wave guides
DESCRIPTION
MicrowaveTRANSCRIPT
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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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