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[parent] Electromagnetic Waves, Antennas, and RF: Huygens Principle and Electromagnetic Field Equivalence - Exercises

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Electromagnetic Waves, Antennas, and RF: Huygens Principle and Electromagnetic Field Equivalence - Exercises and Complete Worked Solutions

This companion to EM29 develops calculation skill with Huygens surfaces, electromagnetic field equivalence, equivalent electric and magnetic surface currents, far-field phase, and finite-aperture superposition. All exercises are stated first. Complete worked solutions follow in Part II. The sign conventions are the same as EM29: the phasor convention is eiωt, outgoing waves contain e−ikr, and the unit Normal n points from region 1 to region 2 [1, 2, 3, 4].

The principal surface jump relations are

^n ×  (H2 − H1 ) = Js,
(1)

and

^n ×  (E2 −  E1) = − Ms.
(2)

For Love exterior equivalence, where the replacement field is zero on the interior side,

Js = ^n × H,      Ms  = − ^n × E.
(3)

Three numerical ideas are used repeatedly in the problems. A far-field phase error is the difference between the exact propagation phase −kR and the phase obtained after replacing R by its far-field approximation. A discrete Huygens-source approximation replaces a continuous surface or line integral by a finite weighted sum. The resulting difference from the continuous-aperture result is the aperture sampling error.

PIC

Figure 1. Region ordering, surface normal, and tangential fields determine the signs of the equivalent electric and magnetic surface currents.

Part I: Exercises

Exercise 1: outgoing Green-function phase and amplitude

At f = 1.00 GHz in free space, compare the outgoing scalar Green function

        e−ikR-
G (R ) =  4πR
(4)

at

R  = 2.00 m,     R  =  2.25 m.
 1                 2
(5)

Find the magnitude ratio |G(R2)∕G(R1)| and the phase change from R1 to R2.

Exercise 2: surface currents from general field jumps

A planar interface has

^n = ^z.
(6)

The tangential fields on the two sides are

H1 = (0.20x + 0.10y) A/m, (7)
H2 = (−0.10x + 0.40y) A/m, (8)
E1 = (10x + 5y) V/m, (9)
E2 = (4x + 11y) V/m. (10)

Find Js and Ms.

Exercise 3: Love currents for a normally incident plane wave

A free-space plane wave propagates in the +z direction with

E  = 2.00^x e−ikz V/m.
(11)

Take the Huygens surface at z = 0 with n = z. Find the Love-equivalent Js and Ms that reproduce the exterior wave and give zero interior field. Use η0 = 376.73 Ω.

Exercise 4: reversing the chosen surface normal

Repeat Exercise 3 with the geometric surface normal reversed:

^n ′ = − ^z.
(12)

Using the corresponding Love formulas with n′, determine how the signed equivalent currents change.

Exercise 5: complementary interior-field equivalence

Let region 1 be the interior of a closed surface and region 2 the exterior, with n pointing outward. Construct a replacement problem that preserves the original fields E,H in region 1 but forces the replacement fields to zero in region 2. Derive the required Js and Ms.

Exercise 6: oblique TE plane wave on a Huygens plane

A free-space plane wave propagates in the x-z plane at

𝜃 = 30∘
(13)

from the +z axis. Its Electric Field is TE-polarized,

E =  3.00 ^ye− ik⋅rV/m,
(14)

with

^k = sin𝜃 ^x + cos𝜃 ^z.
(15)

For the plane z = 0 with n = z, find H, Js, and Ms.

Exercise 7: complex polarization and equivalent currents

At z = 0, a normally propagating +z free-space wave has phasor electric field

E =  (2 ^x + i^y )V/m.
(16)

Find H, Js, and Ms for n = z. Verify the magnitude relation between Ms and Js.

Exercise 8: two discrete Huygens sources

Two equal in-phase elementary Huygens sources are separated by

    λ
d = --
    2
(17)

along the x axis. In the x-z plane, the far-field phase difference is

Δ ϕ = kd sin 𝜃.
(18)

Ignoring the common element pattern, find the magnitude of the coherent sum at 𝜃 = 0∘, 30∘, and 90∘, normalized so that one source alone has amplitude 1.

PIC

Figure 2. Two separated Huygens sources acquire an observation-angle-dependent path difference. Their complex fields must be added before magnitudes or powers are formed.

Exercise 9: quantify a far-field phase approximation

For an aperture point at

x = 2λ,
(19)

an observation distance

r = 100λ,
(20)

and observation angle

𝜃 = 30∘,
(21)

use

         √ ------------------
Rexact =   r2 + x2 − 2rxsin 𝜃
(22)

and

R ff = r − x sin 𝜃.
(23)

Find the path-length error, the associated phase error in degrees, and the ratio between the exact 1∕R amplitude factor and the common far-field factor 1∕r.

Exercise 10: discrete Huygens approximation to a continuous aperture

A uniform line aperture has width

D =  4λ.
(24)

Its normalized continuous aperture factor is

A (𝜃) = sin-u,     u = π D-sin𝜃.
         u              λ
(25)

Approximate the aperture by N equally weighted midpoint Huygens sources,

         1  N∑
AN (𝜃) = ---   eikxnsin𝜃,
         N  n=1
(26)

where the xn are the subinterval midpoints. At 𝜃 = 10∘, evaluate |A| and |A N| for N = 4, 8, 16, 32. Report the percentage magnitude error.

PIC

Figure 3. Midpoint Huygens-source sampling converges toward the continuous uniform-aperture pattern as the number of source samples increases.

Exercise 11: first null of a uniform aperture

A uniformly illuminated one-dimensional aperture has width

D =  6λ.
(27)

Find the first null angle measured from broadside.

Exercise 12: half-power beamwidth of the 4λ aperture

For a uniform aperture with

D =  4λ,
(28)

solve numerically for the positive angle 𝜃hp at which

||sin u ||2   1
|-----| = --,    u =  4πsin 𝜃.
| u   |   2
(29)

Find the full half-power beamwidth

HPBW    = 2𝜃hp.
(30)

Exercise 13: steering a continuous aperture with a phase ramp

A uniform line aperture of width D = 4λ is given the excitation phase

ϕ(x ) = − kx sin𝜃0
(31)

with

𝜃0 = 25∘.
(32)

Show where the main beam points. Find the phase slope dϕ∕dx in rad/λ, the total phase change from x = −D∕2 to x = +D∕2, and the phase step between samples spaced by d = λ∕2.

PIC

Figure 4. A linear phase ramp across an aperture cancels the observation phase in one selected direction and therefore steers the main beam.

Exercise 14: conducting-screen aperture equivalence

An aperture in a large conducting screen has tangential electric field

Ea  = 4.00^x V/m.
(33)

With n = z directed into the radiating half-space, use the common aperture-equivalence convention

M   =  − 2n^× E .
   s           a
(34)

Find Ms and state its units.

Exercise 15: recover tangential fields from Love currents

For Love exterior equivalence with outward normal n, suppose the equivalent currents are known. Starting from

Js = ^n ×  Ht,     Ms  = − ^n × Et,
(35)

derive formulas for the tangential fields Et and Ht in terms of Js and Ms.

Exercise 16: synthesis check from surface fields to a sampled aperture pattern

A 4λ uniform aperture is represented by N = 16 equally spaced midpoint Huygens samples. At 𝜃 = 10∘, compare the sampled magnitude |A 16| with the exact continuous magnitude |A|. Then state what would happen to the sampled pattern if every equivalent current sample were multiplied by the same complex constant Ceiϕ0.

Part II: Complete Worked Solutions

Solution 1: outgoing Green-function phase and amplitude

The free-space wavelength is

    c-   2.99792458-×--108
λ = f =     1.00 × 109     = 0.299792458 m.
(36)

Hence

     2π
k =  λ--≈ 20.9585 rad/m.
(37)

The magnitude of the Green function is 1∕(4πR), so

|      |
||G-(R2)||   R1-   2.00
|G (R1)| = R2  = 2.25 =  0.8889.
(38)

The unwrapped phase change is

Δϕ = −k(R2 − R1) (39)
= −(20.9585)(0.25) (40)
= −5.2396 rad (41)
= −300.21∘. (42)

Modulo 360∘, this is equivalent to

|--------------------------|
|Δϕ =  +59.79 ∘  mod  360∘.|
----------------------------
(43)

Thus increasing range reduces amplitude by the geometric 1∕R factor and advances the negative outgoing-wave phase by 300.21∘.

Solution 2: surface currents from general field jumps

First form the field jumps:

H2 − H1 = (−0.30x + 0.30y) A/m, (44)
E2 − E1 = (−6x + 6y) V/m. (45)

Using

^z × ^x = ^y,     ^z × ^y = − ^x,
(46)

we obtain

Js = z × (−0.30x + 0.30y) (47)
= −0.30x − 0.30y A/m, (48)

and

Ms = −z × (−6x + 6y) (49)
= 6x + 6y V/m. (50)

Therefore

|----------------------------------------------------|
|Js = (− 0.30^x − 0.30 ^y)A/m,    Ms  = (6^x +  6^y)V/m.  |
------------------------------------------------------
(51)

Solution 3: Love currents for a normally incident plane wave

For a +z free-space plane wave,

H =  1-^z × E =  -2.00-y^A/m.
     η0         376.73
(52)

Numerically,

H  = 5.31 × 10−3^y A/m.
(53)

Then

Js = z × H = −5.31 × 10−3x A/m, (54)
Ms = −z × E = −2.00y V/m. (55)

Hence

|--------------------------------------------|
Js =  − 5.31 mA/m  ^x,    Ms  =  − 2.00 V/m ^y.|
----------------------------------------------
(56)

Solution 4: reversing the chosen surface normal

With

^n ′ = − ^z,
(57)

the same algebraic Love formulas give

J ′s = ^n′ × H = − ^z × H  = − Js,
(58)

and

M ′s = − ^n′ × E = + ^z × E =  − Ms.
(59)

Therefore

|-′-------------------------′----------------|
J-s =-+5.31-mA/m---^x,----M--s =-+2.00-V/m--^y.-
(60)

Equivalent-current signs are orientation dependent. A surface normal must always be stated together with the current convention.

Solution 5: complementary interior-field equivalence

For the desired replacement problem,

E1 =  E,     H1 =  H,
(61)

in the interior, while

E2 =  0,    H2  = 0
(62)

in the exterior. The jump conditions give

Js = n × (0 − H) = −n × H, (63)
Ms = −n × (0 − E) = +n × E. (64)

Thus

|----------------------------------|
|Js = − ^n × H,      Ms  = + ^n × E. |
-----------------------------------
(65)

The signs are the reverse of the usual Love exterior-equivalence currents because the prescribed field jump has been reversed.

Solution 6: oblique TE plane wave on a Huygens plane

For a plane wave,

     1
H =  --^k × E.
     η0
(66)

Using

^
k =  sin 𝜃^x + cos 𝜃^z
(67)

and E = E0y,

k ×y = sin 𝜃z − cos 𝜃x. (68)

Hence

|----------------------------------------|
|     -3.00--        ∘          ∘        |
|H =  376.73 (− cos30 ^x + sin30  ^z) A/m. |
------------------------------------------
(69)

The Love electric surface current is

Js = z × H (70)
= −          ∘
3.00-cos30--
  376.73y A/m (71)
= −6.90 × 10−3y A/m. (72)

The equivalent magnetic surface current is

Ms = −z × (3.00y) (73)
= +3.00x V/m. (74)

Therefore

|-------------------------------------------|
Js = − 6.90 mA/m  y^,     Ms  =  3.00 V/m  ^x. |
---------------------------------------------
(75)

Only the tangential part of H contributes to Js on the z = 0 surface.

Solution 7: complex polarization and equivalent currents

For propagation along +z,

H = 1--
η0z × (2x + iy) (76)
= 1
---
η0(−ix + 2y) A/m. (77)

Then

Js = z × H (78)
= −1
---
η0(2x + iy) A/m, (79)

and

Ms = −z × E (80)
= ix − 2y V/m. (81)

The electric-field magnitude is

      √ -2----2   √ --
|E | =  2  + 1  =   5 V/m.
(82)

Therefore

                            √ --
        √ --                --5-
|Ms | =   5V/m,      |Js| =  η  A/m,
                              0
(83)

so

|--------------|
-|Ms--| =-η0|Js|.
(84)

The result holds for normal-incidence free-space plane waves even when the polarization state is complex.

Solution 8: two discrete Huygens sources

With d = λ∕2,

Δ ϕ = kd sin𝜃 = π sin𝜃.
(85)

Two equal phasors separated symmetrically in phase have total magnitude

        ||      ||
|A| = 2 |cos Δ-ϕ|.
        |    2 |
(86)

At broadside,

𝜃 = 0∘ :    |A| = 2.
(87)

At 30∘,

                               --
Δ ϕ =  π,     |A | = 2cos π-= √ 2 ≈ 1.414.
       2                 4
(88)

At endfire,

𝜃 =  90∘ :    Δ ϕ = π,     |A| = 0.
(89)

Thus

|----------------------------------------------|
||A (0∘)| = 2,  |A (30∘)| = 1.414,   |A(90∘)| = 0.|
------------------------------------------------
(90)

Solution 9: quantify a far-field phase approximation

Normalize lengths by λ. With r = 100λ, x = 2λ, and sin 30∘ = 0.5,

Rexact = λ∘ --------------------------
  1002 + 22 − 2(100)(2)(0.5) (91)
= 99.01515λ, (92)

whereas

R  =  100λ − 2λ(0.5) = 99.00000 λ.
 ff
(93)

The path error is

ΔR  = 0.0151504 λ.
(94)

The phase error magnitude is

|Δϕ| = kΔR (95)
= 2π(0.0151504) (96)
= 0.09519 rad (97)
= 5.45∘. (98)

For amplitude,

1∕Rexact=  --r---=  1.00995.
  1∕r      Rexact
(99)

Therefore

|--------------------------------------------------------|
|ΔR  = 0.01515 λ,     |Δ ϕ| = 5.45∘,     --r---= 1.00995. |
|                                       Rexact            |
---------------------------------------------------------
(100)

The approximation is already good in amplitude, while the phase error is the more sensitive quantity.

Solution 10: discrete Huygens approximation to a continuous aperture

At 𝜃 = 10∘ and D = 4λ,

u =  4π sin 10∘ ≈ 2.18213.
(101)

Therefore

      ||sinu ||
|A | = ||----|| ≈ 0.375269.
         u
(102)

Using midpoint samples gives




N |AN| magnitude error



4 0.394550 5.14%
8 0.379963 1.25%
160.376435 0.311%
320.375560 0.0775%



Thus the discrete Huygens sum converges rapidly toward the continuous aperture integral:

|----------------|
||A| ≈ 0.375269. |
-----------------
(103)

The numerical trend also shows why a sampled array and a continuous aperture become closely related when the spatial sampling is sufficiently fine.

Solution 11: first null of a uniform aperture

For a uniform line aperture,

           λ
sin𝜃null =--.
          D
(104)

With D = 6λ,

            (   )
𝜃null = sin−1 1-  = 9.594∘.
              6
(105)

Therefore

|------------|
𝜃null ≈-9.59∘.
(106)

Increasing aperture width narrows the main lobe.

Solution 12: half-power beamwidth of the 4λ aperture

The half-power condition is

(      )
  sin u  2   1
  -u---   = 2-,    u = 4 πsin𝜃.
(107)

The positive root inside the main lobe is

𝜃hp ≈ 6.3578 ∘.
(108)

The pattern is symmetric about broadside, so

HPBW = 2𝜃hp (109)
≈ 12.7156∘. (110)

Thus

|-----------------|
HPBW----≈-12.72∘.--
(111)

Solution 13: steering a continuous aperture with a phase ramp

The aperture integrand contains the product of the imposed excitation phase and the observation phase:

exp [− ikx sin 𝜃0]exp[ikx sin 𝜃] = exp[ikx(sin𝜃 − sin𝜃0)].
(112)

All aperture points are in phase when

sin𝜃 = sin 𝜃0.
(113)

For the main beam near broadside,

|-------------|
𝜃 = 𝜃0 = 25 ∘. |
---------------
(114)

The phase slope is

dϕ-= − k sin 25∘.
dx
(115)

Expressing x in wavelengths,

|--------------------------------------|
|--dϕ--- = − 2π sin 25∘ = − 2.655 rad∕λ. |
-d(x∕-λ)-------------------------------|
(116)

Across D = 4λ,

ΔϕD = −kD sin 25∘ (117)
= −8π sin 25∘ (118)
= −10.622 rad (119)
= −608.6∘. (120)

For d = λ∕2 sampling,

Δϕd = −kd sin 25∘ (121)
= −π sin 25∘ (122)
= −1.328 rad (123)
= −76.1∘. (124)

The phase ramp steers because it precompensates the free-space propagation phase in the selected direction.

Solution 14: conducting-screen aperture equivalence

Using

Ms  = − 2^n × Ea
(125)

with n = z and Ea = 4x V/m,

Ms = −2z × (4x) (126)
= −8y V/m. (127)

Therefore

|--------------------|
|Ms  = − 8.00^y V/m.  |
---------------------
(128)

Magnetic surface-current density has units V/m in this equivalence convention.

Solution 15: recover tangential fields from Love currents

Start with

Js = n^×  Ht.
(129)

Cross both sides with n:

^n × Js =  ^n × (^n × Ht ).
(130)

Because Ht is tangential,

^n ⋅ H  = 0,
     t
(131)

and the vector triple-product identity gives

^n × (^n × H  ) = − H .
           t        t
(132)

Hence

|--------------|
Ht  = − ^n × Js.|
----------------
(133)

Similarly,

Ms  = − ^n × Et.
(134)

Crossing with n gives

^n × Ms  =  −n^×  (^n × Et) = Et.
(135)

Therefore

|---------------------------------|
Et-=--^n ×-Ms,------Ht-=-−-^n-×-Js.--
(136)

This inversion is useful when a numerical or measured equivalent-current distribution is used as the surface data.

Solution 16: synthesis check from surface fields to a sampled aperture pattern

From Solution 10, the continuous 4λ aperture at 10∘ has

|A| = 0.375269.
(137)

For N = 16 midpoint Huygens samples,

|A16| = 0.376435.
(138)

The magnitude difference is

|A16| − |A | = 0.001166,
(139)

corresponding to

|--------|
-0.311%--|
(140)

relative magnitude error.

Now multiply every equivalent source by the same complex constant Ceiϕ0. The whole discrete sum is multiplied by that same factor:

A ′(𝜃) = Cei ϕ0A16 (𝜃).
  16
(141)

Therefore the normalized angular pattern is unchanged. The absolute field magnitude is scaled by |C|, the common phase is shifted by ϕ0, and the power scale changes by |C|2. Beam direction, null angles, and normalized sidelobe structure remain unchanged.

Summary

The exercises demonstrate four central habits for equivalence-principle calculations:

  1. state the surface normal and region ordering before evaluating cross products;
  2. add complex field contributions coherently before taking magnitudes or powers;
  3. treat far-field phase accuracy as more restrictive than amplitude accuracy when large apertures are involved;
  4. recognize a sampled Huygens surface and a discrete antenna array as closely related spatial-superposition models.

EM30 can now use these skills to derive radiation from continuous apertures without treating the surface integral as a purely formal expression.

References

References

[1]   A. E. H. Love, “The integration of the equations of propagation of electric waves,” Philosophical Transactions of the Royal Society of London A, vol. 197, pp. 1–45, 1901.

[2]   S. A. Schelkunoff, “Some equivalence theorems of electromagnetics and their application to radiation problems,” Bell System Technical Journal, vol. 15, no. 1, pp. 92–112, 1936.

[3]   R. F. Harrington, Time-Harmonic Electromagnetic Fields, IEEE Press, 2001 reissue of the 1961 text.

[4]   C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed., Wiley, 2016.

[5]   W. L. Stutzman and G. A. Thiele, Antenna Theory and Design, 3rd ed., Wiley, 2012.


"Electromagnetic Waves, Antennas, and RF: Huygens Principle and Electromagnetic Field Equivalence - Exercises" is owned by bloftin.
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Physics Classification: 41.20.Jb (Electromagnetic wave propagation; radiowave propagation )
 84.40.Ba (Antennas: theory, components and accessories )
 03.50.De (Classical electromagnetism, Maxwell equations )
 41.20.-q (Applied classical electromagnetism)

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