Calculate Shannon channel capacity from bandwidth and SNR in dB or as a ratio using C = B log2(1 + S/N), or find the bandwidth or SNR for a target rate.
Shannon Channel Capacity Formula
The calculator uses the Shannon–Hartley capacity for a band-limited AWGN channel with average received signal power S and noise power N measured over the same bandwidth. At rates below C, suitable coding can make error probability arbitrarily small as block length grows. This is an asymptotic theoretical limit, not a promise of zero errors or delivered throughput from finite hardware. See MIT’s channel-coding notes.
When the SNR is given in decibels, the calculator first converts it to a linear power ratio:
The same relationship gives the theoretical bandwidth or SNR boundary when a target rate is substituted for C. The bandwidth rearrangement holds the supplied S/N fixed. If changing bandwidth changes noise power and therefore S/N, a separate self-consistent power and noise model is needed:
Variables:
- C is the theoretical AWGN channel capacity in bits per second (bps); rates below C can have arbitrarily small error probability with suitable asymptotically long coding
- B is the channel bandwidth in hertz (Hz)
- S/N is average received signal power divided by noise power over bandwidth B, expressed as a linear power ratio rather than dB
- SNRdB is the same ratio expressed in decibels
- C/B is the spectral efficiency in bits per second per hertz (bits/s/Hz)
The solve-for selector switches between capacity, bandwidth, and SNR. Capacity mode uses bandwidth and power SNR in dB or linear form. Bandwidth mode uses target rate and fixed SNR; SNR mode uses target rate and bandwidth. The inverse results are Shannon boundary values, not recommended operating settings. In capacity mode, an optional actual information rate can be compared with the calculated capacity. Use the same channel, bandwidth, and rate definition for that comparison; the percentage is a ratio, not a predicted coding efficiency or a reliability test.
Spectral Efficiency and Illustrative AWGN Capacity Limits
The first table gives the theoretical AWGN spectral efficiency log2(1 + S/N). Multiply it by bandwidth in Hz for capacity in bps. The last column illustrates a 10 MHz bandwidth. It does not assign modulation schemes: constellation bits per symbol alone do not determine coded information bits/s/Hz, which also depend on symbol rate, coding, and waveform overhead.
| SNR (dB) | Linear S/N | Capacity (bits/s/Hz) | Example capacity at 10 MHz (Mbps) |
|---|---|---|---|
| -10 | 0.1 | 0.14 | 1.38 |
| 0 | 1 | 1.00 | 10.00 |
| 5 | 3.16 | 2.06 | 20.57 |
| 10 | 10 | 3.46 | 34.59 |
| 20 | 100 | 6.66 | 66.58 |
| 25 | 316 | 8.31 | 83.09 |
| 30 | 1,000 | 9.97 | 99.67 |
| 40 | 10,000 | 13.29 | 132.88 |
The second table uses illustrative bandwidth and SNR assumptions to calculate single-channel AWGN capacity and spectral efficiency. These are not measurements, typical service rates, or promises for named communication standards. This scalar model does not calculate MIMO capacity, which requires the spatial-channel gains and power allocation.
| Illustrative case | Assumed bandwidth | Assumed SNR | Calculated AWGN capacity | Spectral efficiency (bits/s/Hz) |
|---|---|---|---|---|
| Example A | 3.1 kHz | 37 dB | 38.1 kbps | 12.29 |
| Example B | 6 MHz | 40 dB | 79.7 Mbps | 13.29 |
| Example C | 20 MHz | 25 dB | 166.2 Mbps | 8.31 |
| Example D | 20 MHz | 20 dB | 133.2 Mbps | 6.66 |
| Example E | 100 MHz | 20 dB | 665.8 Mbps | 6.66 |
Example Problems
Example 1: solve for capacity. A wireless link has 10 MHz of bandwidth and an SNR of 20 dB. First convert the SNR to a linear ratio: 10(20/10) = 100. Then apply the formula: C = 10,000,000 * log2(1 + 100) = 10,000,000 * 6.658 = 66.58 Mbps. The spectral efficiency is 6.66 bits/s/Hz. This is an ideal AWGN capacity calculation; practical delivered rate must be evaluated for the chosen code, waveform, hardware, and error target.
Example 2: solve for required SNR. You need 50 Mbps through a 10 MHz channel. The required spectral efficiency is C/B = 50,000,000 / 10,000,000 = 5 bits/s/Hz. The minimum SNR is 25 - 1 = 31 as a linear ratio, which is 10 * log10(31) = 14.91 dB. This is the Shannon boundary. A practical operating SNR must be established from the chosen system and reliability target; the theorem supplies no universal dB margin.
FAQ
How does the Shannon limit relate to real-link rates?
The capacity is an asymptotic coding limit under the stated channel and power assumptions. Finite block length, target error probability, decoding constraints, modulation, pilots, guard intervals, and hardware affect a practical link; fading or interference may require a different channel model. There is no universal delivered-rate percentage, extra-bandwidth percentage, or SNR margin implied by the theorem. Use system-specific measurements or validated performance data for those quantities.
Does adding more bandwidth always increase capacity?
Not without limit. Noise power grows with bandwidth (N = N0 * B, where N0 is the noise power density), so if your signal power is fixed, widening the channel lowers the SNR at the same time. As B goes to infinity, capacity approaches a finite ceiling of about 1.44 * S/N0. This is the origin of the ultimate low-SNR limit: reliable communication requires an energy per bit to noise density ratio (Eb/N0) of at least -1.59 dB. Extra bandwidth helps most when SNR is high; at very low SNR you are power limited, not bandwidth limited.
What is the difference between the Shannon capacity and the Nyquist rate?
For an ideal noiseless real baseband channel, zero intersymbol interference pulse shaping permits a symbol rate of 2B symbols/s. With M distinct levels, each symbol represents log2(M) bits, giving the familiar rate expression 2B log2(M). Shannon–Hartley instead bounds reliable information rate under a band-limited AWGN model and an average-power constraint. These use different assumptions; neither expression by itself predicts the throughput of every real link.
