Link Budgeting: 22-Elec-B3 Digital Communications Systems, December 2014, 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	$P_t$	500 mW
Antenna gain, each end	$G_t = G_r$	9 dB
Receiver losses	$L_{rx}$	6 dB
Receiver noise power spectral density	$N_0$	-168 dBm/Hz
Noise bandwidth	$B$	10 MHz
Fading margin required	$M$	6 dB
Minimum signal-to-noise ratio	$\mathrm{SNR}_{req}$	2 dB
Carrier frequency (part b)	$f$	1.5 GHz
Link distance (part b)	$d$	200 m
Speed of light	$c$	$3.0 \times 10^8$ m/s

Find. (a) the largest path loss $L_p$ the link can tolerate while still delivering the required signal-to-noise ratio with its fading margin intact; (b) whether the free-space loss at 200 m stays inside that allowance; (c) the linear power corresponding to 20 dBm.

Figure 1.1 — the link budget as a chain of decibel gains and losses from transmitter output to receiver detector. The 6 dB fading margin is not drawn: it is an extra reserve subtracted from the allowance so that a fade does not push the received power below the detection threshold.

Approach. Convert every quantity to decibel form, compute the receiver noise floor from $N_0$ and $B$, add the required signal-to-noise ratio and the fading margin to obtain the minimum acceptable received power, and solve the one-line power balance for the path loss; then evaluate the given free-space formula at 200 m and compare.

Part (a) — put the transmitter power on the decibel scale. Power referenced to one milliwatt is $$P_t[\text{dBm}] = 10 \log_{10}\!\left(\frac{P_t}{1\ \text{mW}}\right) = 10 \log_{10}(500) = 26.99\ \text{dBm}.$$ Working the whole budget in dBm and dB turns every multiplication in the link equation into an addition.
Part (a) — compute the receiver noise floor. The quantity quoted as the “receiver noise figure of -168 dBm/Hz” carries units of power per hertz, so it is the receiver’s equivalent input noise power spectral density $N_0$. Integrating it over the noise bandwidth, $$N[\text{dBm}] = N_0[\text{dBm/Hz}] + 10\log_{10}(B) = -168 + 10\log_{10}(10 \times 10^{6}) = -168 + 70.00 = -98.00\ \text{dBm}.$$ The bandwidth contributes $+70$ dB-Hz because 10 MHz is $10^7$ Hz.
Part (a) — set the minimum acceptable received power. The detector needs 2 dB of signal-to-noise ratio, and a further 6 dB must be held in reserve so that a fade does not break the link. Both are added to the noise floor: $$P_{r,\min} = N + \mathrm{SNR}_{req} + M = -98.00 + 2 + 6 = -90.00\ \text{dBm}.$$
Part (a) — solve the power balance for the path loss. With no gains or losses other than the two antennas, the receiver losses and the path loss, the received power is $$P_r = P_t + G_t + G_r - L_{rx} - L_p .$$ Setting $P_r = P_{r,\min}$ and rearranging, $$L_{p,\max} = P_t + G_t + G_r - L_{rx} - P_{r,\min} = 26.99 + 9 + 9 - 6 - (-90.00),$$ $$\boxed{L_{p,\max} = 129.0\ \text{dB}}$$ Any path loss up to about 129 dB still leaves the receiver with 2 dB of signal-to-noise ratio after a 6 dB fade.
Part (b) — evaluate the argument of the given loss formula. The paper supplies $L_p = 30\log_{10}(4\pi d f / c)$, so first form the dimensionless ratio $$\frac{4\pi d f}{c} = \frac{4\pi (200)(1.5\times 10^{9})}{3.0 \times 10^{8}} = 4\pi(1000) = 12\,566.37 .$$ The metres, hertz and metres-per-second cancel, as they must before a logarithm is taken.
Part (b) — compute the free-space path loss at 200 m. Substituting, $$L_p = 30\log_{10}(12\,566.37) = 30 (4.0992) = \boxed{122.98\ \text{dB}}$$ Note that this uses the paper’s own coefficient of 30 rather than the textbook 20; the exam has deliberately made the loss grow as $d^{1.5}$ to represent a channel worse than pure free space, and the question says to use what it gives.
Part (b) — compare against the allowance and confirm the criterion. Since $122.98\ \text{dB} < 129.0\ \text{dB}$, the loss is inside the budget. Carrying the numbers through explicitly, the received power is $$P_r = 26.99 + 9 + 9 - 6 - 122.98 = -83.99\ \text{dBm},$$ so the delivered signal-to-noise ratio is $$\mathrm{SNR} = P_r - N = -83.99 - (-98.00) = \boxed{14.01\ \text{dB}}$$ That comfortably exceeds the 2 dB requirement plus the 6 dB fading reserve, a combined 8 dB. The signal-to-noise criterion is satisfied at $d = 200$ m, with $14.01 - 8 = 6.01$ dB of surplus still in hand — equivalently, the link would still close after a fade of $6 + 6.01 = 12.01$ dB.
Part (c) — invert the dBm definition. Reversing the definition used in step 1, $$P = 1\ \text{mW} \times 10^{P[\text{dBm}]/10} = 10^{20/10} = 100\ \text{mW},$$ $$\boxed{P = 0.100\ \text{W}}$$ The useful mental anchors are 0 dBm = 1 mW, 10 dBm = 10 mW, 20 dBm = 100 mW — each 10 dB is one decade of power.

Quantity	Result
Transmitter power in dBm	26.99 dBm
Receiver noise floor over 10 MHz	-98.00 dBm
Minimum acceptable received power	-90.00 dBm
(a) Maximum allowed path loss	129.0 dB
(b) Free-space path loss at $d = 200$ m	122.98 dB
(b) Received power at 200 m	-83.99 dBm
(b) Delivered signal-to-noise ratio	14.01 dB
(b) Criterion satisfied?	Yes — 6.01 dB of surplus beyond the 2 dB requirement and 6 dB margin
(c) 20 dBm expressed in watts	0.100 W

Check — two readings of “antenna gains of 9 dB”. The plural “gains” with a single value is read here as 9 dB at each end, giving $G_t + G_r = 18$ dB, which is the usual convention in a link budget and the reading the phrase most naturally supports. If instead the 9 dB is meant as the combined gain of the pair, every answer in part (a) shifts down by exactly 9 dB and the maximum allowed path loss becomes 120.0 dB. The conclusion of part (b) is unchanged either way: 122.98 dB would then exceed the allowance by 2.98 dB and the criterion would fail. State whichever reading you adopt, as Note 1 on the cover page invites.

Topic. The decibel link budget: accounting for every gain and loss between a transmitter’s output and a receiver’s detector, and converting a required signal-to-noise ratio into an allowable propagation loss.

Key relations. Received power (Friis, in decibels) $P_r = P_t + G_t + G_r - L_p - L_{\text{other}}$; thermal noise power $N = N_0 B$, or in decibels $N[\text{dBm}] = N_0[\text{dBm/Hz}] + 10\log_{10} B$; signal-to-noise ratio $\mathrm{SNR} = P_r - N$; decibel-milliwatt definition $P[\text{dBm}] = 10\log_{10}(P/1\ \text{mW})$; free-space path loss $L_p = 20\log_{10}(4\pi d f/c)$ in the ideal case, here modified by the paper to a coefficient of 30.

Why this works. The whole point of the decibel scale is that a cascade of multiplicative gains and attenuations becomes a single sum, so a link budget is arithmetic rather than algebra. Thermal noise is white over any practical receiver bandwidth, so its total power is simply the spectral density multiplied by the noise bandwidth — which on the decibel scale is an addition of $10\log_{10}B$. The required signal-to-noise ratio fixes how far above that noise floor the signal must land, and the fading margin is an additional reserve held back so that the inevitable multipath and shadowing fades do not drive the instantaneous received power below the detection threshold. Once the minimum acceptable received power is fixed, the link equation contains exactly one unknown, and the maximum tolerable path loss follows directly. Part (b) then closes the loop by asking whether the propagation physics at the intended range actually fits inside that allowance.

Common pitfall. Two errors dominate. First, treating a noise spectral density in dBm/Hz as though it were already a noise power, which omits the $+70$ dB-Hz bandwidth term and understates the noise floor by seven orders of magnitude. Second, double-counting or omitting the fading margin: it is a reserve to be subtracted from the loss allowance, not a gain, and it is not part of the receiver’s demodulation requirement. A third, subtler slip is silently replacing the paper’s $30\log_{10}$ coefficient with the familiar $20\log_{10}$ — that would give 82 dB instead of 123 dB and change the answer to part (b) by 41 dB.

Source. Haykin, Communication Systems, 5th ed., Ch. 9 (noise in communication systems) and Ch. 6; Lathi & Ding, Modern Digital and Analog Communication Systems, 4th ed., §11.2 (thermal noise, noise figure and equivalent noise temperature).

Question 1 of 5: Link Budgeting

Question 1: Link Budgeting (25 marks)