Decibels – dB

Decibels – dB

The decibel (dB) is a key unit of measurement in RF and microwave engineering

In a typical RF communications system, the signal power transmitted from an antenna may be anything from a few watts up to tens of kilowatts (10kW). In contrast, the signal power at the receiver antenna may only be a few picowatts (10-12 W) in strength.

Using the conventional decimal number system, the ratio of transmitted to received signal power may be of the order 10,000,000,000,000,000:1. Such large ratios are difficult to work with, not to mention conceptualise. The Bel scale was created to simplify working with such large ratio numbers.

The Bel was created by Alexander Graham Bell, often thought of as the father of telephony. The Bel is the logarithm to the base 10 of a power ratio. So the ratio \(P\) of two powers \(P_1\) and \(P_2\) in units of Bel is:

\(P\:(Bel) = \log _{10}\bigg(\displaystyle \frac{P_1}{P_2}\bigg)\)

Some examples expressed in Bels are shown in the following table:

Number Bel
100 2
10 1
2 0.3
1 0
0.5 -0.3
0.1 -1
0.01 -2

Decibel Scale

Although the Bel is useful when working with large power ratios, the increments between Bels are generally too great in practice. For example, one Bel is a factor of 10, but two Bels is a factor of 100. The Bel is not convenient for expressing smaller increments between power levels. The decibel (dB) scale was created to make working with smaller power increments more convenient.

Power Ratios in Decibels

To calculate the ratio of two numbers in decibels, the base 10 logarithm is multiplied by 10. So the ratio \(P\) of two power levels \(P_1\) and \(P_2\) expressed in decibels is:

\(P \:(dB) = 10\log _{10}\bigg(\displaystyle \frac{P_1}{P_2}\bigg)\)

When expressing base 10 logarithms, the 10 subscript is normally dropped and so:

\(P \:(dB) = 10\log\bigg(\displaystyle \frac{P_1}{P_2}\bigg)\)

Voltage Ratios in Decibels

Since power is proportional to voltage squared, the ratio \(V\) in decibels of two voltages \(V_1\) and \(V_2\) for a given impedance level is found by multiplying the base 10 log of the ratio by 20:

\(V\:(dB) = 10\log\bigg(\displaystyle \frac{V_1}{V_2}\bigg)^2 = 2 \times 10\log\bigg(\displaystyle \frac{V_1}{V_2}\bigg) = 20\log\bigg(\displaystyle \frac{V_1}{V_2}\bigg)\)

The above can also be applied to currents by replacing each voltage with the equivalent current.

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2018-09-22T14:36:32+00:00
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