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I am currently looking at the propagation constant $\gamma\in\mathbb{C}$, which is

$$ \gamma = i\omega\sqrt{\mu\epsilon-i\,\frac{\sigma\mu}{\omega}}, $$

where $i^2 = -1$ and all other quantities are real.

The attenuation loss is calculated using the real part of $\gamma$. How do I calculate $\mathfrak{R}(\gamma)$ and, out of interest, $\mathfrak{I}(\gamma)$?

SMA.D
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Harry Smith
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1 Answers1

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There are 2 approximations for $\gamma$

If $\frac{\sigma}{\omega}<<\epsilon$ (low loss) then : $$ \gamma \simeq i\omega\sqrt{\mu\epsilon}+\frac{1}{2}\sqrt\frac{\mu}{\epsilon}\sigma$$ We know $\frac{\sigma}{\omega\epsilon}<<1$ then, $$\gamma =i\omega\sqrt{\mu\epsilon}[1+\frac{\sigma}{i\omega\epsilon}]^{1/2}\simeq i\omega\sqrt{\mu\epsilon}[1+\frac{\sigma}{2i\omega\epsilon}]$$

And if $\frac{\sigma}{\omega}>>\epsilon$ (lossy) then :

$$\gamma\simeq\frac{1}{\sqrt 2} \sqrt{\omega\mu\sigma}(i+1) $$

For proof you can simply omit $\epsilon$, assuming it is small.

$$\require{cancel}\gamma =i\omega\sqrt{\mu}[\cancelto{0}{\epsilon}+\frac{\sigma}{i\omega}]^{1/2}\simeq\frac{1}{\sqrt 2} \sqrt{\omega\mu\sigma}(i+1)$$

Otherwise you can treat it as an ordinary complex nember:

$$\gamma = i\omega\sqrt{\mu\epsilon-i\,\frac{\sigma\mu}{\omega}}= i\omega\sqrt{\sqrt{(\mu\epsilon)^2+(\frac{\sigma\mu}{\omega})^2}\,.e^{-i\tan^{-1}{\frac{\frac{\sigma\mu}{\omega}}{\mu\epsilon}}}} $$ $$ \gamma = i\omega\sqrt[4]{(\mu\epsilon)^2+(\frac{\sigma\mu}{\omega})^2}.e^{-\frac{i}{2}\tan^{-1}{\frac{\sigma}{\omega\epsilon}}} $$

SMA.D
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  • Very nice! Could you perhaps explain the steps of your final calculation a little more? I'm struggling to follow, thank you! – Harry Smith Jul 30 '15 at 15:09
  • I see, very nicely calculated! If it is $1 + x/n$ where $n=1/2$, wouldn't that imply we have $1+2x$? It appears you have treated $n$ as $2$ in your calculations. Also, I am still finding difficulty following the part where you treat $\gamma$ as an ordinary complex number. Is the first step: $\mu\epsilon - i\frac{\sigma\mu}{\omega} = \sqrt{(\mu\epsilon)^2 + \left(\frac{\sigma\mu}{\omega}\right)^2},\exp\left[-i\tan^{-1}\frac{\sigma\mu/ \omega}{\mu\epsilon}\right]$ an identity I'm not recognising? – Harry Smith Jul 31 '15 at 08:27
  • The true formula is $(1+x)^n\simeq 1+nx$ ,$x<<1$ based on Taylor expansion. – SMA.D Jul 31 '15 at 08:37
  • And for the second question, It's the polar representation of the complex number. see link – SMA.D Jul 31 '15 at 08:43
  • Also I should mention that by the word "ordinary" I meant without approximation. – SMA.D Jul 31 '15 at 08:48
  • As an approximation for $(1+x)^n$ we can use the first 2 statements of its series expansion, if $x$ is relatively small to 1. see link – SMA.D Jul 31 '15 at 08:53
  • Oh yes, I completely forgot about including the modulus! I've clearly been using unitary complex numbers too much. Thank you. – Harry Smith Jul 31 '15 at 09:10
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    I'm just going to comment some additional pivotal points in your calculations for anyone else reading. For low-loss media: $\gamma = i\omega\sqrt{\mu\epsilon}\left(1-i\frac{\sigma\mu}{\omega}\frac{1}{\mu \epsilon}\right)$, recall $-i=i^{-1}=1/i$ and then apply the Taylor approximation to obtain the answer provided. For lossy media: $\epsilon\to 0 \implies \gamma = i\omega\sqrt{\mu}\left(\sigma/i\omega\right)^{1/2} = \sqrt{i\omega\sigma\mu}$ and since $\sqrt{i} = (1+i)/\sqrt{2}$ (see proof), the answer provided is obtained. – Harry Smith Jul 31 '15 at 09:20