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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:

The Helmholtz Equation

In a source-free, linear, isotropic, homogeneous region, Maxwell's curl equations in phasor form are $$\nabla \times \bar{E} = -j \omega \mu \bar{H} \tag{1.41a}$$ $$\nabla \times \bar{H} = j \omega \epsilon \bar{E}, \tag{1.41b}$$ and constitute two equations for the unknowns, $\bar{E}$ and $\bar{H}$. As such, they can be solved for either $\bar{E}$ or $\bar{H}$. Taking the curl of (1.41a) and using (1.41b) gives $$\nabla \times \nabla \times \bar{E} = - j\omega \mu \nabla \times \bar{H} = \omega^2 \mu \epsilon \bar{E},$$ which is an equation for $\bar{E}$. This result can be simplified through the use of vector identity (B.14), $\nabla \times \nabla \times \bar{A} = \nabla (\nabla \cdot \bar{A}) - \nabla^2 \bar{A}$, which is valid for the rectangular components of an arbitrary vector $\bar{A}$. Then, $$\nabla^2 \bar{E} + \omega^2 \mu \epsilon \bar{E} = 0, \tag{1.42}$$ because $\nabla \cdot \bar{E} = 0$ in a source-free region. Equation (1.42) is the wave equation, or Helmholtz equation, for $\bar{E}$. An identical equation for $\bar{H}$ can be derived in the same manner: $$\nabla^2 \bar{H} + \omega^2 \mu \epsilon \bar{H} = 0. \tag{1.43}$$ A constant $k = \omega \sqrt{\mu \epsilon}$ is defined and called the propagation constant (also known as the phase constant, or wave number), of the medium; its units are $1/m$.

Plane Waves in a Lossless Medium

In a lossless medium, $\epsilon$ and $\mu$ are real numbers, and so $k$ is real. A basic plane wave solution to the above wave equation can be found by considering an electric field with only an $\hat{x}$ component and uniform (no variation) in the $x$ and $y$ directions. Then, $\partial/\partial{x} = \partial/\partial{y} = 0$, and the Helmholtz equation of (1.42) reduces to $$\dfrac{\partial^2{E_x}}{\partial{z}^2} + k^2 E_x = 0. \tag{1.44}$$ The two independent solutions to this equation are easily seen, by substitution, to be of the form $$E_x(z) = E^+e^{-jkz} + E^-e^{jkz}, \tag{1.45}$$ where $E^+$ and $E^-$ are arbitrary amplitude constants.
The above solution is for the time harmonic case at frequency $\omega$. In the time domain, this result is written as $$\mathcal{E}_x(z, t) = E^+ \cos(\omega t - kz) + E^- \cos(\omega t + kz), \tag{1.46}$$ where we have assumed that $E^+$ and $E^-$ are real constants.

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A complete specification of the plane wave electromagnetic field should include the magnetic field. In general, whenever $\bar{E}$ or $\bar{H}$ is known, the other field vector can be readily found by using one of Maxwell's curl equations. Thus, applying (1.41a) to the electric field of (1.45) gives $H_x = H_z = 0$, and $$H_y = \dfrac{j}{\omega \mu} \dfrac{\partial{E_x}}{\partial{z}} = \dfrac{1}{\eta}(E^+ e^{-jkz} - E^- e^{jkz}), \tag{1.49}$$ where $\eta = \omega \mu / k = \sqrt{\mu/\epsilon}$ is known as the intrinsic impedance of the medium.

I'm confused by this part:

In general, whenever $\bar{E}$ or $\bar{H}$ is known, the other field vector can be readily found by using one of Maxwell's curl equations. Thus, applying (1.41a) to the electric field of (1.45) gives $H_x = H_z = 0$, and $$H_y = \dfrac{j}{\omega \mu} \dfrac{\partial{E_x}}{\partial{z}} = \dfrac{1}{\eta}(E^+ e^{-jkz} - E^- e^{jkz}), \tag{1.49}$$ where $\eta = \omega \mu / k = \sqrt{\mu/\epsilon}$ is known as the intrinsic impedance of the medium.

How is (1.41a) "applied" to (1.45) to get $H_x = H_z = 0$ and $H_y = \dfrac{j}{\omega \mu} \dfrac{\partial{E_x}}{\partial{z}} = \dfrac{1}{\eta}(E^+ e^{-jkz} - E^- e^{jkz})$? If we apply the curl to (1.45), then we get

$$\nabla \times E_x = \hat{\mathbf{j}} \dfrac{\partial{E_x}}{\partial{z}} = (-jk E^+ e^{-jkz} + jk E^- e^{jkz}) \hat{\mathbf{j}} = -jk (E^+ e^{-jkz} - E^- e^{jkz})\hat{\mathbf{j}},$$

so it isn't clear to me how the authors got this.

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  • $\begingroup$ I'm not sure what part is confusing you, because taking the curl of (1.45), along with $E_{z}=0$, and using $\partial/\partial t\rightarrow -i\omega$ in Faraday's Law certainly looks like it will give (1.49). $\endgroup$
    – Buzz
    Apr 23 at 23:25
  • $\begingroup$ @Buzz Taking the curl leads me to $$\nabla \times E_x = \hat{\mathbf{j}} \dfrac{\partial{E_x}}{\partial{z}} = (-jk E^+ e^{-jkz} + jk E^- e^{jkz}) \hat{\mathbf{j}} = -jk (E^+ e^{-jkz} - E^- e^{jkz})\hat{\mathbf{j}}$$ $\endgroup$ Apr 24 at 17:38
  • $\begingroup$ @Buzz If you understand how this works, then I would appreciate it if you would post an answer explaining all of this. $\endgroup$ Apr 24 at 23:55

2 Answers 2

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Faraday's Law, for harmonically varying fields (time dependence $e^{i\omega t}$) in a linear medium is the equation (1.41a), $$\vec{\nabla}\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}=-i\omega\vec{B}=i\omega\mu\vec{H}.$$ Attention is being restricted to an electric field with only its $x$-component $E_{x}$ nonzero and with no dependence on $x$ or $y$. This is a plane wave polarized in the $x$-direction and propagating along the $z$-direction, $$\vec{E}=E_{x}(z)\,\hat{z}.$$ In the curl, $$\vec{\nabla}\times\vec{E}=\left[\begin{array}{ccc} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial x} & \frac{\partial}{\partial z} \\ E_{x} & E_{y} & E_{z} \end{array}\right],$$ therefore only one term can be nonzero, $$\vec{\nabla}\times\vec{E}=\left[\begin{array}{ccc} \hat{x} & \hat{y} & \hat{z} \\ 0 & 0 & \frac{\partial}{\partial z} \\ E_{x} & 0 & 0 \end{array}\right]=\frac{\partial E_{x}}{\partial z}\hat{y}.$$ Plugging this into (1.41a) gives the first equality in (1.49) as its $y$-component, $$\frac{\partial E_{x}}{\partial z}\hat{y}=-i\omega\mu\vec{H}\\ \frac{i}{\omega\mu}\frac{\partial E_{x}}{\partial z}\hat{y}=H_{y}\,\hat{y},$$ along with $0=H_{x}\,\hat{x}$ and $0=H_{z}\,\hat{z}.$

What remains is to insert the plane wave solution (1.45), which includes both left- and right-moving waves (with respective amplitudes $E^{+}$ and $E^{-}$), $$\vec{E}=E_{x}(z)\,\hat{x}=E^{+}e^{-ikz}+E^{-}e^{ikz} \\ \frac{\partial E_{x}}{\partial z}=-ikE^{+}e^{-ikz}+ikE^{-}e^{ikz}.$$ Plugging this $z$-derivative into (1.49) gives $$H_{y}\!=\!\frac{i}{\omega\mu}\left(-ikE^{+}e^{-ikz}+ikE^{-}e^{ikz}\right) =\frac{k}{\omega\mu}\left(E^{+}e^{-ikz}-E^{-}e^{ikz}\right)=\frac{1}{\eta}\left(E^{+}e^{-ikz}-E^{-}e^{ikz}\right),$$ which is just the remainder of (1.49).

As to why $\eta=\omega\mu/k$ is also equal to $\sqrt{\mu/\varepsilon}$, this follows from the fact that the wave speed is $$v_{\mathrm{ph}}=\frac{c}{n}=\frac{1}{\sqrt{\varepsilon\mu}}$$ in the linear material with permittivity $\varepsilon$ and permeability $\mu$. The wave component moving in the positive $z$-direction is proportional to $|E^{+}|e^{i\omega t}e^{-ikz}$ (where we have restored the time dependence that was previously absorbed into the complex amplitude $E^{+}$). Note that the only spacetime coordinate dependence in this functional form for the propagating wave is through $e^{i\omega t}e^{-ikz}=\exp[-i(kz-\omega t)]=\exp\left[-ik\left(z-\frac{\omega}{k}t\right)\right].$ This form shows that this is a traveling wave moving with speed $\omega/k$; equating this to the expression for the phase propagation speed $v_{\mathrm{ph}}$ gives the relation $$\frac{\omega}{k}=\frac{1}{\sqrt{\varepsilon\mu}};$$ multiplying both sides by $\mu$ gives the desired expression for $\eta$.

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(1.41a) states that $\mathbf{\nabla \times E}=-j\omega \mu \mathbf{H}$ and (1.45) tells you that your electric field has only a $x$ component. So $\mathbf{\nabla \times E}=\frac{\partial}{\partial z}E_x \hat{\jmath}=-j\omega \mu H_y \hat{\jmath}$.

So now you know that $H_x=H_z=0$ since you only have $\hat{\jmath}$ in your cross product.

You finally have $H_y=\frac{j}{\omega \mu}\frac{\partial}{\partial z}E_x$

You derived the electric field with respect to $z$, so you just have

\begin{equation} H_y=\frac{j}{\omega \mu}\left[-jk(E^+e^{-jkz}-E^-e^{jkz})\right]=\frac{1}{\eta}(E^+e^{-jkz}-E^-e^{jkz}) \end{equation} Which is the equation (1.49)

Hope this helps ;)

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  • $\begingroup$ I don't see how this answers my question. My question is clearly stated as follows: How is (1.41a) "applied" to (1.45) to get $H_x = H_z = 0$ and $H_y = \dfrac{j}{\omega \mu} \dfrac{\partial{E_x}}{\partial{z}} = \dfrac{1}{\eta}(E^+ e^{-jkz} - E^- e^{jkz})$? As I then clearly state, I began by taking the curl of (1.45): $$\nabla \times E_x = \hat{\mathbf{j}} \dfrac{\partial{E_x}}{\partial{z}} = (-jk E^+ e^{-jkz} + jk E^- e^{jkz}) \hat{\mathbf{j}} = -jk (E^+ e^{-jkz} - E^- e^{jkz})\hat{\mathbf{j}},$$ which is different from what the author has presented. [...] $\endgroup$ Apr 24 at 23:42
  • $\begingroup$ [...] But you are just repeating what the author has written. An answer should be a coherent explanation, logically presented. $\endgroup$ Apr 24 at 23:48
  • $\begingroup$ I just assumed you knew about the cross product formula ;) Buzz does it in details below $\endgroup$ Apr 25 at 9:50

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