AnsweredAssumed Answered

Problems with polar() and zin() Functions

Question asked by GWHITE on Aug 31, 2004
Latest reply on Aug 31, 2004 by mfredrik
I'm trying to create an optimizable impedance that is always at the edge of the Smith chart, but has a variable angle.  I've created a couple of test equations in a Data Display, and the results make no sense.

Eqn Angle=180
Eqn Temp1=rad(Angle)
Eqn Temp2=polar(1,rad(Angle))
Eqn Temp3=zin((polar(1,rad(Angle))),50)

The results are:

Temp1 = 3.1415 (so far, so good)
Temp2 = 0.998 + j0.055 (inaccurate & wrong sign)
                      (expected -1 +j0)
Temp3 = -2.586E-13 + j1.823E3 (nonsense)

What I think I'm doing is taking the angle (in degrees), converting it to radians, and then using the polar() function to make a complex reflection coefficient with magnitude 1.  Then the zin() function should give me the impedance I need, which should be ~zero Ohms.  The real part is close (although how it gets that from the polar conversion is beyond me).

I'm a bit rusty at this stuff, so it's possible I'm missing something fundamental and silly.  I'm certainly not getting what I need.  

As an added bit of excitment, when I put an equation with zin() into a variable block on my schematic, it claims zin is an unresolved reference.

Any suggestions/corrections?

Thanks!

Doug White  

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