I have no time to make something like this or even make a schematic, but if you have some elect experience or know someone who does, here's the basic idea.

Get a 3-terminal regulator, probably a low-dropout regulator like this LT1086CT-12-ND at Digikey. You need low-drop-out because the input voltage from the car's electrical system is so close to the output voltage. "Normal" 3-terminal regulators need about 5v of "overhead" but LDO regulators can work with only 1v of overhead, meaning it will be able to provide a regulated 12v output with as little as 13v input.

3-terminal regulators use the ground pin as a reference, meaning that a 12v regulator will do it's best to keep the output exactly 12v above the ground pin. Now if the ground pin is not really at "ground" then the output won't be at ground either. This method is used to make a normally fixed regulator produce an output that is adjustable. For example, if you connect the ground reference pin to a 1v source, the output of the regulator will be 13v instead of 12v.

How do you adjust the ground reference? The simplest way is to put a resistor in series with it, going to ground instead of a direct connection to ground. See page 1 of the data sheet and look at the diagram at the bottom of the page. You'll see the normal method of connecting the reference terminal to ground. If you put a resistor in this leg, the current consumed by the regulator (not the device it's powering) flows through the resistor and produces a voltage drop. This regulator has a reference current of about 100ua (0.0001 amps) so a 1k resistor in the ground leg will cause a .1v shift in the output voltage.

OK, so we add a 1k resistor in the gound leg to raise the voltage to 12.1v, or take it out (or short it out) to lower it back to 12.0v. Now how do we make it "oscillate" in time with the tach?

Get a N-channel MOSFET transistor (like this one) and put the Drain lead on the ground side of the 1k resistor and the Source lead on the regulator side of the 1k resistor. Connect the Gate lead of the MOSFET through a 100k resistor to the tachometer lead. The tach signal goes from 0v to +12v for every tach pulse. When it's at 0v the MOSFET is off, so the 1k resistor is in line with the ground leg of the regulator and the output is 12.1v. But when the tach signal goes to +12v and turns on the MOSFET, it effectively shorts out the 1k resistor connecting the regulator reference pin directly to ground and lowering the output to 12.0v.

So you end up with a 12v power supply with a guarannteed .1v of noise, and it's noise at the tach frequency. It's only about $5 in parts so it might be worth trying. Make sure the G-tech doesn't draw more than 1 amp though, this regulator won't handle more than that, and at 1 amp it will probably need a heat sink.

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I'm no EE (not even close), but what if you used a long ground cable for the G-tech. Wouldn't that create a ground loop( I think this is the wrong term) and possibly amplify the ignition noise coming through the power system of the G-tech?

 

This guy is trying to make his G-tech's tach work, not use an aftermarket tach. You don't need any dirty power supplies. You can try to connect the g-tech directly to the battery. If that does not work, connect it to the thick wire on the alternator.

An alternator has 3 coils inside that charge the battery. The g-tech is looking for the 3 humps, but your battery acts like a filter and smooths out the humps. After the battery, you also have filter capacitors attached to the ignition system so you won't hear static on the radio every time your spark plug fires.

 

A long ground cable would certainly allow more noise into the system, but that's not necessarily the type of noise we want (unless you wrap that cable around the alternator a few times).