Electronic – Coax cable – capacitive load

how does my signal source see the whole cable?

The characteristic impedance (\$Z_0\$) of any transmission line be it coax or twisted pair (screened or unscreened) is determined by: -

$$Z_0 = \sqrt{\dfrac{R + j\omega L}{G + j\omega C}}$$

Where R, L, G and C are the resistance, inductance, conductance and capacitance of the actual line per unit length. Because it's per unit length, we can choose any length we want to measure those numbers and get the same answer.

At a certain frequency it sees only 50 Ohm ?

If we assume the frequency is quite high (about 1 MHz or more), the above equation can be simplified because \$\omega L\$ and \$\omega C\$ dominate over R and G hence we get: -

$$Z_0 = \sqrt{\dfrac{j\omega L}{j\omega C}} = \sqrt{\dfrac{L}{C}}$$

So, a typical value for L might be 250 nH per metre and C might be 100 pF per metre and this gives: -

$$Z_0 = \sqrt{2500} = 50 \text{ ohms}$$

Above approximately 1 MHz, the characteristic impedance is resistive at some fixed value (quite often 50 ohms) up to the GHz area when other things happen.

How do I calculate that frequency?

It's usually around 1 MHz but, you have the full formula and if you wanted to know how things shape up below 1 MHz, the formula tends to become this for any practical cable with negligible conductance (G) : -

$$Z_0 = \sqrt{\dfrac{R}{j\omega C}}$$

That formula dominates for the majority of the audio spectrum such as a cable like this: -

Picture taken from this wiki site and please note that there is an error in the x axis - it should say "300 k" and not "3 M".

What does my signal source sees at lower frequencies? A Capacitive Load, depending of the length of the cable?

Not quite, the ratio of R to \$j\omega C\$ is easy to understand but the square root of it (and in particular the "j" term) implies a phase angle of 45 degrees.

So, we do Impedance matching to prevent reflections.

Yes, but at low frequencies this is usually pointless because the length of the cable is normally so short compared to the wavelengths of the (audio) signal that reflections are trivial.

What wiki says about coaxial cable R, L, G and C: -

New circuit disclosed by the OP

A picture has appeared showing a differential to single-ended driver with a 50 ohm termination to a length of coax. The coax is terminated in 50 ohm. Given that there is no explanation, I have the following to add: -

• There doesn't need to be a termination resistor at both ends for this to work. It's quite reasonable to use a series driver termination and have the coax far-end open-circuit terminated.
• The advantage of a single driver series termination is that any reflections from the open-circuit load become dissipated at the driver end AND, more importantly, the voltage seen at the receiver is exactly what is transmitted i.e. there is no loss.
• A termination at both ends is missing a trick and the output voltage is 50% of the driver signal voltage.
• Coax cable screens must be grounded, and for best quality, this should be done at both ends BUT....
• If it can't be done at both ends then ground at the sending end and use a differential receiver to adequately avoid noise pick-up AND...
• Grounding should be done using an impedance equal to the impedance used in feeding the inner conductor. This significantly improves EMI susceptibility issues.

This would be my recommendation based on the limited information supplied by the OP: -

But, it might work adequately with 50 ohm on the inner and a hard ground at the shield at the driver end. It’s still unclear what the question addition is all about.