WAVEGUIDES Waveguiding structures are called waveguides and may consist of a single conductor. Transmission lines

tend to guide TEM waves using two or more conductors waves with no field component in the direction

of the propagation. However, waveguides also support propagation of:

Transverse Electric with a longitudinal magnetic field component but no electric field in

the direction of propagation, . They are also known as H waves.

Transverse Magnetic with a longitudinal electric field component , but no magnetic field

component, . They are also known as E waves

Both TE and TM modes have

characteristic cutoff frequencies below

which waves cannot propagate, and

hence power and signal transmission at

those modes is not possible. Thus

waveguides operating in TE and TM

modes are like high pass filters and

only certain patterns of electric and

magnetic fields (modes) can exist for

propagating waves.

In the next part we shall see that these

modes must be solutions to the

governing wave equation while

satisfying appropriate boundary

conditions for the fields.

GENERAL SOLUTIONS FOR ARBITRARY WAVEGUIDE Consider an arbitrary transmission line (waveguide) as shown above situated along the z-direction. We

assume that the fields are time harmonic fields with an dependence. The electric and magnetic fields

can thus be written as the sum of transverse field components and longitudinal

components and :

Which satisfy the source free Maxwells equations:

We then expand the curl operator of the source free Maxwells equations in rectangular coordinates the

derive the derivatives of the transverse field components with respect to z:

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We also note from our definition of the electric fields that:

Then the original curl expansion will yield us:

Similarly with the magnetic field:

We then solve the above equations for and

With (2a) and (1b):

Similarly:

Where we define a the cutoff wavenumber: . We also note that the

wavenumber of the material filling the waveguide is

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To summarise, the following equations of transverse field components are valid for all modes defined

previously:

TEM WAVES TEM waves are characterized by the lack of longitudinal components . We find that

the transverse fields of a TEM mode are non zero only when . Substitution into:

We get:

This indicates that the phase constant of the TEM mode on the guiding structure is equivalent to

the phase constant of a plane wave propagating in a region characterized by the same medium

between the conductors of t he guiding structure. Furthermore, the TEM modes can be

propagated at any non zero frequency since .

We also define the wave impedance of TEM waves from equation (1b) and (2a)

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Combining the two previous results gives us:

We also note that if we consider the Helmholtz wave equation where and

:

Then after calculations using the Laplacian operator and noting that

then:

Which can be expressed in a simpler form where

is the Laplacian operator for two

transverse direction:

The above equations indicate that the transverse electric and magnetic fields satisfy Laplaces equation

with boundary conditions defined by the conductor geometry of the guiding structure.

It should be noted that single conductor waveguides cannot support TEM waves since there are no closed

loops of magnetic field lines in any transverse plane nor longitudinal conduction current .

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TRANSVERSE ELECTRIC In transverse electric waves, and can be supported inside closed conductors and two or more

conductors. Substitution into the general solutions for the four transverse field components yields:

In this case and the propagation constant is given by the relationship

in order to

yield bounded solutions for the transverse field components of TE modes.

From the wave equation we find:

Since then:

This equation represents a reduced Helmholtz equation which can be solved for based on the

boundary conditions of the guiding structure geometry. From the boxed equations we also find that:

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TRANSVERSE MAGNETIC WAVES Starting again with the general solutions of transverse waves:

And noting that will reduce the solutions to:

Again the cutoff wavenumber must be nonzero to yield bounded solutions for the transverse field

component of TM modes so that we must operate the guiding structure above the corresponding cutoff

frequency for the particular TM mode to propagate. The longitudinal field component must satisfy the

wave equation so that:

Since the guided wave electric field is then:

The above reduced Helmholtz equation can be solved for based on the boundary conditions of

the guiding structure geometry. Once is found, the longitudinal magnetic field is known and all of

the transverse field components are found by evaluating the above derivates.

The wave impedance for TM modes is found as:

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ANALYSIS OF TE AND TM WAVEGUIDES In the next section we shall present the analysis for TE and TM waveguides by following the given

procedures:

1. Solve the reduced Helmholtz equation for or which will contain unknown constants

including the unknown

2. Find the transverse fields from or

3. Apply boundary conditions to appropriate field components to find the cutoff wavenumber and

unknown constants

4. Find:

RECTANGULAR WAVEGUIDES The rectangular waveguide can support either TE or

TM modes. The rectangular cross section (a>b) allows

for single mode operation one mode propagates in

the waveguide over a given frequency range.

TE MODE The longitudinal magnetic field of the TE modes within

the rectangular waveguide satisfy:

1. Solve Helmholtz equation

From this the reduced wave equation is:

With the longitudinal component:

Let, . Using method of separation of variables:

The partial derivates become derivates since each term is independent of each other so:

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2. Transverse Field

3. Boundary conditions

The TE boundary conditions are the walls of the waveguide:

From the general TE equations:

The application of the boundary conditions yields:

Thus, the final solution for is:

The four transverse field components for the propagation mode can then be found from

substituting into the general solutions:

is an arbitrary amplitude constant composed of constants A and C

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4. Find other constants and values including the cut off frequency and propagation constant and

wave impedance

Since:

Where is the intrinsic impedance of the filling material. The guide wavelength is:

And phase velocity is:

The notion of dominant mode refers to the propagation mode with the lowest cutoff frequency for any of

the rectangular waveguide modes; in this case it is since we chose a>b. This implies that will

not exist as all the field components are zero. When , is imaginary and hence the wave will decay

exponentially and will not propagate.

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TM MODE The longitudinal electric field of the TM modes within the rectangular waveguide must satisfy the wave

equation . Expansion and simplification leads to:

With . By a similar nature to the TE mode then:

Applying the boundary conditions:

Which gives us:

This allows us to substitute back into the four transverse equations yielding:

Just like in the TE mode we have:

Which is real for the propagating mode but imaginary for the evanescent mode. The cut-off frequency is:

While the guide wavelength and phase velocity are:

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We summarise the information for empty rectangular waveguides:

Parameter

Parameter

Example

Determine the wave impedance and guide wavelength at a frequency equal to twice the cutoff frequency,

in a waveguide for TM and TE modes.

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Example

In AM, the sum of signal frequencies and carrier frequency forms an envelope representing the signal

information. The carrier signal exhibits a phase velocity of

while the envelope propagates at the

group velocity

. Determine the group velocity for a given mode in an air filled rectangular

waveguide.

In an air filled rectangular waveguide, the speed thus:

Note that while the phase velocity within the air filled waveguide can be greater than the speed of light,

the envelope travels at the group velocity less than the speed of light.

Example

Consider a rectangular waveguide with dimensions 1cm X 0.5 cm.

a) What are the cut off frequencies for the first 5 modes?

b) If the waveguide is excited at 20 Ghz, what are the propagation constants for the first five modes

c) If the waveguide is excited at 50 Ghz, how many modes will propagate?

a) Recall that , hence:

Now non-magnetic materials we shall treat , so we can express the cut off frequency as:

For convenience we also assume that , hence this will become an air filled waveguide

For m=0, n=0, there is no mode

For m=1, n=0,

For m=0, n=1,

For m=1, n=1,

For m=2, n=0,

For m=2, n=1,

For m=0, n=2,

For m=1, n=2,

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Example

Mode m n TE 1 0 15

TE 0 1 30 TE 2 0 30

TE 1 1 33.54

TM 1 1 33.54

b) Given that

, with

. Using

the first five modes of propagation as above:

c) Now we have

. In order for propagation to occur, we must have:

We find that:

There are thus 8 possible modes of propagation.

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Example

Write the general instantaneous field expressions for TM and TE modes and deduce those for and

modes.

For

With

For ,

Example

In a rectangular waveguide for which a=1.5 cm, b=0.8cm ,

Determine:

a) The mode of operation

b) The cut-off frequency

c) The phase constant

d) The complex Propagation constant

e) The instrinsic wave impedance .

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a) Given that we have a time dependent function for obtained from , then

comparison with the solutions for rectangular waveguides yields either a mode

b) Suppose we operate in a mode:

Instead, if we consider a mode, the same cutoff frequency is found.

c) The phase constant is:

Where

d) The propagation constant is purely imaginary:

e)

Example

Consider a standard waveguide with a=8cm b=4cm and air dielectric.

a) What wave modes can propagate along this waveguide at

b) What is the dominant frequency range of the waveguide?

a) note that

All other modes have cutoff frequencies higher than the operational frequency.

b) The dominant range in which one wave mode only can propagate (mode TE10) is given by

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Example

A rectangular air filled waveguide has a cross section 90mm X 25 mm. Find

a) Cutoff wavelength for the dominant mode

b) The relative phase velocity (v/c) in guide at 1.6 times the cutoff frequency

c) The cutoff wavelength if the guide is filled with dielectric of

d) The relative phase velocity in guide with dielectric at 1.6 times the cutoff frequency

a)

Or use:

b)

Note that

c)

Or use

d)

Example

Standard air filled waveguides have been designed for radar bands. One type, designated WG-16 is

suitable for X band applications and its dimensions are a=2.29 cm and b=1.02 cm. If it is desired that a

WG-16 waveguide only to operate in mode and that the operating frequency be at least 25% above

the cutoff frequency of mode but no higher than 95% of the next higher cutoff frequency, determine

the allowable operating frequency range.

The two modes having the lowest cutoff frequencies are TE10 and TE20. We thus find:

Thus the operating range is:

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