I am currently studying the textbook Microwave Engineering, fourth edition, by David Pozar. Chapter 1.4 THE WAVE EQUATION AND BASIC PLANE WAVE SOLUTIONS says the following:
Plane Waves in a General Lossy Medium
Now consider the effect of a lossy medium. If the medium is conductive, with a conductivity $\sigma$, Maxwell's curl equations can be written, from (1.41a) and (1.20) as
$$\nabla \times \bar{E} = – j \omega \mu \bar{H}, \tag{1.50a}$$
$$\nabla \times \bar{H} = j \omega \epsilon \bar{E} + \sigma \bar{E}. \tag{1.50b}$$
The resulting wave equation for $\bar{E}$ then becomes
$$\nabla^2 \bar{E} + \omega^2 \mu \epsilon \left( 1 – j \dfrac{\sigma}{\omega \epsilon} \right) \bar{E} = 0, \tag{1.51}$$
where we see a similarity with (1.42), the wave equation for $\bar{E}$ in the lossless case. The difference is that the quantity $k^2 = \omega^2 \mu \epsilon$ of (1.42) is replaced by $\omega^2 \mu \epsilon[1 – j(\sigma/\omega \epsilon)]$ in (1.51). We then define a complex propagation constant for the medium as
$$\gamma = \alpha + j \beta = j \omega \sqrt{\mu \epsilon} \sqrt{1 – j \dfrac{\sigma}{\omega \epsilon}}, \tag{1.52}$$
where $\alpha$ is the attenuation constant and $\beta$ is the phase constant. If we again assume an electric field with only an $\hat{x}$ component and uniform in $x$ and $y$, the wave equation of (1.51) reduces to
$$\dfrac{\partial^2{E_x}}{\partial{z}^2} – \gamma^2 E_x = 0, \tag{1.53}$$
which has solutions
$$E_x(z) = E^+ e^{- \gamma z} + E^- e^{\gamma z}. \tag{1.54}$$
The positive traveling wave then has a propagation factor of the form
$$e^{-\gamma z} = e^{- \alpha z} e^{-j \beta z},$$
which in the time domain is of the form
$$e^{- \alpha z} \cos(\omega t – \beta z).$$
We see that this represents a wave traveling in the $+z$ direction with a phase velocity $v_p = \omega / \beta$, a wavelength $\lambda = 2\pi / \beta$, and an exponential damping factor. The rate of decay with distance is given by the attenuation constant, $\alpha$. The negative traveling wave term of (1.54) is similarly damped along the $-z$ axis. If the loss is removed, $\sigma = 0$, and we have $\gamma = jk$ and $\alpha = 0$, $\beta = k$.
As discussed in Section 1.3, loss can also be treated through the use of a complex permittivity. From (1.52) and (1.20) with $\sigma = 0$ but $\epsilon = \epsilon^\prime – j \epsilon^{\prime\prime}$ complex, we have that
$$\gamma = j \omega \sqrt{\mu \epsilon} = jk = j \omega \sqrt{\mu \epsilon^\prime (1 – j \tan \delta)}, \tag{1.55}$$
where $\tan \delta = \epsilon^{\prime \prime}/\epsilon^\prime$ is the loss tangent of the material.
The associated magnetic field can be calculated as
$$H_y = \dfrac{j}{\omega \mu} \dfrac{\partial{E_x}}{\partial{z}} = \dfrac{-j \gamma}{\omega \mu} (E^+ e^{-\gamma z} – E^- e^{\gamma z}). \tag{1.56}$$
The intrinsic impedance of the conducting medium is now complex,
$$\eta = \dfrac{j \omega \mu}{\gamma}, \tag{1.57}$$
but is still identified as the wave impedance, which expresses the ratio of electric to magnetic field components. This allows (1.56) to be rewritten as
$$H_y = \dfrac{1}{\eta} (E^+ e^{- \gamma z} – E^- e^{\gamma z}). \tag{1.58}$$
Note that although $\eta$ of (1.57) is, in general, complex, it reduces to the lossless case of $\eta = \sqrt{\mu/\epsilon}$ when $\gamma = jk = j \omega \sqrt{\mu \epsilon}$.Plane Waves in a Good Conductor
Many problems of practical interest involve loss or attenuation due to good (but not perfect) conductors. A good conductor is a special case of the preceding analysis, where the conductive current is much greater than the displacement current, which means that $\sigma \gg \omega \epsilon$. Most metals can be categorized as good conductors. In terms of a complex $\epsilon$, rather than conductivity, this condition is equivalent to $\epsilon^{\prime\prime} \gg \epsilon^\prime$. The propagation constant of (1.52) can then be adequately approximated by ignoring the displacement current term, to give
$$\gamma = \alpha + j \beta \simeq j \omega \sqrt{\mu \epsilon} \sqrt{\dfrac{\sigma}{j \omega \epsilon}} = (1 + j) \sqrt{\dfrac{\omega \mu \sigma}{2}}. \tag{1.59}$$
The skin depth, or characteristic depth of penetration, is defined as
$$\delta_s = \dfrac{1}{\alpha} = \sqrt{\dfrac{2}{\omega \mu \sigma}}. \tag{1.60}$$
Thus the amplitude of the fields in the conductor will decay by an amount $1/e$, or $36.8\%$, after traveling a distance of one skin depth, because $e^{-\alpha z} = e^{-\alpha \delta_s} = e^{-1}$. At microwave frequencies, for a good conductor, this distance is very small. The practical importance of this result is that only a thin plating of a good conductor (e.g., silver or gold) is necessary for low-loss microwave components.
The intrinsic impedance inside a good conductor can be obtained from (1.57) and (1.59). The result is
$$\eta = \dfrac{j \omega \mu}{\gamma} \simeq (1 + j) \sqrt{\dfrac{\omega \mu}{2 \sigma}} = (1 + j) \dfrac{1}{\sigma \delta_s} \tag{1.61}$$
Notice that the phase angle of this impedance is $45^\circ$, a characteristic of good conductors. The phase angle of the impedance for a lossless material is $0^\circ$, and the phase angle of the impedance of an arbitrary lossy medium is somewhere between $0^\circ$ and $45^\circ$.
Table 1.1 summarizes the results for plane wave propagation in lossless and lossy homogeneous media.
I'm confused about this part:
The intrinsic impedance inside a good conductor can be obtained from (1.57) and (1.59). The result is
$$\eta = \dfrac{j \omega \mu}{\gamma} \simeq (1 + j) \sqrt{\dfrac{\omega \mu}{2 \sigma}} = (1 + j) \dfrac{1}{\sigma \delta_s} \tag{1.61}$$
Notice that the phase angle of this impedance is $45^\circ$, a characteristic of good conductors. The phase angle of the impedance for a lossless material is $0^\circ$, and the phase angle of the impedance of an arbitrary lossy medium is somewhere between $0^\circ$ and $45^\circ$.
How do we get $\eta = \dfrac{j \omega \mu}{\gamma} \simeq (1 + j) \sqrt{\dfrac{\omega \mu}{2 \sigma}} = (1 + j) \dfrac{1}{\sigma \delta_s}$? Furthermore, what does the author mean when they say that "the phase angle of this impedance is $45^\circ$, a characteristic of good conductors"? This isn't clear to me.
Best Answer
I think, it's just algebra.
First:
$\frac{1}{1+j} = \frac{1-j}{(1+j)(1-j)} = (1-j)/2$
Second : $\gamma =(1+j) \sqrt{\frac{\omega\mu\sigma}{2}} \quad$ then
$$\frac{1}{\gamma }= \frac{1-j}{2}\sqrt{\frac{2}{\omega\mu\sigma}}$$
So:
$$\eta = j\frac{\omega\mu}{\gamma} = (j-j^2) \omega\mu \sqrt{\frac{1}{2\omega\mu\sigma}}= (j+1) \sqrt{\frac{\omega^2\mu^2}{2\omega\mu\sigma}} =(1+j) \sqrt{\frac{\omega\mu}{2\sigma}}$$
The following equal sign is based on the definition of the skin depth which is:
$$\delta_s = \sqrt{\frac{2}{\mu\omega\sigma}}$$
So $(\sigma\delta_s)^{-1}$ is :
$$(1+j) \frac{1}{\sigma\delta_s} = (1+j)\sqrt{\frac{\mu\omega\sigma}{2\sigma^2}} = (1+j)\sqrt{\frac{\mu\omega}{2\sigma}}$$
The $45^\circ$ is related with the fact that
$$\frac{1+j}{\sqrt{2}} = \cos\left( \dfrac{\pi}{4} \right) + j \sin \left( \dfrac{\pi}{4} \right) = \exp\left( j\dfrac{\pi}{4} \right)$$
and that in radians $\pi/4 \equiv 45^\circ$. So one can write:
$\eta = \lvert \eta \rvert e^{i\phi}$ with $\lvert \eta \rvert = \sqrt{\dfrac{\mu\omega}{\sigma}}$ and $\phi = \dfrac{\pi}{4}\equiv 45^\circ$.