Voltage divider with pull up resistor
Have a good understanding of what you are doing, before you do something, and you'll save time, learn something, and speed up the process for the next time.

Before we look at what you did, let's see how the digital input of a raspberry pi works. (You can't add stuff to 'something', without knowing how that 'something' works right ? )

I didn't read the datasheet, I'm too lazy for that, but I found this (you'll have to verify is correctness).



source: http://www.mosaic-industries.com/embedded-systems/microcontroller-projects/raspberry-pi/gpio-pin-electrical-specifications

So when the V_reader > 2.0V, it will register as a HIGH or logic 1. 
When V_reader &lt; 0.54V, it will register as a LOW or logic 0.
Note: That there are 3 values listed on the image that correspond to different ICs on the device. I don't know where you're Reader input is, so you'll have to check the datasheet and confirm what the thresholds actually are.

Also note, the pullup resistor values. This appears to be different than the 1.1kohm you have listed as the internal.



Notice the input has two diodes. Those are protection diodes, and we'll get to that soon. But remember that they are there.

Now let's look at what you did.

You created a voltage divider, such that when when your signal is 0, the output of the divider is 0, and when your signal is -5V, the output is -3.3V. Let's ignore the pullup resistor for now.

So your two input voltages are 0 and -3.3V.

What value gets registered when the voltage coming into the raspberry pi is &lt; 0.54 ? 0. In both cases, because your voltage is less than 0, the raspberry pi would register it as a zero.

Remember those protection diodes from before ? If they were not there, you would have most likely damaged the pin, or damaged the uC itself. Why ? Digitial circuits do not like to go below the 0V or above their Vcc/Vdd. 

The reason you are seeing -600mV, is the voltage drop from the conducting diode (which is saving your raspberry pi). Continuous conduction, will heat up the diode, and it may blow, and now you lost your protection. 

What to do about it ?

One easy way to correct this is  to use an opamp to invert your input signal so that -5 translates to +3.3. This would then feed into your raspberry pi.

Looking at your signal, it seems its a pretty fast signal. So your opamp needs to be able to keep up (slew rate).







simulate this circuit &ndash; Schematic created using CircuitLab

Like the first sentence, its always best to know what you are doing and why before you do it. The protection diodes saved your raspberry pi. Not all ICs will have the, and not all circuits will include protection. Understand what you are doing. If you aren't sure - ask.