3e. Introduction Lecture 3e Rectngulr wveguide So fr in rectngulr coordintes we hve delt with plne wves propgting in simple nd inhomogeneous medi. The power density of plne wve extends over ll spce. Therefore n idel plne wve would crry infinite power nd s such is not physiclly relible. In mny pplictions we need to contin the power of n electromgnetic wve within some specific volume. We cn ccomplish this using wveguide. Exmples of wveguides re coxil cbles, fiberoptics, nd TV ntenn wires. As we will see, when one confines wve the possible solutions form discrete set of modes. Ech mode typiclly hs cutoff frequency below which the mode cnnot propgte, nd the propgtion constnt typiclly hs non-liner dependence on frequency tht leds to the phenomenon of wveguide dispersion. Rectngulr wveguide In this lecture we will consider rectngulr wveguides which guide wves long the xis while confining their power to rectngle of dimensions by b. This is illustrted in the following figure. potentils A, F. We will begin with the cse of field described by F lone. This field will hve no E component, so we will cll it TE mode (the electric field is trnsverse to the direction). TE Modes Using our seprtion of vribles results for the Helmholt eqution in rectngulr coordintes, we look for solutions of the form F ={ cos x sin x x}{ cos y sin y y} e We use the e fctor becuse we re interested in wves propgting long the xis. As before, the brce nottion refers to n rbitrry liner combintion of the enclosed terms. Let's consider the boundry condition E y = t x =,. We hve E y = (2) x F x{ sin x (3) cos x x} Since we re enforcing the boundry conditions t different vlues of x, we need only consider tht fctor of F tht hs x dependence. To get E y = t x = we must use only the sin x x dependence in E y, or we must hve x =. The condition E y = t x = requires x = or sin x = (4) This lst condition gives x =m (5) Figure : Rectngulr wveguide. The guide extends to ± long the xis. We tke b. The interior of the wveguide is x, y b nd. The wveguide surfce is typiclly metllic (idelly PEC), nd the interior is typiclly ir-filled (idelly free spce). However, we could hve PMC or dielectric versions of the sme geometry, nd the interior could, in principle, be filled with n rbitrry mteril. For PEC wveguide, the boundry conditions tht the tngentil electric fields re ero becomes E y = E = E x = E = x =, y=, b As we hve seen previously, ny field cn be represented using ust the components of the mgnetic nd electric vector () where m is n integer: m=,,2,. Since m= gives x = we cn simply sy tht the x dependence of E y must be through fctor sin x x where x is one of the vlues given by (5). Therefore the x dependence of F must be through fctor of cos x x. Note tht if m=, so x =, then E y everywhere. Now consider the boundry conditions E x = t y=, b. We hve E x = y F y{ sin y (6) cos y y} This is nlogous to the E y cse, so we must hve tht F depends on y through fctor of cos y y with y =n b nd n=,,2,. If n= then E x everywhere. We see tht the cse m=n= is trivil becuse we would then hve (7) EE58: Advnced Electromgnetic Theory Scott Hudson 25-3-2
3e.2 E x E y nd there would be no field nywhere. We hve found tht solutions exist only for certin discrete vlues of x, y nd hve the form F =F cos m x/ cosn x/be (8) We refer to this s the TE mn by 2 x 2 y 2 = 2, or mode. The vlue of is fixed = 2 m/ 2 n/ b 2 (9) Using Eqution (24) of Lecture 2c, the electric field is E x = nd the mgnetic field is H x = E y F y cos x xsin y ye E y = F x sin x xcos y ye E = E x H y = F H = 2 x 2 y cos x x cos y y e where the wve impednce is Dominnt mode () () = (2) As we will see, the most importnt, or dominnt mode is the so-clled TE mode with m=, n=. In this cse x = /, y = nd The non-ero field components re = 2 / 2 (3) E y = F sin x x e H x = F sin x x e H = 2 F 2 cos x x e It is convenient to cll E = F /. Then E y = E sin x /e H x = E sin x /e H = E cos x /e (4) (5) with nd Power flow The Poynting vector is = 2 / 2 (6) P = 2 Re E H = (7) = 2 Re [ E x H y E y H x x E y H y E x H ] (8) If the mteril filling the wveguide is lossless so tht, re rel, then the x, y terms re purely imginry due to the fctor in H. In this cse If is rel then is rel nd P= 2 E x 2 E y 2 Re[ ] (9) P= F 2 [ 2 2 2 y cos 2 x xsin 2 y y x 2 sin 2 x xcos 2 y y] (2) If = is imginry then P =. for the TE (ssuming is rel) P= E 2 The totl power crried in the wveguide is For the TE Cutoff mode From the condition 2 sin2 x/ (2) b W = W = E 2 P dx dy (22) 4 b (23) = 2 m/ 2 n/ b 2 (24) we see tht for rel,, if flls below certin vlue then will become imginry, nd the field will no longer propgte long the xis. Insted it will decy s e where =. This is consistent with our result bove tht = gives P =. This frequency is clled the cutoff EE58: Advnced Electromgnetic Theory Scott Hudson 25-3-2
3e.3 frequency. The mode propgtes only bove the cutoff frequency. The cutoff frequency (in H) is f c = 2 m/2 n /b 2 (25) Different modes (m,n vlues) hve different cutoff frequencies. The lowest cutoff frequency for TE mn mode will correspond to m=, n= (since b ). This gives f c = 2 (26) This is the cutoff frequency of the TE mode. The second lowest cutoff will be for the TE mode nd is Over the frequency rnge f c = 2 b 2 f 2b (27) (28) only the TE mode is bove cutoff, nd we sy tht the wveguide hs single-mode opertion. For this reson the mode is sometimes clled the dominnt mode. In the TE next lecture we will consider TM mn modes using the mgnetic vector potentil, nd we will see tht the TE cutoff frequency is lower thn tht of ny TM mn mode. Note tht s we pproch cutoff so. Most generl TE field We hve found some prticulr solutions in the seprtion of vribles form. Wht does this tell us bout n rbitrry TE field in the wveguide described by some function F x, y,? Consider the two-dimensionl function F x, y,. For ny vlue of y we cn represent the x dependence of this over x by cosine series where while for m F x, y, = F m y cos m x / (29) m= F m y= 2 F y= F x, y, dx (3) F x, y,cosm x /dx (3) Likewise, the coefficients F m y cn be expnded in cosine series over y b : F m y = F mn cos n y /b (32) n = Substituting this into the previous expression we rrive t F x, y, = m= n = F mn cos m x / cos n y /b (33) This is nothing more thn liner combintion of our TE mn modes t =. For other vlues of we will hve F x, y, = F mn cos m x / cos n y /b e (34) m= n= Therefore, ny TE field tht cn exist inside the wveguide cn be represented by superposition of the TE mn modes, so we don't need to look for dditionl solutions. If we operte t frequency tht gives us single-mode opertion, then regrdless of the field we strt with t = we will quickly end up with only the TE mode since it is the only mode tht cn propgte. All other mode components will decy exponentilly long the wveguide. Ohmic Losses Our solutions derived bove hve ssumed PEC wveguide surfce. If the surfce is good conductor but not perfect, then we cn pply the surfce resistnce ide to clculte ohmic losses. The surfce current mplitude on PEC is equl to the tngentil mgnetic field, nd the power dissipted over surfce is where W diss = 2 R s H tn 2 ds (35) R s = 2 ' ' = 2 (36) For the dominnt TE mode the fields re given by (5). Consider section of the wveguide extending long the xis. On the surfces x=, only H is tngentil. Since cos 2 =cos 2 =, the power dissipted on these two surfces is 2 2 R s 2 E 2 b dy=r 2 s b 2 E 2 (37) 2 On the surfces y=, b both H x, H re tngentil nd H tn 2 = E 2[ 2 sin2 x / 2 We integrte this over x from to. Since we obtin 2 cos2 x/ ] (38) sin 2 x/ dx= cos 2 x /dx= 2 (39) EE58: Advnced Electromgnetic Theory Scott Hudson 25-3-2
3e.4 2 2 R s E 2 2 [ 2 2 2] (4) 2[ 2 2 Putting the contributions of ll four wlls together we rrive t d W diss = R s 2b E d 2 2 2 (4) ] This is the power lost per unit length for the dominnt TE mode. Wveguide dispersion The dependence of on frequency hs importnt implictions. Suppose we excite wveguide mode with temporl pulse. For exmple, consider Gussin pulse with width nd center frequency The Fourier representtion of this hs the spectrum i t = e 2 t/2 e t (42) i t = A i e t d 2 (43) A i = 2 e 2 2 2 (44) We see tht the spectrl width is / which is the clssic result tht nrrow pulse implies lrge bndwidth nd conversely. Let's investigte wht hppens when this pulse trvels down wveguide of length L. A field component with frequency will experience phse chnge of exp L. Therefore, the output pulse will be o t = A i e L e t d 2 Let's expnd round the center frequency with (45) = 2 2 2 (46) We need to evlute o t =e L 2 = = d d 2 = d 2 d 2 e 2 2 2 e L e 2 2 2 L t d e 2 (47) (48) Clling u= this becomes The integrl is o t =e L 2 e t e 2 2 u 2 e L u e 2 2 L u 2 e u t du (49) e 2 2 2 L u 2 e u t L du (5) Note tht the time vrible t hs been shifted by n mount L. This tells us tht u g =/ is the velocity with which the pulse trvels long the wveguide. This is often clled the group velocity. In ddition, the pulse will be brodened. Using where k 2 = 2 2 L nd e 2 k2 u 2 e ut du= 2 k 2 2 L = 2 2 L 4 2 L 2 e 2 = 2 2 L/ 2 L 2 4 2 L 2 t 2 k 2 (5) we find tht the mplitude of the output pulse vries s o t e The width goes from to 2 t L 2 (52) 2 2 L/ 2 (53) o = 2 2 L/ 2 (54) This is the phenomenon of wveguide dispersion in which pulse widens s it trvels down the wveguide. The nrrower the originl pulse (the smll the vlue of ) the greter the dispersion. Minimiing o 2 = 2 2 L / 2 with respect to we find o,min = 2 2 L (55) for the minimum possible output pulse width. This corresponds to n input pulse width = 2 L. We see tht the 2 nd derivtive (or the curvture) of the curve limits the pulse widths tht cn be sent down the wveguide. Note tht for plne wve in simple medium with =, the 2 nd derivtive of is ero, so dispersion does not occur. However, if = then the 2 nd derivtive of will be non-ero nd dispersion will occur. This is clled mteril dispersion since it rises from the properties of the mteril rther thn from the geometry of wveguide. In fiber-optics, where dielectric wveguides re employed, one typiclly hs both mteril nd wveguide dispersion to contend with. EE58: Advnced Electromgnetic Theory Scott Hudson 25-3-2
3e.5 TM modes In the previous lecture we solved for the TE modes in rectngulr wveguide. We now consider TM modes for which the mgnetic vector potentil hs the form A ={ cos x sin x x}{ cos y sin y y} e (56) The boundry conditions for PEC wveguide re s before E y = E = E x = E = x =, y=, b The electric field is given by (see Lecture 2c) E x = E y = E = x y A A [ 2 x 2 A 2 y 2 A ] (57) (58) Since E x must vnish t y= nd it does not involve y derivtive of A, we must use the sin y y term in our A solution. Likewise, E y must vnish t x = nd it does not involve n x derivtive of A, so we must use the sin x x term. Our solution will therefore hve the form A =A sin x x sin y y e (59) The vnishing of E y t x = nd the vnishing of E x t y= b require x =m y =n b, m=,2,, n=,2, (6) The upper limit is t lest fctor of 2 times the lower limit. We sy tht the wveguide opertes single mode over t lest one octve. For the TM mn mode the electric field is E x = A xcos x xsin y y e E y = A ysin x xcos y y e E = A 2 x 2 y sin x xsin y y e nd the mgnetic field is or H x = H y = H = where the wve impednce is Rectngulr Resontors y A = A y sin x xcos y y x A = A x cos x xsin y y H x = E y E x H y = H = = Consider the sitution in the following illustrtion. (64) (65) (66) (67) Note tht neither m or n cn be ero since tht would led to A nd there would be no field. The vlue of is fixed by the requirement x 2 y 2 2 = 2 = to be = 2 m/ 2 n/ b 2 (6) The cutoff frequency is therefore f c = 2 m/ 2 n/b 2 (62) The lowest cutoff frequency is for the TM mode. Since there is no TM mode we see tht the TE mode is indeed the dominnt mode in the wveguide. The TE nd TE 2 cutoffs will lso be lower thn tht of the TM mode (provided b ). Therefore, the wveguide hs single-mode opertion over the frequency rnge 2 f min 2, /b (63) 2 Figure : Rectngulr resontor. Here we've cut length c of the wveguide nd cpped both ends with PEC pltes. We will no longer be ble to hve power propgting only long the xis. Insted, we will need wves propgting in both the nd directions. Together these should form stnding wve (nonpropgting) field of the type we hve in the x nd y dimensions. Let's consider TE modes. We tke EE58: Advnced Electromgnetic Theory Scott Hudson 25-3-2
3e.6 F =cos x x cos y y F e F 2 e (68) This represents superposition of wves trveling in the the nd directions with mplitudes F, F 2. The electric field will hve E x = / F / y, E y =/ F / x or E x = ycos x xsin y y[ F e F 2 e ] E y = x sin x xcos y y [ F e F 2 e ] (69) Both of these components re tngentil t = nd must therefore be ero. This gives the condition F F 2 =. Therefore nd so F e F 2 e = F e e = F sin (7) F = F cos x x cos y y sin (7) E x = F y cos x xsin y ysin E y = F x sin x xcos y ysin Now, E x = E y = t =c requires sin c= or = p c (72) (73) where p=,2,. Note tht p= would give E. Tken together with the conditions on TE mn wveguide modes we hve x =m y =n b = p c m=,, n=,, p=,2, (74) with the dditionl constnt tht both m nd n cnnot be ero. Therefore so tht 2 m/ 2 n/b 2 p/c 2 = 2 (75) f r = 2 m/ 2 n/b 2 p/c 2 (76) is the resonnt frequency of the TE mnp mode. The lowest resonnt frequency will be tht of the TE mode (provided b ). TM modes From our nlysis bove it's cler tht for TM resontor modes the mgnetic vector potentil will hve the form Since A =A sin x xsin y y{ cos sin } (77) E x = E y = x y A A (78) we must use the cos fctor in our A expression so tht E x, E y re proportionl to sin nd hence vnish t =. We hve A = A sin x xsin y y cos (79) The boundry condition E x = E y = t =c requires = p c (8) where p=,,. Note tht p= is cceptble. Although this gives E x = E y = everywhere, E so there will still be field. The conditions for TM mnp resontor mode re x =m y =n b = p c nd the resonnt frequency is, m=,2,, n=,2,, p=,, (8) f r = 2 m/ 2 n/b 2 p/c 2 (82) The minimum resonnt frequency will be for the TM mode. If cb then the TE frequency will be lower thn the TM frequency. Otherwise the TM frequency will be the lowest. EE58: Advnced Electromgnetic Theory Scott Hudson 25-3-2