Please refer to this example. it is taken from wireless communications book by william stallings.
The whole thing is slightly more complicated: without knowing about the reference magnitude, the dB digits are more or less meaningless.
AND: dB is thought to be an expression of power ratio, but is also used e.g. for voltage ratios.
For power ratios, the above formulas with the factor of 10 apply.
If it is about voltages or currents, the following applies: Vs = Vn * 10dB/20
resp. dB = 20 * log(Vs/Vn)
The factor of 20 adapts voltage, current and similar values to the power 'thinking' of the dB vaues, as power is proportional to the square of voltage or current.
Hope, this helps and does not add to the confusion :)
To add slightly to what Milo said, what he calls "Amp" is the ratio of power amplitudes of the signals you're comparing. So, amplitude of signal power divided by amplitude of noise power, for example.
The math is straightforward. Let's say that "Amp" is signal power divided by noise power. Call that Ps and Pn.
You can solve for Ps.
Ps = Pn * 10dB/10
Pn = Ps / 10dB/10
The whole thing is slightly more complicated: without knowing about the reference magnitude, the dB digits are more or less meaningless.
AND: dB is thought to be an expression of power ratio, but is also used e.g. for voltage ratios.
For power ratios, the above formulas with the factor of 10 apply.
If it is about voltages or currents, the following applies: Vs = Vn * 10dB/20
resp. dB = 20 * log(Vs/Vn)
The factor of 20 adapts voltage, current and similar values to the power 'thinking' of the dB vaues, as power is proportional to the square of voltage or current.
Hope, this helps and does not add to the confusion :)
decibels dB = 20*log(V0/Vi) or simply you can write dB = 20*log( amplitude)
dB is used to quantify the ratio of two values in logarithmic scale which conveniently represent very large or small numbers on a same scale.
For converting the ratio of two power values in dB, we use ans(dB) = 10*log10(ratio) and ratio=10(ans(dB)/10).
For converting the ratio of two voltage or current values in dB, we use ans(dB)=20*log10(ratio) and ratio=10(ans(dB)/20).
You can verify the concept of multiplicative factor of 20 and 10 using P = V2/R = I2*R and taking R=1.
For your information, dBm is used to quantify the value of Power (not a ratio) in logarithmic scale which is generally used in wireless communication and other areas.
Hope you find this answer hepful
Yes and no.
dB is a relative measure, it is the proportional difference between to values in logarithmic scale, if you only has the value in dB you cannot extract the amplitude.
But there are exceptions where the values in dB can be converted to amplitude, they are the cases where is possible to determine the reference value
An example is dBv. In this case the value is measure respect to 1 volt.
For example a value of 10 dBv is 10^(10/20) = 3.1622 volts.
Other example is in sound, in this case the reference value is 10^-2 pascals.
The defining equation for decibels is
A = 10*log10(P2/P1) (dB)
where P1 is the power being measured, and P1 is the reference to which P2 is being compared.To convert from decibel measure back to power ratio:
P2/P1 = 10^(A/10)
For example: To convert 14dB to Magnitude, 14dB = 10^(14/10)=25.12 Hence A dB = 10^(A/10)
For Example: A is Voltage gain then
A in db= 20log(Amplitude)
Amplitude= 10^(A/20)
please refer to standard deviation and variance of signal and noise in the context of wireless communications.
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