example of Electromagnetic Waves, Antennas, and RF: End-to-End RF and GNSS Link-Budget Synthesis

Electromagnetic Waves, Antennas, and RF: End-to-End RF and GNSS Link-Budget Synthesis - Exercises and Complete Worked Solutions

This companion to EM28 turns the complete RF link-budget chain into calculation practice. The problems deliberately vary one physical quantity at a time before combining several changes in a final margin analysis. The sequence covers EIRP, free-space path loss, receiver G∕T, system noise temperature, C∕N₀, finite-bandwidth C∕N, implementation losses, GNSS-scale geometry, and receiver-side interference margin [1, 2, 3, 4, 5].

The principal end-to-end relation is

(1)

For free space,

(2)

while

(3)

and

(4)

Use

(5)

Part I: Exercises

Exercise 1: baseline end-to-end RF link

A telemetry transmitter operates at 2.20 GHz over a 250 km free-space path. The transmitter power is 10.0 dBW, transmit feeder loss is 1.50 dB, transmit antenna gain is 12.0 dBi, and all other propagation/implementation losses total 2.50 dB. The receive antenna gain is 8.00 dBi, the system noise temperature is 500 K, and the receiver noise-equivalent bandwidth is 1.00 MHz.

Find (a) EIRP, (b) free-space path loss, (c) received carrier power, (d) G∕T, (e) C∕N₀, and (f) C∕N.

Exercise 2: vary range while everything else is fixed

Repeat Exercise 1 after the range doubles from 250 km to 500 km. Determine the change in free-space path loss and C∕N₀. If the minimum required C∕N₀ is 75.0 dB-Hz, compute the link margin before and after the range change.

Exercise 3: vary frequency at fixed antenna gains

Return to the 250 km geometry of Exercise 1, but double the carrier frequency from 2.20 GHz to 4.40 GHz. Hold all antenna gains, EIRP, losses, and receiver noise quantities fixed. Find the new FSPL and C∕N₀. Explain why this fixed-gain comparison is not the same as holding physical antenna aperture fixed.

Exercise 4: compensate range loss with EIRP

For the 500 km case of Exercise 2, determine the EIRP required to restore the original 250 km value of C∕N₀. If antenna gain and transmit feeder loss remain unchanged, find the required transmitter power in dBW and watts. Compare with the original 10 W transmitter.

Exercise 5: trade receive gain against system temperature

Starting from Exercise 1, compare two independent receiver improvements:

1. increase receive antenna gain from 8.00 dBi to 11.00 dBi while T_sys = 500 K;
2. keep G_r = 8.00 dBi but reduce T_sys from 500 K to 250 K.

Find the new G∕T and the improvement in C∕N₀ for each case.

Exercise 6: system-temperature degradation

Starting from Exercise 1, suppose environmental or front-end conditions increase T_sys from 500 K to 1000 K while all gains and losses remain fixed. Calculate the new G∕T and C∕N₀. Show that doubling system temperature causes approximately a 3.01 dB penalty.

Exercise 7: bandwidth sweep

The baseline link in Exercise 1 has a fixed C∕N₀. Compute C∕N for noise-equivalent bandwidths of 100 kHz, 1.00 MHz, and 10.0 MHz. Explain why C∕N₀ does not change when the receiver bandwidth is changed in this idealized calculation.

Exercise 8: increase miscellaneous losses

The baseline link uses L_other = 2.50 dB. In an adverse condition suppose the combined atmospheric, polarization, pointing, radome, and implementation losses rise to 6.00 dB. Recalculate C∕N₀ and the margin relative to a 75.0 dB-Hz requirement. How much margin was consumed solely by the additional losses?

Exercise 9: nominal GNSS-style end-to-end budget

Consider an illustrative L1-like link with

(6)

(7)

Find FSPL, received carrier power, G∕T, C∕N₀, and C∕N in a 2.00 MHz front-end bandwidth. If the receiver requires 42.0 dB-Hz, find the nominal C∕N₀ margin.

Exercise 10: GNSS-style range variation

Keep all Exercise 9 quantities fixed except range, which increases from 20,200 km to 26,000 km. Determine the additional path loss, the new C∕N₀, and the new margin relative to 42.0 dB-Hz.

Exercise 11: GNSS-style frequency comparison at fixed gains

Using the Exercise 9 range and all other quantities fixed, compare f₁ = 1.57542 GHz with f₂ = 1.17645 GHz. Calculate the FSPL at the lower frequency and the change in C∕N₀. State clearly the assumption that antenna gains, rather than physical apertures, are being held fixed.

Exercise 12: reference-plane synthesis with feed loss and LNA noise

At the antenna terminals, the antenna noise temperature is 100 K. A 1.00 dB passive feed loss at 290 K precedes an LNA with 0.80 dB noise figure. The receive antenna gain is 25.0 dBi.

(a) Refer the feed and LNA noise temperatures back to the antenna-terminal plane and find T_sys. (b) Find G∕T. (c) For a link with EIRP = 15.0 dBW, L_path = 180 dB, and L_other = 2.00 dB, find C∕N₀ and C∕N in 1.00 MHz. Explain why the 1 dB feed loss must not be subtracted again if its effect has already been included consistently in the quoted G∕T.

Exercise 13: receiver degradation from noise-like interference

For the GNSS-style link of Exercise 9, suppose an admitted noise-like interferer has

(8)

Use

(9)

to find the degradation in dB, the degraded effective C∕N₀, and the remaining margin above a 42.0 dB-Hz threshold.

Exercise 14: allowable interference density from margin

For the same nominal GNSS-style link, use the clean-link margin to determine the maximum J₀∕N₀ that can be admitted before the effective C∕N₀ reaches the 42.0 dB-Hz threshold under the additive noise-like interference model. Use

(10)

where M is the clean-link margin in dB.

Exercise 15: why J∕S is not the same as degradation

For Exercise 9, the carrier power is C and the system temperature is 400 K. Suppose the noise-like interference density is again J₀∕N₀ = −8.00 dB and the receiver bandwidth is 2.00 MHz. Find (a) N₀ in dBW/Hz, (b) J₀ in dBW/Hz, (c) total in-band J in dBW, (d) J∕S in dB, and (e) the additive-power degradation. Explain why a positive J∕S can coexist with less than 1 dB of degradation in this example.

Exercise 16: combined worst-case sensitivity and recovery

Start with the nominal GNSS-style C∕N₀ from Exercise 9. Now apply all of the following changes simultaneously:

• range increases to 26,000 km;
• EIRP decreases by 1.00 dB;
• T_sys rises from 400 K to 600 K;
• miscellaneous losses increase by 0.50 dB;
• noise-like interference has J₀∕N₀ = −6.00 dB.

Find the final effective C∕N₀ and margin relative to 42.0 dB-Hz. If only one scalar budget term can be improved, determine the minimum improvement in EIRP or G∕T required to restore the threshold.

Part II: Complete Worked Solutions

Solution 1: baseline end-to-end RF link

The EIRP is

EIRP	= Pt − Lt + Gt	(11)
	= 10.0 − 1.50 + 12.0	(12)
	= 20.50 dBW.	(13)

Thus

(14)

The free-space path loss is

LFS	= 20 log 10	(15)
	= 20 log 10	(16)
	= 147.255 dB.	(17)

Hence

(18)

The received carrier power at the antenna-terminal reference plane is

C	= EIRP − LFS − Lother + Gr	(19)
	= 20.50 − 147.255 − 2.50 + 8.00	(20)
	= −121.255 dBW.	(21)

Therefore

(22)

The receiver figure of merit is

G∕T	= 8.00 − 10 log 10(500)	(23)
	= −18.990 dB/K.	(24)

Thus

(25)

Now

C∕N0	= 20.50 − 147.255 − 2.50 − 18.990 + 228.599	(26)
	= 80.354 dB-Hz.	(27)

Therefore

(28)

For B_n = 1.00 MHz,

C∕N	= 80.354 − 10 log 10(106)	(29)
	= 20.354 dB.	(30)

Hence

(31)

Solution 2: vary range while everything else is fixed

Doubling range gives

(32)

The new FSPL is

(33)

Because no other term changes, C∕N₀ falls by the same amount:

(34)

The original margin above 75.0 dB-Hz was

(35)

while at 500 km

(36)

Thus

(37)

A factor of two in range consumed just over 6 dB of link margin.

Solution 3: vary frequency at fixed antenna gains

At fixed range,

(38)

Therefore

(39)

and

(40)

This result assumes the quoted transmit and receive gains remain fixed. If physical antenna aperture were held fixed instead, antenna gain would generally increase with frequency, and part or all of the apparent 20 log ₁₀f penalty could be offset. The comparison must therefore state what antenna property is held constant.

Solution 4: compensate range loss with EIRP

The doubled range added

(41)

of path loss. To restore the original C∕N₀, EIRP must increase by the same amount:

(42)

With the original L_t = 1.50 dB and G_t = 12.0 dBi,

Pt,req	= EIRPreq + Lt − Gt	(43)
	= 26.521 + 1.50 − 12.0	(44)
	= 16.021 dBW.	(45)

The corresponding power is

(46)

Thus

(47)

Doubling range requires four times the transmitter power if gain and every other term remain fixed, which is the inverse-square law expressed in power-budget form.

Solution 5: trade receive gain against system temperature

For the gain improvement,

G∕T	= 11.00 − 10 log 10(500)	(48)
	= −15.990 dB/K.	(49)

Compared with the original −18.990 dB/K, the improvement is exactly

(50)

Therefore C∕N₀ also improves by 3.00 dB.

For the temperature improvement,

G∕T	= 8.00 − 10 log 10(250)	(51)
	= −15.979 dB/K.	(52)

The improvement is

Δ(G∕T)	= 10 log 10	(53)
	= 3.010 dB.	(54)

Hence

(55)

A 3 dB gain increase and a factor-of-two temperature reduction are nearly equivalent in this scalar link budget.

Solution 6: system-temperature degradation

With T_sys = 1000 K,

G∕T	= 8.00 − 10 log 10(1000)	(56)
	= −22.00 dB/K.	(57)

Thus

(58)

The new carrier-to-noise-density ratio is

C∕N0	= 20.50 − 147.255 − 2.50 − 22.00 + 228.599	(59)
	= 77.344 dB-Hz.	(60)

Therefore

(61)

The penalty is

(62)

which follows directly from 10 log ₁₀(1000∕500) = 3.010 dB.

Solution 7: bandwidth sweep

The carrier-to-noise-density ratio remains

(63)

For 100 kHz,

(64)

For 1.00 MHz,

(65)

For 10.0 MHz,

(66)

Hence

(67)

for 100 kHz, 1 MHz, and 10 MHz respectively. C∕N₀ is a carrier-power-to-noise-density quantity and therefore does not depend on the later choice of integrated noise bandwidth in this ideal model.

Solution 8: increase miscellaneous losses

The loss increase is

(68)

Thus C∕N₀ decreases by 3.50 dB:

(69)

The remaining margin above 75.0 dB-Hz is

(70)

The additional losses consumed exactly

(71)

of margin because dB losses enter the budget linearly.

Solution 9: nominal GNSS-style end-to-end budget

The free-space path loss is

LFS	= 20 log 10	(72)
	= 182.503 dB.	(73)

Thus

(74)

The carrier power is

C	= 27.0 − 182.503 − 2.00 + 2.00	(75)
	= −155.503 dBW.	(76)

Therefore

(77)

The receiver figure of merit is

G∕T	= 2.00 − 10 log 10(400)	(78)
	= −24.021 dB/K.	(79)

Hence

(80)

The density ratio is

C∕N0	= 27.0 − 182.503 − 2.00 − 24.021 + 228.599	(81)
	= 47.076 dB-Hz.	(82)

Thus

(83)

For B_n = 2.00 MHz,

C∕N	= 47.076 − 10 log 10(2.00 × 106)	(84)
	= 47.076 − 63.010	(85)
	= −15.934 dB.	(86)

Therefore

(87)

The clean-link C∕N₀ margin is

(88)

Solution 10: GNSS-style range variation

The new FSPL at 26,000 km is

(89)

The increase is

ΔLFS	= 20 log 10	(90)
	= 2.192 dB.	(91)

Therefore

(92)

The remaining margin is

(93)

Solution 11: GNSS-style frequency comparison at fixed gains

At 1.17645 GHz and the same 20,200 km range,

(94)

Relative to 1.57542 GHz, the reduction is

ΔLFS	= 20 log 10	(95)
	= 2.536 dB.	(96)

With EIRP, receive gain, losses, and system temperature held fixed, C∕N₀ increases by the same amount:

(97)

Again, this is a fixed-gain comparison. If the physical antenna aperture were fixed, gain would itself vary with frequency.

Solution 12: reference-plane synthesis with feed loss and LNA noise

Convert the 1.00 dB feed loss to a linear power loss:

(98)

The feed equivalent input noise temperature is

Te,f	= (Lf − 1)Tp	(99)
	= (1.25893 − 1)(290)	(100)
	= 75.09 K.	(101)

The LNA noise factor is

(102)

so

(103)

Because the LNA follows the lossy feed, its noise referred to the antenna-terminal plane is multiplied by L_f:

Tsys	= TA + Te,f + LfTe,LNA	(104)
	= 100 + 75.09 + (1.25893)(58.66)	(105)
	= 248.93 K.	(106)

Thus

(107)

The figure of merit is

G∕T	= 25.0 − 10 log 10(248.93)	(108)
	= 1.039 dB/K.	(109)

Hence

(110)

The end-to-end density ratio is

C∕N0	= 15.0 − 180 − 2.00 + 1.039 + 228.599	(111)
	= 62.638 dB-Hz.	(112)

Therefore

(113)

For 1.00 MHz,

(114)

The feed loss has already affected the receiver through the input-referred system temperature used in G∕T. Subtracting the same loss again from the compact C∕N₀ equation would double-count it unless the reference plane were redefined consistently.

Solution 13: receiver degradation from noise-like interference

Convert the interference ratio to linear form:

(115)

The degradation is

DJ	= 10 log 10(1 + 0.15849)	(116)
	= 0.6389 dB.	(117)

Therefore

(118)

The effective density ratio is

(119)

The remaining margin is

(120)

Solution 14: allowable interference density from margin

The clean-link margin is

(121)

Hence

(J0∕N0)max,dB	= 10 log 10	(122)
	= 3.460 dB.	(123)

Therefore

(124)

This is a receiver tolerance result under the stated additive noise-like interference model. It is not a statement that every interference waveform with the same total power will produce identical degradation.

Solution 15: why J∕S is not the same as degradation

At 400 K,

N0	= 10 log 10(kT)	(125)
	= −202.579 dBW/Hz.	(126)

Thus

(127)

Since J₀∕N₀ = −8.00 dB,

(128)

Across 2.00 MHz,

J	= J0 + 10 log 10(2.00 × 106)	(129)
	= −210.579 + 63.010	(130)
	= −147.568 dBW.	(131)

So

(132)

The carrier from Exercise 9 is

(133)

therefore

(134)

Yet the relevant noise-like interference ratio is only J₀∕N₀ = −8 dB, which gives

(135)

The positive J∕S compares integrated interference power with the weak carrier. The degradation model instead compares interference spectral density with thermal-noise density. The bandwidth and spectral distribution therefore matter; J∕S alone is not sufficient to predict receiver degradation.

Solution 16: combined worst-case sensitivity and recovery

Start from

(136)

The range change from 20,200 km to 26,000 km costs

(137)

The EIRP reduction costs

(138)

The temperature increase from 400 K to 600 K costs

ΔT	= 10 log 10	(139)
	= 1.761 dB.	(140)

The extra miscellaneous loss costs

(141)

For J₀∕N₀ = −6 dB,

DJ	= 10 log 10(1 + 10−6∕10)	(142)
	= 0.973 dB.	(143)

The final effective value is therefore

(C∕N0)eff	= 47.076 − 2.192 − 1.000 − 1.761 − 0.500 − 0.973	(144)
	= 40.649 dB-Hz.	(145)

Thus

(146)

Relative to the 42.0 dB-Hz requirement,

(147)

The link is short by 1.351 dB. Therefore a single scalar improvement of at least

(148)

in EIRP or G∕T would restore the threshold under this model. The same deficit could also be recovered by an equivalent reduction in modeled losses or by some combination of smaller improvements.

Summary of sensitivity rules

These problems expose several recurring link-budget scalings:

• doubling range adds 6.02 dB of free-space path loss;
• doubling frequency adds 6.02 dB of FSPL when antenna gains are held fixed;
• +1 dB EIRP produces +1 dB C∕N₀;
• +1 dB G∕T produces +1 dB C∕N₀;
• doubling T_sys reduces G∕T and C∕N₀ by 3.01 dB;
• multiplying bandwidth by ten reduces C∕N by 10 dB while leaving C∕N₀ unchanged;
• ordinary independent losses subtract directly in dB;
• independent powers such as thermal noise and noise-like interference must first be added in linear units, producing 10 log ₁₀(1 + J₀∕N₀) rather than a direct dB subtraction.

References

References