It is common practice in ham radio to use decibels (dB) to describe the ratio of two power levels. Some of the license exam questions require some basic calculations in dB. In this article, we'll describe six Rules of Thumb to assist with these calculations. If you are comfortable with math and using a scientific calculator, you may want to crank out the precise number. However, these Rules of Thumb may still be helpful.
The basic definition of decibels in equation form is:
which simply means we divide the two power levels, take the base 10 log of that result and then multiply by 10 to get the result in dB. This equation translates linear power ratios into the logarithmic scale that we know as decibels. Once a ratio is in decibel form, we can performa gain and loss calculations using just addition and subtraction.
There are a few cardinal values that you may already know concerning decibels. For example, it is well known that a factor of 2 change in power corresponds to a 3 dB change in decibels. A power increase by a factor of 2 is + 3 dB and a decrease by a factor of 2 is - 3 dB. Another rule is that a factor of 10 change in power corresponds to a 10 dB change. (This one is unique in that the ratio and the dB value are the same.)
Let's add two additional Rules of Thumb that will expand our tool kit, especially for some of the questions on the Extra exam. A 1 dB increase in decibels corresponds to multiplying the power level by 1.26. A 1 dB decrease corresponds to multiplying by 0.79. We are showing two digits of precision to the right of the decimal point but you might just want to remember "one and one quarter" for the first value and 0.8 for the second value.
This table summarizes our six Rules of Thumb.
Now let's do a few example calculations to aid our understanding.
Example 1
Q: A transmitter produces 100 watts of power into a transmission line that has 3 dB of loss. How much power is present at the other end of the line?
A: The transmission line has a 3 dB loss, so we multiply the 100 watts by 0.5 to get 50 watts.
Example 2
Q: An amplifier with 20 dB of gain is used to amplify a 5-watt RF signal. What is the power level at the output of the amp?
A: We don't have a 20 dB rule of thumb, so we apply the 10 dB rule twice. That is, we multiply 5 watts times 10 (for the first 10 dB) and again by 10 (for the second 10 dB).
5 x 10 x 10 = 500 watts.
It should be clear from this example that a 20 dB increase equates to multiplying by 100. Perhaps a 7th rule of thumb to throw into the toolkit.
Example 3
Extra Class exam question [E9A02]
Q: What is the effective radiated power (ERP) of a repeater station with 150 watts transmitter power output, 2 dB feed line loss, 2.2 dB duplexer loss, and 7 dBd antenna gain?
A: The transmit power is given as 150 watts, and we need to determine the net
effect of the gains and losses in the system, as follows:
System gain = (-2dB – 2.2 dB + 7 dBd) = +2.8 dB
We don’t have a rule of thumb for +2.8 dB, but +3 dB will provide an answer
that is close enough. A 3 dB increase equates to a power multiplier of 2, or a doubling of the power. Thus: ERP = (2 x 150 watts) = 300 watts.
The precise answer is slightly less than this, or 286 watts. Our use of 3 dB will
be good enough for many computations, including getting the right answer on
the exam.
Example 4
Extra Class exam question [E9A06]
Q: What is the effective radiated power (ERP) of a repeater station with 200 watts transmitter power output, 4 dB feed line loss, 3.2 dB duplexer loss, 0.8 dB circulator loss, and 10 dBd antenna gain?
A: In this question, we have 200 watts of transmit power and three sources of loss: the feed line, the duplexer and the circulator.
The system gain = (-4 dB -3.2 dB -0.8 dB + 10dBd) = 2 dB
Now we need to adjust the 200 watts of power by an increase of 2 dB but we don’t have a rule for 2 dB. However, we can apply the 1 dB rule two times.
ERP = (200 x 1.26 x 1.26) = 317 watts.
Keep these rules of thumb handy in memory, not just for the exam, but for practical application in your everyday ham radio life!