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What are the differences between Decibel (dB) and Decibel milliwatts (dBm)? I'm most interested in dBm with cell phone signal strength. Please give a detailed explanation. I'm not familiar with the basics so I didn't understand what I was reading when I googled it.

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A decibel isn't actually a unit. Decibels tell you a relationship between two things: 10 decibel means something is 10 times more than something else. Specifically, decibels are logarithmic. 10dB means something is 10x something else, 20dB means something is 100x more, 30dB means it is 1000x more, and so on.

So... more than what? That's where the milliwatt comes in. dBm means that you are expressing your signal strength with respect to 1mW. 0dBm means 1mW of signal power, and e.g. 30dBm means 1W. -30dBm means 1µW.

The same goes for all other decibel values. dBµV is often used for line signal voltages, and means decibels with respect to 1µV. Sometimes this suffix is omitted, but that means there is some kind of agreed upon standard value, e.g. dBA (A-weighted sound pressure level), where it's the pressure level with respect to a standardized curve over the audio spectrum.

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  • \$\begingroup\$ Does dBµV express the ratio between a given voltage and 1 microvolt, or between the amount of power that a voltage feeds into a given load compared with what one microvolt signal would do? Would such units only be applicable to resistive loads? \$\endgroup\$
    – supercat
    Jul 23, 2014 at 15:25
  • \$\begingroup\$ The units are always literal: it is the amount of dB compared to 1µV. \$\endgroup\$
    – user36129
    Jul 24, 2014 at 7:43
  • \$\begingroup\$ So would a doubling of voltage represent an increase of 10dB rather than 20? \$\endgroup\$
    – supercat
    Jul 24, 2014 at 13:37
  • \$\begingroup\$ A doubling of voltage would mean about 3dB difference. Note, it's logarithmic. \$\endgroup\$
    – user36129
    Jul 24, 2014 at 21:21
  • \$\begingroup\$ Sorry--I meant a ten-fold increase in voltage. In any case, I infer that you're saying a 3dBµV signal would be 2µV even though that would be four times the power of a 1µV signal? \$\endgroup\$
    – supercat
    Jul 24, 2014 at 21:42
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Decibel is a way of measuring power level ratios. The letter after the dB is what is being measured.

There is a great video on Youtube that explains this relationship.

What is a dB

So to answer your question, dB is the RATIO between two levels and the 'm' is what is being compared. In dBm it is being compared to a milliwatt. It could be WATTS(dBW) VOLTS(dBV), etc..

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Decibel(dB) is a measure of gain beetween two homogeneous quantities. So does not express a precise value, just a logaritmic gain. you can use dB for measure gain beetween: - tension - power ecc

ex:

L(dB) = 10 · log10( P1 / P2) (for power)

L(dB) = 20 · log10( U1 / U2) (for tension)

dBm is the same thing as dB as method of calculation. The difference is that dBm is ALWAYS referred to 1mW. of conseguenzy is a unit of measure of power

P(dBm) = 10 · log10( P(mW) / 1mW)

ex:

0.001 mW = -30 dBm = 10 · log10( 0.001mW / 1mW)

1 mW = 0 dBm = 10 · log10( 1mW / 1mW)

1W = 30 dBm = 10 · log10( 1000mW / 1mW)

1kW = 60 dBm = 10 · log10( 1000000mW / 1mW)

is used to semplify the order of 10. typically in telecommunication.

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If only the ratio of two powers is important, them the decibel is appropriate.

So, for example, if \$p_1\$ is 100 times \$p_2\$ then we can also say \$p_1\$ is larger than \$p_2\$ by 20dB:

$$10\cdot \log\frac{p_1}{p_2} = 10\cdot \log100 = 10 \cdot 2 = 20\mathrm{dB}$$

Now, if we fix the power in the denominator to some standard value and indicate that in some way, we can express the actual value of \$p_1\$. For example, let

$$p_2 = 1\mathrm{mW}$$

and \$p_1 = 100\mathrm{mW}\$, then we say \$p_1 = 20\mathrm{dBm}\$:

$$10\cdot \log\frac{100\mathrm{mW}}{1\mathrm{mW}} = 10\cdot \log100 = 10 \cdot 2 = 20\mathrm{dBm}$$

where we've indicated, by \$\mathrm{dBm}\$ that \$p_1\$ is \$100 \cdot 1\mathrm{mW} = 100\mathrm{mW}\$

To summarize, if a power \$p\$ is expressed in \$\mathrm{dBm}\$, the same power expressed in \$\mathrm{mW}\$ is given by:

$$p\, \mathrm{mW} = 10^{\frac{p\, \mathrm{dBm}}{10}} \mathrm{mW}$$

For example, if

$$p = 25 \mathrm{dBm}$$

then

$$p = 10^{\frac{25}{10}}\mathrm{mW} = 10^{2.5}\mathrm{mW} = 316 \mathrm{mW}$$

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