Waveguides. Parallel plate waveguide Rectangular Waveguides Circular Waveguide Dielectric Waveguide

Transcription

1 Waveguides Parallel plate waveguide Retangular Waveguides Cirular Waveguide Dieletri Waveguide 1

2 Waveguides In the previous hapters, a pair of ond utors was used to guide eletromagne ti wave propagation. This propagatio n was via the transverse eletromagnet i (TEM) mode, meaning both the ele tri and magneti field omponents we re transverse, or perpendiular, to the diretion of propagation. In this hapter we investigate wave-gu iding strutures that support propagati on in non-tem modes, namely in the t ransverse eletri (TE) and transverse magneti (TM) modes. In general, the term waveguide refers t o onstruts that only support non-te M mode propagation. Suh onstruts share an important trait: they are unabl e to support wave propagation below a ertain frequeny, termed the utoff fre queny. Retangular waveguide Dieletri Waveguide Cirular waveguide Optial Fiber

3 Uses To redue attenuation loss High frequenies High power Can operate only above ertain frequenies Ats as a High-pass filter Normally irular or retangular We will assume lossless retangular 3

4 Parallel Plate Waveguide Figure 3. (p. 98) Geometry of a parallel plate waveguide. 4

5 The simplest type of guide that an support TM and TE modes; an also support a TEM mode. TEM Modes TEM mode solution an be obtained by solving Laplae s equation. t Φ ( xy, ) = 0 for 0 x W,0 y d Assume Φ ( x,0) = 0, Φ ( x, d) = V0 Sine there is no variation in x, Φ ( x, y) = A+ By Φ ( x, y) = V0 y / d The transverse E-field from (3.13), e( x, y) = Φ ( x, y) = yv ˆ / d t 0 jk 0 E x y = e x y e = y e (,, ) (, ) ˆ V d jk 5

6 1 V (,, ) ˆ (,, ) ˆ η ηd 0 jk H x y = E x y = x e The voltage of the top plate with respet to the bottom plate d = y = y= 0 V E dy Ve 0 jk The total urrent on the top plate wv ηd w 0 垐 w ( ˆ ) w I = Js dx = y H dx = Hxdx = e x= 0 x= 0 x= 0 Z 0 v p V ηd = = I w ω 1 = = β με jk 6

7 TM Modes H = 0, E satisfies (3.5) with + k (, ) 0 e x y = y e ( x, y) = Asink y + Bosk y / x = 0 B.C. B = 0 and k d = nπ e ( x, y) = 0 at y = 0, d β = k k = k ( nπ / d) nπ y nπ y e( x, y) = Ansin, E( x, y, ) = Ansin e d d jβ 7

8 jωε nπy jβ jβ nπy jβ Hx = Anos e, Ey = Anos e, k d k d E x y = H = 0 TM 0 mode = TEM mode The utoff frequeny f f k n = = π με d με TM 1 mode is the lowest TM mode with a utoff frequeny f 1 = d με 8

9 At frequenies below the utoff frequeny of a given mode, the propagation onstant is purely imaginary, orresponding to a rapid exponential deay of the fields. utoff or evanesent modes. TM n mode propagation is analogous to a high-pass filter response. The wave impedane Ey β βη ZTM = = = H ωε k pure real for f > f, pure imaginary for f < f. The guide wavelength is defined π the distane between equiphase λg = β planes along the -axis. x 9

10 λ g > λ = π/k, the wavelength of a plane wave in the material. The phase veloity and guide wavelength are defined only for a propagation mode, for whih β is real. A utoff wavelength for the TM n mode may be defined as λ = Poynting vetor d n 1 w d 1 w d Po = Re E H dydx ˆ = Re EyH xdydx x= 0 y= 0 x= 0 y= 0 wre( βωε ) nπy = dy k d d A os n y= 0 10

11 wre( βωε ) d A n > = 4k wre( βωε ) d A n k = n for 0 n for 0 Consider the dominant TM 1 mode, whih has a propagation onstant, β 1 = k ( ) π / d, π y jβ A1 E = A1 sin e = e e d j { j[ πy/ d β } 1] j[ πy/ d β1] plane waves traveling obliquely in the y, + and +y, + diretions. 11

12 k π sin θ =, kosθ = β1 d π + β = d 1 k f f : β 1 0: plane waves up and down, no real power flow. Figure 3.3 (p. 10) Bouning plane wave interpretation of the TM 1 parallel plate waveguide mode. 1

13 Condutor loss an be treated using the perturbation method. Pl α = Po where P o : the power flow down the guide in the absene of ondutor loss given by (3.54), P l : the power dissipated per unit length in the lossy ondutors R w s ωε Rsw Pl = Js dx= A 0 n x= k ωε Rs krs α = = for n > 0 βd βηd Rs = for n = 0 ηd 13

14 TE Modes E = 0, H satisfies (3.1) with + k (, ) 0 h x y = y h ( x, y) = Asink y + Bosk y / x = 0 B.C. A = 0 and kd = nπ e (, ) 0 x x y = at y = 0, d jωμ jβ Ex( x, y, ) = [ Aosky Bsin ky] e, k nπ y H( x, y) = Bnos e d jβ 14

15 jωμ nπy jβ jβ nπy jβ Ex = Bnsin e, Hy = Bnsin e, k d k d E y = H = x 0 The utoff frequeny f f k n = = π με d με The wave impedane Z TE E H x = = = y ωμ β β kη 15

16 1 w d 1 w d P = Re o E ˆ Re H dydx = ExHydydx x= 0 y= 0 x= 0 y= 0 wre( βωμ ) = Bn for n> 0 4k If n = 0, E x = H y = 0, P 0 = 0 no TE 0 mode. α kr s kr s = = ωμβ d kβηd 16

17 Attenuation due to ondutor loss for the TEM, TM, and TE 1 modes of a parallel plate waveguide. 17

18 Figure 3.5 (p. 106) Field lines for the (a) TEM, (b) TM1, and () TE 1 modes of a parallel plate waveguide. There is no variation aross the width of the waveguide. 18

19 Retangular Waveguide TE Modes E = 0 H must satisfy the redued wave equation (3.1) x y + + k (, ) 0 h x y = Can be solved by separation of variables h ( x, y) = X( x) Y( y) 1 d X 1 d Y + + k 0 = X dx Y dy 19

20 Figure 3.6 (p. 107) Photograph of Ka-band (WR-8) retangular waveguide omponents. Clokwise from top: a variable attenuator, and E- H (magi) tee juntion, a diretional oupler, an adaptor to ridge waveguide, an E-plane swept bend, an adjustable short, and a sliding mathed load. Courtesy of Agilent Tehnologies, Santa Rosa, CA 0

21 Geometry of a retangular waveguide. 1

22 We define separation onstant k x and k y d X dx d Y dy + k 0, 0, xx = + k yy = kx + ky = k h ( x, y) = ( Aos k x+ Bsin k x)( Cos k y + Bsin k y) x x y y Boundary onditions e ( x, y) = 0 at y = 0, b x e ( x, y) = 0 at x= 0, a y Using (3.19) and (3.19d) jωμ ex = k ( os sin )( sin os ) y A kxx+ B kxx C kyy+ D kyy k jωμ e = k ( Asink x+ Bos k x)( Cosk y+ Dsin k y) y x x x y y k

23 From B.C, D = 0, and k y = nπ/b, B = 0, and k x = mπ/a mπ x nπ y jβ H( x, y, ) = Amnos os e a b The transverse field omponents of TE mn mode jωμnπ mπx nπy jβ Ex = A os sin, mn e kb a b jωμmπ mπ x nπ y jβ Ey = Amnsin os e, ka a b jβmπ mπx nπy jβ Hx = Amnsin os e, ka a b jβnπ mπx nπy jβ Hy = Amnos sin e. kb a b 3

24 mπ nπ β = k k = k a b is real when k mπ nπ > k = + a b f mn k 1 mπ nπ = = + π με π με a b The mode with the lowest utoff frequeny is alled the dominant mode; f 10 = 1 a με 4

25 For f < f, all field omponents will deay exponentially utoff or evanesent modes If more than one mode is propagating, the waveguide is overmoded. The wave impedane E E x y kη ZTE = = = H H β y x The guide wavelength (λ: the wavelength of a plane wave in the filling medium) π π λ = g λ β > k = ω ω v 1/ p = > = με β k 5

26 For the TE 10 mode π x jβ H = A10 os e, a jωμa π x jβ Ey = A10 sin e, π a jβa πx jβ Hx = A10 sin e, π a E = E = H = 0. x y k = π / a, β = k ( π / a) 6

27 The power flow down the guide for the TE 10 mode: 1 a b 1 a b Po = Re E H dydx ˆ = Re EyHxdydx x= 0 y= 0 x= 0 y= 0 ωμa Re( β ) A10 a b π x = sin dydx π x= 0 y= 0 a 3 ωμabre( β ) A10 = 4π Attenuation an our beause of dieletri loss or ondutor loss. Rs Pl = Js dl C 7

28 There are surfae urrents on all 4 walls. The surfae urrent on the x = 0 wall is J = n j 垐 H = x H ˆ = yh 垐 = ya e β s x= 0 x= 0 x= 0 The surfae urrent on the y = 0 wall is ( ) J = n垐 H = y xh + H ˆ s y= 0 x y= 0 y= 0 jβa πx πx = ˆ A sin e + xa ˆ os e π a a jβ jβ b a l = s sy + s sx + s y= 0 x= 0 P R J dy R J J dx a β a = Rs A10 b+ + π

29 The attenuation due to ondutor loss for TE 10 mode Pl π R a β a α = = + + P10 ωμa bβ π Rs 3 = ( bπ + a k ) Np/ m 3 abβkη 3 s 3 b 9

30 TM Modes H = 0 x y + + k (, ) 0 e x y = e ( x, y) = ( Aos k x+ Bsin k x)( Cos k y + Dsin k y) x x y y BC: e (x,y) = x = 0, a and y = 0, b A = 0, k = mπ / a x C = 0, k = nπ / b y mπ x nπ y E( x, y, ) = Bmnsin sin e a b jβ 30

31 jβmπ mπx nπy jβ Ex = Bmnos sin e, ak a b jβnπ mπx nπy jβ Ey = Bmnsin os e, bk a b jωεnπ mπ x nπ y jβ Hx = Bmnsin os e, bk a b jωεmπ mπ x nπ y Hx = Bmnos sin e ak a b mπ nπ β = k k = k a b jβ f 11 1 π π E E x y = +, Z βη TM = = = π με a b H H k y x 31

32 Attenuation of various modes in a retangular brass waveguide with a =.0 m. 3

33 Field lines for some of the lower order modes of a retangular waveguide. Reprinted from Fields and Waves in Communiation Eletronis, Ramo et al, Wiley, 1965) 33

34 Ex a = 1.07 m, b = 0.43 m, f = 15 GH Solution: for Teflon ε r =.08, tan δ = f mn mπ nπ = + π ε a b r Mode m n f (GH) TE TE TE TE, TM TE, TM

35 At 15 GH β α d π = k = s 3 ab k a k tanδ = = np / m = 1.03 db / m β ωμ0 Rs = = 0.03 σ R 3 α = ( bπ + a k ) = np/ m= db/ m β η 35

36 Figure 3.10 (p. 115) Geometry of a partially loaded retangular waveguide. 36

37 Figure on page 117 Referene: Montgomery, et al., Priniples of Mirowave Ciruits, MGraw-Hill, 1948) 37

38 Cirular Waveguide Geometry of a irular waveguide. 38

39 E H ρ ρ j E ωμ H j β E H = β, E + φ = ωμ k ρ ρ φ k ρ φ ρ j ωε E H j E β H = β, H φ = ωε + k ρ φ ρ k ρ ρ φ TE Modes H + k H = E = k (, ) 0 h ρφ = ρ ρ ρ ρ φ h ( ρ, φ) = R( ρ) P( φ) 0 39

40 d R dr d P k = Rdρ ρrdρ ρ Pdφ ρ d R ρ dr 1 d P + + ρ k = R dρ R dρ P dφ 0, 1 d P d P k k P 0 φ φ Pdφ = dφ + = d R dr ρ ρ ρ dρ dρ ( + + k ) 0 kφ R = The general solution is P( φ) = Asink φ + Bosk φ φ φ 40

41 Sine h ( ρ, φ) = h ( ρ, φ± nπ) P( φ) = Asinnφ+ Bosnφ ρ d R dr dρ ρ dρ ρ ( + + k ) 0 n R = R( ρ) = CJ ( k ρ) + DY ( k ρ) But, D = 0 n n h ( ρ, φ) = Asinnφ+ Bos nφ J ( k ρ) BC: Sine E = 0, Eφ ( ρ, φ) = 0 at ρ = ( ) n a jωμ E (,, ) ( sin os ) φ ρφ = A nφ+ B nφ J n( kρ) e k jβ 41

42 J ( k a) = 0 n If the roots of J n ' (x) are defined as p' nm, so that J n '(p' nm ) = 0, where p' nm is the mth root of J n ', then k must have the value. p nm k = a See Table 3.3 The Te mn modes are defined by the utoff wavenumber, k mn = p' nm /a, where n refers to the number of irumferential (φ) variations, and m refers to the number of radial (ρ) variations. 4

43 β p nm k p nm nm = k k = k, fnm = = TE 11 mode: dominant mode a π με π a με jωμn E ( os sin ) ρ = A nφ B nφ Jn( kρ) e k ρ jωμ E ( sin os ) φ = A nφ+ B nφ J n( kρ) e k jβ H ( sin os ) ρ = A nφ+ B nφ J n( kρ) e k jβn H ( os sin ) φ = A nφ B nφ Jn( kρ) e k ρ jβ jβ jβ jβ 43

44 The wave impedane Z TE Eρ Eφ ηk = = = H H β φ ρ Beause of the aimuthal symmetry of the irular waveguide, both sin nφ and os nφ are terms are valid solutions, and an be present in a speifi problem to any degree. The atual amplitudes of these terms will be dependent on the exitation of the waveguide. Consider the dominant TE 11 mode with an exitation suh that B = 0. 44

45 H = Asin φj ( k ρ) e, E = 0 jβ 1 jωμn jωμ Eρ = Aos φj ( k ρ) e, E = Asin nφj ( k ρ) e k ρ jβ jβ 1 φ 1 k jβ jβn Hρ = Asin φj ( k ρ) e, H = Aos nφj ( k ρ) e k The power flow down the guide jβ jβ 1 φ 1 kρ 45

46 1 a π Po = Re E H ˆ d d ρ ρ φ ρ= 0 φ= 0 1 a π Re [ ρ φ φ ρ ] ρ= 0 φ= 0 = E H E H dydx ρd ρdφ ωμ A Re( β ) a π 1 = 4 os J1 ( k ) k sin J1 ( k ) d d k φ ρ φ ρ ρ ρ φ ρ= 0 φ= 0 + ρ ωμ A Re( β ) a 1 = J 4 1 ( kρ) ρkj1 ( kρ) dρ k + ρ = 0 ρ ωμ A β ) = 4Re( ( p 11 1) J1 ( ka ) 4k 46

47 R φ R π s = Hφ + H adφ φ = 0 π s Pl = Js ad φ = 0 A R π s β = os φ sin φ J 4 1 ( kaad ) φ φ = 0 + ka A Rsa β = 1 + J 4 1 ( ka) ka α P R ( k a + β ) R k = = = ( + ) P k a( p 1) k a p 1 4 l s s k o η β 11 η β 11 47

48 TM Modes k (, ) 0 e ρφ = ρ ρ ρ ρ φ e ( ρ, φ) = Asinnφ+ Bos nφ J ( k ρ) E ( ) n ( ρ, φ) = 0 at ρ = a Jn( ka ) = 0 pnm k = a β pnm k pnm nm = k k = k, fnm = = a π με π a με 48

49 jβ E ( sin os ) ρ = A nφ+ B nφ J n( kρ) e k jβ n E ( os sin ) φ = A nφ B nφ Jn( kρ) e k ρ jωεn H ( os sin ) ρ = A nφ B nφ Jn( kρ) e k ρ jωε H ( sin os ) φ = A nφ+ B nφ J n( kρ) e k Z TM Eρ Eφ βη = = = H H k φ ρ jβ jβ jβ jβ 49

50 Attenuation of various modes in a irular opper waveguide with a =.54 m. 50

51 Ex Cutoff frequenies of the first few TE and TM modes of a irular waveguide, relative to the utoff frequeny of the dominant TE 11 mode. 51

52 Field lines for some of the lower order modes of a irular waveguide. Reprinted from Fields and Waves in Communiation Eletronis, Ramo et al, Wiley, 1965) 5