Adapted from notes by ECE 5317-6351 Prof. Jeffery T. Williams Microwave Engineering Fall 2019 Prof. David R. Jackson Dept. of ECE Notes 9 Waveguiding Structures Part 4: Rectangular and Circular Waveguide 1 Rectangular Waveguide . One of the earliest waveguides. Still common for high power and low- loss microwave / millimeter-wave applications. It is essentially an electromagnetic pipe with a rectangular cross-section. Single conductor ⇒ No TEM mode For convenience: . a ≥ b (the long dimension lies along x). 2 Rectangular Waveguide (cont.) y Two types of modes: PEC TEz , TMz 221/2 b ε,, µσ kzc=( kk − ) x z a k=ω µεcr = kj0 ε(1 − tan δ d) σ εε= − j The cutoff wavenumber kc is real. c ω σ We need to solve for kc. =−−εε′jj ′′ ω =εεcc′ − j ′′ ffk>=−: kk22 cz c εc′′ =εc′ 1 − j 22 εc′ f< fcz: k =−− jk c k =εδcd′ (1 − j tan ) =εε0 rd(1 − j tan δ ) 3 TEz Modes For +z propagation: y − jkz z Hzz( xyz,,) = h( xye ,) PEC where b ε,, µσ ∂∂22 ++2 = x 22kcz h( xy,0) ∂∂xy z a 221/2 kcz≡−( kk) From previous field table: Subject to B.C.’s: − j ∂∂EHzz = +ωµ ∂H z Ekxz2 =⇒= kx∂∂ y Ex 00@yb= 0, c ∂y j ∂∂EH Ek=−+zzωµ yz2 ∂H z kyxc ∂∂ E =⇒=00@xa= 0, y ∂x 4 TEz Modes (cont.) ∂∂22 +=−2 22hz( xy,,) kcz h( xy) (eigenvalue problem) ∂∂xy Using separation of variables, let hz ( xy, ) = X( xY) ( y) d22 X dY ⇒Y +=− X k2 XY dx22 dy c Must be a constant This is the 22 11d X dY 2 (If we take one term across the equal sign, we “separation ⇒22 +=−kc have a function of x equal to a function of y.) equation”. X dx Y dy 11d22 X dY ⇒=−kk22and =− X dx22xyY dy 22 2 where kkkxyc+= “separation equation” 5 TEz Modes (cont.) Hence, Xx() Yy() hxyz( ,) =++ (cos A kxB xx sin kxC )(cos kyD y sin ky y ) ∂h z = 0 @yb= 0, A ∂y Boundary Conditions: ∂h z = 0 @xa= 0, B ∂x nπ A ⇒=D0 and kn = =0,1,2,... y b mπ B ⇒=B0 and km= =0,1,2,... x a 22 mxπ ny π 2 mππ n hz ( xy,) = Amn cos cos and kc = + a b ab 6 TEz Modes (cont.) Therefore, 1/2 k= kk22 − ππ zc( ) mn − jkz z Hz ( xyz, ,) = Amn cos x cos y e 221/2 mnππ ab =−−2 k ab 22 From the field table, we obtain the following: mnππ kc = + ab jnωµ π m π n π − = jkz z Ex 2 Amn cos x sin ye Note: kbc a b m = 0,1,2, jmωµ π m π n π − E= − Asin x cos yejkz z y 2 mn n = 0,1,2, kac a b jk mπ mnππ − = z jkz z But m = n = 0 Hx 2 Amn sin x cos ye kac a b is not allowed! jk nπ mnππ − = z jkz z Hy 2 Amn cos x sin ye (non-physical solution) kbc a b − jkz H= zAˆ 00 e;0 ∇⋅ H ≠ 7 TEz Modes (cont.) Reason for non-physical solution Start with the vector wave equation: 2 ∇×( ∇×H) − kH =0 Vector wave equation: from Maxwell’s equations. Take divergence of both sides. 2 ∇⋅( ∇×( ∇×HkH)) − ∇⋅ =0 The divergence of a curl is zero. ∇⋅H =0 Magnetic Gauss law 8 TEz Modes (cont.) Reason for non-physical solution Revisit how we obtained the vector Helmholtz equation: 2 ∇×( ∇×H) − kH =0 Vector wave equation: from Maxwell’s equations. 22 ∇( ∇⋅H) −∇ H − kH =0 From definition of vector Laplacian ∇×∇×H Now use: ∇⋅H =0 Magnetic Gauss law A needed assumption! 22 ∇+H kH =0 Vector Helmholtz equation (what we have solved) 9 TEz Modes (cont.) Reason for non-physical solution Vector wave equation ⇒ magnetic Gauss law Vector Helmholtz equation ⇒ magnetic Gauss law The vector Helmholtz equation does not guarantee that the magnetic Gauss law is satisfied. In the mathematical derivation, we need to assume the magnetic Gauss law in order to arrive at the vector Helmholtz equation. All of the modes that we get by solving the Helmholtz equation should be checked to make sure that they do satisfy the magnetic Gauss law. Note: The TE00 mode is the only one that violates the magnetic Gauss law. 10 TEz Modes (cont.) Lossless case (εc = εε = ′) 221/2 2 1/2 mnππ kmn =−=−− kk22mn k zc( ( ) ) ab ⇒ = mn mn TEmn mode is at cutoff when kkc (k = ωc µε ) 22 mn 1 mn fc = + 2 µε ab Lowest cutoff frequency is for TE10 mode (a > b) We will 10 1 fc = Dominant TE mode revisit this 2a µε (lowest f ) mode later. c 11 TEz Modes (cont.) At the cutoff frequency of the TE10 mode (lossless waveguide): cc c λ =dd = = d = d 10 2a ffc 1 2a µε so a = λd /2 ff= c 12 TEz Modes (cont.) To have propagation: ff> c so 1 f > 2a µε or 11c λ a >==dd 2 f µε 22f or λ a > d 2 Example: Air-filled waveguide, f = 10 GHz. We have that a > 3.0 cm / 2 = 1.5 cm. 13 TMz Modes Recall: y PEC − jkz z Ezz( xyz,,) = e( xye ,) b ε,, µσ x where z a ∂∂22 +=−2 22ez( xy,,) kecz( xy) (eigenvalue problem) ∂∂xy 221/2 kcz=( kk − ) Subject to B.C.’s: E z = 0 @xa= 0, @yb= 0, Thus, following same procedure as before, we have the following result: 14 TMz Modes (cont.) Xx() Yy() exyz( ,) =++ (cos A kxB xx sin kxC )(cos kyD y sin ky y ) Boundary Conditions: ez = 0 @yb= 0, A @xa= 0, B nπ A ⇒=C0 and kn= =0,1,2,... y b mπ B ⇒=A0 and kmx = =0,1,2,... a 22 mπ n π 2 mnππ eBz= mn sin x sin y and kc = + a b ab 15 TMz Modes (cont.) 1/2 =22 − Therefore kzc( kk) 221/2 mnππ − mnππ jkz z =−−k 2 Ez ( xyz, ,) = Bmn sin x sin y e ab ab 22 mnππ kc = + From the field table, we obtain the following: ab jnωεc π mnππ − jk z = = z m 1, 2, 3 Hx 2 Bmn sin x cos ye kbc a b n =1, 2, 3 jmωε π mnππ − = − c jkz z Hy 2 Bmn cos x sin ye kac a b Note: If either m or n is zero, the jk mπ mnππ − = − z jkz z field becomes a trivial one in Ex 2 Bmn cos x sin ye kac a b the TMz case. jk nπ mnππ − = z jkz z Ey 2 Bmn sin x cos ye kbc a b 16 TMz Modes (cont.) ′ Lossless case (εc = εε = ) (same as for 22 mn 22mn 2 mnππ TE modes) kzc=−=−− kk( ) k ab 22 mn 1 mn fc = + 2 µε ab The lowest cutoff frequency is obtained for the TM11 mode 22 11 111 fc = + 2 µε ab Dominant TM mode (lowest fc) 17 Mode Chart ′ y Lossless case (εc = εε = ) PEC Two cases are considered: b ε,, µσ b < a / 2 x Single mode operation z a f TE10 TE20 TE01 TE11 TM11 The maximum bandwidth for single-mode operation is 67%. ff21− BW ≡ ba≤ /2 f0 ( ) b > a / 2 f0 =center frequency Single mode operation f 22 TE TE TE mn 1 mn 10 01 20 TE fc = + 11 2 µε ab TM11 18 Dominant Mode: TE10 Mode For this mode we have y 10 π PEC mn=1, = 0, k = c a b ε,, µσ Hence we have x π − jkz z Hz = A10 cos xe z a a 1/2 2 ka π − 10 2 π z jkz z kk= = k − Hx = j A10 sin xe zz π a a ωµ π π ja − jkz z − jkz z Ey = − A10 sin xe Ey = E10 sin xe π a a E10 −π AE≡ 10 jaωµ 10 EEHxz= = y = 0 19 Dominant Mode: TE10 Mode (cont.) The fields can be put in terms of E : 10 y PEC π − jkz z Ey = E10 sin xe a b ε,, µσ x 1 π − H= − Esin xejkz z x 10 a ZaTE z −ππ − jkz z Hz = E10 cos xe 2 1/2 jaωµ a π kk=10 = k 2 − zz a ωµ Z = EEHxz= = y = 0 TE kz 20 Dispersion Diagram for TE10 Mode Lossless case (εc = εε = ′) 2 π kk=β =2 − > z ω ffc a k 22= ω µε vg = slope ω10 1 c (TEM mode, or “Light line”) µε z β=k = ω µε vp = slope β ω Phase velocity: v = p β Velocities are slopes on dω the dispersion plot. Group velocity: v = g dβ 21 Field Plots for TE10 Mode y PEC Top view b ε,, µσ x z a x E z a H y y b b x a z End view Side view 22 Field Plots for TE10 Mode (cont.) y Top view PEC b ε,, µσ x z a x z a J s H y y b b x a z End view Side view Note: One can cut a narrow z-directed slot in the center of the top wall without disturbing the current. 23 Power Flow for TE10 Mode Time-average power flow in the z direction for +z mode: ab Note: + 1 * P10 =Re (E ×⋅ H) zˆ dydx ab ∫∫ π 2 00 2 x ab ∫∫sin dydx = ab a 2 1 00 = − * Re ∫∫ Eyx H dydx 2 00 π − E= Esin xejkz z y 10 1 ab2 kz −2αz a = ReEe10 1 π − 22 ωµ jkz z Hx = − E10 sin xe ZaTE ωµ Simplifying, we have ZTE = kz +−ab 2 2αz P10 = Re{kEz } 10 e 4ωµ At breakdown: Note: EE= For a given maximum electric field level (e.g., the 10 c breakdown field), the power is increased by increasing the cross-sectional area (ab).