Cable behaving like a low pass filter

In the transmission line theory you can use this circuit to describe a transmission line: So the resistor and the capacitor form together a low pass. So each cable has to have a cutoff frequency where signals no longer get to the other end of the cable, right?

2 Answers 2

No, it doesn't quite work like that. When you derive the characteristic impedance in transmission line theory sure, you use a lumped-model to get the math process started but then, as you approach the solution, you purposely make the "dimensions" of the lump approach zero and this means that what you refer to as the cut-off frequency rises to infinity and we don't care about that. For instance, a simple derivation is to consider this lumped line model: -

Each small section (having an input impedance \$Z_0\$ ) will join on to the next small section. So then, if you do the "fairly straightforward" math you get this impedance looking into the left port: -

$$Z_0 = R + j\omega L + Z_0||\dfrac$$

$$Z_0\cdot \left[1 + Z_0(G+j\omega C)\right] = [R+j\omega L][1 + Z_0(G+j\omega C)]+Z_0$$

$$Z_0 + Z_0^2(G+j\omega C) = R+j\omega L + Z_0[(R+j\omega L)(G+j\omega C)]+Z_0$$

$$Z_0^2(G+j\omega C) = R+j\omega L + Z_0[(R+j\omega L)(G+j\omega C)]$$

The important thing next is to recognize that \$(R+j\omega L)(G+j\omega C)\$ is insignificant as the "lump" approaches zero length and we are left with: -

$$Z_0^2(G+j\omega C) = R+j\omega L$$

If you looked at the equation above for high frequencies, it becomes this: -

And is basically resistive in nature.

But when you take a signal generator with the same impedance as the transmission line and feed it into a transmission line with an open termination you can see that the reflection of the square wave has "slower" edges than the square wave itself

You have to be careful here because any capacitive connection on the t-line output can screw things up - even the oscilloscope input capacitance can cause this when placed at either end. Simulation of 5 pF capacitor loading the end of an unterminated 50 Ω line: -

You can see that the edges are slower but this is wholly because of the 5 pF loading at the end of the 10 metre transmission line.