In the satellite world we frequently work with the decibel scale. Decibels (dB) are a way of expressing the relative difference in signal strength between two sources. It is a logarithmic scale, so increasing numbers represent an exponentional increase in power. In simple terms, they are a convenient way of using addition instead of multiplication.
- I have two carriers, P1 and P2. What is the difference in power (Y) between them? Commonly this question appears as “Customer has capacity P1 and wants P2. How much additional power does he need to support the new carrier?”
- I have carrier P1 of a certain size, and Y dB margin (additional power available). How large can I make my new carrier P2? Or if Y is negative, how small?
This breaks down to fairly simple math. We know two of three variables P1, P2, and Y, and we want to solve for the missing one.
First, the basics. A decibel is the base 10 logarithm of the ratio of the power of two signals, times 10:
dB = 10 * log10 (P2 / P1)
Where P1 and P2 are the two signals to be compared. Since this is base 10, we can be express this without the explicit base:
dB = 10 * log (P2 / P1)
The handy rule of thumb of the decibel scale is this: 3 dB means twice the power. A carrier transmitted at 6 dB is twice as powerful as a carrier transmitted at 3 dB. And a carrier transmitted at 9 dB is four times as powerful as that 3 dB carrier. The scale can run negative as well, so a carrier transmitted at 0 dB is half the power of that 3 dB carrier, and at -3 is one-quarter the power. Remember that 0 dB does not mean “no power”, it is just another point on the scale.
So, if one carrier is 800 kHz in size, and another is 400 kHz in size, what is the difference in power between the two carriers?
P2 = 800, P1 = 400
10 * log (800/400) = 3 dB
Twice the power. Which is logical; if you want to transmit a carrier that is twice as big as your current carrier, you need twice the power.
What if the math isn’t so easy? Let’s say my old carrier is 685 kHz in size, and I want to increase to 976 kHz.
P2 = 976, P1 = 685
10 * log (976/685) = 10 * log (1.425) = 1.54 dB
So we’ll need 1.5 dB more power to transmit the new carrier P2. So if I want to transmit this carrier, I must improve that carrier’s C/N by 1.5 dB somehow. That means a larger antenna or transmitter (BUC) on the transmitting side, or a larger antenna on the receiving side. The same math works for a smaller carrier. Let’s start with a 500 kHz carrier and reduce it to 375 kHz:
P2 = 375, P1 = 500
10 * log (375/500) = -1.25 dB
So I need 1.25 dB less power. We might be able to use a smaller antenna or BUC.
Now we have an answer to the first question – what is the difference in power between two carriers? Now I’ll take a moment to explain a few more things before answering the second question.
Decibels are relative; on their own, they do not express an absolute value. They only tell you how two signals relate to one another. This means you can compare P1 and P2 in any equivalent terms. Let’s use IP rate in kbit instead of Hertz (both carriers must use the same modulation and encoding; we must compare apples to apples). Old carrier is 512 kbit, and new carrier is 768 kbit.
10 * log (768/512) = 1.76 dB
Decibels are sometimes expressed as relative to some reference. When this happens we write them as “dBX”, where X is the reference. I commonly work with dBW (where P2 is expressed in Watts of power and P1 is 1 W) or dBm (where P2 is expressed in milliwatts of power, and P1 is 1 mW). Converting between these two scales is easy; just add or subtract 30 dB.
10 * log (1000/1) = 30 dB
0 dBW = 30 dBm
5 dBW = 35 dBm
-30 dBW = 0 dBm
Which you use – dBW or dBm – does not matter as long as all your inputs are using the same scale. Modems often express transmit power in dBm.
Now we answer the second question. I have a carrier P1 that is 512 kbit in size. My modem is transmitting at -15 dBm, and has a C/N which is exactly what we need for reliable service. My maximum safe transmit power, which we determined early by a 1-dB compression test, is -12 dBm. So we calculate the margin:
Y = -15 - (-12) = 3 dB
How large can I make this carrier if I want to have 0 dB margin?
10 * log (P2 / 512) = 3 dB
If you’ve been paying attention, you know that 3 dB means “double the power”, so the quick answer is P2 = 512 * 2 = 1024 kbit. But let’s do the math anyway. To solve for P2, we need to convert decibels back to a linear scale. The magic formula is:
logax = y
ay = x
Remember that we’re using log base 10, so a = 10. Therefore,
10 * log (P2 / 512) = 3 dB
log10(P2/512) = 3 / 10
log10(P2/512) = 0.3
100.3 = P2 / 512
P2 = 512 * 100.3
P2 = 1021.57
So we can transmit a 1021.57 kbit carrier. Why isn’t that 1024 kbit? Because I wasn’t entirely honest with you. “3 dB equals double the power” is just an estimate. It’s not really equal to 2; it is
3 dB = 100.3 = 1.995. But that is close enough for quick estimates.
Now you can solve the second problem – how to best resize a carrier given Y margin.
- You are transmitting a carrier at 350 kbit, and want to transmit a carrier at 800 kbit. How much additional power will you need?
- You are transmitting a carrier at 700 kHz, and have 7 dB of margin. What is the largest carrier you can transmit with only 2 dB of margin?