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# Using the Decibel - Part 2: Expressing Power as an Audio Level

## 6/4/2008 2:37 PM EDT

You can receive a 15% discount on "Sound System Engineering" by going to www.elsevierdirect.com and typing in the order code 92837 when ordering, or by calling 1-800-545-2522 and mentioning 92837 when ordering. Offer expires 6/30/2008.

[Part 1 introduces the decibel and examines concepts underlying its use in sound systems.]

2.5 Expressing Power as an Audio Level
The reference power is 0.001 W (one milliwatt). When expressed as a level, this power is called 0 dBm (0 dB referenced to 1 mW).

Thus, to express a power level we need two powers - first the measured power W1 and second the reference power W2. This can be written as a power change in dB:

W1/W2 = [(E12/1)/(E22/1)][(1/R1)/(1/R2)]

= (E12/E22)(R2/R1).              (2-12)

This can be written as a power level:

10log[(E12/E22)(R2/R1)] = power change in dB.               (2-13)

or

20log(E1/E2) + 10log(R2/R1) = power change in dB.         (2-14)

Special Circumstance
When R1 = R2 and only then:

Power level in dB = 20log(E1/E2)               (2-15)

where,
E2 is the voltage associated with the reference power.

2.6 Conventional Practice
When calculating power level in dBm, we commonly make E2 = 0.775 V and R2 = 600 Ω.Note that E2 may be any voltage and R2 any resistance so long as together they represent 0.001 W.

Levels in dB
1. The term "level" is always used for a power expressed in decibels.

2. 10log(E12/E22) = 10 log(W1/W2)

when R1 = R2

2 x 10log(E1/E2) = 20log(E1/E2)
= 10 log(W1/W2)

3. Power definitions:
Apparent power = E x I or E2/Z,
The average real or absorbed power is (E2/Z)cosθ,
The reactive power is (E2/Z)sinθ,
Power factor = cosθ

4. The term "gain" or "loss" always means the power gain or power loss at the system's output due to the device under test.

Practical Variations of the dBm Equations
When the reference is the audio standard, i.e., 0.77459 V and 600 Ω, then:

dB level to a reference = 10log[(E12/E22)(R2/R1)]    (2-16)

where,
E2 = 0.77459...V,
R2 = 600 Ω

then:

R2/E22 = 1000

and 1/1000 = 0.001. Note that any E2 and R2 that result in a power of 0.001 W may be used. We can then write:

Level (in dBm) = 10log(E12/0.001R1               (2-17)

and

E1 = √(0.001R1(10dBm/10))               (2-18)

R1 = E2/(0.001(10dBm/10))               (2-19)

See Fig. 2-3.

Figure 2-3. Power in dB across a load versus available input power.

For all of the values in Table 2-2 the only thing known is the voltage. The indication is not a level. The apparent level can only be true across the actual reference impedance. Finally, the presence or absence of an attenuator or other sensitivity control is not known. See Section 2.20 for explanation of VU.

The power output of Boulder Dam is said to be approximately 3,160,000,000 W. Expressed in dBm, this output would be:

10log(3.16 x 109/10-3) = 125 dBm.

Table 2-2. Root Mean Square Voltages Used as Nonstandard References

sharps_eng

5/15/2011 11:57 AM EDT

I am finding this series a useful recap. Although I started in pro-audio I was seduced by embedded processors and digital video, but audio was definitely my first love - it's so damn analog!
The thrill was that to achieve barely acceptable SNRs of -94dB when working with 26dB of headroom took everything we had - the best brains, the best parts and then some. But we did it, and othing much has changed in terms of the net quality delivered, especially in a digital world where we can mix at 96kHz and 24 to 96bits if we want but the compression artefacts in the listeners' source material will reduce the SNR to 12 bits, or worse if we include class H amplification side-effects.
Hopefully the incease in available bandwidth will increase the amount of releases on 16bit linear PCM source, just to get rid of that quantisation weebling and psychacoustic algorithmic breathing!
Hifis still can't reproduce a piano in full voice, let alone a drum kit - maybe a good system a Marshall stack heard from way off.

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