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
and
For Love exterior equivalence, where the replacement field is zero on the interior side,
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.
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
at
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
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
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:
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
from the +z axis. Its electric field is TE-polarized,
with
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
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
along the x axis. In the x-z plane, the far-field phase difference is
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.
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
an observation distance
and observation angle
use
and
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
Its normalized continuous aperture factor is
Approximate the aperture by N equally weighted midpoint Huygens sources,
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.
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
Find the first null angle measured from broadside.
Exercise 12: half-power beamwidth of the 4λ aperture
For a uniform aperture with
solve numerically for the positive angle 𝜃hp at which
Find the full half-power beamwidth
Exercise 13: steering a continuous aperture with a phase ramp
A uniform line aperture of width D = 4λ is given the excitation phase
with
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.
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
With n = z directed into the radiating half-space, use the common aperture-equivalence
convention
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
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
Hence
The magnitude of the Green function is 1∕(4πR), so
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
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
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
Solution 3: Love currents for a normally incident plane wave
For a +z free-space plane wave,
Numerically,
Then
| Js | = z × H = −5.31 × 10−3x A/m, | (54)
|
| Ms | = −z × E = −2.00y V/m. | (55) |
Hence
Solution 4: reversing the chosen surface normal
With
the same algebraic Love formulas give
and
Therefore
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,
in the interior, while
in the exterior. The jump conditions give
| Js | = n × (0 − H) = −n × H, | (63)
|
| Ms | = −n × (0 − E) = +n × E. | (64) |
Thus
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,
Using
and E = E0y,
| k ×y | = sin 𝜃z − cos 𝜃x. | (68) |
Hence
The Love electric surface current is
| Js | = z × H | (70)
|
| = − y 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
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 | = z × (2x + iy) | (76)
|
| = (−ix + 2y) A/m. | (77) |
Then
| Js | = z × H | (78)
|
| = − (2x + iy) A/m, | (79) |
and
| Ms | = −z × E | (80)
|
| = ix − 2y V/m. | (81) |
The electric-field magnitude is
Therefore
so
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,
Two equal phasors separated symmetrically in phase have total magnitude
At broadside,
At 30∘,
At endfire,
Thus
Solution 9: quantify a far-field phase approximation
Normalize lengths by λ. With r = 100λ, x = 2λ, and sin 30∘ = 0.5,
| Rexact | = λ | (91)
|
| = 99.01515λ, | (92) |
whereas
The path error is
The phase error magnitude is
| |Δϕ| | = kΔR | (95)
|
| = 2π(0.0151504) | (96)
|
| = 0.09519 rad | (97)
|
| = 5.45∘. | (98) |
For amplitude,
Therefore
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,
Therefore
Using midpoint samples gives
|
|
|
| N | |AN| | magnitude error |
|
|
|
| 4 | 0.394550 | 5.14% |
| 8 | 0.379963 | 1.25% |
| 16 | 0.376435 | 0.311% |
| 32 | 0.375560 | 0.0775% |
|
|
|
Thus the discrete Huygens sum converges rapidly toward the continuous aperture integral:
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,
With D = 6λ,
Therefore
Increasing aperture width narrows the main lobe.
Solution 12: half-power beamwidth of the 4λ aperture
The half-power condition is
The positive root inside the main lobe is
The pattern is symmetric about broadside, so
| HPBW | = 2𝜃hp | (109)
|
| ≈ 12.7156∘. | (110) |
Thus
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:
All aperture points are in phase when
For the main beam near broadside,
The phase slope is
Expressing x in wavelengths,
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
with n = z and Ea = 4x V/m,
| Ms | = −2z × (4x) | (126)
|
| = −8y V/m. | (127) |
Therefore
Magnetic surface-current density has units V/m in this equivalence convention.
Solution 15: recover tangential fields from Love currents
Start with
Cross both sides with n:
Because Ht is tangential,
and the vector triple-product identity gives
Hence
Similarly,
Crossing with n gives
Therefore
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
For N = 16 midpoint Huygens samples,
The magnitude difference is
corresponding to
relative magnitude error.
Now multiply every equivalent source by the same complex constant Ceiϕ0. The whole discrete sum
is multiplied by that same factor:
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:
- state the surface normal and region ordering before evaluating cross products;
- add complex field contributions coherently before taking magnitudes or powers;
- treat far-field phase accuracy as more restrictive than amplitude accuracy when large
apertures are involved;
- 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.