Cable behaving like a low pass filter
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 &quot;dimensions&quot; 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 &quot;fairly straightforward&quot; math you get this impedance looking into the left port: -
$$Z_0 = R + j\omega L + Z_0||\dfrac{1}{G + j\omega C}$$
$$= R + j\omega L + \dfrac{\frac{Z_0}{G+j\omega C}}{Z_0 + \frac{1}{G+j\omega C}}$$
$$= R + j\omega L + \dfrac{Z_0}{1 + Z_0(G+j\omega C)}$$
$$Z_0\cdot \left[1 + Z_0(G+j\omega C)\right] = [R+j\omega L][1 + Z_0(G+j\omega C)]+Z_0$$
$$Z_0 + Z_0^2(G+j\omega C) = R+j\omega L + Z_0[(R+j\omega L)(G+j\omega C)]+Z_0$$
$$Z_0^2(G+j\omega C) = R+j\omega L + Z_0[(R+j\omega L)(G+j\omega C)]$$
$$\color{red}{\boxed{\text{This next part is the bit where we reduce things to zero}}}$$
The important thing next is to recognize that \$(R+j\omega L)(G+j\omega C)\$ is insignificant as the &quot;lump&quot; approaches zero length and we are left with: -
$$Z_0^2(G+j\omega C) = R+j\omega L$$
hence $$Z_0 = \sqrt{\dfrac{R+j\omega L}{G+j\omega C}}$$
If you looked at the equation above for high frequencies, it becomes this: -
$$Z_0 = \sqrt{\dfrac{L}{C}}$$
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
&quot;slower&quot; 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.