Question 3 of 9: Channel Capacity — Required SNR

Question 3: Channel Capacity — Required SNR (20 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 signal-to-noise ratio (as a ratio, and in dB) needed for the Shannon–Hartley capacity to reach $500\text{ Mbps}$.

Approach. Invert the Shannon–Hartley capacity formula $C = B\log_2(1+\text{SNR})$ to solve for SNR given $C$ and $B$, then convert the ratio to dB.

1. Set up Shannon–Hartley and isolate SNR. $$C = B\log_2(1+\text{SNR}) \;\Rightarrow\; \text{SNR} = 2^{C/B} - 1$$
2. Substitute the given values. $$\frac{C}{B} = \frac{500\times10^{6}}{25\times10^{6}} = 20$$ $$\text{SNR} = 2^{20} - 1 = 1{,}048{,}576 - 1 = \boxed{1{,}048{,}575}$$
3. Convert to decibels. $$\text{SNR(dB)} = 10\log_{10}(\text{SNR}+1) = 10\log_{10}(2^{20}) = 10 \times 20\log_{10}(2) = 200 \times 0.30103$$ $$\text{SNR(dB)} = \boxed{60.21\text{ dB}}$$