|
Decibels |
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.
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:
 |
(Eq. 2) |
where G = Gain in dB Pout = Power output from the device Pin = 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?
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?
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
where P = power, V= 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
 |
(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?
(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):
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!
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. 4: 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.
| ©2005 Jimoid.com |