ECE Notes 4 Functions of a Complex Variable as Mappings. Fall 2017 David R. Jackson. Notes are adapted from D. R. Wilton, Dept.

Transcription

1 ECE 638 Fall 017 Daid R. Jackson Notes 4 Fnctions of a Comple Variable as Mappings Notes are adapted from D. R. Wilton, Dept. of ECE 1

2 A Fnction of a Comple Variable as a Mapping A fnction of a comple ariable, f, is sall ieed as a mapping from the comple plane to the comple plane. f For eample, 3

3 Simple Mappings: Translations Translation: A+ here A is a comple constant. A + A The mapping translates eer point in the b the "ector" A. plane 3

4 Rotation: Simple Mappings: Rotations i( + ) iα iα iθ e e re re α θ here α is a real constant. iα e α The mapping rotates eer point in the plane throgh an angle α. 4

5 Simple Mappings: Dilations Dilation (stretching) : ( iθ ) a a re ar e here a is a real constant. iθ Note: a, a d a d d a d a The mapping magnifies the magnitde of a point in the comple plane b a factor a. 5

6 A General Linear Transformation (Mapping) is a Combination of Translation, Rotation, and Dilation Linear transformation: dilation ( Arg ) i B iθ θ + A + B A + B e re A + B r e here A,B are comple constants. rotation Arg i B translation A+ B Arg B B A Shapes do not change nder a linear transformation! 6

7 Simple Mappings: Inersions Inersion: Im iθ e iθ re r Im 0+ i Re 1 1 Re The straight line 0 + i maps to a circle Inersion circle -presering propert Inersions hae a "circle presering" propert, i.e., circles alas map to circles (straight lines are "circles" of infinite radis). Points otside the nit circle map to points inside the nit circle and ice ersa. 7

8 Circle Propert of Inersion Mapping 1 (This maps circles into circles.) i, + i a Consider a circle: This is in the form + + a+ a + a Hence + + a 1 + a + a J. W. Bron and R. V. Chrchill, Comple Variables and Applications, 9 th Ed., McGra-Hill,

9 Circle Propert of Inersion Mapping (cont.) + + a 1 + a + a Mltipl b + : + a a + a or 1+ a + a + a This is in the form of a circle: b 9

10 Simple Mappings: Inersions (cont.) Geometrical constrction of the inersion : iθ e iθ re r θ 1 1 θ θ Shapes are not presered! Note the circlar bondaries for the region! 10

11 A General Bilinear Transformation (Mapping) Is a Sccession of Translations, Rotations, Dilations, and Inersions Bilinear (Fractional or Mobis) transformation: here A,B,C,D are comple constants. A + B C + D Note that if D 0, A + B A BC D + B D C + D A BC D + C + D C + D C + D BD If e let f C + D, and g ( ξ ) ξ ( A BC D) (translation, dilation, rotation) (inersion, dilation, rotation) then the Mobis transformation ma be ritten as A + B BD+ g( f ) C + D hich is jst a seqence of translations, dilations, rotations, and an inersion. Since each transformation preseres circles, bilinear transformations also hae the circle - presering propert : circles in the plane are mapped into circles in the plane (ith straight lines thoght of as circles of infinite radis). 11

12 Bilinear Transformation Eample: The Smith Chart ( ) ( ) Z Let r + j here Z ( ) R ( ) + jx ( ) is the impedance at - on a Z 0 transmission line of characteristic impedance Z, and Γ is the generalied reflection coefficient : Γ ( ) ( ) 0 ( ) ( ) 0 ( ) Z Z0 1 1 or simpl Γ Z + Z Γ Im Γ Γ r ReΓ Horiontal and ertical ines (contant reactance and resistance) are mapped into circles. For an interpretation of Möbis transformations as projections on a sphere, see 1

13 The Sqaring Transformation i f r e θ ( θ, θ ± π) ( θ, θ ± π) θ 0 The transformation maps half the - plane into the entire - plane The entire - plane coers the - plane tice The transformation is said to be man - to - one (in this case, to - to - one) 13

14 Another Representation of the Sqaring Transformation 70 o 360 o 180 o i f r e θ o o 90 9 o o -180 o -90 o Constant amplitde and phase contors of f 14

15 The Sqare Root Transformation 1 / θ+ πk f re, k 01, i Note: The ale of 1/ on one branch is the negatie of the ale on the other branch. Second branch π < θ π k 1 Principal branch π < θ π k 0 The principal branch is the choice in MATLAB and most programming langages! Re 0 We sa that there are to branches (i.e., ales) of the sqare root fnction. Note that for a gien branch (e.g., the principal branch), the sqare root fnction is not continos on the negatie real ais. (There is a branch ct there.) 15

16 The Sqare Root Transformation (cont.) 1 / 1 / θ ( + kπ ) 01 π < θ π f r e,k,, i 67.5 o 90 o 45 o o o.5 o o 70 o 5 o o o o o -45 o -.5 o 11.5 o 135 o o Top sheet, k 0 Bottom sheet, k 1 16

17 Constant and Contors are Orthogonal and Consider contors in the plane on hich the real qantities,, are constant. The directions normal to these contors are along the gradient direction : ˆ + ˆ ˆ + ˆ constant constant + i f (analtic) The gradients, and therefore the contors, are orthogonal (perpendiclar) b the C. R. conditions : C.R. cond's ˆ + ˆ ˆ + ˆ ˆ + ˆ ˆ + ˆ

18 Constant and Contors are Orthogonal (cont.) Eample: ( ) ( ) + i + i so constant: c constant: c 1 Also, recall that, 0, 0 18

19 Mappings of Analtic Fnctions are Conformal (Angle-Presering) Consider a pair of intersecting paths C, C onto the + i plane. 1 in the plane mapped C 1 γ γ 0 C f f ( 0 ) 0 Γ 1 β 0 β Γ arg arg arg arg arg arg arg( ) arg( 1) arg( ) arg( 1) f f +, along C from f f +, along C from Hence β γ 19

20 Constant and Contors are Orthogonal (Reisited) constant constant f Since the contors constant and constant are (obiosl) orthogonal in the plane, the mst remain orthogonal in the plane. Assmption : d d 0 0

21 Constant and arg() Contors are also Orthogonal iθ If Re the constant R and Θ contors are (obiosl) orthogonal. 1 If f is a mapping back to the plane, the mapping preseres the orthogonalit. Assmption : d d 0 Θconstant R constant 1

22 The Logarithm Fnction ln iθ e e i( θ+ πk ) ( θ π ) ln ln + i + k, k 0, ± 1, ±, There are an infinite nmber of branches (ales) for the ln fnction!

23 Arbitrar Poers of Comple Nmbers a (a ma be comple) Use e ln a ln ln aln + ai( θ+ πk ) aln θ a a ia i πak e e e e e e This has an infinite nmber of ales (branches), nless ak integer for some k, i.e., nless a is real and rational. 3

24 Arbitrar Poers of Comple Nmbers (cont.) Eample : f 3 / 3 ln i θ / e3 e 3 e iπ k 3 ln i θ / k 0 k 0 e e 3 ln i θ i / k 1 k e e e 3 3 ( π) ln i θ i / k k e e e 3 3 ( π) ( 6 3 ln i θ i π) ln i θ / starts k 3 k e3 e 3 e 3 e3 e 3 3 repeating! 8 k 4 k repeats! For p/q the repetition period is k q. For irrational poers, the repetition period is infinite; i.e., ales neer repeat! 4

25 Conformal Mapping This is a method for soling D problems inoling Laplace s eqation. φ, 0 (,) φ constant on C (Dirichlet bondar condition) φ (,) n 0 on C or (Nemann bondar condition) J. W. Bron and R. V. Chrchill, Comple Variables and Applications, 9 th Ed., McGra-Hill,

26 Conformal Mapping (cont.) f The fnction f () is assmed to be analtic in the region of interest. plane C A complicated bondar in the plane is mapped into a simple one in the plane -- for hich e kno ho to sole the Laplace eqation. plane φ (,) ψ (,) Γ φ (,) ψ (, ),(,) 6

27 Conformal Mapping (cont.) The ke to being sccessfl ith the method of conformal mapping is to find a mapping that orks for or problem (i.e., it maps or problem into one the is simple enogh for o to sole). J. W. Bron and R. V. Chrchill, Comple Variables and Applications, 9 th Ed., McGra-Hill, 013. An appendi has man basic conformal mappings. H. Kober, Dictionar of Conformal Representations, Admiralt, Mathematical and Statistical Section, Dept. of Phsical Research, A er thorogh compilation of conformal mappings. 7

28 Conformal Mapping (cont.) Theorem: If ψ (,) satisfies the Laplace eqation in the (,) plane, then φ (,) satisfies the Laplace eqation in the (,) plane. Proof: Assme that (,) ψ 0 or We ant to proe that ψ ψ + φ φ

29 Conformal Mapping (cont.) φ (,) ψ (, ),(,) φ ψ ψ + Using the chain rle: φ ψ ψ ψ ψ ψ ψ ψ + + ψ ψ ψ

30 Conformal Mapping (cont.) φ (,) ψ (, ),(,) φ ψ ψ + Using the chain rle: φ ψ ψ ψ ψ ψ ψ ψ + + ψ ψ ψ

31 Conformal Mapping (cont.) φ φ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ Use Cach-Riemann eqations (red, ble, and black terms): ψ satisfies Laplace s eqation. φ φ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ

32 Conformal Mapping (cont.) φ φ ψ ψ Hence, e hae φ φ + 0 Recall that for an analtic fnction f, 0, 0 (proof complete) 3

33 Conformal Mapping (cont.) Theorem that and are harmonic fnctions (reisited) Choose or (,) ψ (,) ψ so that φ (,) (,) or (,) + i f Note: The mapping fnction is arbitrar here, bt it is assmed to be analtic. Clearl, in either case e hae ψ ψ ψ (,) + 0 Hence, from or theorem, e hae (,) (,) (,) (,) φ 0 φ 0 The fnctions and, the real and imaginar parts of an analtic fnction, satisf Laplace s eqation (as e proed preiosl in Notes ). 33

34 Conformal Mapping (cont.) Eample (,) ψ (This satisfies Laplace s eqation.) Assme: f + i ( + i) (This is an analtic mapping fnction.) φ (,) ψ (,) (,) This mst satisf Laplace s eqation. 34

35 Conformal Mapping (cont.) Eample Assme: ψ (,) f e (This satisfies Laplace s eqation.) (This is an analtic mapping fnction.) φ ( cos sin ) + i i e e i + + e cos e sin (,) ψ (,) (,) (,) so (, ) e cos e sin e cos φ This mst satisf Laplace s eqation. 35

36 Conformal Mapping (cont.) Theorem: If ψ (,) satisfies Dirichlet or Nemann bondar conditions in the (,) plane, then φ (,) satisfies the same bondar conditions in the (,) plane. Proof: Assme that (,) ψ c on Γ Then e immediatel hae that φ (,) c on C since φ (,) ψ (, ),(,) 36

37 Conformal Mapping (cont.) Net, assme that ˆp mapping of n Γ ˆp γ ψ n φ l Γ 0 on Γ ˆp mapping of 0 in direction of ˆp n Γ Γ ˆn Γ π / C Becase of the angle-presering (conformal) propert of analtic fnctions, e hae: γ π / Hence, ˆp φ n nˆc 0 C on C 37

38 Conformal Mapping (cont.) Relation Beteen Charge Densities in the To Planes ρs D nˆc εe nˆc ε φ nˆc ε Φ n ψ ρs D nˆγ εe nˆγ ε ψ nˆγ ε n C Γ so ρ ρ s s n n Γ C Note: φ ψ ˆn C ˆn Γ C Γ 38

39 Conformal Mapping (cont.) Hence, e hae ρ ρ s s n n Γ C Note : f d f d d f d ρ ρ s s f C C n f n Γ ˆn C A ˆn Γ C Γ 39

40 Conformal Mapping (cont.) Theorem: The capacitance (per nit length) beteen to condctie objects remains nchanged beteen the and planes. Proof: C Q V A AB C Q V A AB VAB ΦA ΦB A B A B 40

41 Conformal Mapping (cont.) Q Q A A CA Γ A ρ dl ρ s s C dl Γ ρ ρ dl ρ dl 1 f 1 dlγ ρs dlγ f s C s C Therefore, Q Q C C A A s Note : f d f d d f d A C A ˆn C B A Γ A ˆn Γ B 41

42 Eample Sole for the potential inside of a coa and the capacitance per nit length of a coa. φ (,) φ( ρ) a b ( a) 1[ V] φ ( b) 0 [ V] φ 4

43 Eample (cont.) ln ( i θ ) ln + i ln re r + iθ ln ( r) θ π PMC a b PEC π PEC Assme : π < θ < π PMC ln ( a ) ln ( b) 43

44 Eample (cont.) ψ 0 ψ A + B Bondar conditions ψ ( b) ( ln ) lna lnb π PMC a b PEC π PEC Assme : π < θ < π PMC ln ( a ) ln ( b) 44

45 Eample (cont.) ψ ( b) ( ln ) lna lnb θ ln ( ρ ) (In the final anser e se ρ instead of r.) so φ ( ρ ) ln ( b) ( ln ) lna lnb or φ b ln ρ b ln a a b 45

46 Eample (cont.) C π π ε ε lnb lna b ln a so C πε b ln a [ F/m] π PMC a b PEC π PEC Assme : π < θ < π PMC ln ( a ) ln ( b) 46

47 Eample Sole for the potential inside and otside of a semi-infinite parallel-plate capacitor. φ 1 φ 1 π π E φ (,) Eqipotential contors 1 E φ (,) 47

48 Eample (cont.) φ 1 φ 1 ψ 1 ψ 1 ψ (,) φ (,) π π π π 1 e + e + + i + i e + + i e cos + e sin + e + ± π 48

49 Eample (cont.) π π 1 The corresponding colored dots sho the mapping along the top plate. e + Note: reaches a maimm at 0. π π

50 Eample (cont.) ψ 1 ψ 1 ψ (,) π π This is an ideal infinite parallel-plate capacitor, hose soltion is simple: ψ (,) π 50

51 Eample (cont.) φ 1 φ 1 φ (,) π π 1 The soltion is: φ 1 π (,) (,) here e cos + e sin + For an gien (,), these to eqations hae to be soled nmericall to find (,). 51

52 Eample Sole for the potential srronding a metal strip, and the srface charge densit on the strip. φ (,) φ 1 Note: The potential goes to - as ρ. 5

53 φ (,) Eample (cont.) φ R i Θ e + e Θ i i Re Θ 1 ψ (,) cosθ ψ 1 cosθ The otside of the circle gets mapped into the entire plane. (We don t need to consider the inside of the circle, since the points 0 and 1/ 0 get mapped to the same point.) 0 53

54 Eample (cont.) ψ (,) ψ ( R) 1 ψ 1 Otside the circle, e hae (from simple electrostatic theor): (,) 1 ln ( R) ψ R + Note: To be more general, e cold se (, ) A A ( R ) ψ 1 ln Changing the constant A 1 changes the oltage on the strip. Changing the constant A changes the total charge on the strip. 54

55 Eample (cont.) φ (,) φ φ 1 Hence, e hae: (,) R(,) φ 1 ln 1 + For an gien (,), these to eqations can be soled nmericall to find (R,Θ). iθ 1 + i Re + e R Θ i 1 RcosΘ+ cosθ R 1 RsinΘ sinθ R 55

56 Eample (cont.) Charge densit on strip φ (,) φ φ 1 On the pper srface of the strip, e hae: φ ρs D nˆ ε0e nˆ ε0 φ nˆ ε 0 ε0 φ Also, φ dφ d (along a ertical path aboe strip : d id) 56

57 Eample (cont.) Hence, ρ s ε 0 dφ d The charge densit is no in terms of the normal deriatie. dφ dψ d dψ f d d d d Hence, e hae: On the circle : dφ d d d dψ d f ( R, ) ψ ψ Θ R ( R, ) 1 ln ( R) ψ Θ Hence, on the circle : ψ 1 dψ 1 1 R R d Note: Going normal to the strip in the plane means going normal to the circle in the plane. d dr e Θ i 57

58 Eample (cont.) Hence, e hae: ρ ε s 0 f Net, se 1 + so or so d d f iθ 1 ( 1 i Θ ) f e Re 1 58

59 On the circle: so iθ ( 1 ) f e Eample (cont.) 1 ( 1 cos( ) isin ( )) Θ + Θ 1 (( 1 cos sin ) i ( sin cos )) 1 ( sin i ( sin cos )) Θ+ Θ + Θ Θ Θ+ Θ Θ sin ( sin cos ) f Θ+ i Θ Θ 1 1 ( sin cos ) sin Θ+ i Θ Θ 1 4 sin sin cos 1 Θ+ Θ Θ 1 1 sin Θ sin Θ+ cos Θ 1 1 sinθ 1 59

60 Eample (cont.) f 1 1 sinθ On the strip: so f 1/ 1 cosθ sinθ± 1 Hence, e hae ρ s ε 0 1/ 1 Note: The srface charge densit goes to infinit as e approach the edges. This reslt as first deried b Maell! 60

61 Eample (cont.) Knife-edge singlarit ρ s ρ s Here s is the distance from the edge. 1 s The strip no has a idth of, and the total line charge densit (sm of top and bottom line charge densities) is assmed to be ρ l [C/m]. ρ s / ρl / ρ s ρ d l 1 / π 1 top bot + ρ ρ ρ s s s Note: The normaliation of 1/π corresponds to a nit total line charge densit: / / 1 / π 1 d 1 61

62 Eample (cont.) Microstrip line Js h ε r J s I 1 / π 1 Note: The increased crrent densit near the edges cases increased condctor loss and ssceptibilit to dielectric breakdon. 6

63 Eample Find the capacitance beteen to ires. R ε ε 1 Radii and offset: R1 1 R / 1 + / 1 Hence, e hae: a a a + R 0 C πε 1 ln R0 1 (from Chrchill book) [F/m] J. W. Bron and R. V. Chrchill, Comple Variables and Applications, 9 th Ed., McGra-Hill,

64 Eample (cont.) 1 ε R 1 We therefore hae C ln πε [F/m] here R 1 + R 64

65 Eample (cont.) Smmetrical tin lead transmission line ε r Scale this geometr b 1/a: a h R 1 1 R 1 h/ a h/ a 1 h/ a+ 1 R 1 C πε ε ln h h h a a a 0 r [F/m] 65

66 Eample (cont.) Define: h a C ln πε ε 0 ( )( ) r [F/m] C ln πε ε r [F/m] 66

67 Eample (cont.) C ln πε ε r [F/m] C ln πε ε r [F/m] C ln πε ε r [F/m] 67

68 Eample (cont.) C ln πε ε r [F/m] C ln πε ε r [F/m] C πε ε 0 + ln 1 1 r [F/m] 1 Note : C πε ε 0 ln 1+ 1 r [F/m] 68

69 Eample (cont.) C πε ε 0 ln 1+ 1 r [F/m] C πε ε 0 + ln 1 r [F/m] Note : C πε ε cosh 0 r 1 [F/m] Note : + 1 cosh ln 1 69

70 Eample (cont.) Final Reslt a ε r h C πε0ε r 1 h cosh a [ F/m] Z 0 µε C Z 0 η 0 1 cosh 1 Ω π εr a h [ ] 70

71 Eample Condctor attenation on stripline 71

72 Eample (cont.) Conformal mapping of Bates: 1 / ( 1 p ) 1 ( / 1 ) ( 1 sn ) d dp k p k p a s pc+ C 1 0 s P ln q R sn Jacobi elliptic fnction P Weierstrass elliptic fnction R. H. T. Bates, The characteristic impedance of the shielded slab line, IRE Trans. Microae Theor and Techniqes, ol. MTT-4, pp. 8-33, Jan

73 Eample (cont.) b t ε r 73