## Notes for ece1228 (Electromagnetic Theory) now in book form on Amazon.

September 26, 2020 math and physics play , My notes for ece1228 (Electromagnetic Theory) are now available in book form on Amazon.

This version omits all assigned problem solutions (but includes some self-generated problem solutions.)  As such, it is very short.  I published it so that I could get a copy (of the non-redacted version) for myself , but in the unlikely chance that somebody else is interested I’ve left the redacted version in published state (available very cheaply.)  Feel free to contact me for the full (pdf) version if you are not taking the course (and don’t intend to.)

The official course description at the time was:

Fundamentals: Maxwell’s equations, constitutive relations and boundary conditions, wave polarization. Field representations: potentials, Green’s functions and integral equations. Theorems and concepts: duality, uniqueness, images, equivalence, reciprocity and Babinet’s principles. Plane, cylindrical and spherical waves and waveguides. radiation and scattering.

New material (for me) in this course was limited to:

• dispersion relations.
• Druid-Lorentz model
• magnetic moments, magnetostatic force, and torque (mentioned in class without details, but studied from Jackson)
• matrix representation of transmission and reflection through multiple interfaces

## Posted notes for Electromagnetic Theory (ECE1228H), taught by Prof. M. Mojahedi, fall 2016

February 3, 2017 math and physics play ,

I’ve now posted redacted notes for the Electromagnetic Theory (ECE1228H) course I took last fall, taught by Prof. M. Mojahedi.  This course covered a subset of the following:

• Maxwell’s equations
• constitutive relations and boundary conditions
• wave polarization.
• Field representations: potentials
• Green’s functions and integral equations.
• Theorems and concepts: duality, uniqueness, images, equivalence, reciprocity and Babinet’s principles.
• Plane cylindrical and spherical waves and waveguides.

These notes are fairly compact, only 183 pages, with the full version weighing in at 256 pages.

As always, feel free to contact me for the complete version (i.e. including my problem set solutions) if you interested, but not asking because you are taking or planning to take this course.

## Total internal reflection

From Snell’s second law we have

\begin{equation}\label{eqn:brewsters:20}
\theta_t = \arcsin\lr{ \frac{n_i}{n_t} \sin\theta_i }.
\end{equation}

This is plotted in fig. 3.

For the $$n_i > n_t$$ case, for example, like shining from glass into air, there is a critical incident angle beyond which there is no real value of $$\theta_t$$. That critical incident angle occurs when $$\theta_t = \pi/2$$, which is

\begin{equation}\label{eqn:brewsters:40}
\sin\theta_{ic} = \frac{n_t}{n_i} \sin(\pi/2).
\end{equation}

With
\begin{equation}\label{eqn:brewsters:340}
n = n_t/n_i
\end{equation}

the critical angle is
\begin{equation}\label{eqn:brewsters:60}
\theta_{ic} = \arcsin n.
\end{equation}

Note that Snell’s law can also be expressed in terms of this critical angle, allowing for the solution of the transmission angle in a convenient way
\begin{equation}\label{eqn:brewsters:360}
\begin{aligned}
\sin\theta_i
&= \frac{n_t}{n_i} \sin\theta_t \\
&= n \sin\theta_t \\
&= \sin\theta_{ic} \sin\theta_t,
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:brewsters:380}
\sin\theta_t = \frac{\sin\theta_i}{\sin\theta_{ic}}.
\end{equation}

Still for $$n_i > n_t$$, at angles past $$\theta_{ic}$$, the transmitted wave angle becomes complex as outlined in , namely

\begin{equation}\label{eqn:brewsters:400}
\begin{aligned}
\cos^2\theta_t
&=
1 – \sin^2 \theta_t \\
&=
1 –
\frac{\sin^2\theta_i}{\sin^2\theta_{ic}} \\
&=
-\lr{
\frac{\sin^2\theta_i}{\sin^2\theta_{ic}}
-1
},
\end{aligned}
\end{equation}

or
\begin{equation}\label{eqn:brewsters:420}
\cos\theta_t =
j \sqrt{
\frac{\sin^2\theta_i}{\sin^2\theta_{ic}}
-1
}.
\end{equation}

Following the convention that puts the normal propagation direction along z, and the interface along x, the wave vector direction is
\begin{equation}\label{eqn:brewsters:440}
\begin{aligned}
\kcap_t
&= \Be_3 e^{ \Be_{31} \theta_t } \\
&= \Be_3 \cos\theta_t + \Be_1 \sin\theta_t.
\end{aligned}
\end{equation}

The phase factor for the transmitted field is

\begin{equation}\label{eqn:brewsters:460}
\begin{aligned}
\exp\lr{ j \omega t \pm j \Bk_t \cdot \Bx }
&=
\exp\lr{ j \omega t \pm j k \kcap_t \cdot \Bx } \\
&=
\exp\lr{ j \omega t \pm j k \lr{ z \cos\theta_t + x \sin\theta_t } } \\
&=
\exp\lr{
j \omega t
\pm j k \lr{ z j \sqrt{ \frac{\sin^2\theta_i}{\sin^2\theta_{ic}} -1 } + x \frac{\sin\theta_i}{\sin\theta_{ic}} }
} \\
&=
\exp\lr{
j \omega t \pm k
\lr{
j x \frac{\sin\theta_i}{\sin\theta_{ic}}
– z \sqrt{ \frac{\sin^2\theta_i}{\sin^2\theta_{ic}} -1 }
}
}.
\end{aligned}
\end{equation}

The propagation is channelled along the x axis, but the propagation into the second medium decays exponentially (or unphysically grows exponentially), only getting into the surface a small amount.

What is the average power transmission into the medium? We are interested in the time average of the normal component of the Poynting vector $$\BS \cdot \ncap$$.

\begin{equation}\label{eqn:brewsters:480}
\begin{aligned}
\BS
&= \inv{2} \BE \cross \BH^\conj \\
&= \inv{2} \BE \cross \lr{ \inv{\eta} \kcap_t \cross \BE^\conj } \\
&= -\inv{2 \eta} \BE \cdot \lr{ \kcap_t \wedge \BE^\conj } \\
&= -\inv{2 \eta} \lr{
(\BE \cdot \kcap_t) \BE^\conj

\kcap_t \BE \cdot \BE^\conj
} \\
&=
\inv{2 \eta}
\kcap_t \Abs{\BE}^2.
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:brewsters:500}
\begin{aligned}
\kcap_t \cdot \ncap
&= \lr{ \Be_3 \cos\theta_t + \Be_1 \sin\theta_t } \cdot \Be_3 \\
&= \cos\theta_t \\
&=
j \sqrt{
\frac{\sin^2\theta_i}{\sin^2\theta_{ic}}
-1
}.
\end{aligned}
\end{equation}

Note that this is purely imaginary. The time average real power transmission is

\begin{equation}\label{eqn:brewsters:520}
\begin{aligned}
\expectation{\BS \cdot \ncap}
&=
\textrm{Re} \lr{
j \sqrt{
\frac{\sin^2\theta_i}{\sin^2\theta_{ic}}
-1
}
\frac{1}{2 \eta} \Abs{\BE}^2
} \\
&= 0.
\end{aligned}
\end{equation}

There is no power transmission into the second medium at or past the critical angle for total internal reflection.

## Brewster’s angle

Brewster’s angle is the angle for which there the amplitude of the reflected component of the field is zero. Recall that when the electric field is parallel(perpendicular) to the plane of incidence, the reflection amplitude ( eq. 4.38)

\begin{equation}\label{eqn:brewsters:80}
r_\parallel
=
\frac
{
\frac{ n_t }{\mu_t} \cos \theta_i
-\frac{ n_i }{\mu_i} \cos \theta_t
}
{
\frac{ n_t }{\mu_t} \cos \theta_i
+\frac{ n_i }{\mu_i} \cos \theta_t
}
\end{equation}
\begin{equation}\label{eqn:brewsters:100}
r_\perp
=
\frac
{
\frac{ n_i }{\mu_i} \cos \theta_i
-\frac{ n_t }{\mu_t} \cos \theta_t
}
{
\frac{ n_i }{\mu_i} \cos \theta_i
+\frac{ n_t }{\mu_t} \cos \theta_t
}
\end{equation}

There are limited conditions for which $$r_\perp$$ is zero, at least for $$\mu_i = \mu_t$$. Using Snell’s second law $$n_i \sin\theta_i = n_t \sin\theta_t$$, that zero is found at

\begin{equation}\label{eqn:brewsters:120}
\begin{aligned}
n_i \cos \theta_i
&= n_t \cos \theta_t \\
&= n_t \sqrt{ 1 – \sin^2 \theta_t } \\
&= n_t \sqrt{ 1 – \frac{n_i^2}{n_t^2} \sin^2 \theta_i },
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:brewsters:140}
\frac{n_i^2}{n_t^2} \cos^2 \theta_i = 1 – \frac{n_i^2}{n_t^2} \sin^2 \theta_i,
\end{equation}

or
\begin{equation}\label{eqn:brewsters:160}
\frac{n_i^2}{n_t^2} \lr{ \cos^2 \theta_i + \sin^2 \theta_i } = 1.
\end{equation}

This has solutions only when $$n_i = \pm n_t$$. The $$n_i = n_t$$ case is of no interest, since that is just propagation, so naturally there is no reflection. The $$n_i = -n_t$$ case is possible with the transmission into a negative index of refraction material that is matched in absolute magnitude with the index of refraction in the incident medium.

There are richer solutions for the $$r_\parallel$$ zero. Again considering $$\mu_1 = \mu_2$$ those occur when

\begin{equation}\label{eqn:brewsters:180}
\begin{aligned}
n_t \cos \theta_i
&= n_i \cos \theta_t \\
&= n_i \sqrt{ 1 – \frac{n_i^2}{n_t^2} \sin^2 \theta_i } \\
&= n_i \sqrt{ 1 – \frac{n_i^2}{n_t^2} \sin^2 \theta_i }
\end{aligned}
\end{equation}

Let $$n = n_t/n_i$$, and square both sides. This gives

\begin{equation}\label{eqn:brewsters:200}
\begin{aligned}
n^2 \cos^2 \theta_i
&= 1 – \inv{n^2} \sin^2 \theta_i \\
&= 1 – \inv{n^2} (1 – \cos^2 \theta_i),
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:brewsters:220}
\cos^2 \theta_i \lr{ n^2 + \inv{n^2}} = 1 – \inv{n^2},
\end{equation}

or
\begin{equation}\label{eqn:brewsters:240}
\begin{aligned}
\cos^2 \theta_i
&= \frac{1 – \inv{n^2}}{ n^2 – \inv{n^2} } \\
&= \frac{n^2 – 1}{ n^4 – 1 } \\
&= \frac{n^2 – 1}{ (n^2 – 1)(n^2 + 1) } \\
&= \frac{1}{ n^2 + 1 }.
\end{aligned}
\end{equation}

We also have

\begin{equation}\label{eqn:brewsters:260}
\begin{aligned}
\sin^2 \theta_i
&=
1 – \frac{1}{ n^2 + 1 } \\
&=
\frac{n^2}{ n^2 + 1 },
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:brewsters:280}
\tan^2 \theta_i = n^2,
\end{equation}

and
\begin{equation}\label{eqn:brewsters:300}
\tan \theta_{iB} = \pm n,
\end{equation}

For normal media where $$n_i > 0, n_t > 0$$, only the positive solution is physically relevant, which is

\begin{equation}\label{eqn:brewsters:320}
\boxed{
\theta_{iB} = \arctan\lr{ \frac{n_t}{n_i} }.
}
\end{equation}

# References

 E. Hecht. Optics. 1998.

 JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975.

## Fresnel angular sum and difference formulas

November 22, 2016 math and physics play , ,

In  are some sum and angle difference formulations for the Fresnel formulas given a $$\mu_1 = \mu_2$$ constraint. The proof of these trig Fresnel equations is left to an exercise, and will be derived here.

\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:20}
\begin{aligned}
\sin(a + b)
&=
\textrm{Im}\lr{ e^{j(a + b)} } \\
&=
\textrm{Im}\lr{
e^{ja} e^{+ jb}
} \\
&=
\textrm{Im}\lr{
(\cos a + j \sin a) (\cos b + j \sin b)
} \\
&=
\sin a \cos b + \cos a \sin b.
\end{aligned}
\end{equation}

Allowing for both signs we have

\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:240}
\begin{aligned}
\sin(a + b) &= \sin a \cos b + \cos a \sin b \\
\sin(a – b) &= \sin a \cos b – \cos a \sin b.
\end{aligned}
\end{equation}

The mixed sine and cosine product can be expressed as a sum of sines

\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:40}
2 \sin a \cos b = \sin(a + b) + \sin(a – b).
\end{equation}

With $$2 x = a + b, 2 y = a – b$$, or $$a = x + y, b = x – y$$, we find

\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:60}
\begin{aligned}
2 \sin(x + y) \cos (x – y) &= \sin( 2 x ) + \sin( 2 y ) \\
2 \sin(x – y) \cos (x + y) &= \sin( 2 x ) – \sin( 2 y ).
\end{aligned}
\end{equation}

Returning to the problem. When $$\mu_1 = \mu_2$$ the Fresnel equations were found to be

\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:100}
\begin{aligned}
r^{\textrm{TE}} &= \frac { n_1 \cos\theta_i – n_2 \cos\theta_t } { n_1 \cos\theta_i + n_2 \cos\theta_t } \\
r^{\textrm{TM}} &= \frac{n_2 \cos\theta_i – n_1 \cos\theta_t }{ n_2 \cos\theta_i + n_1 \cos\theta_t } \\
t^{\textrm{TE}} &= \frac{ 2 n_1 \cos\theta_i } { n_1 \cos\theta_i + n_2 \cos\theta_t } \\
t^{\textrm{TM}} &= \frac{2 n_1 \cos\theta_i }{ n_2 \cos\theta_i + n_1 \cos\theta_t }.
\end{aligned}
\end{equation}

Using Snell’s law, one of $$n_1, n_2$$ can be eliminated, for example

\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:120}
n_1 = n_2 \frac{\sin \theta_t}{\sin\theta_i}.
\end{equation}

Inserting this and proceeding with the application of the trig identities above, we have

\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:160}
\begin{aligned}
r^{\textrm{TE}}
&= \frac { n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i – n_2 \cos\theta_t } { n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i + n_2 \cos\theta_t } \\
&=
\frac {
\sin\theta_t \cos\theta_i – \cos\theta_t \sin\theta_i
} {
\sin\theta_t \cos\theta_i + \cos\theta_t \sin\theta_i
} \\
&=
\frac {
\sin( \theta_t – \theta_i )
} {
\sin( \theta_t + \theta_i )
}
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:180}
\begin{aligned}
r^{\textrm{TM}}
&= \frac{n_2 \cos\theta_i – n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_t }{ n_2 \cos\theta_i + n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_t } \\
&= \frac{
\sin\theta_i \cos\theta_i – \sin\theta_t \cos\theta_t
}{
\sin\theta_i \cos\theta_i + \sin\theta_t \cos\theta_t
} \\
&= \frac{\inv{2} \sin(2 \theta_i) – \inv{2} \sin(2 \theta_t) }{ \inv{2} \sin(2 \theta_i) + \inv{2} \sin(2 \theta_t) } \\
&= \frac
{\sin(\theta_i – \theta_t)\cos(\theta_i + \theta_t) }
{\sin(\theta_i + \theta_t)\cos(\theta_i – \theta_t) } \\
&=
\frac
{\tan(\theta_i -\theta_t)}
{\tan(\theta_i +\theta_t)}
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:200}
\begin{aligned}
t^{\textrm{TE}}
&= \frac{ 2 n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i } { n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i + n_2 \cos\theta_t } \\
&= \frac{ 2 \sin\theta_t \cos\theta_i } { \sin\theta_t \cos\theta_i + \cos\theta_t \sin\theta_i } \\
&= \frac{ 2 \sin\theta_t \cos\theta_i }
{ \sin(\theta_i + \theta_t) }
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:fresnelSumAndDifferenceAngleFormulas:220}
\begin{aligned}
t^{\textrm{TM}}
&= \frac{2 n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_i }{ n_2 \cos\theta_i + n_2 \frac{\sin\theta_t}{\sin\theta_i} \cos\theta_t } \\
&= \frac{2 \sin\theta_t \cos\theta_i }{ \sin\theta_i \cos\theta_i + \sin\theta_t \cos\theta_t } \\
&= \frac{2 \sin\theta_t \cos\theta_i }
{ \inv{2} \sin(2 \theta_i) + \inv{2} \sin(2 \theta_t) } \\
&= \frac{2 \sin\theta_t \cos\theta_i }
{ \sin(\theta_i + \theta_t) \cos(\theta_i – \theta_t) }
\end{aligned}
\end{equation}

# References

 E. Hecht. Optics. 1998.

## Normal transmission and reflection through two interfaces

November 21, 2016 math and physics play , , , ,

### Motivation

In class an outline of normal transmission through a slab was presented. Let’s go through the details.

### Normal incidence

The geometry of a two interface configuration is sketched in fig. 1.

Given a normal incident ray with magnitude $$A$$, the respective forward and backwards rays in each the mediums can be written as

[I]

1. \begin{equation}\label{eqn:twoInterfaceNormal:20}
\begin{aligned}
A e^{-j k_1 z} \\
A r e^{j k_1 z} \\
\end{aligned}
\end{equation}
2. \begin{equation}\label{eqn:twoInterfaceNormal:40}
C e^{-j k_2 z} \\
D e^{j k_2 z} \\
\end{equation}
3. \begin{equation}\label{eqn:twoInterfaceNormal:60}
A t e^{-j k_3 (z-d)}
\end{equation}

Matching at $$z = 0$$ gives
\begin{equation}\label{eqn:twoInterfaceNormal:80}
\begin{aligned}
A t_{12} + r_{21} D &= C \\
A r &= A r_{12} + D t_{21},
\end{aligned}
\end{equation}

whereas matching at $$z = d$$ gives

\begin{equation}\label{eqn:twoInterfaceNormal:100}
\begin{aligned}
A t &= C e^{-j k_2 d} t_{23} \\
D e^{j k_2 d} &= C e^{-j k_2 d} r_{23}
\end{aligned}
\end{equation}

We have four linear equations in four unknowns $$r, t, C, D$$, but only care about solving for $$r, t$$. Let’s write $$\gamma = e^{ j k_2 d }, C’ = C/A, D’ = D/A$$, for

\begin{equation}\label{eqn:twoInterfaceNormal:120}
\begin{aligned}
t_{12} + r_{21} D’ &= C’ \\
r &= r_{12} + D’ t_{21} \\
t \gamma &= C’ t_{23} \\
D’ \gamma^2 &= C’ r_{23}
\end{aligned}
\end{equation}

Solving for $$C’, D’$$ we get

\begin{equation}\label{eqn:twoInterfaceNormal:140}
\begin{aligned}
D’ \lr{ \gamma^2 – r_{21} r_{23} } &= t_{12} r_{23} \\
C’ \lr{ \gamma^2 – r_{21} r_{23} } &= t_{12} \gamma^2,
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:twoInterfaceNormal:160}
\begin{aligned}
r &= r_{12} + \frac{t_{12} t_{21} r_{23} }{\gamma^2 – r_{21} r_{23} } \\
t &= t_{23} \frac{ t_{12} \gamma }{\gamma^2 – r_{21} r_{23} }.
\end{aligned}
\end{equation}

With $$\phi = -j k_2 d$$, or $$\gamma = e^{-j\phi}$$, we have

\begin{equation}\label{eqn:twoInterfaceNormal:180}
\boxed{
\begin{aligned}
r &= r_{12} + \frac{t_{12} t_{21} r_{23} e^{2 j \phi} }{1 – r_{21} r_{23} e^{2 j \phi}} \\
t &= \frac{ t_{12} t_{23} e^{j\phi}}{1 – r_{21} r_{23} e^{2 j \phi}}.
\end{aligned}
}
\end{equation}

### A slab

When the materials in region I, and III are equal, then $$r_{12} = r_{32}$$. For a TE mode, we have

\begin{equation}\label{eqn:twoInterfaceNormal:200}
r_{12}
=
\frac{\mu_2 k_{1z} – \mu_1 k_{2z}}{\mu_2 k_{1z} + \mu_1 k_{2z}}
= -r_{21}.
\end{equation}

so the reflection and transmission coefficients are

\begin{equation}\label{eqn:twoInterfaceNormal:220}
\begin{aligned}
r^{\textrm{TE}} &= r_{12} \lr{ 1 – \frac{t_{12} t_{21} e^{2 j \phi} }{1 – r_{21}^2 e^{2 j \phi}} } \\
t^{\textrm{TE}} &= \frac{ t_{12} t_{21} e^{j\phi}}{1 – r_{21}^2 e^{2 j \phi}}.
\end{aligned}
\end{equation}

It’s possible to produce a matched condition for which $$r_{12} = r_{21} = 0$$, by selecting

\begin{equation}\label{eqn:twoInterfaceNormal:240}
\begin{aligned}
0
&= \mu_2 k_{1z} – \mu_1 k_{2z} \\
&= \mu_1 \mu_2 \lr{ \inv{\mu_1} k_{1z} – \inv{\mu_2} k_{2z} } \\
&= \mu_1 \mu_2 \omega \lr{ \frac{1}{v_1 \mu_1} \theta_1 – \frac{1}{v_2 \mu_2} \theta_2 },
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:twoInterfaceNormal:260}
\inv{\eta_1} \cos\theta_1 = \inv{\eta_2} \cos\theta_2,
\end{equation}

so the matching condition for normal incidence is just

\begin{equation}\label{eqn:twoInterfaceNormal:280}
\eta_1 = \eta_2.
\end{equation}

Given this matched condition, the transmission coefficient for the 1,2 interface is

\begin{equation}\label{eqn:twoInterfaceNormal:300}
\begin{aligned}
t_{12}
&= \frac{2 \mu_2 k_{1z}}{\mu_2 k_{1z} + \mu_1 k_{2z}} \\
&= \frac{2 \mu_2 k_{1z}}{2 \mu_2 k_{1z} } \\
&= 1,
\end{aligned}
\end{equation}

so the matching condition yields
\begin{equation}\label{eqn:twoInterfaceNormal:320}
\begin{aligned}
t
&=
t_{12} t_{21} e^{j\phi} \\
&=
e^{j\phi} \\
&=
e^{-j k_2 d}.
\end{aligned}
\end{equation}

Normal transmission through a matched slab only introduces a phase delay.