High voltage transmission line with ABCD complex equations

MrAl

Joined Jun 17, 2014
7,849
Hi All,

I am still struggling to understand how the left series element = (A-1)/C, Shunt element = 1/C , Right series element = (D-1)/C are obtained.
I found in my notes that ABCD matrix formed from the T-network looks like this:


What should be done next? Should i Plug for A in the matrix 0.8698+j0.03542 and so on and then equate A with?
Please help, I am completely stuck!
Hi,

Looks like you could just subtract 1 from A, then divide by C, that gives you
(1/2)*(R+jwL)*dx

Is that what you wanted?
 

The Electrician

Joined Oct 9, 2007
2,801
If you consider the T network, letting the left series element be ZL, then the a matrix for just that element will be:
[ 1 ZL ]
[ 0 1 ]

Let the center shunt element be G (this is an admittance rather than an impedance), then the a matrix for that element will be:
[ 1 0 ]
[ G 1 ]

Let the right series element be ZR, then the a matrix for that element will be:
[ 1 ZR ]
[ 0 1 ]

If you multiply those three matrices in the order shown you will get:
Code:
[ G*ZL+1  (G*ZR+1)*ZL+ZR ]
[   G          G*ZR+1    ]
If we consider this to be equivalent to:
[ A B ]
[ C D ]

We can determine, for example, that ZL = (A-1)/C, and so on for the rest of them as shown in earlier posts. It's important to remember that the determinant A*D-B*C is equal to 1 for a reciprocal network, which this network is.
 

MrAl

Joined Jun 17, 2014
7,849
Hello there,

An even simpler method is to just handle this as a mixed hybrid parameter problem where we just have to find those parameters. We can relate the different branches to the parameters with a little calculation later.

As a quick example, if we start with the three element network made of z1, z2, and z3, where z1 is the left (input) branch and z2 is the right (output) branch and z3 is the center branch that goes to ground, and the two equations again:
Vs=Vr*(A1+j*A2)+Ir*(B1+j*B2)
Is=Vr*(C1+j*C2)+Ir*(D1+j*D2)

when we set Ir=0 we get for A and C
A=z1/z2+1
C=1/z2

and then it follows that z1=(A-1)/C.

What else we could do later is compare the results using the lumped T network model with results using a distributed network model derived from the Telegraphers Equation and see how well they correlate.
 
Last edited:

MrAl

Joined Jun 17, 2014
7,849
You made two mistakes here. The correct result is:

A=z1/z3+1
C=1/z3

The two errors compensate when finding z1, but you get a wrong result for z2.
Hi there,

Yes thanks for catching that, where actually what happened was i worded it wrong. The impedances were supposed to be ordered from left to right schematically so that would make z2 the bottom resistor and z3 the right side resistor, but since you calculated it that way we can use that orientation also.

With z1 left and z2 bottom and z3 right, the write-up would be:
As a quick example, if we start with the three element network made of z1, z2, and z3, where z1 is the left (input) branch, and z2 is the center branch that goes to ground, and z3 is the right (output) branch and the two equations again:
Vs=Vr*(A1+j*A2)+Ir*(B1+j*B2)
Is=Vr*(C1+j*C2)+Ir*(D1+j*D2)

we have:
A=z1/z2+1
C=1/z2

and instead with z3 on the bottom and z2 on the right, then we have:
A=z1/z3+1
C=1/z3

This was meant to be a simple way to look at this because it works like other types of two port network parameters, like a mixed hybrid set. In fact, the parameters A, B, C, D should be related to the hybrid parameters:
A=h12-(h11*h22)/h21
B=h11/h21
C=-h22/h21
D=1/h21

although there may be a discrepancy in one or more of the signs because sometimes they show the current leaving the output node for transmission lines whereas for the hybrid parameters they show it entering the output node (like the input current enters the input node). Someone can double check this set as well as the signs if they care to.
 

MrAl

Joined Jun 17, 2014
7,849
Wikipedia's Two-port page was incorporated into this thread by reference in post #6. :)

Hi,

Well that's good, but i see the reference was also to a matrix form whereas the discrete form i was giving is still a little simpler and straightforward.

I do see one thing interesting there though, and that is that they have I2 entered into the vector as -I2 which illustrates the difference in output current flow between the usual set of parameters and the ABCD parameters.

I think it would be more interesting is to compare the results we get here to the more realistic non lumped calculations. Would be interesting to see how much they differ. I've also used finite element models which give some pretty good results, but then again these T networks are more common in these classroom type problem sets.
 
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