TUTORIAL: dB & dBm (REVISED)
Paul H. Bock
Author's Note: This tutorial was originally written for digital
& software engineers and non-engineers, but it may be of value to
others who wish 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.
Comments and criticisms may be e-mailed to:
* Paul H. Bock, Jr. K4MSG * Principal Systems Engineer
* E-Systems/Melpar Div. * Telephone: (703) 560-5000 x2062
* Internet: pb...@melpar.esys.com
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
USING AND UNDERSTANDING DECIBELS
by
Paul H. Bock, Jr. K4MSG
*General*
The decibel, or dB, is a means of expressing either 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). The decibel was
developed by the telephone company to conveniently express the gain
or loss in telephone transmission systems. The decibel 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 indicted, 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 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 decided that it might be convenient to
handle gains and losses this way, so they invented a unit of gain
measurement called a "Bel," named after Alexander Graham Bell.
They defined the Bel 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 quickly occurred to the 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 negative.
There are two common methods of using the decibel. The first
is to express a known 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 not covered in this discussion. 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
*Uses of the Decibel with a Defined Reference*
The most common "defined reference" use of the decibel
is the dBm, or decibel relative to 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
either whatever reference a particular user chooses or 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:
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 = Power level referenced to 1 watt
dBk = Power level referenced to 1 kilowatt (1000 watts)
One other common usage is dBc, which is essentially a relative
term with a varible reference, like dB alone. It means "dB
referenced to a carrier level" and is most commonly seen in
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").
Bruce A. Black
Bock) writes:
>
> ...
>
> * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
>
> USING AND UNDERSTANDING DECIBELS
>
> by
>
> Paul H. Bock, Jr. K4MSG
>
Thanks for the lucid tutorial. Do you plan to follow it up with a
discussion of decibels used for voltage and current ratios? That is the
part people tend to find confusing!
I have always been under the impression that the logarithmic measure of
power was introduced because it mimics the response of the human ear.
Bruce Black, WB9Q
bruce...@rose-hulman.edu