Decade to Octave Conversion

[Tesla23]

THen please explain to me what this extract is saying, because it seems to say that Gdb is power gain, not voltage gain. What I'm basically looking for is an explanation of dB that refers explicitly specifically to amplitude gain in dB rather than power gain.

Someone said earlier than until the advent digital signal processing, power gain has almost always been more important and that's why definitions are the way they are. So far it seems that the 10log power and 20log amplitude formulas only allow me to say that the the power gain is X dB from the input and output power of X, or that the amplitude gain has resulted in a power increase of X dB. I'm looking for the formula that would let me say the voltage gain is X dB.

GdB is simply the gain.

What it actually means does, however, depend somewhat on the system. An RF amplifier measured in a matched 50 ohm system with 20dB of gain will show a power gain of 100 and an amplitude gain (voltage or current) of 10. An op-amp with high impedance input and low impedance output configured to have a voltage gain of 10 still has 20dB of gain. If the source was low impedance this would supply 100x the power to a load, compared to connecting that same load to the input. However (not wishing to confuse you) you could clearly obtain a power gain of >> 100 simply by lowering the impedance of the load. If the source and load impedances change then you need to be careful, for example a 2:1 transformer can drop the voltage by 2 without changing the power available.

However in most cases, the simple 10 log power ratios or 20 log amplitude ratios works a treat.

 

[Sceadwian]

Tesla.... umm you couldn't be more wrong.

I'll quote from Wikipedia for a quick reference.

The decibel (dB) is a logarithmic unit of measurement that expresses the magnitude of a physical quantity (usually power or intensity) relative to a specified or implied reference level. Since it expresses a ratio of two quantities with the same unit, it is a dimensionless unit. A decibel is one tenth of a bel (B).

Please carefully note it is a 'dimensionless unit' which means it has no inherent value in and of itself, only in relation to something else. There db volts, db watts, db current, db ANYTHING. The decibel itself is not a unit of anything, it is the description of a refrence value. Comonly decibel/watts or decibel/volts. Which because power increases logrhythmicaly in respect to voltage they are VERY much not the same.

 

[dougy83]

Which because power increases logrhythmicaly in respect to voltage they are VERY much not the same.

Oh really? I would have thought power increased exponentially with respect to voltage.

Also, you may notice that 10log10(P1/P0) is used for power ratios, while 20log10(V1/V0) is used for amplitude ratios. Have you any idea why this is? Maybe because the same dB value is obtained for a given gain, whether you express the ratio as relative amplitude or relative power.

 

[Sceadwian]

Let me please insert my foot in my mouth, thanks dougy =)
As tesla stated though a DB is not a DB is not a DB.

 

[Tesla23]

Let me please insert my foot in my mouth, thanks dougy =)
As tesla stated though a DB is not a DB is not a DB.

I assume you have examples where a dB isn't a dB, such as where the dB power gain is different from the dB voltage gain etc.

The point I was trying to make is that the 10log and 20log thing is simply an artifice to measure power ratios - you still get the same dBs. I think if you consult any authoritive reference you will find the bel is defined as log10 of a power ratio. It is then consistent to define it also as 2log10 of an amplitude ratio.

Of course it is a ratio, and there are many actual quantities measured as dBs relative to a reference level, dBV, dBu, dBm etc.... But the original issue - what is a dB - still only has one answer, and a dB is a dB is a dB in the sense that it doesn't matter whether you are measuring relative amplitudes or powers.

 

[Sceadwian]

Tesla, simple example, decibel sound pressure vs decibel power. No device is for acoustical reproduction is completely linear.

 

[dknguyen]

I assume you have examples where a dB isn't a dB, such as where the dB power gain is different from the dB voltage gain etc.

The point I was trying to make is that the 10log and 20log thing is simply an artifice to measure power ratios - you still get the same dBs. I think if you consult any authoritive reference you will find the bel is defined as log10 of a power ratio. It is then consistent to define it also as 2log10 of an amplitude ratio.

The bel is DEFINED as a power ratio? Okay, that's the type of thing I was looking for. So dB is always the power ratio for something or resulting power from whatever ratio of whatever amplitudes, and there is no such thing as dB to represent an amplitude ratio (though the dB can be converted with the 20log formula to produce the voltage ratio that would caues it). THen it makes sense to use 20log for amplitudes the resultant dB always represents a power.

But dBV? I've personally never seen a dBV. But the main thing I was concerned about is if you ran across a dB that was calculated using an amplitude, but you weren't sure whether the dB represented a power gain or an amplitude gain. If dB is always defined as a power gain, then you always know which formula to use (10log if you want to calculate the linear power gain, or 20log if you want to calculate the amplitude gain that produces that power gain).

EDIT: Okay, so it would seem dBV is using 1V as a reference, so I would think this technically means it is using the power in a 1V signal as the reference...with the actual power depending on the details of the system.

I suppose it also removes some confusion if dB is always a power ratio...you just say dB and the user can use whatever formula they want to convert that power ratio to a resultant linear power ratio or linear amplitude ratio. If there are power dB and amplitude dB floating around then if someone gets lazy and just writes dB rather than dB_there would be more confusion than there already is in practical matters.

 

[Tesla23]

Tesla, simple example, decibel sound pressure vs decibel power. No device is for acoustical reproduction is completely linear.

Sorry, I still don't understand where the difficulty lies. I'm actually interested, can you give me an example where you get different dB values by considering powers and amplitudes.

 

[Tesla23]

The bel is DEFINED as a power ratio? Okay, that's the type of thing I was looking for. So dB is always the power ratio for something or resulting power from whatever ratio of whatever amplitudes.

YES!

Just to back up what I've been saying, from Fink and Christiansen "Electronic Engineers Handbook":

The bel, named in honour of Alexander Graham Bell, is defined as the common logarithm of the ratio of two powers P1 and P2. Thus the number of bels Nb is
Nb = log10(P2/P1)

for dB add a factor of 10.

It then goes on to say that in terms of electrical quantities that as P=V^2/R = I^2R then
dB = 20log10(V2/V1) + 10log(R1/R2)

and so on.

So for physical quantities where the impedance level is constant, use 20log amplitude.

This particular book predates DSP, but I think the analogies with amplitude and power is pretty clear.

 

[Sceadwian]

Is this okay Tesla?

To calculate the ratio of 1 kW (one kilowatt, or 1000 watts) to 1 W in decibels, use the formula **broken link removed**

• To calculate the ratio of 1 mW (one milliwatt) to 10 W in decibels, use the formula

**broken link removed**

• To find the power ratio corresponding to a 3 dB change in level, use the formula

**broken link removed**

 

[Tesla23]

Is this okay Tesla?

Are you saying that in any of those cases that you would get different results by calculating the ratio of amplitudes?

 

[Bob Scott]

I also used to be confused about dB, power and voltage relationships.

Now repeat carefully after me:

a dB is a dB is a dB

This is exactly correct, and it made me instantly understand it years ago.

If I raise the volume of my stereo by 6dB, I've doubled the voltage going to the speakers and quadrupled the power going to the speakers, and my ears hear a 6dB increase. Power is up by 6dB. Voltage is up by 6dB.

There are no such things as power dBs or amplitude dBs.

The thing that makes it possible to measure dBs using either amplitude or power ratios is that the number of dBs is 10 log a power ratio or 20 log an amplitude ratio. It's that simple.

It sounds like Matlab presents this in some confusing way, I don't have it so can't test it.

dB's are handy for plotting filter response graphs, as all we technical types know. If you also have a logarithmic frequency axis, just like the layout of a piano keyboard, it makes your response graph linear. You draw straight lines.

Also,
Logarithmic = Exponential.
There is no difference.

 

[dougy83]

I also used to be confused about Db, power and voltage relationships. If I raise the volume of my stereo by 6Db, I've doubled the voltage going to the speakers and quadrupled the power going to the speakers, and my ears hear a 6Db increase. Power is up by 6Db. Voltage is up by 6Db.

And you're still confused with how to spell it =)

Also,
Logarithmic = Exponential.
There is no difference.

You will find that the logarithmic and exponential functions are inverses of each other. They are related, but very much different.

 

[Tony Stewart]

Anything with a controlled impedance is generally defined in dB ratios of power such as all devices, components , filters, amplifiers , Spectrum Analyzers with 50 Ohms.

- everything else is usually defined in dB for ratios of voltage due to changing impedances in the circuit from high to low.

- exceptions might include 300 or 600 Ohm phone lines because the impedance changes with the number of phones connected.

absolute dB readings always have a suffix like dBmV, dBuV, for voltage and dBm or dBW or dBuW for power

Relative readings of dB are all the same in the context of ratios , but the context is usually implied by a controlled impedance application for inputs and outputs to mean power.

Otherwise electronics with unknown or differing impedance implies Amplitude and not Power ratio.

Voltage Ratio in dB differs by a factor of Power dB ratios x2.

Half voltage is -6dB is not the same as half power @ -3dB = .707V/1V

So a dB is any log10 ratio. Since deci (-bel) means base 10 logarithm.
In the context of a controlled impedance like 50 Ohm it is always power, if not specified otherwise.

 

[Ian Rogers]

6 year old thread!!! I doubt any of the above ( apart from Dougy ) come here anymore..

 

[steveB]

Lol, my wife does the same thing. - always bringing up issues from the past!

 

[AnalogKid]

For example, the cutoff frequency is defined as the -3dB point for POWER...not amplitude.

Don't think so. In classic active filter design terminology, for example for 1-pole and 2-pole Butterworths, both cutoff frequencies are where the amplitude response is down 3 dB with respect to the passband amplitude, while the attenuation slope is 6 dB or 12 dB per octave depending on the number of poles.

ak

 

[DerStrom8]

AK, did you not see Post #35?

 

[AnalogKid]

Historically, I am blind to thread start dates. duh.

 

[Chad Hidalgo]

This thread went all kinds of crazy. I don't have MATLAB, but because power is measured in Watts (or milliWatts as the case may be), if you want the dBm value that tells you the attenuation or gain of the signal, you have to use 10 log(Output/Input). If you are calculating using voltage or current, the formula is 20 log (Output/Input) because power is proportional to the square of the signal's amplitude (voltage). The logarithmic squares law applies.