# From nodal admittance matrix to an electrical circuit

Bear with me on this one. I have a one port circuit whose impedance has been measured from input port 'x' (the circuit is a wide complicated electrical circuit run in LTspice)

I was able to make it a two port circuit and extract the nodal admittance matrix of the form using LTspice (excite one port while short circuit the other and vice-versa ok?)

$$Y = \begin{bmatrix} Y_{11} & Y_{12} \\ Y_{21} & Y_{22}\end{bmatrix}$$

Now we all know that I can deduce the equivalent circuit of the now two port circuit (input nodes named 'x' and 'y') to look like this

where

$$Z_1 = \frac 1 {Y_{11}+Y_{12}}$$

$$Z_2 = -Y_{12} = -Y_{21}$$

$$Z_3 = \frac 1 {Y_{21}+Y_{22}}$$

The two ports circuit's impedance equals: Z = 1/(1/(Z2+Z3)+1/Z1) which eventually is the same as the one seen in the first figure.

However when I compare the two impedances (of the first figure) and the one recomputed using the previous expression I have a problem in the phase (I have a -1 that I don't know where it's coming from?) as seen below:

This has taken a me while so feedback will be very much appreciated.

• The units of your formula for Z2 look wrong. You have impedance = negative admittance. Are you sure that's right? Jan 12, 2023 at 19:54
• @AdamHaun Yes it is, check this paper ieeexplore.ieee.org/abstract/document/… Jan 12, 2023 at 20:04
• The paper is not free and costs \$33 USD. I made a snarky comment earlier but I guess the point went over the moderator's head. Jan 12, 2023 at 23:01
• @SteKulov I found your comment unnecessary. There are other ways to find the paper. Jan 13, 2023 at 8:44
• Oh, OK. Thanks! Jan 13, 2023 at 13:16

The equivalent impedances should be the inverse of the admittances, but your $$\Z_2\$$ formula is $$\Z_2 = -Y_{12}\$$, which is still an admittance. Try using:
$$Z_2 = \frac 1 {-Y_{12}}$$