Converting dB to Voltage Ratio (V/V) and Reverse: Understanding the Formulas

As in the question whether it is possible to convert the logaytmic scale to linear and vice versa (V / V to dB and dB to V / V). If there is such a possibility, please provide which formulas how to convert it.

X [dB] = 20log (V1 [V] / V2 [V]) - it is one way and the other way:
V1 [V] / V2 [V] = 20 ^ (X [dB] / 20)

The logarithm is decimal and with the power before the logarithm it is 10, not 20.

Hello
The logarithmic scale is different from the linear distribution of the points on the axis, in the linear you have the sequence points at the same intervals, e.g. 1,2, ..., 9,10,11, ..., 20,21, ... etc and in the scale logarithmic in these male intervals you have 1,2, ..., 9,10,20, ..., 90,100,200, ... etc.
When it comes to dB, in them you express some ratio of two signals, e.g. gain or attenuation. If it is a ratio of two voltages, e.g. Uwej / Uwyj, then notice that you have V / V and if you want in dB, you have to log it 20log (Uwej / Uwyj).

I think so, if I am wrong, please correct me :)

greetings

Colleague Consul was wrong.
It has to be like this
V1 [V] / V2 [V] = 10 ^ (X [dB] / 20)

So we have, for example, Ku = 40dB, we calculate it
10 ^ (40/20) = 10 ^ 2 = 10? = 100V / V
Ku [db] = 70dB
ku [V / V] = 10 ^ (70/20) = 10 ^ 3.5 = 3162V / V

Ku [db] = - 30dB
Ku [V / V] = 10 ^ (- 30/20) = 10 ^ -1.5 = 0.0315V / V

And if we have a gain of 20V / V then
we recalculate as given by a colleague above.
Ku [V / V] = 20 * logKu = Ku [dB]
so we count
Ku [V / V] = 20V / V
Ku [db] = 20 * log20V / V = 20 * 1.3 = 26dB

Ku [V / V = 450V / V
Ku [db] = 20log450V / V = 20 * 2.65 = 53dB

Ku [V / V] = 0.06V / V
Ku [db] = 20log0.06 = 20 * -1.22 = -24dB

And when we calculate the power, we divide and multiply by 10.

Hello !
If you need it, you have ready-made level conversion tables:

Jony is right. I wanted to write back quickly and I was wrong. I do not like the way of writing at the very bottom, because the written equations are contradictory, especially when there are no dB units next to them. It seems to me that you should not create such records at all and clearly separate linear units from dB by writing, for example:
Ku = 100V / V
Ku [db] = 20log100V / V = 40dB
Writing such equations on one line gives the appearance of the possibility of making two-way actions on these expressions, which is of course a mistake.

I heard that there is still a dBA volume unit. How is it converted between dB and dBA?

The dBA units are used to measure loudness (noise), and precisely it is the averaged power of the sound pressure.

The basis is the value 2 * 10 ^ -5 Pa, level 0 dB, denoted p0.

Basically the loudness is L = 10 log10 (p / p0) ^ 2.

Only that due to the properties of human hearing, the pressure p is averaged over the audible frequency range according to one of the characteristics A, C or Z (well, I was sure it was B, but I read that it was not). Hence the dBA (dBC and dBZ).

And if someone wishes in more detail, I recommend http://www.sonopan.com.pl/pliki/pawod/definicje_a.pdf