Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

ABCD parameters matrix of unsymmetric coupled lines

PlanarMetamaterials

Advanced Member level 4
Joined
Jun 13, 2012
Messages
1,152
Helped
356
Reputation
712
Reaction score
308
Trophy points
1,363
Location
Edmonton, Canada
Activity points
7,649
Checking through equation (7.109) and earlier part of chapter 7 could not locate the proof to derive the modal terminal transformation (which is equivalent to the expression for ABCD) in post #4

Please correct me if I miss anything.
The derivation is not in Paul's text, but it may be readily found in the MTL literature. I'll outline it here:

To start with, the terminal-domain ABCD matrix is defined as:


\[
\left[ \begin{array}{c} \vec{V}_T(0) \\ \vec{I}_T(0) \end{array} \right] = \left[ \begin{array}{cc} \left[A_T\right] & \left[B_T\right] \\ \left[C_T\right] & \left[D_T\right] \\ \end{array} \right] \left[ \begin{array}{c} \vec{V}_T(l) \\ \vec{I}_T(l) \end{array} \right]
\]

where \(l\) is the length of the transmission-line section and in general, each of the ABCD sub-matrices are fully populated. Alternative, we can look at the modal domain, where by definition the various modes are decoupled from each other, and therefore their properties may be described with diagonal matrices:

\[
\left[ \begin{array}{c} \vec{V}_M(0) \\ \vec{I}_M(0) \end{array} \right] = \left[ \begin{array}{cc} \left[A_M\right] & \left[B_M\right] \\ \left[C_M\right] & \left[D_M\right] \\ \end{array} \right] \left[ \begin{array}{c} \vec{V}_M(l) \\ \vec{I}_M(l) \end{array} \right]
\]

Now, we know that the voltages and current between domains are related by Paul's equations 7.8:

\[ \vec{V}_T = \left[T_V\right] \vec{V}_M \]
\[ \vec{I}_T = \left[T_I\right] \vec{I}_M \]

Such that we may take the terminal-domain expressions and write:

\[
\left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right] \left[ \begin{array}{c} \vec{V}_M(0) \\ \vec{I}_M(0) \end{array} \right] = \left[ \begin{array}{cc} \left[A_T\right] & \left[B_T\right] \\ \left[C_T\right] & \left[D_T\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right] \left[ \begin{array}{c} \vec{V}_M(l) \\ \vec{I}_M(l) \end{array} \right]
\]

Multiplying by the inverse of the first matrix:

\[
\left[ \begin{array}{c} \vec{V}_M(0) \\ \vec{I}_M(0) \end{array} \right] = \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right]^{-1} \left[ \begin{array}{cc} \left[A_T\right] & \left[B_T\right] \\ \left[C_T\right] & \left[D_T\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right] \left[ \begin{array}{c} \vec{V}_M(l) \\ \vec{I}_M(l) \end{array} \right]
\]

Then, compare this with the second equation above:

\[
\left[ \begin{array}{cc} \left[A_M\right] & \left[B_M\right] \\ \left[C_M\right] & \left[D_M\right] \\ \end{array} \right] = \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right]^{-1} \left[ \begin{array}{cc} \left[A_T\right] & \left[B_T\right] \\ \left[C_T\right] & \left[D_T\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right]
\]

We can recast this as:

\[
\left[ \begin{array}{cc} \left[A_T\right] & \left[B_T\right] \\ \left[C_T\right] & \left[D_T\right] \\ \end{array} \right] = \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[A_M\right] & \left[B_M\right] \\ \left[C_M\right] & \left[D_M\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right]^{-1}
\]

Now, we know from Paul 7.16 that:

\[
\left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right]^{-1} = \left[ \begin{array}{cc} \left[T_I\right] & \left[0\right] \\ \left[0\right] & \left[T_V\right] \\ \end{array} \right]^{T}
\]

And, as such:

\[
\left[ \begin{array}{cc} \left[A_T\right] & \left[B_T\right] \\ \left[C_T\right] & \left[D_T\right] \\ \end{array} \right] = \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[A_M\right] & \left[B_M\right] \\ \left[C_M\right] & \left[D_M\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[T_I\right] & \left[0\right] \\ \left[0\right] & \left[T_V\right] \\ \end{array} \right]^{T}
\]

At which point, I must apologize unreservedly for inverting the transformation matrix on the left-hand side in post #4, instead of the one on the right-hand side, which you can verify for yourself is the correct solution.
 

    promach

    points: 2
    Helpful Answer Positive Rating

    andre_teprom

    points: 2
    Helpful Answer Positive Rating

promach

Advanced Member level 4
Joined
Feb 22, 2016
Messages
1,028
Helped
1
Reputation
2
Reaction score
3
Trophy points
38
Activity points
9,314
As for Paul's equation (7.8) and its phasor MTL equations (7.4) , I am not quite sure why the second term inside the equation (3.9) disappear/optimized away ?



 

PlanarMetamaterials

Advanced Member level 4
Joined
Jun 13, 2012
Messages
1,152
Helped
356
Reputation
712
Reaction score
308
Trophy points
1,363
Location
Edmonton, Canada
Activity points
7,649
As for Paul's equation (7.8) and its phasor MTL equations (7.4) , I am not quite sure why the second term inside the equation (3.9) disappear/optimized away ?
Equations 7.8 and 7.4 are not directly related, did you perhaps mean to reference a different equation?

The second term inside of 3.9 is included in Z. As per equation 7.6, Z = R + jwL. (where the time derivative is replaced by jw)
 

    promach

    points: 2
    Helpful Answer Positive Rating

promach

Advanced Member level 4
Joined
Feb 22, 2016
Messages
1,028
Helped
1
Reputation
2
Reaction score
3
Trophy points
38
Activity points
9,314
1. May I know why exactly is time derivative is being replaced by jw ?

2. As for this paper: Decoupling the multiconductor transmission line equations , how does Paul's text "We will use a change of variables to decouple the
second-order differential equations in (7.7) by putting them into the form of n separate equations describing n isolated two-conductor lines." actually work ?
 

PlanarMetamaterials

Advanced Member level 4
Joined
Jun 13, 2012
Messages
1,152
Helped
356
Reputation
712
Reaction score
308
Trophy points
1,363
Location
Edmonton, Canada
Activity points
7,649
1. May I know why exactly is time derivative is being replaced by jw ?
We are assuming the steady-state, time-harmonic case. That is, V(z,t) = V(z)*exp(jwt).

2. As for this paper: Decoupling the multiconductor transmission line equations , how does Paul's text "We will use a change of variables to decouple the
second-order differential equations in (7.7) by putting them into the form of n separate equations describing n isolated two-conductor lines." actually work ?
You are getting more into the philosophy of MTL theory; I commend you for making it this far.

What Paul is describing is the terminal to modal conversion process, which is a change of basis. We convert from describing the system in terms of the terminal-domain voltages and currents -- which are coupled with one another -- to the system where the voltages and currents are decoupled from one another -- the modal domain.
 

    promach

    points: 2
    Helpful Answer Positive Rating

promach

Advanced Member level 4
Joined
Feb 22, 2016
Messages
1,028
Helped
1
Reputation
2
Reaction score
3
Trophy points
38
Activity points
9,314
For Paul's MTL text, how does jw substitution/transformation actually work for equations (3.27) to become equations (3.32) or (7.4) ?
I mean I do not quite understand how the transformation actually works.
 

PlanarMetamaterials

Advanced Member level 4
Joined
Jun 13, 2012
Messages
1,152
Helped
356
Reputation
712
Reaction score
308
Trophy points
1,363
Location
Edmonton, Canada
Activity points
7,649
For Paul's MTL text, how does jw substitution/transformation actually work for equations (3.27) to become equations (3.32) or (7.4) ?
Assuming time-harmonic excitations is a standard topic in many engineering fields; I'm surprised you haven't come across it before. You can simply insert the expression into the equation with the derivative, and reduce, e.g.:

\[ \frac{\partial}{\partial z}V(z,t) = - R I(z,t) - L \frac{\partial}{\partial t}I(z,t) \]

If we assume time-harmonic dependence, then

\[ V(z,t) = V(z)e^{j\omega t} \]
\[ I(z,t) = I(z)e^{j\omega t} \]

Inserting these into the above equation yields:

\[ \frac{\partial}{\partial z} V(z)e^{j\omega t} = - R I(z)e^{j\omega t} - j \omega L I(z)e^{j\omega t} \]

So the exponentials all cancle:

\[ \frac{\partial}{\partial z} V(z) = - R I(z) - j \omega L I(z) \]

Now, as in Paul's (7.6), we define Z as R + jwL, such that

\[ \frac{\partial}{\partial z} V(z) = - Z I(z) \]

And the same can be done with the expression for \( \frac{\partial}{\partial z} I(z) \)

I mean I do not quite understand how the transformation actually works.
You'll have to be ask a more specific question, sorry. I would suggest thinking about Paul's equations (7.8).
 

    promach

    points: 2
    Helpful Answer Positive Rating

promach

Advanced Member level 4
Joined
Feb 22, 2016
Messages
1,028
Helped
1
Reputation
2
Reaction score
3
Trophy points
38
Activity points
9,314
What Paul is describing is the terminal to modal conversion process, which is a change of basis. We convert from describing the system in terms of the terminal-domain voltages and currents -- which are coupled with one another -- to the system where the voltages and currents are decoupled from one another -- the modal domain.
May I know why in terminal-domain, voltages and currents are coupled with one another , while
voltages and currents are decoupled from one another in the modal domain ?

Besides, how exactly is equations (7.8) which is described as terminal to modal conversion process being done ?
 

PlanarMetamaterials

Advanced Member level 4
Joined
Jun 13, 2012
Messages
1,152
Helped
356
Reputation
712
Reaction score
308
Trophy points
1,363
Location
Edmonton, Canada
Activity points
7,649
May I know why in terminal-domain, voltages and currents are coupled with one another , while
voltages and currents are decoupled from one another in the modal domain ?
Because that is how we choose to define them :)

The terminal domain is simply just expressing the quantities on each conductor, so I am sure you can appreciate that (in most cases), they all exhibit capacitive and inductive coupling with each other.

The modes we're are interested in, on the other hand, are the individual "waves" which travel through the transmission line. Similar to most waves in passive linear media, the waves don't couple with one another.

Besides, how exactly is equations (7.8) which is described as terminal to modal conversion process being done ?
Probably, it is best to view 7.8 alongside equations 7.10, which can be seen to be eigenmode equations, or a change of basis. You can see that the transformation matrices work to diagonalize [Z][Y] (or [Y][Z]), where the diagonalized matrix is the square of the modal propagation constants.

So (7.8) is a change of basis which converts the coupled voltages and currents in the terminal domain, to the decoupled voltages and currentsof the modal domain (since the propagation constants form a diagonal matrix, they are unrelated to each other).
 

    promach

    points: 2
    Helpful Answer Positive Rating

promach

Advanced Member level 4
Joined
Feb 22, 2016
Messages
1,028
Helped
1
Reputation
2
Reaction score
3
Trophy points
38
Activity points
9,314
So (7.8) is a change of basis which converts the coupled voltages and currents in the terminal domain, to the decoupled voltages and currents of the modal domain (since the propagation constants form a diagonal matrix, they are unrelated to each other).
Wait, why does eigenmode equation or diagonal matrix imply unrelated/decoupled voltages and currents in modal domain ?

Besides, how to obtain equations (7.7) from equations (7.4) ?
 

PlanarMetamaterials

Advanced Member level 4
Joined
Jun 13, 2012
Messages
1,152
Helped
356
Reputation
712
Reaction score
308
Trophy points
1,363
Location
Edmonton, Canada
Activity points
7,649
Wait, why does eigenmode equation or diagonal matrix imply unrelated/decoupled voltages and currents in modal domain ?
If we look at the modal-domain wave equations (7.9), we see that \( [T_V]^{-1}[Z][Y][T_V] \) is the matrix of diagonal modal propagation constants (squared). Since the matrix is diagonal, the propagation of a single mode's voltage or current does not depend on the voltages or currents of the other modes; i.e., it is decoupled from the others.

Besides, how to obtain equations (7.7) from equations (7.4) ?
One substitutes 7.4a into 7.4b, or vice-versa:

We can express 7.4a as:

\[ \vec{I}(z) = -\frac{d}{dz}[Z]^{-1}\vec{V}(z) \]

and insert this into 7.4b to get:


\[ \frac{d}{dz} \frac{d}{dz}[Z]^{-1}\vec{V}(z) = -[Y]\vec{V}(z) \]

which re-arranges to:

\[ \frac{d^2}{dz^2} \vec{V}(z) = -[Z][Y]\vec{V}(z) \]

Not that [Y] and [Z] are assumed to be z-invariant, so they don't need to be inside the derivative.
 

    promach

    points: 2
    Helpful Answer Positive Rating

promach

Advanced Member level 4
Joined
Feb 22, 2016
Messages
1,028
Helped
1
Reputation
2
Reaction score
3
Trophy points
38
Activity points
9,314
The following questions are derived from the last paragraph just before section 7.2 of Paul's MTL book.

1. May I know why "the per-unit-length parameter matrices Ẑ and Ŷ do not commute, that is,
ZY = YZ" ?

2. Why is equation (7.4) called "first-order coupled" forms while equation (7.7) is called "second order uncoupled" form ?

3. As for the diagonal matrix issue, I am checking on reference text [B.25] : Decoupling the multiconductor transmission line equations . How to obtain equations (5a) and (5b) in the paper ?
 

PlanarMetamaterials

Advanced Member level 4
Joined
Jun 13, 2012
Messages
1,152
Helped
356
Reputation
712
Reaction score
308
Trophy points
1,363
Location
Edmonton, Canada
Activity points
7,649
The following questions are derived from the last paragraph just before section 7.2 of Paul's MTL book.

1. May I know why "the per-unit-length parameter matrices Ẑ and Ŷ do not commute, that is,
ZY = YZ" ?
In general, matrices do not commute; is there some reason you would expect them to? In this particular case, since both [Z] and [Y] are symmetric, [Z][Y] is actually just the transpose of [Y][Z] -- enabling the modal domain equivalents to the be the same diagonal matrix.

2. Why is equation (7.4) called "first-order coupled" forms while equation (7.7) is called "second order uncoupled" form ?
While I'm not entirely sure here, I would assume he's referring to coupling between voltages and currents. Since the wave equations only contain either voltages or currents, we can say they are decoupled. Since they are still terminal-domain matrices, though, the voltages are still coupled with one another. This does seem like a bit confusing terminology to use, now that you point that out.

3. As for the diagonal matrix issue, I am checking on reference text [B.25] : Decoupling the multiconductor transmission line equations . How to obtain equations (5a) and (5b) in the paper ?
Simply insert equations 4 into equations 1 (In the textbook, equations 7.8 into 7.4), premultiplying (a) by the inverse of [Tv] and (b) by the inverse of [Ti].
 

    promach

    points: 2
    Helpful Answer Positive Rating

promach

Advanced Member level 4
Joined
Feb 22, 2016
Messages
1,028
Helped
1
Reputation
2
Reaction score
3
Trophy points
38
Activity points
9,314
In this particular case, since both [Z] and [Y] are symmetric, [Z][Y] is actually just the transpose of [Y][Z] -- enabling the modal domain equivalents to the be the same diagonal matrix.
Wait, why are both [Z] and [Y] symmetric ?
 

PlanarMetamaterials

Advanced Member level 4
Joined
Jun 13, 2012
Messages
1,152
Helped
356
Reputation
712
Reaction score
308
Trophy points
1,363
Location
Edmonton, Canada
Activity points
7,649
Wait, why are both [Z] and [Y] symmetric ?
Paul discusses this on page 106 of his text, starting just before equations 3.45. In my opinion, he over-complicates things, and it can be simply stated that the symmetric forms of the matrices are a result of reciprocity.
 

    promach

    points: 2
    Helpful Answer Positive Rating

PlanarMetamaterials

Advanced Member level 4
Joined
Jun 13, 2012
Messages
1,152
Helped
356
Reputation
712
Reaction score
308
Trophy points
1,363
Location
Edmonton, Canada
Activity points
7,649
You may have missed my link to the Wikipedia article on reciprocity, it is attached to the word "reciprocity" in my post, but it's a bit hard to see on this forum.

Regardless, essentially what Lorentz reciprocity means in this case (in my understanding) is that is if we have a voltage applied to one conductor, it induces some current in any other conductor in some proportion, call it Y. The Reciprocity theorem tells us that if we excite the same voltage on that second conductor, what happens is that a current with the same proportion Y is excited on the first conductor. Alternatively, in terms of our MTL notation, \(Y_{ij} = Y_{ji}\); hence, the matrix is symmetric. The same applies to the per-unit-length impedance.
 

    promach

    points: 2
    Helpful Answer Positive Rating

promach

Advanced Member level 4
Joined
Feb 22, 2016
Messages
1,028
Helped
1
Reputation
2
Reaction score
3
Trophy points
38
Activity points
9,314
How are expressions (6a) and (6b) obtained ?

 

PlanarMetamaterials

Advanced Member level 4
Joined
Jun 13, 2012
Messages
1,152
Helped
356
Reputation
712
Reaction score
308
Trophy points
1,363
Location
Edmonton, Canada
Activity points
7,649
How are expressions (6a) and (6b) obtained ?
By comparing equations (7) with (5). This type of comparison is commonly used to derive relations between the modal and terminal domain representations of various properties.

Please note, as a I have tried to emphasize in my previous posts: for most modal quantities such as per-unit-length impedances and admittances, the general consensus of experts in the field at the moment is that due to scaling of the eigenvectors (that is, the columns of [Tv] and [Ti]), the modal values cannot be uniquely determined.
 

    promach

    points: 2
    Helpful Answer Positive Rating

promach

Advanced Member level 4
Joined
Feb 22, 2016
Messages
1,028
Helped
1
Reputation
2
Reaction score
3
Trophy points
38
Activity points
9,314
1. What do you mean by This type of comparison is commonly used to derive relations between the modal and terminal domain representations of various properties. ?

2. Why the matrices inside equations (6a) and (6b) only contain entries at the diagonal ?
 

Toggle Sidebar

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top