Link budgeting: 22-Elec-B3 Digital Communications Systems, December 2015, Q1

Question 1: 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 single free-space hop whose only gains and losses are the two antennas, a lumped receiver loss and the path loss itself.

Given data
Quantity	Symbol	Value
Transmitter power	$P_{t}$	20 W
Antenna gain, each end	$G_{t}=G_{r}$	6 dB
Receiver losses	$L_{rx}$	9 dB
Receiver noise density	$N_{0}$	$-174$ dBm/Hz
Noise bandwidth	$B$	20 MHz
Fading margin	$M$	6 dB
Required signal-to-noise ratio	$\mathrm{SNR}_{req}$	6 dB
Carrier frequency (part b)	$f$	1.5 GHz
Range (part b)	$d$	200 m
Speed of light	$c$	$3.0\times10^{8}$ m/s

Find. (a) the largest path loss the link can absorb and still meet the signal-to-noise requirement with its fading margin intact, (b) whether the free-space law as printed satisfies that requirement at 200 m, and (c) 20 dBm expressed in watts.

Figure 1.1 — the link budget as a ladder of decibel levels. Every term in the budget is one rung; the received level must finish above the required level, which already contains the fading margin.

Approach. Convert every quantity to a decibel-referenced level, build the thermal noise power in the stated bandwidth, add the required signal-to-noise ratio and the fading margin to obtain the minimum usable receive level, and then solve the one-line budget equation for the path loss — first as an allowance in part (a), then as the actual free-space value in part (b).

Check: two readings of the question statement, both stated explicitly.

First, the paper calls $-174$ dBm/Hz a “receiver noise figure”. A noise figure is a dimensionless ratio and would be quoted in plain dB; a quantity in dBm per hertz is a noise power spectral density. The value is also exactly the thermal floor $kT = -174$ dBm/Hz at $T\approx 290$ K, so it is treated here as $N_{0}$ and integrated over the 20 MHz bandwidth. Reading it as a 174 dB noise figure is physically impossible.

Second, “antenna gains of 6 dB” is plural with a single value, which is taken to mean 6 dB at each end, so $G_{t}+G_{r}=12$ dB. If instead the 6 dB is the combined figure, the allowance in (a) becomes 129.0 dB and the achieved ratio in (b) becomes 18.02 dB — still above the 12 dB the link needs, so the pass/fail verdict in part (b) is the same under either reading. Both numbers are given below.

Part (a) — put the transmit power on the dBm scale. Power in dBm is referred to one milliwatt, $P[\text{dBm}] = 10\log_{10}(P/1\ \text{mW})$, so with $P_{t}=20\ \text{W}=20{,}000\ \text{mW}$, $$P_{t} = 10\log_{10}(20{,}000) = 43.01\ \text{dBm}.$$ The same figure is $13.01$ dBW, which is a useful check: adding 30 dB converts dBW to dBm.
Build the noise power in the channel bandwidth. A flat noise density integrated over $B$ hertz gives $N = N_{0} + 10\log_{10}B$. With $B = 20$ MHz, $10\log_{10}(20\times10^{6}) = 73.01\ \text{dB-Hz}$, so $$N = -174 + 73.01 = -100.99\ \text{dBm}.$$ In absolute terms that is $7.96\times10^{-14}\ \text{W}$, about 80 femtowatts — the reason a receive level of $-77$ dBm is comfortable rather than marginal.
Convert the requirement into a minimum receive level. The detector needs $\mathrm{SNR}_{req}$ above the noise floor, and the fading margin $M$ must sit on top of it so that a fade of up to 6 dB still leaves the ratio intact. Hence $$P_{r,\min} = N + \mathrm{SNR}_{req} + M = -100.99 + 6 + 6 = -88.99\ \text{dBm}.$$ Note the margin is inside this level, not an extra subtraction later.
Solve the budget for the path-loss allowance. The budget in decibels is $P_{r} = P_{t} + G_{t} + G_{r} - L_{rx} - L_{path}$. Setting $P_{r}=P_{r,\min}$ and rearranging, $$L_{path,\max} = P_{t} + G_{t} + G_{r} - L_{rx} - P_{r,\min} = 43.01 + 12 - 9 - (-88.99),$$ which gives the allowance $$\boxed{L_{path,\max} = 135.0\ \text{dB}}$$ (129.0 dB if the 6 dB is read as the total antenna gain).
Part (b) — evaluate the path-loss law printed in the question at 200 m. The argument is dimensionless: $$\frac{4\pi d f}{c} = \frac{4\pi(200)(1.5\times10^{9})}{3.0\times10^{8}} = 4\pi(1000) = 12{,}566.37 .$$ The ratio $df/c$ is exactly 1000 here, which makes the arithmetic easy to check by hand. Applying the coefficient the paper gives, $$L_{path} = 30\log_{10}(12{,}566.37) = \boxed{122.98\ \text{dB}}.$$
Received power and the achieved ratio. Substituting back into the same budget, $$P_{r} = 43.01 + 12 - 9 - 122.98 = -76.97\ \text{dBm},$$ and subtracting the noise floor found in step 2 gives $$\mathrm{SNR} = P_{r} - N = -76.97 - (-100.99) = \boxed{24.02\ \text{dB}}.$$
Compare with the criterion and answer the question. The link must deliver the 6 dB the detector needs plus the 6 dB fading margin, i.e. 12 dB. Since $24.02\ \text{dB}\ge 12\ \text{dB}$, the signal-to-noise criterion is satisfied at 200 m, with $24.02-12 = 12.02$ dB to spare. The same conclusion follows directly from part (a): the actual loss of 122.98 dB is 12.02 dB below the 135.0 dB allowance, and the two routes must agree because they are the same equation rearranged. Under the alternative 6 dB total-gain reading the ratio is 18.02 dB, still 6.02 dB clear, so the verdict does not change. Solving $30\log_{10}(4\pi d f/c)=135.0$ shows the budget is exhausted only at $d = 503$ m.
Part (c) — invert the dBm definition. $P[\text{mW}] = 10^{P[\text{dBm}]/10}$, so $$P = 10^{20/10} = 100\ \text{mW} = \boxed{0.100\ \text{W}}.$$ The anchors are worth remembering for an exam without a calculator: 0 dBm is 1 mW, every $+10$ dB is one decade and every $+3$ dB roughly doubles the power, so $+20$ dBm is two decades above 1 mW.

Final results
Quantity	Result
Transmit power, $P_{t}$	43.01 dBm
Noise floor in 20 MHz, $N$	$-100.99$ dBm
Minimum receive level, $P_{r,\min}$	$-88.99$ dBm
(a) Maximum allowed path loss	135.0 dB (129.0 dB on the 6 dB total-gain reading)
Free-space loss at 200 m, 1.5 GHz	122.98 dB
Received power at 200 m	$-76.97$ dBm
(b) Achieved signal-to-noise ratio	24.02 dB — criterion satisfied, 12.02 dB spare
Range at which the budget is exhausted	503 m
(c) 20 dBm expressed in watts	0.100 W

Topic. Decibel link budgets and the thermal noise floor.

Key relations. Budget in decibels, $P_{r}=P_{t}+G_{t}+G_{r}-L_{rx}-L_{path}$; noise power from a density, $N = N_{0}+10\log_{10}B$; thermal density $N_{0}=kT$, which is $-174$ dBm/Hz at 290 K; free-space loss $L_{fs}=20\log_{10}(4\pi d/\lambda)$; the dBm definition $P[\text{dBm}]=10\log_{10}(P/1\ \text{mW})$.

Why this works. Every multiplicative effect in a radio path — amplification, aperture gain, spreading loss, feeder loss — becomes an addition once expressed in decibels, so a chain of a dozen effects collapses to a single sum that can be checked term by term. The noise side works the same way: because thermal noise is white over any practical channel, its power is the density times the bandwidth, which is an addition in decibels. The signal-to-noise ratio is then a difference of two levels, and a design requirement stated as a ratio becomes a requirement on that difference. Writing the fading margin into the required receive level rather than carrying it separately keeps the comparison a single inequality.

Common pitfall. Two errors recur. The first is double-counting the fading margin — once inside the required level and again as a subtraction from the received level — which makes the link look 6 dB worse than it is. The second is using the textbook $20\log_{10}$ free-space coefficient when a paper prints $30\log_{10}$: at 200 m and 1.5 GHz the two differ by 40.99 dB (81.98 dB against 122.98 dB), enough to reverse the verdict, so the printed law governs. A third is treating a density in dBm/Hz as a noise figure and never integrating it over the bandwidth.

Source. Rappaport, Wireless Communications: Principles and Practice, 2nd ed., §3.2 (free-space propagation) and §4.2; Proakis and Salehi, Communication Systems Engineering, 2nd ed., §5.1 (noise in receivers).

Question 1 of 5: Link budgeting

Question 1: Link budgeting (25 marks)