I see that it requires at least 5V, and outputs 4.7V minimum pulses. Are you using a voltage divider so that you are not putting more than 3.3V into the GPIO?
Voltage Source (VS): 5 V
Resistance 1 (R1): 1 kohms (use the drop down list)
Resistance 2 (R2) 1.8 kohms
Press the calculate button to calculate the output voltage. (3.2V) This is your worst case high logic 1
Now drop the voltage source to 4.7V and calculate again. (3.0V) This is your worst case low logic 1 output
We’ll assume that the logic 0 is going to be so close to 0 that there is no need to check that (it will be lower that what was already working with no divider).
Thus with the resistors connected as shown in the accompanying picture, where the source voltage is your sensor and the output voltage is connected to your GPIO, you have converted a 5V signal to one that is compatible with 3.3V logic.
There’s a bit of finesse to it.
You have to look up the actual values of resistors that exist (search for an E24 series table) . If you ever need an odd non-existent value you can use series / parallel combinations (look for another calculator) but in this case we are ok with commonly available values.
We must size the series resistance of the two resistors (R1+R2, 1+1.8 = 2.8KOhm) so that they don’t draw too much current from your sensor. I think I saw it could supply 10mA.
5V / 2.8K = 1.7mA (assuming that the GPIO contributes very little load, which it does). So all ok there.
We also have to check that 1.7mA is enough to supply the GPIO (it is). Making both resistors 10x larger would still work but would be more susceptible to induced noise as you would only have a 170uA signal.