On Sat, Dec 21, 2019 at 06:55 PM, Gary O'Neil wrote:
it is the objective of calibration to normalize the measured standards to
those boundaries.
Hi Gary,
Actually, it is the objective of calibration to normalize the measured standards to their imperfect definitions, not to the ideal boundaries.
For example, let's say the actual short standard has a Gamma of -0.99 + j0.01. After calibration, if that short is then attached to the VNA's port, the VNA should display -0.99+j0.01 as its Gamma.
And if this is the result that the new algorithm would generate, great! But if the new equation normalizes it to -1,0, there's a problem.
One of the beauties of the traditional one-port, 3-term error model is that the three standards used for calibration could be anything and do not need to be S.O.L. -- as long as you know the Gammas of the standards, you can use these "actual" Gammas, plus their three measured Gammas, to error-correct your "Device Under Test" measurements.
In other words, if you substitute the equations for e00, e11, and delta_e into the traditional equation Gamma(DUT,actual) = (Gamma(DUT,measured - e00)/(Gamma(DUT,measured)*e11 - delta_e), you will get a single function in terms of 7 parameters. I.e:
Gamma(DUT, actual) = function(Gamma(DUT, measured), Gamma(S1, measured), Gamma(S2, measured), Gamma(S3, measured), Gamma(S1, actual), Gamma(S2, actual), Gamma(S3, actual), where S1, S2, and S3 are the three calibration standards.
But this is a very messy equation, and usually its calculation is broken up into several steps, i.e. the separate calculation of e00, e11, and delta_e, as e00 and e11 are also used in the 12-term 2-port correction. But, if just doing 1-port S11 measurements, there's no reason why Gamma(DUT,actual) can't be expressed, and solved, as a single equation of seven parameters.
If the new equation is also a function of the same seven parameters, but if it looks different from the traditional equation "expanded out", then it should be possible, through equation manipulation, to show that the two are equivalent. If this cannot be done, then I would suspect an error or incorrect assumption.
Anyway, this is not to say that the new equation is not correct, but I'm personally hesitant to declare it "finished" until I see more information.
- Jeff, k6jca
