Complex permittivity

Status
Not open for further replies.

polarmystery

Diamond Member
Aug 21, 2005
3,888
8
81
Hey guys,

Anyone have any idea how you would model an equation for the complex permittivity of a substance given it's loss tangent and relative permittivity graphs? I'm a little stuck on this one. Cheers.

graphsq.jpg
 

Farmer

Diamond Member
Dec 23, 2003
3,334
2
81
By frequency dependent dielectric model, do they mean frequency-dependent permittivitty and conductance?

I'm not an EE, but the definition for the loss is from Wiki:

96500035ca5369537a03f97dd599bb7b.png


And from what I understand the Wiki its some measure of the time-damping of the EM field in a dielectric (both from resistance and from bound charge).

Well you already know the real part of permittivity (e' in the above) and you also know tan delta for the frequency range given. You can back out imaginary part of permittivity given you have some information about the conductance (or maybe you guys work with impedance).

I definitely am not an expert, but if this is a school problem, this seems like what they mean.
 

Born2bwire

Diamond Member
Oct 28, 2005
9,840
6
71
By frequency dependent dielectric model, do they mean frequency-dependent permittivitty and conductance?

I'm not an EE, but the definition for the loss is from Wiki:

96500035ca5369537a03f97dd599bb7b.png


And from what I understand the Wiki its some measure of the time-damping of the EM field in a dielectric (both from resistance and from bound charge).

Well you already know the real part of permittivity (e' in the above) and you also know tan delta for the frequency range given. You can back out imaginary part of permittivity given you have some information about the conductance (or maybe you guys work with impedance).

I definitely am not an expert, but if this is a school problem, this seems like what they mean.

If you have the real/imaginary part then you can get the associated imaginary/real part of the permittivity or permeability using the Kramers-Kronig relation (which is really a Hilbert Transform). Technically, this requires you to know the entire bandwidth for the known part which is never known. What most people do is know the behavior over a limited bandwidth and then assume the limiting cases and make a smooth transition to those limiting cases. This gives us an approximation of the complimentary part over a similar limited bandwidth (like in the Debye model or Cole-Cole).

But in regards to the OP, it would seem that the relative permittivity is the coefficient for the real part and you use the loss tangent to calculate the imaginary part. The loss tangent is the ratio of the imaginary to the real part of the permittivity. So just scale the loss tangent by your real part to get the imaginary part. Thus,

\epsilon = \epsilon_0 \epsilon_r ( 1 + \tan \delta )

EDIT: Regarding the Wiki equation. We really can ignore their inclusion of the conductivity. That's a rather confusing equation because the imaginary part of the permittivity INCLUDES the conductivity of the material. It seems that this would imply double counting the conductivity. However, I think what they are doing is allowing for you to take into account any conductance current that would be included in the relation that they are using to derive the loss tangent. They get it by taking the curl of the magnetic field which would include a contribution from the time varying electric flux density (which gives us the factor of permittivity) and any source currents (which can be taken as the product of the conductivity and the electric field via Ohm's Law). But for the most part, we are not looking at a point where source currents exist and so the contribution from that conductivity term would not be included.
 
Last edited:

polarmystery

Diamond Member
Aug 21, 2005
3,888
8
81
I don't believe we are supposed to use any form of conductance to this equation we are supposed to come up with. In the book, the only thing they mention is the Debye model as Born said to show permittivity behavior decreasing as the loss tangent increases.

They show a few examples where 'poles' of the medium (when the oscillating frequency is equal to the harmonic frequency of the dielectric) cause the loss tangent to evaporate over time, but they do not include how they arrived at that conclusion. I'm under the assumption one of these poles arrives at 17GHz according to the loss tangent graph when it switches directions, but beyond that I'm not sure using the Debye model to approximate epsilon_r would be accurate in this case.
 

Born2bwire

Diamond Member
Oct 28, 2005
9,840
6
71
I don't believe we are supposed to use any form of conductance to this equation we are supposed to come up with. In the book, the only thing they mention is the Debye model as Born said to show permittivity behavior decreasing as the loss tangent increases.

They show a few examples where 'poles' of the medium (when the oscillating frequency is equal to the harmonic frequency of the dielectric) cause the loss tangent to evaporate over time, but they do not include how they arrived at that conclusion. I'm under the assumption one of these poles arrives at 17GHz according to the loss tangent graph when it switches directions, but beyond that I'm not sure using the Debye model to approximate epsilon_r would be accurate in this case.

The Debye model only has a single resonance. The resonance manifests itself as a sign change in the imaginary part and a peak in the real part. So looking at your data, the loss tangent is just a scaled imaginary part (albeit it is not a constant scaling). So we see that the first data point represents a smooth peak in your real part and a zero in your imaginary part. If we are to apply the Debye model then I would say that your first data point represents the resonant frequency of your Debye equation. Take a look at how the Debye model looks:

f3.gif


So it looks like we have our resonance at zero (if that is possible) and then you mention that we need another pole so that the imaginary part diminishes. As to how that happens I do not know off-hand. However, the relationship between the real and imaginary parts is that of a Hilbert Transform. So one can probably make use of the property of the transform to give a rough model for the behavior in such the same way as we do for circuit filters via the Fourier Transform. That is, a circuit filter is defined by the location of the poles and zeroes and these translate to knees and so forth in the frequency domain. I would imagine that similar relationships can be found for the Hilbert Transform that would allow you to construct a rough estimation using a similar system of poles and zeros in the transfer function.

One thing though is that I am not used to looking at the loss tangent as it reflects the behavior of both the real and imaginary parts. So I am not sure how to directly relate the behavior of the loss tangent here.
 

polarmystery

Diamond Member
Aug 21, 2005
3,888
8
81
The Debye model only has a single resonance. The resonance manifests itself as a sign change in the imaginary part and a peak in the real part. So looking at your data, the loss tangent is just a scaled imaginary part (albeit it is not a constant scaling). So we see that the first data point represents a smooth peak in your real part and a zero in your imaginary part. If we are to apply the Debye model then I would say that your first data point represents the resonant frequency of your Debye equation. Take a look at how the Debye model looks:

f3.gif


So it looks like we have our resonance at zero (if that is possible) and then you mention that we need another pole so that the imaginary part diminishes. As to how that happens I do not know off-hand. However, the relationship between the real and imaginary parts is that of a Hilbert Transform. So one can probably make use of the property of the transform to give a rough model for the behavior in such the same way as we do for circuit filters via the Fourier Transform. That is, a circuit filter is defined by the location of the poles and zeroes and these translate to knees and so forth in the frequency domain. I would imagine that similar relationships can be found for the Hilbert Transform that would allow you to construct a rough estimation using a similar system of poles and zeros in the transfer function.

One thing though is that I am not used to looking at the loss tangent as it reflects the behavior of both the real and imaginary parts. So I am not sure how to directly relate the behavior of the loss tangent here.

The Hilbert transform section of our book doesn't happen for a few chapters later, so I am under the impression that we aren't supposed to use that.

I'm already to the point where I have no desire to become a signal integrity engineer, that's for sure...
 

PsiStar

Golden Member
Dec 21, 2005
1,184
0
76
I'm already to the point where I have no desire to become a signal integrity engineer, that's for sure...
Oh please, this is just mathematical curve fitting. There are also the Drude-Lorentz and Debye-Lorentz material models. Were you awake in class for those 15 minutes?
 

polarmystery

Diamond Member
Aug 21, 2005
3,888
8
81
Oh please, this is just mathematical curve fitting. There are also the Drude-Lorentz and Debye-Lorentz material models. Were you awake in class for those 15 minutes?

I am always awake in class (Considering it cost me $3,000). Our instructor did not go over either of those models. The only models mentioned in our class was the Debye model. Normally I'd agree with the assessment that I might not have a firm grasp of the material (which I do not), but when 71% of your students fail the midterms (5 out of 7 students), something is up. But I digress...

I've never even heard of the Drude model before. For reference, here is our books where this came from:

http://books.google.com/books/about/Advanced_signal_integrity_for_high_speed.html?id=J8zE7u_C26oC
 

PsiStar

Golden Member
Dec 21, 2005
1,184
0
76
I was obnoxious in my comment. You don't get to these classes by snoozing.

That book is a keeper for referencing in the future. I have it somewhere. I am self employed in EMC/SI RF work & have a lot of work. I'm even on the East Coast & not the Bay Area.

Keep with it ... if I can't challenge then I'll encourage.
 

polarmystery

Diamond Member
Aug 21, 2005
3,888
8
81
Well luckily, it's over.

Almost nobody in our class got the right answer. Our professor had to use a curve-fitting program to get it work as well. Furthermore, he said that was the most difficult homework assignment for the semester...but color me skeptical :hmm:
 

PsiStar

Golden Member
Dec 21, 2005
1,184
0
76
I suppose that I should have said that a friend of mine at cst.com & the VP of Engineering, produced one of the 1st curve fitting utilities for Debye material models.

If you talk your prof into extra credit because of your insatiable interest & personable challenge for extra credit,:whiste: maybe we can get some very useful info from him. Companies as this really like to help academia.
 

polarmystery

Diamond Member
Aug 21, 2005
3,888
8
81
I suppose that I should have said that a friend of mine at cst.com & the VP of Engineering, produced one of the 1st curve fitting utilities for Debye material models.

If you talk your prof into extra credit because of your insatiable interest & personable challenge for extra credit,:whiste: maybe we can get some very useful info from him. Companies as this really like to help academia.

Oh don't get me wrong, I find the material highly fascinating, but I think this class hits my limits of academia. I will however, let him know anything you want to tell him...I have never had a problem asking him questions and giving him information, he knows this.
 

sm625

Diamond Member
May 6, 2011
8,172
137
106
Just use a lookup table. Calculus is for nerds. Flash and RAM are cheap.
 
Status
Not open for further replies.