Waveguide notes 018 Electromagnetic waves in free space We start with Maxwell s equations for an LIH medum in the case that the source terms are both zero. = =0 =0 = = Take the curl of Faraday s law, then use Ampere s law: = = = Use the first Maxwell equation (the "H" in LIH assures us that spatial derivatives of are zero 1 ), we obtain the wave equation with wave speed =1 = = A similar derivation gives the same equation for wave solution: = 0 exp = 0 exp + Now let s look at a plane where = the wave phase speed. By including the phase constant in the expression for we allow for a possible phase shift bewteen Inserting these expressions into Maxwell s equations, we have = 0 0 =0 (1) = 0 0 =0 Thus both are perpendicular to the direction of propagation. Faraday s law 0 exp = 0 exp + From 1 Here we also assume that is independent of 1
Since this relation must be true for all is real, we have =0( oscillate in phase) 0 = 0 = 1 ˆ 0 () Thus is also perpendicular to its magnitude is If the waves propagate in a vacuum, the derivation goes through in the same way the only difference is that the wave speed is =1 0 0 In an LIH medium, = where the refractive index = p 0 0 ' p 0 Electromagnetic fields in a wave guide A wave guide is a region with a conducting boundary inside which EM waves are caused to propagate. In this confined region the boundary conditions create constraints on the wave fields. We shall idealize, assume that the walls are perfect conductors. If they are not, currents flowing in the walls lead to energy loss. See Jackson Ch 8 for a discussion of this case, especially sections 1 5. The boundary conditions at the walls of our perfectly conducting guide are (see notes 1 eqns 10,1,13 15): where Σ is the free surface charge density on the wall, ˆ = Σ (3) ˆ =0 (4) ˆ =0 (5)
ˆ = (6) where isthefreesurfacecurrentdensity. (Notethatwehaveassumed =0 inside the conducting material. This is true here because all our fields are assumed to be time-dependent, then non-zero implies non-zero by Faraday s law. Non-zero is not allowed inside a perfect conductor, so must be zero too. ) Since we do not know Σ or equations (4) (5) will be most useful. Now we use cylindrical coordinates with ˆ along the guide in the direction of wave propagation. The transverse coordinates will be chosen to match the cross-sectional shape of the guide Cartesian for a rectangular guide polar for a circular guide. Next we assume that all fields may be written in the form = 0 () We are not making any special assumptions about the time variation, because we can always Fourier transform the fields to get combinations of terms of this form. Then Maxwell s equations in the guide take the form: = =0 =0 = Taking the curl of Faraday s law, inserting from Ampere s law, we get: = = = = (7) This equation is the same as we obtained for free space. Note that is usually a function of Next we look for solutions that take the form of waves propagating in the direction, that is: 0 () = ( ) Thewaveequation(7)thenbecomes: t + =0 (8) where t is the Laplacian operator in the two transverse coordinates ( or for example.) Thus equation (8) is an equation for the function of the two transverse coordinates. Since we were able to simplify the equations by separating the function into its dependence on the coordinates along transverse to the guide, we 3
nowtrytodothesamethingwiththecomponents. Atfirst glance, based on equation (1), you might want to jump to the conclusion that there is no component of a wave propagating in the direction, but in general there is. The waves are propagating between conducting boundaries, we have to allow for the possibility that waves travel at an angle to the guide center-line, bounce back forth off the walls as they travel. Since is perpendicular tothewavevector, in such a bouncing wave has a component. The total electromagnetic disturbance in the guide is a sum of such waves. The sum is a combination of waves that interfere constructively. Thus we take = ˆ + t similarly = ˆ + t This decomposition simplifies the boundary conditions, since the normal ˆ on the boundary has no component. Then eqn (5) becomes ˆ t =0on (9) However equation (4) has two components. The transverse component gives while the component gives =0on (10) ˆ t =0 on (11) Next we put these components into Maxwell s equations. The divergence equations are scalar equations, so let s start with them: µ =0= + t t evaluating the derivative, we get Similarly: + t t =0 (1) + t t =0 (13) We separate the curl equations into transverse components. Take the dot product of Faraday s law with ˆ: ˆ = =ˆ t t (14) also the cross product: ˆ =ˆ 4
Let s investigate the triple cross product on the left. Since ˆ is a constant, we may move it through the operator in the BAC-CAB rule: ˆ = ˆ ˆ =ˆ t The derivative times ˆ appears in both terms in the middle, so cancels, leaving: t t = ˆ t evaluating the derivative, we get t t = ˆ t (15) Similarly, from Ampere s law, we have the transverse component: t t = ˆ t (16) the component =ˆ t t (17) Equations (1), (13), (17) (14) show that the longitudinal components act as sources of the transverse fields t t. Now we can simplify by looking at the normal modes of the system. Transverse Electric (TE) (or magnetic) modes. In these modes there is no longitudinal component of : 0 everywhere Thus boundary condition (10) is automatically satisfied. The remaining boundary conditions are (9) (11), we can find the version that we need by taking the dot product of ˆ with equation (16): ˆ t ˆ t = ˆ ˆ t The second term is zero on (eqn 9), we rearrange the triple scalar product on the right, leaving: = ˆ ˆ t =0on (18) where we used the other boundary condition (11) for Transverse Magnetic (TM) (or electric) modes. In these modes there is no longitudinal component of : 0 everywhere 5
Thus the boundary condition (18) is trivially satisfied, we must impose the remaining condition (10) =0on S. Since the Maxwell equations are linear, we can form superpositions of these two sets of modes to obtain fields in the guide with non-zero longitudinal components of both Thesemodesaretheresultoftheconstructiveinterference mentioned above. Transverse electromagnetic (TEM) modes In these modes both are zero everywhere. Then from (1) (14), t t t t are zero everywhere. This means we can express t as the gradient of a scalar function Φ that satisfies Laplace s equation in two dimensions. The boundary condition (11) becomes µ ˆ Φ Φ =ˆ ˆ =0 on where is a coordinate parallel to the surface Since ˆ ˆ are perpendicular, ˆ ˆ is not zero, so Φ = constant on therefore is constant everywhere inside the volume making t =0 Thus these modes cannot exist inside a hollow guide. They may exist, in fact become the dominant modes, inside a guide with a separate inner boundary, like a coaxial cable. We will not consider them further here. Now let s see how the equations simplify for the TE TM modes.. TM modes We start by findinganequationfor Since 0 equation (16) simplifies to: t = ˆ t t = ˆ t (19) we substitute this result back into equation (15). t + t = ˆ ˆ t t = t t = µ1 t (0) finally we substitute this result for back into equation (1): t + t 1 = 0 t + = 0 6
or t + =0 (1) with () Equation (1) is the defining differential equation for Once we have solved for we can find t from equation (0) then t from equation (19). TE modes The argument proceeds similarly. We start with equation (15) with =0, to get: t = ˆ t (3) substitute into equation (16) t + t = ˆ ˆ t µ t = t (4) Then finally from equation (13) we have: + t t = 0 t + = 0 which is the same differential equation that we found for in the TM modes. The solutions are different because the boundary conditions are different. Thus thesolutionforthetwomodesproceedsasfollows: TM modes TE modes assumed 0 0 differential equation t + =0 t + =0 This is an eigenvalue/ boundary condition =0on S =0on S eigenfunction problem. next find = t = t then find t = ˆ t t = ˆ t The differential equation plus boundary condition is an eigenvalue problem that produces a set of eigenfunctions (or ) a set of eigenvalues The wave number is then determined from equation (): This equation is the Helmholtz equation. = (5) 7
Clearly if is greater than = where is the wave phase speed in unbounded space, becomes imaginary the wave does not propagate. There is a cut-off frequency for each mode, given by = If is the lowest eigenvalue for any mode, the corresponding frequency is the cutoff frequency for the guide, waves at lower frequencies cannot propagate in the guide. A few things to note: the wave number is always less than the freespace value thus the wavelength is always greater than the free-space wavelength. The phase speed in the waveguide is = = 1 1 p 1 1 = we can differentiate eqn (5) to get = Thenthegroupspeedintheguideis = = = 1 = 1 r 1 1 = Thus information travels more slowly than if the wave were to propagate in free space. TM modes in a rectangular wave guide We use Cartesian coordinates with origin at one corner of the guide. Let the guide have dimensions in the -direction by in the direction. Let the interior be full of air so 0 = 0 ' 1 Then the differential equation for is (1) µ + + =0 As usual we look for a separated solution, choosing = () () to obtain: 00 + 00 + =0 Each term must separately be constant, so we have: 00 = 00 = + =0 8
The boundary condition is =0on so: =0at =0 = =0at =0 = Thus the appropriate solutions are =sin =sin with eigenvalues chosen to fit the second boundary condition in each coordinate: = = Thus, putting back the dependence on we have = sin sin (6) = + (7) Notice that the lowest possible values of are 1 in each case, since taking or =0would render identically zero. Thus the lowest eigenvalue is r 1 11 = + 1 the cutoff frequency for the TM modes is: r 1 c,tm = 11 = + 1 = r 1+ Jackson solves for the TE modes (pg 361). The eigenvalues are the same, but in this case it is possible for one (but not both) of to be zero, leadingtoalowercutoff frequency for the TE modes: TE = assuming This would be the cutoff frequency for the guide. IntheTMmode,theremainingfields are (eqns 0 6): t = t sin sin = ˆ cos sin + ˆ sin cos (eqn 19) t = ˆ ˆ cos sin + ˆ sin cos = ˆ cos sin ˆ sin cos 9 (8) (9)
where (eqns 5 7) r = As usual, the physical fields are given by the real part of each mathematical expression, so that, for example, t sin ( ). You should verify that these fields satisfy the boundary conditions at =0 at =0 Power The power transmitted by the waves in the guide is: () = 1 0 where here we must take the real,physical fields. Usually we are interested in the time-averaged Poynting flux, which is given by =Re 1 0 (30) where the fields on the right are the complex functions we have just found. Proof of this result: If = 0 =ˆ 0 similarly for where 0 0 are real, then = 1 Re Re 0 the time average is = ˆ ˆ 0 0 cos ( )cos( ) 0 = ˆ ˆ 0 0 (cos cos +sin sin )(cos cos +sin sin ) 0 = ˆ ˆ 0 0 cos cos cos + 0 (cos sin +cos sin )cos sin +sin sin sin = ˆ ˆ 0 0 0 (cos cos +sin sin )=ˆ ˆ 0 0 0 cos ( ) = Re 1 0 so the two results are the same. (See also Notes page 11.) 10
Using the solution (8, 9), the components of are: =Re 1 0 = Re 1 0 ( ) cos sin + sin cos = 0 4 cos sin + sin cos p = 0 4 cos sin + sin cos q = 0 i cos sin + sin cos h + Check the dimensions! is positive for all values of showing that power is propagating continuously along the guide in the positive -direction. The transverse component is: =Re 1 µ 0 0 sin sin ( ) = Re 1 0 = 0 1 =Re 0 0 cos sin Because there is no real part, the time averaged power flowing across the guide is zero. Power sloshes back forth, but there is no net energy transfer. Fields in a parallel plate wave guide By simplifying the shape of the guide even more, we can demonstrate how thewavemodesareformedbyreflection of waves at the guide walls. Let this guide exist in the region 0 with infinite extent in For the TM modes, theequationtobesatisfied is: t + =0 with =0at =0 = Because the region is infinite in the direction, the appropriate solution has no dependence: = sin (31) with = Thus the cutoff frequency for these modes is TM = 1 = (3) 11
Then the other components of the fields are: = = cos ˆ = cos ˆ (33) = ˆ = cos ˆ = cos ˆ (34) Let s look at the electric field first. We write the sine cosine as combinations of complex exponentials. From (31), µ = = µ + from (33) = µ + Thus we can write the electric field as a superposition total = 1 1 + where the two superposed fields are Similarly: with t = ˆ Now define the four vectors 1 = (ˆ ˆ) = (ˆ + ˆ) µ + exp ( + ) exp ( ) = 1 1 + 1 = ˆ exp [ ( ± )] 1 = ˆ ˆ; 1 = ˆ + ˆ; = ˆ + ˆ = ˆ + ˆ 1
Then for =1 thevectorsareperpendicular: =0 1 1 =(ˆ + ˆ) (ˆ ˆ) = ˆ + = ˆ = where we used equation (5). The vectors are shown in the diagram below. The two electric field components are then: while 1 = 1 exp ( 1 ) = exp ( ) 1 1 = ˆ exp [ ( + )] = 1 consistent with () (34). Each of these sets of fields ( 1 1 ) has the form of a free-space wave propagating in the direction given by the vectors 1 respectively with wave number 1 = = p + =. These waves are moving across the guide at an angle given by tan = = ± = ± q that is, the waves are reflecting off the plates at =0asshowninthe diagram. When the angle becomes the wave ceases to propagate along the guide, but just bounces back forth. This happens when tan or = This gives the cut-off frequency (3) we found before. See also Lea Burke pages 1058-1060. 13