# Attenuation constant (low-loss dielectrics)

1. Nov 10, 2006

### big man

$$\alpha =\frac {\omega \epsilon \prime \prime} {2} \sqrt {\mu \epsilon \prime}$$

My question is what exactly does the $$\epsilon \prime \prime$$ term signify??

I see sections of my notes that say $$\epsilon \prime \prime << \epsilon$$, 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 $$( \epsilon )$$ I don't know what to do with the $$\epsilon \prime \prime$$ 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.

Last edited: Nov 10, 2006
2. Nov 10, 2006

### big man

Sorry I was looking up how to correctly write up the primes using Latex. It's fixed now.

3. Nov 10, 2006

### OlderDan

Last edited: Nov 10, 2006
4. Nov 10, 2006

### quasar987

Dan is probably right.

The basic equation for the absorption $\alpha$ is

$$\alpha = 2\Im(k) = 2\omega \frac{\mu}{2}\sqrt{\sqrt{\Re(\epsilon)^2+\Im(\epsilon)^2}-\Re(\epsilon)}$$

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. Nov 10, 2006

### OlderDan

$$\epsilon = \epsilon \prime + i \epsilon \prime \prime$$

6. Nov 10, 2006

### quasar987

That is the most sensible guess, but then the expression e''<<e doesn't make sense since e is not real.

7. Nov 10, 2006

### big man

The $$\epsilon$$ in my notes is the permittivity given by this expression:

$$\epsilon = \epsilon_r \epsilon_0$$

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:

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

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:

$$\epsilon \prime \prime = \frac {\sigma} {\omega}$$

Am I right in doing this??
,

Last edited: Nov 10, 2006
8. Nov 11, 2006

### OlderDan

That apperas to be consitent with the approach taken in that link I posted earlier