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. The largest path loss the link can tolerate and still deliver the required SNR with the stated margin, the range that path loss corresponds to at 2.4 GHz, the path-loss exponent implied by the given expression, and −30 dBm expressed in watts.

Approach. Convert every quantity to decibels referred to one milliwatt, build the required receive power upward from the thermal noise floor, and let the path loss absorb whatever is left over in the link equation; then invert the given path-loss formula for distance.

1. Put the transmit power in dBm. Decibel-milliwatts are referred to 1 mW, so a power $P$ in watts becomes $$P_t\,[\mathrm{dBm}] = 10\log_{10}\!\left(\frac{P_t}{1\ \text{mW}}\right) = 10\log_{10}(4000) = 36.02\ \text{dBm}.$$ Everything downstream is now an addition or a subtraction rather than a multiplication.
2. Find the thermal noise power in the receiver bandwidth. The figure quoted as “−174 dBm/Hz” is a noise power spectral density $N_0$, not a noise figure — a noise figure is a dimensionless ratio in dB, whereas this quantity carries units of dBm per hertz. It is exactly $kT$ at the standard reference temperature $T_0 = 290$ K. Integrating it over the noise bandwidth, $$N = N_0 + 10\log_{10} B = -174 + 10\log_{10}(10^{6}) = -174 + 60 = -114\ \text{dBm}.$$
3. Build the minimum acceptable receive power. The detector needs the signal to sit $\mathrm{SNR}_{req}$ above that noise floor, and the fading margin is extra headroom that must be present in the median case so that fades do not drop the link below threshold. Both are additive in dB: $$P_{r,\min} = N + \mathrm{SNR}_{req} + M = -114 + 10 + 6 = -98\ \text{dBm}.$$
4. Apply the link equation and solve for the path loss. Along the chain of Figure 1.1 the received power is $$P_r = P_t + G_t + G_r - L_{rx} - L .$$ Setting $P_r = P_{r,\min}$ and rearranging gives the largest loss the budget can absorb: $$L_{\max} = P_t + G_t + G_r - L_{rx} - P_{r,\min} = 36.02 + 6 + 6 - 3 - (-98)$$ $$\boxed{L_{\max} = 143.02\ \text{dB}}$$ Under the alternative reading of the gain statement (6 dB in total) the same arithmetic gives $L_{\max} = 137.02$ dB.
5. Part (b) — invert the path-loss expression for distance. The paper supplies $$L = 30\log_{10}\!\left(\frac{4\pi d f}{c}\right)\ \text{dB},$$ so setting $L = L_{\max}$ and solving, $$d_{\max} = \frac{c}{4\pi f}\,10^{\,L_{\max}/30}.$$ The leading factor is $\lambda/4\pi$ with $\lambda = c/f = 3.0\times10^{8}/2.4\times10^{9} = 0.125$ m, giving $\lambda/4\pi = 9.947\times10^{-3}$ m. The exponential factor is $10^{143.02/30} = 10^{4.7674} = 5.854\times10^{4}$. Multiplying, $$\boxed{d_{\max} = 9.947\times10^{-3} \times 5.854\times10^{4} = 582\ \text{m}}$$ Substituting 582 m back into the given expression returns 143.02 dB, which confirms the inversion. With the 6 dB-total reading the range falls to 367 m — a useful reminder of how sharply range depends on the budget when the exponent is large.
6. Part (c) — read off the path-loss exponent. The standard log-distance model is written $$L = 10\,n\log_{10} d + \text{constant},$$ where $n$ is the path-loss exponent. Expanding the given expression separates the distance dependence from the frequency dependence: $$L = 30\log_{10}\!\left(\frac{4\pi f}{c}\right) + 30\log_{10} d .$$ Matching coefficients, $10n = 30$, hence $$\boxed{n = 3}$$ Free space would give $n = 2$; the value 3 is a mildly obstructed outdoor or suburban channel, which is why the range in part (b) is a few hundred metres rather than several kilometres.
7. Part (d) — convert −30 dBm to watts. Inverting the definition of the dBm scale, $$P = 1\ \text{mW}\times 10^{\,-30/10} = 10^{-3}\times 10^{-3}\ \text{W}$$ $$\boxed{P = 1\times10^{-6}\ \text{W} = 1\ \mu\text{W}}$$ The useful mental anchor is that every 10 dB is one decade and 0 dBm is 1 mW, so −30 dBm is three decades below a milliwatt.

Final Results