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I have an ABCD matrix represenataion which is defined as:

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For such a problem, what approach should I follow in order to organize the new ( reversed) network ports voltages and currents properly. I have tried to keep the currents of ports same and making voltages switching places but this did not make any sense in the substituted-general form calculations.

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  • \$\begingroup\$ Is there a restriction on the type of network? Like it's a network composed only of passive components? \$\endgroup\$ – The Photon Dec 5 '18 at 23:17
  • \$\begingroup\$ Not sure if this helps, but at least you can see the basics from this:$$\begin{align*}\left[\begin{matrix}V_2\\ I_2\end{matrix}\right]&=\left[\begin{matrix}A&B\\ C&D\end{matrix}\right]\left[\begin{matrix}V_1\\ I_1\end{matrix}\right]\\\\ \left[\begin{matrix}A&B\\ C&D\end{matrix}\right]^{-1}\left[\begin{matrix}V_2\\ I_2\end{matrix}\right]&=\left[\begin{matrix}A&B\\ C&D\end{matrix}\right]^{-1}\left[\begin{matrix}A&B\\ C&D\end{matrix}\right]\left[\begin{matrix}V_1\\ I_1\end{matrix}\right] \end{align*}$$ \$\endgroup\$ – jonk Dec 6 '18 at 1:23
  • \$\begingroup\$ and:$$\begin{align*} \left[\begin{matrix}A&B\\ C&D\end{matrix}\right]^{-1}\left[\begin{matrix}V_2\\ I_2\end{matrix}\right]&=\left[\begin{matrix}V_1\\ I_1\end{matrix}\right]\\\\ \frac{1}{A\cdot D-B\cdot C}\left[\begin{matrix}D&-B\\ -C&A\end{matrix}\right]\left[\begin{matrix}V_2\\ I_2\end{matrix}\right]&=\left[\begin{matrix}V_1\\ I_1\end{matrix}\right] \end{align*}$$ \$\endgroup\$ – jonk Dec 6 '18 at 1:23
  • \$\begingroup\$ This is very straightforward. I do not think that few manipulations in linear algebra will solve the problem. I am worried more about the reconstruction of the network ports, currents, ...etc. No specific restriction on the type of the network. But due to RF purposes, it can be assumed to contain only passive elements if that can simplify the issue. \$\endgroup\$ – utdlegend Dec 6 '18 at 6:46
  • \$\begingroup\$ Different lengths of the forward and back parts of the network may point to asymmetric network ! \$\endgroup\$ – utdlegend Dec 6 '18 at 6:49
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No full solutions to homeworks, only guidance!

The new matrix is in math called the inverse matrix of the original. You can find it by solving V1, I1 from the equations which the matrix present. In numerical cases one can use software. For ex. Excel can find the inverses of numerical matrices. Complex numbers are accepted. But your case is not numerical because A,B,C,D are variables.

The inverse does not always exist. That's the case when several inputs can cause the same output.

The inverse of a 2x2 matrix is so simple and often used that many people remember the inversion result as a general formula. It's shown also here:

https://support.office.com/en-us/article/minverse-function-11f55086-adde-4c9f-8eb9-59da2d72efc6 ADD: the general 2x2 inversion formula seems to be given already in a comment, too.

If you compare the general inversion formula and what the inversion should be in your case, you can decide what A,B,C,D values are possible. There of course are several possible circuits which fullfill the resulted equation, but the simplest possible do not need many lines to draw.

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    \$\begingroup\$ But in general the inverse of a 2x2 matrix isn't just the same entries rearranged, so how do we get the result OP asked for? \$\endgroup\$ – The Photon Dec 6 '18 at 2:21
  • \$\begingroup\$ I am not seeking for full answer. I could not understand the "reversing" term of the problem, if A and D switch places how is it possible for B and C stay same. I think that ports still constant with same voltages and currents but switching the network elements if possible is required. \$\endgroup\$ – utdlegend Dec 6 '18 at 13:27
  • \$\begingroup\$ However, is it possible to get a general prove in this case? I tried to solve the same question with a general T-network but I could not get the same results! \$\endgroup\$ – utdlegend Dec 6 '18 at 13:28
  • \$\begingroup\$ @utdlegend try to find those numbers which satisfy equation -X=X. \$\endgroup\$ – user287001 Dec 6 '18 at 13:38
  • \$\begingroup\$ @user287001 unfunny joke. I think I solved the problem by defining a semi-general network elements though. \$\endgroup\$ – utdlegend Dec 7 '18 at 22:17

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