电磁场与电磁波第25讲矩形波导的传播特性课件.ppt

1、Electromagnetic Field and Wave电磁场与电磁波电磁场与电磁波2011.5.212Review1.General Wave Behaviors along Uniform Guiding Structures()()zj tj tzzjtzeeeee2222220 0 xyxyE(k)EandH(k)H00021jzzxEHEhxy00021jzzyEHEhyx00021zzxHEHjhxy00021jzzyHEHhyx0()(,;)Re(,)j tzE x y z tEx y e321 (rad/m).cfkfg222211ccffkff211cggfuuu.d/d

2、f21pcuuuff2gpu uu.21 ()cTMfZf2 ()1TEcZff222hk2222220 0 xyxyE(k)EandH(k)H22 or 2cchhf 4Main topic1.Rectangular Waveguides 51.Rectangular Waveguides Select the rectangular coordinate system and let the broad side be placed along the x-axis,the narrow side along the y-axis,and the propagating direction

3、 be along the z-axis.azyxb,For TM waves,Hz=0,and according to the method of longitudinal fields,the component Ez should first be solved,and from which the other components can be derived.The z-component of the electric field intensity can be written as0ezzzE(x,y,z)E(x,y)1.1 TM waves6It satisfies the

4、 following scalar Helmholtz equation,i.e.2220220z(h)E(x,y)xyIn order to solve the above equation,the method of separation of variables is used.Let0zE(x y)X(x)Y(y),We obtain2XYhXY where X denotes the second derivative of X with respect to x,and Y denotes the second derivative of Y with respect to y.2

5、222220 0 xyxyE(k)EandH(k)H7The only way the equation can be satisfied is that both terms on the left side are constants.2XYhXY Now let 2xXkX 2ykYY where k x and k y are called the separation constants,and they can be found by using the boundary conditions.222xyhkkObviously22221200cossinxxxxxxXkXk XX

6、kjkXCk xCk x 812cossinxxXCk xCk x34cossinyyYCk yCk ywhere 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 The z-component of the electric field intensity can be w

7、ritten as01234(,)e(cossin)(cossin)ezzzzxxyyEEx yCk xCk x Ck yCk yBoundary conditions:212()0naEEazyxb,0000000000E00000yzxyzx axzytxzy bEEEEEEEE910301,2,3.000()000(1,2),3.zxzxxx azyzyyy bECmEk amkmaECnEk bnknb01234(,)e(cossin)(cossin)ezzzzxxyyEEx yCk xCk x Ck yCk y01234240(,)e(cossin)(cossin)esin()sin

8、()esin()sin()ezzzzxxyyzzEEx yCk xCk x Ck yCk ymnCCxyabmnExyab10 1 2 3ynk,nb,1,2,3,xmkmaAnd all the field components are0ezzmnEE sinxsinyab02ezxmmnEE cosx sinyhaab02ezynmnEE sinx cosyhbab02ezxjnmnHE sinx cosyhbab02ezyjmmnHE cosx sinyhaabwhere22222xyjj khj(kk)1122222xyjj khj(kk)The cutoff of a particu

9、lar mode is the condition that makes vanish.For the TMmn mode the cutoff frequency is 221 (Hz)2cmnmnfabAlternatively,we may write 222 (m)cmnmnabWhere c is the cutoff wavelength.21 (rad/m).cfkf21 ()cTMfZf12 1，电磁波的相位仅与变量，电磁波的相位仅与变量 z 有关，而振幅与有关，而振幅与 x,y 有关。因此，在有关。因此，在Z方方向上为向上为行波行波，在，在 X 及及 Y 方向上形成方向上形成

10、驻波驻波。2，z 等于常数的平面为波面。但振辐与等于常数的平面为波面。但振辐与 x,y 有关，因此上述有关，因此上述TM波为波为非均匀非均匀的平面波；的平面波；0sinsinezzmnEExyab02cjcossinezzxk EmmnExykaab 02cjsincosezzyk EnmnExykbab 02cjsincosezxEnmnHxykbab02cjcossinezyEmmnHxykaab 3，当，当 m 或或 n 为零时，上述各个分量均为零，因此为零时，上述各个分量均为零，因此 m 及及 n 应为应为非零非零的整数。的整数。m 及及 n 具有明显的物理意义，具有明显的物理意义，m

11、 为宽壁上的为宽壁上的半个驻波半个驻波的数目，的数目，n 为窄壁上为窄壁上半个驻波半个驻波的数目。的数目。4，由于，由于m 及及 n 为多值，因此场结构均具有为多值，因此场结构均具有多种模式多种模式。m 及及 n 的每一的每一种组合构成一种模式，以种组合构成一种模式，以TMmn表示。表示。例如例如 TM11表示表示 m=1,n=1 的场结的场结构，具有这种场结构的波称为构，具有这种场结构的波称为TM11波。波。5，数值大的，数值大的 m 及及 n 模式称为模式称为高次模高次模，数值小的称为，数值小的称为低次模低次模。由于。由于 m 及及 n 均不为零，故矩形波导中均不为零，故矩形波导中TM波的

12、波的最低模式最低模式是是TM11波。波。13 Select the rectangular coordinate system and let the broad side be placed along the x-axis,the narrow side along the y-axis,and the propagating direction be along the z-axis.azyxb,For TE waves,Ez=0,and according to the method of longitudinal fields,the component Hz should firs

13、t be solved,and from which the other components can be derived.The z-component of the magnetic field intensity can be written as0ezzzH(x,y,z)H(x,y)1.2 TE waves14It satisfies the following scalar Helmholtz equation,i.e.2220220z(h)H(x,y)xyIn order to solve the above equation,the method of separation o

14、f variables is used.Let0zH(x y)X(x)Y(y),We obtain2XYhXY where X denotes the second derivative of X with respect to x,and Y denotes the second derivative of Y with respect to y.01234(,)e(cossin)(cossin)ezzzzxxyyHHx yCk xCk x Ck yCk y15 Similarly,we can derive all the components of a TE wave in the re

15、ctangular waveguide,as given by0ezzmnHH cosx cosyab02ezxmmnHH sinx cosyhaab02ezynmnHH cosx sinyhbab02ezxjnmnEH cosx sinyhbab02ezyjmmnEH sinx cosyhaab where ,but both should not be zero at the same time.0 1 2,.m n.,Boundary conditions:00000000000000000000E00000yxyx axzzxzzx azzyzzyxybbyEHExHExHEyHyEE

16、EE16The cutoff of a particular mode is the condition that makes vanish.For the TEmn mode the cutoff frequency is 221 (Hz)2cmnmnfabAltermatively,we may write 222 (m)cmnmnabWhere c is the cutoff wavelength.21 (rad/m).cfkf2 ()1TEcZff TE wave has the multi-mode characteristics as the TM wave.The lowest

17、order mode of TE wave is the TE01 or TE10 wave.17 Example.The inside of a rectangular metal waveguide is vacuum,and the cross-section is 25mm10mm.What modes can be transmitted if an electromagnetic wave of frequency enters the waveguide?Will the modes be changed if the waveguide is filled with a per

18、fect dielectric of relative permittivity?MHz104f4r Solution:Due to the inside is vacuum,the operating wavelength is mm 30fcand the cutoff wavelength is 2222c25.6502nmbnamThen the cutoff wavelength of TE10 wave is ,that of TE20 wave is ,and that of TE01 wave is .The cutoff wavelength of the higher mo

19、des will be even shorter.In view of this,only TE10 wave can be transmitted in this waveguide.mm50cmm25cmm20c18 If the waveguide is filled with a perfect dielectric of ,then the operating wavelength is4rr15mmHence,TE10 and TE20 waves can be transmitted,and some other modes TE01，TE30，TE11，TM11，TE21，TM

20、21 can exist.191.3 TE10 Wave in Rectangular Waveguides Let ,we find0 ,1nm0ezzHH cosxa02ezxHH sinxhaa02ezyjEH sinxhaa 0zH(,t)H cosx cos(tz)ar022xH(,t)H sinx cos(tz)haar022yE(,t)H sinx cos(tz)haa rThe corresponding instantaneous values areAnd .0zxyEEH20Let m=1,n=0,we find the cutoff wavelength of TE10

21、 mode asa2c It means that the cutoff wavelength of the TE10 wave is independent of the narrow side.The phase velocity and the guide wavelength can be found asp212uua2g21athe energy velocity as2ec1uu21 Example.A rectangular wave-guide is filled with dielectric(perfect)medium(r=1,r=9),.and operates at

22、 a frequency 3GHz.If the dimensions of the wave-guide is a=2cm and b=1cmSolution：(1)Show that the 10TE can propagate at this frequency811 10prrcu 899101 102.5 103 1022 0.02pcufHzHza m/s 10TESo the can propagate at this frequency in the waveguide.221 (Hz)2cmnmnfab222 (m)cmnmnab222101116cff210101czpfk

23、uf2998923 102.5 10110 111 103 10(2)Determine phase constant:rad/m 21TEmncmnZff002101120240 113111116rrcff (3)Determine wave impedance21 (rad/m).cfkf2 ()1TEcZff23102101ppcuuff86 11 10/11m s(4)Determine the phase velocity21cmngmnpfuuf81110/6m s(5)Determine the group velocity21pcuuuff211cggfuuu.d/df241

24、.Rectangular Waveguides the method of separation of variablesazyxb,the method of longitudinal fieldssummary2220220z(h)E(x,y)xy0zE(x y)X(x)Y(y),2XYhXY 0000000000E00000yzxyzx axzytxzy bEEEEEEEE00000000000000000000E00000yxyx axzzxzzx azzyzzyxybbyEHExHExHEyHyEEEE25 1 2 3ynk,nb,1,2,3,xmkma,0,1,2,ynknb,0,

25、1,2,xmkmaTM waveTE wave0ezzmnEE sinxsinyab02ezxmmnEE cosx sinyhaab 02ezynmnEE sinx cosyhbab 02ezxjnmnHE sinx cosyhbab02ezyjmmnHE cosx sinyhaab 0ezzmnHH cosx cosyab02ezxmmnHH sinx cosyhaab02ezynmnHH cosx sinyhbab02ezxjnmnEH cosx sinyhbab02ezyjmmnEH sinx cosyhaab 26221 (Hz)2cmnmnfab222 (m)cmnmnab21 (rad/m).cfkf21 ()cTMfZf2 ()1TEcZffg222211ccffkff211cggfuuu.d/df2=1pcuuuff2gpu uu.27homework10-14;10-16;10-18Thank you!Bye-bye!Thank you!Bye-bye!答疑安排答疑安排时间：时间：地点：地点：