Everyone is an expert in Computer music ...... Hey, that classmate! Don't be nervous ~ You are so excited that the experts are not really amazing ...... Believe it? Let me ask you a few questions. You can answer them. You are amazing! When we use computers to make music, we often come into contact with a variety of tables. No matter what tables are measured, they cannot be separated from a single unitDB ), my problem is related to it. I have heard of it:
1. What is the difference between 20 dB and 60 dB? (Do not answer me 6020 = 40 (DB), I will smoke you! Do you tell me how loud the 40 dB sound is? Are you using your fingers to measure the distance in the peak table ?)
2. How much noise does the sound of 72db and 66db sound together? (STOP! I know your profile  138db, right? Please ~ This is equivalent to a jet fighter flying one meter away from you! Are you nuts? The two values I mentioned are equivalent to playing with a drummer and a guitar player. Do you think a band is as noisy as flying fighters in the Air Force Base ?)
3. I often hear about various device metrics10dbv and + 4dbu. Are you familiar with this? They said that the + 4dbu devices belong to the "Professional level" and the10dbv devices belong to the "civil level". Do you know why?
4. Why do some articles say that digital devices cannot exceed 0 dB, but analog devices can exceed it?
5. What is the dynamic range of 16bit digital audio? What about 24bit? If you want to say 21bit, can you say it?
6. How much can 100watt guitar speakers sound better than 50watt guitar speakers?
If you think it is a trivial matter for you, you don't have to read this article. You are really a high person! If you try to dig your head ...... Please, you can't solve the problem. You need at least one scientific calculator! What do you think is simple?
I know that everyone can use computers to make songs, but this is due to advanced technology and silly operations. If you put your work in the 1930s s, do you think you can do what you know today? I have seen many "Masters". They may not even know what the decibels are like! Of course, some people disagree and think that the result is important, not the process. It doesn't matter if you don't understand the truth. Just do it. What I want to say is that those real recording engineers will never think like this, because they really know the art of recordingit's not just about pulling a bunch of buttons to completeyou have to understand all the mysteries if you want to create an unprecedented sound. So I said that the "Masters" who just want to imitate or even imitate are nothing remarkable.
I'm glad you can stick to this. It means you are definitely not a guy that can easily satisfy your needs. Your head is full of endless desires for knowledge. You may read the instruction and help documents of all the devices you can find, and you will often see: dBSPL, dbu, DBV, DBM, dbvu, dbfs, and other terms related to decibels. But unfortunately, there is almost no detailed explanation in this regard. You often get confused: Who are they? What is the relationship between them? Don't blame those vendors for not explaining them in the manual, because they just want you to be the "master" I just mentioned ", in this way, you will purchase their products/software from generation to generation. If you gradually understand it, you may not need it. ^ Of course, these are really not easy to understand, because they involve professional knowledge related to mathematics and physics. (As I said, it is not so exaggerated. You can understand it as long as you pass the exam in high school.) let's take a look at what dB is?
Decibel: the unit that represents the relative difference between two sound or power signals in terms of power or strength, equivalent to 10 times the ratio of the common logarithm of the two levels. "
This is a scientific general definition of decibels I found in a professional dictionary. This is the case with DB! "...... Wait! 'Relative difference ...... Ratio of two levels ...... Common logarithm ...... 'Here ...... What is this? I cannot understand it !" Well, don't worry. Of course I won't let you get such a conclusion after reading it for a long time. Please let me talk about it slowly.
First, based on the above definition, we can find the subject, the predicate, and the object, and omit the other parts first. We can get that "DB is a unit". Is this clear conclusion? Our common sense tells us that the unit is used for measurement. We can obtain the specific value of this unit using an instrument or a formula. So what is dB measurement? Practice tells us that the peak table can be used to measure it, but we don't know what the measurement data means to us, even an abstract one! Therefore, we need mathematics to help us with this problem. Scientists chose logarithm.
Why use logarithm? Because they are lazy ...... I'm not kidding! When you go deep into the mysteries of decibels, you will find that you need to deal with a lot of headaches, scientistsa bit like an instrumental performerare characteristic of using every possible means to make the problem easier. Let's take a look at the complexity and simplicity of decibels (please, I have already seen it here. Please give me some patience and support, and I will be right here ):
The sound intensity refers to the sum of the energy in a specified area per unit time (do you know this? But it doesn't matter if you don't know, hey ):
Response = energy/(time * Area)
We know that the ratio of energy to time is power (should we know this? Do not know yet? I rely on ...... I really gave it back to my dear teacher), so:
Response = power/Area
The unit of power is watt, and we use square meters. The unit of sound is: watt/meter ^ (it is difficult to write special symbols on the forum. I use ^ to replace the square, the same below)
Now let's assume that you know that the smallest sound that ordinary people can hear is. 000000000001 watt/meter ^, and the first painful sound is 1 watt/meter ^, then we will get a lot of values between the two numbers, for example. 000792710162 watt/meter ^, and. 000006288415 watt/meter ^, etc. Try to quickly compare the two numbers and calculate their difference! How do I get dizzy? Can you imagine that our peak table uses this unit for representation? Oh, my God ......
Our lovely scientists did not do such stupid things, so they wrote the formula:
Log (. 000792710162) =3.1
Log (. 000006288415) =5.2
Is this much worse? It's 2.1 ...... Ah? What do you mean? What is this 2.1? It's just a poor volume. If you're smart, you may suddenly think about what it is calledyes, it's bell!
However, this is not a DB because Bell's later scientists inherited his tradition and carried it forward (what tradition? Lazy !)...... This time, they don't even want to see the decimal point, so they multiply 10 again and become like this:
10 * log (. 000792710162) =31
10 * log (. 000006288415) =52
The answer has changed from 2.1 to 21. This "21" is the main character of todaydecibels.
How are scientists smart? Students, you need to learn how they can use various formulas ...... Er, I mean: Dare to explore! They are really lazy, right? It's more lazy! The logarithm has a feature that converts subtraction to Division. Therefore, we can make it simpler:
10 * log (x)10 * log (y) = 10 * log (x/y)
In this way, we don't need to separate the problem just now. We can solve the problem by using a formula:
10 * log (. 000792710162/. 000006288415) = 21 DB
This is why we need to use the logarithm. With this simple method, we can finally conduct more indepth research on decibels.
There is another small problem. What should we do if the measurement data we get is not all measured in the unit of sound intensity? If the two data units are different, will the formula be ruined? Think about it. What method do we usually use to calculate the values of different units and get the results of the same unit? In fact, we only need to find a fixed constant and bring it into this formula to solve this problem. We call this constant "reference number ". What is used as the reference number? We just mentioned that the smallest sound that ordinary people can hear is. 000000000001 watt/meter ^. Let's use this! (The same is true for other numbers. We just want to unify the unit.) We use the letter "N" to represent this constant, so:
10 * log (x/N)10 * log (y/N)
= 10 * log [(x/N)/(y/n)]
= 10 * log (x/y)
For the sake of insurance, let's check if there is any problem with this formula, or use the example just now:
10 * log (. 000792710162/0 .000000000001) = 89 DB
10 * log (. 000006288415/0 .000000000001) = 68 DB
89 DB68 DB = 21 DB
OK, all done! This method allows us to compare the values of different units. (The two data units in this example are the same, so it seems that "Reference Number" is useless)
Frequently Used measurement units include the Sound Power (Watt), the sound (W/meter ^), and the sound pressure (Pascal)  hey! You should pay attention to what I will talk about next. This is the easiest place to confuse decibels.
We can use the above formula to calculate the data measured in units of power or sound. However, when people talk about "decibels", it usually refers to pressure. After all, it is the pressure of sound waves that compress our eardrum to let us tell how many sounds sound. Therefore, we generally talk about dBSPL (sound pressure levels ).
Pressure is the force acting on the unit area, and the Force Unit is Newton (seeing your strong nod, I am really helpless ......), Therefore, the unit of pressure is ox/meter ^. Another commonly used unit is Pascal. 1 Pa equals 1 ox/meter ^.
The relationship between sound (I) and sound pressure (P) can be expressed using the following formula:
I = P ^/P
P is a Greek letter, read as: "meat", it represents the air resistance, is a constant. This value depends on factors such as atmospheric pressure and air temperature. Generally, at room temperature, the air resistance is about 400. Therefore, the minimum audible sound that ordinary people can hear is converted into sound pressure:
. 000000000001 W/m2 = (. 00002 Pa) ^/400
However, in the formula just now, there is another square behind P. That is to say, the sound pressure is doubled, and the sound is quadrupled. The sound pressure is quadrupled, and the sound is sixteen times ...... In this case, when we use sound pressure as the unit of measurement, will the previous formula not be faulty? Let's calculate a little bit: DB = 10 * log (x/y)  at this time, X and Y use sound as the measurement unit. We bring P ^/P into the formula, then: DBSPL = 10 * log [(PX ^/P)/(PY ^/P)] = 10 * log (PX ^/Py ^) = 10 * log (PX/Py) ^ = 20 * log (PX/Py) In this way, the problem is solved. Unlike the previous formula, it is multiplied by 20. This is the dBSPL formula. When we talk about "decibels", 99% is about it. The DB we see on various scales is actually dBSPL, no one said this, but it always had three SPL letters. (Some may be in trouble, but most of them may not know ...... But now you know) When we use sound pressure as the measurement unit, we choose the "reference number. 00002 Pa, close to the minimum sound that ordinary people can hear. Let's take a look at the formula we just obtained: DBSPL = 20 * (p/. 00002 PA) Because log1 = 0, so: 20 * log (. 00002 PA/. 00002 Pa) = 0 dB SPL Please note that if we take the same value as the reference number, we will always get "0 dB", no matter what type it is  dBm, dbu, DBV, dbfs... that's all! Also, you may have questions. Isn't 00002 Pa nearly impossible to hear? How is it 0db? Yes! Is zero equal to none? Oh, I understand what you mean. in the computer, you often see that 0 dB represents the highest value of the peak table, right? The digital circuit is different from what we are talking about now. Don't worry. I'll talk about it later. The strongest sound pressure we can endure is about 20 Pa. Try using decibels for a look? It should be as follows: 20 * log (20 PA/. 00002 Pa) = 120 dB What, remember what I said in the physics class? We can't bear the sound of over 120 decibels. This is the value. Here, we should review it. I believe that a lot of formulas and calculations have left you dizzy? No way. To make it clear, I can only do this, but you only need to understand it. What you need to remember is the following two: DB = 10 * log (x/y)  the formula used to calculate the decibel when measured by sound. The unit should be W/m ^ DB = 20 * log (x/y)  the formula used to calculate the decibel when the sound pressure is used as the measurement unit. The unit is Pa. So far, you already know what dB is. However, this lesson is not over yet, because we do not know dbu, DBV, DBV, DBM, dbvu, and dbfs. However, with the above foundation, you understand that these little things are only a matter of time. Let's start with the principle: We have understood the meaning of decibels. We should pay special attention to the following:Decibels indicate the ratio between two data types.(The type should be the same. This is very important. You cannot compare them directly with Watt ). In the two data types, one of them is called the reference number ", the measured value and reference number are substituted into the formula to calculate the corresponding decibel value. For example, we have used sound pressure as the measurement unit before, and that is the reference number we selected is. 00002 Pa. The DB value we finally get is called "dBSPL ". That is to say, the different letters behind the DB indicate what we use as the measurement unit to get the decibel value. Sound pressure is the SPL (sound pressure levels ). Should this be clearly explained? If you understand this, I will explain other DBrelated units one by one. DBM and dbvu [/color] We have discussed how to use the power measurement to obtain the decibel value. At that time, we were talking about the sound power, measured in watt. However, we know that apart from sound, there are many other phenomena that can generate power, such as electricity. Long ago, engineers relied on a device called vu to complete their work in the "ancient times" that led to the birth of light emitting diodes and LCD screens. Vu looks like a speedometer in the cab, and uses a pointer to indicate the increment of current through this question clockwise. VU is short for "volume unit", meaning: volume measurement unit. The problem with vutable is that every vutable is different! It was not until the end of the 1930s s that a group of engineers sat together and decided to unify the VUV table metering specifications. They determined the standard: when the current power is 1 MW (1 MW), vuvu 0 dB. In other words: 0dbm = 0 dbvu. The m behind the DB represents the milliwatt. DBM is also measured in units of power, and the reference number is 1 MW. DBM = 10 * log (power/1 MW) In this way, we can easily use DBM to represent changes in current power. Remember? When the measured value is equal to the reference object, is the DB value always 0? So: 10 * log (1 MW/1 MW) = 10 * log (1) = 0 dBm When the vu Pointer Points to + 3dbm, the power is doubled. How can this problem be solved? In this way: 10 * log (2 mW/1 MW) = 10 * log (2) = 3 dBm  as I said, at least you have to prepare a scientific calculator, and the logarithm is hard to calculate. What if it points to6dbm? 10 * log (. 25 MW/1 MW) = 10 * log (. 25) =6 dBm Dbu (also called DBV)[/Color] Recall the high school physics. Power (p) can also be expressed by the relationship between voltage (V) and resistance (R: P = V ^/R  the unit of resistance is Ohm (Ω) When talking about dBm, the reference number is 1 MW. This standard was set up in the 1930s S. At that time, the input impedance of all audio devices was 600 ohm, the tape recorder, the mixer, and the frontend power amplifier ...... If there is a plug, the resistance from the FireWire to the ground is 600 ohm. So when the resistance is 600 ohm, how much voltage is required to generate 1 MW of power? Use the formula just now to calculate: P = V ^/R . 001 W = V ^/600 Ω V2 =. 001 w * 600 Ω V = SQRT (. 001 w * 600 Ω)  SQRT is square. I don't know how to use this symbol. V =. 775 v The answer is 0.775 volts. So, when the input impedance of all devices is 600 ohm, the reference number used for dbu calculation is. 775 V, that is, dbu is the decibel value calculated based on the voltage. However, we noticed that the voltage in the formula just now is the number of voltages. Based on the previous experience, we know how to solve this problem: Dbu = 20 * log (tested Voltage/. 775 V) If you are very careful, you may find it strange: Why is it dbu instead of DBV? In fact, people used DBV for a long time, but later people found that DBV and DBV are too confusing, the lowercase letter "U" is used to replace the lowercase letter "V. If you still can see DBV, it means the dbu we talked about today  unless the DBV writer cannot figure out what he wants to say! So what is the difference between DBV and DBV?

For a long time, the input impedance of all the audio devices used is 600 ohm. Today, we will see some devices with higher impedance, such as 10000 Ω. The higher the resistance, the lower the power consumption of the circuit. (According to the above formula, we know that the power is inversely proportional to the resistance) Remember that the reference number for dbu is. 775v? Many engineers think this number is too troublesome, but because all the devices were at a fixed input impedance at that time, the use of. 775v as a reference number is also logical. If the device is not improved, the reference number cannot be changed. However, for convenience, a new reference number is developed rapidlyA New DB unit DBV is generated along with the new reference number. The reference number is 1 V: DBV = 20 * log (tested Voltage/1 V) In fact, DBV and dbu are very similar, but the reference numbers are different. Now, let's take a look at the differences between "Professionallevel" and "userlevel" devices. You may have known that professional devices are + 4dbu and userlevel devices are10dbv. Of course, this is ridiculous, haha. We have seen that both dbu and DBV calculate the DB value by comparing the voltage. There is no difference except the reference number. The socalled professional level means that most users of these devices are "uncles" (because the standards are early, most of the users are "qualified ). In fact, it is too arbitrary to determine the "level" of the device based on the two parameters. In any case, the two types of devices can fulfill their work requirements well. I think we should give full play to people's subjective initiative in this regard. We know the hard differences between devices, but how to use our knowledge to maximize the potential of the devices in your hands is what we should pursue. Poor equipment is a matter of money. If you have a good device or a good music, it is a human problem. The money problem can be solved, and the human problem cannot be solved! There is a small island across the Strait. Although there are not many people on it, there are many people engaged in music. We admit that their music has developed well, however, they have a very high level of people playing music, and there is a bird forum in them, in the above, some birds say nothing about the "professional" and "user" devices! Let me look at cainiao on the other side of the Strait (by the way, many people think that people on the other side of the Strait are far behind them ...... This is the root user ~ But who makes this happen now? In order to make the comrades in the Strait do not look as "professional" as they do, they are actually very "fucking", so I wrote this paragraphit should be said that I wrote this article, this is also a big reason! Okay. Let's take a look at the difference between + 4dbu and10dbv: + 4 dbu = 20 * log (tested Voltage/. 775 V) Tested voltage = 1.228 v 10 DBV = 20 * log (tested Voltage/1 V) Tested voltage = 0.3162 v 20 * log (1.228 V/0.3162 v) = 11.79 DB If you have these two devices, you can perform a test: connect10 DBV output to + 4 dbu input, and then read the vutable of + 4 dbu. Is it 11.79 dbvu? Dbfs[/Color] Finally, let's take a look at the dbfs that is most closely related to us. Dbfs stands for "decibels full scale" (full decibel scale)a decibel value representation method created for digital audio devices. This guy is not the same as several other brothers. His reference number is not the smallest, nor the middle, but the largest! That is to say, "0 dbfs" is the highest level that a digital device can reach. In addition, all values are smaller than this valueall are negative. This is why the highest scale of the peak table we see on the computer is "0", and the pointer will never read a higher number. But why? To solve this problem, let's briefly talk about the principle of digital audio storage. We use 16bit digital audio as an example: "16bit" means that the sampling signal is stored in 16bit binary numbers. There are two binary numbers: "0" and "1 ". Therefore, the maximum value is 1111 1111 1111 1111 (binary, converted to decimal is 65536). Therefore, the formula for calculating dbfs is: Dbfs = 20 * log (sample signal/1111 1111 1111 1111) In this way, it is easy to explain why it cannot exceed "0", because the reference number of dbfs is the maximum value, so: 20 * log (1111 1111 1111 1111/1111 1111 1111 1111) = 0 dbfs What about the smallest? Except for 0, the minimum 16bit binary number is 0000 0000 0000, so: 20 * log (0000 0000 0000 0001/1111 1111 1111 1111) =96 dbfs Do you know why the peak tables you see are from 0 dB to96 dB? Next, you can calculate the dynamic range of 24bit and 32bit digital audio. I will tell you that the dynamic range of 24bit digital audio is 144db. Try it on your own? (Do not forget to convert the binary into decimal. I will not use the binary to calculate the logarithm! ^) At this point, the content of this article has almost been written, and the time has been too short. omissions are inevitable. Thank you for your correction ...... However, when I look back at the previous content, I always feel that there are still some things that can be written, but they cannot be too hasty. It is true that this article is not very easy to understand, but I hope you can read it with some thoughts. I can assure you that there is no harm! If you think you have understood the problem, please try to solve the first few questions in the article. If you can solve it all, it means that I have written it clearly, I don't have to explain it more. If there are many problems, my worries are still reasonable. I will write another article about decibels to solve these problems, it's just a supplement. (What is the problem? I will not talk about it first, so that you will not be lazy and will not discover your own problems) Finally, I would like to thank the bird Forum I just mentioned and some of the above birds for giving me the motivation to write this article. I would also like to thank some effective tool (forgot, it seems to be PSP vintage. This document is not fully explained, so I have the opportunity to read it.Lionel Dumond(You can go to prorec to search for this article, eArticle). Finally, I would like to thank you (this is a real Thank you)Lionel DumondWithout your good words, I would not know what dB is! Zookeeper ~~~ When I was writing this article, I used textbooks and extracurricular books for high school physics, Acoustic physics, and advanced algebra ...... Comrades! 12 years of hard reading is still useful, and never be lost ~~~ : P 