Thread: Matching impedance with coax

"Jeff Liebermann" <jeffl@cruzio.com> wrote in message news:h60ff55vd3ctrjopff7m879a8i3th148ah@4ax.com... > On Sun, 8 Nov 2009 19:06:49 -0600, "amdx" <amdx@knology.net> wrote: > >>Ok, here are some more pictures. If anyone is so interested that they want >>to >>model the antenna I'll post picures or dimensions or both of...

On Sun, 8 Nov 2009 18:17:57 -0800, dplatt@radagast.org (Dave Platt) wrote: >> Odd multiples of 1/4 wavelength will neatly transform the endpoint >> impedances according to: >> Zcoax = sqrt (Zin * Zout) >> or >> Zcoax^2 = Zin * Zout >> So, with a 50 ohm load, 75 ohm coax, and 3/4 wavelengths of coax: >> Zout = 112.5 ohms >> which is...

On Sun, 08 Nov 2009 21:09:27 -0800, Jeff Liebermann <jeffl@cruzio.com> wrote: >Well, I previous guestimated that the 6 mm of exposed center conductor >at the coax connector was good for about 3 nH or about 45 ohms at >2.4Ghz. If the balun represents 50 ohms from the antenna, then the RF >power is roughly split evenly between being radiated by the 6 mm >"leak" and going to the...

On Sat, 07 Nov 2009 06:32:25 +0800, who where <noone@home.net> wrote: >you seem to need the limelight, posting a name and callsign. Yeah, as a longstanding convention for thousands of posters here, it is a strange thing about being public and open in this world isn't it? If you can't put your name to it, then any posting is only vacant spam. "No one at home" informs us all about quality. On the...

On Fri, 06 Nov 2009 20:55:19 -0600, tom <news4792@taring.org> wrote: >who where wrote: >> >> Whatever - and whoever Art is. I wonder why people like you carry on >> at a personal level towards posters whose views you don't share. And >> you seem to need the limelight, posting a name and callsign. >> >> I'm describing how the...

who where wrote: > > Whatever - and whoever Art is. I wonder why people like you carry on > at a personal level towards posters whose views you don't share. And > you seem to need the limelight, posting a name and callsign. > > I'm describing how the matching IS done commercially. You can crap on > forever if you wish about how you might do it. Fini. The "ways it's done...

On Fri, 06 Nov 2009 20:55:19 -0600, tom <news4792@taring.org> wrote: >who where wrote: >> >> Whatever - and whoever Art is. I wonder why people like you carry on >> at a personal level towards posters whose views you don't share. And >> you seem to need the limelight, posting a name and callsign. >> >> I'm describing how the...

who where wrote: > > Whatever - and whoever Art is. I wonder why people like you carry on > at a personal level towards posters whose views you don't share. And > you seem to need the limelight, posting a name and callsign. > > I'm describing how the matching IS done commercially. You can crap on > forever if you wish about how you might do it. Fini. The "ways it's done...

> Odd multiples of 1/4 wavelength will neatly transform the endpoint > impedances according to: > Zcoax = sqrt (Zin * Zout) > or > Zcoax^2 = Zin * Zout > So, with a 50 ohm load, 75 ohm coax, and 3/4 wavelengths of coax: > Zout = 112.5 ohms > which is a bit closer to what I would expect to see with a folded > dipole antenna. Another thing to note: based on the pictures posted...

Jeff Liebermann wrote: > On Mon, 09 Nov 2009 10:11:42 +1000, atec7 7 <"atec 77"@hotmail.com> > wrote: > >>> One wavelength at 2.4Ghz is 12.5cm. Guessing from the photo, there's >>> a total of about 15mm of exposed conductor. That's about 1/8th >>> wavelenth, which will still radiate rather badly, but not as badly as...