Like Patrick said, in terms of electronics, "high" and "low" simply denote one of the two possible states. You could just as easily call it on, and off. 1 and 0. Energized, de-energized, so on and so forth.
In this case it's a 12v output that sits on the line. If the PCM doesn't act it remains 12v+. The PCM then yanks them to ground. It then releases the grounds and the lines jump back to 12v. For example for FDCS it leaves the wire in the high state for the duration of the pw, then pulls it back to ground perfectly defining a "Square". Up, to the right, and back down. A series of this forms the FDCS signal.
It is a pair of modulated grounds by the PCM that forms this digital signaling.
Now I think the PCM/IDM com link is getting confused with the IDM/Injector circuits.
On the output side, there are two high side wires. One per bank. Now when someone says high side, they mean positive (using conventional current notation). It takes both a positive and a negative for electron flow, so the injector needs two wires to fire. The high side is common to all 4 injectors on each bank. It's the center wire in the UVC. This is your 120+volt wire. The next 2 wires to each side are your injector wires (in order of appearance) for that bank. The next two on each side outside that are the GP wires.
For instance. To fire injector number 1, the IDM goes high on bank one (center wire on your passenger's side UVC) and also grounds the #1 injector wire (second wire to the front from the center wire on your passenger's side UVC). The circuit is complete, and #1 fires.
The leading edge of the FDCS into the IDM determined when this would happen, and the width of that wave determined how long it would last. To fire number 3 you would do the same thing except ground the #3 injector wire instead of the number one. This would be the adjacent wire, just forward of the center wire in the passenger's side UVC.
That is the output side of the IDM. High voltage.
Might as well explain IPR while we're at it.
The IPR duty cycle (I'm rusty on this, it's been a while) is like 440hz? I think it is, so lets just pretend for now that we know that it is.
That means that 440 times per second, (that's what "Hz" means...per second) a square wave of variable duration is sent down the line. Meaning about every 2.3ms a command is given. That's the frequency. The timeframe for this pulse is also 2.3ms long. So at 100% duty cycle, the pulse would go high, and remain high for 2.3ms. As you might have guessed, that would mean that at the time it would normally go back low, it's time for the next pulse (remember it's a constant frequency) so in reality, it never actually would turn back off, and instead would simply remain high. The result would be full power output with no modulation.
In contrast a 0% duty cycle would mean that when it came time for each pulse (440 times every second) nothing would happen. Because it's 0%. It would never turn on. Meaning that the output would be 0 volts. Nothing. 440 times per second you would get nothing.
Now everything in between 0 and 100% DC is a proportion of available on time, vs off time.
For instance. Take a 50% DC. That means it's on 50% of the time and off 50% of the time. So out of our 2.3ms pulse window, it would go high at the start (every 2.3ms it does this, 440 times a second) and stay high for 1.15ms. It would then go low for the remaining 1.15ms of the possible 2.3 before the next pulse. At which point it would again go high for 1.15ms, then go low for the remaining 1.15 again. Over and over again, 440 times each second until the DC was changed.
A 25% DC would be an on time of .575ms and an off time of 1.725ms. 75% DC would be an on time of 1.725ms and an off time of .575ms, exactly the opposite of the 25% DC.
I hope that makes sense.