Electromagnetic Waves, Antennas, and RF: Friis Transmission Equation, Free-Space Path Loss, and RF Link Budgets - Exercises

Electromagnetic Waves, Antennas, and RF:
Friis Transmission Equation, Free-Space Path Loss, and RF Link Budgets - Exercises and Complete Worked Solutions

This companion to EM25 turns the Friis transmission equation and RF link-budget notation into calculation tools. The exercises begin with the linear-power form of Friis, build the free-space path-loss equation, introduce logarithmic Power units, and then add polarization and impedance mismatch. The final problems assemble complete terrestrial and GNSS-style links in preparation for thermal noise and C∕N₀ in EM26 [1, 2, 3, 4, 5].

The central ideal-link relation is

(1)

with

(2)

In decibel form,

(3)

and a practical power budget is

(4)

Unless otherwise stated, the antennas are reciprocal, mutually far-field, polarization matched, and used in homogeneous free space. Gain values refer to the propagation directions of interest. Additional losses are power losses and are therefore subtracted in a dB link budget.

Part I: Exercises

Exercise 1: Friis in linear units

A transmitter delivers P_t = 2.00 W of accepted power to an antenna having directional gain G_t = 8.00. A receiving antenna 5.00 km away has directional gain G_r = 3.00. The frequency is 2.40 GHz. Find (a) wavelength, (b) received power in watts, (c) received power in dBW, and (d) received power in dBm.

Exercise 2: free-space path loss from first principles

For an isotropic link at 915 MHz and range 3.00 km, calculate (a) wavelength, (b) the dimensionless free-space path loss, and (c) free-space path loss in dB. Verify the result using the engineering form

(5)

Exercise 3: distance scaling of path loss

At 2.40 GHz, calculate free-space path loss at 1.00 km, 2.00 km, and 10.0 km. From the results, verify that doubling distance adds approximately 6.02 dB of path loss and multiplying distance by ten adds 20 dB.

Exercise 4: frequency scaling at fixed antenna gains

At a fixed range of 1.00 km, calculate free-space path loss at 1.00 GHz and 2.00 GHz. Explain why doubling frequency adds approximately 6.02 dB when the antenna gains are held fixed. Relate the result to the receive-aperture law

(6)

Exercise 5: dB, dBm, dBW, and watts

Convert each quantity as requested:

1. 20.0 dBm to watts and dBW;
2. −100 dBm to watts and dBW;
3. 5.00 W to dBW and dBm.

Then state the distinction between a quantity expressed in dB and an absolute power expressed in dBm or dBW.

Exercise 6: a basic logarithmic link budget

A transmitter produces 30.0 dBm. The transmit feed line loses 1.50 dB, the transmitting antenna gain is 12.0 dBi, the receiving antenna gain is 8.00 dBi, and the receive feed line loses 2.00 dB. An additional 1.00 dB implementation loss is specified. The link operates at 5.80 GHz over 2.00 km. Find (a) free-space path loss and (b) received power after all listed losses, in dBm and watts.

Exercise 7: linear-polarization mismatch

Two linearly polarized antennas have polarization axes separated by 40.0^∘. Find (a) the polarization loss factor, (b) polarization loss in dB, and (c) the received power if all other link effects would have produced −85.0 dBm under perfect polarization alignment.

Exercise 8: circular-to-linear polarization mismatch

An ideal circularly polarized wave is received by an ideal linearly polarized antenna. Find the polarization loss factor and polarization loss in dB. If the matched-polarization link prediction was −90.0 dBm, find the available power after this polarization mismatch alone.

Exercise 9: terminal mismatch loss

At the receiving antenna terminals, the magnitude of the reflection coefficient is

(7)

Find (a) the accepted-power fraction 1 −|Γ_A|², (b) mismatch loss in dB, and (c) delivered power if the available matched power would have been −80.0 dBm. Explain why the same mismatch loss must not be subtracted again if the receive antenna specification already uses realized gain.

Exercise 10: complete microwave link budget

A 10.0 GHz point-to-point link spans 15.0 km. The transmitter output is 43.0 dBm. The transmit feed loss is 2.00 dB and transmit antenna gain is 18.0 dBi. The receive antenna gain is 25.0 dBi and receive feed loss is 1.50 dB. Include a 0.70 dB polarization loss, 1.20 dB atmospheric loss, and 1.00 dB pointing loss. Calculate (a) free-space path loss, (b) received power in dBm, and (c) received power in watts.

Exercise 11: maximum range from a receiver threshold

A 2.40 GHz radio transmits 20.0 dBm. Its transmit feed loss is 1.00 dB, transmit antenna gain is 5.00 dBi, receive antenna gain is 2.00 dBi, and all other losses total 3.00 dB. The receiver threshold is −100 dBm. Neglect fading margin. Determine (a) the maximum allowable free-space path loss and (b) the corresponding free-space range.

Exercise 12: field-strength and Friis cross-check

A transmitter has EIRP 100 W. At r = 1.00 km, a receiving antenna operates at 900 MHz and has gain 6.00 dBi toward the transmitter. Using

(8)

and

(9)

find (a) incident power density, (b) RMS electric-field magnitude, (c) receive effective aperture, and (d) received power. Verify the same received power directly from Friis using P_tG_t = EIRP.

Exercise 13: GNSS-style received carrier power

Consider an intentionally simplified GPS-L1-like link with

(10)

Assume transmitter EIRP 27.0 dBW in the receiver direction, receive gain 0 dBi, and 2.00 dB of additional propagation and implementation loss. Find (a) free-space path loss, and (b) received carrier power in dBW, dBm, and watts. This is a link-budget exercise, not a specification for a particular GNSS satellite or receiver.

Exercise 14: GNSS-style link with receive gain and polarization loss

Use the same L1 frequency and range as Exercise 13, but now let the transmitter EIRP be 26.5 dBW, receive antenna gain be 3.00 dBi, polarization loss be 1.50 dB, and all other losses total 1.00 dB. Find the received power in dBW, dBm, and watts. Identify which terms are transmitter-side, propagation-side, and receiver-side.

Exercise 15: compare L1-like and L5-like path loss

At the same 20,200 km range, compare the ideal free-space path loss at

(11)

and

(12)

Find the path-loss difference in dB and the corresponding linear power ratio for links having the same EIRP and the same receive gain. Explain why this comparison assumes fixed gain rather than fixed physical receive aperture.

Exercise 16: Julia sweep of free-space link behavior

Write a Julia program that

1. computes L_FS from 100 MHz to 10 GHz at ranges of 1, 10, and 100 km;
2. verifies the 20 dB-per-decade distance scaling;
3. verifies the 20 dB-per-decade frequency scaling at fixed gains;
4. calculates received power for a user-defined P_t, G_t, G_r, and loss term;
5. reproduces the simplified GNSS-style calculation of Exercise 13.

State the expected logarithmic slopes before writing the code.

Part II: Complete Worked Solutions

Solution 1: Friis in linear units

The wavelength is

λ	=	(13)
	=	(14)
	= 0.124914 m.	(15)

Friis gives

Pr	= PtGtGr2	(16)
	= (2.00)(8.00)(3.00)2	(17)
	= 1.897 × 10−10 W.	(18)

Therefore

(19)

In dBW,

Pr,dBW	= 10 log 10(1.897 × 10−10)	(20)
	= −97.22 dBW.	(21)

Since

(22)

we obtain

(23)

Solution 2: free-space path loss from first principles

At 915 MHz,

λ	=	(24)
	= 0.327642 m.	(25)

Then

LFS	= 2	(26)
	= 2	(27)
	= 1.324 × 1010.	(28)

Thus

(29)

In decibels,

LFS,dB	= 10 log 10LFS	(30)
	= 101.22 dB.	(31)

The engineering form gives

LFS,dB	≈ 32.45 + 20 log 10(3.00) + 20 log 10(915)	(32)
	≈ 101.22 dB,	(33)

with the small last-digit difference determined by the rounded constant 32.45.

Solution 3: distance scaling of path loss

At 2.40 GHz,

LFS(1 km)	= 100.05 dB,	(34)
LFS(2 km)	= 106.07 dB,	(35)
LFS(10 km)	= 120.05 dB.	(36)

Therefore doubling range changes the loss by

(37)

and multiplying range by ten changes it by

(38)

This follows directly from

(39)

Solution 4: frequency scaling at fixed antenna gains

At r = 1.00 km,

LFS(1.00 GHz)	= 92.45 dB,	(40)
LFS(2.00 GHz)	= 98.47 dB.	(41)

The difference is

(42)

The same result follows from

(43)

At fixed receive gain,

(44)

Thus a gain-normalized receiving antenna has one-quarter the effective aperture when frequency doubles, corresponding to 6.02 dB less received power. This does not mean that ideal free space absorbs more energy at high frequency.

Solution 5: dB, dBm, dBW, and watts

For dBm,

(45)

Therefore

(46)

Similarly,

(47)

For 5.00 W,

PdBW	= 10 log 10(5.00)	(48)
	= 6.990 dBW,	(49)

so

(50)

A value in dB is ordinarily a logarithmic ratio. dBm and dBW are absolute power levels because their reference powers are fixed at 1 mW and 1 W, respectively.

Solution 6: a basic logarithmic link budget

First calculate free-space path loss:

LFS,dB	= 20 log 10	(51)
	= 113.74 dB.	(52)

Now perform the power bookkeeping:

Pr	= 30.0 − 1.50 + 12.0 + 8.00 − 113.74 − 2.00 − 1.00	(53)
	= −68.24 dBm.	(54)

Converting to watts,

Pr	= 10(−68.24−30)∕10	(55)
	= 1.50 × 10−10 W.	(56)

Thus

(57)

Solution 7: linear-polarization mismatch

For two linear polarizations separated by angle ψ,

(58)

At ψ = 40.0^∘,

PLF	= cos 2(40.0∘)	(59)
	= 0.5868.	(60)

The polarization loss is

Lpol	= −10 log 10(0.5868)	(61)
	= 2.315 dB.	(62)

Therefore

(63)

Solution 8: circular-to-linear polarization mismatch

An ideal circularly polarized wave has equal average power in two orthogonal linear components. An ideal linear antenna selects one of those components, so

(64)

The loss is

Lpol	= −10 log 10(1∕2)	(65)
	= 3.010 dB.	(66)

Therefore the predicted receive power becomes

(67)

Solution 9: terminal mismatch loss

The fraction of available power accepted by the mismatched load is

ηmis	= 1 −|ΓA|2	(68)
	= 1 − (0.350)2	(69)
	= 0.8775.	(70)

Thus

(71)

The mismatch loss is

Lmis	= −10 log 10(0.8775)	(72)
	= 0.568 dB.	(73)

Therefore

(74)

If the antenna gain in the link budget is already a realized gain, mismatch has already been included in that gain definition. Subtracting L_mis again would double-count the same physical effect.

Solution 10: complete microwave link budget

The free-space path loss is

LFS,dB	= 20 log 10	(75)
	= 135.97 dB.	(76)

The complete budget is

Pr =	43.0 − 2.00 + 18.0 + 25.0 − 135.97	(77)
	− 0.70 − 1.20 − 1.00 − 1.50	(78)
=	− 56.37 dBm.	(79)

In watts,

Pr	= 10(−56.37−30)∕10	(80)
	= 2.31 × 10−9 W.	(81)

Hence

(82)

Solution 11: maximum range from a receiver threshold

Rearrange the link budget so the maximum allowable path loss is the remaining power margin:

LFS,max =	Pt − Lt + Gt + Gr − Lother − Pr,min	(83)
=	20.0 − 1.00 + 5.00 + 2.00 − 3.00 − (−100)	(84)
=	123.0 dB.	(85)

Thus

(86)

From

(87)

solve for range:

(88)

At 2.40 GHz,

r	= 10123∕20	(89)
	= 1.404 × 104 m.	(90)

Therefore

(91)

A practical design would normally reserve additional fading, implementation, and environmental margin rather than operate exactly at this threshold.

Solution 12: field-strength and Friis cross-check

The incident power density is

S	=	(92)
	= 7.958 × 10−6 W/m2.	(93)

The RMS electric field is

Erms	=	(94)
	=	(95)
	= 5.475 × 10−2 V/m.	(96)

At 900 MHz,

(97)

A gain of 6.00 dBi corresponds to

(98)

Hence

Ae	=	(99)
	= 3.515 × 10−2 m2.	(100)

The receive power is

Pr	= SAe	(101)
	= (7.958 × 10−6)(3.515 × 10−2)	(102)
	= 2.797 × 10−7 W	(103)
	= −35.53 dBm.	(104)

Now use Friis with P_tG_t = 100 W:

(105)

which gives the same 2.797 × 10⁻⁷ W. The field-strength, power-density, effective-aperture, and Friis descriptions are therefore mutually consistent.

Solution 13: GNSS-style received carrier power

For f = 1.57542 GHz and r = 20,200 km,

LFS,dB	= 20 log 10	(106)
	= 182.50 dB.	(107)

The link budget is

Pr	= 27.0 + 0 − 182.50 − 2.00	(108)
	= −157.50 dBW.	(109)

Therefore

(110)

In watts,

Pr	= 10−157.50∕10	(111)
	= 1.78 × 10−16 W.	(112)

The extremely small carrier power is why GNSS receiver analysis is normally expressed relative to noise spectral density rather than by received power alone.

Solution 14: GNSS-style link with receive gain and polarization loss

The L1 free-space path loss is unchanged:

(113)

The link budget is

Pr	= 26.5 + 3.00 − 182.50 − 1.50 − 1.00	(114)
	= −155.50 dBW.	(115)

Therefore

(116)

In watts,

(117)

Here 26.5 dBW is the transmitter-side EIRP. The 182.50 dB FSPL and 1.50 dB polarization term describe propagation/coupling effects. The 3.00 dBi receive gain is receiver-side directional collection. The final 1.00 dB term represents other specified implementation or propagation losses according to the chosen reference planes.

Solution 15: compare L1-like and L5-like path loss

At 20,200 km,

LFS,L1	= 182.503 dB,	(118)
LFS,L5	= 179.966 dB.	(119)

Thus

ΔLFS	= 182.503 − 179.966	(120)
	= 2.536 dB.	(121)

For otherwise identical links,

(122)

Thus, under the stated fixed-gain assumptions, the lower-frequency link has about 1.79 times the received power purely from the wavelength-dependent Friis factor.

The fixed-gain qualification matters. If instead the same physical aperture were used efficiently at both frequencies, gain itself would scale approximately as 1∕λ², and the apparent frequency dependence could change. Friis must always be interpreted together with how the antennas are being held fixed.

Solution 16: Julia sweep of free-space link behavior

Because

(123)

the expected slope is 20 dB per decade in either range or frequency when the other variables and antenna gains are fixed.

One Julia implementation is:

using Printf

const c = 299_792_458.0

fspl_db(r, f) = 20*log10(4*pi*r*f/c)
received_dbm(pt_dbm, gt_dbi, gr_dbi, r, f, loss_db=0.0) =
    pt_dbm + gt_dbi + gr_dbi - fspl_db(r,f) - loss_db

freq = 10 .^ range(log10(100e6), log10(10e9), length=201)
ranges = [1e3, 10e3, 100e3]

for r in ranges
    vals = fspl_db.(r, freq)
    @printf("r = %.0f km: FSPL %.2f to %.2f dB\n",
            r/1e3, vals[1], vals[end])
end

# Verify decade scaling.
f0 = 1.0e9
@printf("range decade change = %.6f dB\n",
        fspl_db(10e3,f0) - fspl_db(1e3,f0))
@printf("frequency decade change = %.6f dB\n",
        fspl_db(1e3,10e9) - fspl_db(1e3,1e9))

# Example user-defined link.
pt_dbm = 30.0
gt_dbi = 8.0
gr_dbi = 3.0
loss_db = 2.0
r = 5e3
f = 2.4e9
pr_dbm = received_dbm(pt_dbm,gt_dbi,gr_dbi,r,f,loss_db)
@printf("example Pr = %.3f dBm\n", pr_dbm)

# Simplified GNSS-style Exercise 13.
f_l1 = 1.57542e9
r_gnss = 20_200e3
eirp_dbw = 27.0
                                                                                         
                                                                                         
other_loss_db = 2.0
lfs = fspl_db(r_gnss,f_l1)
pr_dbw = eirp_dbw - lfs - other_loss_db
@printf("GNSS-style FSPL = %.3f dB\n", lfs)
@printf("GNSS-style Pr = %.3f dBW = %.3f dBm\n",
        pr_dbw, pr_dbw + 30)

The expected checks are approximately

(124)

and

(125)

For the simplified L1-like case, the script should reproduce approximately

(126)

and

(127)

What EM25E1 adds to the series

EM25 derived the Friis equation and free-space path loss from electromagnetic power density and receive effective aperture. EM25E1 makes that theory operational. The worked problems connect linear power, logarithmic units, polarization and mismatch factors, practical loss bookkeeping, inverse range calculations, field strength, and satellite-scale received powers in one consistent reference-plane framework.

The calculation chain is

(128)

The next article can therefore introduce thermal-noise power and spectral density,

(129)

so that the received carrier power obtained here can be converted into C∕N₀.

References