Electromagnetic Waves, Antennas, and RF: Antenna Directivity, Radiation Efficiency, Gain, EIRP, and Effective Aperture - Exercises and Complete Worked Solutions

Electromagnetic Waves, Antennas, and RF: Antenna Directivity, Radiation Efficiency, Gain, EIRP, and Effective Aperture - Exercises and Complete Worked Solutions

This companion to EM24 turns the antenna definitions into calculation tools. The exercises begin with radiation intensity and directivity, then add efficiency and gain, move into EIRP and far-field power density, and finally develop effective aperture and receive power. The last problems reconnect the transmit and receive viewpoints in preparation for the Friis transmission equation of EM25 [1, 2, 3, 4].

The central relations are

(1)

(2)

(3)

and

(4)

Unless a problem states otherwise, gains are power ratios, propagation is in free space, polarization is matched, the receiver is conjugately matched, and quoted electric-field phasor magnitudes are peak amplitudes.

Part I: Exercises

Exercise 1: radiation intensity and directivity of a Hertzian-type pattern

An antenna has radiation intensity

(5)

independent of ϕ. Find (a) the total radiated power, (b) the maximum directivity, and (c) the maximum directivity in dBi.

Exercise 2: beam solid angle

An antenna has beam solid angle

(6)

Find its maximum directivity and maximum directivity in dBi.

Exercise 3: radiation efficiency and gain

An antenna accepts 20.0 W at its terminals and radiates 14.0 W. In the direction of peak radiation its directivity is D_max = 8.00. Find (a) radiation efficiency, (b) dissipative loss power, (c) peak gain, and (d) peak gain in dBi.

Exercise 4: realized gain and antenna mismatch

The antenna of Exercise 3 has an input reflection coefficient with magnitude

(7)

Using

(8)

find the realized gain in linear form and dBi. Also find the mismatch loss in dB.

Exercise 5: EIRP, power density, and electric-field strength

A transmitter supplies 10.0 W of accepted power to an antenna having gain G = 12.0 in the observation direction. At a range of 2.00 km, find (a) EIRP, (b) far-field power density, and (c) RMS electric-field magnitude in free space.

Exercise 6: dB link-budget bookkeeping at the transmitting antenna

A transmitter output is 37.0 dBm. A feed cable introduces 2.0 dB loss before an antenna whose gain is 8.0 dBi. Find the EIRP in dBm and watts. State clearly which power reaches the antenna terminals.

Exercise 7: effective aperture at GPS L1

For GPS L1,

(9)

Find the maximum effective aperture of (a) an ideal 0 dBi antenna and (b) an antenna with 5.00 dBi gain.

Exercise 8: received power from incident power density

At 1.20 GHz a receiving antenna has gain 6.00 dBi in the source direction. A matched-polarization plane wave arrives with power density

(10)

Find the antenna effective aperture and the maximum available receive power in watts and dBm.

Exercise 9: derive the Hertzian-dipole effective aperture

For a Hertzian dipole oriented with the incident electric field, use

(11)

(12)

(13)

and

(14)

to derive A_e = P_av∕S_inc. Show that the result is identical to Gλ²∕(4π) for an ideal Hertzian dipole.

Exercise 10: polarization mismatch

Two linearly polarized antennas have polarization axes separated by 35.0^∘. Find the polarization loss factor and polarization loss in dB. If the received power would be −90.0 dBm under matched polarization, what power is available after polarization mismatch alone?

Exercise 11: physical aperture, aperture efficiency, and dish gain

A circular aperture antenna has diameter

(15)

aperture efficiency η_ap = 0.650, and operates at 10.0 GHz. Find its physical area, effective aperture, gain, and gain in dBi.

Exercise 12: infer gain from measured effective aperture

A receiving antenna intercepts

(16)

from an incident power density

(17)

at 2.00 GHz. Find A_e, then infer antenna gain in linear units and dBi.

Exercise 13: numerical check of the Hertzian-dipole aperture derivation

At 300 MHz, a Hertzian dipole has length ℓ = λ∕50. A matched-polarization plane wave has peak electric-field phasor magnitude |E| = 1.00 V/m. Calculate (a) R_rad, (b) open-circuit voltage, (c) maximum available receive power, (d) incident time-average power density, and (e) effective aperture. Verify the result against

(18)

Exercise 14: frequency scaling of effective aperture

An antenna has constant gain G = 10.0 at both 1.00 GHz and 2.00 GHz. Find the effective aperture at each frequency and their ratio. Explain why a fixed gain does not imply a fixed effective area.

Exercise 15: transmit-to-receive bridge

A transmitter has accepted antenna power P_t = 2.00 W and directional gain G_t = 10.0. A receiver 10.0 km away has gain G_r = 4.00 toward the transmitter. The frequency is 1.00 GHz. Compute the received power in two stages:

1. find the incident power density from P_tG_t∕(4πr²);
2. find A_e,r = G_rλ²∕(4π) and then P _r = SA_e,r.

Finally combine the expressions algebraically and identify the Friis form that EM25 will derive in detail.

Exercise 16: Julia sweep of gain, effective aperture, and EIRP

Write a Julia program that

1. sweeps frequency from 0.5 to 5 GHz for fixed gain G = 6 dBi and evaluates A_e;
2. verifies the A_e ∝ 1∕f² scaling using endpoint ratios;
3. computes EIRP for accepted power P_acc = 5 W and gains from 0 to 15 dBi;
4. calculates far-field power density at 1 km for each gain value.

State the expected scaling before writing the code.

Part II: Complete Worked Solutions

Solution 1: radiation intensity and directivity

The radiated power is

Prad	= ∫ 02π ∫ 0π1.50 sin 2𝜃 sin 𝜃 d𝜃 dϕ	(19)
	= (1.50)(2π) ∫ 0π sin 3𝜃 d𝜃.	(20)

Using

(21)

we obtain

(22)

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

(23)

Therefore

Dmax	=	(24)
	= = 1.50.	(25)

Thus

(26)

In dBi,

(27)

This reproduces the ideal Hertzian-dipole directivity derived in EM23 and EM24.

Solution 2: beam solid angle

The beam-solid-angle identity is

(28)

Hence

Dmax	=	(29)
	≈ 10.472.	(30)

Therefore

(31)

In logarithmic form,

(32)

A small beam solid angle corresponds to high directivity because the radiated power is concentrated into a smaller set of directions.

Solution 3: radiation efficiency and gain

Radiation efficiency is

(33)

Thus

(34)

The dissipative loss is

(35)

Gain combines efficiency and directivity:

Gmax	= ηradDmax	(36)
	= (0.700)(8.00) = 5.60.	(37)

Therefore

(38)

and

(39)

Directivity describes how the radiated power is shaped; gain additionally remembers that 30% of accepted power was dissipated rather than radiated.

Solution 4: realized gain and mismatch

The mismatch efficiency is

(40)

Hence

Grealized	= ηmG	(41)
	= (0.910)(5.60)	(42)
	= 5.096.	(43)

Thus

(44)

with

(45)

The mismatch factor itself corresponds to

(46)

Thus the mismatch loss magnitude is

(47)

Solution 5: EIRP, power density, and field strength

The EIRP is

(48)

Thus

(49)

At r = 2000 m,

S	=	(50)
	=	(51)
	≈ 2.387 × 10−6 W/m2.	(52)

Therefore

(53)

For a free-space plane wave,

(54)

so

Erms	=	(55)
	≈ 2.999 × 10−2 V/m.	(56)

Hence

(57)

Using η₀ ≈ 120π Ω gives the familiar equivalent form

(58)

Solution 6: dB transmitter bookkeeping

The cable reduces the 37.0 dBm transmitter output to

(59)

Thus 35.0 dBm, not 37.0 dBm, reaches the antenna terminals.

Adding antenna gain,

EIRPdBm	= 35.0 + 8.0	(60)
	= 43.0 dBm.	(61)

Therefore

(62)

Converting to watts,

(63)

Hence

(64)

The result is an isotropic-equivalent directional power, not the actual total power radiated in all directions.

Solution 7: effective aperture at GPS L1

The wavelength is

λ	=	(65)
	=	(66)
	≈ 0.190294 m.	(67)

For 0 dBi, G = 1, so

Ae	=	(68)
	≈ 2.882 × 10−3 m2.	(69)

Thus

(70)

For 5.00 dBi,

(71)

Therefore

Ae	=	(72)
	≈ 9.113 × 10−3 m2.	(73)

Hence

(74)

Solution 8: received power from incident power density

The wavelength is

(75)

The linear gain corresponding to 6.00 dBi is

(76)

Hence

Ae	=	(77)
	≈ 1.9773 × 10−2 m2.	(78)

Thus

(79)

The maximum available receive power is

Pav	= SincAe	(80)
	= (2.00 × 10−10)(1.9773 × 10−2)	(81)
	≈ 3.955 × 10−12 W.	(82)

Therefore

(83)

In dBm,

PdBm	= 10 log 10	(84)
	≈−84.03 dBm.	(85)

Solution 9: derive the Hertzian-dipole effective aperture

Start from the available receive power under conjugate match:

(86)

The incident time-average power density is

(87)

Therefore

Ae	=	(88)
	=	(89)
	= .	(90)

Now substitute

(91)

Then

Ae	=	(92)
	= .	(93)

Using

(94)

we get

Ae	=	(95)
	= .	(96)

For an ideal Hertzian dipole,

(97)

Therefore

(98)

which is the same result. The receive calculation and the transmit directivity therefore meet at the reciprocity relation

(99)

Solution 10: polarization mismatch

For two linear polarizations separated by angle Δα,

(100)

Thus

PLF	= cos 2(35.0∘)	(101)
	≈ 0.6710.	(102)

Therefore

(103)

The loss magnitude is

Lpol	= −10 log 10(0.6710)	(104)
	≈ 1.733 dB.	(105)

Hence the received power becomes

(106)

This is a receive-orientation loss, not a free-space propagation loss.

Solution 11: physical aperture, aperture efficiency, and dish gain

The physical aperture area is

Aphys	= π2	(107)
	= π(0.300)2	(108)
	≈ 0.28274 m2.	(109)

Thus

(110)

The effective aperture is

Ae	= ηapAphys	(111)
	= (0.650)(0.28274)	(112)
	≈ 0.18378 m2.	(113)

At 10.0 GHz,

(114)

Therefore

G	=	(115)
	≈ 2.570 × 103.	(116)

Hence

(117)

with

(118)

Solution 12: infer gain from measured effective aperture

First,

Ae	=	(119)
	=	(120)
	= 0.0250 m2.	(121)

Thus

(122)

At 2.00 GHz,

(123)

Using

(124)

we obtain

(125)

In dBi,

(126)

Solution 13: numerical Hertzian-dipole aperture check

At 300 MHz,

(127)

so

(128)

Using the exact vacuum-impedance form,

(129)

Thus

(130)

The peak open-circuit phasor voltage is

(131)

The maximum available power is

Pav	=	(132)
	≈ 1.5820 × 10−4 W.	(133)

The incident time-average power density is

Sinc	=	(134)
	≈ 1.3272 × 10−3 W/m2.	(135)

Therefore

Ae	=	(136)
	≈ 0.11920 m2.	(137)

Thus

(138)

The gain relation predicts

(139)

which agrees exactly apart from rounding.

Solution 14: frequency scaling of effective aperture

For fixed gain,

(140)

so

(141)

At 1.00 GHz,

(142)

At 2.00 GHz,

(143)

Their ratio is

(144)

Doubling frequency halves wavelength, and squaring that change reduces effective aperture by a factor of four. Gain is dimensionless directional performance; effective aperture is an area and therefore retains the wavelength scale.

Solution 15: transmit-to-receive bridge

The wavelength at 1.00 GHz is

(145)

The transmitter EIRP in the receiver direction is

(146)

At r = 10.0 km,

S	=	(147)
	=	(148)
	≈ 1.5915 × 10−8 W/m2.	(149)

The receiver effective aperture is

Ae,r	=	(150)
	=	(151)
	≈ 2.8608 × 10−2 m2.	(152)

Thus

Pr	= SAe,r	(153)
	≈ (1.5915 × 10−8)(2.8608 × 10−2)	(154)
	≈ 4.553 × 10−10 W.	(155)

Therefore

(156)

Combining the two stages algebraically gives

Pr	=	(157)
	= PtGtGr2 .	(158)

This is the ideal free-space Friis transmission relation. EM25 will separate this result into antenna terms, propagation loss, and practical link losses.

Solution 16: Julia sweep

Before computation, the expected scalings are

(159)

for fixed gain, and

(160)

for fixed accepted power.

A direct Julia implementation is provided with this article as

Its core calculations are

c = 299792458.0
Pacc = 5.0
Gfixed_dBi = 6.0
Gfixed = 10.0^(Gfixed_dBi/10.0)

f = range(0.5e9, 5.0e9, length=200)
lambda = c ./ f
Ae = Gfixed .* lambda.^2 ./ (4*pi)

ratio_numeric = Ae[1] / Ae[end]
ratio_expected = (f[end] / f[1])^2

gain_dBi = 0.0:1.0:15.0
gain = 10.0 .^ (gain_dBi ./ 10.0)
EIRP = Pacc .* gain
r = 1000.0
S = EIRP ./ (4*pi*r^2)

Because the frequency changes by a factor of ten, the endpoint effective-aperture ratio should be

(161)

The gain sweep should increase EIRP and far-field power density by the same linear gain factor.

What EM24E1 adds to the series

EM24 established the conceptual bridge from radiation pattern to receive aperture. EM24E1 makes that bridge operational through direct calculations and inverse problems. The complete chain is now

(162)

The final worked problem already exposes the ideal Friis equation. EM25 can therefore focus on the physical meaning of free-space path loss, logarithmic link budgets, system losses, and the assumptions under which Friis is valid.

References

References