1
$\begingroup$

When deriving the speed of light in vacuum, one usually starts from Maxwell's equations, does some calculus and finds a wave solution with the phase velocity $c = 1/\sqrt{\mu_0 \varepsilon_0}$. This is clear to me.

When analyzing transmission lines, the approach is not quite the same. One starts with a model of distributed inductance, capacitance, resistance and admittance along the transmission line from which one obtains the telegrapher's equations.

Distributed element model of a transmission line

Solving these equations (with a sinusoidal ansatz) again leads to a wave solution of the form $\exp(\mathrm i \omega t - \gamma x)$ where $$ \gamma = \sqrt{(R + \mathrm i \omega L) (G + \mathrm i \omega C)} . $$ For a lossless transmission line ($R = 0 = G$) we thus find $v_{\mathrm{ph}} = \omega / \operatorname{Im}(\gamma) = 1 / \sqrt{L C}$.

Now, in the literature I've been looking at (on calculating $v_{\mathrm{ph}}$ for coplanar waveguides, but this is unimportant, I suspect) it is stated as an obvious fact that in the absence of any dielectric except vacuum $v_{\mathrm{ph}} = c$ in the transmission line. This is not obvious to me at all. Shouldn't the waveguide geometry (which determines $L$ and $C$, after all) be able to affect this value? The idea seems to be that only the space in which the fields propagate determine this velocity and that the conductors, carrying merely currents and charges, can't affect it.

Can you justify this fact and make it intuitive for me?

Does this still hold with a lossy transmission line? That is, if I "switch on" the resistance $R > 0$ without changing the geometry of an existing transmission line or any nearby dielectrics, will $L$ and $C$ "magically" adjust in order to keep $\operatorname{Im}(\gamma)$ (and thus $v_{\mathrm{ph}}$) constant?

Improve this question
$\endgroup$
1
  • $\begingroup$ The geometry matters. Look up "slow wave structures". $\endgroup$
    – John Doty
    Jan 11 at 2:01

1 Answer 1

Reset to default
1
$\begingroup$

This is an excellent question but you must restrict it for the case of a homogeneous transmission line, otherwise it is not true. The line must consist of a pair of parallel conductors that individually may have different but otherwise arbitrary cross sections, one may contain other as in a coaxial line, and the lines are embedded in propagating medium that is homogeneous both in cross-section and along the axis.

If you have that then the fundamental mode of propagation is essentially the static $E$ and $H$ fields along the line and because of the assumed homogeneity that field has no longitudinal component, that is a TEM mode. The fields can be derived from a scalar potential $\Phi$ so that with $\kappa = \omega \sqrt{\epsilon \mu}$: $$ \mathbf E_t =\nabla_t \Phi e^{-\mathfrak j \kappa z} \qquad E_z=0\\ \mathbf H_t = \pm \sqrt{\frac{\epsilon}{\mu}} \mathbf {\hat z} \times \mathbf E_t \qquad H_z=0 \\ \nabla^2_t \Phi = 0 \\ \Phi({\mathcal{C}_1}) =\Phi_1 \qquad \Phi({\mathcal{C}_2}) =\Phi_2$$ where $\mathcal{C_1}$ and $\mathcal{C_2}$ are the contours of the conductors at which the scalar potential $\Phi$ are given constants. If the medium is lossy then the $\epsilon$ and $\mu$ are complex quantities.

The intuition you are asking for is in the static nature of this propagating field it being a slight modification, with restriction to the conductors, of a plane wave whose phase velocity then depends only on the medium. This does not mean that a different, not TEM, mode may not propagate at a geometry dependent speed, and, in fact, the phase velocity of all the other modes, be it TE or TM, does depend on the shape of the contours and their separation.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Ah, I see, that makes a lot of sense! A lossy medium (i.e. complex $\epsilon$ and $\mu$) correspond only to a nonzero $G$ though, wouldn't it? A nonzero $R$ means that the losses happen in the conductor, after all, not in the dielectric. Or am I missing something? $\endgroup$
    – schtandard
    Jan 11 at 10:48
  • 1
    $\begingroup$ Exactly, but note that while $R$ usually models the losses in the conductor that I completely neglect in the above one can have both $\epsilon$ and $\mu $ be lossy and then result in a complex $\kappa$ and wave impedance $\sqrt{\frac{\mu}{\epsilon}}$. Thus media contribute to $R$ as well as to $G$, but transmission lines are rarely used on magnetic material (except in a few ferrite-based applications) so only then only $\Im[ {\epsilon}]$ would show up in modeling $G$. To include conductor losses formally you would need to use the so-called "impedance boundary" approximation with the TEM waves. $\endgroup$
    – hyportnex
    Jan 11 at 14:34
  • $\begingroup$ Alright, that seems a lot less straight-forward then. To be clear: Does this mean that conductor losses will reduce the phase velocity of a travelling wave? Or will it just lead to attenuation, leaving the phase velocity unchanged from the case without conductor losses? $\endgroup$
    – schtandard
    Jan 11 at 14:52
  • 1
    $\begingroup$ neglect $G$, keep $R<<\omega L$, then $\gamma = \sqrt{-\omega^2 LC} \sqrt{1-\mathfrak j R/(\omega L)} \approx \mathfrak j \omega \sqrt{LC} (1-\mathfrak j R/(2\omega L)$, so in the propagation term $e^{-\gamma z} = e^{-\mathfrak j \kappa z} e^{-\alpha z}$ where $\alpha = R/(2Z_0)$, $Z_0=\sqrt{L/C}$ showing that $R$ contributes only to the exponential amplitude decay and not to the phase velocity. $\endgroup$
    – hyportnex
    Jan 11 at 15:46

Your Answer

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

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