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I know that $\text{dB}$ actually represents dimensionless value, meaning, it is the ratio of two values that have the same dimension, for example:

$$p_1=10\log\frac{P_1}{P_0} $$

This can be expressed in decibels if $P_0=1\text{ W}$, however, if $P_0=1\text{ mW}$ then unit we use is $\text{dBm}$.

I am wondering, are we allowed to subtract and add values if one of them is in $\text{dB}$ and another is $\text{dBm}$ ?

The reason i am asking this is because i've found an example in one workbook where two values are summed up where the first one was in $\text{dB}$ and another $\text{dBm}$, however, i don't understand how's that possible.

For example, if we say $p_1=10\text{ dB}$ and $p_2=20\text{ dBm}$ if we sum these two up, we end up with $30$ but how is this possible? It's $30$ what? $\text{dB}$ or $\text{dBm}$? I hope someone could clarify me this. Any help appreciated!

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Short answer: $p_\textrm{dB} =\alpha\textrm{ dB}$ means multiplication factor (no unit) while $P_\textrm{dBm} =A\textrm{ dBm}$ is truly a power, nothing more!

Then $p_\textrm{dB} + P_\textrm{dBm} = 10\log_{10}(10^{\alpha/10} \times 10^{A/10})$ is a power, hence has unit $\textrm{dBm}$ in logarithm scale.

$p_\textrm{dB} + q_\textrm{dB} = 10\log_{10}(10^{\alpha/10} \times 10^{\beta/10})$ is factor, hence no unit (or $\textrm{dB}$ in logarithm scale).

$P_\textrm{dBm} + Q_\textrm{dBm} = 10\log_{10}(10^{A/10} \times 10^{B/10})$ is a power square ($\textrm{mW}^2$), unit undefined in logarithm scale.

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  • $\begingroup$ Isn't dBm a ratio of two powers, too? $\endgroup$ – cdummie Nov 29 '18 at 16:51
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    $\begingroup$ @cdummie dBm is the unit of power expressed in logarithm scale, i.e. 1dBm = f_log(1mW) where f_log() is an invertible function. It is like 1kW = 1000x1W = f_k(1W) which is power written in linear scale, again f_k() is an invertible function. $\endgroup$ – AlexTP Nov 29 '18 at 17:48
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Gain (or loss) is taken as a simple ratio of powers. So if your system doubles the signal amplitude, it has a gain of 6dB.

Your 2 examples are ratios relative to a reference value.

The first example should have been expressed as dB re 1 watt not just dB.

You would add 6 dB to an input power regardless of the reference.

The use of dB can be confusing at first because it is taken in context

There are many conversations that depend on the application such as dB re 1 micro pascal per Hz.

The use of dBs is partly historical. A vestige of the days of slide rules. They also have some physical significance. Some scales are logarithmic like loudness perception.

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  • $\begingroup$ Ok, so dB ratio of two powers (input power and output power, or gain power noise power) in case where none of them is fixed. On the other hand dBm tells us about power with respect to 1mW, but again i don't understand how come that we can add up or subtract those two values. I must add that i don't understand why would i have to add 6dB to an input power in my examples. $\endgroup$ – cdummie Nov 29 '18 at 16:48
  • $\begingroup$ log(ab)=log(a) + log (b) $\endgroup$ – Stanley Pawlukiewicz Nov 29 '18 at 16:51
  • $\begingroup$ Ok, so if b=1mW, log(b)=0 and if b=1W then log(B)=0 again, but how is this related to any of my two questions. I am sorry but i don't get it. $\endgroup$ – cdummie Nov 29 '18 at 16:58
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    $\begingroup$ if you apply a signal that is 0dB re 1 watt to a system that doubles the amplitude the output will be 6dB re 1 watt. $\endgroup$ – Stanley Pawlukiewicz Nov 29 '18 at 17:04
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    $\begingroup$ repeating log(ab)=log(a) + log (b) or 10log(ab)=10log(a)+10log(b). I don't know how to make it more simple. write out 2 times a in dB and assume a has units of dorks. $\endgroup$ – Stanley Pawlukiewicz Nov 29 '18 at 17:38
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Decibels, with no suffix, always refers to a ratio of powers.

Your idea that $dB$ refers to $P_0 = 1\rm{W}$ is wrong. If this were the case, then you would denote it as $dB_W$.

In a chain of subsystems, each with a gain expressed in $dB$, the total gain is the sum of all the individual gains expressed in $dB$.

If you have the absolute input power expressed in $dB_W$, then add the gain (in $dB$) to get the output power in $dB_W$. Same for $dB_m$.

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  • $\begingroup$ How come that value expressed in dBW is power? Isn't that ratio of powers too? $\endgroup$ – cdummie Nov 29 '18 at 16:50
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    $\begingroup$ @cdummie dBW is not power. The W, however, is a reminder that the reference is one watt, and that, when you convert back to linear scale, you'll get a value in watts. $\endgroup$ – MBaz Nov 29 '18 at 17:36

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