Introduction / Context:
Shannon’s channel capacity quantifies the theoretical maximum data rate that can be achieved with arbitrarily small error over a band-limited, noisy channel. Broadcasters and communication engineers use this bound to gauge feasibility and to compare modulation/coding schemes against physical limits.
Given Data / Assumptions:
- Channel bandwidth B = 6 MHz.
- Signal-to-noise ratio SNR = 35 dB.
- Shannon capacity formula: C = B * log2(1 + SNR_linear).
- No constraints from implementation loss, roll-off, or spectral shaping are considered; this is an ideal bound.
Concept / Approach:
First convert the decibel SNR to a linear ratio. Then substitute into C = B log2(1 + SNR). Finally, convert the result to megabits per second for comparison with the options.
Step-by-Step Solution:
Convert SNR: 35 dB ⇒ SNR_linear = 10^(35/10) = 10^3.5 ≈ 3162.3.Compute the log term: log2(1 + 3162.3) ≈ log2(3163.3) = ln(3163.3) / ln(2) ≈ 8.058 / 0.693 ≈ 11.63 bits/s/Hz.Compute capacity: C = 6 × 10^6 × 11.63 ≈ 69.8 × 10^6 bps ≈ 69.8 Mbps.Round and match: closest option is 72 Mbps.
Verification / Alternative check:
Back-of-envelope: with SNR ≈ 3.16 × 10^3, the log2 term ~ 11.6; times 6 MHz gives just under 70 Mbps, consistent with the calculation.
Why Other Options Are Wrong:
36 Mbps: roughly half the bound; too low for 35 dB SNR.50 Mbps and 28 Mbps: also underestimate capacity.100 Mbps: exceeds the calculated bound for the given SNR and bandwidth.
Common Pitfalls:
Using 35 (not 10^(3.5)) directly in the formula; mixing natural log and base-2 log without conversion; forgetting to add 1 to SNR in the log term.
Final Answer:
72 Mbps
Discussion & Comments