FUCKING SUPER DIODES WHAT THE FUCK

Linear relationship. Like I said, I want to use the signal as a power switch. In the final circuit there will be no physical switch between the signal source and the op amp. The presence of the signal will indicate power. I don't care about the data on the signal.
Signal present at op amp = 2.8VDC out
Signal not present at op amp = 0VDC out
That's it.

That's brillaint, actually, but
1. Parts count uuughh
2. does it work on lower frequencies as well?

this

>does it work on lower frequencies as well?
Yes. The original purpose was to drive a motor from the Doppler frequency output of a 9 GHz motion detector which could go down to a few Hertz on slow motion while the amplitude is a measure of distance. That's why the (relatively) big 10uF capacitor at the input. The speed of the motor followed the 0..3 V output connected to the analog input of a PWM power amp.
>Parts count uuughh
Not really. Normally you count the pins (solder points), not the parts. Leave off the last stage and compare to your (incomplete) original circuit. There is also a single supply version, LM358 instead of TL082.

Naturally, the problem arises from using diodes to rectify a small signal because of their voltage drop, so by the time it's rectified there's barely anything. One way to solve it is to amplify it first.

Pic related is pretty crude but you get the idea.

A gain of 20 may look impressive but doesn't change that much and now you have two diodes in series. The main drawback is that the output is no longer ground-referenced, it floats. This configuration can be used to drive a moving coil indicator, but not a meter. The idea behind the precision rectifier is to use the open-loop gain of an opamp to (almost) fully eliminate the diode voltage as shown in the picture. The current through R2 (and the instrument) precisely follows the input signal, even if it is small.

>The main drawback is that the output is no longer ground-referenced

I guess you could do half wave then, and you'd get rid of the two diodes in series problem. The ripples would be all sorts of fucked tho.

Here it is.

(4.5^-1 + 4.5^-1 + 9^-1)^-1 = 5.4

I dont know much about EE, but from a basic view your scheme looks like a AC to DC converter. When your switch isn't rectifying why do you assume current isn't flowing? The idea is to control the direction of the current by the loop and diodes, not to cut it off.

(4.5^-1 + 4.5^-1 + 9^-1)^-1 x 3 = 5.4

Yes, that will certainly work. The next step would be to get rid of the inconvenient (+/-) dual supply requirement (which is possible) and pretty soon you realize that a simple transistor circuit can do all you need. The real challenge is a single supply (5V), dc-coupled full wave precision rectifier circuit..