What is the ABCD matrix of the following configuration, where a transmission line with characteristic impedance \$Z_0\$ is connected to another line with characteristic impedance \$Z_1\$ (if we consider \$Z_0\$ as port 1 and \$Z_1\$ as port 2)?

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The \$S\$ matrix given in this reference is

$$ S = \begin{pmatrix} \frac{Z_1 - Z_0}{Z_1 + Z_0} & \frac{2\sqrt{Z_0Z_1}}{Z_0 + Z_1}\\ \frac{2\sqrt{Z_0Z_1}}{Z_0 + Z_1} & \frac{Z_0 - Z_1}{Z_0 + Z_1} \end{pmatrix}, $$

which I converted to the ABCD matrix as

$$ \text{ABCD} = \begin{pmatrix} \sqrt{Z_1/Z_0} & 0\\ 0 & \sqrt{Z_0/Z_1} \end{pmatrix} $$

by the formulas from for example here. I wonder if there is a way to see the form of the ABCD matrix directly.

  • 1
    \$\begingroup\$ Where are you wanting to measure the impedance and what does this actually mean: "There is no further electrical length on either side" - please be clearer. \$\endgroup\$
    – Andy aka
    Jul 4, 2021 at 17:53
  • 1
    \$\begingroup\$ The questioner has obviously an idea: He uses ABCD presentations for chained parts and calculates the total effect as a matrix product. Only the ABCD matrix of one element is missing: That part is the zero length piece of transmission line which has only an impedance step. He already has ABCD matrices for parts in both sides of the step. \$\endgroup\$
    – user287001
    Jul 4, 2021 at 18:11

1 Answer 1


The ABCD matrix is defined by the equation

$$\begin{bmatrix}V_1\\I_1\end{bmatrix}=\begin{bmatrix}A & B\\C & D\end{bmatrix} \begin{bmatrix}V_2\\-I_2\end{bmatrix}$$

What equations apply at infinitesimal point of transition between transmission lines with different characteristic impedance?

$$V_1 = V_2$$ $$I_1 = -I_2$$

So now all you have to do is work out what values of \$A\$, \$B\$, \$C\$ and \$D\$ make the ABCD equation match the equation for your network.

  • \$\begingroup\$ If \$V_1 = V_2\$, \$I_1 = -I_2\$, isn't the ABCD matrix just the identity matrix? But I think the answer should be $$\begin{pmatrix}\sqrt{Z_1/Z_0} & 0 \\ 0 & \sqrt{Z_0/Z_1}\end{pmatrix}$$ \$\endgroup\$
    – snsunx
    Jul 4, 2021 at 20:39
  • \$\begingroup\$ How is KCL satisfied in that case? \$\endgroup\$
    – The Photon
    Jul 4, 2021 at 20:44
  • \$\begingroup\$ Actually I think the answer should be $$\begin{pmatrix} 1 & 0 \\ 0 & Z_0 / Z_1 \end{pmatrix}.$$ I am getting this answer by converting the \$S\$ matrix shown in the reference to the ABCD matrix. \$\endgroup\$
    – snsunx
    Jul 4, 2021 at 21:27
  • \$\begingroup\$ That's exactly where I'm confused about. I understand that by KCL \$I_1 = -I_2\$ should hold. But if \$V_1 = V_2, I_1 = -I_2\$, doesn't it imply both sides have the same impedance, which is not necessarily the case? \$\endgroup\$
    – snsunx
    Jul 4, 2021 at 21:30
  • \$\begingroup\$ Because the thing you're characterizing with this ABCD matrix is not the transmission lines, but the connection between them. Which is essentially just an ideal wire in circuit theory terms. \$\endgroup\$
    – The Photon
    Jul 4, 2021 at 21:53

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