# Permittivity and Permeability in shielded microstrip

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.
âœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
spam snipped-for-privacy@yahoo.com wrote:

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
âœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
snipped-for-privacy@ic.ac.uk wrote:

The "always" is misleading -- I mean: "hence mu_r is 1"
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
âœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
On Oct 15, 1:47 am, snipped-for-privacy@ic.ac.uk wrote:

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.
âœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
On Oct 15, 5:47 am, snipped-for-privacy@ic.ac.uk wrote:

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!
âœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Benj

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.
âœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
On Oct 18, 6:51 pm, spam snipped-for-privacy@yahoo.com wrote:

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.
âœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
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.
âœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
On Oct 20, 8:38 pm, spam snipped-for-privacy@yahoo.com wrote:

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.
âœ–
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

## Site Timeline

• ### 1500 mile e-car for Austin in UK non-toxic fuel cell

• - the site's last updated thread. Posted in ⏚ Electrical Engineering
• Share To

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.