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Chap. 4 Plane Wave FunctionsThe Helmholtz equation in rectangular coordinates is

By separation of variables, assume . We have

.

The only possible solution of the above is

,

where , and are constants of , and satisfying

.The solutions of the above second order differential equations areharmonic functions of the form

Therefore, the final solution for a give set of , and can beexpressed as

The exact values of , , and the form of the harmonic functionsare determined by the boundary conditions.

General solutions1. Discrete eigenvalues (bounded regions)

2. Continuous eigenvalues (unbounded regions)

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Plane Waves

Consider the elementary wave function of the form

Let

,

then

Note in general, can be a complex vector expressed as following:

where and are real vectors. must satisfy

1. is complex ( )a. Uniform plane wave only when and are in the same

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direction.b. In general, and

2. is real ( ).a. : Uniform plane wave.b. : (ex: evanescent field in total reflection)

Let (TM to z), then

1. real: TM to z, , TEM to , the propagationdirection.

2. complex: TM to z, , TEM to but not thepropagation direction unless .

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Rectangular Waveguides

TM to z: Boundary conditions:

,,,

therefore

Similarly,TE to z:

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Note that excluded.

Also, the following equation must be satisfied.

Cut-off frequency: the minimum possible frequency for propagationmode (m,n).

The complex propagation constant is

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Wave impedance

1. TE:

a. For propagating modes ( ), .

b. For nonpropagating modes, is inductive.

2. TM:

a. For propagating modes ( ),

b. For nonpropagating modes, is capacitive.

3. .

4. It is always true that .

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Alternative Mode SetsTM to x:

TE to x:

Solutions are

Note:1. Modes are .2. Modes are 3. All others are hybrid of and .

Characteristic impedance

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Partially Filled Waveguides

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TM to xLet , where is as follow

for , and

Note that at , the tangential and must continue.

We can derive

Continuity of and at requires that

Similarly,

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Continuity of and at requires that.

From the two boundary conditions, we have

, a function of , solved

numerically.Similarly, for TE to x,

, a function of , solved

numerically.Note:1. Mode patterns are the distorted versions of , .

Concentrate in the material of higher and .2. Cutoff frequencies lie between those of waveguides filled by the

two materials.3. Cutoff frequencies of the corresponding and modes

are different.4. Knowledge of the

cutoff frequencies isnot sufficient todetermine at otherfrequencies.

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Dielectric-slab Waveguides

Assume no variation in y direction and z-traveling waves.

TM to zConsider separately two cases: (1) an odd function of , denotedby , and (2) an even function of , denoted by . For case (1),we have

where

satisfying

Evaluating the field components tangential to the air-dielectricinterface, we have

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Continuity of and at requires that

The ratio of the two equation to the second gives the characteristicequation

Similarly for even mode,

Homework #5, prove the even mode characteristic equation.

For TE to z case, we have

odd mode:

even mode:

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The phase constant of an unattenuated mode lies between the intrinsicphase constant of the dielectric and that of air; that is,

Cut-off occurs when the wave in the air is unattenuated, that is and . Thus for both TE and TM

Solving for the cut-off frequencies

Note that the lowest-order of TE and TM modes are odd modesand have zero cut-off frequency.

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Surface-Guided Waves

Equivalent to slab waveguide modes with zero tangential electricfield at , that is, and

For thick coatings, tightly bound,

For thin coatings, loosely bound,

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Corrugated Waveguides

Assume the teeth are infinitely thin and that there are many slots perwavelength such that at the surface.

Choosing TM modes of slab waveguides, we have for

The wave impedance looking into the corrugated surface is

(Inductive)

In the slot ( ), parallel-plate transmission-line mode.

(Inductive for )

Equating both, we have

1. Only approximate solution.2. Loosely bound for small .3. Tightly bound for large .

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Discontinuities and Modal Analysis (Pozar, 4.6)

Let the modes existing in a waveguide be

Assuming two waveguides and are connected by anaperture located at . Let the remaining areas atwaveguide a and b be and respectively. Assume onlythe first mode incident from waveguide , we have the totaltangential fields in

Likewise in waveguide

At the aperture , the fields at both sides must be the same,that is

(178)

(179)

And the electric fields at and must equal zero.

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Integrate the above electric field equation with the mode pattenof mode in waveguide over surface , we have

Due to the orthogonal properties between the modes in awaveguide, the above equations lead to

(186)

where

Note that is the normalization constant of mode inwaveguide . Rewriting the above Eq. (186) in matrix form, wehave

(191)

where

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(192)

Likewise, integrate the magnetic field equation (Eq. 179) withthe mode pattern of mode of waveguide only overaperture , we have

which leads to

(197)

where

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Rewriting the above Eq. (197) in matrix form, we have(199)

where

(200)

From Eq. (191) and Eq. (199), we have

(201)

Thus is solved. Using Eq. (191), we have

(203)

Thus is solved.

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Excitation of Waveguides (Pozar, 4.7)

Assume sources and exist in a waveguide between and . The fields outside this region can be expressed as

where and only have x and y components. Assume ,from reciprocity theorem, we have

Let , then and are the fields generated by , which

are , , and .

Let and , we have

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Likewise, let and , we have

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Probe-Fed Rectangular Waveguide

for

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Modal Expansions of Fields

Assume and at . Use to expand the field for .We have

.

Then,

(Double Fourier series)

Therefore,

,

where

.

For and , use . Also, the complex power

Example: waveguide step junction. Assume only the fundamentalmode propagates.

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Note: the first term is positive real, the second is positive imaginary.Let the total admittance be

.

Choose the voltage at the center. Then we have , capacitive.

Example: An inductive waveguide junction

, for

and ,

, inductive.

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Properties of Cylindrical Waveguides

Suppose a cylindrical waveguide extending in z direction. The vectorpotential can be expressed as

and satisfies

where

For TE mode, , then

The tangential component of electric field at the boundary is

where is the tangential unit vector at the boundary.Similarly, for TM mode, , then

The tangential component of electric field at the boundary is

If the boundary is PEC, the required boundary condition is

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for TE modes

for TM modes.

Obviously, the transverse electric and magnetic fields areperpendicular to each other.

In a microwave network, define mode functions and ,mode voltages and mode currents according to thetransverse components of the electric and magnetic fields

.

Furthermore, we normalize the mode vectors according to

The integration is over the cross section of the waveguide.

Prove that all the eigenvalues are real. Consider two-dimensionaldivergence theorem

Let , then

.The divergence theorem becomes

Since or over the boundary, therefore

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.

Orthogonality for TE and TM modes

For TE mode, let and be the mode patterns of mode and .From

also

We have

Similarly, for TM.Homework #6, For one TE one TM, prove the above equation holds.Hint: use .

Currents in Waveguides

Assume in a rectangular waveguide, at .Due to symmetry and B. C., . Therefore, use

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modes.Then,

Continuity of and leads to . The boundarycondition of leads to

By Fourier Series,

All the supplied power is

Example: coaxial-fed waveguides.Assume . Then,

Input impedance

, where .

If only the first mode propagates, then

Note:1. diverges because of the delta function . Can be

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avoided if the current is models according to the real sizeof the conductor.

2. For small a, capacitive.

3. Near , . Again, must model the

current correctly to avoid.

Apertures in Ground Planes (plane wave expansion)

Assume the field at the aperture is only x-directed, no z component Use TE to z.

Then,

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At ,

where

Note: to satisfy out-going wave condition

Example: Parallel-plate waveguide TEM mode incident.

Assume

Then,

Also, the aperture admittance can be computed as

where the power can be computed by Parseval’s theorem

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Plane Current Sheets

Assume at . Then use ,

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From boundary conditions at ,

where

Example: . Show that

where

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Rectangular Cavities

For , the first mode is .