example of Electromagnetic Waves, Antennas, and RF: Radiation from Time-Varying Currents and the Hertzian Dipole

Electromagnetic Waves, Antennas, and RF: Radiation from Time-Varying Currents and the Hertzian Dipole - Exercises and Complete Worked Solutions

EM23 derived the fields of the Hertzian dipole and separated the surrounding electromagnetic field into terms proportional to 1∕r³, 1∕r², and 1∕r. This companion article turns those formulas into calculations. The emphasis is on reading physical meaning from the equations: retardation, electrical distance kr, near-to-far-field transition, angular radiation pattern, outward Poynting flux, total radiated Power, radiation resistance, and directivity [1, 2, 3, 4].

The phasor convention is

(1)

so an outward wave carries the radial phase factor e^−ikr.

For a z-directed Hertzian element with current moment I₀ℓ in a lossless medium,

(2)

(3)

and

(4)

In the radiation zone,

(5)

with

(6)

For the ideal Hertzian element,

(7)

(8)

(9)

and in free space

(10)

How to use this problem set

Attempt all problems in Part I before consulting Part II. For numerical problems in vacuum, use

(11)

Keep a clear distinction among field amplitude, power density, radiation intensity, total power, and radiation resistance.

Part I: Exercises

Exercise 1: retarded time and propagation phase

A sinusoidal source operates at

(12)

An observation point is one quarter wavelength from the source in vacuum.

1. Find the wavelength.
2. Find the source-to-observer distance.
3. Find the propagation delay R∕c.
4. Find the phase lag kR in radians and degrees.
5. Explain how the time delay and phasor factor e^−ikR encode the same physics.

Exercise 2: is the element electrically short?

A straight current element has physical length

(13)

Evaluate ℓ∕λ at 1.00 GHz and 10.0 GHz. Discuss at which frequency the ideal Hertzian approximation is more credible and why the condition ℓ ≪ λ matters.

Exercise 3: comparing the 1∕r, 1∕r², and 1∕r³ terms

Inside E_𝜃, compare the magnitudes of the induction and reactive terms with the radiation term. Show that

(14)

and

(15)

Evaluate both ratios for

(16)

Interpret the trend physically.

Exercise 4: exact Hertzian-dipole fields in the transition region

In vacuum let

(17)

At

(18)

calculate the complex phasors E_r, E_𝜃, and H_ϕ. Report magnitudes and phases. Explain why the simple far-field relation E∕H = η₀ should not yet be expected for the total local fields.

Exercise 5: exact field versus far-field approximation

Use the same 300 MHz Hertzian element as Exercise 4, but evaluate the broadside field at

(19)

1. Calculate the exact E_𝜃 and H_ϕ magnitudes from the complete expressions.
2. Calculate the far-field E_𝜃^ff magnitude.
3. Find the percentage amplitude error of the far-field electric-field approximation.
4. Evaluate |E_𝜃∕H_ϕ| and compare it with η₀.

Exercise 6: radiation pattern and half-power beamwidth

For a Hertzian dipole,

(20)

Find the directions of the nulls, the direction of maximum radiation, the two half-power angles in the principal plane, and the half-power beamwidth.

Exercise 7: power density and radiation intensity

For the 300 MHz element of Exercise 4, evaluate the far-field time-average power density and radiation intensity at

(21)

Verify that the radiation intensity is independent of r in the far field.

Exercise 8: derive the total radiated power

Starting from

(22)

with

(23)

derive the total power by integrating over solid angle:

(24)

Show every integration step and recover the EM23 result.

Exercise 9: radiated power and radiation resistance

For the 300 MHz element with I₀ = 1.00 A and ℓ = λ∕50:

1. calculate P_rad;
2. calculate R_rad from P_rad = (1∕2)|I₀|²R_rad;
3. verify the free-space formula 80π²(ℓ∕λ)².

Exercise 10: what fraction of the power lies near broadside?

Find the fraction of the total Hertzian-dipole radiated power that leaves through the angular band

(25)

The band includes all azimuth angles 0 ≤ ϕ < 2π.

Exercise 11: derive directivity from the pattern

Use

(26)

to derive the maximum directivity of a Hertzian dipole. Express the result as a dimensionless ratio and in dBi.

Exercise 12: frequency scaling at fixed physical length

A physically fixed Hertzian current element is driven with the same current amplitude at 100 MHz and 300 MHz. Assume its length remains electrically short at both frequencies.

1. By what factor does R_rad change?
2. By what factor does P_rad change if I₀ is unchanged?
3. Explain the result from R_rad ∝ (ℓ∕λ)².

Exercise 13: distance required for small nonradiative terms

For the E_𝜃 expression, determine the electrical distance kr and physical distance in wavelengths required so that

1. the induction term has magnitude only 10% of the radiation term;
2. the induction term has magnitude only 1% of the radiation term;
3. state the corresponding reactive-term ratios at those same distances.

Explain why the boundary of the “far field” is not a single universal radius for every source and every accuracy requirement.

Exercise 14: Hertzian element versus a physically short dipole

A Hertzian element assumes uniform current along its infinitesimal length. A physically short center-fed dipole of total length ℓ is more realistically approximated by a triangular current distribution that falls from I₀ at the feed to zero at the ends.

If the triangular distribution has effective current moment I₀ℓ∕2, use the Hertzian result to derive the radiation resistance of the physically short dipole. Compare it with the uniform-current Hertzian value for the same total length and feed current.

Exercise 15: spherical spreading and conservation of radiated power

At a fixed far-field angle, show that the Hertzian-dipole power density satisfies

(27)

Compare the power density at r = 10λ and r = 20λ. Then explain why the total power through a sphere remains constant even though the local power density decreases.

Exercise 16: Julia sweep of field zones and radiation pattern

Write a Julia program that:

1. sweeps kr from 0.05 to 30 and evaluates the relative induction and reactive terms 1∕(kr) and 1∕(kr)²;
2. sweeps 𝜃 from 0 to π and evaluates sin ²𝜃;
3. numerically integrates sin ³𝜃 to verify the angular integral 4∕3;
4. computes R_rad versus ℓ∕λ for 10⁻³ ≤ ℓ∕λ ≤ 10⁻¹.

State the numerical checks the program should reproduce.

Part II: Complete Worked Solutions

Solution 1: retarded time and propagation phase

The wavelength is

(28)

One quarter wavelength is

(29)

The propagation delay is

(30)

Thus

(31)

The Wavenumber is

(32)

so

(33)

Therefore

(34)

In time language, the observation point sees the source 2.50 ns late. In phasor language, that same delay multiplies the source amplitude by

(35)

The delay and phase shift are two representations of the same finite propagation time.

Solution 2: is the element electrically short?

At 1.00 GHz,

(36)

Hence

(37)

At 10.0 GHz,

(38)

so

(39)

Thus the element is much more plausibly approximated as electrically short at 1 GHz. At 10 GHz its length is about one third wavelength, so phase and current variation along the Conductor can no longer be ignored.

The requirement ℓ ≪ λ allows the source current to be treated as nearly in phase over the entire element and permits the current distribution to be collapsed into the single current moment I₀ℓ.

Solution 3: comparing the three radial families

The radiation contribution inside the E_𝜃 bracket has magnitude proportional to

(40)

The induction contribution has magnitude 1∕r², so

(41)

The reactive contribution has magnitude 1∕(kr³), giving

(42)

For r = λ∕100,

(43)

so

(44)

For r = λ∕(2π),

(45)

and both ratios equal unity.

For r = λ,

(46)

so

(47)

For r = 10λ,

(48)

so

(49)

The faster-decaying 1∕r² and 1∕r³ contributions dominate close to the source but become negligible compared with the 1∕r radiation term as kr grows.

Solution 4: exact fields in the transition region

At 300 MHz,

(50)

(51)

and

(52)

The observation radius is

(53)

so

(54)

Also,

(55)

Substituting into the complete field expressions gives

(56)

(57)

and

(58)

The ratio of the transverse fields is therefore complex:

(59)

Its magnitude is not η₀. This is expected because kr < 1 and the local field still contains strong induction and reactive contributions. The plane-wave-like ratio E∕H = η₀ applies to the surviving radiation terms in the far zone, not to the complete transition-region field.

Solution 5: exact field versus far-field approximation

At broadside, sin 𝜃 = 1 and E_r = 0. At

(60)

we have

(61)

Using the complete formulas gives

(62)

and

(63)

The far-field expression gives

|E𝜃ff|	=	(64)
	≈ 0.376991 V/m.	(65)

The percentage amplitude error is

(66)

Thus the far-field approximation is already very accurate at 10λ for this ideal element.

The exact transverse ratio is

(67)

which is extremely close to

(68)

Solution 6: radiation pattern and half-power beamwidth

The power pattern is

(69)

Nulls occur where

(70)

so

(71)

The maximum occurs at

(72)

Half power means

(73)

Therefore

(74)

which gives

(75)

Hence the half-power beamwidth in that principal plane is

(76)

Solution 7: power density and radiation intensity

The far-field power density is

(77)

For 𝜃 = 30^∘,

(78)

At r = 10λ the numerical result is

(79)

Radiation intensity is defined by

(80)

Hence

(81)

Because the far-field power density contains 1∕r², multiplying by r² removes the radial dependence:

(82)

Thus radiation intensity records angular power flow without depending on which far-field sphere is used.

Solution 8: total radiated power

Start with

(83)

Since U_max is constant,

(84)

The azimuth integral is

(85)

For the polar integral, write

(86)

Let

(87)

Then

∫ 0π sin 3𝜃 d𝜃	= ∫ −11(1 − u2) du	(88)
	= −11	(89)
	= .	(90)

Therefore

(91)

Substituting

(92)

gives

(93)

Solution 9: radiated power and radiation resistance

For the numerical example,

(94)

Thus

(95)

Radiation resistance is defined by

(96)

For I₀ = 1 A,

(97)

The closed-form formula gives

Rrad	= 80π22	(98)
	= 0.31583 Ω,	(99)

The familiar coefficient 80π² uses the standard approximation η ₀ ≈ 120π Ω. Using η₀ = 376.7303 Ω directly gives 0.31561 Ω; using 120π Ω gives 0.31583 Ω. The distinction is only about 0.07% and does not change the antenna physics.

Solution 10: fraction of power near broadside

Because all azimuth angles are included, the required fraction is

(100)

The denominator is 4∕3.

An antiderivative is

(101)

Therefore

∫ π∕32π∕3 sin 3𝜃 d𝜃

= .

(102)

Hence

(103)

Thus

(104)

of the Hertzian-dipole power lies between 60^∘ and 120^∘.

Solution 11: directivity

The maximum radiation intensity occurs at 𝜃 = 90^∘:

(105)

Directivity is

(106)

Substituting the expressions for U_max and P_rad,

Dmax	= 4π	(107)
	= .	(108)

Therefore

(109)

In dBi,

(110)

Solution 12: frequency scaling at fixed physical length

For fixed physical length ℓ,

(111)

Since

(112)

we have

(113)

Increasing frequency from 100 to 300 MHz multiplies frequency by three, so

(114)

For unchanged current amplitude,

(115)

so

(116)

The physical element becomes electrically longer at higher frequency, making it a more effective radiator as long as the Hertzian approximation remains valid.

Solution 13: distance required for small nonradiative terms

The induction-to-radiation ratio is

(117)

For a 10% induction term,

(118)

so

(119)

Since

(120)

we obtain

(121)

At the same distance, the reactive-to-radiation ratio is

(122)

or 1%.

For a 1% induction term,

(123)

so

(124)

and

(125)

At that distance the reactive ratio is

(126)

These estimates concern the ideal Hertzian field terms. A real finite antenna also has an aperture-size far-field criterion such as r ≳ 2D²∕λ, and the acceptable error depends on the application. Therefore there is no single universal far-field radius that applies to every radiator and accuracy requirement.

Solution 14: Hertzian element versus a physically short dipole

For the Hertzian element,

(127)

A triangular current distribution with peak feed current I₀ has effective current moment

(128)

Since the far field is proportional to current moment, the radiated power and radiation resistance are proportional to the square of effective length. Therefore

Rrad,short	= 80π22	(129)
	= 20π22.	(130)

Thus

(131)

For the same total physical length and feed current,

(132)

This is why one must distinguish an ideal uniform-current Hertzian element from a physically short center-fed dipole.

Solution 15: spherical spreading and conservation of radiated power

In the far field,

(133)

Therefore the time-average Poynting flux satisfies

(134)

Doubling radius from 10λ to 20λ therefore gives

(135)

But a sphere has area

(136)

When radius doubles, area grows by a factor of four. Thus the quarter-sized local power density is spread across four times the area, leaving

(137)

unchanged. The 1∕r² law is therefore precisely what conservation of radiated power requires for a freely expanding spherical wave.

Solution 16: Julia sweep of field zones and radiation pattern

One implementation is supplied with the bundle as EM23E1_hertzian_sweep.jl. Its essential calculations are

using Printf

kr = 10 .^ range(log10(0.05), log10(30.0), length=300)
induction_ratio = 1.0 ./ kr
reactive_ratio  = 1.0 ./ kr.^2

theta = range(0.0, pi, length=2001)
pattern = sin.(theta).^2

# trapezoidal integral of sin^3(theta)
y = sin.(theta).^3
integral = sum((y[1:end-1] .+ y[2:end]) .* diff(theta) ./ 2)

ell_over_lambda = 10 .^ range(-3.0, -1.0, length=200)
Rrad = 80*pi^2 .* ell_over_lambda.^2

@printf("integral sin^3(theta) dtheta = %.8f\n", integral)
@printf("expected                         = %.8f\n", 4/3)
@printf("Rrad at ell/lambda=1/50         = %.6f ohm\n",
        80*pi^2*(1/50)^2)

The main checks are

(138)

and

(139)

The numerical sweep should also show the slopes implied by 1∕(kr), 1∕(kr)², and the quadratic scaling R_rad ∝ (ℓ∕λ)².

What EM23E1 adds to the series

EM23 established the radiation formulas. EM23E1 makes them operational. The central calculation chain is

(140)

The exercises also make the field-zone hierarchy quantitative through the single dimensionless distance kr.

The next antenna article can now treat radiation intensity as the starting point for antenna directivity, efficiency, gain, EIRP, and receive effective aperture.

References

References