How to plot constant Q curves on the ADS datadisplay smith chart

Do you mean locuses of Q=X/R or Q=B/G on smith chart ?

Yes, I mean locuses of Q = X/R ( Q isn't the same value for both X/R for circuit in serie or B/G for circuit in parallel ?! and, isnt the same for both Z chart or Y chart or ZY chart?)

Thank you,

No.

Gamma=(Z/Z0-1)/(Z/Z0+1)=-(Y/Y0-1)/(Y/Y0+1)

Yes, of course there is a shift of 180 degree between Gamma calaculated according to Z and that calculated with Y, I didn't say the opposite.

But, the locus of constante Q in the Z, Y, and ZY chat is the same, for example, for Q=1, the locus ploted in the Z, Y and ZY chart is the same.

Anyways, how to plot the constant Q circle in the Z chart

Thanks,

Draw partial circle based on definition.
Center is (0, +1/Q) or (0, -1/Q).
Radius is sqrt(1+1/Q^2).

I don't know how to access to the resistance and reactance of the smith chart on the ADS datadisplay to plot the Q
And, I don't know how to plot the equation of a circle on ADS datadisplay

I want to notice that is not as simple as we could think, I searched but didnt find, this is why I am asking here

Thanks

we ?
Only you.

It's very easy and simple.

I'm also interested. Please tell us about the simple method how to draw these circles in ADS data display.

There is one solution on the attached document,

If needed, create custom AEL function.

Actually I have custom Skill function "Constat_Q_Circle(Q)" which returns Gamma for Q in Cadence dfII.

It's too unefficient since S is bilnear transformation of z, S=(z-1)/(z+1).

I think a following is good than Smith chart utility tool of ADS.

http://www.fritz.dellsperger.net/smith.html

Update....

Qserie=X/R=Qparallel=B/G => So, it results on the same

It is simple to demonstrate by taking a serie R-L circuit and calulate the parallel equivalent R-L...

"Constant_Q_Circle.ael"

defun Constant_Q_Circle(Q)
{
  decl radius, S0p, th1, th2, dth, thp, nlen, Sp, Sm, Sq;

  radius = sqrt( 1 + 1/(Q*Q) );
  S0p = -j * (1/Q);

  th1 = acos(1 / radius);
  th2 = pi - th1;
  dth = (th2 - th1) / 50;

  thp = [th1::dth::th2];
  nlen = sweep_size(thp);
  
  Sp = S0p + radius * exp(j*thp);
  Sp[0] = 1.0;
  Sp[nlen-1] = -1.0;
  
  Sm = conj(Sp);

  Sq = [ Sp, Sm[nlen-2::-1::0] ];

  return Sq;
}

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