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Background - transmission line

$\newcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ A transmission line can be modeled as an infinite sequence of inductors and capacitors:

     z   z+dz
     |   |
     v   v
  ---L---L--- <-- signal
   |   |   |
.. C   C   C ..
   |   |   |
  ----------- <-- ground

Each $L$ and $C$ is the inductance and capacitance per length. The current and voltage in this circuit is represented by the so-called telegrapher's equations \begin{align} V(z + dz) = V(z) - L \, dz \, \dot{I}(z) \quad \text{and} \quad I(z + dz) = I(z) - C \, dz \, \dot{V}(z) \end{align} which can be rearranged, in the limit $dz \rightarrow 0$, to $$ \frac{\partial V}{\partial z} = - L \frac{\partial I}{\partial t} \quad \text{and} \quad \frac{\partial I}{\partial z} = - C \frac{\partial{V}}{\partial t} \, . $$ If we take $\partial/\partial z$ on the left equation, and $\partial/\partial_t$ on the right equation, and substitude the right into the left, we get $$ \left( \frac{\partial^2 V}{\partial z^2} \right) - LC \left(\frac{\partial^2 V}{\partial t^2} \right) = 0 $$ which is the wave equation for waves moving with phase velocity $v^2 = 1 / LC$ (remember $L$ and $C$ are per-length quantities).

Context - coupled transmission lines

Now suppose we have coupled transmission lines

    z      z+dz
    |      |
    v      v
  ------------------- <-- ground
    |      |      |
..  Ca     Ca     Ca ..
    |      |      |
  -----La-----La----- <-- signal a
    |      |      |
..  Cm Lm  Cm Lm  Cm ..
    |      |      |
  -----Lb-----Lb----- <-- signal b
    |      |      |
..  Cb     Cb     Cb ..
    |      |      |
  ------------------- <-- ground

We have two lines $a$ and $b$, each with inductance per length $L_a$ or $L_b$, and capacitance per length $C_a$ or $C_b$. The lines are coupled through $C_m$ and $L_m$; the mutual inductance is hard to draw in ASCII art, so think of $L_m$ as a mutual inductance between the inductors $L_a$ and $L_b$. It's pretty easy to show that the equations for these coupled lines are \begin{align} \frac{\partial V_a}{\partial z} &= - L_a \frac{\partial I_a}{\partial t} - L_m \frac{\partial I_b}{\partial t} \\ \frac{\partial I_a}{\partial z} &= -(C_a + C_m) \frac{\partial V_a}{\partial t} + C_m \frac{\partial V_b}{\partial t} \end{align} and similarly for line $b$.

The equations for the coupled lines can be expressed in a matrix form \begin{align} \frac{\partial}{\partial z} \ket{V} &= - [L] \frac{\partial}{\partial t} \ket{I} \\ \frac{\partial}{\partial z} \ket{I} &= - [C] \frac{\partial}{\partial t} \ket{V} \end{align} where $\ket{V}$ is the vector of voltages, $\ket{I}$ is the vector of currents, and $[L]$ and $[C]$ are matrices. The matrices are $$ [L] = \begin{bmatrix} L_a & L_m \\ L_m & L_b \end{bmatrix} \quad \text{and} \quad [C] = \begin{bmatrix} (C_a + C_m) & -C_m \\ -C_m & (C_b + C_m) \end{bmatrix} \, . $$

Question

In the book Fundamentals of Microwave Transmission Lines by Jon C. Freeman, it is written

It turns out that the velocity of propagation is $$ [L][C] = \frac{1}{v^2} [I]$$ where $$ v = \frac{1}{\sqrt{\mu \epsilon}} $$ (here $[I]$ is the identity matrix)

Why is the product of the $L$ and $C$ matrices diagonal, and why are the diagonal entries equal to each other ($1/v^2$)? It seems like the author must be making an assumption and I haven't figured out what that assumption is.

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  • $\begingroup$ If you just multiply together the general matrices, what are the componentwise expressions for the claimed matrix equation $LC = I/v^2$? Are they really messy? $\endgroup$
    – tparker
    Commented Jan 4, 2023 at 5:47
  • $\begingroup$ It’s weird if the author suddenly jumps from only using abstract lumped-circuit elements to the microscopic physical material parameters $\epsilon$ and $\mu$, which is a pretty different “picture”. $\endgroup$
    – tparker
    Commented Jan 4, 2023 at 5:52

1 Answer 1

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See Chapter 12 in Orfanidis: Electromagnetic Waves and Antennas, especially equation 12.1.35 where he shows that for a symmetrical pair of coupled lines in a homogeneous medium you $$(L_0+L_m)(C_0-C_m)= (L_0-L_m)(C_0+C_m)=\mu\epsilon$$and $$\frac{L_0}{C_0}=\frac{L_m}{C_m}=\frac{\mu\epsilon }{C_0^2-C_m^2}$$ from which the $\mathbf{LC}=\frac{1}{\mu\epsilon}\mathbf{I}$ follows. (In your notation this would be $L_a=L_b=L_0$ and $C_a=C_b=C_0$)

Note that in this derivation symmetry with medium homogeneity is only sufficient and not necessary for the equation to hold.

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