[600MRG] Scopematch question

[Robert Grizzard]

You are converting from polar coordinates to rectangular for this
problem. You have some impedance Z, made up of a real part R and an
imaginary part j that are at some angle theta. If you multiply sine
theta by Z you will get the imaginary part of the impedance j. If theta
is positive then the impedance is inductive, while if theta is negative
the impedance is capacitive.

To get the real part you can either multiply Z by cosine theta or square
Z, square j, and subtract j^2 from Z^2 then take the square root.

HTH

de kg7yy

On 12/06/2017 12:17 PM, N1BUG wrote:
> I'm embarrassed that I haven't yet found (or figured out on my own) an
> answer to this, but am going to ask...
> 
> Assuming one is using a scopematch and there is some reactance so one
> trace leads the other... is there some math that will convert from
> degrees leading or lagging to ohms capacitive or inductive reactance? It
> seems that should be readily possible but I can't quite get my mind
> around it.
> 
> I guess this could be complicated or inaccurate if the waveforms are
> distorted, but let's assume for now they're not.
> 
> Thanks, 73,
> Paul N1BUG
> 
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