Re: Discharge impedance questions. (RG-8, matching)

• To: tesla-at-pupman-dot-com
• Subject: Re: Discharge impedance questions. (RG-8, matching)
• From: "Jim Lux" <jimlux-at-jpl.nasa.gov> (by way of Terry Fritz <twftesla-at-uswest-dot-net>)
• Date: Sun, 12 Mar 2000 08:56:06 -0700
• Approved: twftesla-at-uswest-dot-net
• Delivered-To: fixup-tesla-at-pupman-dot-com-at-fixme

> >Hi Terry,
> >
> >
> >Thanks for the preliminary info!
> >
> >
> >I'm afraid I don't understand the 'through about 10 feet of RG-8' part.
> >Remember normally I speak dutch, so for us Europeans it's a little bit
> >harder to understand the 'Tesla language'.
> 
> The generator is connected to the primary coil through a 10 foot long
> length of type RG-8 coaxial cable.  A common 50 ohm transmitter cable
used
> for Ham radio and such.  Since I could tune the coil to almost no
reflected
> power, this cable should not have had any effect on anything.

The fact that you got a  match was achieved at the generator end doesn't
mean the load is matched at the other end.  A match at one end only implies
a match at the other end if the cable is a multiple of a half wavelength.
Otherwise, you'll want to use the formula:

Zin = Z0 * (Zload*cosh(gamma) + Z0*sinh(gamma))/
(Zload*sinh(gamma)+Z0*cosh(gamma))
gamma is the complex propagation constant. gamma = alpha + j beta where:
alpha is attenuation in nepers (1 neper = 8.688 dB)
beta is length in radians

(http://home.earthlink-dot-net/~jimlux/radio/tleqn.htm for more)

Fortunately, your data is probably reasonably valid.  10 feet is a quarter
wave at (18.47 meters = 16.23 MHz)... you're so much lower in frequency,
that the cable just acts like some capacitance in parallel (160 pF I
think.. 16 pF/foot for RG-8?  maybe 30 pF?  Don't have the chart in front
of me).

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