# Signal Speed in Coaxial Cable

I want to calculate the signal propagation speed in a real life coaxial cable. I have the data sheet which tells me its impedance and capacitance per unit length.

For long parallel wires carrying currents in opposite directions (i.e. transmission lines), we can use $$Z=\sqrt{\frac{L}{C}}$$ (where $$Z$$ is characteristic impedance, $$L$$ is inductance/unit length, $$C$$ is capacitance/unit length) to get $$L$$ given $$Z$$ and $$C$$. Then the signal speed is simply $$c=\frac{1}{\sqrt{LC}}$$ (see e.g. Griffith's, fourth edition, problem 7.62)

and boom, problem solved. How can I carry out a similar calculation for long, coaxial wires?

• Unclear. You say "boom, problem solved" and then ask how to solve it. Commented Dec 14, 2022 at 14:35
• Actually the same method works whether you are dating with parallel or coaxial cables. You'll just need to adapt the value of $L,C$ accordingly, but if the coaxial cables are separated by the same homogenous, isotropic medium as the parallel cables, then it turns out that $c$ will be the same.
– LPZ
Commented Dec 14, 2022 at 15:17
• L and C are values for unit length, so the calculation is the same for short or long wires. You should basically compare to wavelength of signal, that gives you answer if the line is short or long. Anyway, I advice you to read about telegraphers equations for better understanding. Also, it is very well explained in Pozar's book "microwave engineering". There is certain duality between (current, voltage) <-> (H-field, E-field), the solutions for these fields are used in order to obtain impedance and phase speed. Commented Dec 14, 2022 at 16:57
• This is an everyday problem in radio engineering and its answer can be gotten at the amateur radio stack exchange. Commented Dec 15, 2022 at 6:52

The datasheet of the coaxial cable gives $$Z$$ (the impedance) and $$C$$ (the capacitance per unit length). From $$Z=\sqrt{\frac{L}{C}}$$ you can calculate the inductance per unit length: $$L=Z^2C$$
Then you can calculate the signal speed $$c=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{Z^2 C C}}=\frac{1}{ZC}$$