Nyquist Bit Rate

Nyquist’s result gives the highest symbol rate a noiseless channel can support without intersymbol interference, based only on its bandwidth. Each transmitted symbol can encode \(\log_2 M\) bits when there are \(M\) distinct signal levels, so multiplying the maximum symbol rate \(2B\) by bits per symbol gives the bit rate \(C = 2B\log_2 M\). The key idea is that a limited bandwidth channel cannot change arbitrarily fast, so the faster you want to send data, the more carefully you must choose the number of voltage or phase levels.

Use this when analyzing an ideal or noiseless baseband channel and you know its bandwidth and number of discrete signal levels. It is common in digital communications and line coding problems where you want the theoretical maximum bit rate before noise or practical hardware limits are considered.

• Using the formula for a noisy channel instead of a noiseless one; Shannon’s formula is needed when noise is present
• Forgetting that \(M\) is the number of distinct signal levels, so \(\log_2 M\) is bits per symbol
• Mixing up bandwidth units and data rate units, such as plugging in kHz but interpreting the result without converting to bits per second