Decibel Conversion
Presented by: Dr Eman Morsi
 The use of decibels is widespread throughout the electronics
industry. Many electronic instruments are calibrated in
decibels, and it is common for data sheets describing various
instruments and devices to give the specifications in terms of
decibels. For these reasons a study of their use is essential.
 Let us begin by considering the situation depicted by Fig. 1. A
voltage generator is driving an amplifier. The input resistance
of the amplifier is R3.
Fig. 1 Developing the decibel concept
 The amplifier delivers an output voltage to a load resistor
whose value is R2. In describing the gain of the amplifier we
may speak of either the power gain or the voltage gain. The
power gain G is-defined as the power delivered to R2 divided
by the power delivered to the amplifier resistance R1.That is,
2
P  / R 2   2  R1
G  2  22
  
P1  1 / R1   2  R 2
2
(1)
 The voltage gain A is defined simply as the output voltage
divided by the input voltage . That is
2
A
1
(2)
It will be recalled from basic electronics courses that the power
gain in decibels is defined as
Gdb = 10 log G
(3)
where G is the power gain and, Gdb is the decibel equivalent
of G. Note that the base 10 is used in Eq. (3).
 If we like, we can obtain an alternate formula for Gdb by
substituting Eq. (1) into Eq. (3) to obtain
2
G db  10 log G  10 log 
 1
2
 R1

 R2
(4)
2
R1
Gdb  20 log
 10 log
1
R2
(5)
 The properties of logarithms allow us to rewrite this
equation as
 Historically, decibels were defined strictly for use with power
ratios. However, modern usage of decibels has developed to
the point where it is now common to speak of either the
power gain in decibels or the voltage gain in decibels. The
voltage gain in decibels is the first term of Eq. (5), that is,
Adb = 20 log A
(6)
where A is the voltage gain, , and Adb is the decibel equivalent
of A. Thus, we can rewrite Eq. (5) simply as
Gdb  Adb  10 log
R1
R2
(7)
 Note carefully that if R1/R2 is unity, then Gdb = Adb. If R1/R2
does not equal unity, then Gdb Adb. It is very important,
therefore, when using decibel equivalents to specify voltage
gain or power gain. In other words, it is not enough to say
that the gain is so many decibels. We must specify that the
voltage gain is so many decibels or that' the power gain is so
many decibels, whichever the case may be.
 Let us summarize the use of decibels up to this point. In the
modern usage of decibels we may speak of either the decibel
equivalent of a power ratio or the decibel equivalent of a
voltage ratio. The decibel equivalent of a power ratio is given
by
G db  10 log G
(8)
where G is the ratio of two powers, p2/p1. The decibel
equivalent of a voltage ratio is given by
Adb = 20 log A
(9)
 where A is the ratio of two voltages, . In speaking of the gain
of an amplifier, Gdb = Adb only if the load resistance R2 equals
the input resistances R1. If the two resistances are not equal,
it is essential to specify whether the gain in decibels is for the
voltage gain or for the power gain.
 In the remainder of this section we develop some shortcuts




that allow rapid conversion between ordinary numbers and
decibel equivalents. Adb is a function of A, meaning that each
value of A selected, only one value of Adb can be calculated.
For instance,
When A = 1
Adb = 20 log 1 = 0 db
When A = 2
Adb = 20 log 2 = 20(0.3) = 6 db
When A = 4
Adb = 20 log 4 = 20(0.6) = 12 db
When A = 10
Adb = 20 log 10 = 20(1) = 20 db
 We can continue in this fashion until there are enough pairs
of A and Adb to tabulate as shown in Table 1.
To bring out certain properties of decibels more clearly, let us
graph the data of Table 1, using semi-logarithmic paper to
compress the A values .
Table 1
Voltage Gain and Its Decibel Equivalent
A
Adb
1
100
1
10
1
8
1
4
½
-40
-20
-18
-12
-6
1
0
2
6
4
12
8
18
10
20
100
40
1000
60
 First, note in Fig. 2 a than each time A is increased by a factor
of two.
A = 1, 2, 4, S, …
Adb increases by 6 db
Adb = 0, 6, 12, 18, …
Conversely, each time that A decreases by a factor of two
A = 1, ½ , ¼ , , …
Adb decreases by 6 db
Adb = 0, -6, -12, -18, …
 Adb increases by 20db
Adb = 0, 20, 40, 60, ….
Fig. 2Graphs of Adb VS. A.
,…
 Conversely, each time A is decreased by a factor of te
1
1
1
A = 1, 10 , 100 , 1000
 Adb decreases by 20 db
Adb = 0, -20, -40, 60, …
These properties of decibels make the conversion from
ordinary numbers into decibels a simple matter. We need
only express the ordinary number in factors of two and ten
and convert according to the decibel properties described. As
an example, let us convert A= 4000 into its decibel
equivalent.
A = 4000 = 2 . 2 . 10 . 10 .10
Adb = 6 + 6 + 20 + 20 + 20 = 72db
 We have factored A into twos and tens and added 6 or 20 db
for each factor of two or ten to obtain the total of 72 db.
As another example, consider A = 0.004. We write this as a
fraction and then factor into twos and tens
A  0.004 
4
2.2

1000 10 .10 .10
Adb  6  6  20  20  20   48db
 In this case we add 6 db for the numerator factors and
subtract 20 db for each denominator factor.
 When the value of A is not exactly factorable into twos and
tens, we can obtain an approximate answer by interpolation.
For instance, if A = 60, we observe that this is a number
between A = 40 and A = 80. Hence,
Since A = 40 = 2 - 2 - 10
Adb = 32 db
Since A = 80 =2-2-2-10
Adb = 38 db
 A = 60 is halfway between 40 and 80, so that Adb 35 db. We
have interpolated to find the approximate decibel equivalent
of A = 60. using the exact formula Adb = 20 log 60, yields a
value of Adb = 35.56db. The error in our approximate answer
is only about 0.5db. Usually, errors of less than 1 db are
acceptable in practice. Only in those situations where the
greatest is required must we use the exact formula,
Adb = 20 log A
 Let us summarize our procedure for finding decibel equivalents.
 1-For any ratio A of voltages or currents, express the number in
factors of two and ten If the number is not exactly factorable into
twos and tens, bracket between the next lower and higher
numbers that are factorable into twos and tens.
 2-Add 6 db for every factor of two in the numerator and 20 db for
every factor of ten. Subtract 6 db for every factor of two in the
denominator and 20 db for every factor of ten.
 3-Interpolate, if necessary, to obtain the decibel equivalent.
 4-When dealing with a power ratio G, proceed as in steps 1 to 3
but -divide the result by 2 to obtain Gdb.
EXAMPLE 1
 Find the decibel equivalent of A = 2000.
SOLUTION
A = 2000 = 2.10 . 10 . 10
Adb = 6 + 20 + 20 + 20 = 66 db
EXAMPLE 2
 Find the decibel equivalent of A = 3000
SOLUTION
 This number is not factorable into twos and tens, but it lies
between 2000 and 4000, numbers which are so factorable.
A = 2000 = 2 . 10 . 10 . 10
Adb = 6 + 20 + 20 + 20 = 66db
A = 4000 implies that we add 6 db to obtain Adb = 72 db.
For A = 3000, we interpolate to obtain Adb = 69 db.
(The exact answer is 69.5 db. Whenever we interpolate the
maximum error possible is about 0.5 db.)
EXAMPLE 3
 Find the decibel equivalent of P2/P1 = 2000
SOLUTION
This is a ratio of two powers. The decibel equivalent of a
power ratio is one-half the decibel equivalent of a voltage
ratio of the same numerical value. We need only proceed in
our usual manner and divide the answer by 2.
2000 = 2 . 10 . 10 . 10
6 + 20 + 20 + 20 = 66db
Hence,
Gdb = 33 db
EXAMPLE 4
 An amplifier has an input voltage of 1 my and an output
voltage of 1.6 volts. Express the voltage gain of the amplifier
in decibels.
 SOLUTION
The voltage gain of the amplifier is the output voltage divided
by the input voltage.
A
1.6
 1600  2 . 2 . 2 . 2 .10 .10
3
10
Adb = 6+6+6+6+20+20 = 64db
EXAMPLE 5
 Find the decibel equivalent of A =
1
200
 SOLUTION
A
1
1

200 2 .10 .10
Adb = -6 –20 –20 = 46 db
EXAMPLE 6
 Voltages are often expressed in decibel equivalents by
comparing their value to a reference voltage. In Fig.3,
suppose that we use a reference voltage of 0.5 volt. Form the
ratio of each given voltage to 0.5 volt and find the decibel
equivalent of these ratios.
Fig. 3
SOLUTION:
 For 0.1 volt, we have
0 .1 2

0.5 10
 which has a decibel equivalent of -14 db. Hence, we would
say that the first voltage, 0.1 volt, is -14 db with respect to
0.5 volt.
 In a similar way, the ratio of the second voltage to the
reference voltage is
1 .5
3
0 .5
 which has a decibel equivalent of about 9 db.
Finally, the ratio of 10 volts to 0.5 volt is
10
 20
0.5
which has a decibel equivalent of 26 db.
Hence, our system can be labeled with the decibel equivalents
as shown in Fig. 3b. It is important to realize that these
decibel values have the correct meaning only for a reference
voltage of 0.5 volt. Had we chosen a different reference
voltage, the decibel equivalents would all be different from
those shown.
Decibel Gain of a System
 One important reason for the use of decibels is that for a
system consisting of many stages, the overall gain in decibels
is the sum of the stage gains expressed in decibels.
Fig.4 The decibel gain of a cascade of stages.
A1, A2, and A3 are the voltage gains of each stage expressed in
ordinary numbers, that is, as ratios. For instance, the first stage
may have a voltage gain of 100, so that A1 is 100, meaning that the
output voltage divided by the input voltage is 100.
Decibel Gain of a System
 To find the ordinary voltage gain of the entire system we
already know that the gains are multiplied
A = A1 A 2 A 3
where A is the overall gain.
Let us find the decibel equivalent of the overall gain.
Adb = 20 log A = 20 log A1 A2 A3
Recall that the logarithm of a product of numbers is equal to,
the sum of the logarithms of each number.
 Adb = 20 (log A1 + log A2 + log A3)
= 20 log A1 + 20 log A2 + 20 log A3
Each term on the right-hand side of the last equation is merely the
decibel gain of each stage. Hence,
 Adb = A1 (db) + A2(db) + A3(db)
(10)
Equation (10) tells us that the overall decibel gain is the sum of the
decibel gains of the individual stages. This property is another reason for
the popularity of decibels. If we work with decibel gains, we add the
stage gains to find the overall gain. This is considerably easier than
working with ordinary gains, where it is necessary to multiply to find
the overall gain.
 In practice, we will find that voltmeters often have a decibel
scale, so that the gain of a stage can be measured in decibels.
For instance, on some voltmeters a reference voltage of 0.77
volt is used. A decibel scale is provided on the meter face, so
that all voltages can be. read in decibels with respect to 0.775
volt.
 We might find, for example, that the input to a stage reads -
10 db and the output reads +20 db. The gain of the stage is
the algebraic difference between these two values, or 30 db.
In this way, the decibel gains of different stages are easily
found. Once they are known, they can be added to find the
overall gain of a system in decibels.
EXAMPLE 7
 Find the overall gain for the system of the following Fig.
SOLUTION
Adb = 20 – 10 + 35 = 45 db
EXAMPLE 8
 A data sheet for an amplifier specifies that the voltage gain is
40 db. If we cascade three amplifiers of this type, what is the
overall gain expressed as an ordinary number?
SOLUTION
Adb = 40 + 40 + 40 = 120db
For every 20db we know that there is a factor of ten in A.
Hence,
Adb = 20 + 20 + 20 + 20 + 20 + 20
A = 10 . 10 .10 . 10 . 10 . 10 = 106
EXAMPLE 8
 A voltmeter is calibrated in decibels with a reference voltage
of 0.775 volt. What does the voltmeter read in decibels for a
voltage of 3.1 volts?
 SOLUTION

3.1

4
 ref 0.775
The voltmeter will read 12 db, meaning that given voltage is
four times greater than the reference of 0.775 volt.