far field

Updated notes for ece1229 antenna theory

March 16, 2015 ece1229 No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

I’ve now posted a first update of my notes for the antenna theory course that I am taking this term at UofT.

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 which go by faster than I can easily take notes for (and some of which match the textbook closely). In class I have annotated my copy of textbook with little details instead. This set of notes contains 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), as well as some notes Geometric Algebra formalism for Maxwell’s equations with magnetic sources (something I’ve encountered for the first time in any real detail in this class).

The notes compilation linked above includes all of the following separate notes, some of which have been posted separately on this blog:

Reciprocity theorem: background

February 19, 2015 ece1229 No comments , , , , , , ,

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The class slides presented a derivation of the reciprocity theorem, a theorem that contained the integral of

\begin{equation}\label{eqn:reciprocityTheorem:360}
\int \lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} } \cdot d\BS = \cdots
\end{equation}

over a surface, where the RHS was a volume integral involving the fields and (electric and magnetic) current sources.
The idea was to consider two different source loading configurations of the same system, and to show that the fields and sources in the two configurations can be related.

To derive the result in question, a simple way to start is to look at the divergence of the difference of cross products above. This will require the phasor form of the two cross product Maxwell’s equations

\begin{equation}\label{eqn:reciprocityTheorem:100}
\spacegrad \cross \BE = – (\BM + j \omega \mu_0 \BH) % \BM^{(a)} + j \omega \mu_0 \BH^{(a)}
\end{equation}
\begin{equation}\label{eqn:reciprocityTheorem:120}
\spacegrad \cross \BH = \BJ + j \omega \epsilon_0 \BE, % \BJ^{(a)} + j \omega \epsilon_0 \BE^{(a)}
\end{equation}

so the divergence is

\begin{equation}\label{eqn:reciprocityTheorem:380}
\begin{aligned}
\spacegrad \cdot
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} }
&=
\BH^{(b)} \cdot \lr{ \spacegrad \cross \BE^{(a)} } -\BE^{(a)} \cdot \lr{ \spacegrad \cross \BH^{(b)} } \\
&-\BH^{(a)} \cdot \lr{ \spacegrad \cross \BE^{(b)} } +\BE^{(b)} \cdot \lr{ \spacegrad \cross \BH^{(a)} } \\
&=
-\BH^{(b)} \cdot \lr{ \BM^{(a)} + j \omega \mu_0 \BH^{(a)} } -\BE^{(a)} \cdot \lr{ \BJ^{(b)} + j \omega \epsilon_0 \BE^{(b)} } \\
&+\BH^{(a)} \cdot \lr{ \BM^{(b)} + j \omega \mu_0 \BH^{(b)} } +\BE^{(b)} \cdot \lr{ \BJ^{(a)} + j \omega \epsilon_0 \BE^{(a)} }.
\end{aligned}
\end{equation}

The non-source terms cancel, leaving

\begin{equation}\label{eqn:reciprocityTheorem:440}
\boxed{
\spacegrad \cdot
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} }
=
-\BH^{(b)} \cdot \BM^{(a)} -\BE^{(a)} \cdot \BJ^{(b)}
+\BH^{(a)} \cdot \BM^{(b)} +\BE^{(b)} \cdot \BJ^{(a)}
}
\end{equation}

Should we be suprised to have a relation of this form? Probably not, given that the energy momentum relationship between the fields and currents of a single source has the form

\begin{equation}\label{eqn:reciprocityTheorem:n}
\PD{t}{}\frac{\epsilon_0}{2} \left(\BE^2 + c^2 \BB^2\right) + \spacegrad \cdot \inv{\mu_0}(\BE \cross \BB) = -\BE \cdot \BJ.
\end{equation}

(this is without magnetic sources).

This suggests that the reciprocity theorem can be expressed more generally in terms of the energy-momentum tensor.

far field integral form

Employing the divergence theorem over a sphere the identity above takes the form

\begin{equation}\label{eqn:reciprocityTheorem:480}
\int_S
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} } \cdot \rcap dS
=
\int_V \lr{
-\BH^{(b)} \cdot \BM^{(a)} -\BE^{(a)} \cdot \BJ^{(b)}
+\BH^{(a)} \cdot \BM^{(b)} +\BE^{(b)} \cdot \BJ^{(a)}
}
dV
\end{equation}

In the far field, the cross products are strictly radial. That surface integral can be written as

\begin{equation}\label{eqn:reciprocityTheorem:500}
\begin{aligned}
\int_S
\lr{ \BE^{(a)} \cross \BH^{(b)} – \BE^{(b)} \cross \BH^{(a)} } \cdot \rcap dS
&=
\inv{\mu_0}
\int_S
\lr{ \BE^{(a)} \cross \lr{ \rcap \cross \BE^{(b)}} – \BE^{(b)} \cross \lr{ \rcap \cross \BE^{(a)}} } \cdot \rcap dS \\
&=
\inv{\mu_0}
\int_S
\lr{ \BE^{(a)} \cdot \BE^{(b)} – \BE^{(b)} \cdot \BE^{(a)}
}
dS \\
&= 0
\end{aligned}
\end{equation}

The above expansions used \ref{eqn:reciprocityTheorem:540} to expand the terms of the form

\begin{equation}\label{eqn:reciprocityTheorem:560}
\lr{ \BA \cross \lr{ \rcap \cross \BC } } \cdot \rcap
= \BA \cdot \BC -\lr{ \BA \cdot \rcap } \lr{ \BC \cdot \rcap },
\end{equation}

in which only the first dot product survives due to the transverse nature of the fields.

So in the far field we have a direct relation between the fields and sources of two source configurations of the same system of the form

\begin{equation}\label{eqn:reciprocityTheorem:580}
\boxed{
\int_V \lr{
\BH^{(a)} \cdot \BM^{(b)} +\BE^{(b)} \cdot \BJ^{(a)}
}
dV
=
\int_V \lr{
\BH^{(b)} \cdot \BM^{(a)} +\BE^{(a)} \cdot \BJ^{(b)}
}
dV
}
\end{equation}

Application to antenna

This is the underlying reason that we are able to pose the problem of what an antenna can recieve, in terms of what the antenna can transmit.

More on that to come.

Identities

Lemma: Divergence of a cross product.

\begin{equation}\label{thm:polarizationReview:400}
\spacegrad \cdot \lr{ \BA \cross \BB } =
\BB \lr{\spacegrad \cross \BA}
-\BA \lr{\spacegrad \cross \BB}.
\end{equation}

Proof.

\begin{equation}\label{eqn:reciprocityTheorem:420}
\begin{aligned}
\spacegrad \cdot \lr{ \BA \cross \BB }
&=
\partial_a \epsilon_{a b c} A_b B_c \\
&=
\epsilon_{a b c} (\partial_a A_b )B_c

\epsilon_{b a c} A_b (\partial_a B_c) \\
&=
\BB \cdot (\spacegrad \cross \BA)
-\BB \cdot (\spacegrad \cross \BA).
\end{aligned}
\end{equation}

Lemma: Triple cross product dotted
\begin{equation}\label{thm:polarizationReview:520}
\begin{aligned}
\lr{ \BA \cross \lr{ \BB \cross \BC } } \cdot \BD
=
\lr{ \BA \cdot \BC } \lr{ \BB \cdot \BD }
-\lr{ \BA \cdot \BB } \lr{ \BC \cdot \BD }
\end{aligned}
\end{equation}

Proof.

\begin{equation}\label{eqn:reciprocityTheorem:540}
\begin{aligned}
\lr{ \BA \cross \lr{ \BB \cross \BC } } \cdot \BD
&=
\epsilon_{a b c} A_b \epsilon_{r s c } B_r C_s D_a \\
&=
\delta_{[a b]}^{r s}
A_b B_r C_s D_a \\
&=
A_s B_r C_s D_r
-A_r B_r C_s D_s \\
&=
\lr{ \BA \cdot \BC } \lr{ \BB \cdot \BD }
-\lr{ \BA \cdot \BB } \lr{ \BC \cdot \BD }.
\end{aligned}
\end{equation}

Notes for ece1229 antenna theory

February 4, 2015 ece1229 No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

I’ve now posted a first set of notes for the antenna theory course that I am taking this term at UofT.

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.)

The notes linked above include:

  • Reading notes for chapter 2 (Fundamental Parameters of Antennas) and chapter 3 (Radiation Integrals and Auxiliary Potential Functions) of the class text.
  • Geometric Algebra musings.  How to do formulate Maxwell’s equations when magnetic sources are also included (those modeling magnetic dipoles).
  • Some problems for chapter 2 content.

On Tai and Pereira’s half power beamwidth approximations

January 31, 2015 ece1229 No comments , , , , , ,

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E and H plane directivities

In [2] directivities associated with the half power beamwidths are given as

\begin{equation}\label{eqn:taiAndPereira:20}
D_1 = \frac{\Abs{E_\theta}^2_{\textrm{max}}}{\inv{2} \int_0^\pi \Abs{E_\theta(\theta, 0)}^2 \sin\theta d\theta}
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:40}
D_2 = \frac{\Abs{E_\phi}^2_{\textrm{max}}}{\inv{2} \int_0^\pi \Abs{E_\phi(\theta, \pi/2)}^2 \sin\theta d\theta},
\end{equation}

whereas [1] lists these as

\begin{equation}\label{eqn:taiAndPereira:60}
\inv{D_1} = \inv{2 \ln 2} \int_0^{\Theta_{1 r}/2} \sin\theta d\theta
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:80}
\inv{D_2} = \inv{2 \ln 2} \int_0^{\Theta_{2 r}/2} \sin\theta d\theta.
\end{equation}

where the total directivity is given by the associated arithmetic mean formula

\begin{equation}\label{eqn:taiAndPereira:160}
\inv{D_0} = \inv{2}\lr{\inv{D_1} + \inv{D_2}}.
\end{equation}

This should follow from the far field approximation formula for \( U \). I intended to derive that result, but haven’t gotten to it. What follows instead are a few associated notes from a read of the paper, which I may revisit later to complete.

Short horizontal electrical dipole

Problem

In [2] a field for which directivities can be calculated exactly was used in comparisons of some directivity approximations

\begin{equation}\label{eqn:taiAndPereira:140}
\BE = E_0 \lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap }.
\end{equation}

(Observe that an inverse radial dependence in \(E_0\) must be implied here for this to be a valid far-field representation of the field.)

Show that Tai & Pereira’s formula gives \( D_1 = 3 \), and \( D_2 = 1 \) respectively for this field.

Calculate the exact directivity for this field.

Answer

The field components are

\begin{equation}\label{eqn:taiAndPereira:180}
E_\theta = E_0 \cos\theta \cos\phi
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:200}
E_\phi = -E_0 \sin\phi
\end{equation}

Using \ref{eqn:taiAndPereira:10} from the paper, the directivities are

\begin{equation}\label{eqn:taiAndPereira:220}
D_1 = \frac{2}{\int_0^\pi \cos^2 \theta \sin\theta d\theta}
= \frac{2}{\evalrange{-\inv{3}\cos^3\theta}{0}{\pi}}
= 3,
\end{equation}

and

\begin{equation}\label{eqn:taiAndPereira:240}
D_2
= \frac{2}{\int_0^\pi \sin\theta d\theta}
= \frac{2}{\evalrange{-\cos\theta}{0}{\pi}}
= 1.
\end{equation}

To find the exact directivity, first the Poynting vector is required. That is

\begin{equation}\label{eqn:taiAndPereira:260}
\begin{aligned}
\BP
&= \frac{
\Abs{E_0}^2
}{2 c \mu_0}
\lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap }
\cross
\lr{ \rcap \cross \lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap } } \\
&= \frac{
\Abs{E_0}^2
}{ 2 c \mu_0}
\lr{ \cos\theta \cos\phi \thetacap – \sin\phi \phicap }
\cross
\lr{ \cos\theta \cos\phi \phicap + \sin\phi \thetacap } \\
&= \frac{
\Abs{E_0}^2 \rcap
}{2 c \mu_0}
\lr{ \cos^2\theta \cos^2\phi + \sin^2\phi },
\end{aligned}
\end{equation}

so the radiation intensity is

\begin{equation}\label{eqn:taiAndPereira:280}
U(\theta, \phi) \propto \cos^2\theta \cos^2\phi + \sin^2\phi.
\end{equation}

The \( \thetacap \), and \( \phicap \) contributions to this intensity, and the total intensity are all plotted in fig. 1, fig. 2, and fig. 3 respectively.

TaiAndPereiraSampleFieldThetaCapComponentFig1pn

fig 1. The theta direction contribution to the radiation intensity.

 

TaiAndPereiraSampleFieldPhiCapComponentFig2pn

fig 2. The phi direction contribution to the radiation intensity.

 

TaiAndPereiraSampleFieldAllComponentsFig3pn

fig 3. Radiation intensity (both theta and phi direction contributions).

 

Given this the total radiated power is

\begin{equation}\label{eqn:taiAndPereira:300}
P_{\textrm{rad}} = \int_0^{2 \pi} \int_0^\pi
\lr{ \cos^2\theta \cos^2\phi + \sin^2\phi } \sin\theta d\theta d\phi
= \frac{8 \pi}{3}.
\end{equation}

Observe that the radiation intensity \( U \) can also be decomposed into two components, one for each component of the original \( \BE \) phasor.

\begin{equation}\label{eqn:taiAndPereira:320}
U_\theta = \cos^2 \theta \cos^2 \phi
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:340}
U_\phi = \sin^2 \phi
\end{equation}

This decomposition allows for expression of the partial directivities in these respective (orthogonal) directions

\begin{equation}\label{eqn:taiAndPereira:360}
D_\theta = \frac{4 \pi U_\theta}{P_{\textrm{rad}}} = \frac{3}{2} \cos^2 \theta \cos^2 \phi
\end{equation}
\begin{equation}\label{eqn:taiAndPereira:380}
D_\phi = \frac{4 \pi U_\phi}{P_{\textrm{rad}}} = \frac{3}{2} \sin^2 \phi
\end{equation}

The maximum of each of these partial directivities is both \( 3/2 \), giving a maximum directivity of

\begin{equation}\label{eqn:taiAndPereira:400}
D_0 =
\evalbar{D_\theta}{{\textrm{max}}}
+\evalbar{D_\phi}{{\textrm{max}}} = 3,
\end{equation}

the exact value from the paper.

References

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

[2] C-T Tai and CS Pereira. An approximate formula for calculating the directivity of an antenna. IEEE Transactions on Antennas and Propagation, 24:235, 1976.

Fundamental parameters of antennas

January 22, 2015 ece1229 No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

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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

ParametricPlotListing

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.

CoSineAndCoSineSqFig1pn

fig 1. Cosinusoidal radiation intensities

SineAndSinSqFig3pn

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.

SineSq3DFig4pn

fig 3. Square sinusoidal radiation intensity

 

CoSineSq3DFig2pn

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.

trigIntegralsUpToPiBy2

trigIntegralsUpToPi

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}

RCSsquareGeometryFig1

fig. 6. Square geometry for RCS example.

 

Sphere.

In the optical limit the radar cross section for a sphere

RCSsphereGeometryFig3

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.

RCScylinderGeometryFig1

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

trihedralCornerReflectorFig6

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}
sphericalScatteringFig5

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: \( \BE(x,y,z,t) = \ecap E(x,y) e^{j \lr{\omega t – k z}}, \boldsymbol{\mathcal{E}}(x, y, z, t) = \textrm{Re} \BE \).
greekStyleOmegaFig1

fig. 11. Prof. handwriting decoder ring: Omega

sigmaFig1

fig 12. Prof. handwriting decoder ring: sigma

 

References

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

[2] digi.com. Antenna Gain: dBi vs. dBd Decibel Detail, 2015. URL http://www.digi.com/support/kbase/kbaseresultdetl?id=2146. [Online; accessed 15-Jan-2015].

[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.