Channel Capacity

• We measure data rate in bits per second (bps)
• Bandwidth, as before, the difference in frequencies.
  • Stallings seems to drop down to bandwidth means primary frequency
• Rule of thumb: The greater the bandwidth, the more expensive the system.
• But we know the greater the bandwidth the greater the bps.
• Nyquist models maximum bandwidth
  • C = 2*B*log₂M
  • B is the bandwidth in Hz.
  • M is the number of voltage levels being detected.
  • For us, M has been high and low, but it can be more if we wish.
  • This is really voltage levels.
• Example: What is the data rate for a system with a bandwidth of 4kHz that uses 8 levels of signaling.

  C = 2*B*log2M
     M = 8
     B = 4x103
  C = 2 * 4x103 * lg (8)
    = 24 kbps
     

• Notice for 2 levels, Nyquist's is what we have been using.
• the problem with M>2 is distinguishing between signals.
• A Second formula, Shannon's, relates data rate to noise.
• SNR - signal to noise ratio
• SNR_dB = 10 log₁₀(signal power)/(noise power)
• SNR_dB = 10 log SNR
• Shannon's states: C = B lg(1+SNR), where B is the bandwidth in Hz.
• Stallings does an exmaple:

  Suppose that the spectrum of a channel is between 3MHz and 4MHz.
  and SNRdB = 24dB.
     B = 4MHz - 3MHz = 1MHz
     24dB = 10 log (SNR)
         2.4dB = log (SNR)
         102.4 = SNR
         251 = SNR
  
     Shannon's  Tells us that 
         C = 1MHz*lg(251+1) = 8x106bps
  
     But by Nyquest's we have
         8x106 =2* 1x106lg(M)
         4 = lg(M)
         M = 16
  
      So we would need to use 16 levels to achieve this data rate.