Rectangular Waveguide Notes

ECE 3317 Applied Electromagnetic Waves Notes 20 Rectangular Waveguides 1 Rectangular Waveguide Rectangular Waveguide y b ε,µ a x Cross section  We assume that the boundary is a perfect electric conductor (PEC). No TEMz mode can exist! 2 Rectangular Waveguide (cont.) Why is there no TEMz mode? y E b z=0 x a TEMz mode: k z= k= ω µε Rectangular waveguide mode (m, n): k z( m ,n ) ( m,n ) kz ≠k 2  mπ   nπ  =k 2 −   −  a    b  2 ( m, n ) ≠ ( 0, 0 ) 3 Rectangular Waveguide (cont.) Rectangular Waveguide  Two types of modes can exist independently: TMz: Ez only TEz: Hz only 4 Rectangular Waveguide (cont.) Rectangular Waveguide y b ε,µ a x Cross section  We analyze the problem to solve for Ez or Hz (all other fields come from these). TMz: Ez only TEz: Hz only 5 TMz Modes = H z 0, E z ≠ 0 ∇2 E z + k 2 E z = 0 (Helmholtz equation) E z = 0 on boundary Guided-wave assumption: (PEC walls) E z ( x, y , z ) = E z 0 ( x, y ) e − jk z z  ∂ 2 Ez ∂ 2 Ez ∂ 2 Ez  2 0  ∂x 2 + ∂y 2 + ∂z 2  + k E z =    ∂ 2 Ez ∂ 2 Ez  2 2 0  ∂x 2 + ∂y 2 − k z E z  + k E z =   6 TMz Modes (cont.) ∂ 2 Ez ∂ 2 Ez 2 2 0 k k + + − ( z ) Ez = 2 2 ∂x ∂y Define: kc2 ≡ k 2 − k z2 ∂ 2 Ez ∂ 2 Ez 2 Note that kc is an We then have: 0 + + k E = c z unknown at this point. ∂x 2 ∂y 2 Dividing by the exp(-j kz z) term, we have: ∂ 2 Ez 0 ∂ 2 Ez 0 2 k + + 0 c Ez 0 = 2 2 ∂x ∂y We solve the above equation by using the method of separation of variables. Please see Appendix A for the solution. 7 TMz Modes (cont.) Solution from separation of variables method: TMmn mode  mπ x   nπ y  − jk z(m ,n ) z E z ( x, y , z ) = Amn sin   sin  e  a   b  k z( m ,n ) = k 2 − kc2 = 2  mπ   nπ  k2 −   −   a   b  2  mπ   nπ  kc2  =  +  a b     m = 1, 2, n = 1, 2, 2 2 = k ω= µε k0 µrε r k= ω µ0ε= 0 0 ω 2π = c λ0 Note: If either m or n is zero, the entire field is zero. 8 TMz Modes (cont.) Cutoff Frequency for Lossless Waveguide We start with Set k z( m,n ) k z( m n ) , =0 2  mπ   nπ  2 =k −   −  a b     Note: Cutoff frequency only has a clear meaning in the lossless case (k is real). 2 This defines the cutoff frequency. TM m ,n k f = f 2π f c = µε = c cd TM m , n fc = 2π 2 2  mπ   nπ    +   a   b   mπ   nπ    +   a   b  2 cd = 2 c εr ( nonmagnetic material ) 9 TMz Modes (cont.) Summary of TMz Solution: TMmn mode  mπ x   nπ y  − jk z(m ,n ) z E z ( x, y , z ) = Amn sin   sin  e  a   b  k ( m ,n ) z = k 2 − kc2 = 2  mπ   nπ  k2 −   −   a   b  2 m = 1, 2, n = 1, 2, Note: If either m or n is zero, the entire field is zero. cd TM f c m ,n = 2π 2  mπ   nπ    +   a   b  2 (lossless waveguide) 10 TEz Modes = E z 0, H z ≠ 0 We now start with ∂2H z ∂2H z 2 2 0 + + k − k ( z ) Hz = 2 2 ∂x ∂y Guided-wave assumption: Define: H z ( x, y, z ) = H z 0 ( x, y ) e − jkz z kc2 ≡ k 2 − k z2 ∂2 H z0 ∂2 H z0 2 0 + + k H z0 = c 2 2 ∂x ∂y Please see Appendix B for the solution. 11 TEz Modes (cont.) Summary of TEz Solution: TEmn mode  mπ x   nπ y  − jkz( m ,n) z H z ( x, y, z ) = Amn cos   cos  e  a   b  k z( m ,n ) 2  mπ   nπ  =k 2 −   −   a   b  m = 0,1, 2 n = 0,1, 2 cd TE m , n fc = 2π 2 2 ( m, n ) ≠ ( 0, 0 )  mπ   nπ    +   a   b  2 (lossless waveguide) Note: Same formula for cutoff frequency as the TMz case! 12 Summary for Both Modes TMmn  mπ x   nπ y  − jk z(m ,n ) z E z ( x, y , z ) = Amn sin   sin  e  a   b  TEmn  mπ x   nπ y  − jk z(m ,n ) z H z ( x, y , z ) = Amn cos   cos  e  a   b  k z( m ,n ) cd f c( m ,n ) = 2π 2  mπ   nπ  =k 2 −   −  a b     2  mπ   nπ    +  a    b  2 cd = c εr (lossless waveguide) TMz m = 1, 2, n = 1, 2, TEz 2 same formula for both modes same formula for both modes m = 0,1, 2, n = 0,1, 2, ( m, n ) ≠ ( 0, 0 ) 13 Field Plots Color denotes magnitude, arrows show direction of electric field. 14 Wavenumber kz TMz or TEz mode: = 2 k −k 2 c 2 kc = with  mπ   nπ    +   a   b  2 Note: The (m,n) notation is suppressed here on kz. Lossless waveguide: Above cutoff: Below cutoff: kz = β k z = − jα 2  mπ   nπ  β =k 2 −   −   a   b  α= 2 2 2  mπ   nπ  2   +  −k  a   b  (Recall the general formula for k z : k z= β − jα ) 15 Wavenumber Plot β 2 kc  mπ   nπ    +   a   b  2 α kc “Light line” β= k= 2π f µε f fc 2 2  mπ   nπ  β =k −   −  ,  a   b  2 α= 2 f > fc 2  mπ   nπ  2   +  −k ,  a   b  cd f c f= = c 2π ( m,n ) f < fc 2  mπ   nπ    +   a   b  cd = 2 c εr 16 Guided Wavelength Recall: The guided wavelength λg is the distance z that it takes for the wave to repeat itself. λg = 2π β (This assumes that we are above the cutoff frequency – otherwise guided wavelength makes no sense.) After some algebra (see next slide): λg = λd 1 − ( fc / f ) (Note : λ g 2  λ0   λd =   ε r   (lossless waveguide) > λd ) 17 Guided Wavelength (cont.) Derivation of wavelength formula (lossless waveguide): = λg λg = 2π = β 2π = 2 2 m n π π     k2 −   −   a   b  2π k 2 − kc2 2π 2π 2π = = = 2 2 2 π 2 2 k − kc k 1 − ( kc / k ) 1 − ( kc / k ) λd = k ω= µε 2π f µε 1 − ( kc / k ) 2 kc / k = f c / f = kc ω= 2π f c µε c µε λg = λd λd 1 − ( fc / f ) 2 18 Dominant Mode The "dominant" mode is the one with the lowest cutoff frequency. Assume b < a cd = fc 2π 2  mπ   nπ    +  a    b  cd = y 2 c (lossless waveguide) b ε,µ εr a Lowest TMz mode: TM11 Lowest TEz mode: TE10 The dominant mode is the TE10 mode. TMz m = 1, 2, n = 1, 2, x TEz m = 0,1, 2, n = 0,1, 2, ( m, n ) ≠ ( 0, 0 ) 19 Dominant Mode (cont.) Summary (TE10 Mode)  π x  − jk z z H z ( x, y , z ) = A10 cos  e  a  π  k2 −  a = kz fc = 2 cd 2a cd = k = k0 ε r c εr 2 π  β= k2 −  , a f > fc 2 π  α =   − k2 , a f < fc 20 Dominant Mode (cont.) Fields of the Dominant TE10 Mode  π x  − jk z z H z ( x, y , z ) = A10 cos  e  a  Find the other fields from these equations (Appendix A of Notes 19):  − jωµ  ∂H z  jk z  ∂E z = − 2 Ex  2 2  2  − ∂ − k k y k k z  z  ∂x    jωµ  ∂H z  jk z  ∂E z = − 2 Ey  2 2  2  − ∂ − k k x k k z  z  ∂y    jωε  ∂E z  jk z  ∂H z = − 2 Hx  2 2  2  − ∂ − k k y k k z  z  ∂x    − jωε  ∂E z  jk z  ∂H z = − 2 Hy  2 2  2  − ∂ − k k x k k z  z  ∂y   21 Dominant Mode (cont.) Summary of fields for TE10 mode:  π x  − jk z z H z ( x, y , z ) = A10 cos  e  a   π x  − jkz z E y ( x, y, z ) = E10 sin  e  a   kz   π x  − jkz z H x ( x, y , z ) = −  e  E10 sin   a   ωµ  where  jωµ   π  = E10  2 −  A10 2   k − kz   a  22 Dominant Mode (cont.) TE10 Mode y z=0 E b y x a x Length of arrows denotes magnitude of field Color denotes magnitude of field y E H b a x Spacing between arrows denotes magnitude of field 23 Dominant Mode (cont.) TE10 Mode 3D View 24 Dominant Mode (cont.) What is the mode with the next highest cutoff frequency? fc cd 2π 2  mπ   nπ    +   a   b  2 (1,0 ) fc c = d 2a y Assume b < a / 2 2 cd  2π  cd  1  = = fc     2π  a  2 a/2 ( 2,0 ) 2 cd  π  cd  1  = = fc     2π  b  2 b ( 0,1) Then the next highest is the TE20 mode. cd = b f c( 2,0) = 2 f c(1,0) ε,µ a A 2:1 operating band! c εr x (lossless waveguide) Useful operating region TE01 TE10 TE20 fc 25 Dominant Mode (cont.) What is the mode with the next highest cutoff frequency? 2 cd  2π  cd  1  = = fc     2π  a  2 a/2 ( 2,0 ) cd = Assume b > a / 2 2 cd  π  cd  1  = = fc     2π  b  2 b ( 0,1) c y εr (lossless waveguide) b Then the next highest is the TE01 mode. ε,µ The useable bandwidth is now lower than before. Useful operating region TE10 x a TE01 TE20 fc 26 Dominant Mode (cont.) b a Power flow in lossless waveguide (f > fc): 1 E × H * ) ⋅ zˆ dxdy ( 2 0 0 = Pz Re ∫ ∫ b a = Re ∫ ∫ − 1 E y H x* ) dxdy ( 2  π x  − jkz z E y ( x, y, z ) = E10 sin  e 1 = − b Re ∫ E H dx  a  2  ab  2 Pz =   β E10 [ W ] (watts flowing down the waveguide)  4ωµ  0 0 a y * x 0 (The derivation is omitted, but please see the formula box above.) Note: Above cutoff, there is only watts flowing (no vars). Below cutoff there is no watts flowing (only vars).  Make b larger to get more power flow for a given value of a.  Keep b smaller than a/2 to get maximum bandwidth. The optimum dimension for b is a/2 (gives maximum power flow without sacrificing bandwidth). 27 Dominant Mode (cont.) Plane wave interpretation of TE10 mode  π x  − jk z z − jk z z E y ( x, y , z ) E= e E k x e sin sin ( ) 10 10 x    a   e jk x x − e − jk x x  − jk z z = E10  e j 2   π  k ≡  x  a   e jz − e − jz  Note : sin z =    2j  E y ( x, y , z ) E10′ e − jk x x e − jk z z + E10′′ e jk x x e − jk z z = PW #1 x (E, H) θ PW #1 PW #2 PW #2 tan= θ  E10′ ≡ − E10 / ( 2 j )    ′′ E ≡ + E / 2 j ( ) 10  10  kx = kz π /a π  k −  a 2 2 z 28 Dominant Mode (cont.) Losses in Waveguide (f > fc) α ≈ αd + αc Dielectric loss: π  2 β − jα d kz = k02ε r (1 − j tan δ d ) −   = a π  αd = − Im k ε (1 − j tan δ d ) −   a 2 0 r  2  Conductor loss: αc εr Rs bη0 1 − ( f / f )2 c ε rc ε r (1 − j tan δ ) Recall : = Note: If we are below cutoff, attenuation is mainly due to evanescence, so we don’t worry about conductor and dielectric loss then.  2b  f  2  1 +  c   [np/m]   a f     (This is derived in ECE 5317.) 29 Example Find the single-mode operating frequency region for air-filled X-band waveguide. Standard X-band* waveguide: a = 0.900 inches (2.286 cm) b = 0.400 inches (1.016 cm) Use Hence, we have: (1,0 ) fc Note: b < a / 2 c = 2a f c(1,0) = 6.56 [GHz] f c( 2,0) = 13.11 [GHz] X-band waveguide * X-band: from 8.0 to 12 GHz. 6.56 < f < 13.11 [GHz] 30 Example (cont.)  Find the phase constant of the TE10 mode at 9.00 GHz.  Find the attenuation in dB/m at 5.00 GHz 1,0 Recall: f c( ) = 6.56 [GHz] X-band waveguide 2 π  β= k2 −  , a f > fc 2 π  α =   − k2 , a f < fc = k k= ω µ0ε= 2π f / = c 2π / λ0 0 0 At 9.0 GHz : k = 188.62 [rad/m] At 5.0 GHz : k = 104.79 [rad/m] = kc π= / a 137.43 [rad/m] At 9.00 GHz: β = 129.13 [rad/m] At 5.00 GHz: α = 88.91 [nepers/m] 31 Example (cont.) At 5.0 GHz: α = 88.91 [ nepers/m ] Recall: dB/m = 8.68589 α X-band waveguide Attenuation = 772 dB/m This is a very rapid attenuation! 32 Waveguide Components Straight sections Waveguide adapters Flexible waveguides Waveguide couplers Waveguide bends Waveguide terminations https://www.pasternack.com 33 Waveguide Modes in Transmission Lines  A transmission line normally operates in the TEMz mode, where the two conductors have equal and opposite currents.  At high frequencies, waveguide modes can also propagate on transmission lines.  This is undesirable, and it limits the high-frequency range of operation for the transmission line. 34 Waveguide Modes in Coax Dominant waveguide mode in coax (derivation omitted): TE11 mode: f TE11 c  1  1  ≈    a εr  π   1 + b / a  c Note: In this notation, the “11” subscript refers to the angular and radial variation. Example: RG 142 coax = a 0.035 inches = 8.89 × 10−4 [m] = b 0.116 inches = 29.46 × 10−4 [m] ε r = 2.2 b= / a 3.31 ⇒ Z 0 = 48.4 [Ω] εr b z f cTE11 ≈ 16.8 [GHz] This coax cannot be used above 16.8 [GHz] a = Z0 η0 L b = ln   C 2π ε r  a  35 Appendix A: TMz Modes We want to solve: Define: ∂ 2 Ez ∂ 2 Ez 2 2 + + − k k 0 ( z ) Ez = 2 2 ∂x ∂y kc2 ≡ k 2 − k z2 ∂ 2 Ez ∂ 2 Ez 2 Note that kc is an We then have: 0 + + k E = c z unknown at this point. ∂x 2 ∂y 2 Dividing by the exp(-jkzz) term, we have: ∂ 2 Ez 0 ∂ 2 Ez 0 2 k + + 0 c Ez 0 = 2 2 ∂x ∂y We solve the above equation by using the method of separation of variables. We assume: E z 0 ( x, y ) = X ( x ) Y ( y ) 36 Appendix A (cont.) ∂ 2 Ez 0 ∂ 2 Ez 0 2 + + k 0 c Ez 0 = 2 2 ∂x ∂y E z 0 ( x, y ) = X ( x ) Y ( y ) Hence X ′′Y + X Y ′′ = −kc2 XY Divide by XY : X ′′ Y ′′ + = −kc2 X Y Hence This has the form X ′′ Y ′′ 2 = − kc − X Y F ( x) = G ( y) Both sides of the equation must be a constant! 37 Appendix A (cont.) X ′′ Y ′′ 2 = − kc − = constant X Y Denote X ′′ = −k x2 = constant X General solution: = X ( x) A sin(k x x) + B cos(k x x) y Boundary conditions: X (0) = 0 X (a ) = 0 (1) ( 2) b ε,µ a (1) B= 0 ⇒ X ( x) = A sin(k x x) (2) sin(k x a ) = 0 x 38 Appendix A (cont.) From the last slide: sin(k x a ) = 0 This gives us the following result: k x a m= π , m 1, 2 = mπ kx = a Hence  mπ x  X ( x) = A sin   a   Now we turn our attention to the Y (y) function. 39 Appendix A (cont.) We have X ′′ Y ′′ 2 = − kc − = −k x2 X Y Hence Denote Y ′′ = k x2 − kc2 Y 2 2 2 k= k − k y c x Then we have General solution: = Y ( y) Y ′′ = −k y2 Y C sin(k y y ) + D cos(k y y ) 40 Appendix A (cont.) = Y ( y ) C sin(k y y ) + D cos(k y y ) y Boundary conditions: Y (0) = 0 (3) Y ( b) = 0 ( 4) b ε,µ a (3) (4) x D= 0 ⇒ Y ( y) = C sin(k y y ) sin(k y b) = 0 Equation (4) gives us the following result: = kb y n= π , n 1, 2 nπ ky = b 41 Appendix A (cont.) The Y(y) function is then  nπ y  Y ( y ) = C sin   b   Therefore, we have  mπ x   nπ y  E z 0 ( x, y ) X= = ( x ) Y ( y ) AC sin   sin    a   b  New notation:  mπ x   nπ y  E z 0 ( x, y ) = Amn sin   sin   a b     The Ez field inside the waveguide thus has the following form:  mπ x   nπ y  − jk z(m ,n ) z E z ( x, y , z ) = Amn sin   sin  e  a   b  42 Appendix A (cont.) Recall that 2 2 2 k= k − k y c x 2 2 2 k= k + k c x y Hence, Therefore, the solution for kc is given by 2  mπ   nπ  2 kc  =  +   a   b  2 2 2 2 Next, recall that k= k − k c z 2 2 2 k − kc Hence k= z ( m,n ) z kz = k 2  mπ   nπ  =k 2 −   −  a    b  2 43 Appendix B: TEz Modes = E z 0, H z ≠ 0 We now start with ∂2H z ∂2H z 2 2 0 + + k − k ( z ) Hz = 2 2 ∂x ∂y Using the separation of variables method again, we have H z 0 ( x, y ) = X ( x ) Y ( y ) where = X ( x) A sin(k x x) + B cos(k x x) = Y ( y ) C sin(k y y ) + D cos(k y y ) and 2 2 2 k= k + k c x y 2 2 2 k= k − k z c 44 Appendix B (cont.) Boundary conditions: E x ( x,0) = 0 E y (0, y ) = 0 E x ( x , b) = 0 E y (a, y ) = 0 y b ε,µ a x The result is  mπ x   nπ y  H z 0 ( x, y ) = Amn cos   cos    a   b  This can be shown by using the following equations:  − jωµ  ∂H z  jk z  ∂Ez = − 2 Ex  2 2  2  − ∂ − k k y k k   z  z  ∂x ∂H z 0,= = y 0, b ∂y  jωµ  ∂H z  jk z  ∂Ez = − 2 Ey  2 2  2  − ∂ k k x   k − k z  ∂y z  ∂H z = 0,= x 0, a ∂x 45 Appendix B (cont.) The Hz field inside the waveguide thus has the following form:  mπ x   nπ y  − jkz( m ,n) z H z ( x, y, z ) = Amn cos   cos  e  a   b  k z( m ,n ) 2  mπ   nπ  2 =k −   −  a    b  m = 0,1, 2 n = 0,1, 2 2 Same formula for cutoff frequency as the TEz case! ( m, n ) ≠ ( 0, 0 ) Note: The (0,0) TEz mode is not valid, since it violates the magnetic Gauss law: H ( x, y , z ) = zˆ A00 e − jkz ∇ ⋅ H ( x, y , z ) ≠ 0 46