Line impedance versus shunt capacitance

[Landru]

Hello, I'm having the damnedest time wrapping my head around a concept of audio electronics:

http://whirlwindusa.com/support/tech-articles/high-and-low-impedance-signals/

High impedance lines are more adversely affected by the inherent capacitance that is present in the cable itself. This capacitance combines with the impedances of the source and destination to set up a filter. As the impedance increases and/or the capacitance per foot increases, the active frequency at which the filter comes into play gets lower.

How exactly does a higher impedance bring down the frequency at which capacitance dominates? My current understanding is that the frequency at which capacitance will dominate is dependent on the capacitance value. Through what mechanism of physics does the a increase in source or input impedance decrease the frequency at which capacitance dominates?

Thanks!

 

[.Scott]

Well, if there was no impedance at all, there would only be a fixed load on the signal - no low pass filter at all.
At low frequencies, the resistance dominates. At high frequencies, the capacitor acts as a low resistance element compared to the actual resistors.
Here is a link that will allow you to experiment:
https://www.circuitlab.com/editor/#?id=7xn74dzye4qa

Try a frequency sweep from 1Hz to 1MHz.

 

[Paul Colby]

Neat website. So R1 in the simulation is meant to be the internal impedance of the source, right? Increasing the load impedance, R2, has little effect on the simulation as I would expect.

 

[.Scott]

@Paul Colby That's right.

 

[tech99]

Neat website. So R1 in the simulation is meant to be the internal impedance of the source, right? Increasing the load impedance, R2, has little effect on the simulation as I would expect.

For the case of short cables at low frequencies:-
The low frequency attenuation is determined by source and load resistances in series, forming a potential divider.
The turn over frequency, where the slope commences, is determined by source and load resistances in parallel, so they have equal effect. At the -3dB point, the two resistances in parallel equal the capacitive reactance of the cable.

 

[Landru]

Well, if there was no impedance at all, there would only be a fixed load on the signal - no low pass filter at all.
At low frequencies, the resistance dominates. At high frequencies, the capacitor acts as a low resistance element compared to the actual resistors.
Here is a link that will allow you to experiment:
https://www.circuitlab.com/editor/#?id=7xn74dzye4qa

Try a frequency sweep from 1Hz to 1MHz.

Thanks for making that demo. Between your response and and tech99's post it makes good sense now. So the issue is that a high impedance causes the capacitance to become a "path of least resistance" as frequency rises, where as a low impedance deprives the capacitance of that opportunity to become a path of least resistance.

I recreated it in LTSpice so I could see multiple resistances in a single plot. The green line is the lowest resistance value, and each line below it corresponds to higher values for the resistors.

 

[.Scott]

Thanks for making that demo.

You're welcome - but I did not need to create that demo. It was already there.