Attenuation constant (low-loss dielectrics)

  • Thread starter big man
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  • #1
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[tex]\alpha =\frac {\omega \epsilon \prime \prime} {2} \sqrt {\mu \epsilon \prime} [/tex]

My question is what exactly does the [tex]\epsilon \prime \prime[/tex] term signify??

I see sections of my notes that say [tex]\epsilon \prime \prime << \epsilon [/tex], but what is it?

I mean when I am trying to find the attenuation constant and the material happens to be a low-loss dielectric and you are given the actual permittivity [tex]( \epsilon )[/tex] I don't know what to do with the [tex] \epsilon \prime \prime[/tex] term. It's the only value I don't know in the problem?

I would REALLY appreciate any help on this because I really need to understand this before my exam on Monday.
 
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  • #2
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Sorry I was looking up how to correctly write up the primes using Latex. It's fixed now.
 
  • #4
quasar987
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Dan is probably right.

The basic equation for the absorption [itex]\alpha[/itex] is

[tex]\alpha = 2\Im(k) = 2\omega \frac{\mu}{2}\sqrt{\sqrt{\Re(\epsilon)^2+\Im(\epsilon)^2}-\Re(\epsilon)}[/tex]

And my guess is that your expression is some approximation of that.

But are there really 3 epsilons?! a non prime, a primed and a double primed?!?
 
  • #5
OlderDan
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quasar987 said:
Dan is probably right.

The basic equation for the absorption [itex]\alpha[/itex] is

[tex]\alpha = 2\Im(k) = 2\omega \frac{\mu}{2}\sqrt{\sqrt{\Re(\epsilon)^2+\Im(\epsilon)^2}-\Re(\epsilon)}[/tex]

And my guess is that your expression is some approximation of that.

But are there really 3 epsilons?! a non prime, a primed and a double primed?!?
[tex] \epsilon = \epsilon \prime + i \epsilon \prime \prime[/tex]
 
  • #6
quasar987
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That is the most sensible guess, but then the expression e''<<e doesn't make sense since e is not real.
 
  • #7
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The [tex]\epsilon[/tex] in my notes is the permittivity given by this expression:

[tex]\epsilon = \epsilon_r \epsilon_0 [/tex]

Thanks for the quick replies guys and I understand this a little better, but could you just tell me if I have this right.

OK let's say that you are wanting to find the attenuation constant for a low-loss dielectric with the following properties:

[tex]\sigma=5.80*10^-^2 (S/m)[/tex]
[tex]\omega = 100 GHz[/tex]
[tex]\epsilon_r = 1[/tex]
[tex]\mu = \mu_0[/tex]

You first test to see if it is a low-loss dielectric or a good conductor.
Then finding that at that the material is a low-loss dielectric you can proceed to use the previsously stated equation for the attenuation constant (in first post). Do you say that it doesn't behave like a perfect dielectric, which means that the conduction current is not negligible and this means that you can reexpress the magnitude of the complex part as follows:

[tex]\epsilon \prime \prime = \frac {\sigma} {\omega}[/tex]

Am I right in doing this??
,
 
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  • #8
OlderDan
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big man said:
[tex]\epsilon \prime \prime = \frac {\sigma} {\omega}[/tex]

Am I right in doing this??
That apperas to be consitent with the approach taken in that link I posted earlier
 

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