In a thesis, [1] I found the following statement:

To determine [characteristic impedance and] dispersion relationships for an arbitrary reciprocal network having the transmission matrix

$$ \begin{bmatrix} A & B \\ C & D \end{bmatrix}, $$

one sets the determinant of

$$ \begin{bmatrix} A-e^{-\gamma d} & B \\ C & D-e^{-\gamma d} \end{bmatrix} $$

to zero where gamma is the complex propagation constant and d is the physical length of the whole network.

I can calculate this, and use it to verify the propagation constant of an element of an artifical transmission line, for example.

But can someone give a compelling, concise explanation why do we subtract $$\mathbf{I} e^{-\gamma d}$$ from the ABCD matrix and set the determinant to zero? What is it about the determinant that gives us the propagation constant? What is the intuition.

(I definitely see a similarity to determining eigenvalues via the characteristic polynomial, for example, but I don't see how this plays here)

[1] Michael Garth Case, Nonlinear Transmission Lines for Picosecond Pulse, Impulse and Millimeter-Wave Harmonic Generation


That's an eigenvalue problem: The goal is not to calculate the determinant, the goal is to calculate the eigenvalues and -vectors of that matrix.

If you go back to your "higher math 1" or 2" course, then you'll find that you always find the eigenvalues by subtracting \$\lambda\mathbf I\$ from a matrix, then writing down the determinant of that (which is a polynomial in \$\lambda\$), and finding the zeros of that.

Here, we're looking for eigenvalues that are some \$e^{-\gamma d}\$ to determine these parameters. That's why we directly subtract \$e^{-\gamma d}\mathbf I\$.

The intuition to finding the eigenvalue: For an eigenfunction of the system (and this we're expecting single tones \$e^{j\pi f t}\$ to be eigenfunctions), the relationship between in- and output can be described as a single complex number. That complex number is literally what we're looking for.

  • 1
    \$\begingroup\$ Great! One useful step to more insight is to repeat the exercise with \$\lambda\$ (which is a more obvious eigenvalue problem). Turns out calculating magnitude and argument of \$\lambda\$ gives the same result (\$|\lambda|=1, \operatorname{arg}\lambda=-\gamma d\$); the exponential just simplifies things. The intuition is now clear as well: \$\gamma d\$ describes the phase shift of the signal through the circuit \$\endgroup\$ – divB Dec 3 '20 at 20:43

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