Peeter Jootpeeter.joot@gmail.com

Linear wire antennas

These are notes for the UofT course ECE1229, Advanced Antenna Theory, taught by Prof. Eleftheri-ades, covering ch.4 [1] content.

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

1.1 Magnetic Vector Potential.

In class and in the problem set A was referred to as the Magnetic Vector Potential. I only recalledthis referred to as the Vector Potential. Prefixing this with magnetic seemed counter intuitive to mesince it is generated by electric sources (charges and currents). This terminology can be justified dueto the fact that A generates the magnetic field by its curl. Some mention of this can be found in [4],which also points out that the Electric Potential refers to the scalar φ. Prof. Eleftheriades points outthat Electric Vector Potential refers to the vector potential F generated by magnetic sources (becausein that case the electric field is generated by the curl of F.)

1.2 Plots of infinitesimal dipole radial dependence.

In §4.2 of [1] are some discussions of the kr < 1, kr = 1, and kr > 1 radial dependence of the fields andpower of a solution to an infinitesimal dipole system. Here are some plots of those kr dependence,along with the kr = 1 contour as a reference. All the θ dependence and any scaling is left out.

The CDF notebook visualizeDipoleFields.cdf is available to interactively plot these, rotate the plotsand change the ranges of what is plotted.

FIXME: nbref.A plot of the real and imaginary parts of Hφ = jk

r e−jkr(

1− jkr

)can be found in fig. 1.1 and fig. 1.2 .

A plot of the real and imaginary parts of Er = 1r2

(1− j

kr

)e−jkr can be found in fig. 1.3 and fig. 1.4.

Finally, a plot of the real and imaginary parts of Eθ = jkr

(1− j

kr −1

k2r2

)e−jkr can be found in fig. 1.5

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Figure 1.1: Radial dependence of Re Hφ.

Figure 1.2: Radial dependence of Im Hφ.

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Figure 1.3: Radial dependence of Re Er.

Figure 1.4: Radial dependence of Im Er.

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and fig. 1.6.

Figure 1.5: Radial dependence of Re Eθ .

Figure 1.6: Radial dependence of Im Eθ .

1.3 Electric Far field for a spherical potential.

It is interesting to look at the far electric field associated with an arbitrary spherical magnetic vectorpotential, assuming all of the radial dependence is in the spherical envelope. That is

(1.1)A =e−jkr

r(rar(θ, φ)

+ θaθ

(θ, φ)

+ φaφ

(θ, φ))

.

The electric field is

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(1.2)E = −jωA− j1

ωµ0ε0∇ (∇ · A) .

The divergence and gradient in spherical coordinates are

(1.3a)∇ · A =1r2

∂

∂r(r2Ar

)+

1r sin θ

∂

∂θ(Aθ sin θ) +

1r sin θ

∂Aφ

∂φ

(1.3b)∇ψ = r∂ψ

∂r+

θ

r∂ψ

∂θ+

φ

r sin θ

∂ψ

∂φ.

For the assumed potential, the divergence is

(1.4)

∇ · A =ar

r2∂

∂r

(r2 e−jkr

r

)+

1r sin θ

e−jkr

r∂

∂θ(sin θaθ) +

1r sin θ

e−jkr

r∂aφ

∂φ

= are−jkr(

1r2 − jk

1r

)+

1r2 sin θ

e−jkr ∂

∂θ(sin θaθ) +

1r2 sin θ

e−jkr ∂aφ

∂φ

≈ −jkar

re−jkr.

The last approximation dropped all the 1/r2 terms that will be small compared to 1/r contributionthat dominates when r → ∞, the far field.

The gradient can now be computed

(1.5)

∇ (∇ · A) ≈ −jk∇( ar

re−jkr

)= −jk

(r

∂

∂r+

θ

r∂

∂θ+

φ

r sin θ

∂

∂φ

)ar

re−jkr

= −jk(

rar∂

∂r

(1r

e−jkr)

+θ

r2 e−jkr ∂ar

∂θ+ e−jkr φ

r2 sin θ

∂ar

∂φ

)= −jk

(−r

ar

r2

(1 + jkr

)+

θ

r2∂ar

∂θ+

φ

r2 sin θ

∂ar

∂φ

)e−jkr

≈ −k2rar

re−jkr.

Again, a far field approximation has been used to kill all the 1/r2 terms.The far field approximation of the electric field is now possible

(1.6)

E = −jωA− j1

ωµ0ε0∇ (∇ · A)

= −jωe−jkr

r(rar (θ, φ) + θaθ (θ, φ) + φaφ (θ, φ)

)+ j

1ωµ0ε0

k2rar

re−jkr

= −jωe−jkr

r(���

��rar (θ, φ) + θaθ (θ, φ) + φaφ (θ, φ))

+��

������

�

jc2

ω

(ω

c

)2r

ar

re−jkr

= −jωe−jkr

r(θaθ (θ, φ) + φaφ (θ, φ)

).

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Observe the perfect, somewhat miraculous seeming, cancellation of all the radial components ofthe field. If AT is the non-radial projection of A, the electric far field is just

Eff = −jωAT. (1.7)

1.4 Magnetic Far field for a spherical potential.

Application of the same assumed representation for the magnetic field gives

B = ∇ × A

=r

r sin θ∂θ

(Aφ sin θ

)+

θ

r

(1

sin θ∂φ Ar − ∂r

(rAφ

))+

φ

r(∂r (rAθ)− ∂θ Ar)

=r

r sin θ∂θ

(e−jkr

raφ sin θ

)+

θ

r

(1

sin θ∂φ

(e−jkr

rar

)− ∂r

(r

e−jkr

raφ

))+

φ

r

(∂r

(r

e−jkr

raθ

)− ∂θ

(e−jkr

rar

))=

rr sin θ

e−jkr

r∂θ

(aφ sin θ

)+

θ

r

(1

sin θ

e−jkr

r∂φar − ∂r

(e−jkr

)aφ

)+

φ

r

(∂r

(e−jkr

)aθ −

e−jkr

r∂θar

)≈ jk

(θaφ − φaθ

) e−jkr

r

= −jkr×(θaθ + φaφ

) e−jkr

r

=1c

Eff.

(1.8)

The approximation above drops the 1/r2 terms. Since

1µ0c

=1µ0

√µ0ε0 =

√ε0

µ0=

1η

, (1.9)

the magnetic far field can be expressed in terms of the electric far field as

H =1η

r× E. (1.10)

1.5 Plane wave relations between electric and magnetic fields

I recalled an identity of the form eq. (1.10) in [3], but didn’t think that it required a far field approx-imation. The reason for this was because the Jackson identity assumed a plane wave representationof the field, something that the far field assumptions also locally require.

Assuming a plane wave representation for both fields

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E(x, t) = Eej(ωt−k·x) (1.11a)

B(x, t) = Bej(ωt−k·x) (1.11b)

The cross product relation between the fields follows from the Maxwell-Faraday law of induction

0 = ∇× E +∂B∂t

, (1.12)

or

(1.13)

0 = er × E∂rej(ωt−k·x) + jωBej(ωt−k·x)

= −jerkr × Eej(ωt−k·x) + jωBej(ωt−k·x)

= (−k× E + ωB) jej(ωt−k·x),

or

(1.14)H =

kkcµ0

k× E

=1η

k× E,

which also finds eq. (1.10), but with much less work and less mess.

1.6 Transverse only nature of the far-field fields

Also observe that its possible to tell that the far field fields have only transverse components usingthe same argument that they are locally plane waves at that distance. The plane waves must satisfythe zero divergence Maxwell’s equations

(1.15a)∇ · E = 0

(1.15b)∇ · B = 0,

so by the same logic

(1.16a)k · E = 0

(1.16b)k · B = 0.

In the far field the electric field must equal its transverse projection

(1.17)E = ProjT

(−jωA− j

1ωµ0ε0

∇ (∇ · A))

.

Since by eq. (1.5) the scalar potential term has only a radial component, that leaves

(1.18)E = −jω ProjT A,

which provides eq. (1.7) with slightly less work.

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1.7 Vertical dipole reflection coefficient

In class a ground reflection scenario was covered for a horizontal dipole. Reading the text I was sur-prised to see what looked like the same sort of treatment §4.7.2, but ending up with a quite differentresult. It turns out the difference is because the text was treating the vertical dipole configuration,whereas Prof. Eleftheriades was treating a horizontal dipole configuration, which have different re-flection coefficients. These differing reflection coefficients are due to differences in the polarizationof the field.

To understand these differences in reflection coefficients, consider first the field due to a verticaldipole as sketched in fig. 1.7, with a wave vector directed from the transmission point downwards inthe z-y plane.

Figure 1.7: vertical dipole configuration.

The wave vector has direction

k = zezxθ = z cos θ + y sin θ. (1.19)

Suppose that the (magnetic) vector potential is that of an infinitesimal dipole

A = zµ0 I0l4πr

e−jkr (1.20)

The electric field, in the far field, can be computed by computing the normal projection to the wavevector direction

(1.21)

E = −jω(

A ∧ k)· k

= −jωµ0 I0l4πr

(z ∧ (z cos θ + y sin θ)) (z cos θ + y sin θ)

= −jωµ0 I0l4πr

(zy sin θ) (z cos θ + y sin θ)

= −jωµ0 I0l4πr

sin θ (−y cos θ + z sin θ)

= jωµ0 I0l4πr

sin θyezyθ .

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This is directed in the z-y plane rotated an additional π/2 past k. The magnetic field must then bedirected into the page, along the x axis. This is sketched in fig. 1.8.

Figure 1.8: Electric and magnetic field directions

Referring to [2] (eq.4.40) for the coefficient of reflection component

(1.22)R =nt cos θi − ni cos θt

ni cos θi + nt cos θt

This is the Fresnel equation for the case when that corresponds to E lies in the plane of incidence,and the magnetic field is completely parallel to the plane of reflection). For the no transmission case,allowing vt → 0, the index of refraction is nt = c/vt → ∞, and the reflection coefficient is 1 as claimedin §4.7.2 of [1]. Because of the symmetry of this dipole configuration, the azimuthal angle that thewave vector is directed along does not matter.

1.8 Horizontal dipole reflection coefficient

In the class notes, a horizontal dipole coming out of the page is indicated. With the page representingthe z-y plane, this is a magnetic vector potential directed along the x-axis direction

A = xµ0 I0l4πr

e−jkr. (1.23)

For a wave vector directed in the z-y plane as in eq. (1.19), the electric far field is directed along

(1.24)

(x ∧ k

)· k = x−

(x · k

)k

= x−(((((

((((((

x · (z cos θ + y sin θ))

k

= x.

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The electric far field lies completely in the plane of reflection. From [2] (eq.4.34), the Fresnel reflec-tion coefficients is

(1.25)R =ni cos θi − nt cos θt

ni cos θi + nt cos θt,

which approaches −1 when nt → ∞. This is consistent with the image theorem summation thatProf. Eleftheriades used in class.

Azimuthal angle dependency of the reflection coefficient Now consider a horizontal dipole directedalong the y-axis. For the same wave vector direction as above, the electric far field is now directedalong

(1.26)

(y ∧ k

)· k = y−

(y · k

)k

= y− (y · (z cos θ + y sin θ)) k= y− k sin θ

= y− sin θ (z cos θ + y sin θ)= y cos2 θ − sin θ cos θz= cos θ (y cos θ − sin θz)= cos θyezyθ .

That is

(1.27)E = −jωµ0 I0l4πr

e−jkr cos θyezyθ .

This far field electric field lies in the plane of incidence (a direction of θ rotated by π/2), not in theplane of reflection. The corresponding magnetic field should be directed along the plane of reflection,which is easily confirmed by calculation

(1.28)k× (y cos θ − sin θz) = (z cos θ + y sin θ)× (y cos θ − sin θz)

= −x cos2 θ − x sin2 θ

= −x.

The far field magnetic field is seen to be

(1.29)H = jωI0l

4πre−jkr cos θx,

so a reflection coefficient of 1 is required to calculate the power loss after a single ground reflectionsignal bounce for this relative orientation of antenna to the target.

I fail to see how the horizontal dipole treatment in §4.7.5 can use a single reflection coefficientwithout taking into account the azimuthal dependency of that reflection coefficient.

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Bibliography

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley & Sons, 3rd edition, 2005.1, 1.2, 1.7

[2] E. Hecht. Optics. 1998. 1.7, 1.8

[3] JD Jackson. Classical Electrodynamics. John Wiley and Sons, 2nd edition, 1975. 1.5

[4] Wikipedia. Magnetic potential — Wikipedia, The Free Encyclopedia, 2015. URL http://en.

wikipedia.org/w/index.php?title=Magnetic_potential&oldid=642387563. [Online; accessed5-February-2015]. 1.1

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