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[TowerTalk] CATV length

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Subject: [TowerTalk] CATV length
From: k6ll@juno.com (David O Hachadorian)
Date: Fri, 04 Jul 1997 13:55:52 EDT
On Fri, 4 Jul 1997 10:03:48 -0500 ROBERT SAUVAN <cabmas@bankpds.com>
writes:
>Towertalkians:
>    As you know,I raised a question about CATV hardline connectors 
>which
>prompted other questions about matching 75 ohm to 50 ohms.I am now
>trying to calculate the electrical length I need to cut the 75 ohm 
>CATV so
>as to eliminate using 50-75 ohm transformers.Both towers will accept 
>up to
>250 ft of feedline each.The formula I used to calculate electrical 
>wavelength
>is:983.6 divided by freq.X velocity factor.The VF I used was 81%.On 
>one of
>my towers I have a 3 ele 40 mtr beam,160 inv L,and a 80 mtr inv 
>L.Using the
>formula,I came up with the following:
>               
>               1/2 wave at 1810 = 220.20 ft
>               1 wave at 3550 = 224.42 ft
>               2 waves at 7100 = 224.42
>
>Should I cut this feedline at 224.42 ft to operate all the antennas on 
>this 
>tower?Will that length work OK at the 1810 frequency with acceptable 
>swr?

The first thing I would do is MEASURE the velocity factor of
the cable. Short one end, then use a grid dip meter or
antenna analyzer to find the lowest frequency at which an
impedance null occurs. This is the electrical length of a half wave.
You can then calculate the actual vf:


       (physical length in feet)x(frequency in mHz)
vf =  ----------------------------------------------
                   491.8


Then I would redo all of your calculations above, using the
actual measured vf. The calculations are all correct for a
vf of .81.

Assuming your antenna is exactly 50 ohms,
the impedance presented to the transmitter at 1810 is
51.1 +j2.5. The swr seen by a 52 ohm transmitter at
1810 is 1.06.




>
>   My second tower has a KT34XA tribander,80 mtr dipole,40 mtr sloper.
>I calculate the following lengths:
>
>               1 wave at 3550 = 224.42
>               2 waves at 7100 = 224.42
>               4 waves at 14100 = 226.00
>               6 waves at 21100 = 226.50
>               8 waves at 28.100 = 226.80
>
>The tribander being the most important of the antennas on this 
>tower,should 
>I cut the feedline at 226.00 to operate all of the antennas on this 
>tower?Will
>that length be close enough to be acceptable with the other antennas 
>also?

Assuming that all your antennas are a flat 50 ohms across the
band, half inch 75 ohm hardline, and vf=.81, the TL program
comes up with the following swr's for a 52 ohm transmitter:

freq    swr

3.5     1.05
4.0     1.81
7.0     1.08
7.3     1.41
14.0    1.16
14.35   1.4
21.0    1.24
21.45   1.47
28.0    1.31
28.6    1.63

In all cases, an swr of 1 occurs somewhere in the band. Not bad!

>
>Both towers have remote antenna switches where the hardline 
>terminates.
>My next question is this:I will be needing to run flexible coax to the 
>antennas
>from the remote coax switches.Do I run 50 ohm cable or continue with a 
>75 ohm.
>And,depending on which impedence I am told to use, are these 
>additional lengths to be considered in my total electrical length 
>calculations?

Assuming once again that the antennas are about 50 ohms, you
can use any convenient length of 52 ohm coax.

The TL program is pretty handy for this kind of stuff. It
came packaged with my 1994 ARRL Antenna Handbook. I don't
know if it's available elsewhere.

Dave Hachadorian, K6LL
k6ll@juno.com

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