Shannon capacity of a 3 kHz channel at 30 dB SNR
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A channel has a bandwidth of $3\ \text{kHz}$ and a signal-to-noise ratio of $30\ \text{dB}$.
Its Shannon capacity, in kbps, is ________.
Show the step-by-step derivation
Answer
29.9 kbps (accept 29.8 to 30.0) kbps (accept 29.8 to 30.0)
Step-by-step derivation
- Convert the SNR from decibels to a power ratio. Since SNR is a power quantity, $$30 = 10\log_{10}\!\left(\frac{S}{N}\right) \;\Longrightarrow\; \frac{S}{N} = 10^{3} = 1000.$$
- Apply the Shannon-Hartley theorem: $$C = B\log_2\!\left(1 + \frac{S}{N}\right).$$
- Substitute the values: $$C = 3000 \times \log_2(1 + 1000) = 3000 \times \log_2(1001).$$
- Evaluate the logarithm by change of base: $$\log_2(1001) = \frac{\ln 1001}{\ln 2} = \frac{6.9088}{0.6931} = 9.967.$$
- Multiply out: $$C = 3000 \times 9.967 = 29\,901\ \text{bits/s} \approx \mathbf{29.9\ kbps}.$$
- The rule of thumb. Every $3$ dB of SNR buys roughly one extra bit per second per hertz, because $\log_2$ of a doubling is $1$. So capacity grows only logarithmically with power but linearly with bandwidth - which is why modern systems chase spectrum rather than transmit power.
The trap this question is built around
Using the dB value $30$ directly inside the logarithm. $\log_2(1+30) = 4.95$ gives $14.9$ kbps, half the true capacity. Shannon's formula takes a linear power ratio, never decibels.
The idea behind this question
Shannon's theorem gives the highest error-free bit rate a channel can carry: $C = B\log_2(1 + S/N)$. Capacity grows in proportion to bandwidth but only logarithmically with signal-to-noise ratio - at high SNR, doubling the signal power (3 dB) buys roughly one more bit per second per hertz.
Try a variation
What is the capacity of a $4$ kHz channel at $20$ dB SNR, in kbps?
Show the answer
Answer: $26.6$ kbps
$20$ dB is a ratio of $100$, and $4000 \times \log_2 101 = 26.6$ kbps.
Other mistakes to avoid
- Converting decibels with 20 log instead of 10 log. SNR is a power ratio.
- Using the natural logarithm instead of $\log_2$.