Let’s turn now to explore more of the mathematics of concepts related to acoustics.
In Section 2, Table 4.2 lists some general guidelines regarding sound perception, and Table 4.5 gives some rules of thumb regarding power or voltage changes converted to decibels. We can’t mathematically prove the relationships in Table 4.2 because they’re based on subjective human perception, but we can prove the relationships in Table 4.5.
First let’s prove that if we double the power in watts, we get a 3 dB increase. As you work through this example, you see that you don’t always use decibels related to the reference points in Table 4.3. (That is, the standard reference point is not always the value in the denominator.) Sometimes you compare one wattage level to another, or one voltage level to another, or one sound pressure level to another, wanting to know the difference between the two in decibels. In those cases, the answer represents a difference in two wattage, voltage, or sound pressure levels, and it is measured in dB.
In general, to compare two power levels, we use the following:
[equation caption=”Equation 4.15″]
The difference in decibels between power $$P_{0}$$ and power $$P_{1}$$ = $$10\log_{10}\left ( \frac{P_{0}}{P_{1}} \right )$$
[/equation]
If $$P_{1} = 2P_{0}$$ then we have
$$!10\log_{10}\left ( \frac{2P_{0}}{P_{0}} \right )=10\log_{10}2\approx 3\: dB\: increase$$
You can illustrate this rule of thumb with two specific wattage levels – for example 1000 W and 500 W. First, convert watts to dBm. Table 4.3 gives the reference point for the definitions of dBm, dBW, dBV, an dBu. The table shows that dBM uses 0.001 W as the reference point, which means that it is in the denominator inside the log.
$$!10\log_{10}\left ( \frac{1000}{0.001} \right )=60\: dBm$$
Thus, 1000 W is 60 dBm.
What is 500 W in dBM? The standard reference point for dBm is 0.001 W. This yields.
$$!10\log_{10}\left ( \frac{500}{0.001} \right )\approx 57\: dBm$$
We see that 500 W is about 57 dBm, confirming that doubling the wattage results in a 3 dB increase, just as we predicted. We get the same result if we compute the increase in decibels based on dBW. dBW uses a reference point of 1 W in the denominator.
$$!10\log_{10}\left ( \frac{1000}{1} \right )= 30\: dBW$$
1000 W is about 30 dBW.
$$!10\log_{10}\left ( \frac{500}{1} \right )\approx 27\: dBW$$
500 W is about 27 dBW. Again, doubling the wattage results in a 3 dB increase, as predicted.
Continuing with Table 4.5, we can show that if we multiply power by 10, we have a 10 dB increase in power.
$$!10\log_{10}10= 10\: dB \:increase \:in \:power$$
If we divide the power by 10, we get a 10 dB decrease in power.
$$!10\log_{10}\left ( \frac{1}{10} \right )= -10\: dB \:decrease \:in \:power$$
For voltage, we use the formula $$20\log_{10}\left ( \frac{V_{1}}{V_{0}} \right )$$, as shown in Table 4.3. From this we can show that if we double the voltage, we have a 6 dB increase.
$$!20\log_{10}2\approx 6\: dB\: increase\: in\: voltage$$
If we multiply the voltage times 10, we get a 20 dB increase
$$!20\log_{10}10= 20\: dB\: increase\: in\: voltage$$
Don’t be fooled into thinking that if we multiply the voltage by 5, we’ll get a 10 dB increase. Instead, multiplying voltage times 5 yields about 14 dB increase in voltage.
$$!20\log_{10}5\approx 14\: dB\: increase\: in\: voltage$$
The rest of the rows in the table related to voltage can be proven similarly.