RGBHV video signal voltage level question

[Electronizer]

Now that I have my Atari Jaguar working with the Sony KV-20XBR TV I got recently, I'd like to try programming an FPGA to generate some simple video signals. I'm using the book Designing Video Game Hardware in Verilog.

In the book, they present several examples that generate simple one-bit RGBHV output (i.e. RGB either on or off). I'm trying to figure out how to interface this to my Sony TV's RGB input. I'd like to limit the luminance to 75% of maximum, which if I understand correctly, is 0.75 x 0.7V = 0.525V.

The FPGA I'm using puts out 3.3V. So, if the input to the TV is terminated at 75 ohms, I need a series resistor that will drop 3.3 - 0.525V = 2.775V. This resistor will form a voltage divider with the 75 ohm termination at the TV, so that means I need something around 390 ohms.

Now, here's where I started getting confused. Because I don't want to fry anything on the Jaguar or in the TV, I took some measurements. I measured about 800 ohms on the red input to the TV. Definitely not 75 ohms!

Since the Jaguar is working nicely with the TV, I decided to look at its RGB output. I used an oscilloscope to measure the red signal while triggering on the composite sync. The red signal maxed out at about 1V, not 0.7V as I've seen listed online in various discussions about the standard. Also, there was a +1VDC offset for the red signal.

Jaguar RGB signals.jpg

So, now I'm not sure what to do. The sync signal output from the FPGA can probably be used with a 75ohm series resistor to match the impedance of the VGA cable I'm using since the Jaguar was putting out around 3V on the composite sync line. However, the RGB signals definitely need to be stepped down from 3.3V. I have two questions:

1. What R, G, B voltage should I aim for at the input to the TV in order to get 100% luma?

2. Once I know the voltage, should I recalculate my voltage divider resistor using 800 ohms termination at the TV?