Link Budgeting: 22-Elec-B3 Digital Communications Systems, May 2018, 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.

Quantity	Symbol	Value
Transmitter power	$P_t$	8 W
Antenna gain, each end	$G_t = G_r$	9 dB
Receiver losses	$L_{rx}$	6 dB
Receiver noise power spectral density	$N_0$	−174 dBm/Hz
Noise bandwidth	$B$	10 MHz
Fading margin requirement	$M$	6 dB
Required signal-to-noise ratio	$\mathrm{SNR}_{req}$	10 dB
Carrier frequency (part b)	$f$	2.4 GHz
Range (part b)	$d$	200 m
Speed of light	$c$	$3.0\times10^{8}$ m/s

Find. (a) the largest path loss $L_p$ the link can absorb and still deliver 10 dB of SNR with 6 dB of fading margin; (b) whether that budget is met at a 200 m free-space range; (c) the path-loss exponent implied by the model used in (b).

Figure 1.1 — The link as a chain of dB terms. Everything to the left of the receiver adds power; everything to the right of the path removes it. The maximum allowable path loss is the gap between the two ends.

Approach. Put every quantity into dBm/dB, build the receiver noise floor from the noise density and the bandwidth, add the required SNR and the fading margin to obtain the minimum acceptable received power, and take the maximum path loss as whatever is left over in the link equation.

Check — two readings of the question, stated as assumptions (cover-page Note 1).

“Receiver noise figure of −174 dBm/Hz” is dimensionally a noise power spectral density, not a noise figure (a noise figure is a unitless dB ratio). The value quoted is exactly the thermal noise density $kT_0$ at $T_0 = 290$ K, so it is treated here as $N_0$ and integrated over the bandwidth. Treating it as a 0 dB noise figure on a room-temperature front end gives the same noise floor, so the reading is not in dispute.
“Antenna gains of 9 dB” (plural noun, single value) is read as 9 dB at each end, i.e. $G_t + G_r = 18$ dB total. The alternative reading (9 dB total) lowers every answer below by exactly 9 dB — the maximum path loss becomes 130.03 dB and the achieved SNR becomes 59.96 dB. The part (b) verdict is unchanged either way, which is the useful thing to note: the conclusion is not sensitive to the ambiguity.

Part (a) — Convert the transmitter power to dBm. The dBm scale is referenced to 1 mW, so $$P_t[\mathrm{dBm}] = 10\log_{10}\!\left(\frac{P_t}{1\ \mathrm{mW}}\right) = 10\log_{10}(8000) = 39.03\ \mathrm{dBm}.$$ A useful sanity anchor: 1 W is +30 dBm, and 8 W is three doublings above that, i.e. $30 + 3\times3.01 = 39.03$ dBm.
Build the receiver noise floor from the noise density and the bandwidth. Noise power is the density integrated over the noise bandwidth, which in dB is an addition: $$N = N_0 + 10\log_{10}B = -174 + 10\log_{10}(10\times10^{6}) = -174 + 70 = -104\ \mathrm{dBm}.$$ Every 10× of bandwidth costs 10 dB of noise, so a 10 MHz channel sits 70 dB above the per-hertz floor.
Set the minimum acceptable received power. The detector needs $\mathrm{SNR}_{req}$ above the noise floor, and the fading margin is extra signal held in reserve so that a fade of up to 6 dB does not break the link: $$P_{r,\min} = N + \mathrm{SNR}_{req} + M = -104 + 10 + 6 = -88\ \mathrm{dBm}.$$
Solve the link equation for the path loss. Collecting the fixed gains and losses, $$P_r = P_t + G_t + G_r - L_{rx} - L_p,$$ and setting $P_r = P_{r,\min}$ gives the largest loss the budget can absorb: $$L_{p,\max} = P_t + G_t + G_r - L_{rx} - P_{r,\min} = 39.03 + 9 + 9 - 6 - (-88)$$ $$\boxed{L_{p,\max} = 139.03\ \mathrm{dB}}$$ (With the 9 dB–total reading of the antenna gains this becomes 130.03 dB.)
Part (b) — Evaluate the free-space path loss at 200 m. Using the model the question supplies, $$L_p = 20\log_{10}\!\left(\frac{4\pi d f}{c}\right),\qquad \frac{4\pi d f}{c} = \frac{4\pi (200)(2.4\times10^{9})}{3.0\times10^{8}} = 2.0106\times10^{4},$$ $$L_p = 20\log_{10}(2.0106\times10^{4}) = 86.07\ \mathrm{dB}.$$ The same number follows from the wavelength form: $\lambda = c/f = 0.125$ m, so $4\pi d/\lambda = 4\pi(200)/0.125 = 2.0106\times10^{4}$.
Compare against the budget and quantify the surplus. Since $86.07\ \mathrm{dB} \ll 139.03\ \mathrm{dB}$, the link closes. Carrying the numbers through explicitly, the received power is $$P_r = 39.03 + 18 - 6 - 86.07 = -35.04\ \mathrm{dBm},$$ so the achieved signal-to-noise ratio is $$\mathrm{SNR} = P_r - N = -35.04 - (-104) = 68.96\ \mathrm{dB}.$$ Against the 10 dB requirement plus the 6 dB fading margin, $$\boxed{\mathrm{SNR} = 68.96\ \mathrm{dB} \gg 16\ \mathrm{dB\ required} \Rightarrow \text{criterion satisfied, with } 52.96\ \mathrm{dB\ to\ spare.}}$$ The surplus is enormous because 200 m at 2.4 GHz is a very short link; the same budget would still close at roughly 90 km of pure free space, which is exactly why real 2.4 GHz systems are limited by walls, foliage and interference rather than by free-space spreading.
Part (c) — Read the path-loss exponent off the model. A path-loss exponent $n$ is defined by $L_p \propto d^{\,n}$, i.e. $$L_p[\mathrm{dB}] = 10\,n \log_{10} d + \text{constant}.$$ The model given carries $d$ inside a $20\log_{10}(\cdot)$, so $L_p = 20\log_{10}d + \text{const}$, which matches $10n\log_{10}d$ only for $$\boxed{n = 2}$$ In one sentence: the distance appears once inside a $20\log_{10}$, and $20 = 10n$, so the exponent is 2 — the free-space value, as expected for a model derived from the inverse-square spreading of power over a sphere.

Result	Value
(a) Transmitter power in dBm	39.03 dBm
(a) Receiver noise floor, $N = N_0 + 10\log_{10}B$	−104 dBm
(a) Minimum acceptable received power	−88 dBm
(a) Maximum allowed path loss	139.03 dB (130.03 dB if 9 dB is the total antenna gain)
(b) Free-space path loss at d = 200 m, f = 2.4 GHz	86.07 dB
(b) Received power	−35.04 dBm
(b) Achieved SNR / verdict	68.96 dB — criterion satisfied (52.96 dB surplus)
(c) Path-loss exponent	n = 2

Topic. Decibel link-budget accounting and the free-space (Friis) path-loss model.

Key relations. $P[\mathrm{dBm}] = 10\log_{10}(P/1\ \mathrm{mW})$; noise floor $N = N_0 + 10\log_{10}B$ with $N_0 = kT$ ($kT_0 = -174$ dBm/Hz at 290 K); link equation $P_r = P_t + \sum G - \sum L - L_p$; Friis free-space loss $L_p = 20\log_{10}(4\pi d/\lambda) = 20\log_{10}(4\pi df/c)$; generalised loss $L_p = 10\,n\log_{10}d + C$.

Why this works. A link budget is nothing more than a conservation statement written in logarithms: power leaves the amplifier, is multiplied by a chain of gains and divided by a chain of losses, and must arrive above a threshold set by the receiver's own noise. Because multiplication becomes addition in dB, the whole chain collapses to one line of arithmetic and any single term can be isolated — here the path loss. The noise floor is fixed by physics rather than by design: a matched resistive source at temperature $T$ delivers $kT$ watts per hertz, which is $-174$ dBm/Hz at room temperature, and widening the receiver bandwidth admits proportionally more of it. The fading margin is not part of the physics; it is a design reserve held back so that the multipath and shadowing variations of a real channel, which the deterministic Friis model does not represent at all, do not push the instantaneous SNR below threshold.

Common pitfall. Two dominate. First, adding the fading margin to the path loss rather than to the required received power — it must make the budget harder, not easier, so it is subtracted from the allowance. Second, forgetting that the given −174 dBm/Hz is a density: candidates who use it directly as the noise power obtain a noise floor 70 dB too low and a path-loss allowance 70 dB too generous. A third, subtler one is mixing dB and dBm — gains and losses are dB (ratios), powers are dBm (absolute), and the two may only be added in the pattern dBm ± dB.

Source. B. Sklar, Digital Communications: Fundamentals and Applications, 2nd ed., Ch. 5 (Communications link analysis); S. Haykin, Communication Systems, 5th ed., §8.3 (noise figure and equivalent noise temperature).

Question 1 of 5: Link Budgeting

Question 1: Link Budgeting (25 marks)