Presented by: Dr Eman Morsi Decibel Conversion
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.depicted by Fig. 1 Developing the decibel concept
The amplifier delivers an output voltage to a load resistor whose value is R 2. 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 R 2 divided by the power delivered to the amplifier resistance R 1. That is, (1)
The voltage gain A is defined simply as the output voltage divided by the input voltage. That is (2) It will be recalled from basic electronics courses that the power gain in decibels is defined as (3) where G is the power gain and, G db is the decibel equivalent of G. Note that the base 10 is used in Eq. (3). G db = 10 log G
If we like, we can obtain an alternate formula for G db by substituting Eq. (1) into Eq. (3) to obtain (4) The properties of logarithms allow us to rewrite this equation as (5)
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, A db = 20 log A (6) where A is the voltage gain,, and A db is the decibel equivalent of A. Thus, we can rewrite Eq. (5) simply as (7)
Note carefully that if R 1 /R 2 is unity, then G db = A db. If R 1 /R 2 does not equal unity, then G db A db. 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 (8) where G is the ratio of two powers, p 2 /p 1. The decibel equivalent of a voltage ratio is given by A db = 20 log A (9)
where A is the ratio of two voltages,. In speaking of the gain of an amplifier, G db = A db only if the load resistance R 2 equals the input resistances R 1. 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. A db is a function of A, meaning that each value of A selected, only one value of A db can be calculated. For instance, When A = 1A db = 20 log 1 = 0 db When A = 2A db = 20 log 2 = 20(0.3) = 6 db When A = 4A db = 20 log 4 = 20(0.6) = 12 db When A = 10A db = 20 log 10 = 20(1) = 20 db
We can continue in this fashion until there are enough pairs of A and A db 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.
AA db ½ Table 1 Voltage Gain and Its Decibel Equivalent
First, note in Fig. 2 a than each time A is increased by a factor of two. A = 1, 2, 4, S, … A db increases by 6 db A db = 0, 6, 12, 18, … Conversely, each time that A decreases by a factor of two A = 1, ½, ¼,, … A db decreases by 6 db Adb = 0, -6, -12, -18, …
A db increases by 20 db A db = 0, 20, 40, 60, …. Fig. 2Graphs of A db VS. A.
Conversely, each time A is decreased by a factor of te A db decreases by 20 db A db = 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 = 1,, …
A = 4000 = A db = = 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 = We write this as a fraction and then factor into twos and tens
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 = A db = 32 db Since A = 80 = A db = 38 db
A = 60 is halfway between 40 and 80, so that A db 35 db. We have interpolated to find the approximate decibel equivalent of A = 60. using the exact formula A db = 20 log 60, yields a value of A db = 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, A db = 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 G db.
EXAMPLE 1 Find the decibel equivalent of A = SOLUTION A = 2000 = A db = = 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 = A db = = 66db A = 4000 implies that we add 6 db to obtain A db = 72 db. For A = 3000, we interpolate to obtain A db = 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 P 2 /P 1 = 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 = = 66db Hence,G db = 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 db = = 64db
EXAMPLE 5 Find the decibel equivalent of A = SOLUTION A db = -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 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
which has a decibel equivalent of about 9 db. Finally, the ratio of 10 volts to 0.5 volt is 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. A 1, A 2, and A 3 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 A 1 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 = A 1 A 2 A 3 where A is the overall gain. Let us find the decibel equivalent of the overall gain. A db = 20 log A = 20 log A 1 A 2 A 3 Recall that the logarithm of a product of numbers is equal to, the sum of the logarithms of each number.
A db = 20 (log A 1 + log A 2 + log A 3 ) = 20 log A log A log A 3 Each term on the right-hand side of the last equation is merely the decibel gain of each stage. Hence, A db = A 1 (db) + A 2 (db) + A 3 (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 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 A db = 20 – = 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 A db = = 120db For every 20db we know that there is a factor of ten in A. Hence, A db = A = = 10 6
EXAMPLE 8 A voltmeter is calibrated in decibels with a reference voltage of volt. What does the voltmeter read in decibels for a voltage of 3.1 volts? SOLUTION The voltmeter will read 12 db, meaning that given voltage is four times greater than the reference of volt.