If you want to learn a subject well, first of all, the concept must be very clear. If you don't understand it well, it's best to go back and forth to understand those concepts as you learn. I think the reason why understanding theory is sometimes confusing is that the concept itself is not clear. Look at some questions on Zhihu. If the concept is clear, you will not ask such unprofessional questions.
Let's first look at the two basic concepts of data transmission rate, symbol (symbol), and symbol rate.
1. Data transfer rate
Also known as code rate, bit rate or data bandwidth, it describes the number of bits of data code transmitted per second in communication, in bps. This is easy to understand, it is "just needed", how many bits of data are transmitted per second.
2. Symbol (symbol)
Also called a symbol. Through different modulation methods (such as FSK, QAM, etc.), multiple bits of information can be loaded on one symbol symbol. For example, the figure below shows all four symbol symbols modulated by 4QAM (ie QPSK), and one symbol can carry two bits of information.
3. Symbol rate
The symbol rate is the symbol rate, in Baud/s or sym/s, and represents the number of symbols transmitted per second. Symbol rate is also called baud rate or symbol rate. The symbol rate determines the communication efficiency. Obviously, the more symbol states of a modulation method (4QAM in the above example), the larger the symbol rate value, and the more bit information can be transmitted per second. obviously have
Data transmission rate = symbol rate x number of bits in a symbol
The serial port we usually use does not have any modulation at all. The high and low levels sent directly represent 1 and 0, that is, a bit is a symbol, so its baud rate is the transmission rate. The serial port baud rate we are talking about is 115200, that is, under this setting, the transmission rate can reach 115200bit/s.
After talking about the above three concepts, we can talk about bandwidth.
Bandwidth is actually a physical concept, it refers to the width of the occupied spectrum. When designing a communication system, bandwidth is actually a quantity determined by design. It's very important to understand that a system, what data rate are you going to support? What modulation method is used? What encoding is used? After taking it all into consideration, these indicators determine how much bandwidth your channel needs. Various encoding methods (various purposes, check, error correction, etc., with only one purpose, to improve the reliability of transmission) determine the total amount of information you ultimately transmit (the data to be transmitted + other necessary information), modulation The method determines the symbol rate at which these data are ultimately transmitted.
So the question is, what is the relationship between bandwidth and bandwidth? The relationship between channel bandwidth and data transmission rate can be described by Shannon's theorem and Nyquist's criterion.
Shannon's theorem:
Cmax=Wlog2(1+S/N)(b/s) S is the average power of the signal transmitted in the channel, N is the Gaussian noise power inside the channel
That is to say, if the channel has no noise, the bandwidth supported by the channel is infinite. Of course, it is impossible to actually have no noise.
Shannon's theorem gives the theoretical upper limit of the channel capacity, but it looks a bit illusory, because it seems to have nothing to do with baud rate, code rate, etc., and the relationship between them is given by the Nyquist criterion.
Nyquist criterion: For a noise-free low-pass channel with a bandwidth of W (Hz), the highest symbol transmission rate Bmax:
Bmax=2W (Baud), that is, the maximum symbol transmission rate of an ideal low-pass channel per Hertz bandwidth is 2 symbols per second.
According to the previous definition of the Baud unit, if the number of symbol states of the encoding method is M, the limit information transmission rate (channel capacity) Cmax is obtained:
Cmax=2Wlog2(M) (b/s) (comments point out that this is a low-pass case, but it does not affect understanding)
What Nyquist wants to tell us is that if each symbol transmits a certain bit, if my channel only supports a bandwidth of W (Hz), you can only give me Cmax (bit) information per second at most, I can't eat any more. Conversely, when the bandwidth is known and the channel capacity Cmax has been determined by Shannon's theorem, the Nyquist criterion actually gives the maximum number of bits (such as the number of QAMs) transmitted per symbol in the system.
Going back to the sentence above, bandwidth is a quantity that is determined by design. I want to transmit so much data, and the maximum signal-to-noise ratio of the channel can basically have a predicted value. You at least have to create a channel for me that satisfies Shannon's theorem. Needless to say, the bandwidth is less, and there is more waste. You need to know the spectrum. Resources are often very precious. Also, your RF circuit, hardware design, and filter must meet this bandwidth for me. If it is less, it will not work. If the bandwidth is too much, the interference signal outside may leak in, and the anti-interference will not work.
Finally, let's talk about the carrier. As the name implies, the carrier is the carrier of signal modulation and transmission. It has only one center frequency and has nothing to do with the bandwidth itself. For example, the 11n protocol stipulates that it can work in the 2G frequency band or in the 5G frequency band, and other factors are the same. Assuming 20M bandwidth, the carrier frequency is 2.4GHz when working in the 2G frequency band, then the spectrum resources it actually occupies are 2.390GHz-2.410GHz. When working in the 5G frequency band, the carrier frequency is 5GHz, and the spectrum resources it actually occupies are 4.990GHz-5.010GHz.
At the end of this article, how do I understand the relationship between data bandwidth, carrier frequency and carrier bandwidth in wireless communication systems with my answer in this link? as an end. Why the signal occupies the bandwidth is really basic, because the spectrum occupied by the digital signal (isn't it just aperiodic like a square wave) is actually infinitely wide after the Fourier transform.
