Wednesday, November 26, 2014

Decibel Deathmatch (or how I learned to love logarithms)

I have noticed that one of the most confusing topics in ham radio is the decibel.  I believe a lot of the confusion relates to the complex look of the formula:

 That “log” really freaks people out, especially if you've never taken a math class that included logarithms in the curriculum.  Logarithms are a way of multiplying or dividing large numbers by adding smaller ones.  Another area of confusion is in the descriptions of the inverse operation, that is, taking a decibel number and working it backwards to get to the number that was multiplied or divided by.

Decibel calculations are not intuitive.  You have to practice and pay attention to what the calculations mean.  The mechanics of the formula are fairly straightforward and today’s scientific calculators do all the heavy lifting for you.  You literally just have to push a button to get the answers.  Unfortunately, if you  don't understand what’s going on you won’t perform the steps in the correct order.

First you should take a look at your calculator and notice that the key labeled “log” has a second function above it.  That second function is the inverse operation of “log” and is labeled as “10X.”  You pronounce that “ten to the X” and all that means is you are raising the number 10 to the power of the number in the calculator’s display.  If the number in the display is 2, when you press the 10X key you will get the result of 102 (ten squared, or 10 to the second power) which is 100 (10 times 10 is ten squared).  Logarithms take that relationship and put it into a different form.  In that 10 to the 2nd power equals 100, the base is 10.  There are as many bases for logs as there are numbers, but the one on your calculator, and the one that doesn’t specify a base, is the base 10 logarithm, or log10.  So the log10 of 100 is 2 because 10 to the 2 power is 100.  All you really have to understand is that the 10X and log key are inverses of each other.

So try it.  Enter 100 into your calculator and press the log button.  The calculator should now show the number 2.  Now press the 2ND button on your calculator (to use the second function of a key) and press log again.  This time because you pressed the 2ND key first you are accessing the 10X function and should now see 100 on your screen.  See?  We’re back to our starting point.  This is an important point to recognize as we go forward.

Another point of confusion is in which number to put on top in the Pmeasured/Preference part of the equation.  If you are looking for a gain you put the bigger number on top and divide by the smaller number.  E.g., if you have a 50 watt radio and your amplifier puts out 100 watts, what is the gain?  Well, in plain numbers your gain is 50 watts.  That’s a doubling of your input power.  If you plug those numbers into the formula as 100/50 you will get 2.  That’s the multiplier.  The log of 2 is the power you raise 10 to in order to get 2 as the answer.  If you accidentally put the numbers the other way, 50/100, then you’d get 0.5 or 1/2.  100 watts is not half of 50 so you know right away that you did something wrong.  So what decibel value represents that doubling of power?  Well, with the original answer of 2 in the display (from dividing 100 by 50) press the log key and you should get a number like 0.3010299 etc.  We’re almost done at this point, all that’s left is to multiply this answer by 10 to get, rounded off, 3.  So a gain of twice our input power is 3dB.  If you had instead used the numbers backwards, so 50/100 resulting in 1/2 (0.5) then when you pressed the log key you would have seen -0.3010299, which when multiplied by 10 would give you minus 3dB!

Eventually most people come to terms with the formula for calculating dB.  The most important thing to remember is that you are dividing 2 power values (or voltages if you are working with voltages) to get the ratio of them.  Gain is what you have when you divide the bigger number by the smaller, and loss (minus dB) is what you get when you divide the smaller by the larger.

Where a lot of people have the most trouble is when given a decibel value and either asked to calculate the multiplication/division factor, or given a dB value and a power and asked to calculate either the starting power or ending power.  For instance, if I tell you that I have an amplifier that provides 7dB of gain no matter how much power is put into it and ask you to tell me how much power out I’d get if I fed it with 50 watts on its input how would you calculate that?  You do it by working the dB formula backwards.  When you calculate the dB value of a pair of power numbers, you first divided the 2 powers, pressed the log key to get the log of that answer, then multiplied by 10 for the final decibel figure.  Here we start with the decibel figure so the first thing we do is divide that dB number by 10.  So we have 7dB divided by 10 equals 0.7 (zero point seven).  The second to last operation was pressing log, but when we divide by 10 to get 0.7 that IS the log of the ratio of the 2 powers!  We have to find out what that ratio is by using the inverse of the log function: 10X.  So with 0.7 showing on the calculator press the 2nd key and log (to access the 10X function) and you should see 5.011872…, basically just 5.  5 is the ratio of the input to output power.  If you are given the input power, as we were, you just multiply that by 5 to get the answer.  Here we were given 50 watts in, so 50 times 5 equals 250 watts output!  If you were given the output power and asked to calculate the input power to a 7dB amplifier, you would divide the output power by 5.  Using the same example, we have a 7dB amplifier with 250 watts output.  What’s the input power?  It’s just 250 divided by 5 equals 50!

If you really want to get into logarithms and exponents, check out this page:  As always I invite discussion and questions.  I really want you to understand this stuff so you can impress all your ham friends with your mastery of decibels.

Friday, November 21, 2014

To the ARRL: How to alienate faithful members.

One of the things I try to do is to support organizations that further my interests. For instance, I love to hike and appreciate the beauty of nature and so I belong to several hiking organizations that maintain trails and advocate for conservation of the natural environs.

I also like ham radio and as such am a member of the ARRL (the American Radio Relay League), which is the premier advocate for supporting amateur radio activities.  They also advocate for legislation that helps further the hobby.  When I renew, I usually do so as soon as I get my notice that my membership is going to expire soon, and I normally renew for more than 1 year at a time.

Good member, right?  Well you know the old saying, "the squeaky wheel gets the grease?"  I was hanging out one day with a fellow ham and I noticed that he had a really nice collection of ARRL publications.  Since I have many of those same books myself I know they are worth a good bit of change.  I remarked to him that he must spend as much as I do on books, but he said not really.  It turns out that if you don't renew your ARRL membership when it comes do they will send you an incentive in the form of a book.  So my friend (lots of them it turns out) doesn't renew until they offer the free book.  He does this every year and so HIS membership is apparently worth a lot more than mine!

This makes me feel like a chump.  Why should I renew for multiple years, on time or ahead of schedule, for the full membership fee, when I can just wait a few weeks and get a free book.  Most of these books average in the $20 to $30 range, the median being around $25.  With a one-year membership currently going for $39, they are effectively paying only $14 a year for their membership.

I know the ARRL wants to keep member roles up, but they should offer something to those of us who renew right away too.  At least that's my opinion on the matter.  What do you think?

Kevin, AB2ZI