You can google that exact question to find several variations of this answer:
A noiseless channel can carry an arbitrary large amount of information, no matter how
often it is sampled.
Just send a lot of data per sample.
For 4KHz channel, make 1000 samples/sec. If each sample is 16 bits, the channel can send 16 Kbps.
If each sample is 1024 bits, the channel can send 1000 samples/sec * 1024 bits = 1024 Mbps.
The key word here is “noiseless”. With a normal 4 KHz channel, Shannon limit would
not allow this.
For the 4 KHz channel we can make 8000 samples/sec. In this case if each sample is 1024 bits this channel can send 8.2 Mbps.
Could anyone explain in more detail the meaning of the case factor c? In practice, what does it mean to say that the signal rate depends on the data pattern?
The explanation in the text isn't very clear, and this term is not used in other texts I know of. I think what it's saying is that different messages might produce different signal spectra. For example, in a an 2-level FSK system, a message composed of all 1's or all 0's would just be single tone, and have a very narrow bandwidth; while a message composed of alternating 1's and 0's would contain both the one-level tone and the zero-level tone (as well as a spread of frequency content related to switching between them) and produce a broader spectrum if measured on a spectrum analyzer.
Why does the minimum bandwidth for a digital signal equal the signal rate?
This is not correct. The minimum bandwidth for a digital signal is given by the Shannon-Hartley theorem,
\$ C = B\log_2\left(1+\frac{S}{N}\right)\$
Turned around,
\$B = \frac{C}{\log_2\left(1+{S}/{N}\right)}\$.
Approaching this bandwidth minimum depends on making engineering trade offs between encoding scheme (which would relate to the number of bits per symbol), equalization, and error correcting codes (actually sending extra symbols to include redundant information that allows recovering the signal even if a transmission error occurs).
A typical rule of thumb used for on-off coding in my industry (fiber optics) is that the channel bandwidth in Hz should be at least 1/2 of the baud rate. For example, a 10 Gb/s on-off-keyed transmission requires at least 5 GHz of channel bandwidth. But that is specific to the very simple coding and equalization methods used in fiber optics.
If we set c to 1/2 in the formula for the minimum bandwidth to find Nmax (the maximum data rate for a channel with bandwidth B), and consider r to be log2(L) (where L is the number of signal levels), we get Nyquist formula. Why? What is the meaning of setting c to 1/2?
Choosing between L signal levels is equivalent to a \$\log_2(L)\$-bit digital-to-analog conversion. So it's not surprising Nyquist's formula is lurking in the shadows somewhere.
Best Answer
The difference between the two formulas arises from the fact that the Nyquist formula uses the number of encoding levels that was explicitly given (16 levels implies 4 bits/baud), while the Shannon formula is the theoretical maximum based on the SNR of the channel (40 dB implies about 6.64 bits/baud).
3000 Hz × 2 baud/cycle × 4 bits/baud = 24000 bits/sec
3000 Hz × 2 baud/cycle × 6.64 bits/baud = 39840 bits/sec