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 Posted: Wed Oct 14, 2009 11:29 am |
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Guest
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Hi
I have a question about how the Finite Element Method
is implemented in EM. Specifically, when you are dealing
with a shielded microstrip, whose diagram looks like:
----------------------------
| |
| | air
| |
| _______ | microstrip
|-------------------------|
|/////////////////////////|
|/////////////////////////| dielectric
|/////////////////////////|
|/////////////////////////|
|/////////////////////////|
In the books I am using:
Jin, Jianming, The Finite Element Method in Electromagnetics, John
Wiley & Sons,
New York, 1993.
Volakis, John L., et. al., Finite Element Method for Electromagnetics,
IEEE Press, 1998.
the values of the Relative Permittivity, er = 4.0,
and Relative Permeability, ur = 1.0. This seems
to be the same for both the air and the dielectric.
Can someone tell me why there is no difference in
values between the air and the dielectric. I am
trying to solve the waveguide problem.
I will say my specialties are in mechanical
engineering and numerical methods rather than EE,
so I hope the answer isn't too simplistic.
Also, could someone tell me what are the boundary
conditions for the electric field on the perimeter,
on the microstrip, and the dielectric?
Thanks. |
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| Posted: Thu Oct 15, 2009 3:47 am |
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Guest
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spam_list at (no spam) yahoo.com wrote:
[quote:477409d0b1]the values of the Relative Permittivity, er = 4.0,
and Relative Permeability, ur = 1.0. This seems
to be the same for both the air and the dielectric.
Can someone tell me why there is no difference in
values between the air and the dielectric.
[/quote:477409d0b1]
Are you asking why the Permeability is the same in both
air and dielectric? That's because the Permeability
represents the magnetic response of the medium, and
neither air nor an (ordinary) dielectric have a magnetic
response; hence mu_r is always 1.
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (Photonics) (ph) +44-20-759-47734 (fax) 47714
Imperial College London, Dr.Paul.Kinsler at (no spam) physics.org
SW7 2AZ, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/ |
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| Posted: Thu Oct 15, 2009 4:42 am |
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Guest
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p.kinsler at (no spam) ic.ac.uk wrote:
[quote:b6b9249715]hence mu_r is always 1.
[/quote:b6b9249715]
The "always" is misleading -- I mean: "hence mu_r is 1"
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (Photonics) (ph) +44-20-759-47734 (fax) 47714
Imperial College London, Dr.Paul.Kinsler at (no spam) physics.org
SW7 2AZ, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/ |
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| Posted: Thu Oct 15, 2009 7:22 am |
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Guest
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On Oct 15, 1:47 am, p.kins... at (no spam) ic.ac.uk wrote:
[quote:a74eb21f33]the values of the Relative Permittivity, er = 4.0,
and Relative Permeability, ur = 1.0. This seems
to be the same for both the air and the dielectric.
Are you asking why the Permeability is the same in both
air and dielectric? That's because the Permeability
represents the magnetic response of the medium, and
neither air nor an (ordinary) dielectric have a magnetic
response; hence mu_r is always 1.
[/quote:a74eb21f33]
Thanks. I am treating this purely as a math problem and
basic facts like in your response elude me. The books I
mentioned don't(and can't) explain things on the level needed
for non-EE people to do the calculation. Still, I wish the books
could provide a complete data set of a problem. |
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| Benj... |
| Posted: Thu Oct 15, 2009 9:36 pm |
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Guest
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On Oct 15, 5:47 am, p.kins... at (no spam) ic.ac.uk wrote:
[quote:ff4bc8a909]spam_l... at (no spam) yahoo.com wrote:
the values of the Relative Permittivity, er = 4.0,
and Relative Permeability, ur = 1.0. This seems
to be the same for both the air and the dielectric.
Can someone tell me why there is no difference in
values between the air and the dielectric.
Are you asking why the Permeability is the same in both
air and dielectric? That's because the Permeability
represents the magnetic response of the medium, and
neither air nor an (ordinary) dielectric have a magnetic
response; hence mu_r is always 1.
[/quote:ff4bc8a909]
This is correct. However, the relative permittivity (actually the
dielectric constant...the real situation is much more complex
including a loss tangent term etc) is obviously not the same above and
below the stripline. It is 1 in the air and 4 in the dielectric.
(ignore losses in dielectric). If you are looking for examples in
books you may find them where the dielectric constant is the same
above and below the line. That would be because a standard way to
build such circuits is to sandwich two circuit cards together. The
stripline is etched on one side and a ground plane is the unetched
other side of the board. Two of these board together forms a system
known as a "triplate" stripline. K is the same then (4) above and
below the strip line. The example you are using is hard to build
because of the difficulty in providing a stable air gap above the
line.
Good luck! |
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| Posted: Sun Oct 18, 2009 12:51 pm |
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Guest
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Benj
[quote:5a9e42a7bf]This is correct. However, the relative permittivity (actually the
dielectric constant...the real situation is much more complex
including a loss tangent term etc) is obviously not the same above and
below the stripline. It is 1 in the air and 4 in the dielectric.
(ignore losses in dielectric). If you are looking for examples in
books you may find them where the dielectric constant is the same
above and below the line. That would be because a standard way to
build such circuits is to sandwich two circuit cards together. The
Good luck!
[/quote:5a9e42a7bf]
Thanks. Since you have experience with these devices, I'm wondering
if you or anyone could answer a few more questions.
1) I set the boundary of the shielded microstrip to be:
electrical field = 0.0
on the perimeter. Is this correct and are there other BC's I
should
use. In the book I am using, they also solve based on the
voltage(potential), and have potential = 0.0 on the boundary and
potential = 1.0 on the strip. How does this translate into a
boundary condition for the electric field.
2) In mechanical engineering, when I solve for the vibration of a
structure,
I can rely on the mathematical linear system being positive
definite
(all positive eigenvalues/frequencies) if I prescribe the problem
correctly(no rigid body movement). When I look at the mathematical
systems that arise from the waveguide problem for the microstrip,
I can see that individual matrices are at least positive-semi-
definite
(frequencies >= 0.0).
Ultimately, my books direct me to solve the waveguide to get the
propagation constant, kz, for a given wavenumber:
ko = (omega)*sqrt(permittivity*permeability)
Because these are related to the frequencies, shouldn't they always
be > 0.0?
3) The authors do say that in the general eigenvalue system that
arises:
[A][x] = lambda[B][x]
[A] is symmetric and [B] is symmetric and positive definite. Even
more
generally, I think [A] and [B] may be Hermitian, and [B] is
positive
definite since this is the type of matrix solved by LAPACK. Does
this
fit the description of the linear systems for what others have
encountered?
Thanks for any insight you can provide on any one of my questions. |
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| Benj... |
| Posted: Mon Oct 19, 2009 2:16 pm |
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Guest
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On Oct 18, 6:51 pm, spam_l... at (no spam) yahoo.com wrote:
[quote]Benj
Good luck!
Thanks. Since you have experience with these devices, I'm wondering
if you or anyone could answer a few more questions.
Thanks for any insight you can provide on any one of my questions.
[/quote]
The boundary conditions on the copper are the usual ones where the
metal is assumed to be a perfect conductor. The electric field is
always perpendicular to those surfaces. And note that all electric
fields parallel to the metal must be zero at the boundary. Also note
that the system isn't exactly "shielded" since the sides are open, but
instead relies on the rapid attenuation of the fields away from the
strip in lateral direction for isolation.
I am not sure of the solutions to the stripline as I haven't dealt
with these for many years, and I don't recall if TEM solutions are
possible or not. Generally in waveguides they are not since the TEM
electric field must be derivable as the transverse gradient of a
scalar potential that satisfies Laplace's equation. However the
triplate line is actually sort of a hybrid between a waveguide and a
standard transmission line. So I don't really know off the top of my
head.
And yes you do want to calculate the propagation of the waveguide
because you'd be interested in the cutoff wavelength of the system
which relate to the propagation constant and both the phase velocity
and waveguide wavelength, both of which are greater than that of free
space. Nevertheless, of course, it does not mean that superluminal
information transfer can occur since information is transmitted by
group velocity.
That's about what I know right now. |
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| ... |
| Posted: Tue Oct 20, 2009 2:38 pm |
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Guest
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Benj
[quote]The boundary conditions on the copper are the usual ones where the
metal is assumed to be a perfect conductor. The electric field is
always perpendicular to those surfaces. And note that all electric
fields parallel to the metal must be zero at the boundary. Also note
that the system isn't exactly "shielded" since the sides are open, but
instead relies on the rapid attenuation of the fields away from the
strip in lateral direction for isolation.
[/quote]
Thank you very much for your response. I'm beginning to get
a handle on the physics now and the reasoning behind what is
given in my books. Your description of boundary conditions
is especially helpful. I envy your knowledge of this subject. |
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| Benj... |
| Posted: Tue Oct 20, 2009 9:33 pm |
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Guest
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On Oct 20, 8:38 pm, spam_l... at (no spam) yahoo.com wrote:
[quote]Benj
Thank you very much for your response. I'm beginning to get
a handle on the physics now and the reasoning behind what is
given in my books. Your description of boundary conditions
is especially helpful. I envy your knowledge of this subject.
[/quote]
Glad I could help. What you should envy, however, is that I somewhere
(lost in my house) have a book that was put out by some manufacturer
way back when, that had just about everything you needed to know in
it. Unfortunately these days that esoteric knowledge seems to have
become lost like knowing how to make a decent mummy etc. |
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