transformer coupling co efficient in agilent ADS momentum

hi,

I am trying to make a transformer in ADS momentum..
i am using following equations to calculate L and Q of individual inductors
L = imag(1/y(1,1))/(2*pi*f)
Q = imag(y(1,1))/real(y(1,1))

replace y(1,1) by y(2,2) for second inductor

Qustion:(1) does this formula hold true when both inductors are modeled as differential inductors?

(2) how do i calculate coupling co-efficient k of the transformer?

thanks

Z=stoz(S,50)
Y=stoy(S,50)
k=mag(sqrt((1/Y(1,1)-Z(1,1))*Z(2,2)/imag(Z(1,1))/imag(Z(2,2))))

Can you show the source of this formulae ?

See my ppt http://www.home.agilent.com/upload/c...x_FrontEnd.pdf page 13.

My question was that.. How could you extract this relation between s-parameters and coupling coefficient ?
To my knowledge, coupling coefficient is a mechanical parameter and can not be pre-determined before.
How could you get this correlation ? Which theory supports this relation behind ?
Thanks for your answers in advance..

@bigboss/Zander_ua: can any of you tell me how to assign a ground to a transformer in ADS momentum? if i assign port 1 to one end of primary and ground the other end, ie not make a differential circuit, the analysis of equivalent circuit becomes very easy..i'll assign port 2 to one end of secondary and ground the other end of secondary too..thanks in advance..

申明：网友回复良莠不齐，仅供参考。如需专业帮助，请学习易迪拓培训专家讲授的ADS视频培训课程。