A Brief Treatise on the Decibel

Contributed by George T. Baker, W5YR

Stated simply, the problem is that decibels never change, but people do - and therein lies the problem. They tend to memorize formulas and then blindly apply them without thinking through what they are doing.

Enter lecture mode: press BACK on your browser now to move on . . .

The decibel first, foremost and always is and always has been a logarithmically based representation of a power ratio. Period. Yet we find people using the classic formula

dB = 10 log10 (X2/X1)

for expressing just about any ratio X2/X1 regardless of what X2 and X1 are physically. This leads to confusion. The above formula yields decibels only if X2 and X1 are power values. This goes back to Bell Labs' original definition of the the Bel as 10 times the common logarithm (log10) of the ratio between two levels.

Now, nothing says that I cannot use that formula to express the ratio logarithmically of any two quantities; the problem comes in calling the results "decibels" (dB) which in the engineering world always connote a power ratio, regardless of how they are calculated. Once something has been expressed in dB, correctly or not, those "dB" tend to be treated just like "real dB", and that is where the problems come in. For example, we are so accustomed to the notion that doubling power is equivalent to increasing power by 3 dB, that we can fall into the trap of thinking that any increase by a factor of two corresponds to that same 3 dB. But, it may not.

The decibel can be calculated from either voltage or current ratios, and most commonly is, but doing so implies a common value of resistance (not impedance) for the voltages or currents involved. This requirement is often overlooked or simply taken for granted.

Errors then result . . .

For example, what is the power gain of an audio amplifier when 100 millivolts of input signal produces 5 volts of output signal?

Well, the formula says:

dB = 20 log10 (V2/V1)

where V2 is the output voltage (5 volts) and V1 is the input voltage (0.1 volt). The answer is clearly:

dB = 20 log10 (5/0.1)

which my calculator shows to be just a hair under 34 dB.

Is that correct? Only if the resistances across which the input and output voltages are measured are the same. The input resistance might be 10,000 ohms while the output resistance might be 8 ohms. For the sake of clarity, let's do this one step at a time.

First the actual input power is (0.1)2/10,000 which is 1 X 10-6 watts or one microwatt.

The output power is (5)2/8 which is 3.125 watts.

The actual power ratio then is 3.125/0.000001 which is 3,125,000.

So the gain of the amp in dB is:

dB = 10 log10 (3,125,000)

which is just under 65 dB. Quite a respectable gain and far different from the 34 dB that the "formula" gave us. The error is about 31 dB or a factor of a little more than 1000.

Let's compare that with the overall gain of a modern amateur receiver which can produce 2 watts of audio from the speaker when presented with a 0.1 µV (microvolt) signal across 50 ohms at the antenna input. The input power is:

{(0.1 X 10-6)2}/50 = 2 X 10-16 watts

The output power is given by the manufacturer as 2 watts of audio.

The overall gain of the receiver then is:

dB = 10 log10 2/(2 X 10-16) = 160 dB,

not an unreasonable result compared to the specs of current radios which can start with a signal 3 dB above a noise floor of -130 dbm and produce 2 watts of audio noise output. That performance amounts to an overall gain of 160 dB to go from 127 dB below one milliwatt to 33 dB above one milliwatt (2000 mW, or 2W, output).

On the other hand, if we assume that the 2 watts was measured across an 8-ohm resistive load, the voltage calculates to be:

P = (V2)/R or V = (PR)-2

So, the output voltage across 8 ohms is:

V = (2 X 8)-2 = 4 volts

The "blind" formula finds the receiver gain to be:

dB = 20 log10 4/(1 X 10-7) = 152 dB

which is in error by about 8 dB with respect to the correct gain figure. In this case, the erroneous use of the formula produced a fairly close answer only because the input resistance of 50 ohm is not all that far from the output resistance of 8 ohms.

So where did that 8 dB of error come from? Let's look into this mystery.


dB = 10 log10 (50/8) = 7.96 dB

which is exactly the difference between our approximate formula application neglecting different value of resistances and the correct result calculated from the power ratio.

Similarly, if we look back at the first example, we were in error by about 31 dB by neglecting the actual resistances as being different. Now look at the result of:

dB = 10 log10 (10,000/8) = 30.97 dB

which is our missing "about 31 dB."

What magic is this? First you tell us that we cannot just use voltage or current values alone to get a correct answer, yet if we do, we get a result that can be corrected by doing a second calculation involving only a resistance ratio. Yet, you say that only voltage, current or power can be used to calculate dB.

The answer, of course, is no mystery. I will let the gentle reader bring forth his rusty algebra talents to show how dealing separately with the voltage ratios and the resistance ratios produces two dB values which, when properly combined, produce the correct answer.

But don't let mathematical cuteness obscure the basic truth that a dB always represents a power ratio regardless of how it is calculated.

This little demonstration probably doesn't warrant the length of this treatise, but it seems to me that every now and then we need to look back at the fundamentals a bit and refresh our memories. Those who disagree will have already hit BACK at the start, so nobody should be offended.

Copyright 2002, George T. Baker, W5YR

Page created by A. Farson. Last updated: 11/27/2015

Microsoft SharePoint Designer 2007.