Hello all,
in designing a monopole should i use the velocity factor of the wire in
the equation or not? and if i want its impedance to be 50ohm should i
use a matching network or is there another way?

On 11 Jun 2005 14:04:37 -0700, "redhat" wrote:

Hello all,
in designing a monopole should i use the velocity factor of the wire in
the equation or not? and if i want its impedance to be 50ohm should i
use a matching network or is there another way?

Are you sure you want 50-ohm impedance and not 37-ohms?

Danny, K6MHE

yes, i want to match it to a 50ohm amplifier,why do you think it should
be 37ohm, i have simulated it using EZNEC and the source impedance is
1.684 - J 4592, is there something wrong in that? it is 1/8 wavelength
monopole.
Regards

Well, undoubtedly...

the 1.684 is the "pure resistance value", and this would best be at 50
ohms...

if "- J 4592" is a real minus 4592 ohms cap reactance--it would be good
to insert a +Jx of 4592 ohms (an equal and opposite inductive
reactance)... so that Jx ends up zero...

.... at that time you will find the finals in the amp start lasting
indefinitely--and you will not be bothered replacing them every few
minutes or so...

Warmest regards,
John

"redhat" wrote in message
oups.com...
yes, i want to match it to a 50ohm amplifier,why do you think it
should
be 37ohm, i have simulated it using EZNEC and the source impedance is
1.684 - J 4592, is there something wrong in that? it is 1/8 wavelength
monopole.
Regards

redhat wrote:
yes, i want to match it to a 50ohm amplifier,why do you think it should
be 37ohm, i have simulated it using EZNEC and the source impedance is
1.684 - J 4592, is there something wrong in that? it is 1/8 wavelength
monopole.

Try a loading coil in the center of the 1/8 wavelength.
--
73, Cecil http://www.qsl.net/w5dxp


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what about the velocity factor of the wire, should i use it in the
equation of the antenna length or not?

On 11 Jun 2005 18:03:05 -0700, "redhat" wrote:

what about the velocity factor of the wire, should i use it in the
equation of the antenna length or not?

Hi OM,

Your problems with modeling go much further. The impedance you
reported was seriously low, even for an 1/8th wave. What others were
trying to communicate to you is that even the best 1/4th wave (a
standard sized whip or monopole) will only give you 35 to 37 Ohms, not
50 Ohms. However, this is rarely an issue in achieving good
performance. A 1 Ohm antenna, on the other hand, seriously begs close
examination, patience, and care in feeding.

73's
Richard Clark, KB7QHC

yes, a wire in free space has a velocity factor... if the formula for
the antenna in question does not all ready take that velocity factor
into consideration, you will need to consider the velocity...

.... there is also "end effect"...

.... simplest thing to do would be to use the formula which takes this
all into consideration...

.... first, I would google for "shortened antennas" or "shortened
radiators" and get a bit more familiar with "loading" the antenna to
resonance...

Warmest regards,
John
"redhat" wrote in message
ps.com...
what about the velocity factor of the wire, should i use it in the
equation of the antenna length or not?

On 11 Jun 2005 16:29:47 -0700, "redhat" wrote:

yes, i want to match it to a 50ohm amplifier,why do you think it should
be 37ohm, i have simulated it using EZNEC and the source impedance is
1.684 - J 4592, is there something wrong in that? it is 1/8 wavelength
monopole.
Regards

You hadn't stated that you were using a 1/8-wavelength monopole in
your original posting and I had assumed that you were discussing a
*standard* 1/4-wave monopole.

Danny, K6MHE

i have made a modification to it, it is now 1/4 wavelength monopole.
the simulation output is : source voltage= 3092 at -89.79 deg. and
impedance= 11.08-j3092 ohm . what is the meaning of this source
voltage? i have placed a source with amplitude 1v at 0% from E1 because
it is a monopole, the ground type is free space.