From a theoretical point of view, the maximum capacity of a channel affected by AWGN Noise (Additive Gaussian Gaussian Noise) is determined by the Shannon–Hartley theorem:
$$ C\leq log_2(1+\frac{S}{N_0B})$$
This means you can't put more than that information on a channel with a band (B= \$f_{MAX}-f_{MIN}\$) without making the communication unreliable.
Then we go on the modulations: every modulation has a particular spectrum efficiency and an erroneous bit probability. More levels you use (QPSK vs 16-QAM, p.e.), more bit for each symbol (= more efficiency) but more erroneous symbols (similar to the bit error rate, with a Gray code).
The spectrum is directly related to the shaping impulse used by the modulation. A very common one is the raised cosine impulse (cause it has no Inter-Symbol Interference), that decreases the efficiency of a factor \$ (1+ \alpha) \$
Again we go on codes, that could give a huge gain, especially using concatenated codes like Reed-Solomon + Viterbi, using Turbo codes or LDPC.
Every effort is done to approach the Shannon capacity limit.
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
Light is part of the electro-magnetic spectrum: -
To the left is AC power and to the right are gamma rays and some cosmic rays. Somewhere in the middle is visible light. This is called the electro-magnetic spectrum and the optical spectrum for an optical device refers to the bandwidth (or range of colours) that it produces or is most sensitive to.
For an LED, how quickly you can "modulate" the light level with an electrical signal is related to the electrical bandwidth of the device. How fast a photodiode responds to modulations of light is also related to the electrical spectrum of the device.
Here is an expanded slice of the radio part of the spectrum: -
Somewhere approximately in this region is where the electical bandwidth limits for a given device will lie.