Physical Layer - Data Rate Limits
The channel capacity can be defined in two different ways:
- Data Rate:
- In bits per second.
- Rate at which data can be communicated.
- Bandwidth:
- The frequency width of the transmitted signal.
- In cycles per second or Hz.
- Constrained by the transmitter and medium.
Nyquist Bandwidth
If the bandwidth is $B$, then the highest signal transmission (baud) rate is $2B$:
- For a binary signal the data rate supported by $B$ Hz is $2B$ bps.
This can be increased by using $M$ signal states:
- Each state encodes multiple bits.
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With no noise the maximum data rate is:
\[2B\log_2(M)\]This is what QAM techniques use to transmit more bits per state.
To find the number of bits from the number of states $M$ you can then calculate:
\[\log_2(M)\]This is as we can encode the states in binary.
Shannon’s Capacity Formula
The amount of thermal noise present is measured by the ratio of signal power to noise power.
This is called the signal to noise ratio: $\frac SN$.
The ratio itself is not usually quoted but is represented is dB:
\[10\log_{10}(\frac SN)\]Therefore for $\frac SN=100$ then this 20 dB.
Typical analogue voice telephone usually gives 30dB of signal to nose.
Shannon’s Law
The maximum rate of a noisy channel with bandwidth $B$ and signal to noise ratio of $\frac SN$ is:
\[B\log_2(1+\frac SN)\]Shannon’s Law Example
For a channel bandwidth of 3.1 KHz and a signal to noise ratio of 30 dB ($\frac SN=1000$), the maximum bits per second is:
\[3100\log_2(1+1000)=30894\text{ bps}\]We can combine this with the Nyquist Bandwidth formula to find the maximum data rate of a noisy channel with multiple signal states.