = 0 k (Schaum s Outline Mathematical Handbook, Eq. (24.9)) 1 2 1 3 1 p ( ) p ( ) 0 = + + + + = where 1 ( 0) p 0 19

Bessel Functions (cont.) Example ( 1) = n ( ) x ( ) J J x Prove: n n ( ( ) k + 2 n k 1 + x = ( ) x J ) n ! ! 2 k n k ) = 0 k ( + k + 2 n k 1 n x = ( ) x J ( ) n ! ! 2 k k = 0 k k = + k n Denote: ( + ) ) ( ) ! n k + + 2 n k 1 k x = ( ) x J ( n ! 2 n k = k n 20

Bessel Functions (cont.) Example (cont.) ( ) n k + + 2 n k 1 n x = ( ) x J ( ) ( k ) n + ! ! 2 k = k n Plot of function (from Wikipedia) Note that ( ) = + = 1, 2, 3... = ! 1 , 21

Bessel Functions (cont.) Example (cont.) Hence ( ) n k + + 2 n k 1 n x = ( ) x J ( ) ( k ) n + ! ! 2 k = k n ( ) n k + + 2 n k 1 n x = ( ) x J ( ) ( k ) n + ! ! 2 k = 0 k 22

Bessel Functions (cont.) Example (cont.) Hence, we have ( ( ) k + 2 n k 1 + x = ( ) x J ) n ! ! 2 k n k = 0 k ( ! ) + k + 2 n k 1 n x ( ) n = ( ) x 1 J ( ) ( k ) n ! 2 k = 0 k so ( 1) = n ( ) x ( ) J J x n n 23

Bessel Functions (cont.) 0 J x as From the Frobenius solution and the symmetry property, we have that 1 x 1, 2, 3,... ( ) ~ x J 2 ! 1 n = n ( ) ~ J x 0,1,2,.... x n n 2 ! n ( 1) = n ( ) x ( ) J J x n n 1 n = n n ( ) ~ ( 1) x 0,1,2,.... J x n n 2 ! n 24

Bessel Functions (cont.) 0 Y x as + 2 x = ( ) ~ Y x ln , 0.5772156 0 2 n 1 2 x 1 2 x ( ) ~ ( 1)! ( ) ~ Y x ( 1)! , 0 n Y x n = 1,2,3,..... n ( 1) = n cos sin 1 x ( ) x ( ) Y Y x n n ( ) ~ Y x , 0 ! 2 n 1 2 x ( ) n ( ) ~ x 1 ( 1)! Y n n + (2 1)/ 2 n n = 1,2,3,..... n 25

Bessel Functions (cont.) Asymptotic Formulas x 2 ( ) ~ x cos J x 2 4 x 2 ( ) ~ Y x sin x 2 4 x 26

Hankel Functions ( ) 1 + ( ) x ( ) x ( ) H J jY x ( ) 2 ( ) x ( ) x ( ) H J jY x x As 2 + ( ) j x ( ) 1 Incoming wave ( ) ~ x H e 2 4 x 2 ( ) j x ( ) 2 ( ) ~ x H e Outgoing wave 2 4 x These are valid for arbitrary order . 27

Fields In Cylindrical Coordinates 1 1 ( ) = E A F j 1 1 ( ) = + H A F j = = z A z F A F or z z We expand the curls in cylindrical coordinates to get the following results. 28

TMzFields = z A TMz: 2 1 = + 2 z E k H = 0 2 j z z 1 2 1 = H = E j z 2 1 1 = = E H j z 29

TEzFields TEz: = z F 2 1 1 = H = E j z 2 1 1 = H = E j z 2 1 z E = = + 2 0 H k z 2 j z 30