How can I transform a T matrix to become a Z matrix for a two-port network?

I found the following conversion to get the T matrix from Z matrix from here: Two-Port Parameter Conversions

Step 7 - Therefor, the T parameters matrix is
$$\begin{bmatrix} A & B\\ C & D \end{bmatrix} = \begin{bmatrix} \frac{Z_{11}}{Z_{21}} & \frac{Z_{11}Z_{22}-Z_{12}Z_{21}}{Z_{21}}\\ \frac{1}{Z_{21}} & \frac{Z_{22}}{Z_{21}} \end{bmatrix}$$

But I would like to get the opposite of this. To go from T matrix to Z matrix

  • 1
    \$\begingroup\$ Just do some algebra. If you look at it for a minute you should be able to figure this out on your own. \$\endgroup\$
    – The Photon
    Commented Jun 29, 2019 at 14:41
  • \$\begingroup\$ This page clearly provides the definitions of each type. Since they already provide you with the answer for \$Z\rightarrow T\$ and from the page I've linked you know the independent definitions of \$T\$ and \$Z\$, you should be easily able to see how they arrived at the transformation they provided on the link you gave. Knowing that much, it should be a walk in the park to do the reverse. Do you follow how they derived \$Z\rightarrow T\$? If so, there are no problems. If not, focus on their derivation first. \$\endgroup\$
    – jonk
    Commented Jun 30, 2019 at 5:18

1 Answer 1


table of conversion from T to Z and Z to T

This document from ieee has a bunch of tables for such conversions, and

  • 1
    \$\begingroup\$ It looks like you started a useful answer but didn't finish. Can you complete the answer, and include a summary of anything from key links? \$\endgroup\$
    – colintd
    Commented Nov 28, 2023 at 23:55

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