X".Y--keXYThe second term on the left side of the above equationis a functionof y only, while the right side is a constant. The only way the equationcan be satisfiedis that both terms on the left side are constants"--k.YNow letXwherek,andk,are calledthe separation constants,andtheycan befound by using the boundary conditions.Obviouslyk? =k?+k?The two equations are second order ordinary differentialequations.and thegeneral solutions,arerespectivelyX =C, cosk,x +C2 sin k,xY = C, cosk,y+C4 sin k,Jwhere all the constants Cr, C2 , C3 , C4, and k x, k y, depend on theboundaryconditions.u7
The second term on the left side of the above equation is a function of y only, while the right side is a constant. The only way the equation can be satisfied is that both terms on the left side are constants. 2 c k Y Y X X = − + Now let 2 x k X X = − 2 y k Y Y = − where k x and k y are called the separation constants, and they can be found by using the boundary conditions. 2 2 2 c x y Obviously k = k + k X C k x C k x x x = 1 cos + 2 sin Y C k y C k y y y = 3 cos + 4 sin where all the constants C1 , C2 , C3 , C4 , and k x , k y , depend on the boundary conditions. The two equations are second order ordinary differential equations, and the general solutions, are respectively
Since the component E.is parallelto the walls, we have E,= O atthe boundariesx = O, a and y = O, b . Using these results we findn元m元k,=m = 1,2,3, ..n = 1,2,3, .k.=6hAnd allthefieldcomponentsarenm元Je-jk.:E, = E,sinxsnba.k.Eommn元le-ik.-EsinCOSxVkebaak.Eon元m元n元e-jk,-E.sinCOSxkbbaOE.n元mn元-jk.H.一sinCOSk?bbaWEmmTnle-jk.H,=sinCOSke6aaKV
Since the component Ez is parallel to the walls, we have Ez = 0 at the boundaries x = 0, a and y = 0, b . Using these results we find , 1,2,3, π = n = b n k y , 1,2,3, π = m = b m kx And all the field components are k z z z y b n x a m E E j 0 e π sin π sin − = z k z x z y b n x a m a m k k E E j 2 c 0 e π sin π cos π j − = − z k z y z y b n x a m b n k k E E j 2 c 0 e π cos π sin π j − = − k z x z y b n x a m b n k E H j 2 c 0 e π cos π sin π j − = k z y z y b n x a m a m k E H j 2 c 0 e π sin π cos π j − = −
mTn元Je-ik.-E, = E, sinxsinbak.EommnTe-jk.E.sincOSkehaak.Eon元m元n元-jk.E.sincOskehbaOEn元 mn元-jk.HcossinYkabham元WE.m元le-jk.H,=sinCOSV11k2baa(e) The modes with larger m and n are calledthe modes of higherorder orthe higher modes, and that with less m and n are calledthe modesoflower order or the lower modes. Since both m and n are not zero, and thelowest mode of TM wave is TM, in the rectangular waveguide
(a) The phase of the electromagnetic wave is related to the variable z only, while the amplitude to the variables x and y. Hence, a traveling wave is formed in the z-direction, and a standing wave is in the xdirection and y-direction. (b) The plane z = 0 is a wave front. Because the amplitude is related to x and y, the TM wave is a non-uniformplane wave. k z z z y b n x a m E E j 0 e π sin π sin − = z k z x z y b n x a m a m k k E E j 2 c 0 e π sin π cos π j − = − z k z y z y b n x a m b n k k E E j 2 c 0 e π cos π sin π j − = − k z x z y b n x a m b n k E H j 2 c 0 e π cos π sin π j − = k z y z y b n x a m a m k E H j 2 c 0 e π sin π cos π j − = − (c) If m or n is zero, then ( for TM wave), and all components will be zero. Thus the m and n are non-zero integrals, and they have clear physical meanings. The value of m stands for the number of half-cycle variations of the fields along the broad side, while n denotes that for the narrow side. E z = 0 H z = 0 (d) Since m and n are multi-valued, the pattern of the field has multiple forms, also called multiple modes. A pair of m and n lead to a mode, and it is denoted as the TMmn mode. For instance, TM11 denotes the pattern of the field for m = 1 , n = 1 , and the wave with this character is called TM11 wave or mode. (e) The modes with larger m and n are called the modes of higher order or the higher modes, and that with less m and n are called the modes of lower order or the lower modes. Since both m and n are not zero, and the lowest mode of TM wave is TM11 in the rectangular waveguide
Similarly,we can derive all the components of a TE wave in therectangularwaveguide,as given bym元le-jk.2H, = HcoscOsXVbak.H.m元nm元e-jk.sHxsincOS=1kebaak.Hn元m-jk.H.sinCOS二kebbaoμH.n元mn元-jk.E, =sinCOSk?bbaouHmTnTm元jk.EsinCOSk2hOawhere m,n - O, 1, 2,..., but both should not be zero at the same timeTE wave has the multi-mode characteristics as the TM wave. Thelowest order mode of TE wave is the TEo or TEo wave
Similarly, we can derive all the components of a TE wave in the rectangular waveguide, as given by k z z z y b n x a m H H j 0 e π cos π cos − = z k z x z y b n x a m a m k k H H j 2 c 0 e π cos π sin π j − = z k z y z y b n x a m b n k k H H j 2 c 0 e π sin π cos π j − = k z x z y b n x a m b n k H E j 2 c 0 e π sin π cos π j − = k z y z y b n x a m a m k H E j 2 c 0 e π cos π sin π j − = − where , but both should not be zero m,n = 0,1, 2, at the same time. TE wave has the multi-mode characteristics as the TM wave. The lowest order mode of TE wave is the TE01 or TE10 wave
3.CharacterizationofElectromagneticWavesinRectangularWaveguidesSince k? = k2 - k2, ork2 = k2 - k3, if k = k。, then k, = 0 . This meansthat the propagation of the wave is cut off, and k, is called the cutoffpropagationconstant→ α (_) ()k2 =k? +k,Fromk = 2πf eu, we can find the cutoff frequencyf correspondingtothe cutoffpropagationconstantk.,as givenbyk.())12元/u2/8U7
3. Characterization of Electromagnetic Waves in Rectangular Waveguides Since , or , if , then . This means that the propagation of the wave is cut off, and is called the cutoff propagation constant. 2 2 2 c z k = k − k 2 c 2 2 k k k z = − c k = k kz = 0 c k From , we can find the cutoff frequency corresponding to the cutoff propagation constant , as given by k = 2πf c k c f 2 2 c c 2 1 2π + = = b n a k m f 2 2 2 c x y k = k + k 2 2 2 c π π + = b n a m k