1
\$\begingroup\$

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

\$\endgroup\$
1

1 Answer 1

1
\$\begingroup\$

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.

\$\endgroup\$
1
  • 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
    Commented Dec 3, 2020 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.