Link budgeting: 22-Elec-B3 Digital Communications Systems, May 2016, Q2

Question 2: Link budgeting (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A one-hop wireless link whose only propagation loss is the stated path-loss law, with the budget entries below.

Given data — link parameters
Parameter	Symbol	Value
Transmitter power	$P_t$	10 W
Antenna gain (each end)	$G_t = G_r$	4 dB
Receiver losses	$L_{rx}$	6 dB
Receiver noise density	$N_0$	−174 dBm/Hz
Noise bandwidth	$B$	10 MHz
Fading margin	$M$	6 dB
Required signal-to-noise ratio	$\mathrm{SNR}_{req}$	6 dB
Carrier frequency, range, light speed	$f,\ d,\ c$	2.4 GHz, 200 m, $3.0\times10^{8}$ m/s

Find. (a) the largest path loss the budget can absorb; (b) whether the stated law at $d=200$ m keeps the link inside that allowance; (c) the path-loss exponent implied by the given law.

Check: two readings of the data are worth stating, per cover-page Note 1. First, "receiver noise figure of −174 dBm/Hz" carries the units of a noise power spectral density, not of a noise figure (which is dimensionless, in dB). The value is exactly $kT_0$ at $T_0 = 290$ K, so it is read here as the receiver's input-referred noise density, integrated over the bandwidth to obtain the noise floor. Second, "antenna gains of 4 dB" is plural with a single value, read here as 4 dB at each end (+8 dB total). Both readings are carried explicitly through the arithmetic; the alternative reading of 4 dB in total is also evaluated below, and the part (b) verdict is unchanged by it.

Link-budget level diagram: power in dBm from the transmitter power amplifier to the demodulator input, against the noise floor and the sensitivity that already includes the 6 dB fade margin.

Approach. Work the whole budget in decibel units: convert the transmitter power to dBm, build the noise floor from the noise density and the bandwidth, add the required signal-to-noise ratio and the fading margin to obtain the minimum acceptable received power, and let the path loss be whatever the remaining budget permits. Then evaluate the given path-loss law at 200 m and compare.

Part (a) — put the transmitter power on the dBm scale. With 0 dBm defined as 1 mW, $$P_t\,[\text{dBm}] = 10\log_{10}\!\left(\frac{10\ \text{W}}{1\ \text{mW}}\right) = 10\log_{10}(10\,000) = 40\ \text{dBm}.$$ The anchors worth memorising for the exam room are $0\ \text{dBm}=1\ \text{mW}$, $+10$ dB per decade and $+3$ dB per doubling, so 10 W reads off directly as $30+10 = 40$ dBm.
Build the receiver noise floor. Thermal noise is white across the receiver bandwidth, so its total power is the density multiplied by the bandwidth, which is an addition in decibels: $$N = N_0 + 10\log_{10}B = -174 + 10\log_{10}(10\times10^{6}) = -174 + 70 = -104\ \text{dBm}.$$ The 70 dB is worth sanity-checking on its own: 10 MHz is seven decades above 1 Hz.
Set the receiver sensitivity. The demodulator needs 6 dB of signal-to-noise ratio, and the link must additionally hold 6 dB in reserve against fading, so the received power may never fall below $$P_{r,\min} = N + \mathrm{SNR}_{req} + M = -104 + 6 + 6 = -92\ \text{dBm}.$$ Adding the fade margin here (rather than subtracting it from the path-loss allowance later) is the bookkeeping convention that keeps the margin visible in the sensitivity figure.
Solve the link equation for the path loss. Every gain adds and every loss subtracts along the chain from power amplifier to demodulator input: $$P_r = P_t + G_t + G_r - L_{rx} - L_{path}.$$ Setting $P_r = P_{r,\min}$ and rearranging, $$L_{path,\max} = P_t + G_t + G_r - L_{rx} - P_{r,\min} = 40 + 4 + 4 - 6 - (-92)$$ $$\boxed{L_{path,\max} = 134\ \text{dB}}$$ Under the alternative reading of 4 dB of antenna gain in total, the same arithmetic gives 130 dB.
Part (b) — evaluate the argument of the given path-loss law. The frequency and the speed of light combine first, since $f/c = 2.4\times10^{9}/3.0\times10^{8} = 8\ \text{m}^{-1}$ exactly, which keeps the rest of the arithmetic clean: $$\frac{4\pi d f}{c} = 4\pi (200)(8) = 6400\pi = 2.0106\times10^{4}.$$
Apply the 30 log law. With $\log_{10}(2.0106\times10^{4}) = 4.3033$, $$L_{path} = 30\log_{10}\!\left(\frac{4\pi d f}{c}\right) = 30(4.3033)$$ $$\boxed{L_{path}(200\ \text{m}) = 129.1\ \text{dB}}$$ Note this is not the textbook free-space figure: the ordinary Friis law would use $20\log_{10}$ and give 86.1 dB. The paper prescribes the 30 log form, so the 43 dB difference is intended, and using the textbook expression instead is the single most costly error available in this question.
Compare against the allowance and state the verdict. The link needs 129.1 dB of the 134 dB available, so $$134 - 129.1 = 4.9\ \text{dB of headroom} \Rightarrow \text{the criterion is satisfied.}$$ It is worth restating this as a received power and a signal-to-noise ratio, because that is what a commissioning measurement would actually show: $$P_r = 40 + 8 - 6 - 129.1 = -87.1\ \text{dBm}, \qquad \mathrm{SNR} = P_r - N = -87.1 - (-104) = 16.9\ \text{dB}.$$ The requirement including the fade margin is $6+6 = 12$ dB, and 16.9 dB clears it by the same 4.9 dB.
Confirm the verdict survives the ambiguous antenna reading. With 4 dB of gain in total the received power falls to $-91.1$ dBm and the signal-to-noise ratio to 12.9 dB, still above the 12 dB needed. The answer to part (b) is therefore $$\boxed{\text{Yes — the criterion is satisfied at } d = 200\ \text{m under either reading}}$$ which is the useful thing to report: the ambiguity in the question does not change the engineering decision, only the size of the margin.
Part (c) — identify the path-loss exponent. A path-loss model with exponent $n$ is written $L = 10\,n\log_{10}(d) + \text{constant}$, because received power falls as $d^{-n}$. Matching that form to the given law, $$30\log_{10}\!\left(\frac{4\pi d f}{c}\right) = 30\log_{10}(d) + 30\log_{10}\!\left(\frac{4\pi f}{c}\right) \equiv 10\,n\log_{10}(d) + \text{const},$$ so $10n = 30$ and $$\boxed{n = 3}$$ The physical reading is that the model is not free space at all: true free-space propagation has $n=2$, while $n=3$ represents a moderately obstructed environment, roughly the value measured in urban microcells and in-building corridors. The extra 10 dB per decade of range is the price of that obstruction.

Question 2 — final results
Quantity	Result
Transmitter power	40 dBm
Noise floor over 10 MHz	−104 dBm
Receiver sensitivity (with 6 dB fade margin)	−92 dBm
(a) Maximum allowed path loss	134 dB (130 dB if the 4 dB gain is a total)
(b) Path loss at $d=200$ m	129.1 dB
(b) Received power and SNR	−87.1 dBm, 16.9 dB
(b) Verdict	Criterion satisfied, 4.9 dB of headroom
(c) Path-loss exponent	$n = 3$

Topic. Decibel link budgeting: thermal noise floors, receiver sensitivity, fade margin, and empirical path-loss exponents.

Key relations. $P[\text{dBm}]=10\log_{10}(P/1\,\text{mW})$; noise floor $N = N_0 + 10\log_{10}B$ with $N_0 = kT_0 F \approx -174\ \text{dBm/Hz}$ at 290 K for $F=1$; sensitivity $P_{r,\min}=N+\mathrm{SNR}_{req}+M$; link equation $P_r = P_t + G_t + G_r - L_{rx} - L_{path}$; generalised path loss $L = 10\,n\log_{10}d + \text{const}$, with $n=2$ free space and $n \approx 3$ to 4 obstructed.

Why this works. The budget is a single conservation statement about power, written in logarithmic units so that the cascade of multiplicative gains and losses becomes a sum. Because both the wanted signal and the noise floor are expressed in dBm, their difference is directly the signal-to-noise ratio in dB, and the whole design question collapses to one subtraction. The $-174$ dBm/Hz figure is not arbitrary: $kT_0 = 1.38\times10^{-23}\times290 = 4.0\times10^{-21}$ W/Hz, which is $-174$ dBm/Hz, and it is the floor any receiver at room temperature must live above. The fade margin is held separately from the demodulator's requirement because they answer different questions — one is a property of the modulation and coding, the other a statistical allowance for the channel.

Common pitfall. The dominant error here is substituting the familiar Friis $20\log_{10}$ expression for the $30\log_{10}$ law the paper prescribes, which understates the loss by 43 dB at 200 m and inverts nothing about the verdict but everything about the margin. A close second is mixing units in the noise floor, either by leaving $N_0$ in dBW/Hz or by multiplying rather than adding after converting to dB. A third is treating "antenna gains" as a single end; state the reading explicitly and carry both if the verdict could turn on it.

Source. Rappaport, Wireless Communications: Principles and Practice, 2nd ed., ch. 4 (free-space and log-distance path-loss models, path-loss exponent tables); Proakis and Salehi, Communication Systems Engineering, 2nd ed., ch. 5 (noise figure, noise temperature and link budgets).

Question 2 of 5: Link budgeting

Question 2: Link budgeting (25 marks)