# Linear 2-port network representation

• cianfa72
In summary, the conversation discusses the existence of an implicit linear equations based representation for linear two-port networks. If the network contains only passive one-port components with no independent current/voltage generators, the representation always exists. However, the question arises if this representation also exists when controlled generators are included in the internal structure of the network. It is suggested that treating the controlled generators as external components may lead to a linear network with nonlinear boundary conditions.
cianfa72
TL;DR Summary
about the existence of implicit representation for a linear 2-port network
Hi,

in the context of linear two-ports networks my (italian) textbook says that if its internal structure consists of passive one-port components with no independent current/voltage generators then the following (implicit) linear equations based representation always exist:

##\left[A\right] V + \left[B\right] I = 0## where ##V## is the column vector of port voltages and ##I## that of port currents

Now, if we take in account just only resistors as passive one-port components than the reason behind is clear: just close the 2-port network with 2 resistor and apply the Tellegen theorem to the overall circuit. All voltages and currents have to be null so the complete set of linear equations have to be linear independent, thus by elimination the two linear equations representing the 2-port network are independent themselves.

On the other hand, what if we include (voltage and/or current) controlled generators as one-port component inside the internal structure of the 2-port network ? Starting from the complete set of linear equations can we always obtain that linear representation by elimination ?

If I understand right, you are describing a linear network with nonlinear boundary conditions. The controlled generators make the nonlinear boundary conditions.

Do you realize that is a description of the power grid? You could model the entire European power grid as you describe. A linear network with nonlinear boundary conditions. Because of the nonlinearities, solutions must be iterative.

anorlunda said:
If I understand right, you are describing a linear network with nonlinear boundary conditions. The controlled generators make the nonlinear boundary conditions.
As far as I can tell maybe my point is not clear...

Controlled generators we're talking about are inside the 2-port network (inside the 'black-box' accessible from 2 ports). Just as an example consider a 2-port accessible 'black-box' with a (current) controlled voltage generator inside. Its value depends linearly on the current flowing through another component inside the network itself.

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If the controlled generator responds with anything other than constant impedance (for example constant power), it becomes a nonlinear device. In general, you can not eliminate them with linear equations.

You can make a piecewise linear representation and solve that, but the solution is valid only at one operating point.

If you have trouble visualizing that, just imagine the network having no components other than the controlled generator. Would that fit your linear equations?

The general way to handle those problems is to treat the generators as external components, not part of the network. That leads you to linear network with nonlinear boundary conditions as I described before.

DaveE and berkeman
anorlunda said:
If the controlled generator responds with anything other than constant impedance (for example constant power), it becomes a nonlinear device. In general, you can not eliminate them with linear equations.
Consider the following two-port network in which the controlled generator voltage ##V_g## is equal to ##K*I_3## (##I_3## is the current through ##R_3##)

Here the 2-port network has a simple external representation (a 2 equations homogeneous linear system), don't you ?

##\begin{bmatrix} V_a\\V_b\end{bmatrix} + \left[B\right]\begin{bmatrix} I_a\\I_b\end{bmatrix} = 0##

I see two networks. IA is not dependent on anything on the B side. Solve for IA first, then get VG, then IB. The equations are not simultaneous. Giving the B side an external dependency on A, does not pull A into a set of simultaneous equations unless A depends on B.

Sorry, but I'm not able to catch your point...

Just to rephrase the original question: consider the following schema - a box internally made of linear two-pole elements with no independent current/voltage generators, externally accessible from 2 ports each one closed on a two-pole element.

Does the following linear equations system representation always exist for it ?

##\left[A\right] V + \left[B\right] I = 0##; ##V=\begin{bmatrix} V_a\\V_b\end{bmatrix}## and ##I=\begin{bmatrix} I_a\\I_b\end{bmatrix}##

cianfa72 said:
Summary: about the existence of implicit representation for a linear 2-port network

Hi,

in the context of linear two-ports networks my (italian) textbook says that if its internal structure consists of passive one-port components with no independent current/voltage generators then the following (implicit) linear equations based representation always exist:

##\left[A\right] V + \left[B\right] I = 0## where ##V## is the column vector of port voltages and ##I## that of port currents

Now, if we take in account just only resistors as passive one-port components than the reason behind is clear: just close the 2-port network with 2 resistor and apply the Tellegen theorem to the overall circuit. All voltages and currents have to be null so the complete set of linear equations have to be linear independent, thus by elimination the two linear equations representing the 2-port network are independent themselves.

On the other hand, what if we include (voltage and/or current) controlled generators as one-port component inside the internal structure of the 2-port network ? Starting from the complete set of linear equations can we always obtain that linear representation by elimination ?
I believe this question is equivalent to the question, "Can you diagonalize a matrix?"

And I'm pretty sure the answer to that will always be "yes" for a properly-defined network. In particular, if the matrix is symmetric it is always diagonalizable.

I think that the matrix must be invertible to be diagonalizable.

kimbyd said:
I believe this question is equivalent to the question, "Can you diagonalize a matrix?"

And I'm pretty sure the answer to that will always be "yes" for a properly-defined network. In particular, if the matrix is symmetric it is always diagonalizable.
I tried to reason as follows. Consider the complete circuit obtained closing each 'box' port with a two-pole linear element. Assume the box inside consists of connected two-pole linear elements with no independent voltage/current generators.

Starting from it, write down KCL, KVL and two-pole linear costitutive equations (including controlled generators if any inside) getting an homogeneous linear system having the number of linear equations the same as the number of unknowns (square system resolution matrix)

Take for instance the network in post #7 having a resolving matrix with 14 rows and 14 unknowns (14x14)
$$\begin{pmatrix} * & * & * & * & * & * & * & * & * & * & * & * & * & * & \\ * & * & * & * & * & * & * & * & * & * & * & * & * & * & \\ * & * & * & * & * & * & * & * & * & * & * & * & * & * & \\ * & * & * & * & * & * & * & * & * & * & * & * & * & * & \\ \vdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a & b & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & c & d \\ \end{pmatrix} \begin{pmatrix} * \\ * \\ * \\ * \\ \vdots \\ V_{a} \\ I_{a} \\ V_{b} \\ I_{b} \\ \end{pmatrix}$$
By means of Gauss elimination we can transform it into raw echelon form up to the last 4 raws (basically eliminating 14-4=10 voltage/current unknowns 'inside' the box). At that point we find ourselves with a set of linear equations involving just only the port voltage/current unknowns. From those we can obtain up to to 4 independent linear equations (last 2 are independent each other for sure).

This way we can conclude '2 linear equation representation" always exist with the given conditions even if we cannot assume, in general, they are independent though...

I think the representation you're looking for: "[A]V+I=0 where V is the column vector of port voltages and I that of port currents "

is called "ABCD" parameters on that page.

Often, a network in a black box can be characterized by any of the 6 parameter sets, and any particular one can be transformed to any other one using the table "Interrelation of parameters" at the bottom of the Wikipedia page.

It is possible to have a black box with a network of two terminal elements inside that cannot be represented by every one of the 6 parameter sets on the Wikipedia page. For example, the network consisting of a single resistor from the a (input) port to the b (output) port can be represented with admittance parameters, but not with impedance parameters.

If there is no coupling either in the forward or the reverse direction, which is lacking in your network of post #7, there probably won't be a representation in ABCD parameters.

The network in post #5 does have coupling from the input port to the output port (but not in the reverse direction), so it does have an ABCD parameter representation.

The Electrician said:

I think the representation you're looking for: "[A]V+I=0 where V is the column vector of port voltages and I that of port currents "

is called "ABCD" parameters on that page.

from wikipedia link [A]V+I=0 -> I = - [A]V should be actually the y-parameters model.

I think the main point is that, starting from let me say the 'implicit' representation ##\left[A\right] V + \left[B\right] I = 0## we assume always exist (see my previous post #10), the 6 models result from a different selection of the 2 independent port variable (combination without repetition of 2 element taken in 4 is actually 6)

The Electrician said:
If there is no coupling either in the forward or the reverse direction, which is lacking in your network of post #7, there probably won't be a representation in ABCD parameters.
ok that's right.

When I "copied and pasted" from post #1, part of what I copied wasn't bold, and in the process of fixing that, something got lost. I should have had:

This gives ABCD parameters.

I think it's easier to just use the nodal method of analysis to get the two-port Y parameters in a couple of steps.
The admittance matrix for your network in post #5 can be written by inspection like this:

There are 4 nodes if a non-essential node is placed between the controlled source and R4.

Then all the (internal) nodes except the two port nodes are eliminated using the Schur complement technique: https://en.wikipedia.org/wiki/Schur_complement (as described under the heading "Application to solving linear equations").

Can you use the procedure you describe in post #10 to find two equations for what is probably the simplest case with no coupling:

The Electrician said:
When I "copied and pasted" from post #1, part of what I copied wasn't bold, and in the process of fixing that, something got lost. I should have had:

View attachment 246811

This gives ABCD parameters.
The representation AV + BI = 0 is actually ##\begin{pmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{pmatrix}\begin{pmatrix} V_{1}\\V_{2}\end{pmatrix}+\begin{pmatrix}b_{11} & b_{12}\\b_{21} & b_{22}\end{pmatrix}\begin{pmatrix} I_{1}\\I_{2}\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}##

Depending on which minors of the combined rectangular matrix A|B = ##\begin{pmatrix}a_{11} & a_{12} & b_{11} & b_{12}\\a_{21} & a_{22} & b_{21} & b_{22}\end{pmatrix}## are ##\neq 0## we are able to 'extract' one or another of the 6 different models (possibly all of them).

To me it seems ABCD model you're talking about is just one of the 6 possible models so it is not the general 2-port 'implicit' representation AV + BI = 0 , are you agree ?

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marcusl said:
I think that the matrix must be invertible to be diagonalizable.
Not necessarily. A matrix with more than one eigenvalue that is zero can still often be diagonalized. But any matrix with one or more zero eigenvalues cannot be inverted. Nearly all matrices are diagonalizable if you permit complex-valued diagonalization matrices. See here:
https://en.wikipedia.org/wiki/Diagonalizable_matrix#Characterization

As a rule of thumb, over
almost every matrix is diagonalizable. More precisely: the set of complex
matrices that are not diagonalizable over
, considered as a subset of
, has Lebesgue measure zero.

Of course, if a matrix isn't square, diagonalization doesn't even make sense.

Real matrices associated with real problems like this, however, are almost always both diagonalizable and invertible, for the simple reason that if they weren't, then the solution would be ill-defined. And if the solution were ill-defined, then the physical system couldn't behave in a sensible manner. I can't guarantee that the matrix will always be diagonalizable and invertible, but it probably is as long as the problem is well-formed.

marcusl
cianfa72 said:
The representation AV + BI = 0 is actually ##\begin{pmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{pmatrix}\begin{pmatrix} V_{1}\\V_{2}\end{pmatrix}+\begin{pmatrix}b_{11} & b_{12}\\b_{21} & b_{22}\end{pmatrix}\begin{pmatrix} I_{1}\\I_{2}\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}##

Depending on which minors of the combined rectangular matrix A|B = ##\begin{pmatrix}a_{11} & a_{12} & b_{11} & b_{12}\\a_{21} & a_{22} & b_{21} & b_{22}\end{pmatrix}## are ##\neq 0## we are able to 'extract' one or another of the 6 different models (possibly all of them).

To me it seems ABCD model you're talking about is just one of the 6 possible models so it is not the general 2-port 'implicit' representation AV + BI = 0 , are you agree ?

You are showing a more general description of AV + BI = 0 than you did in earlier posts. I didn't think that A was a 2x2 matrix of values; I thought it was only a 2x1 vector.

You speak of a "resolving matrix", which is a term I'm not familiar with. It appears that you are studying topological analysis of circuits. I would be more familiar with terms such as "incidence matrix", etc.

I wonder if it would be possible to extract one or more of the 6 standard parameter sets from your very general formulation? It might be instructive to solve the system in post #14 with your general method and see whitch of the 6 parameter sets you can extract.

The Electrician said:
You are showing a more general description of AV + BI = 0 than you did in earlier posts. I didn't think that A was a 2x2 matrix of values; I thought it was only a 2x1 vector.
In that case multiplication operation involved should be in the form column vector x column vector, for instance: ##\begin{pmatrix} a_{1}\\a_{2}\end{pmatrix} \begin{pmatrix} V_{1}\\V_{2}\end{pmatrix} ## ; does it actually exist ? I'm not aware of...

The Electrician said:
You speak of a "resolving matrix", which is a term I'm not familiar with. It appears that you are studying topological analysis of circuits. I would be more familiar with terms such as "incidence matrix", etc.
Sorry, that name is not a well-known/standard name, it was just used by me to indicate the complete matrix associated to the linear system.

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cianfa72 said:
The representation AV + BI = 0 is actually ##\begin{pmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{pmatrix}\begin{pmatrix} V_{1}\\V_{2}\end{pmatrix}+\begin{pmatrix}b_{11} & b_{12}\\b_{21} & b_{22}\end{pmatrix}\begin{pmatrix} I_{1}\\I_{2}\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}##

Depending on which minors of the combined rectangular matrix A|B = ##\begin{pmatrix}a_{11} & a_{12} & b_{11} & b_{12}\\a_{21} & a_{22} & b_{21} & b_{22}\end{pmatrix}## are ##\neq 0## we are able to 'extract' one or another of the 6 different models (possibly all of them).

To me it seems ABCD model you're talking about is just one of the 6 possible models so it is not the general 2-port 'implicit' representation AV + BI = 0 , are you agree ?
So, the thing is: this representation is over-specifying the problem. Typically either A or B will be invertible (or both), in which case the problem simplifies to:
$$V + A^{-1}BI = 0$$
or
$$B^{-1}AV + I = 0$$

Both of these equations only have four unknowns each. If neither is invertible, the problem is poorly-defined (i.e., the solution is ambiguous).

kimbyd said:
So, the thing is: this representation is over-specifying the problem. Typically either A or B will be invertible (or both), in which case the problem simplifies to:
$$V + A^{-1}BI = 0$$
or
$$B^{-1}AV + I = 0$$

Both of these equations only have four unknowns each. If neither is invertible, the problem is poorly-defined (i.e., the solution is ambiguous).
Consider the following formal "singular" example:

the 'resolving' linear system for it is ##\begin{bmatrix}1 & -1\\0 & 0\end{bmatrix}\begin{bmatrix} V_{a}\\V_{b}\end{bmatrix}+\begin{bmatrix}0 & 0\\1 & 1\end{bmatrix}\begin{bmatrix} I_{a}\\I_{b}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}##
because neither ##A## or ##B## are invertible we are 'forced' to choose columns taken from both matrix as independent variables (for instance the couple ##V_{a}## and ##I_{b}##)

Take another example depicted in the picture:

The complete linear system is:

##\begin{bmatrix}0 & 0 & 1 & 0 & 1 & 0 & 0 & 0\\ -1 & 0 & -1 & 0 & 0 & 0 & 1 & 0\\0 & -1 & 0 & 1 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 1 & 0 & 1 & 0 & -1\\-1 & 0 & 0 & 0 & 0 & 0 & 0 & k\\0 & 0 & -R_{1} & 1 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & R_{a} & 1 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & R_{b} & 1\end{bmatrix}\begin{bmatrix} I_{0}\\V_{0}\\I_{1}\\V_{1}\\I_{a}\\V_{a}\\I_{b}\\V_{b}\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\\0\\0\\0\end{bmatrix}##

swapping first row with second:
##\begin{bmatrix}-1 & 0 & -1 & 0 & 0 & 0 & 1 & 0\\0 & 0 & 1 & 0 & 1 & 0 & 0 & 0\\0 & -1 & 0 & 1 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 1 & 0 & 1 & 0 & -1\\-1 & 0 & 0 & 0 & 0 & 0 & 0 & k\\0 & 0 & -R_{1} & 1 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & R_{a} & 1 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & R_{b} & 1\end{bmatrix}\begin{bmatrix} I_{0}\\V_{0}\\I_{1}\\V_{1}\\I_{a}\\V_{a}\\I_{b}\\V_{b}\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\\0\\0\\0\end{bmatrix}##

zeroes the elements under the first column pivot
##\begin{bmatrix}-1 & 0 & -1 & 0 & 0 & 0 & 1 & 0\\0 & 0 & 1 & 0 & 1 & 0 & 0 & 0\\0 & -1 & 0 & 1 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 1 & 0 & 1 & 0 & -1\\0 & 0 & 1 & 0 & 0 & 0 & -1 & k\\0 & 0 & -R_{1} & 1 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & R_{a} & 1 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & R_{b} & 1\end{bmatrix}\begin{bmatrix} I_{0}\\V_{0}\\I_{1}\\V_{1}\\I_{a}\\V_{a}\\I_{b}\\V_{b}\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\\0\\0\\0\end{bmatrix}##

swapping second row with third
##\begin{bmatrix}-1 & 0 & -1 & 0 & 0 & 0 & 1 & 0\\0 & -1 & 0 & 1 & 0 & 1 & 0 & 0\\0 & 0 & 1 & 0 & 1 & 0 & 0 & 0\\0 & 0 & 0 & 1 & 0 & 1 & 0 & -1\\0 & 0 & 1 & 0 & 0 & 0 & -1 & k\\0 & 0 & -R_{1} & 1 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & R_{a} & 1 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & R_{b} & 1\end{bmatrix}\begin{bmatrix} I_{0}\\V_{0}\\I_{1}\\V_{1}\\I_{a}\\V_{a}\\I_{b}\\V_{b}\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\\0\\0\\0\end{bmatrix}##

zeroes the elements under the third column pivot
##\begin{bmatrix}-1 & 0 & -1 & 0 & 0 & 0 & 1 & 0\\0 & -1 & 0 & 1 & 0 & 1 & 0 & 0\\0 & 0 & 1 & 0 & 1 & 0 & 0 & 0\\0 & 0 & 0 & 1 & 0 & 1 & 0 & -1\\0 & 0 & 0 & 0 & -1 & 0 & -1 & k\\0 & 0 & 0 & 1 & R_{1} & 0 & 0 & 0\\0 & 0 & 0 & 0 & R_{a} & 1 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & R_{b} & 1\end{bmatrix}\begin{bmatrix} I_{0}\\V_{0}\\I_{1}\\V_{1}\\I_{a}\\V_{a}\\I_{b}\\V_{b}\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\\0\\0\\0\end{bmatrix}##

zeroes the elements under the fourth column pivot
##\begin{bmatrix}-1 & 0 & -1 & 0 & 0 & 0 & 1 & 0\\0 & -1 & 0 & 1 & 0 & 1 & 0 & 0\\0 & 0 & 1 & 0 & 1 & 0 & 0 & 0\\0 & 0 & 0 & 1 & 0 & 1 & 0 & -1\\0 & 0 & 0 & 0 & -1 & 0 & -1 & k\\0 & 0 & 0 & 0 & R_{1} & -1 & 0 & 1\\0 & 0 & 0 & 0 & R_{a} & 1 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & R_{b} & 1\end{bmatrix}\begin{bmatrix} I_{0}\\V_{0}\\I_{1}\\V_{1}\\I_{a}\\V_{a}\\I_{b}\\V_{b}\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\\0\\0\\0\end{bmatrix}##

So fare we have eliminated the 'black-box' internal voltages/currents obtaining an homogeneous system of 4 linear equations with the first 2 linearly independent.

The complete system of 4 equations admit infinite solutions when ##k = - \frac {R_{1}+R_{a}+R_{b}} {R_{b}(R_{1}+R_{a})}## as expected

cianfa72 said:
Take another example depicted in the picture:
View attachment 247133

I get the same value for K by doing this:

A controlled current source whose output current is proportional to the voltage across it is equivalent to a resistor. Your dependent source is equivalent to a resistor of 1/K ohms.

Instead of eliminating the internal voltages and currents, how would you eliminate the external voltages and currents, and finally, obtain the two linear equations representing the two-port alone?

In post #1, you say:

"Now, if we take in account just only resistors as passive one-port components than the reason behind is clear: just close the 2-port network with 2 resistor and apply the Tellegen theorem to the overall circuit. All voltages and currents have to be null so the complete set of linear equations have to be linear independent, thus by elimination the two linear equations representing the 2-port network are independent themselves. "

The ABCD parameters for your two-port are:

Is there another pair of linear equations representing your two-port that are linearly independent?

The Electrician said:
Is there another pair of linear equations representing your two-port that are linearly independent?

Sure, starting from the last row take the third and fourth equations and use second and third columns as indipendent variables:

##\begin{bmatrix}-1 & 0 & -1 & k\\R_{1} & -1 & 0 & 1\end{bmatrix}\begin{bmatrix} I_{a}\\V_{a}\\I_{b}\\V_{b}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}##

##\begin{bmatrix}0 & -1\\-1 & 0\end{bmatrix}\begin{bmatrix} V_{a}\\I_{b}\end{bmatrix}+\begin{bmatrix}-1 & k\\R_{1} & 1\end{bmatrix}\begin{bmatrix} I_{a}\\V_{b}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\Rightarrow \begin{bmatrix} V_{a}\\I_{b}\end{bmatrix}=\begin{bmatrix}R_{1} & 1\\-1 & k\end{bmatrix}\begin{bmatrix} V_{b}\\I_{a}\end{bmatrix}##

we get the Hybrid model (h-parameters)

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As last example consider the following:

here performing Gauss elimination of the "internal" variables ##I_0,V_0,I_1,V_1## we get just one homogeneous equation ##I_a + I_b = 0## (the other one results in identically null).

Thus, with this example in mind, I believe we can conclude that any two-port linear network with no internal independent voltage/current generators has got a representation made of up to two independent homogeneous linear equations.

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## What is a linear 2-port network representation?

A linear 2-port network representation is a mathematical model used to describe the behavior of a 2-port network. It consists of two input variables and two output variables, and is often represented using matrices.

## How is a linear 2-port network represented mathematically?

A linear 2-port network is typically represented using a scattering matrix or S-matrix. This matrix relates the input and output variables of the network through a set of linear equations.

## What are the advantages of using a linear 2-port network representation?

One of the main advantages of using a linear 2-port network representation is that it allows for easy analysis and calculation of the network's behavior. It also provides a simplified way to model complex networks and their interactions with other components.

## What types of networks can be represented using a linear 2-port network representation?

Linear 2-port network representations can be used for a wide range of networks, including electrical circuits, mechanical systems, and even biological systems. As long as the network can be described using two input and two output variables, it can be represented using this model.

## Are there any limitations to using a linear 2-port network representation?

While a linear 2-port network representation is useful for many applications, it does have some limitations. It assumes that the network is linear, meaning that the output is directly proportional to the input. This may not always be the case in real-world systems.

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