Can we break the Shannon capacity?

Shannon capacity and nyquist.each places an upper limit on the bit rate ofchannel based on 2 differentapproach

• the difference between shannon and nyquist on channel capacity. each places an upper limit on the bit rate of a channel based on two different approaches. How are the two related

• Answer:

  I see you are doing homework from the 7th edition of the William Stallings "Data and Computer Communications" book, problem 3.18 on page 89. So now that you're busted :-) maybe I can help you out, haha. The upper limits of both Nyquist and Shannon are related to the noise. The Shannon limit gives a theoretical maximum limit based on the signal to noise ratio, while the Nyquist formula can be used to determine how many signal levels are required to achieve that limit based on how much bandwidth is available. But the higher the number of signal levels, the more precise (and expensive) the receiver circuitry becomes, because it is that much more sensitive to noise. Again, noise is the related, limiting factor here. See example 3.3 on page 85 of the 7th edition text.