Low Pass Filter With Hall Effect and ADC Help
The browser doesn't like that link to the datasheet, so I couldn't read it.  Provide the link to just the PDF file, not a page with all kinds of fluff around it.

In any case, the reason for the low pass filter is that the motor current can have short term spikes and other noise, but what you care about is more of a recent "average", or more precisely, you care only about the low frequencies of the current signal.  Since you only want readings at 10 Hz (a reasonable rate for looking at motor current), you should filter out frequencies above 5 Hz at least.

A simple way to accomplish this is with a R-C low pass filter:



The rolloff frequency of such a filter is

F = 1 / 2&pi;RC

When R is in Ohms and C in Farads, then F is in Hz.  In this example, the rolloff frequency is 4.4 Hz.  That's the frequency at which it roughly starts to attenuate, with the attenuation being 3 dB at that point.  Much below that frequency, the amplitude is unchanged.  Much above that frequency and the amplitude falls off 6 dB per octave above the rolloff frequency, which is also the ratio of the rolloff frequency to the frequency being passed.  For example, 100 Hz is 23 times the rolloff frequency, so this filter will attenuate a 100 Hz signal by 23 in voltage.  If you stick in a 100 Hz at 10 V, you will get out a 100 Hz at 440 mV.

You also have to consider loading of the current sensor output and what maximum impedance the A/D input requires.  The above is fine if the current sensor can drive a 1.2 k&Omega; load, and if the A/D is OK with its signal having 1.2 k&Omega; impedance.  You can adjust this by changing the resistor but keeping the R*C product the same.  For example, R1 = 12 k&Omega; and C1 = 3 &micro;F would give you the same frequency response, load the current sensor output less, but also present a higher impedance signal to the A/D.