[TowerTalk] Dollars per dB

[DavidC]

I am interested in reactions to this presentation (below)by K2UVG.
It sounds logical on the face of it ... though I'll have to rerun his
numbers for used solid state and tube amplifiers and see how
they look.  Since I recently acquired a used Cushcraft A3S w/40m
I appreciate that illustration ... used is more cost efficient!  :-)
73 & Happy New Year to All!  DavidC  K1YP

Dollars per dB
or...or so you want to improve your station! 
------------------------------------------------------------------------
If I held up a dollar bill, you all would recognize it for what it is. If i
wrote "dB" on the blackboard, you would probably tell me that you herd
about it but that you don't quite understand it. Well let's work on that. 

When I talk about station improvement, I am really talking about signal
inprovement. It is useful to appraise signal strengths in terms of relative
loudness as registered by the human ear. For example, if a person estimates
that a signal is "twice as loud" when the transmitter power is increased
from 10 watts to 40 watts, he will also estimate that a 400 watt signal is
twice as loud as a 100 watt signal. 

The human ear has a logarithmic response. Dont let that word throw you.
Well have it licked in short order. This fact is the basis for the use of a
relative power unit called the decibel (dB). A change of one decibel in the
power level is just dectable as a change in loudness under ideal
conditions. The number of decibels corresponding to to a given power ratio
is given by the equation: 

db = 10 log P2/P1 

There is that nasty word log again.... dont worry we will slay that dragon.
Lets create some rules of thumb. First lets calculate a power ratio of 2/1
.. 

db = 10 log P2/P1 = 10 log 2/1 = 10 log 2 

It turns out the log of 2 = .3010299957. So the power ratio of 2 is equal
to 3.010299957 dB. Well just call it 3 dB. Another ratio to remember is
P2/P1 = 10. 

dB = 10 log 10 

The log of 10 is 1. So a power ratio of 10 is also 10 dB. Knowing these two
rules and the fact that since by the definition of logs we are talking
about exponents... and when we multiply numbers with exponents we just add
exponents... so when multiplying power ratios we just add decibles. Lets
look at an example: 

Suppose we have a power ratio of 8. How many dB is that? Well we know from
our 2 rules that its greater than 3dB and less than 10 db. We also know
that a power ratio of 8 is also equal to 2 X 2 X 2. Using the last rule we
have 3 dB + 3 dB + 3 dB = 9 dB. Every time you double the power you add 3
dB. Evey time you reduce the power by 2 (half) you loose 3 dB. 

The S-unit and dB are used as references on receiver signal-strength
meters. No particular standard has been adopted by the industry at this
time, as a S meter is a relative reading instrument on most amateur radios.
However, during WWII at least one receiver manufacturer used 50 micro volts
for S9 and each S unit below S9 was supposed to be equal to 6 dB. The units
above S9 are already in dB. I will use this "standard" convention that a S
unit is equal to 6 dB. 

Pop quiz...what is the power ratio associated with 6 dB ? Well 3 dB is 2
and another 3 dB is another 2... thats 6 dB and 2 time 2 or 4. 

Lets go back to $ per dB. You all knew what a $ was. Now you are dB experts
and "per" is just the arithmetic operator of division. Miles per hour
...miles divided by hours...$ per dB ... dollars divided by dB. This gives
us a Figure of Merit or Measure (FOM) that we can use to look at several
ways of improving our station. 

Lets now use this new found knowledge to improve a typical start-up
station. This station will consis of the following: 

Power out - 100 watts
Antenna - dipole
Coax - 100 ft of RG-58

You notice that when I described this station, I am only talking about the
business end. I make certain assumptions like you are not using a crystal
radio to receive. We will also make use of the fact that an improvement in
the antenna system (coax and antenna) on transmit will also improve the
received signal. 

Lets tackle the transmitter. How do we improve the power out. Well 100
watts makes a good exciter for a linear amp. Lets use that approach. We are
also going to compare apples to apples or use CW figures of power out. From
a recent catalog: 

Ameritron ALS-500M - solid state linear - 400 watts CW - Price $679.95 

To go from 100 watts to 400 watts...you double 100 to 200 (3dB) and you
double 200 to 400(another 3 dB). Adding the dB we have a 6dB improvement.
Now just divide dollars by dB or in this case 679.95/6 = $113.33/dB. For
every $113 we get a 1 dB improvement in our signal. What is this total
improvement going to do at the received end ? Remember that 6 dB is equal
to 1 S-unit. If your are S8 with 100 watts your now S9. If you were S9 you
are S9+6dB. Note: To keep it simple, I did not include the 12 volt supply
in this calculation. When you consider cost it should be TOTAL cost. 

Lets look at a bigger amp. 

Alpha ETO - 91 Beta - 1500 watts - Price $2300 

Double 100 = 200, double 200 = 400, double 400 = 800, double 800 = 1600.
Each double is 3 dB or 1600 watts is 12 db. We also know that 10 times 100
or 1000 watts is 10 dB. If we guess 11 dB for 1500 watts we would be right
on. Our FOM would be $2300/11 dB or $209.09/dB. Note that the improvement
on the received side would be almost 2 S-units. Note: Again I did not
include the cost of putting in 220 volts to the shack. 

Now lets look at the antenna and take down that dipole and put up a beam. 

Cushcraft A3S - 3 elm tri-bander - Gain 8 dBd - $349.95 

Whoops, what is that "d" behind the dB ? Well in antennas they use 2
references when describing the gain of an antenna. The ideal or perfect
antenna that radiates equally in all directins is called an isotropic
antenna. Gain measured against this fictious antenna is labeled dBi. Gain
measured relative to a dipole is dBd. The difference between them is that a
dipole has a gain of 2.14 dBi or 0 dBd. Since we have a dipole and the new
beam is rated in dBd, we are all set. All we have to do is take the cost
and divide by the gain. $349.95/8 = $43.74/db. Note: Here again I did not
include a rotor or tower/mast. 

Finally lets look at the coax. Here we must introduce attenuation. Coax
never gives us anything like the linear or antenna did. It always takes
something away from us. To improve our system we must insure that it takes
away the least amount. The amount it takes away is called attenuation. You
can go to the Handbook or Antenna Book and look up the attenuation factors
of various coaxial cables. They are measured in db per 100 ft at a
particular frequency. The higher the frequency the higher the attenuation.
Here is a sample table: 

Coax4 MHz30 MHz150 MHzRG-58.82.77RG-8 foam.27.927/8 in hardlineless than
.1.25.7

Since we have been talking about HF in our past examples, we will stick to
HF here and use the 30 MHz figures. This is a worst case for HF. Presently
we have an attenuation of 2.7 dB if we go to RG-8 foam filled coax, we can
improve that by 1.8 dB (2.7 - .9 = 1.8). The cost of 100 foot of Belden
8214 which is RG-8 foam is $49.95. Doing the now familar math $49.95/1.8 =
$27.75/dB. 

An aside...42 feet of RG-58 at 2 meters takes that 50 watt rig and makes it
a 25 watt rig! 

By now it should be obvious that I chose the order of improvements to prove
a point. Use the best coax you can, use the best antenna you can and then
worry about the power or $113/dB > $43/dB > $27/dB. 

The Moral of this story is NEVER use RG-58!
------------------------------------------------------------------------
If you would like to use this document, feel free to do so. All I ask is
that you give proper credit and drop me an e-mail telling me what you are
doing with it.
K2UVG

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