In Elements of Electromagnetics (Sadiku, 3rd edition, Section 10.8), the author says to consider two lossless dielectric materials joined at an interface $z=0$. Here two lossless dielectric materials can be translated to mean that the conductivity of medium 1 (located to the left of $z=0$, i.e. for $z<0$) is equal to the conductivity of medium 2 (located to the right of $z=0$, i.e. for $z>0$) which is equal to 0 Siemens/meter.

Recall that intrinsic impedance is given by $$\eta = \sqrt{\frac{j\omega\mu}{\sigma+j\omega\varepsilon}}$$So if $\sigma=0$, then $$\eta = \sqrt{\frac{j\omega\mu}{j\omega\varepsilon}} = \sqrt{\frac{\mu}{\varepsilon}}$$This means that the intrinsic impedances $\eta_1$ and $\eta_2$ are real numbers (with units) since $\sigma_1=\sigma_2=0$ S/m and $\mu$ and $\varepsilon$ are taken to be just real constants (with units), i.e. they do not change with space or time, except at $z=0$ where there is a transition from one lossless dielectric to another of course.

Also note that the reflection coefficient $\Gamma$ is given by $$\Gamma = \frac{E_{r0}}{E_{i0}} = \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}$$ where $E_{r0}$ is the amplitude of the reflected wave at $z=0$ (the interface) and $E_{i0}$ is the amplitude of the reflected wave at $z=0$ (the interface).

The author considers the case where $\eta_2 > \eta_1$ (meaning $\Gamma > 0$). A normally incident electromagnetic wave (normal to the interface) reaches the interface and sends some EM energy through to medium 2 in the form of a transmitted electromagnetic wave and reflects some EM energy in the form of a reflected electromagnetic wave. The incident and reflected waves have different amplitudes. The electric field in medium 1 is the sum of the incident and the reflected waves (described mathematically below) $$ \vec{E}_{1} = \text{Re}\left[\underbrace{\left[\vec{E}_{i0}e^{-\gamma_1z} + \vec{E}_{r0}e^{\gamma_1 z}\right]}_{\text{Spatial variation}}\underbrace{\left[e^{j\omega t}\right]}_{\text{Temporal variation}}\right]$$ Here, $\gamma_1$, is the propagation constant (most generally a complex number) in medium 1 given by $$ \gamma = \underbrace{\alpha}_{\omega\sqrt{\frac{\mu\varepsilon}{2}\left[\sqrt{1+\left[\frac{\sigma}{\omega\varepsilon}\right]^2}-1\right]}} + j\underbrace{\beta}_{\omega\sqrt{\frac{\mu\varepsilon}{2}\left[\sqrt{1+\left[\frac{\sigma}{\omega\varepsilon}\right]^2}+1\right]}} $$ But since $\sigma_1=0$ S/m, $\gamma_1=j\beta=j\left[\omega\sqrt{\mu\varepsilon}\right]$. Since $e^{j\theta}=\cos\theta+j\sin\theta$, taking the real part of a complex exponential is a compact way of writing a sinusoidal wave, which is very relevant here as there are three sinusoidal waves involved here -- the incident wave representing the incident electromagnetic wave's electric field, the reflected wave representing the reflected electromagnetic wave's electric field, and the transmitted wave representing the transmitted electromagnetic wave's electric field.

With this background, here is what I do not understand.

Problem: The author claims, just before Eqs. 10.88 and 10.89, that the amplitude of $\vec{E}_{1}$, $\left|\vec{E}_{1}\right|$ is maximum for when $$-\beta_1 z_{\text{max}} = n\pi$$Further, the author claims the amplitude of $\vec{E}_{1}$ is minimum for when $$-\beta_1 z_{\text{min}} = \left(2n+1\right)\frac{\pi}{2}$$I cannot seem to replicate this result. Here is my attempt at a proof.

Partial proof: Assuming the electric field oscillates in the $\mathbf{a}_x$ direction (alternately called $\mathbf{e}_x$ or $\hat{x}$ direction) and $E_{r0}=\left|\vec{E}_{r0}\right|$, $E_{i0}=\left|\vec{E}_{i0}\right|$ Proof up to phasor envelope representation

To determine the magnitude of this, it may be helpful to consider the geometry of this situation by using a phasor diagram. The phasor diagram would plot two vectors (phasors) – one of magnitude $E_{i0}$ and the other of magnitude $\Gamma E_{i0}$ (remember $\Gamma>0$ in this case but $0\le\left|\Gamma\right|\le 1$ holds in general). If the reference phasor is $\cos\left(\beta_1 z\right)$, then $\cos\left(\beta_1 z-\omega t\right)$ will be located a certain distance in the clockwise direction away from the reference and $\cos\left(\beta_1 z\right)$ will be located the same distance in the counterclockwise direction. The resulting vector (or phasor) sum of the two will have a magnitude of $\sqrt{E_{i0}^2 + \left(\Gamma E_{i0}\right)^2}$. I am trying to figure out the phase angle for this resulting vector. I am currently stuck here, hoping someone can help me out. Thanks in advance!

EDIT: Here's the picture I drew.Phasor diagram

It seems like this all simplifies to finding $\phi$ using the law of sines / law of cosines. The goal is to see for what values of $\phi$ will $\left|\vec{E}_1\right|$ be maximum. Hopefully, it'll end up validating what the book says.

  • $\begingroup$ What is $z_{max},z_{min}$? $\endgroup$
    – akhmeteli
    Jun 23, 2013 at 3:03
  • $\begingroup$ First, thanks for reading. I was worried this question would get buried by all the homework questions. $z_{\text{max}}$ are the $z$ values for when $\left|\vec{E}_1\right|$ is at maximum. $z_{\text{min}}$ are the $z$ values for when $\left|\vec{E}_1\right|$ is at minimum. As a simple example, if the goal was to find $x_{\text{max}}$ for which $f(x)=-x^2$ would be maximum, it would be when $\frac{df(x)}{dx}=0=-2x$. So $x_{\text{max}}$ would be when $x=0$. Similarly, for $g(y)=y^2$, $y_{\text{min}}$ would be when $y=0$. $\endgroup$ Jun 23, 2013 at 3:30

3 Answers 3


Your notation is quite confusing, so let me reformulate and partly simplify your problem. If my reformulation/simplification is wrong... Well, tough luck...

Let us consider the following magnitude: $E=\Re[E_i\exp(i(\delta z+\omega t))+E_r\exp(i(-\delta z+\omega t))]$, where $\delta$ is real, $E_i$ and $E_r$ are real and positive. We are trying to find out when the time-averaged $E^2$ is maximum and minimum (depending on $z$). $E=\frac{1}{2}(E_i\cos(\delta z+\omega t)+E_r\cos(-\delta z+\omega t)$), so $$E^2=\frac{1}{4}(E_i^2\cos^2(\delta z+\omega t)+E_r^2\cos^2(-\delta z+\omega t)+2 E_i E_r\cos(\delta z+\omega t)\cos(-\delta z+\omega t))=\frac{1}{4}(E_i^2\cos^2(\delta z+\omega t)+E_r^2\cos^2(-\delta z+\omega t)+2 E_i E_r\frac{1}{2}(\cos(2\omega t)+\cos(2\delta z)))$$ (http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities ). Let us average $E^2$ over time (we denote time average by a bar, so, e.g., time-averaged $E^2$ is denoted as $\overline{E^2}$. As $\overline{\cos^2(\delta z+\omega t)}=\overline{\cos^2(-\delta z+\omega t)}=\frac{1}{2}$, and $\overline{\cos(2\omega t)}=0$, we obtain: $\overline{E^2}=\frac{1}{4}(\frac{1}{2}E_i^2+\frac{1}{2}E_r^2+E_i E_r\cos(2\delta z))$. Therefore, $\overline{E^2}$ is maximum when $\cos(2\delta z)=1$ and minimum when $\cos(2\delta z)=-1$.

  • $\begingroup$ Thanks for your response. First, why is the magnitude of $E$ considered the time-averaged $E^2$? Second, when the book says to find the maxima of the amplitude $\left|\vec{E}_1(z)\right|$ (or $\left|E\right|$), it means to find values of $z$ will $\left|\vec{E}_1\right|$ (or $\left|E\right|$) be the highest, but when it says to find minima, it means find values of $z$ when the magnitude is $0$. I incorrectly explained that previously. Not a big deal. It would just mean your answer would change to say minima are found at $\cos\left(2\delta z\right)=0$. But, I am confused about the first point. $\endgroup$ Jun 23, 2013 at 21:25
  • $\begingroup$ Because the frequency is very high, and any instrument will typically measure some averaged magnitude. However, magnitude of $E$ is not the time-averaged $E^2$, it is a square root of this time averaged $E^2$. The averaging is typically done for $E^2$ because this magnitude has better analytical properties than, say, $|E|$. However, we may assume that, say, $|E|$ has maximum when time-averaged $E^2$ has maximum. $\endgroup$
    – akhmeteli
    Jun 23, 2013 at 22:00
  • $\begingroup$ I disagree that "when it says to find minima, it means find values of z when the magnitude is 0." If $\Gamma<1$, you won't get 0, and you get minimum only if $\cos(2\delta z)=-1$, not 0. By the way, this seems consistent with the statements in the book. $\endgroup$
    – akhmeteli
    Jun 23, 2013 at 22:06
  • $\begingroup$ I would also like to add that en.wikipedia.org/wiki/Standing_wave_ratio#Further_analysis seems relevant to this discussion. Also, $\Gamma > 0$ in this particular case. $\endgroup$ Jun 23, 2013 at 22:32
  • $\begingroup$ I don't see any contradiction with my answer or comments there. $\endgroup$
    – akhmeteli
    Jun 23, 2013 at 23:06

$$ E_1 = Re\left[\left(E_{i0}e^{-j\beta z} + \Gamma E_{i0}e^{+j\beta z}\right)e^{j\omega t}\right]$$

$$ E_1 = E_{i0}\cos(\omega t - \beta z) + \Gamma E_{i0}\cos(\omega t +\beta z) $$

$$ E_1 = E_{i0}\left(\cos(\omega t)\cos (\beta z) + \sin(\omega t)\sin(\beta z) \right) + \Gamma E_{i0}\left(\cos(\omega t)\cos (\beta z) - \sin(\omega t)\sin(\beta z) \right) $$

$$ = E_{i0}\left(1 + \Gamma \right)\cos (\beta z)\cos(\omega t) + E_{i0}\left( 1 - \Gamma \right)\sin(\beta z)\sin(\omega t) $$

the following is true: $$A \cos(\omega t) + B \sin(\omega t) = \sqrt{A^2 + B^2} \sin(\omega t + \phi)$$ the value of phi is irrelevant for the locations of the maxima and minima along z.

$$ E_1 = E_{i0}\sqrt{\left(1 + \Gamma \right)^2\cos^2(\beta z) + \left( 1 - \Gamma \right)^2\sin^2(\beta z)}\sin(\omega t + \phi)$$

$$ E_1 = E_{i0}\sqrt{\left(1 + 2\Gamma + \Gamma^2 \right)\cos^2(\beta z) + \left( 1 - 2\Gamma+ \Gamma^2 \right)\sin^2(\beta z)}\sin(\omega t + \phi)$$ factoring out $\cos^2(\beta z) + \sin^2(\beta z) = 1 $ from some of the terms $$ E_1 = E_{i0}\sqrt{\left(1 + \Gamma^2 \right) + 2\Gamma\left(\cos^2(\beta z) - \sin^2(\beta z) \right)}\sin(\omega t + \phi)$$

$$\sin^2(\beta z) = 1 - \cos^2(\beta z)$$

$$ E_1 = E_{i0}\sqrt{1 + \Gamma^2 + 2\Gamma\left(2\cos^2(\beta z) - 1 \right)}\sin(\omega t + \phi)$$

$$ E_1 = E_{i0}\sqrt{4\Gamma\cos^2(\beta z) + \Gamma^2 - 2\Gamma + 1}\sin(\omega t + \phi)$$

$$ E_1 = E_{i0}\sqrt{4\Gamma\cos^2(\beta z) + \left(1 - \Gamma \right)^2}\sin(\omega t + \phi)$$

Now to find the location of the maxima and minima, you need only find the maxima and minima of $\cos^2(\beta z) $. The maxima will be where $\cos(\beta z) = \pm 1 $ and the minima will be where $\cos(\beta z) = 0 $

  • $\begingroup$ I forgot about the very useful equation you included after saying "the following is true:" and that was something I was trying to recall from my communication systems knowledge. Thanks! $\endgroup$ Jun 27, 2013 at 6:44

No such thing as a lossless dielectric. This violates causality by Kramers-Kronig. I can easily measure the loss in monocrystalline quartz, teflon, mono-layer graphene, you name it. Well, OK, there is one. Vacuum.


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