Speed of EM waves from the point of view of an electrical engineer

A loss-less coaxial cable has inductance and capacitance and, their associated formulas introduce $\epsilon$ and $\mu$. But, to home-in on the velocity of propagation, the characteristic impedance must also be determined. We can avoid the complexity of the telegrapher's equations by looking at a short piece of loss-less cable in the frequency domain: -

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$$Z_0\hspace{1cm} = \hspace{1cm}sL + \dfrac{\frac{1}{sC}\cdot Z_0}{\frac{1}{sC} + Z_0}\hspace{1cm}=\hspace{1cm} sL + \dfrac{Z_0}{1 + sCZ_0}$$$$Z_0 + sCZ_0^2\hspace{1cm} =\hspace{1cm} sL + s^2LCZ_0 + Z_0$$$$sCZ_0^2 \hspace{1cm}=\hspace{1cm} sL + s^2LCZ_0$$

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When s, L and C are small, we can ignore the $s^2LCZ_0$ term, hence: -

$$sCZ_0^2 = sL$$

And, $\hspace{7cm}Z_0 = \sqrt{\dfrac{L}{C}}\hspace{1cm}$ (a well known formula).

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Having $Z_0$, we can analyse the circuit's transfer function. This allows us to find the output phase lag and, with a little more manipulation, the time delay for 1 metre of cable: -

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A general bode plot using lumped-values for L and C (per metre) is of interest: -

Here's a close-up of the phase response plotted linearly against frequency up to 1 MHz: -

• At 1 MHz, the output phase lag is 1.8°

• As a fraction of the period (1 μs) it's 0.005 hence, it's a time lag of 5 ns.

• At 100 kHz, the phase lag is 0.18° but, it's still a time lag of 5 ns

Because 1 metre lengths of capacitance and inductance are modelled, the equivalent velocity of propagation is 200 million metres per second (the inverse of the time lag).

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A simulator is great as a demonstrator but, a phase angle formula is needed so that we know what the dependencies are. It's a simple 2nd order low pass filter and, if you went through the derivation (omitted to keep the answer shorter) you find this: -

$$\text{Phase lag} = \arctan\left(\dfrac{2\zeta\dfrac{\omega}{\omega_n}}{1 - \dfrac{\omega^2}{\omega_n^2}}\right)$$

Because we are making $\omega$ a lot smaller than $\omega_n$ we can simplify: -

• The arctan of a small number is the small number because $$\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - ...$$
• The denominator equals 1

$$\text{Phase lag} = {2\zeta\dfrac{\omega}{\omega_n}}$$

But, we can also determine $\zeta$ for the circuit. Wikipedia RLC circuits shows it is this: -

And, because we know that R is $Z_0$ $\left(\sqrt{\frac{L}{C}}\right)$, $\zeta$ must equal 0.5. The phase lag now becomes: -

$$\text{Phase lag} = {\dfrac{\omega}{\omega_n}} = \omega\sqrt{LC}$$

_{Anyone studying telegrapher's equations will recognize this as the imaginary part of the propagation constant.$$$$}

Dividing the phase lag by $\omega$ gives us the time lag: -

$$\text{Time lag} = \dfrac{1}{\omega_n} = \sqrt{LC}$$

And, the velocity of propagation is the reciprocal of the time lag (for a 1 metre length of cable): -

$$\text{Velocity of propagation} = \dfrac{1}{\sqrt{LC}}$$

Nearly there!

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Right at the start I mentioned coaxial cable and its inductance and capacitance per unit length. The formulas are: -

$$L = \dfrac{\mu}{2\pi}\ln\left(\dfrac{b}{a}\right)$$

Taken from Inductance of a Coaxial Structure

$$C = \dfrac{2\pi\epsilon}{\ln\left(\dfrac{b}{a}\right)}$$

Taken from Capacitance of a coaxial structure

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If we multiply L and C we get $\epsilon\cdot\mu$ hence,

$$\text{Velocity of propagation} = \dfrac{1}{\sqrt{\epsilon\cdot\mu}}$$