How to derive intrinsic impedance of parallel plate transmission line?

Question

I've been Googling variations of this for hours but I can't seem to find any information on it and the formula my professor provided just seems to come out of nowhere and he provides no explanation or derivation of it.
His PowerPoint says that

assum[ing] $E_y$ and $H_x$ are almost 'unperturbed', i.e. same as for perfectly conducting walls except with attenuation in the $z$ direction:

$E = -\hat{y}E_0 \exp(-(\alpha z + i \beta z))$

$H = \hat{x}\frac{E_0}{\eta} \exp(-(\alpha z + i \beta z))$

and $E_z$ is a small perturbation in the field given by

$E_z = \pm \eta_c H_x$

the intrinsic impedance of the conducting plates is given by

$\eta_c = (1+i)\sqrt{\frac{\pi f \mu_\text{conductor}}{\sigma_\text{conductor}}}$

where $\sigma$ is conductivity, not charge density, and $f$ is the operating frequency.

I've got no idea where this formula comes from though. I know that, conceptually, the real part of impedance is just resistance to current flow and the imaginary part is resistance to changes in current, or at least that's how it was explained when I took circuits a few years ago. In regards to electromagnetics, which is the subject of my current class, I know it's given by the ratio of the magnitude of the E field over the magnitude of the H field, but just dividing the expressions he gave for $E$ and $H$ doesn't result in the formula he gives, and I can't seem to find anything anything online about it. I don't even know where to start in trying to derive the formula myself -- about the only part of it that even looks similar to expressions for impedance used previously in the course is the factor of $\mu$ in the numerator inside the square root. Where does this formula come from?

Answer 1

Since $\text{curl} \mathbf E = -\mathfrak j \mu \mathbf H$ and you assumed that $\mathbf E =-\hat y E_0 e^{-\gamma z}$, $\gamma = \alpha + \mathfrak j \beta$, thus $\partial_x \mathbf E =\partial_y \mathbf E = 0$ and you also have
$$\text{curl} \mathbf E = \hat z \times \partial_z \mathbf E=-\hat x \partial_z E_0 e^{-\gamma z}\\ =\hat x E_0\gamma e^{-\gamma z} =-\mathfrak j \mu \mathbf H$$ In other words $E_y = -\mathfrak j \frac{\mu}{\gamma} H_x$ and $\zeta_1=\frac{E_0}{H_0}=-\mathfrak j \frac{\mu}{\gamma}$.

Now do the same for $\text{curl} \mathbf H = \mathfrak j \epsilon \mathbf E$ and $\mathbf H = \hat x H_0 e^{-\gamma z}$: $$ \text{curl} \mathbf H = \hat z \times \partial_z \mathbf H\\ =-\hat y \gamma H_0 e^{-\gamma z}=\mathfrak j \epsilon \mathbf E.$$

That is $\mathbf E= -\hat y \mathfrak j \frac{\gamma}{\epsilon } H_0 e^{-\gamma z}$ and $\zeta_2=\frac{E_0}{H_0}=-\mathfrak j\frac{\gamma}{\epsilon}.$

The two "zetas" are equal, therefore $\gamma^2 = \mu \epsilon$. For lossy propagation medium $\epsilon = \epsilon'+\mathfrak j \epsilon''$ and $\epsilon = \mu'+\mathfrak j \mu''$ with $\gamma = \pm \sqrt{(\epsilon'+\mathfrak j \epsilon'')(\mu'+\mathfrak j \mu'')}$. The sign before the "sqrt" is to be selected so that the real part of $\gamma = \alpha + \mathfrak j \beta$ be positive for propagation in the positive $\hat z$ direction.