I'm interested to understand what affect the capacitance (between cable core and shielding) and the inductance have on a signal transmitting down a coaxial cable.
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\$\begingroup\$ Have a look here. \$\endgroup\$– Vladimir CraveroMay 21, 2014 at 7:07
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2 Answers
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The equation which answers your questions is that of the characteristic impedance of the transmission line, which arises from both inductance and capacitance: $$Z_0=\sqrt{\frac{R+j\omega L}{G+j\omega C}}$$
For a lossless transmission line this is simply: $$ Z_0 = \sqrt{\frac{L}{C}} $$
You should also check out the Telegrapher's Equations. They are one of the first steps in to transmission lines, a scary and wonderful topic.
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\$\begingroup\$ Thank you, I'll check those out. Do you have any info in more simple terms? such as 'a higher the capacitance would affect the signal propagation in this way..' \$\endgroup\$ May 21, 2014 at 7:33
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\$\begingroup\$ @Jenny_the_Argonian Well, we can see it what it does from the equation for characteristic impedance. For instance, if you hold the inductance constant and increase the capacitance the characteristic impedance will decrease. As for signal propagation, that depends on a lot more, like what the impedance is of the signal source and the load. I can't make an accurate statement about the propagation without knowing more about the system. \$\endgroup\$– SamuelMay 21, 2014 at 7:51
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Since the op asked for simple infos:
- high C means that higher frequencies get cut before, i.e. the cable act as a low pass filter, the higher the C the lower the corner frequency.
- high L means pretty much the same thing, higher frequencies are cut before.
That is because the transmission line is modeled as a second order LP filter (series L, parallel C), so the corner pulsation is \$\omega_0=\frac{1}{\sqrt{LC}}\$.
nb
a transmission line is actually modelled with distributed components, so it's not exactly an LP filter
answered May 21, 2014 at 7:50