Friday, March 11, 2016

TMRS: Too Many Radios Syndrome

Have you ever found yourself involved in an ongoing QSO for an extended period of time  on a repeater only to find out later that you were inadvertently cross-band the whole time?

If so you may be suffering from TMRS­—Too Many Radios Syndrome.

Sure, it all started innocently enough with the purchase of your fist dual VFO rig for the car.  You thought to yourself, hey, this is great!  I can monitor more than one frequency or band at the same time!  I can even scan multiple bands at the same time—neat!  Then one day it happened... You were driving along and heard someone throw their call out into the ether.  You picked up your microphone without looking down and replied with your callsign and soon you were having a nice QSO with a fellow ham.

What you and the other ham didn’t realize, however, was you were talking on a 440 repeater and the ham you were talking to was replying on a 2 meter machine.  How can this happen?  It’s simple really, you were both probably monitoring 2 of the club’s repeaters on dual VFO rigs.  You had your transmit selection set for the 440 side and the other ham had theirs set on 2 meters when they threw their call looking for a QSO.  When you answered the call you were  transmitting on the 440 side, which the other ham was also listening to, and he, assuming you were answering his call on 2 meters, replied to you on the other repeater which you were also listening to.

We have dual-band, tri-band and quad-band rigs to choose from.  We have mixed mode rigs (FM analog/D-Star for example).  Now there’s talk of rigs that will combine even more modes!

No big deal for you guys, but to anyone listening in you probably both sounded like you were crazy people talking to yourselves!

Now add in some more complications like a D-Star radio, maybe a 900 Mhz transceiver, DMR, Fusion and maybe MotoTurbo for good measure.  Also, let’s not forget those of you out there with GMRS licenses too, and there’s probably no room for your XYL (or even a napkin) left in your vehicle.

Hams are nothing if not collectors of technology.  We love our toys—essential emergency communications infrastructure is the excuse we use to try to convince ourselves and others that we’re not suffering from something that can only be diagnosed by a visit to a professional and a deep search of  the DSM-V.

Oh, and don’t get me started on the emergency service fanatics!  Some of these hams are carrying so many HTs and spare batteries that they wouldn’t be allowed on a cruise ship for fear they may cause it to take on excess water or at the least cause it to ride dangerously low in the water and risk running aground on a reef or sandbar.

Just how many different modes do we need for FM repeaters? We’re approaching a tipping point where it will soon be possible for every ham on the planet to have their own individual personalized mode at which time they won’t have anyone else to talk to.

Don’t think it could happen?  Next time you go to a large event with ARES support look around at some of the people who are there to help.  I can guarantee you will see hams there with 4 or more radios operating with at least as many different modes.

We have to stop this insanity while we still can.

Wednesday, April 1, 2015

Hams to lose frequency allocations above 2 Meters

The FCC and Homeland Security have been conducting a systematic and on-going survey of amateur radio frequency use in response to requests from communications industry lobbyists.  The survey has consisted in frequency monitoring and statistical analysis of usage rates and has covered the the past 7 years.

Communications industry lobbyists have complained that amateur frequency allocations in the UHF and higher have caused security concerns related to RF infrastructure and had made the claim that radio amateurs were not making appropriate use of the spectrum.

A new classified report confirms accounts of malicious and childish behavior, not only on the frequencies in the UHF and above, but also on the VHF, HF and LF spectrum allocations.  Recordings of amateurs using profanity, rebroadcasting copyrighted music, engaging in hate speech, etc., reinforced those claims.

A spokesperson for the FCC said, "The very valuable RF spectrum allocated to radio amateurs is being either misused or not used at all."  Amateur radio repeaters, which are used to extend the range of low powered radios, are often idle for days at a time, and in instances when they are in use are mostly used as a chat room for commuters.

The FCC had pointed out the role amateurs play in emergency communications, however the director of Homeland Security made assurances that local, state and federal emergency management agencies, were more than up to the task and not in need of assistance from civilian amateurs.  One official told congress, "let the professionals do their jobs, that's what we pay them for and it's also the reason we give them all that money to buy equipment.  In these modern times of wi-fi, smartphones, Twitter and Facebook there's more than enough ways for people to communicate."

The FCC and Homeland Security have identified the radio frequencies from 148 MHz and up as those the amateurs will no longer have access to beginning within the week.  "The amateurs will need to immediately cease operation on those frequencies and be subject to arrest, fines and imprisonment.
Amateurs will be given until April 30th to remove all equipment capable of transmitting on frequencies in excess of 148 MHz including walkie-talkies, mobile radios, base stations and all associated repeater systems.

The spectrum above 148 MHz will be sold at auction in the coming months and is estimated to be worth over a billion dollars in revenue to the government.

Sunday, January 25, 2015

Using Algebra to Derive New Formulas

Using Algebra to Derive New Formulas and Solve others

To follow this you are going to need to know some algebra.  If you need a refresher, I recommend  I am first going to show how to use algebra to find the temperature at which both a Fahrenheit and Celsius thermometer would be showing the same number.  That is, what temperature is the same in degrees Fahrenheit and Celsius.  You can think of this like using 2 different rulers to measure your height.  One in feet and one in meters.  Your height is the same no matter which you use, just the number will be different, i.e. 6 feet or 182.88 centimeters.

First we start with the conversion formulas.  To convert from degrees Celsius (C) to Fahrenheit (F), we multiply the temperature by 1.8 (also regularly stated as 9/5) then add 32.  So 20 degrees C = 68 degrees F.

Converting from F to C the formula is degrees F minus 32 then that result divided by 1.8:

To calculate the temperature where these are equal to each other, we set the 2 formulas equal to each other, use just one variable for temperature, e.g. “t” and solve for that variable:

The idea is to get the t all by itself and we do this by performing the inverse operations on various parts of the equation, remembering to do the same operation on BOTH sides!  First we’ll get rid of the 1.8 on the bottom of the right side by multiplying both sides by 1.8:

 The 1.8’s on the right side cancel leaving just the t - 32 and we multiply the 2 terms inside the parenthesis on the left side each by 1.8, giving us:

Now subtract t from both sides:

Doing the math this leaves us with:

Next we subtract 57.6 from both sides:

This gives us:

Next we divide both sides by 2.24 to get the t by itself (this is the same as multiplying by 1/2.24):

The result of this is:

You can check this by inserting minus 40 into the 2 equations above and you’ll see the conversions both come out to minus 40.

Now as they say, I showed you that so I can show you this, specifically, how to derive a new formula.

In the license books we’ve seen examples, usually poorly explained, where a new formula is derived by substituting values.  In the case of the power formula, P = E I, we are told to substitute the ohms law equivalents of E and I to get 2 new formulas.

The example I’m going to show you is from the ARRL Extra Class license manual which is only shown worked through superficially.  The assumption being that either the student is well versed in algebra, or will go to one of the many resources available online or elsewhere to bone up on it.

However, I have found that many who come to ham radio, especially those who come to it after a prolonged time away from high school or college, may have forgotten most, if not all that they’ve learned.  So I’d like to build on the previous example and work through the derivation of the formula for calculating resonant frequency of LC circuits.

The definition of LC circuit resonance is given as the frequency where both inductive reactance (XL) and capacitive reactance (XC) are equal.  Each reactance being 180 degrees out of phase with the other means that their reactances will cancel out leaving only a purely resistive load.  Let’s first look at the formulas for each, then we’ll set them equal to each other and solve for frequency to get the resonant frequency formula.

Setting them equal:

We want to find “f” so we’ll multiply both sides by f to remove it from the bottom of the right side.

Combining the f’s on the left and cancelling on the right leaves us with:

Next I’m going to get rid of the 2πL on the left by dividing both sides by 2πL (this is the same as multiplying by the reciprocal 1/L)

When we do the math we cancel the 2πL on the left side and multiply the top and bottom on the right, combining the like terms to get:

We now take the square root of both sides…
 This gives us:

…and that’s the formula for resonant frequency!  Not too bad, right?

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