"Tricking" a Capacitor?

[johnywhy]

With an OOK-modulated, high frequency, high current carrier, if you filter out the high-frequency carrier, will the modulator frequency remain, providing the same current as the original carrier (with losses)?

[Andy Chee]

Definitely not the same current.  A certain amount of current will be dissipated as heat within the filter. 

AM crystal receiver is essentially what you seem to be replicating:

https://en.wikipedia.org/wiki/Envelope_detector

....except you seem to want a "power" version.

[johnywhy]

Definitely not the same current.  A certain amount of current will be dissipated as heat within the filter. 

You mean losses? So, Isupply - filterLosses = Ioutput.

[Andy Chee]

Definitely not the same current.  A certain amount of current will be dissipated as heat within the filter. 

You mean losses? So, Isupply - filterLosses = Ioutput.

Correct.  Filter losses are unavoidable.

[f4eru]

if you use a linear low pass filter, the output will be zero (except on abrupt changes).
You probably first need a non-linear element, like a diode bridge (also with losses)

[johnywhy]

if you use a linear low pass filter, the output will be zero (except on abrupt changes).

The "abrupt changes" is the modulation happening below the cutoff freq of the filter.

You probably first need a non-linear element, like a diode bridge (also with losses)

You mean a rectifier? Won't that convert AC to DC? I don't want DC output, i want AC output.

[Andy Chee]

You mean a rectifier? Won't that convert AC to DC? I don't want DC output, i want AC output.

Not necessarily.  As you can see in this diagram, a diode by itself will create a pulsed output which is definitely not DC:

https://en.wikipedia.org/wiki/Pulsed_DC#/media/File:Rectification.svg