Question 1 of 5: Link Budgeting

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.

Find. (a) the largest path loss the link can tolerate and still deliver the required SNR with the fading margin intact; (b) whether the printed free-space model at 200 m stays inside that allowance; (c) the path-loss exponent implied by the model.

Approach. Put every quantity on the decibel scale, build the receiver noise floor from the noise density and the bandwidth, add the required SNR and the fading margin to get the minimum acceptable received power, then rearrange the link equation to release the path-loss allowance; compare the printed free-space model against that allowance at 200 m and read the exponent off the coefficient of the logarithm.

1. Part (a) — Express the transmitter power in dBm. The decibel-milliwatt scale is referenced to 1 mW, so $$P_t = 10\log_{10}\!\left(\frac{10\ \text{W}}{1\ \text{mW}}\right) = 10\log_{10}(10\,000) = 40\ \text{dBm}.$$ The anchors worth memorising are 0 dBm = 1 mW, +10 dB = one decade and +3 dB = a doubling; 10 W is four decades above 1 mW, hence 40 dBm.
2. Build the receiver noise floor. Thermal noise power is the density integrated over the noise bandwidth, which on the decibel scale is an addition: $$N = N_0 + 10\log_{10} B = -174 + 10\log_{10}(10\times10^{6}) = -174 + 70 = -104\ \text{dBm}.$$ Ten megahertz is 70 dB above 1 Hz, so a receiver of this bandwidth sits 70 dB above the −174 dBm/Hz floor.
3. Set the minimum acceptable received power. The demodulator needs $\mathrm{SNR}_{req}$ above the noise floor, and the fading margin is extra power held in reserve so that a fade of up to 6 dB still leaves the demodulator above threshold: $$P_{r,req} = N + \mathrm{SNR}_{req} + M = -104 + 6 + 6 = -92\ \text{dBm}.$$ The margin is not a loss in the chain; it is a floor raised deliberately.
4. Rearrange the link equation for the path-loss allowance. With only antenna gains, receiver losses and path loss present, $$P_r = P_t + G_t + G_r - L_{rx} - L_{path} \;\ge\; P_{r,req},$$ so that $$L_{path,\max} = P_t + G_t + G_r - L_{rx} - P_{r,req} = 40 + 6 + 6 - 9 - (-92)$$ $$\boxed{L_{path,\max} = 135\ \text{dB}}$$ Under the alternative reading of the antenna data (6 dB total) the allowance would be 129 dB.
5. Part (b) — Evaluate the argument of the printed loss model. The bracket is dimensionless, which is the quickest check that the substitution is right: $$\frac{4\pi d f}{c} = \frac{4\pi (200)(1.5\times10^{9})}{3.0\times10^{8}} = 4\pi(1000) = 12\,566.4.$$ Physically this is $2\pi d/\lambda$ with $\lambda = c/f = 0.20$ m, i.e. the range measured in radians of carrier phase.
6. Evaluate the free-space loss. Using the coefficient the paper prints — 30, not the textbook 20 — $$L_{fs} = 30\log_{10}(12\,566.4) = 30(4.0992)$$ $$\boxed{L_{fs} = 122.98\ \text{dB at } d = 200\ \text{m}}$$ Had the classical $20\log_{10}$ model been used the loss would be only 81.99 dB, so the printed model must be taken at face value.
7. Compare against the allowance and report the received SNR. Since $122.98\ \text{dB} < 135\ \text{dB}$, the link closes with 12.02 dB to spare. Carrying the numbers through the chain confirms it directly: $$P_r = 40 + 6 + 6 - 9 - 122.98 = -79.98\ \text{dBm},$$ $$\mathrm{SNR} = P_r - N = -79.98 - (-104) = 24.02\ \text{dB}.$$ The requirement is 6 dB of SNR plus a 6 dB fading margin, i.e. 12 dB, so the criterion is satisfied with 12.02 dB in hand. Under the 6-dB-total reading the received SNR is 18.02 dB and the criterion is still satisfied, by 6.02 dB.
8. Part (c) — Read the path-loss exponent off the coefficient. A path-loss exponent $n$ is defined by $L \propto d^{\,n}$, i.e. $L = 10\,n\log_{10} d + \text{constant}$; the printed model expands to $30\log_{10} d + 30\log_{10}(4\pi f/c)$, so matching coefficients gives $$10\,n = 30 \quad\Longrightarrow\quad \boxed{n = 3}$$ Equivalently, doubling the range costs $10n\log_{10}2 = 9.03$ dB here rather than the 6.02 dB of true free space — the model is a free-space form carrying a cubic decay law, of the kind fitted to obstructed or ground-reflected paths.