The Decibel is a subunit of a larger unit called the bel. As originally used, the bel representedthe power ratio of 10 to 1 between the strength or intensity i.e., power, of two sounds, and wasnamed after Alexander Graham Bell. Thus a power ratio of 10:1 = 1 bel, 100:1 = 2 bels, and 1000:1 = 3bels. It is readily seen that the concept of bels represents a logarithmic relationship since thelogarithm of 100 to the base 10 is 2 (corresponding to 2 bels), the logarithm of 1000 to the base10 is 3 (corresponding to 3 bels), etc. The exact relationship is given by the formula
 Bels = log(P2/P1)
where P2/P1 represents the power ratio.
Since the bel is a rather large unit, its use may prove inconvenient. Usually a smaller unit, theDecibel or dB, is used. 10 decibels make one bel. A 10:1 power ratio, 1 bel, is 10 dB; a 100:1ratio, 2 bels, is 20 dB. Thus the formula becomes
 Decibels (dB) = 10 log(P2/P1)
The power ratio need not be greater than unity as shown in the previous examples. Inequations  and , P1 is usually the reference power. If P2 is lessthanP1, the ratio is less then 1.0 and the resultant bels or decibels are negative. Forexample, if P2 is one-tenth P1, we have
bels = log(0.1/1) = -1.0 bels
and dB = 10 log(0.1/1) = -10 dB.
It should be clearly understood that the term decibel does not in itself indicate power, butrather is a ratio or comparison between two power values. It is often desirable to express powerlevelsin decibels by using a fixed power as a reference. The most common references in the world ofelectronics are the milliwatt (mW) and the watt. The abbreviation dBm indicates dB referenced to1.0 milliwatt. One milliwatt is then zero dBm. Thus P1 in equations  or becomes 1.0 mW. Similarly, The abbreviation dBW indicates dB referenced to 1.0 watt, withP2 being 1.0 watt, thus one watt in dBW is zero dBW or 30 dBm or 60 dBuW. Forantenna gain, thereference is the linearly polarized isotropic radiator, dBLI. Usually the `L' and/or `I' is understoodand left out.
dBc is the power of one signal referenced to a carrier signal, i.e. if a second harmonicsignal at 10GHz is 3 dB lower than a fundamental signal at 5 GHz, then the signal at 10 GHz is -3 dBc.
P2/P1 = 100/0.1 = 1000
or a gain of:
10 log(P2/P1) = 10 log(100/0.1) = 30 dB.
(notice the3 in 30 dB corresponds to the number of zeros in the power ratio)
The ability of an antenna to intercept or transmit a signal is expressed in dB referenced to anisotropic antenna rather than as a ratio. Instead of saying an antenna has an effective gain ratio of7.5, it has a gain of 8.8 dB (10 log 7.5).
A ratio of less than 1.0 is a loss, a negative gain, or attenuation. For instance, if 10 watts ofpower is fed into a cable but only 8.5 watts are measured at the output, the signal has beendecreased by a factorof
8.5/10 = 0.85
10 log(0.85) = -0.7 dB.
This piece of cable at the frequency of the measurement has a gain of -0.7 dB. This is generallyreferred to as a loss or attenuation of 0.7 dB, where the terms "loss" and "attenuation" imply thenegative sign. Anattenuator which reduces its input power by factor of 0.001 has an attenuation of 30 dB. Theutility of the dB isvery evident when speaking of signal loss due to radiation through the atmosphere. It is mucheasier to work with aloss of 137 dB rather than the equivalent factor of 2 x 10-14.
Instead of multiplying gain or loss factors as ratios we can add them as positive or negativedB. Suppose we have a microwave system with a 10 watt transmitter, and a cable with 0.7 dBloss connected to a 13 dBgain transmit antenna. The signal loss through the atmosphere is 137 dB to a receive antennawith a 11 dB gainconnected by a cable with 1.4 dB loss to a receiver. How much power is at the receiver? First,we must convertthe 10 watts to milliwatts and then to dBm:
10 watts = 10,000 milliwatts
10 log (10,000/1) = 40 dBm
40 dBm - 0.7 dB + 13 dB - 137 dB + 11 dB - 1.4 dB = -75.1 dBm.
dBm may be converted back to milliwatts by solving the formula:
mW = 10(dBm/10)
giving 10(-75.1/10) = 0.00000003 mW
Voltage and current ratios can also be expressed in terms of decibels, provided the resistanceremains constant. First we substitute for P in terms of either voltage, V, or current, I. Since P=VI and V=IR wehave:
P = I2R = V2/R
Thus for a voltage ratio we have
dB = 10 log[(V22/R)/(V12/R)] = 10log(V22/V12) = 10log(V2/V1)2 = 20log(V2/V1)
Like power, voltage can be expressed relative to fixed units, so one volt is equal to 0 dBV or 120dBuV.
Similarly for current ratio dB = 20 log(I2/I1)
Like power,amperage can be expressed relative to fixed units, so one amp is equal to 0 dBA or 120dBŠA.
Decibel Formulas (where Z is the general form of R, including inductance andcapacitance)
When impedances are equal:
When impedances are unequal:
|Voltage orCurrent Ratio|
|Voltage orCurrent Ratio|
An easy way to remember how to convert dB values that are a multiple of 10 to the absolutemagnitude of the power ratio is to place a number of zeros equal to that multiple value tothe right of the value 1.
i.e. 40 dB = 10,000 : 1 (for Power)
Minus dB moves the decimal point that many places to the left of 1.
i.e. -40 dB =0.0001 : 1 (for Power)
For voltage or current ratios, if the multiple of 10 is even, then divide the multiple by2, and apply the above rules. i.e. 40 dB = 100 : 1 (for Voltage)
-40 dB = 0.01 :1
If the power in question is not a multiple of ten, then some estimation is required. Thefollowing tabulation lists some approximations, some of which would be useful to memorize.
|DB RULES OFTHUMB|
|Multiply Current / VoltageBy||.||Multiply Power By:|
|if -dB||dB||if +dB||if -dB|
You can see that the list has a repeating pattern, so by remembering just three basic valuessuch as one, three, and 10 dB, the others can easily be obtained without a calculator by additionand subtraction of dB values and multiplication of corresponding ratios.
A 7 dB increase in power (3+3+1) dB is an increase of (2 x 2 x 1.26) = 5 times whereas
A 7dB decrease in power (-3-3-1) dB is a decrease of (0.5 x 0.5 x 0.8) = 0.2.
Example2: Assume you know that the ratio for 10 dB is 10, and that the ratio for 20 dB is 100(doubling the dB increases the power ratio by a factor of ten), and that we want to find someintermediate value.
We can get more intermediate dB values by adding orsubtracting one to the above, for example, to find the ratio at 12 dB we can:
work up fromthe bottom; 12 = 1+11 so we have 1.26 (from table) x 12.5 = 15.75
alternately, workingdown the top 12 = 13-1 so we have 20 x 0.8(from table) = 16
The resultant numbers are not an exact match (as they should be) because the numbers in thetable are rounded off. We can use the same practice to find any ratio at any other given value ofdB (or the reverse).
dB AS ABSOLUTE UNITS
Power in absolute units can be expressed by using 1 Watt (or 1 milliwatt) as the reference powerinthe denominator of the equation for dB. We then call it dBW or dBm. We can then build a tablesuch as the adjoining one.
|dB AS ABSOLUTE UNITS|
|60||30||1 W (1000mW)||0|
From the above, any intermediate value can be found using the same dB rules and memorizingseveral dB values i.e. for determining the absolute power, given 48 dBm power output, wedetermine that 48 dBm = 50 dBm - 2 dB so we take the value at 50 dB which is 100W anddivide by the value 1.58 (ratio of 2 dB) to get: 100 watts/1.58 = 63 W or 63,291mW.
Because dBW is referenced to one watt, the Log of the power in watts times 10 is dBW. TheLogarithm of 10 raised by any exponent is simply that exponent. That is: Log(10)4= 4. Therefore, a power that can be expressed as any exponent of 10 can also be expressed indBW as that exponent times 10. For example, 100 kw can be written 100,000 watts or105 watts. 100 kw is then +50 dBW. Another way to remember this conversion isthat dBW is the number of zeros in the power written in watts times 10. If the transmitter power in question is conveniently a multiple of ten (it often is) the conversion to dBW is easy andaccurate.