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

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*[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 *W*_{1} and second the reference power *W*_{2}. This can be written as a power change in dB:

*W*_{1}/*W*_{2} = [(*E*_{1}^{2}/1)/(*E*_{2}^{2}/1)][(1/*R*_{1})/(1/*R*_{2})]

= (*E*_{1}^{2}/*E*_{2}^{2})(*R*_{2}/*R*_{1}). (2-12)

This can be written as a power level:

10log[(*E*_{1}^{2}/*E*_{2}^{2})(*R*_{2}/*R*_{1})] = *power change in dB*. (2-13)

or

20log(*E*_{1}/*E*_{2}) + 10log(*R*_{2}/*R*_{1}) = *power change in dB*. (2-14)

*Special Circumstance*

When *R*_{1} = *R*_{2} and *only* then:

*Power level in dB* = 20log(*E*_{1}/*E*_{2}) (2-15)

where,

*E*_{2} is the voltage associated with the reference power.

**2.6 Conventional Practice**

When calculating power level in dBm, we commonly make *E*_{2} = 0.775 V and *R*_{2} = 600 O.Note that *E*_{2} may be any voltage and *R*_{2} 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(*E*_{1}^{2}/*E*_{2}^{2}) = 10 log(*W*_{1}/*W*_{2})

when *R*_{1} = *R*_{2}

2 x 10log(*E*_{1}/*E*_{2}) = 20log(*E*_{1}/*E*_{2})

= 10 log(*W*_{1}/*W*_{2})

3. Power definitions:

Apparent power = *E* x *I* or *E*^{2}/*Z*,

The average real or absorbed power is (*E*^{2}/*Z*)cos?,

The reactive power is (*E*^{2}/*Z*)sin?,

Power factor = cos?

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 O, then:

*dB level to a reference* = 10log[(*E*_{1}^{2}/*E*_{2}^{2})(*R*_{2}/*R*_{1})] (2-16)

where,

*E*_{2} = 0.77459...V,

*R*_{2} = 600 O

then:

*R*_{2}/*E*_{2}^{2} = 1000

and 1/1000 = 0.001. Note that any *E*_{2} and *R*_{2} that result in a power of 0.001 W may be used. We can then write:

*Level (in dBm)* = 10log(*E*_{1}^{2}/0.001*R*_{1} (2-17)

and

*E*_{1} = v(0.001*R*_{1}(10^{dBm/10})) (2-18)

*R*_{1} = *E*^{2}/(0.001(10^{dBm/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 10^{9}/10^{-3}) = 125 dBm.

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