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)

One-port network with a single impedance, Z

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

Two-port pi network with impedances Z1 (left), Z2 (middle), and Z3 (right)


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

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

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

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:

enter image description here

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

  • 1
    \$\begingroup\$ The units of your formula for Z2 look wrong. You have impedance = negative admittance. Are you sure that's right? \$\endgroup\$
    – Adam Haun
    Jan 12, 2023 at 19:54
  • \$\begingroup\$ @AdamHaun Yes it is, check this paper ieeexplore.ieee.org/abstract/document/… \$\endgroup\$
    – Wallflower
    Jan 12, 2023 at 20:04
  • \$\begingroup\$ 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. \$\endgroup\$
    – Ste Kulov
    Jan 12, 2023 at 23:01
  • \$\begingroup\$ @SteKulov I found your comment unnecessary. There are other ways to find the paper. \$\endgroup\$
    – Wallflower
    Jan 13, 2023 at 8:44
  • \$\begingroup\$ Oh, OK. Thanks! \$\endgroup\$
    – Ste Kulov
    Jan 13, 2023 at 13:16

1 Answer 1


I can't access the paper you linked to, but according to Wikipedia and other sources, there's an error in your formula. The pi equivalent circuit for your admittance parameters is:

Pi network of admittances

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:

$$Z_2 = \frac 1 {-Y_{12}}$$

  • \$\begingroup\$ Yes you are right about that, I made a mistake writing the question, but my script is still correct \$\endgroup\$
    – Wallflower
    Jan 14, 2023 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.