jack8int.pdf

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Solutions to Problems in Jackson,Classical Electrodynamics , Third Edition

Homer Reid

May 24, 2002

Chapter 8: Waveguide Derivations

Before starting the problems, I thought it would be useful to run through

my own derivations of some of the formulas from this chapter.

Waveguides and cavities: basic pedagogy

The unifying feature of waveguide and cavity problems is that we single outone spatial coordinate and announce from the start that the elds will havesinusoidal dependence on that coordinate. Taking the special coordinate to be z,this means that all components of all elds have the functional form f (x, y )e ikzfor some wavevector k.

Assuming harmonic time dependence, we write explicitly

E (x ) = E x (x, y )i + E y (x, y ) j + E z (x, y )k ei (kz t )

B (x ) = Bx(x, y )i + B

y(x, y ) j) + B

z(x, y )k ei (kz

t )

(1)

We have here a total of six functions f (x, y ) that we must nd to satisfyMaxwells equations with the relevant boundary conditions. At rst this wouldappear tough since the six elds are all coupled by Maxwells equations, butafter a little algebra we obtain the following simplied situation: The z di-rection elds E z (x, y ) and B z (x, y ) turn out to satisfy (separately) simple one-dimensional differential equations, which may be readily solved upon specifyingthe boundary conditions for a particular situation. Meanwhile, the remainingelds (E x , E y , B x , B y ) can be expressed simply as linear combinations of E z andBz and their derivatives, so once we obtain the z elds we have everything. Inwhat follows well derive the differential equations satised by E z and Bz andthe equations giving the remaining elds in terms of them.

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Homer Reids Solutions to Jackson Problems: Chapter 8 2

The differential equations for E z and B z

The Maxwell curl equations are

E = Bt

, B = 1c2m

Et

where cm is the speed of light in the medium. We begin by applying the rstcurl equation to our ansatz (1), obtaining

y E z ikE y = iB x (2) x E z + ikE x = iB y (3)

x E y y E x = iB z , (4)and we pause to solve the rst two of these for E x and E y :

E x = k

By ik

x E z , E y = k

Bx ik

y E z . (5)

Next we apply the second curl equation to our ansatz, obtaining

y B z ikB y = ic2m E x (6)

x B z + ikB x = i c2m

E y (7)

x By y Bx = i c2m

E z . (8)

But in (5) we solved for E x and E y , and if we then plug those solutions into (6)and (7) we can solve for B x and By in terms of B z and E z :

Bx = ikc2m

2 k2c2m x B z +

c2m k

y E z (9)

By = ikc2m

2

k2c2m

y B z c2m k

x E z . (10)

Finally, with the ansatz (1) the equation B = 0 readsB xx

+ By

y = ikB z .

When we plug (9) and (10) into this, the terms involving E z elds cancel, andwe obtain an equation involving B z alone:

2

x 2 +

2

y 2B z +

2

c2m k2 B z = 0

or

2

x 2 + 2

y 2 B z + 2B z = 0 (11)


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where

= 2c2 k2 .If we had carried out this derivation in the reverse order we would have obtainedthe same equation for E z :

2x 2

+ 2y 2

E z + E z = 0 . (12)

We can think of equations (11) and (12) as eigenvalue equations that have solu-tions only for certain values of the parameter , which depend on the boundaryconditions.

Armed with equations (11) and (12) and the boundary conditions appropri-ate to our problem we can now solve for Bz and E z and then use (9) and (10)to nd the remaining components of the B eld. The remaining components of the E eld are given by analogous equations:

E x = ikc2m

2 k2c2m x E z +

k y B z (13)

E y = ikc2m

2 k2c2m y E z

k

x B z . (14)

Boundary Conditions; TE, TM, TEM Modes

The boundary conditions on the elds at the surfaces of the waveguide or cavityare that E and B

be continuous, where denotes the component of thevector normal to the boundary surface and includes all other components of the vector. This means that the two eigenvalue equations (11) and (12) must be

solved subject to different boundary conditions, which means in general theireigenvalues will be different. If we have a solution of (12) for some value of (i.e. for some combination of values of and k), then there will be no nonzerosolution of (11) for that value of , and hence we must have B z = 0 identicallyfor the mode at that frequency and wavevector. Since in this case the magneticeld has nonzero components only transverse to the direction of propagation,this is called a transverse magnetic mode. Similarly, if (11) can be solved withnonzero B z at some , then E z = 0 and we have a transverse electric mode. Amode for which both E z and B z are zero is called a transverse electromagneticmode, and can only exist in the region between two conducting surfaces, notwithin a single conductor as is possible for TE and TM modes.

Since either E z or B z is zero, we can simplify some of the equations aboveand collect results appropriate to the two cases.


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TM Modes TE Modes

Bz 0

2

tE z + 2E z = 0 , E n

S = 0

E x = ikc2m

2 k2c2m x E z

E y = ikc2m

2 k2c2m y E z

Bx = i

2 k2c2m y E z

By = i

2 k2c2m x E z

E z 0

2t B z +

2

B z = 0 , B n

n S = 0

E x = ic2m

2 k2c2m y Bz

E y = ic2m

2 k2c2m x B z

Bx = ikc2m

2 k2c2m x B z

By = ikc2m

2 k2c2m y Bz

(A factor of ei (kz t ) is understood in all of these expressions.)

For TM modes, the boundary condition is E = 0, and E z is always perpen-dicular to the boundary surfaces, so the boundary condition for the eigenvalueequation is E z = 0 . For the case of TE modes, the boundary condition is B = 0.Suppose one boundary surface is the yz plane. The normal to this plane is thex direction, so B x must vanish at this surface; but we just saw that in the TEcase Bx x B z , i.e. the derivative of B z normal to the boundary surface mustvanish. This is general: the boundary condition for the eigenvalue equation inthe TM case is Bz /n = 0 .

Power ow; Energy Loss

The ow of power down a waveguide is described by the z component of thePoynting vector S = E H = 1 E B . Using the boxed expressions above, forthe two types of modes we obtainS TMz =

1

(E x By E y Bx )

= kc2m

(2 k2c2m )2( x E z )2 + ( y E z )2 e2i (kz t )

or, in the time average,

= kc2m

2(2 k2c2m )2( x E z )2 + ( y E z )2

= k2 4 ( x E z )2 + ( y E z )2 . (15)


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Similarly, for TE modes we obtain

S TEz = k2 4

( x B z )2 + ( y B z )2 . (16)

To address the issue of dissipation in the boundaries, we solve Maxwells

equations within the boundary surfaces. The two curl equations are

E = iB (17)B = J i E

= ( i ) EE (18)

since in most cases. (For example, for a copper waveguide with airor vacuum interior we have we have 6 107 1 m1 , while 9 1012 1 m1 ( in rad/sec), so the approximation is good up to frequencies1019 rad/sec.)Now we assume that the elds are only changing signicantly in the direc-tion normal to the boundary surface (i.e., as we go deeper and deeper into theboundary surface the elds die out rapidly, whereas as we move along parallelto the boundary surface the elds dont change much) and keep only the normalderivative in the curl equations. If measures the depth of penetration into thesurface, the curl equations become

E

= iB

B

= E

Differentiating the rst of these, taking the cross product with of both sides,and substituting in the second equation yields

2E

2 = iE

or, using the bac-cab rule,

2E 2

2E2

= iE .

Evidently the component of the LHS vanishes here, so E = 0; the electriceld within the conducting boundary has no component normal to the surface.For the remaining components we obtain

2E

2 + iE = 0


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with solution

E = e i E 0= e(1+ i )

E 0

where =

2/ is the skin depth and E 0 is the eld just at the surface of

the boundary. To keep the solution from blowing up as we penetrate into theconductor we take the negative sign in the exponent. From (17) we then obtain

B = i 1

( E 0)e(1+ i )

.

Evaluating this at the surface yields the modied boundary condition on theelds in the cavity or waveguide:

B 0 = i 1

( E 0). (19)

From this equation we can work out the power loss per unit length in thecavity or waveguide. The power dissipated in a volume dV is

(J

E ) dV =

E 2 dV. We integrate over the volume occupied by the boundary surfaces ina length dz :dP = dz 0 E 20 e2(1+ i )

ddl e2i (kz t )

= dz

2(1 + i) E 20 dl e2i (kz t )or, taking the time average,

dP dz

= 4 2 E 20 dl (20)

where the line integral is over the cross section of the surface boundary at axed value of z.

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