Demystifying Decibels

When I took my foundation licence, I never really understood decibels. It only finally clicked at advanced level, and even then never from the book. When I help out at training sessions I see a lot of people struggling with this topic, yet it's actually quite easy to understand with the right teaching.

What's a decibel?

Decibels are a unit, like Volts and Amps are units. Volts are a unit of Potential Difference, Amps are a unit of Current, Decibels are a unit of ratio.

Ooooh, ratio, sounds scary

Nope. Usually we use decibels to quantify a power ratio. For example, if I have an amplifier that takes 4 watts from my transmitter in and produces 200 watts out to my loft-mounted antenna, I have an amplifier with a gain of:

$gain_{\times} = \frac{power\_out}{power\_in}$ $gain_{\times} = \frac{200}{4}$ $gain_{\times} = 50$

That is, I have fifty times the power coming out as I'm putting in. I also have a rather warm loft.

Ratios like this one $(200/4)$ can quickly get large and difficult to work with. We can also express the ratio in decibels.

$db = 10 \times \log_{10}\left(\frac{power\_out}{power\_in}\right)$ $db = 10 \times \log_{10}\left(\frac{200}{4}\right)$ $db = 16.99\mathrm{dB}$

We could then say my amplifier has a gain of 17dB. It means the same as "fifty times."

If something produces less power out than it takes in (for example a filter or an attenuator), we say it has a loss not a gain. For example, something that takes in 200 watts and produces 4 watts (fifty times less) has a loss of 17dB. You could say it has a gain of -16.99dB, but nobody does.

Why would you do that?

Decibels are a logarithmic unit rather than a linear one. Voltage is linear: 12 volts is twice as many volts as 6 volts, and joining two 6 volt batteries in series will sure enough give you a 12 volt battery.

12 decibels is not double the ratio of 6 decibels.

Actually, as the ratios get larger, the decibels pack tighter together, meaning that really big differences are still easy to work with. A gain of million times is still only 60dB.

$1,000_{\times} = 10 \times \log_{10}\left(\frac{1,000}{1}\right) = 30\mathrm{dB}$ $2,000_{\times} = 10 \times \log_{10}\left(\frac{2,000}{1}\right) = 33\mathrm{dB}$ $1,000,000_{\times} = 10 \times \log_{10}\left(\frac{1,000,000}{1}\right) = 60\mathrm{dB}$ $1,001,000_{\times} = 10 \times \log_{10}\left(\frac{1,001,000}{1}\right) = 60.004\mathrm{dB}$

Once you convert big ratios into decibels, you can work with them in your head.

I hate that table

At foundation and intermediate level, you're shown a table to convert decibels to 'times.' You're then usually taught that you can multiply combinations of these numbers together and somehow deduce ratios for 'unusual' numbers of decibels.

It's true, you can learn this and it will work, but for me all this did was confuse the issue. The table was rounded, the method didn't explain why the scale worked, and I always felt that there must be some kind of formula for this. There is, and it's really easy.

First off, I'm going to replicate the table using the decibels formula we saw above, giving a more precise value for each.

Gain (dB)	Gain (times)	Gain (times)
3	2	1.9953
6	4	3.9812
9	8	7.9433
10	10	10.0000

The teaching goes: When you add decibels, you multiply the ratios. If you have a gain of 13dB and you remember this table, 13dB is 10dB plus 3dB…

$13\mathrm{dB} = 10\mathrm{dB} + 3\mathrm{dB}$ $13\mathrm{dB} = 10_{\times} \times 2_{\times}$ $13\mathrm{dB} = 20_{\times}$

It holds true, and 13dB is actually 19.95 times, but my the mental agility required to get there.

What's 17dB using this table? Well we know it has to be more than 16dB; 10dB plus 6dB (10 times $\times$ 4 times = 40 times), but we're pretty stuck. The answer is 50 times, as per our example earlier.

So, the table is there because the maths alternative must be pretty horrific right? Wrong.

$ratio = 10^{\left(\frac{db}{10}\right)}$

I've never seen this taught in amateur radio, I had to work it out by transposing the formula for decibels just to prove to myself I wasn't going insane. Yet, it's so simple to convert decibels back to times. Ten to the power of the figure over ten.

$gain_{\times} = 10^{\left(\frac{db}{10}\right)}$ $gain_{\times} = 17\mathrm{dB} = 10^{\left(\frac{17}{10}\right)}$ $gain_{\times} = 17\mathrm{dB} = 10^{1.7}$ $gain_{\times} = 17\mathrm{dB} = 50$

No seriously, why?

Right, so we can convert ratio (times) to decibels and back, and we know that when we add decibels, it's equivalent to multiplying the ratios they represent. We also learnt that decibels aren't linear, so they can represent a big range of values using smaller numbers that are easier to work with.

Decibels are all about never having to do any of the maths I've outlined on this page.

Practical example: Decibels for a radio system

I have a VHF CW transmitter which outputs 9.2W of power. It's connected to 32 metres of coax cable that has a loss of 15dB per 100m at VHF. The cable connects to a yagi antenna with a gain of 9.5dBd.

What's the total gain or loss of this system?

$coax\_loss\_per\_meter = \frac{15\mathrm{dB}}{100\mathrm{m}} = 0.15\mathrm{dB/m}$ $coax\_loss = 0.15\mathrm{dB/m} \times 32\mathrm{m} = 4.8\mathrm{dB}$ $system\_gain = antenna\_gain - coax\_loss = 9.5\mathrm{db} - 4.8\mathrm{dB} = 4.7\mathrm{dB}$

So, even though the coax is losing two thirds of the power, the antenna gain overcomes this, and we still get a net power gain from this setup. What's the power ratio?

$system\_gain\_times = 4.7\mathrm{dB} = 10^{\left(\frac{4.7}{10}\right)} = 2.95$

We get just shy of three times out what we put in, in the direction the antenna is focussed. We can calculate the effective radiated power for our transmitter:

$erp = system\_gain \times transmit\_power$ $erp = 4.7\mathrm{dB} \times 9.2\mathrm{W}$ $erp = 2.95 \times 9.2 = 27.14\mathrm{W}$

Not bad. In a more complex system, you might have lots of gains and losses; you can even calculate loss through the air to work out what power a receiving station might experience from your signal.

Working with decibels in this way is actually very useful. It is far easier to take 4.8 from 9.5 than to calculate what an 8.91 times increase less a $1/3.02$ times decrease would be, yet the answers will be the same.

Bonus point: Effective radiated power

If you put 10W into an antenna with a gain of 3dBd, your E.R.P will be 20W. Your antenna does not however, magic up another 10 watts of power like some kind of aluminium free-energy hippie.

What it does, is take the power you put into it, and efficiently radiate it in a given direction, which makes it throw twice as much power in that direction than a dipole would.

The key here is that antenna gain is always relative to some other antenna. 3dBd means "twice as much power compared to a dipole." 6dBi means "four times as much power compared to an isotropic radiator" and most always 'in a specific direction'.

When we calculate antenna gain, we do so relative to a dipole, and gain figures are assumed dBd unless otherwise specified. If they were presented as dBi, they may be converted by subtracting 2.15dB.

A yagi antenna has very high gain figures, not because it magically creates extra power, but because it reflects and focusses power into a high power beam which would otherwise be radiated all over the place or be lost to space or ground.

Gotcha: Decibels aren't always for power ratios

These examples have been working with power, and so I should have qualified my units as dBW, or decibels relative to 1 watt.

Voltage ratios and sound pressures use 20 in the formulae instead of 10, because their basis is the square of the ratio and twice the log of a number is the same as the log of the number squared. ¯\(ツ)/¯

This article isn't aimed at any particular level: It's beyond what's needed for foundation or intermediate and a starting point for advanced revision. I hope though that it demystifies the subject just enough to help with your own learning, and shows that there is a mathematical alternative if juggling with tables isn't your thing.