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?  

1. 2 ns
2. 3.3 ns
3. 6.6 ns
4. 10 ns

Answer

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

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

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

$T_{PD}=\frac{\mathcal{l}}{v}=\frac{2\text{m}}{198\times {10}^8\text{m/s}}\approx 10\text{ns.}$

These measurements also provide us with the characteristic impedance of the cable, $Z_0=\sqrt{\frac{L}{C}}=\sqrt{\frac{510\times {10}^{-9}\text{H/2m}}{200\times {10}^{-12}\text{F/2m}}}=50\Omega .$

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