# Relationship between bel and decibel

Bel is a unit of $$log_{10}$$ of ratio of two quantities.

1 Bel = $$\log_{10}\frac{P_1}{P_2}$$

On Wikipedia it says: 1 decibel = $$\frac{1}{10}$$ bel

According to this definition then,

1dB = $$\frac{1}{10}$$ $$log_{10}\frac{P_1}{P_2}$$

But in the book they have written:

1dB = 10 $$log_{10}\frac{P_1}{P_2}$$

Please explain what am I missing here? Is

1 dB = $$\frac{1}{10}$$ bel

or

1 dB = 10 bel

Here is a link for the definitions of bel and dB.

• A random commercial website is not necessarily the best source for this kind of information. Nov 15, 2023 at 18:38
• They're not saying "1 dB = 10 * log...", they're saying that "counted in dB, the power is = 10 * log ...". It's a confusing way to put it, though. Nov 15, 2023 at 19:10

1 decibel = 0.1 bel.

According to this definition then, 1dB = $$\frac{1}{10}$$ $$log_{10}\frac{P_1}{P_2}$$

This is incorrect. Say for example that $$P_1=100 P_2$$. Then we would have

$$\log_{10}\left(\frac{P_1}{P_2}\right) = 2 \text{ B}$$ Since $$1$$ bel = $$10$$ decibels,

$$\log_{10}\left(\frac{P_1}{P_2}\right) = 2 \text{ B} = 20 \text{ dB}$$

The lesson is that $$\log_{10}\frac{P_1}{P_2}$$ expresses the relative intensity of $$P_1$$ with respect to $$P_2$$ in units of bels. Each bel is equal to 10 decibels, so to express the relative intensity of $$P_1$$ with respect to $$P_2$$ in units of decibels, you'd need to multiply this expression by 10.

What you have to remember is what a length $$\ell\,m$$ means.

The length is $$\ell \times (1\rm m)$$ where $$(1\rm m)$$ is a unit of length.

$$\rm 1\, m = 1\times (1m)= 100\times (1cm)=100\,\rm cm$$

In general $$\rm \ell \,m = \ell\times (1m)= \ell\times 100\times (1cm)= 100\ell \,cm$$

$$\rm 1\, B = 1\times (1B) = 10\times (1dB) = 10\,dB$$

$$\rm\log_{10}\frac{P_1}{P_2}\,B= \log_{10}\frac{P_1}{P_2} \times (1B) = \log_{10}\frac{P_1}{P_2} \times 10\times (1dB) = 10\log_{10}\frac{P_1}{P_2}\times (1dB) = 10\log_{10}\frac{P_1}{P_2}\,dB$$

I wouldn't take an "equation" like bel = $$\log_{10}\frac{P_1}{P_2}$$ too seriously. It is confusing at best.

Here is another, arguably neater way to think about things. Suppose $$x$$ is a ratio of two powers. It is sometimes useful to represent $$x$$ in logarithmic scale as an equivalent quantity $$x_{\text{log}}$$, expressed in the units B or dB. The defining relations are $$x_{\text{log}} =\log_{10}x\text{ B}=10\log_{10}x\text{ dB}.$$

In the second equation we have used $$1\text{ B} = 10\text{ dB}$$. This is no different from "usual" units with the deci- prefix, e.g. $$1\text{ m} = 10\text{ dm}.$$

If $$x$$ is the ratio of a "field" quantity instead, like voltage, current or E-field, whose squares are proportional to some sort of power or intensity, the relations are instead

$$x_{\text{log}} =\log_{10}(x^2)\text{ B}=10\log_{10}(x^2)\text{ dB}$$ or $$x_{\text{log}} =2\log_{10}x\text{ B}=20\log_{10}x\text{ dB}.$$

• What exactly is "confusing" about the equation in the first sentence? Seems like the asker's confusion is more about what it means mathematically that 1 dB = 0.1 B. Nov 16, 2023 at 3:20
• @ToddWilcox it's confusing because given all those information, we have 1 dB = 0.1 B, and B is bel, and bel = log_10 (p_1/p_2). If instead we have X = log_10 (p_1/p_2) B (with B as a unit), then it's less confusing. Nov 16, 2023 at 3:28
• @justhalf Oh yeah that makes sense. Nov 16, 2023 at 4:00

A millimeter is 1/1000 of a meter. So if we measure your foot and it is 0.28 meters long, would the length in millimeters be 0.00028 mm or 280 mm?

The formula is telling you how to convert a given power measurement (with reference) to bels or decibels. Just like with millimeters and meters, when you measure with the smaller unit, you need to come out with a larger numerical value.

The extract that you have copied is very confusing.

$$\log_{10}(P_1/P_2)$$ is not the bel, but the relative power expressed in bel(s). If you wanted a definition of the bel (hardly ever sought) it would be

1 bel = logarithm to the base 10 of a powers ratio, $$P_1/P_2$$, of 10 .

The first and second lines of maths in the extract are clearly inconsistent with each other. The second line starts correctly with the relationship between units:

dB=$$\frac{\text{bel}}{10}$$

But then, with incorrect mathematics (the mysterious movement of the 10 from divisor to multiplier), it substitutes for bel from the first line. The two mistakes cancel out (sort of) provided it is realised that what the second line ends up giving you is not the unit dB, but the logarithmic relative power in dB.

Since the unit dB is a tenth the size of the bel, to express a given logarithmic powers ratio you need 10 times more dB than bel(s). That's the gist of it.

I hope that the rest of the source you are quoting from has fewer mistakes and muddles ...

It's very simple. The textbooks are WRONG, when they write a variable result as a unit: dB= . . . or 1dB= . . . it should say either VoltageGain(in dB)= . . . or PowerGain(in dB)= . . . (which can be any quantity as per the FORMULA, with the RESULT being in dB). They are NOT equating fixed values with different SI prefixes as they appear to do!!