ECE Spring Prof. David R. Jackson ECE Dept. Notes 33

Transcription

1 ECE 6341 Spring 16 Prof. David R. Jackson ECE Dept. Notes 33 1

2 Complex Integral: Steepest-Descent Method Ω g z I f z e dz Ω = C This is an extension of Laplace s method to treat integrals in the complex plane. The method was published b Peter Debe in 199. Debe noted in his work that the method was developed in a unpublished note b Bernhard Riemann (1863). Peter Joseph William Debe (March 4, 1884 November, 1966) was a Dutch phsicist and phsical chemist, and Nobel laureate in Chemistr. Georg Friedrich Bernhard Riemann (September 17, 186 Jul, 1866) was an influential German mathematician who made lasting contributions to analsis and differential geometr, some of them enabling the later development of general relativit.

3 Complex Integral: Ω g z I f z e dz Ω = C The functions f (z) and g(z) are analtic (except for poles or branch points), so that the path C ma be deformed if necessar (possibl adding residue contributions or branch-cut integrals). Saddle Point (SP): g ( z ) = g = x g = 3

4 Path deformation: If the path does not go through a saddle point, we assume that it can be deformed to do so. If an singularities are encountered during the path deformation, the must be accounted for (e.g., residue of captured poles). C = + z C C' C p z p X C p = π Res ( p ) F z dz j F z C p x C 4

5 Denote g( z) = u( z) + jv( z) Cauch Reimann eqs.: u x u v = v = x Hence u u v v = = = x x x x x u = 5

6 or u x u + = If u xx <, then u > Near the saddle point: 1 u( x, ) u( x, ) + uxx ( x x) 1 + u + u x x x Ignore (rotate coordinates to eliminate). 6

7 1 1 u x u x u x x u u x x (, ) (, ) + ( ) + ( ) + ( )( ) xx x In the rotated coordinate sstem: 1 1 u x u x u x x u (, ) (, ) + ( ) + ( ) xx Assume that the coordinate sstem is rotated so that u < u > xx 7

8 The u (x, ) function has a saddle shape near the saddle point: ux (, ) z x 8

9 ux (, ) x x z Note: The saddle does not necessaril open along one of the principal axes (onl when u x (x, ) = ). 9

10 Along an descending path (where the u function decreases): or Ω Ω g( z) g( z ) I ~ f z e e dz Ω g z C jω vz vz Ω uz uz I Ω ~ f z e e e dz Ω g z C Behaves like a delta function Both the phase and amplitude change along an arbitrar descending path C. If we can find a path along which the phase does not change (v is constant), then the integral will look like that in Laplace s method. 1

11 Choose path of constant phase: C : v z = v z = constant z C x 11

12 Gradient Propert (proof on next slide): ux, C is parallel to Hence C is either a path of steepest descent () or a path of steepest ascent (SAP). (Of course, we want to choose the.) : u(x,) decreases as fast as possible along the path awa from the saddle point. SAP: u(x,) increases as fast as possible along the path awa from the saddle point. 1

13 Proof of gradient propert g( z) = u( z) + jv( z) u u = xˆ + ˆ x v v = xˆ + ˆ x u v Hence, u v u v u v u v= + x x u u u u = + x x = v C Also, v C v ( is constant on C ) u x Hence u C 13

14 Recall: jω vz vz Ω uz uz I Ω ~ f z e e e dz Ω g z C Because the v function is constant along the, we have or Ω u( z) u( z ) I ~ f z e e dz Ω Ω Ω g z Ω g( z) g( z ) I ~ f z e e dz Ω g z This is real on the. 14

15 Local behavior near the saddle point 1 g z g z g z z z g z z z + ( ) + ( ) so 1 g z g z g z z z ( ) Denote g z = Re z z = re jφ jα z r φ x Then we have 1 ( + ) j g z g z Rr e α φ 15

16 1 g z g z Rr e α φ j( ) + 1 u z u z Rr α φ cos( + ) SAP: : α + φ = + πn α α φ =, + π α + φ = π + πn φ = α π, α π + π 1 uz uz ( ) = Rr vz vz = 1 uz uz ( ) = Rr vz vz = Note: The two paths are 9 o apart at the saddle point. 16

17 SAP SP u 9 o u increases u u decreases x 17

18 ux (, ) x z SAP 18

19 Ω Ω u( z) u( z ) I f z e e dz ~ Ωg z Define s u( z ) u( z) This defines z= zs () ux (, ) s x s > s < z z 19

20 Ωg ( z + ) Ωs dz Ω ~ I f z e e ds ds ~ dz Ω s Ω g z Ω ds s= I f z e e ds + (This gives the leading term of the asmptotic expansion.) Hence ~ I f z e Ωg z dz ds Ω s= π Ω

21 To evaluate the derivative: = ( ) g( z ) s u z u z = g z (Recall: v is constant along.) ds s = g z dz At the saddle point this gives =. Take one more derivative: ds ds d s s = g z dz dz dz 1

22 s = ds g ( z ) At : dz = Z 1/ so dz = ds g z s= 1/ Hence, we have 1/ Ωg( z ) π I( Ω) ~ f ( z ) e g ( z ) Ω

23 Note: There is an ambiguit in sign for the square root: dz = ds g z s= 1/ To avoid this ambiguit, define z θ arg dz ds s= = θ x 3

24 The derivative term is therefore dz = ds g z s= jθ e Hence π I Ω f z e e Ω g z Ω g z jθ ~ 4

25 To find θ : Denote: g z = Re z z = re jα jθ α = arg g z 1 1 g z g z g z z z Rr e α θ ( ) j( + ) = α + θ =± π θ α π = ± 5

26 θ α π θ = ± α = arg g z SAP x Note: The direction of integration determines The sign. The user must determine this. 6

27 Summar π I Ω f z e e Ω g z Ω g z jθ ~ α π θ = ± Ω g z I f z e dz Ω = C ( ) α = arg g z 7

28 Example J 1 Ω = π π / π / cos Ωcos z dz 1 = Re I Ω π where π / π / ( jcos z) Ω I e dz Ω = Hence, we identif: f ( z) = 1 g z = jcos z g z = jsin z = z = ± nπ 8

29 Example (cont.) π z = π x g( z) = jcos z g z = jcos z = j π α = arg g ( z ) = α π π π θ = ± = ± 4 π 3π θ = or 4 4 9

30 Example (cont.) Identif the and SAP: = cos( + ) = j[ cos xcosh jsin xsinh ] g z j x j ux, = sin xsinh vx, = cos xcosh and SAP: v( z) = v( z ) = constant cos xcosh = constant = cos cosh = 1 3

31 Example (cont.) and SAP: cos xcosh = 1 SAP Examination of the u function reveals which of the two paths is the. π C π x ux, = sin xsinh 31

32 Example (cont.) Vertical paths are added so that the path now has limits at infinit. = C+ Cv + Cv 1 It is now clear which choice is correct for the departure angle: θ π = 4 C v1 π π C C v x ( Ω ) = v1 v I I I I 3

33 Example (cont.) Ω g( z ) π I f ( z ) e e Ω g jθ ~ ( z ) This is the answer for I(Ω) if we ignore the contributions from the vertical paths. Hence, π I( Ω) ~ ( 1) e e Ω j π Ω j cos( j ) 4 or I so 4 ~ e Ω J π π j Ω Ω π 1 π j Ω Ω π Ω 4 ~ Re e 33

34 Example (cont.) Hence J ( Ω) π ~ cos Ω π Ω 4 34

35 Example (cont.) Examine the path C v1 (the path C v is similar). C v1 π π x I v1 π / = π / + j e Ω ( jcos z) dz Let z π = + j 35

36 Example (cont.) v1 π jωcos + j Ωsinh Ωsinh I = j e d = j e d = j e d since π cos + j = sin( j) = j sinh Use integration b parts (we can also use Watson s Lemma): 1 d ( Ω sinh I = j e ) d v1 Ωcosh d 1 j Ωcosh ( Ω sinh e ) 1 I j v1 Ω 36

37 Example (cont.) Hence, 1 I = O v1 Ω Note: I v1 is negligible compared to the saddle-point contribution as Ω. However, if we want an asmptotic expansion that is accurate to order 1/ Ω, then the vertical paths must be considered. 37

38 Example (cont.) Alternative evaluation of I v1 using Watson s Lemma (alternative form): v1 Ω sinh = Use s = sinh ( ) I j e d s d Ω I = j e ds v1 ds ds = cosh d ( ) = = = Ωs 1 j e ds cosh ( ) Ωs 1 j e ds 1+ sinh Ωs 1 j e ds 1+ s ( ) Appling Watson s Lemma: so 1 1+ s ( 1) s 1 + Ωs Ωs s I j e ds j e ds v1 + 38

39 Example (cont.) Ωs Ωs s ( 1) I j e ds j e ds v1 + so Ωs j Ωs I j e ds + e s ds + v1 Recall: αn s e Ωs ds 1 = Γ αn + 1 n + Ω ( α 1) Hence, I x 1 ( 1! ) t x x t e dt Γ = ( 3) 1 j Γ j + + Ω Ω v1 3 39

40 Complete Asmptotic Expansion B using Watson's lemma, we can obtain the complete asmptotic expansion of the integral in the steepest-descent method, exactl as we did in Laplace's method. Define: b Ωg( z) I Ω = f z e dz a s g z g z dz s h( s) f ( z( s) ) ds Assume: Then we have n h s ~ as as s n=,1,... g( z ) a n + 1 Ω Γ ( 1) n=,,4 Ω n I ~ e Ω n+ n 4