## EMC Question of the Week: August 5, 2024

With a shorted termination, the inductance measured at the input of a 2-meter coaxial cable is 510 nH. With an open termination, the capacitance measured at the input is 200 pF. How long does it take a signal to propagate from one end of the cable to the other?

- 2 ns
- 3.3 ns
- 6.6 ns
- 10 ns

## Answer

The best answer is “d.” The velocity of propagation in the cable can be calculated as,

$v=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{\left(\frac{510\times {10}^{-9}\text{\hspace{0.33em}}\text{H}}{2\text{\hspace{0.33em}}\text{m}}\right)\left(\frac{200\times {10}^{-12}\text{\hspace{0.33em}}\text{F}}{2\text{\hspace{0.33em}}\text{m}}\right)}}=1.98\times {10}^{8}\text{\hspace{0.33em}}\text{m/s.}$

The propagation delay is the length of the cable divided by the velocity of propagation,

${T}_{PD}=\frac{\mathcal{l}}{v}=\frac{2\text{\hspace{0.33em}}\text{m}}{198\times {10}^{8}\text{\hspace{0.33em}}\text{m/s}}\approx 10\text{\hspace{0.33em}}\text{ns.}$

These measurements also provide us with the characteristic impedance of the cable, ${Z}_{0}=\sqrt{\frac{L}{C}}=\sqrt{\frac{510\times {10}^{-9}\text{\hspace{0.33em}}\text{H/}\left(\text{2m}\right)}{200\times {10}^{-12}\text{\hspace{0.33em}}\text{F/}\left(\text{2m}\right)}}$.

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