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I am trying to derive a expression for calculating the input impedance of a 3 port network to use as direct calculating code and avoid SPICE/simulator solving of the same.

I am able to solve the input impedance of a 2 port system with a load, \$Z_{load}\$, connected to port 2. The impedance looking into port 1 would be (by solving basic 2 port theory):

$$ Z_{in} = Z_{11} - \frac{Z_{21}Z_{12}}{Z_{22}+Z_{load}} $$

If I have a 3 port network with 2 ports connected to different loads, \$Z_{load1}\$ and \$Z_{load2}\$, how can I generate an expression for \$Z_{in}\$ looking in from port 1 starting from only the Z-matrix, \$Z_{load1}\$ and \$Z_{load2}\$?

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  • \$\begingroup\$ Can you use LaTex to type in your equation? It looks better. \$\endgroup\$ – Niteesh Shanbog Nov 5 '18 at 7:15
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    \$\begingroup\$ Didn't you think it relevant to actually show the circuit of your 3 port network? \$\endgroup\$ – Andy aka Nov 5 '18 at 11:02
  • \$\begingroup\$ @Andyaka I believe he's referring to the 3-port version of a generic two-port network instead of a specific circuit. \$\endgroup\$ – Sven B Nov 5 '18 at 13:47
  • \$\begingroup\$ @Andyaka Im sorry if my question wasnt too clear, but as Sven mentioned, it is for a generic network \$\endgroup\$ – Sum_one Nov 6 '18 at 6:48
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The 3-port can be described in 3 equations, using

$$\left(\begin{matrix} V_1 \\ V_2 \\ V_3 \end{matrix}\right) = \left(\begin{matrix} Z_{11} & Z_{12} & Z_{13} \\ Z_{21} & Z_{22} & Z_{23} \\ Z_{31} & Z_{32} & Z_{33} \end{matrix}\right)\left(\begin{matrix} I_1 \\ I_2 \\ I_3\end{matrix}\right)$$

I will now load ports 1 and 2, while using port 3 as the input. Adding two loads adds two new equations

$$\begin{align} V_1 &= -Z_{L1}\cdot I_1 \\ V_2 &= -Z_{L2}\cdot I_2 \end{align}$$

This can be inserted into the matrix notation:

$$\left(\begin{matrix} Z_{11} & Z_{12} & Z_{13} \\ Z_{21} & Z_{22} & Z_{23} \\ Z_{31} & Z_{32} & Z_{33} \end{matrix}\right)\left(\begin{matrix} I_1 \\ I_2 \\ I_3\end{matrix}\right) = \left(\begin{matrix} -Z_{L1}\cdot I_1 \\ -Z_{L2}\cdot I_2 \\ V_3 \end{matrix}\right)$$

Which is the same as

$$\left(\begin{matrix} Z_{11} + Z_{L1} & Z_{12} & Z_{13} \\ Z_{21} & Z_{22} + Z_{L2} & Z_{23} \\ Z_{31} & Z_{32} & Z_{33} \end{matrix}\right)\left(\begin{matrix} I_1 \\ I_2 \\ I_3\end{matrix}\right) = \left(\begin{matrix} 0 \\ 0 \\ V_3 \end{matrix}\right)$$

Since we want to solve for \$Z_{in} = \frac{V_3}{I_3}\$, we can use Cramer's rule to find

$$ I_3 = \frac{\left|\begin{matrix} Z_{11} + Z_{L1} & Z_{12} & 0 \\ Z_{21} & Z_{22} + Z_{L_2} & 0 \\ Z_{31} & Z_{32} & V_3 \end{matrix}\right|}{\left|\begin{matrix} Z_{11} + Z_{L1} & Z_{12} & Z_{13} \\ Z_{21} & Z_{22} + Z_{L2} & Z_{23} \\ Z_{31} & Z_{32} & Z_{33} \end{matrix}\right|}\ $$

Or also

$$ Z_{in} = \frac{V_3}{I_3} = \frac{\left|\begin{matrix} Z_{11} + Z_{L1} & Z_{12} & Z_{13} \\ Z_{21} & Z_{22} + Z_{L2} & Z_{23} \\ Z_{31} & Z_{32} & Z_{33} \end{matrix}\right|}{\left|\begin{matrix} Z_{11} + Z_{L1} & Z_{12} & 0 \\ Z_{21} & Z_{22} + Z_{L_2} & 0 \\ Z_{31} & Z_{32} & 1 \end{matrix}\right|} $$

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  • \$\begingroup\$ Thank you very much for the very clear derivation of the expression. Which position of the resultant 3x3 matrix generated would be used as the Zin term to be plotted? I tried implementing this expression in Python( taking inverse of the denominator matrix and multipying with numerator array to generate a 3x3 matrix) but could not match the result with simulator output ( sorry but unable to reproduce code due to company restrictions) \$\endgroup\$ – Sum_one Nov 6 '18 at 6:53
  • \$\begingroup\$ Cramer's rule involves calculating determinants, ie. scalars. The result is not a matrix. \$|A| = det(A)\$. \$\endgroup\$ – Sven B Nov 6 '18 at 9:21
  • \$\begingroup\$ Thank you for the clarification. Really appreciate your help in understanding this complex problem \$\endgroup\$ – Sum_one Nov 6 '18 at 23:51

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