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The reason the second matrix is incorrect is because you've applied the equation for a TL stub, and not the general expression for a TL. You might want the matrix for an unterminated TL here (hint: it has a function of Bl in every term).
Good Luck!
Ok, I found the general expression for the impedance of a transmission line, which is the first equation I put down on the piece of paper as shown in the attached picture below, is this what you meant? When you say a matrix with an unterminated TL, then that would mean there is no load; meaning infinite resistance. So I took the limit as ZL --> infinity and get the same expression for an open transmission line = -jZo*cot(Bl). So I am very lost
This is for a series transmission line by itself, which I showed on the top right corner of the matrix, would this be right? I am kind of thinking in terms of capacitors, inductors, and resistors, where the impedance of one of those elements would be at the top right corner of the matrix and the shunt would be on the bottom left of the matrix (which would be in terms of admittance).
Ok, I found the general expression for the impedance of a transmission line, which is the first equation I put down on the piece of paper as shown in the attached picture below, is this what you meant? When you say a matrix with an unterminated TL, then that would mean there is no load; meaning infinite resistance. So I took the limit as ZL --> infinity and get the same expression for an open transmission line = -jZo*cot(Bl). So I am very lost
What you did is correct, but an unterminated TL does not mean an open circuit. You need to find the ABCD expressions for a TL section (fairly simple via google).
As far as I understand skatefast's math, he calculated the input impedance of an open ended line, using the general expresssion for line input impedance and inserting infinite load impedance. The result is the open ended line input impedance that can be used with pancho_hideboo's ABCD matrix from post 2.
Yes, that is correct, but the OP was looking to solve the problem via a matrix multiplication, which at some point needs to be reduced to a single impedance value.
In fairness, it is more simple without matrices, in this case.