Link Budgeting: 22-Elec-B3 Digital Communications Systems, December 2019, Q3

Question 3: 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.

Parameter	Symbol	Value
Transmit power	$P_t$	8 W
Antenna gains (one value, both ends)	$G_t,\ G_r$	3 dB each
Receiver losses	$L_{rx}$	6 dB
Receiver noise power spectral density	$N_0$	−174 dBm/Hz
Noise bandwidth	$B$	10 MHz
Required signal-to-noise ratio	$\mathrm{SNR}_{\text{req}}$	3 dB
Fading margin	$M$	3 dB
Path-loss model / carrier	—	$30\log_{10}(4\pi d f/c)$, $f = 5$ GHz, $c = 3.0\times10^8$ m/s

Find. (a) the maximum path loss the link can tolerate while still meeting the SNR requirement with the stated margin; (b) the range at which that loss is reached; (c) the path-loss exponent implied by the model; (d) −30 dBm converted to watts.

Check: two readings of the source, both answered. The paper prints “antenna gains of 3 dB” — plural noun, single value. The natural engineering reading, and the one used for the boxed answers below, is 3 dB at each end, i.e. $G_t + G_r = 6$ dB of antenna gain in the budget. A reader who instead takes the 3 dB as the combined figure obtains a maximum path loss of 134.03 dB and a range of 140.2 m; both variants are carried through the working so either marking scheme can be followed. Note also that the quantity labelled “receiver noise figure of −174 dBm/Hz” is dimensionally a noise power spectral density, not a noise figure — −174 dBm/Hz is exactly $kT_0$ at $T_0 = 290$ K — so it is integrated over the bandwidth rather than added as a dB penalty. Cover-page Note 1 invites exactly this kind of stated assumption.

Approach. Convert everything to dB units referred to 1 mW, form the noise power in the 10 MHz bandwidth, add the SNR requirement and the fading margin to obtain the minimum acceptable received power, and take the difference between transmitted and required received power after gains and losses. Then invert the given path-loss law for distance, read the exponent off its coefficient, and convert dBm to watts from the definition.

Part (a) — put the transmit power in dBm. The dBm scale is referred to 1 mW, so $P_t = 8$ W $= 8000$ mW gives $$P_t\,[\text{dBm}] = 10\log_{10}(8000) = \boxed{39.03\ \text{dBm}}$$ A useful mental check: 1 W is $+30$ dBm and a factor of 8 is $3 \times 3 = 9$ dB, so 8 W is about 39 dBm.
Compute the receiver noise floor. The noise power in a bandwidth $B$ follows from the power spectral density by integration, which on a logarithmic scale is an addition: $$N\,[\text{dBm}] = N_0\,[\text{dBm/Hz}] + 10\log_{10} B = -174 + 10\log_{10}(10\times10^{6}) = -174 + 70 = -104\ \text{dBm}.$$ This is the thermal noise floor of a 10 MHz channel at room temperature, and it is the reference against which the SNR requirement is measured.
Set the minimum acceptable received power. The receiver must sit at least $\mathrm{SNR}_{\text{req}}$ above the noise floor, and the fading margin is extra headroom held in reserve for the deepest fades the link must survive: $$P_{rx,\min} = N + \mathrm{SNR}_{\text{req}} + M = -104 + 3 + 3 = -98\ \text{dBm}.$$
Close the budget and solve for the allowed path loss. Following the signal from the transmitter to the detector, every gain adds and every loss subtracts: $$P_{rx} = P_t + G_t + G_r - L_{rx} - L_{\text{path}} \;\ge\; P_{rx,\min}.$$ Rearranging for the largest loss the link can absorb and substituting the primary reading $G_t = G_r = 3$ dB, $$L_{\text{path,max}} = P_t + G_t + G_r - L_{rx} - P_{rx,\min} = 39.03 + 3 + 3 - 6 - (-98)$$ $$\boxed{L_{\text{path,max}} = 137.03\ \text{dB}}$$ Under the alternative reading (3 dB of antenna gain in total) the same arithmetic gives 134.03 dB.
Part (b) — invert the path-loss law for distance. Setting the model equal to the allowance, $$30\log_{10}\!\left(\frac{4\pi d f}{c}\right) = 137.03 \quad\Longrightarrow\quad \frac{4\pi d f}{c} = 10^{137.03/30} = 10^{4.5677} = 3.696\times10^{4}.$$ The frequency-dependent factor is $$\frac{4\pi f}{c} = \frac{4\pi (5\times10^{9})}{3.0\times10^{8}} = 209.44\ \text{m}^{-1},$$ so the maximum range is $$d_{\max} = \frac{3.696\times10^{4}}{209.44} = \boxed{176.5\ \text{m}}$$ Substituting back, $30\log_{10}(209.44 \times 176.46) = 137.03$ dB, which closes the loop. Under the 3-dB-total reading the range falls to 140.2 m — a 20 % reduction from a 3 dB change, which is the practical point of the exercise: at an exponent of 3 the range is only weakly sensitive to link-budget errors.
Part (c) — read the path-loss exponent from the coefficient. The general empirical path-loss law is written $$L_{\text{path}}\,[\text{dB}] = 10\,n\log_{10} d + \text{constant},$$ where $n$ is the path-loss exponent. Expanding the given expression separates the distance dependence from the frequency dependence: $$30\log_{10}\!\left(\frac{4\pi d f}{c}\right) = 30\log_{10} d + 30\log_{10}\!\left(\frac{4\pi f}{c}\right),$$ so matching coefficients gives $10n = 30$ and $$\boxed{n = 3}$$ Equivalently, doubling the range costs $10n\log_{10} 2 = 9.03$ dB, against 6.02 dB for free space. An exponent of 3 sits between ideal free space ($n=2$) and a cluttered urban or indoor environment ($n \approx 3.5$–4), and describes a link with partial obstruction and ground reflection.
Part (d) — convert −30 dBm to watts. By the definition of the dBm scale, $P\,[\text{dBm}] = 10\log_{10}(P/1\,\text{mW})$, so inverting, $$P = 10^{-30/10}\ \text{mW} = 10^{-3}\ \text{mW} = 10^{-3} \times 10^{-3}\ \text{W}$$ $$\boxed{P = 1\times10^{-6}\ \text{W} = 1\ \mu\text{W}}$$ The anchors make this a mental calculation: 0 dBm is 1 mW and every $-10$ dB is one decade, so $-30$ dBm is three decades below 1 mW, i.e. 1 µW.

Quantity	Result (3 dB per antenna)	Alternative (3 dB total)
Transmit power in dBm	39.03 dBm	39.03 dBm
Noise power in 10 MHz	−104 dBm	−104 dBm
Minimum received power	−98 dBm	−98 dBm
(a) Maximum allowed path loss	137.03 dB	134.03 dB
(b) Maximum range at 5 GHz	176.5 m	140.2 m
(c) Path-loss exponent	$n = 3$
(d) −30 dBm in watts	$1\times10^{-6}$ W (1 µW)

Topic. Wireless link power budgets: dB arithmetic, thermal noise floors, fade margins 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 $N_0 = kT_0 = -174$ dBm/Hz at 290 K; link equation $P_{rx} = P_t + G_t + G_r - L_{rx} - L_{\text{path}}$; required receive power $P_{rx,\min} = N + \mathrm{SNR}_{\text{req}} + M$; path loss $L = 10n\log_{10}d + \text{const}$, with free space $n = 2$ giving the Friis form $20\log_{10}(4\pi d f/c)$.

Why this works. A link budget is a single conservation statement written on a logarithmic scale, where cascaded multiplicative gains and losses become additions. Its two ends are fixed by physics rather than design: the transmitted power at one end, and at the other the thermal noise floor $kTB$, below which no amount of receiver engineering can see. Everything in between — antenna gain, feed and implementation losses, propagation loss — is bookkeeping between those two limits. The fade margin is the only genuinely statistical entry: it converts an instantaneous power balance into one that holds for a specified fraction of time in a fading channel, which is why it is added to the requirement rather than subtracted from the allowance.

Common pitfall. Treating a figure quoted in dBm/Hz as a noise figure and adding it as a dB penalty, which is dimensionally wrong and produces a nonsensical budget; it is a spectral density and must be integrated over the bandwidth. Equally common is applying a single antenna gain when the paper's plural implies one at each end, and using the free-space exponent 2 when the question has explicitly printed a coefficient of 30, not 20 — at 200 m the difference between the two models exceeds 40 dB.

Source. Sklar, Digital Communications, 2nd ed., Ch. 5 (Communications Link Analysis — link budget, noise figure and effective noise temperature); Lathi & Ding, Modern Digital and Analog Communication Systems, 4th ed., Ch. 12.

Question 3 of 5: Link Budgeting

Question 3: Link Budgeting (25 marks)