Steepest-Descent Path Physics Analysis

ECE 6341 Spring 2016 Prof. David R. Jackson ECE Dept. Notes 37 1

Line Source on a Grounded Slab y x r ( ( ) ) ( ( ) ) + TE in TE in TE Z Z k k Z Z k k = j A E ( ) 0 = x x TE k z z x TE 0 x x 0 0 I 1 + ( ) jk y = + jk x TE 1 A k e e dk 0 y ( ) x ( (even function of ky1) ) = TE in TE z x x tan Z k jZ k h 4 j k 1 1 x y 0 y ( ) ( ) 1/2 1/2 = = 2 0 2 x 2 2 x k k k k k k = TE 0 Z 0 1 1 y y 0 k 0 y = TE 0 Z = xk k There are branch points only at 1 k 0 1 y 2

Steepest-Descent Path Physics Steepest-descent transformation: = = sin cos k k k k 0 0 0 x y There are no branch points in the plane (cos is analytic). Both sheets of the kx plane get mapped into a single sheet of the plane. 3

Steepest-Descent Path Physics Examine ky0 to see where the plane is proper and improper: ( ) = + cos k k j 0 0 y r i = cos cosh sin sinh k j 0 r i r i = Im sin sinh k k 0 0 y r i Proper Improper :Im 0 k 0 y :Im 0 k 0 y 4

SDP Physics (cont.) = Im sin sinh k k 0 0 y r i Proper Improper :Im 0 k P: proper I: improper 0 y i :Im 0 k 0 y C I I P P r 2 2 P I I P 5

SDP Physics (cont.) Mapping of quadrants in kx plane = = + sin sin cosh cos sinh k k k j 0 0 x r i r i k xi 0k i 2 0k 1 C k 3 4 P I xr 3 4 I P 2 1 SWP r LWP 2 2 3 4 P I 1 2 P I 6

SDP Physics (cont.) Non-physical growing LW poles (conjugate solution) also exist. i C P I 3 4 I P 2 1 SWP r LWP 2 2 3 4 P I = * LW x k k 1 2 P xp I ( ) = + = conj r /2 /2 The conjugate pole is symmetric about the /2 line: r r = conj i i 7

SDP Physics (cont.) ( ) = cos cosh 1 SDP: r i i C A leaky-wave pole is considered to be physical if it is captured when deforming to the SDP (otherwise, there is no direct residue contribution). SWP SDP r 2 LWP 2 = 0 8

SDP Physics (cont.) Comparison of Fields on interface ( = / 2): ( ) LW x = jk x 2 Res z E j e LWP: (exists if pole is captured) = LW xk j jk x e 0 E A SDP: (from higher-order steepest-descent method) z 3/2 x The leaky-wave field is important if: 1) The pole is captured (the pole is said to be physical ). 2) The residue is strong enough. 3) The attenuation constant is small. 9

SDP Physics (cont.) LWP captured: b i SDP The angle b represents the boundary for which the leaky-wave pole is captured (the leaky-wave field exists). b Note: rp r b rp LWP = + j p rp ip 10

SDP Physics (cont.) Behavior of LW field: + ( ) jk y = jk x E F k e e dk 0 y x z x x ( ) ( ) ( ) cos j k = 0cos F e k d 0 C ( ) ( )( ) ( ) cos j k = LW z 2 Res cos E j F k e p 0 p p In rectangular coordinates: jk x jk y = LW z E Ae e 0 xp y p (It is an inhomogeneous plane-wave field.) = = LW x k k j where xp 11

SDP Physics (cont.) ( ) ( ) cos j k = Examine the exponential term: e 0 p ( ) ( ) = + j cos cos p rp ip ( ) ( ) = cos cosh sin sinh j rp ip rp ip Hence ( ) ( ) sin sinh k = e 0 rp ip ( ) ( ) 0 sinh sin k since = e 0 ip rp ip 12

SDP Physics (cont.) ( ) ( ) sinh sin k = e 0 ip rp Radially decaying: rp LW decays radially y LW exists: rp b b Also, recall that LW exists b rp x Line source r 13

Power Flow Power flows in the direction of the vector. = = + xk yk Re Re( ) k 0 x y ( ) ( ) ( ) x k y k = + + + Re sin cos j j 0 0 rp ip rp ip ( ) = + x y sin cosh cos cosh k 0 rp ip rp ip y 0 x 14

Power Flow (cont.) ( sin cosh rp ip k x ) = + 0 y cos cosh rp ip ( ) = + x y Also, sin cos 0 0 Note that = = y rp b = = x tan tan 0 rp y Hence Note: There is no amplitude change along the rays ( is perpendicular to in a lossless region). = 0 rp 0 x r 15

ESDP (Extreme SDP) The ESDP is the SDP for = / 2. The ESDP is important for evaluating the fields on the interface (which determines the far-field pattern). = = / 2 0 i ESDP We can show that the ESDP divides the LW region into slow-wave and fast-wave regions. r Fast 2 Slow 16

ESDP (cont.) ( ) = = cos cosh cosh 1 1 (SDP) To see this: r i sin (ESDP) r i = sin k k Recall that 0 xp p ( ) = + j sin k 0 rp ip = = Re k Hence xp sin cosh k 0 rp ip 17

ESDP (cont.) k = sin cosh Hence rp ip 0 k 1 sin cosh 1 Fast-wave region: rp ip 0 k sin cosh 1 1 Slow-wave region: rp ip 0 = sin cosh 1 Compare with ESDP: r i 18

ESDP (cont.) The ESDP thus establishes that for fields on the interface, a leaky-wave pole is physical (captured) if it is a fast wave. i ESDP = /2 SWP r LWP captured 2 Fast Slow LWP not captured 19

SDP in kx Plane We now examine the shape of the SDP in the kx plane. k = = sin sin k k 0 x ( ) + j 0 r i so that = = sin cos cosh sinh k k k k 0 xr r i 0 xi r i ( ) = cos cosh 1 SDP: r i The above equations allow us to numerically plot the shape of the SDP in the kx plane. 20

SDP in kx Plane (cont.) k xi = 0sin xk k C k xr 0k SW LW SDP = 2 (Please see the appendix for a proof.) 21

Fields on Interface k xi The SDP is now a lot simpler (two vertical paths)! 2 0k SW k xr LW The leaky-wave pole is captured if it is in the fast-wave region. ESDP fast- wave region 22

Fields on Interface (cont.) k xi = + SW z CS z E E E E z 2 ( ) = + + SW z LW z RW z E E 0k SW k xr LW The contribution from the ESDP is called the space-wave field or the residual-wave (RW) field. ESDP (It is similar to the lateral wave in the half-space problem.) 23

Asymptotic Evaluation of Residual-Wave Field ( ) = = jk x RW z ( 0) E F k e dk y x x x EDSP k xi Use = k dk k js 0 x = jds 0k x k xr s - + 24

Asymptotic Evaluation of Residual-Wave Field (cont.) + ( ) + = jk x RW z sx E je F k js e ds 0 0 0 ( ) 0 ( ) + jk x sx je F k js e ds 0 0 + ( ) + = jk x RW z sx E je F k js e ds 0 0 0 ( ) ( ) jk x sx je F k js e ds 0 0 0 ( ) ( ) s ( ) s + H s F F Define 25

Asymptotic Evaluation of Residual-Wave Field (cont.) Then ( ) = jk x RW sx z E je H s e ds 0 0 = x x for ( ) ~ H s 0 As s Assume as We then have ( ) + + 1 A Watson s lemma (alternative form): jk x RW z ~ E je 0 x 1 26

Asymptotic Evaluation of Residual-Wave Field (cont.) It turns out that for the line-source problem at an interface, = 1/ 2 Hence jk x 3 2 e 0 RW z ~ E j A 3/2 x Note that the wavenumber is that of free space. jk e 0 RW z E A Note: For a dipole source we have 1 2 27

Discussion of Asymptotic Methods We have now seen two ways to asymptotically evaluate the fields on an interface as x for a line source on a grounded substrate: 1) Steepest-descent ( ) plane There are no branch points in the steepest-descent plane. The function f ( ) is analytic at the saddle point 0= = /2, but is zero there. The fields on the interface correspond to a higher- order saddle-point evaluation. 2) Wavenumber (kx) plane The SDP becomes an integration along a vertical path that descends from the branch point at kx = k0. The integrand is not analytic at the endpoint of integration (branch point) since there is a square-root behavior at the branch point. Watson s lemma is used to asymptotically evaluate the integral. 28

Summary of Waves y Continuous spectrum LW SW RW x jk x e 0 SW x e = jk x RW z SW z LW x ~ E A e E A = jk x LW z E A RW SW 3/2 LW x = SW SW xk = LW LW LW xk j 29

Interpretation of RW Field The residual-wave (RW) field is actually a sum of lateral-wave fields. y x c 30

Appendix: Proof of Angle Property Proof of angle property: = ( ) k k k k 0 = xr xr tan k xi xi ( ) 2 tan r i = = sin cos cosh sinh k k k k The last identity follows from k k 0 xr r i xr tan r xi 0 xi r i ~ ~ Hence or r r 31

Proof (cont.) i On SDP: u 0 u 0 SDP As u = 0 i 2 2 r + r + 2 2 2 (the asymptote) u SAP 0 u 0 + + ~ ~ Hence or 2 2 = 2 32

Proof (cont.) To see which choice is correct: = ESDP: 2 = 0 In the kx plane, this corresponds to a vertical line for which = Hence 2 33