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In [1] section 3.2 is a demonstration of the required (curl) form for the magnetic field, and potential form for the electric field.

I was wondering how this derivation would proceed using the Geometric Algebra (GA) formalism.

Maxwell’s equation in GA phasor form.

Maxwell’s equations, omitting magnetic charges and currents, are

\begin{equation}\label{eqn:phasorMaxwellsGA:20}
\spacegrad \cross \boldsymbol{\mathcal{E}} = -\PD{t}{\boldsymbol{\mathcal{B}}}
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:40}
\spacegrad \cross \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{J}} + \PD{t}{\boldsymbol{\mathcal{D}}}
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:60}
\spacegrad \cdot \boldsymbol{\mathcal{D}} = \rho
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:80}
\spacegrad \cdot \boldsymbol{\mathcal{B}} = 0.
\end{equation}

Assuming linear media \( \boldsymbol{\mathcal{B}} = \mu_0 \boldsymbol{\mathcal{H}} \), \( \boldsymbol{\mathcal{D}} = \epsilon_0 \boldsymbol{\mathcal{E}} \), and phasor relationships of the form \( \boldsymbol{\mathcal{E}} = \textrm{Re} \lr{ \BE(\Br) e^{j \omega t}} \) for the fields and the currents, these reduce to

\begin{equation}\label{eqn:phasorMaxwellsGA:100}
\spacegrad \cross \BE = – j \omega \BB
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:120}
\spacegrad \cross \BB = \mu_0 \BJ + j \omega \epsilon_0 \mu_0 \BE
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:140}
\spacegrad \cdot \BE = \rho/\epsilon_0
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:160}
\spacegrad \cdot \BB = 0.
\end{equation}

These four equations can be assembled into a single equation form using the GA identities

\begin{equation}\label{eqn:phasorMaxwellsGA:200}
\Bf \Bg
= \Bf \cdot \Bg + \Bf \wedge \Bg
= \Bf \cdot \Bg + I \Bf \cross \Bg.
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:220}
I = \xcap \ycap \zcap.
\end{equation}

The electric and magnetic field equations, respectively, are

\begin{equation}\label{eqn:phasorMaxwellsGA:260}
\spacegrad \BE = \rho/\epsilon_0 -j k c \BB I
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:280}
\spacegrad c \BB = \frac{I}{\epsilon_0 c} \BJ + j k \BE I
\end{equation}

where \( \omega = k c \), and \( 1 = c^2 \epsilon_0 \mu_0 \) have also been used to eliminate some of the mess of constants.

Summing these (first scaling \ref{eqn:phasorMaxwellsGA:280} by \( I \)), gives Maxwell’s equation in its GA phasor form

\begin{equation}\label{eqn:phasorMaxwellsGA:300}
\boxed{
\lr{ \spacegrad + j k } \lr{ \BE + I c \BB } = \inv{\epsilon_0 c}\lr{c \rho – \BJ}.
}
\end{equation}

Preliminaries. Dual magnetic form of Maxwell’s equations.

The arguments of the text showing that a potential representation for the electric and magnetic fields is possible easily translates into GA. To perform this translation, some duality lemmas are required

First consider the cross product of two vectors \( \Bx, \By \) and the right handed dual \( -\By I \) of \( \By \), a bivector, of one of these vectors. Noting that the Euclidean pseudoscalar \( I \) commutes with all grade multivectors in a Euclidean geometric algebra space, the cross product can be written

\begin{equation}\label{eqn:phasorMaxwellsGA:320}
\begin{aligned}
\lr{ \Bx \cross \By }
&=
-I \lr{ \Bx \wedge \By } \\
&=
-I \inv{2} \lr{ \Bx \By – \By \Bx } \\
&=
\inv{2} \lr{ \Bx (-\By I) – (-\By I) \Bx } \\
&=
\Bx \cdot \lr{ -\By I }.
\end{aligned}
\end{equation}

The last step makes use of the fact that the wedge product of a vector and vector is antisymmetric, whereas the dot product (vector grade selection) of a vector and bivector is antisymmetric. Details on grade selection operators and how to characterize symmetric and antisymmetric products of vectors with blades as either dot or wedge products can be found in [3], [2].

Similarly, the dual of the dot product can be written as

\begin{equation}\label{eqn:phasorMaxwellsGA:440}
\begin{aligned}
-I \lr{ \Bx \cdot \By }
&=
-I \inv{2} \lr{ \Bx \By + \By \Bx } \\
&=
\inv{2} \lr{ \Bx (-\By I) + (-\By I) \Bx } \\
&=
\Bx \wedge \lr{ -\By I }.
\end{aligned}
\end{equation}

These duality transformations are motivated by the observation that in the GA form of Maxwell’s equation the magnetic field shows up in its dual form, a bivector. Spelled out in terms of the dual magnetic field, those equations are

\begin{equation}\label{eqn:phasorMaxwellsGA:360}
\spacegrad \wedge \BE = – j \omega \BB I
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:380}
\spacegrad \cdot \lr{ -\BB I } = \mu_0 \BJ + j \omega \epsilon_0 \mu_0 \BE
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:400}
\spacegrad \cdot \BE = \rho/\epsilon_0
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:420}
\spacegrad \wedge (-\BB I) = 0.
\end{equation}

Constructing a potential representation.

The starting point of the argument in the text was the observation that the triple product \( \spacegrad \cdot \lr{ \spacegrad \cross \Bx } = 0 \) for any (sufficiently continuous) vector \( \Bx \). This triple product is a completely antisymmetric sum, and the equivalent statement in GA is \( \spacegrad \wedge \spacegrad \wedge \Bx = 0 \) for any vector \( \Bx \). This follows from \( \Ba \wedge \Ba = 0 \), true for any vector \( \Ba \), including the gradient operator \( \spacegrad \), provided those gradients are acting on a sufficiently continuous blade.

In the absence of magnetic charges, \ref{eqn:phasorMaxwellsGA:420} shows that the divergence of the dual magnetic field is zero. It it therefore possible to find a potential \( \BA \) such that

\begin{equation}\label{eqn:phasorMaxwellsGA:460}
\BB I = \spacegrad \wedge \BA.
\end{equation}

Substituting this into Maxwell-Faraday \ref{eqn:phasorMaxwellsGA:360} gives

\begin{equation}\label{eqn:phasorMaxwellsGA:480}
\spacegrad \wedge \lr{ \BE + j \omega \BA } = 0.
\end{equation}

This relation is a bivector identity with zero, so will be satisfied if

\begin{equation}\label{eqn:phasorMaxwellsGA:500}
\BE + j \omega \BA = -\spacegrad \phi,
\end{equation}

for some scalar \( \phi \). Unlike the \( \BB I = \spacegrad \wedge \BA \) solution to \ref{eqn:phasorMaxwellsGA:420}, the grade of \( \phi \) is fixed by the requirement that \( \BE + j \omega \BA \) is unity (a vector), so a \( \BE + j \omega \BA = \spacegrad \wedge \psi \), for a higher grade blade \( \psi \) would not work, despite satisifying the condition \( \spacegrad \wedge \spacegrad \wedge \psi = 0 \).

Substitution of \ref{eqn:phasorMaxwellsGA:500} and \ref{eqn:phasorMaxwellsGA:460} into Ampere’s law \ref{eqn:phasorMaxwellsGA:380} gives

\begin{equation}\label{eqn:phasorMaxwellsGA:520}
\begin{aligned}
-\spacegrad \cdot \lr{ \spacegrad \wedge \BA } &= \mu_0 \BJ + j \omega \epsilon_0 \mu_0 \lr{ -\spacegrad \phi -j \omega \BA } \\
-\spacegrad^2 \BA – \spacegrad \lr{\spacegrad \cdot \BA} &=
\end{aligned}
\end{equation}

Rearranging gives

\begin{equation}\label{eqn:phasorMaxwellsGA:540}
\spacegrad^2 \BA + k^2 \BA = -\mu_0 \BJ – \spacegrad \lr{ \spacegrad \cdot \BA + j \frac{k}{c} \phi }.
\end{equation}

The fields \( \BA \) and \( \phi \) are assumed to be phasors, say \( \boldsymbol{\mathcal{A}} = \textrm{Re} \BA e^{j k c t} \) and \( \varphi = \textrm{Re} \phi e^{j k c t} \). Grouping the scalar and vector potentials into the standard four vector form \( A^\mu = \lr{\phi/c, \BA} \), and expanding the Lorentz gauge condition

\begin{equation}\label{eqn:phasorMaxwellsGA:580}
\begin{aligned}
0
&= \partial_\mu \lr{ A^\mu e^{j k c t}} \\
&= \partial_a \lr{ A^a e^{j k c t}} + \inv{c}\PD{t}{} \lr{ \frac{\phi}{c} e^{j k c t}} \\
&= \spacegrad \cdot \BA e^{j k c t} + \inv{c} j k \phi e^{j k c t} \\
&= \lr{ \spacegrad \cdot \BA + j k \phi/c } e^{j k c t},
\end{aligned}
\end{equation}

shows that in \ref{eqn:phasorMaxwellsGA:540} the quantity in braces is in fact the Lorentz gauge condition, so in the Lorentz gauge, the vector potential satisfies a non-homogeneous Helmholtz equation.

\begin{equation}\label{eqn:phasorMaxwellsGA:550}
\boxed{
\spacegrad^2 \BA + k^2 \BA = -\mu_0 \BJ.
}
\end{equation}

Maxwell’s equation in Four vector form

The four vector form of Maxwell’s equation follows from \ref{eqn:phasorMaxwellsGA:300} after pre-multiplying by \( \gamma^0 \).

With

\begin{equation}\label{eqn:phasorMaxwellsGA:620}
A = A^\mu \gamma_\mu = \lr{ \phi/c, \BA }
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:640}
F = \grad \wedge A = \inv{c} \lr{ \BE + c \BB I }
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:660}
\grad = \gamma^\mu \partial_\mu = \gamma^0 \lr{ \spacegrad + j k }
\end{equation}
\begin{equation}\label{eqn:phasorMaxwellsGA:680}
J = J^\mu \gamma_\mu = \lr{ c \rho, \BJ },
\end{equation}

Maxwell’s equation is

\begin{equation}\label{eqn:phasorMaxwellsGA:700}
\boxed{
\grad F = \mu_0 J.
}
\end{equation}

Here \( \setlr{ \gamma_\mu } \) is used as the basis of the four vector Minkowski space, with \( \gamma_0^2 = -\gamma_k^2 = 1 \) (i.e. \(\gamma^\mu \cdot \gamma_\nu = {\delta^\mu}_\nu \)), and \( \gamma_a \gamma_0 = \sigma_a \) where \( \setlr{ \sigma_a} \) is the Pauli basic (i.e. standard basis vectors for \R{3}).

Let’s demonstrate this, one piece at a time. Observe that the action of the spacetime gradient on a phasor, assuming that all time dependence is in the exponential, is

\begin{equation}\label{eqn:phasorMaxwellsGA:740}
\begin{aligned}
\gamma^\mu \partial_\mu \lr{ \psi e^{j k c t} }
&=
\lr{ \gamma^a \partial_a + \gamma_0 \partial_{c t} } \lr{ \psi e^{j k c t} }
\\
&=
\gamma_0 \lr{ \gamma_0 \gamma^a \partial_a + j k } \lr{ \psi e^{j k c t} } \\
&=
\gamma_0 \lr{ \sigma_a \partial_a + j k } \psi e^{j k c t} \\
&=
\gamma_0 \lr{ \spacegrad + j k } \psi e^{j k c t}
\end{aligned}
\end{equation}

This allows the operator identification of \ref{eqn:phasorMaxwellsGA:660}. The four current portion of the equation comes from

\begin{equation}\label{eqn:phasorMaxwellsGA:760}
\begin{aligned}
c \rho – \BJ
&=
\gamma_0 \lr{ \gamma_0 c \rho – \gamma_0 \gamma_a \gamma_0 J^a } \\
&=
\gamma_0 \lr{ \gamma_0 c \rho + \gamma_a J^a } \\
&=
\gamma_0 \lr{ \gamma_\mu J^\mu } \\
&= \gamma_0 J.
\end{aligned}
\end{equation}

Taking the curl of the four potential gives

\begin{equation}\label{eqn:phasorMaxwellsGA:780}
\begin{aligned}
\grad \wedge A
&=
\lr{ \gamma^a \partial_a + \gamma_0 j k } \wedge \lr{ \gamma_0 \phi/c + \gamma_b A^b } \\
&=
– \sigma_a \partial_a \phi/c + \gamma^a \wedge \gamma_b \partial_a A^b – j k
\sigma_b A^b \\
&=
– \sigma_a \partial_a \phi/c + \sigma_a \wedge \sigma_b \partial_a A^b – j k
\sigma_b A^b \\
&= \inv{c} \lr{ – \spacegrad \phi – j \omega \BA + c \spacegrad \wedge \BA }
\\
&= \inv{c} \lr{ \BE + c \BB I }.
\end{aligned}
\end{equation}

Substituting all of these into Maxwell’s \ref{eqn:phasorMaxwellsGA:300} gives

\begin{equation}\label{eqn:phasorMaxwellsGA:800}
\gamma_0 \grad c F = \inv{ \epsilon_0 c } \gamma_0 J,
\end{equation}

which recovers \ref{eqn:phasorMaxwellsGA:700} as desired.

Helmholtz equation directly from the GA form.

It is easier to find \ref{eqn:phasorMaxwellsGA:550} from the GA form of Maxwell’s \ref{eqn:phasorMaxwellsGA:700} than the traditional curl and divergence equations. Note that

\begin{equation}\label{eqn:phasorMaxwellsGA:820}
\grad F
=
\grad \lr{ \grad \wedge A }
=
\grad \cdot \lr{ \grad \wedge A }
+
\grad \wedge \lr{ \grad \wedge A }
=
\grad^2 A – \grad \lr{ \grad \cdot A },
\end{equation}

however, the Lorentz gauge condition \( \partial_\mu A^\mu = \grad \cdot A = 0 \) kills the latter term above. This leaves

\begin{equation}\label{eqn:phasorMaxwellsGA:840}
\begin{aligned}
\grad F
&=
\grad^2 A \\
&=
\gamma_0 \lr{ \spacegrad + j k }
\gamma_0 \lr{ \spacegrad + j k } A \\
&=
\gamma_0^2 \lr{ -\spacegrad + j k }
\lr{ \spacegrad + j k } A \\
&=
-\lr{ \spacegrad^2 + k^2 } A = \mu_0 J.
\end{aligned}
\end{equation}

The timelike component of this gives

\begin{equation}\label{eqn:phasorMaxwellsGA:860}
\lr{ \spacegrad^2 + k^2 } \phi = -\rho/\epsilon_0,
\end{equation}

and the spacelike components give

\begin{equation}\label{eqn:phasorMaxwellsGA:880}
\lr{ \spacegrad^2 + k^2 } \BA = -\mu_0 \BJ,
\end{equation}

recovering \ref{eqn:phasorMaxwellsGA:550} as desired.

References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley & Sons, 3rd edition, 2005.

[2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[3] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.

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This is my first set of notes for the UofT course ECE1229, Advanced Antenna Theory, taught by Prof. Eleftheriades, covering ch. 2 [1] content.

Unlike most of the other classes I have taken, I am not attempting to take comprehensive notes for this class. The class is taught on slides that match the textbook so closely, there is little value to me taking notes that just replicate the text. Instead, I am annotating my copy of textbook with little details instead. My usual notes collection for the class will contain musings of details that were unclear, or in some cases, details that were provided in class, but are not in the text (and too long to pencil into my book.)

Poynting vector

The Poynting vector was written in an unfamiliar form

\begin{equation}\label{eqn:chapter2Notes:560}
\boldsymbol{\mathcal{W}} = \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}}.
\end{equation}

I can roll with the use of a different symbol (i.e. not \(\BS\)) for the Poynting vector, but I’m used to seeing a \( \frac{c}{4\pi} \) factor ([6] and [5]). I remembered something like that in SI units too, so was slightly confused not to see it here.

Per [3] that something is a \( \mu_0 \), as in

\begin{equation}\label{eqn:chapter2Notes:580}
\boldsymbol{\mathcal{W}} = \inv{\mu_0} \boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{B}}.
\end{equation}

Note that the use of \( \boldsymbol{\mathcal{H}} \) instead of \( \boldsymbol{\mathcal{B}} \) is what wipes out the requirement for the \( \frac{1}{\mu_0} \) term since \( \boldsymbol{\mathcal{H}} = \boldsymbol{\mathcal{B}}/\mu_0 \), assuming linear media, and no magnetization.

Typical far-field radiation intensity

It was mentioned that

\begin{equation}\label{eqn:advancedantennaL1:20}
U(\theta, \phi)
=
\frac{r^2}{2 \eta_0} \Abs{ \BE( r, \theta, \phi) }^2
=
\frac{1}{2 \eta_0} \lr{ \Abs{ E_\theta(\theta, \phi) }^2 + \Abs{ E_\phi(\theta, \phi) }^2},
\end{equation}

where the intrinsic impedance of free space is

\begin{equation}\label{eqn:advancedantennaL1:480}
\eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} = 377 \Omega.
\end{equation}

(this is also eq. 2-19 in the text.)

To get an understanding where this comes from, consider the far field radial solutions to the electric and magnetic dipole problems, which have the respective forms (from [3]) of

\begin{equation}\label{eqn:chapter2Notes:740}
\begin{aligned}
\boldsymbol{\mathcal{E}} &= -\frac{\mu_0 p_0 \omega^2 }{4 \pi } \frac{\sin\theta}{r} \cos\lr{w t – k r} \thetacap \\
\boldsymbol{\mathcal{B}} &= -\frac{\mu_0 p_0 \omega^2 }{4 \pi c} \frac{\sin\theta}{r} \cos\lr{w t – k r} \phicap \\
\end{aligned}
\end{equation}
\begin{equation}\label{eqn:chapter2Notes:760}
\begin{aligned}
\boldsymbol{\mathcal{E}} &= \frac{\mu_0 m_0 \omega^2 }{4 \pi c} \frac{\sin\theta}{r} \cos\lr{w t – k r} \phicap \\
\boldsymbol{\mathcal{B}} &= -\frac{\mu_0 m_0 \omega^2 }{4 \pi c^2} \frac{\sin\theta}{r} \cos\lr{w t – k r} \thetacap \\
\end{aligned}
\end{equation}

In neither case is there a component in the direction of propagation, and in both cases (using \( \mu_0 \epsilon_0 = 1/c^2\))

\begin{equation}\label{eqn:chapter2Notes:780}
\Abs{\boldsymbol{\mathcal{H}}}
= \frac{\Abs{\boldsymbol{\mathcal{E}}}}{\mu_0 c}
= \Abs{\boldsymbol{\mathcal{E}}} \sqrt{\frac{\epsilon_0}{\mu_0}}
= \inv{\eta_0}\Abs{\boldsymbol{\mathcal{E}}} .
\end{equation}

A superposition of the phasors for such dipole fields, in the far field, will have the form

\begin{equation}\label{eqn:chapter2Notes:800}
\begin{aligned}
\BE &= \inv{r} \lr{ E_\theta(\theta, \phi) \thetacap + E_\phi(\theta, \phi) \phicap } \\
\BB &= \inv{r c} \lr{ E_\theta(\theta, \phi) \thetacap – E_\phi(\theta, \phi) \phicap },
\end{aligned}
\end{equation}

with a corresponding time averaged Poynting vector

\begin{equation}\label{eqn:chapter2Notes:820}
\begin{aligned}
\BW_{\textrm{av}}
&= \inv{2 \mu_0} \BE \cross \BB^\conj \\
&=
\inv{2 \mu_0 c r^2}
\lr{ E_\theta \thetacap + E_\phi \phicap } \cross
\lr{ E_\theta^\conj \thetacap – E_\phi^\conj \phicap } \\
&=
\frac{\thetacap \cross \phicap}{2 \mu_0 c r^2}
\lr{ \Abs{E_\theta}^2 + \Abs{E_\phi}^2 } \\
&=
\frac{\rcap}{2 \eta_0 r^2}
\lr{ \Abs{E_\theta}^2 + \Abs{E_\phi}^2 },
\end{aligned}
\end{equation}

verifying \ref{eqn:advancedantennaL1:20} for a superposition of electric and magnetic dipole fields. This can likely be shown for more general fields too.

Field plots

We can plot the fields, or intensity (or log plots in dB of these).
It is pointed out in [3] that when there is \( r \) dependence these plots are done by considering the values of at fixed \( r \).

The field plots are conceptually the simplest, since that vector parameterizes
a surface. Any such radial field with magnitude \( f(r, \theta, \phi) \) can
be plotted in Mathematica in the \( \phi = 0 \) plane at \( r = r_0 \), or in
3D (respectively, but also at \( r = r_0\)) with code like that of the
following listing

Intensity plots can use the same code, with the only difference being the interpretation. The surface doesn’t represent the value of a vector valued radial function, but is the magnitude of a scalar valued function evaluated at \( f( r_0, \theta, \phi) \).

The surfaces for \( U = \sin\theta, \sin^2\theta \) in the plane are parametrically plotted in fig. 2, and for cosines in fig. 1 to compare with textbook figures.

fig 1. Cosinusoidal radiation intensities

fig 2. Sinusoidal radiation intensities

Visualizations of \( U = \sin^2 \theta\) and \( U = \cos^2 \theta\) can be found in fig. 3 and fig. 4 respectively. Even for such simple functions these look pretty cool.

fig 3. Square sinusoidal radiation intensity

fig 4. Square cosinusoidal radiation intensity

dB vs dBi

Note that dBi is used to indicate that the gain is with respect to an “isotropic” radiator.
This is detailed more in [2].

Trig integrals

Tables 1.1 and 1.2 produced with tableOfTrigIntegrals.nb have some of the sine and cosine integrals that are pervasive in this chapter.

Polarization vectors

The text introduces polarization vectors \( \rhocap \) , but doesn’t spell out their form. Consider a plane wave field of the form

\begin{equation}\label{eqn:chapter2Notes:840}
\BE
=
E_x e^{j \phi_x} e^{j \lr{ \omega t – k z }} \xcap
+
E_y e^{j \phi_y} e^{j \lr{ \omega t – k z }} \ycap.
\end{equation}

The \( x, y \) plane directionality of this phasor can be written

\begin{equation}\label{eqn:chapter2Notes:860}
\Brho =
E_x e^{j \phi_x} \xcap
+
E_y e^{j \phi_y} \ycap,
\end{equation}

so that

\begin{equation}\label{eqn:chapter2Notes:880}
\BE = \Brho e^{j \lr{ \omega t – k z }}.
\end{equation}

Separating this direction and magnitude into factors

\begin{equation}\label{eqn:chapter2Notes:900}
\Brho = \Abs{\BE} \rhocap,
\end{equation}

allows the phasor to be expressed as

\begin{equation}\label{eqn:chapter2Notes:920}
\BE = \rhocap \Abs{\BE} e^{j \lr{ \omega t – k z }}.
\end{equation}

As an example, suppose that \( E_x = E_y \), and set \( \phi_x = 0 \). Then

\begin{equation}\label{eqn:chapter2Notes:940}
\rhocap = \xcap + \ycap e^{j \phi_y}.
\end{equation}

Phasor power

In section 2.13 the phasor power is written as

\begin{equation}\label{eqn:chapter2Notes:620}
I^2 R/2,
\end{equation}

where \( I, R \) are the magnitudes of phasors in the circuit.

I vaguely recall this relation, but had to refer back to [4] for the details.
This relation expresses average power over a period associated with the frequency of the phasor

\begin{equation}\label{eqn:chapter2Notes:640}
\begin{aligned}
P
&= \inv{T} \int_{t_0}^{t_0 + T} p(t) dt \\
&= \inv{T} \int_{t_0}^{t_0 + T} \Abs{\BV} \cos\lr{ \omega t + \phi_V }
\Abs{\BI} \cos\lr{ \omega t + \phi_I} dt \\
&= \inv{T} \int_{t_0}^{t_0 + T} \Abs{\BV} \Abs{\BI}
\lr{
\cos\lr{ \phi_V – \phi_I } + \cos\lr{ 2 \omega t + \phi_V + \phi_I}
}
dt \\
&= \inv{2} \Abs{\BV} \Abs{\BI} \cos\lr{ \phi_V – \phi_I }.
\end{aligned}
\end{equation}

Introducing the impedance for this circuit element

\begin{equation}\label{eqn:chapter2Notes:660}
\BZ = \frac{ \Abs{\BV} e^{j\phi_V} }{ \Abs{\BI} e^{j\phi_I} } = \frac{\Abs{\BV}}{\Abs{\BI}} e^{j\lr{\phi_V – \phi_I}},
\end{equation}

this average power can be written in phasor form

\begin{equation}\label{eqn:chapter2Notes:680}
\BP = \inv{2} \Abs{\BI}^2 \BZ,
\end{equation}

with
\begin{equation}\label{eqn:chapter2Notes:700}
P = \textrm{Re} \BP.
\end{equation}

Observe that we have to be careful to use the absolute value of the current phasor \( \BI \), since \( \BI^2 \) differs in phase from \( \Abs{\BI}^2 \). This explains the conjugation in the [4] definition of complex power, which had the form

\begin{equation}\label{eqn:chapter2Notes:720}
\BS = \BV_{\textrm{rms}} \BI^\conj_{\textrm{rms}}.
\end{equation}

Radar cross section examples

Flat plate.

\begin{equation}\label{eqn:chapter2Notes:960}
\sigma_{\textrm{max}} = \frac{4 \pi \lr{L W}^2}{\lambda^2}
\end{equation}

fig. 6. Square geometry for RCS example.

Sphere.

In the optical limit the radar cross section for a sphere

fig. 7. Sphere geometry for RCS example.

\begin{equation}\label{eqn:chapter2Notes:980}
\sigma_{\textrm{max}} = \pi r^2
\end{equation}

Note that this is smaller than the physical area \( 4 \pi r^2 \).

Cylinder.

fig. 8. Cylinder geometry for RCS example.

\begin{equation}\label{eqn:chapter2Notes:1000}
\sigma_{\textrm{max}} = \frac{ 2 \pi r h^2}{\lambda}
\end{equation}

Tridedral corner reflector

fig. 9. Trihedral corner reflector geometry for RCS example.

\begin{equation}\label{eqn:chapter2Notes:1020}
\sigma_{\textrm{max}} = \frac{ 4 \pi L^4}{3 \lambda^2}
\end{equation}

Scattering from a sphere vs frequency

Frequency dependence of spherical scattering is sketched in fig. 10.

Low frequency (or small particles): Rayleigh\begin{equation}\label{eqn:chapter2Notes:1040}
\sigma = \lr{\pi r^2} 7.11 \lr{\kappa r}^4, \qquad \kappa = 2 \pi/\lambda.
\end{equation}
Mie scattering (resonance),\begin{equation}\label{eqn:chapter2Notes:1060}
\sigma_{\textrm{max}}(A) = 4 \pi r^2
\end{equation}
\begin{equation}\label{eqn:chapter2Notes:1080}
\sigma_{\textrm{max}}(B) = 0.26 \pi r^2.
\end{equation}
optical limit ( \(r \gg \lambda\) )\begin{equation}\label{eqn:chapter2Notes:1100}
\sigma = \pi r^2.
\end{equation}

fig 10. Scattering from a sphere vs frequency (from Prof. Eleftheriades’ class notes).

FIXME: Do I have a derivation of this in my optics notes?

Notation

Time average.
Both Prof. Eleftheriades
and the text [1] use square brackets \( [\cdots] \) for time averages, not \( <\cdots> \). Was that an engineering convention?
Prof. Eleftheriades
writes \(\Omega\) as a circle floating above a face up square bracket, as in fig. 1, and \( \sigma \) like a number 6, as in fig. 1.
Bold vectors are usually phasors, with (bold) calligraphic script used for the time domain fields. Example: \( \BE(x,y,z,t) = \ecap E(x,y) e^{j \lr{\omega t – k z}}, \boldsymbol{\mathcal{E}}(x, y, z, t) = \textrm{Re} \BE \).

fig. 11. Prof. handwriting decoder ring: Omega

fig 12. Prof. handwriting decoder ring: sigma

References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley \& Sons, 3rd edition, 2005.

[3] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.

[4] J.D. Irwin. Basic Engineering Circuit Analysis. MacMillian, 1993.

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

[6] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980. ISBN 0750627689.

linear media

Maxwell’s (phasor) equations in Geometric Algebra

Maxwell’s equation in GA phasor form.

Preliminaries. Dual magnetic form of Maxwell’s equations.

Constructing a potential representation.

Maxwell’s equation in Four vector form

Helmholtz equation directly from the GA form.

References

Like this:

Fundamental parameters of antennas

Poynting vector

Typical far-field radiation intensity

Field plots

dB vs dBi

Trig integrals

Polarization vectors

Phasor power