[SOLVED] how to get the inductance from the impedance in ADS？

[orz]

i use the Zin function .
now i can get its real and imag parts.
THX

by the way:
what is the unit of "imag(Zin1)" ? I guess it's H if it is positive and F if it's negative. Am i right?

 

[WimRFP]

the unit of Imag(Z) is Ohm, as it puts out "B" from Z = A+jB. When it is positive, it represents the inductive part, negative means capacitive part.

When positive: L [H] = Imag(Zin)/(2*pi*freq).

When negative C [F] = -1/(2*pi*freq*imag(Zin) ).

 

[orz]

THX!
now I am studying active inductors.
I simulated the schematics of the paper about active induct .
but no matter how I tuning the values of the devices, I still could not get the right results.
I am wondering if anyone simulated active inductors before, and it would be nice if you could give me some advice!