**by**

Paul H. Bock, Jr. K4MSG

**Hamilton, VA U.S.A.**

Author’s Note: This tutorial was first written in 1992 and was originally intended for digital & software engineers and non-engineers, in the belief that the concepts and explanations might prove useful to anyone wishing to develop a better understanding of the decibel and its value in electronics work.

All references to "telephone company" and "telephone company engineers" are based on anecdotal evidence rather than historical fact, so the author apologizes for any inaccuracies.

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**General.** The decibel, or dB, is a means of expressing the gain of an active device (such as an amplifier) or the loss in a passive device (such as an attenuator or length of cable). It is simply the ratio of output to input expressed in logarithmic form. The decibel was developed by the telephone company to express the gain or loss in telephone transmission systems, and is best understood by first discussing the rationale for its development.

If we have two cascaded amplifiers as shown below, with power gain factors A1 and A2 as indicated, the total gain is the product of the individual gains, or A1 x A2.

Input >——– Amp #1 ——— Amp #2 ——> Output

A1 = 275 ………… A2 = 55

In the example, the total gain factor At = 275 x 55 = 15,125.

Now, imagine for a moment what it would be like to calculate the total gain of a string of amplifiers. It would be a cumbersome task at best, and especially so if there were portions of the cascade which were lossy and reduced the total gain, thereby requiring division as well as multiplication.

It was primarily for the reason stated above that Bell Telephone developed the decibel. Thinking back to the rules for logarithms, we recall that rather than multiplying two numbers we can add their logarithms and then take the antilogarithm of this sum to find the product we would have gotten had we multiplied the two numbers.

Mathematically, log (A x B) = log A + log B

If we want to divide one number into another, we subtract the logarithm of the divisor from the logarithm of the dividend, or in other words

log (A/B) = log A – log B

The telephone company apparently decided that it might be convenient to handle gains and losses this way since it would simplify transmission path calculations. It is also possible that the fact that the human ear’s response to sound is logarithmic may have influenced this decision. In any event, a new unit of gain measurement was invented and called the "Bel," after Alexander Graham Bell. The Bel was defined as

Gain in Bels = log A

where A = Power amplification factor

Going back to our example, we find that log 275 = 2.439 and log 55 = 1.740, so the total gain in our cascade is

2.439 + 1.74 = 4.179 Bels

It probably occurred rather quickly to telephone company engineers that using Bels meant they would be working to at least two decimal places. They couldn’t just round things off to one decimal place, since 4.179 bels is a power gain of 15,101 while 4.2 bels is a power gain of 15,849, yielding an error of about 5%. At that point it was decided to express power gain in units which were equal to one-tenth of a Bel, or in deci-Bels. This simply meant that the gain in Bels would be multiplied by 10, since there would be ten times more decibels than Bels. This changes

the formula to

Gain in decibels (dB) = 10 log A (Eq. 1)

Again using our example, the gain in the cascade is now

24.39 + 17.40 = 41.79 decibels

The answer above is accurate, convenient to work with, and can be rounded off to the first decimal place will little loss in accuracy; 41.79 dB is a power gain of 15,101, while 41.8 dB is a power gain of 15,136, so the error is only 0.23%.

What if the power gain factor is less than one, indicating an actual power loss? The calculation is performed as shown above using Equation 1, but the result will be different. Suppose we have a device whose power gain factor is 0.25, which means that it only outputs one-fourth of the power fed into it? Using Equation 1, we find

G = 10 log (0.25)

G = 10 (-0.60)

G = -6.0 dB

The minus sign occurs because the logarithm of any number less than 1 is always negative. This is convenient, since a power loss expressed in dB will always be a negative number.

**Using the Decibel**

There are two common methods of using the decibel. The first is to express a power gain factor in dB as just described; the second is to determine the power gain factor and convert it to dB, which can all be done in one calculation. The formula for this operation is as follows:

Po

G = 10 log —- (Eq. 2)

Pi

where G = Gain in dB Po = Power output from the device Pi = Power input to the device

Both Po and Pi should be in the same units; i.e., watts, milliwatts, etc. Note that Equation 2 deals with power, not voltage or current; these are handled differently when converted to dB and are covered later. Below are two examples of the correct application of Equation 2:

Ex. 1: An amplifier supplies 3.5 watts of output with an input of 20 milliwatts. What is the gain in dB?

3.5 watts G = 10 log ----------- 0.02 watts

G = 10 log (175)

G = 10 (2.24)

G = 22.4 dB

Ex. 2: A length of coaxial transmission line is being fed with 150 watts from a transmitter, but the power measured at the output end of the line is only 112 watts. What is the line loss in dB?

112 watts G = 10 log ----------- 150 watts

G = 10 log 0.747

G = 10 (-0.127)

G = -1.27 dB

**Alternative Calculation Methods**

It is also possible to calculate gain or loss in decibels using voltage or current measurements rather than power measurements. Power is the product of voltage and current, and is also defined as the square of the current times the circuit resistance, or the square of the voltage divided by the circuit resistance. The mathematical expressions for power are

P = EI P = I2R P= E2/R

where P = power E = voltage I = current R = resistance

Power will thus vary as the square of the voltage or current, so if either voltage or current ratios are measured in a circuit the decibel representation of that ratio will be found from

dB = 20 log E1/E2 *OR* dB = 20 log I1/I2 (Eq. 3)

CAUTION: The circuit resistance must be the same for the two voltage or current measurements, otherwise calculating a dB value will give an erroneous answer.

The following is an example of the same kind of calculation shown previously except that a voltage ratio is used instead of a power ratio (Equation 3):

Ex. 3. An audio engineer measures the RMS voltage level of a test tone at the output of an amplifier running at maximum gain setting and finds it to be 6 volts. He then measures the RMS level of the input signal and finds it to be 135 millivolts. If both the input and output impedances of the amplifier are 600 ohms, what is the gain of the amplifier?

G = 20 log 6volts/0.135 G = 20 log (44.444)

G = 20 (1.65)

G = 32.9 dB

(NOTE: As an exercise, the reader is encouraged to convert the output and input voltages to power in watts using the stated output/input impedance, and then to calculate the amplifier gain in dB using Equation 2. If the mathematics are done correctly the answer will be the same.)

To further expound on the need for maintaining the same reference impedance when calculating decibel gain or loss using voltage (or current), let’s consider what will happen in the example above if the input impedance is 600 ohms as stated but the output impedance is 16 ohms.

If we do the calculation using Equation 3 we arrive at the same answer shown in Example 3; if, however, we first convert to power levels using the impedances just stated we find that the input power is 3.04E-5 watts, while the output power is 2.25 watts. Substituting these two values in Equation 2 (for power):

2.25 watts G = 10 log -------- 3.04E-5 watts

G = 10 log (74,074.1)

G = 10 (4.87)

G = 48.7 dB

Big difference! If the impedances are not equal, the answer found by using Equation 3 is in error by almost 16 dB! This should point out the need for caution in ensuring that impedance levels are constant when using voltage (or current) for decibel calculations. Remember that an audio (or RF) voltmeter doesn’t know what the impedance of the circuit is; it simply registers voltage. In addition, some voltmeters are designed to operate at a certain impedance level, so if the circuit impedance is some other value the voltage reading itself may be erroneous, throwing

yet more confusion into the calculation process. Caution is ever the watchword!

**Some Rules of Thumb**

As with many other mathematical or technical concepts, there are some easy-to-remember "rules of thumb" which can make calculating with decibels much easier. Here are several which will make working with decibels simpler when you don’t have a calculator handy:

a) A power ratio of 2 is equal to 3 dB. Conversely, a power ratio of 1/2 is equal to -3 dB. (NOTE: The -3 dB point is often referred to as a "half-power point," as when describing, say, a frequency response curve.)

b) A voltage or current ratio of 2 is equal to 6 dB. Conversely, a voltage or current ratio of 1/2 is equal to -6 dB.

c) A power ratio of 10 is equal to 10 dB. Conversely, a power ratio of 1/10 is equal to -10 dB.

d) A voltage or current ratio of 10 is equal to 20 dB. Conversely, a voltage or current ratio of 1/10 is equal to -20 dB.

Here is an example of how the rules of thumb can be useful:

Ex. 5: A spurious signal specification for a transmitter states that the maximum level of any spurious signal shall not exceed -36 dB relative to the output power. If the output power is 5 watts, what is the maximum permissible level for spurs?

(a) Since -10 dB is 1/10th, then -20 dB is 1/100th (or 1/10th of 1/10th) and -30 dB is 1/1000th (or 1/10th of 1/10th of 1/10th). So 30 dB below our 5 watt output power becomes 1/1000 x 5 = 0.005 watts.

(b) For the additional -6 dB, remember that -3 dB = 1/2, so -6 dB = 1/4th (or 1/2 of 1/2). Thus, our 0.005 watts previously calculated becomes .005 x 1/4 = 0.00125 watts, or 1.25 milliwatts. This is the maximum power level any spur can have if the transmitter is to meet the stated specification.

**Uses of the Decibel with a Defined Reference**

The most common "defined reference" use of the decibel is the dBm, or decibels relative to a power level of one milliwatt. It is different from the dB because it uses the same specific, measurable power level as a reference in all cases, whereas the dB is relative to whatever reference a particular user chooses or even to no reference at all.

The difference between "relative" and "defined reference" can be understood easily by considering temperature. For example, if I say that it is "20 degrees colder now than it was this morning," it’s a relative measurement; unless the listener knows how cold it was this morning, there is no reference for comparison.

If, however, I say, "It was 20 degrees C this morning, but it’s 20 degrees colder now," then the listener knows exactly what is meant; it is now 0 degrees C. This can be measured on a thermometer and is referenced to a defined temperature scale. So it is with dB and dBm. A dB has no particular defined reference while a dBm is referenced to a specific quantity: the milliwatt (1/1000 of a watt).

{NOTE: The IEEE definition of dBm is "a unit for expression of power level in decibels with reference to a power of 1 milliwatt." Note that no mention is made of the value of circuit impedance; the dBm is merely an expression of power present in a circuit relative to a known fixed amount (i.e., 1 milliwatt) and the circuit impedance is irrelevant.}

We can apply this concept to Equation 1 as follows:

dBm = 10 log (P) (1000 mW/watt)

where dBm = Power in dB referenced to 1 milliwatt P = Power in watts

For example, take the case where we have a power level of 1 milliwatt:

dBm = 10 log (0.001 watt) (1000 mW/watt)

dBm = 10 log (1)

dBm = 10 (0)

dBm = 0

Thus, we see that a power level of 1 milliwatt is 0 dBm. This makes sense intuitively, since our reference power level is also 1 milliwatt. If the power level was 1 watt, however, we find that

dBm = 10 log (1 watt) (1000 mW/watt)

dBm = 10 (3)

dBm = 30

The dBm can also be negative, just like the dB; if our power level is 1 microwatt, we find that

dBm = 10 log (1 x 10E-6 watt) (1000 mW/watt)

dBm = -30 dBm

Since the dBm has a defined reference it can be converted back to watts if desired. Since it is in logarithmic form it may also be conveniently combined with other dB terms, making system analysis easier. For example, suppose we have a signal source with an output power of -70 dBm, which we wish to connect to an amplifier having 22 dB gain through a cable having 8.5 dB loss. What is the output level from the amplifier? To find the answer, we just add the gains and losses as follows:

(signal) (amp. gain) (cable loss)

Output = -70 dBm + 22 dB + (-8.5 dB)

Output = -70 dBm + 22 dB – 8.5 dB

Output = -56.5 dBm

As a final note, power level may be referenced to other quantities and expressed in dB form. Below are some examples:

dBW = Decibels referenced to a power level of 1 watt

dBk = Decibels referenced to a power level of 1 kilowatt (1000 watts)

dBm0 = Decibels with respect to 1 mW at the zero (0) Transmission Level Point

dBrnC = Decibels above the reference noise level for a C-weighted communications circuit

Another common usage is dBc, which is essentially a relative term with a variable reference, like dB alone. It is usually taken to mean "dB referenced to a carrier level" and is most commonly seen in radio receiver specifications regarding spurious signals or images. For example, "Spurious signals shall not exceed -50 dBc" means that spurious signals will always be at least 50 dB less than some specified carrier level present (which could mean "50 dB less than the desired signal").

For Further Study Readers wishing to pursue other applications of the decibel are encouraged to consult any of the various handbooks dealing with specific disciplines such as Bellcore Telecommunications Transmission Engineering, Volumes 1 through 3; IEEE Press Subscriber Loop Signaling and Transmission Handbook (Analog); ITT Radio Engineer’s Handbook; NAB Engineering Handbook (broadcasting); ARRL The Radio Amateur’s Handbook; etc.

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About the author Paul H. Bock, Jr. has worked as an electronics professional since 1960. He served 14 years in the U.S. Navy as an Electronics Technician, rising to the rank of Chief Petty Officer (enlisted) and thence to Chief Warrant Officer. After resigning his Warrant Officer commission he spent 9 years with the Naval Reserve, retiring as a Master Chief Petty Officer in 1983.

After leaving active duty, Bock worked for four years as Chief Engineer of a commercial AM & FM radio station complex, then spent 19 years as a design and systems engineer in the defense industry. Since 1996 he has been employed as a design engineer in the telecommunications industry, specializing in digital loop carrier (DLC) systems.

First licensed in 1957, Bock holds an Extra Class amateur radio license (K4MSG) as well as a General Radiotelephone Operator License. He is a life member of ARRL, and also holds memberships in IEEE, SOWP, and VWOA. His other primary hobby is astronomy and he holds memberships in the Royal Astronomical Society of Canada, Association of Lunar & Planetary Observers, and the William Herschel Society.