Fun with 2.4 GHz on the cheap (May 2004)
With the demise of the Galaxy 2.4GHz MDS pay TV system, grid dish antenna's very similar to this one have become available for nothing, they are fitted to many houses but are no longer used - just ask and someone will let you have theirs for nothing.

Now what to do with them, well recently 2.4GHz TV transmitter modules have become available at very cheap prices, especially if several club members can come together and make a bulk purchase, prices like $22 for a TX and $27 for an RX module can be achieved.

Another possible use is to connect them to wireless network cards in your PC, and share files and live 2 way video/audio contacts across town.

The biggest problem is getting the signals to the antenna, the secret to this is to put the transmitter/receiver modules on the back of the antenna's (in a water proof box of course) then run the video/data signals back to your shack.

To run 2.4 GHz any distance up a co-axial cable requires something like helix coax, however if you put, say a video transmitter module (about the size of a box of matches) at the antenna, then video can be run up to the antenna using cheap 75 ohm TV coax, with very little to no signal loss problems. Power can be run using figure 8 flex as the modules consume very little power (at 5V), so voltage drop is not a problem.

In the first picture, the grid dish antenna was mounted on a light duty camera tripod for field work. This is a good way to experiment with simple short hop links before one gets too ambitions.

The feed to the dish consists of a simple dipole antenna with a reflector element. The one above was made on printed circuit board, however you can use bits or brass rod, or heavy copper wire, just soldered together.

This one is made from brass rod, housed in PVC water pipe.

Come to this Friday's meeting and talk to John VK3YTV who made them.

Talk to Peter VK3KCG about how to get the video modules for a good price

Paul VK3TGX

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Decibels Made Easy (Without Logarithms)(April 2004)
Bruce Spratling, Jr.

Suppose your station has a transmitter that puts out 50 watts of power, a feed line that loses all but 40% of the power, a duplexer that loses half the power, and an antenna that produces enough gain to make the signal 4 times as strong as a dipole antenna.  To find the effective radiated power of your station you need to multiply 50 watts times .4 (feed line loss) times .5 (duplexer loss) times 4 (antenna gain) = 40 watts.

Rather than multiply all these factors together, someone decided it would be good to represent them in a way that allows us to add them.  An increase of a factor of 10 is defined to be a 10 decibel increase.  Decibels are abbreviated dB.  Two 10dB increases produces a 20dB increase, because we add decibel increases.

Two increases by a factor of 10 results in an increase of 10 X 10 = 100; therefore 20dB = 100.  Three 10dB increases is 30dB, which is 10 X 10 X 10 = 1000. 

Because we want to be able to add the dB increases, 0 dB is a factor of 1.  Multiplying something by 1 does not change it, just as adding 0 to something results in no change.

It’s easy to figure out the meaning of 10, 20, 30, 40... dB, but how about decibels less than 10?

Note that if we multiply 2 by itself 10 times, it generates the following: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024.

2 times itself 10 times is 1024, which is slightly more than 1000, which is 30dB.  3 added to itself 10 times is 30, and 30dB is a factor of 1000.  Therefore, 3dB represents a factor of (approximately) 2.

Another way to derive that 3dB represents a factor of 2 is this:  If we add 3dB to itself 3 times we get 9dB, which is less than, but close to, 10dB.  If we multiply 2 times itself 3 times we get 2X2X2=8, which is close to, but less than 10.  This is good because 9dB should be less than 10 (since 10 dB is 10).  We have to multiply 8 by 5/4 to equal 10, and we have to add 1dB to 9dB to equal 10dB.  Having 3dB = 2  will work if 1dB = 5/4.  Does 1 dB represent a factor of 5/4?  If we apply 1dB 3 times, we’ll have 3dB, which = 2.  (5/4)X(5/4)X(5/4) = 125/64, which is close to 2.  Therefore, a 3dB increase represents a 2 times increase, and 1 dB represents an increase of 1.25.

6dB = 3dB + 3dB.  A 3dB increase is a factor of 2, so 6dB = 2 X 2 = 4.  Similarly, 9 dB = 2 X 2 X 2 = 8.

So far we have:

Let’s consider negative decibels.  -1dB changes 10dB to 9dB, which changes 10 to 8.  Therefore, -1 dB is .8, because .8 times 10 = 8.

To find 8dB, think of 8dB as 9dB – 1dB = 8 X .8 = 6.4.  To find 5dB, think of 5dB as 6dB - 1dB = 4 times .8 = 3.2.  Note that 5dB + 5dB = 10dB = 10.  Using 5dB = 3.2:  3.2 X 3.2 = 10.24, which is about 10.  To find 2dB, use 2dB = 3dB – 1dB = 2 X .8 = 1.6

You’ll notice from the chart that the results are all within 1.5%.

To summarize the system:

Realize that 3dB = 2, so 6dB = 4, and 9dB = 8.

You know the value for 3, 6, and 9dB (2, 4, 8).

To find 2, 5, or 8dB (1 less than 3dB, 6dB, or 9dB), use -1dB = .8.

If you want to know 4 or 7dB (1 more than 3dB or 6dB), use 1dB = 1.25.

To find fractions of 1dB, realize that 1dB represents a 25% increase, so .1dB is a 2.5% increase, .2dB is a 5% increase, .4dB is a 10% increase.  (This interpolation method isn’t exact, but it’s fairly close when dealing with such small values).

You can carry this further for hundredths of a decibel.  .01dB is one tenth of .1dB.  .1dB is 2.5%, so .01dB is .25%.  .04 dB is 1%.

An example:  Suppose an antenna has a gain of 17.68 dB.  How much increase is this?  Note that 17.68 = 10 + 7 + .68.  A 10 dB increase is a factor of 10.  A 7 dB increase is a factor of 5.  Therefore, a 17 dB increase is a factor of 10 X 5 = 50.  .6 dB is 15%, .08 dB is 2%, so .68 dB is about 17%.  50 X 1.17 = 58.5, so the antenna increases the signal strength by a factor of 58.5, the signal is 58.5 times as strong (the exact value is 58.61).

Another example:  An antenna’s signal is 25 times stronger than a dipole antenna.  How many decibels is this?  This is actually quite easy.  A 10 dB increase is a factor of 10.  Because 25 = 10 X 2.5, we still have an increase of 2.5 to account for.  An increase of 2.5 is 4 dB, so the total increase = 10 dB + 4 dB = 14 dB (the exact value is 13.98 dB).

Suppose we have an increase of a factor of 30.  How many decibels is this?  30 = 10 X 3.  To multiply by 10 requires 10dB.  But, we need to add the decibels needed to multiply by 3.   4dB is 2.5, but we need 20% more (3 is 20% more than 2.5).  20% is about .8dB, so 30 = 14.8dB (the exact value is 14.77).

Well, that’s my little system for decibels, and as promised, I didn’t mention logarithms!

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GGREC IRLP/70cm Repeater Project

John VK3XJW

Two weeks after the repeater had been installed, LIPD interference started to be a problem, so the tone access had to be enable at 88.5Hz. This stopped most of the interference but one LIPD produced some low frequency modulation that sometimes would trigger the repeater. What we are going to do about this problem I am not sure, anybody got any ideas? Last weekend we tried some new antennas on the repeater made by John VK3YTV. Both are collinear with about 6 to 8DB of gain. The antenna without the top hat preformed the best out of all the antenna’s tried and was installed on the mast.

LEFT Collinear with top hat for beam down tilt. RIGHT Standard Collinear

Testing Antenna's before installing on mast.

Standard Collinear Beam width. Top Hat Collinear Beam width

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Up And Comming Projects

In an effort to build funds for the club and also start the grey matter working, several projects have been proposed.

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