For a channel with a bandwidth of 20 MHz and an S/N (signal-to-noise ratio) of 30 dB, which is closest to the channel capacity (the theoretical upper limit on the error-free transmission rate) given by Shannon's theorem?

Answer: B. 200 Mbit/s

By Shannon's theorem, channel capacity is C = B·log₂(1 + S/N), with S/N as a linear (power) ratio. 30 dB is a power ratio of 10³ = 1000, so C = 20×10⁶ × log₂(1001) ≈ 20×10⁶ × 9.97 ≈ 200 Mbit/s. Doubling the bandwidth at the same S/N doubles the capacity, but when S/N is large, doubling S/N (about +3 dB) adds only about 1 bit/s per Hz. Practical systems combine error-correcting codes and multilevel modulation to approach this limit.

• A. This takes 30 dB as a power ratio of 30, giving 20×10⁶ × log₂(31) ≈ 99 Mbit/s. dB values must be converted to linear ratios before use in the formula.
• B. Correct. 20×10⁶ × log₂(1001) ≈ 199 Mbit/s.
• C. This multiplies the bandwidth by the dB value directly (20 MHz × 30). Capacity is proportional to the logarithm of S/N.