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# Decibels

**Decibel** is a logarithmic expression used in filter applications as a
unit of gain or loss. Characteristic of logarithmic operations, processes
involving multiplication or division are reduced to additions or subtractions,
that in many but not all cases is advantageous because the calculation is
simpler. Decibels have come to be used in a manner that is not strictly
correct according to the original definition; however, the massive use of
the incorrect form on manufacturers' data sheets, in books, and throughout
the electronics industry in general has virtually forced a redefinition of
the term decibel.

The *original* definition of decibels is embodied in the figure above.
The box represents an amplifier, filter, or other device that may be passive or
active. Input power *P*_{1} across input impedance
*Z*_{1} with a phase angle *φ*_{1} results in an
input voltage *V*_{1}. Correspondingly, at the output there
appears power *P*_{2} in *Z*_{2} (with a phase angle
*φ*_{2}) along with *V*_{2}. The ratio of
output power to input power of the circuit is

The power ratio computed by this equation has no units. The number of decibels denoting the power ratio is defined as

The power gain or loss in decibels (dB) is therefore found by taking the logarithm to the base 10 of the power ratio and multiplying it by 10. Combining the previous two equations gives

If the impedances are equal, the last two terms can be dropped. The remaining term can be manipulated to become

Note that this equation is an expression for the *voltage* gain or loss
of the circuit expressed in dB. Strictly speaking, it should be used only
when the input and output impedances are equal; for example, in a 600-ohm
telephone system; however, this expression for voltage gain is used
frequently even when the impedances are not equal.
For simplicity, and because it is used extensively in literature,
this equation is used to define voltage gain also in this textbook.

**Example 1**:

An amplifier has a voltage gain of 2; express this in dB.

**Solution**:

*N* = 20 log_{10} 2 = 6.02 dB

Accordingly, an amplifier with a gain of 2 is usually said to have a gain of 6 dB.

**Example 2**:

A circuit attenuates an input voltage by a factor of 2; express its gain in dB.

**Solution**:

*N* = 20 log_{10} 1/2 = -6.02 dB

Note that for a gain or loss of 2, the numerical answer is the same, namely 6 dB, but the gain is indicated by a plus sign while a loss or attenuation is indicated by a minus sign.

**Example 3**:

An amplifier has a 3-position gain switch to provide gains of 10, 1.0, and 0.1;
express these gains in dB.

**Solution**:

Gain = 10: *N* = 20 log_{10} 10 = 20 dB

Gain = 1: *N* = 20 log_{10} 1 = 0 dB

Gain = 0.1: *N* = 20 log_{10} 0.1 = -20 dB

The decibel is useful for representing large gains or losses. For example,
to say that a voltage signal has increased 100 dB is much easier than to
say the signal has increased 100,000 times.
The value of using decibels in *all* cases to describe gain is highly
questionable. For example, it is probably better to say that an amplifier has
a gain of 100 than to state that its gain is 40 dB. Moreover, the use of dB
to simplify cascaded-stage gain calculations can result in complication rather
than simplification. This is illustrated by the following example:

**Example 4**:

Three amplifiers with gains of 3, 4, and 5 are connected in series. Express
the overall gain as a number and in dB.

**Solution**:

Gain = 3×4×5 = 60

*N* = 9.5424 + 12.0412 + 13.9794 = 35.563 dB

In the above example, it is obvious that computations have not been simplified by adding dB's instead of multiplying gains, in fact, it is clear that expressing performance in decibels does not enhance comprehension of what the amplifier chain is doing in this case. Nevertheless, decibels will be used extensively in this textbook because they are widely used in filter design and because their use provides convenient graphs of filter performance.