Hi guys,
Winterstone:
He's probably just trying to test this filter to see if it does what he wants it to do, or something like that
Mech:
Connect channel A probe to Vin, connect channel B probe to Vo.
Using a DC input of 1v, drive the filter input Vin. After about a second measure Vo and make a note of that call it VoDC.
Now using a sine wave generator connected to Vin with frequency of 1Hz and with a peak of 1v do the same, measure the output, call it Vo1.
Now using the sine generator again set to 10Hz do the same again, measure the output, call it Vo10.
Next set the sine to 100Hz, do the same again, call the output Vo100.
Again for 1000Hz, call the output Vo1k.
Again for 10000Hz, call the output Vo10k
Now you have readings at DC,1Hz, 10Hz, 100Hz, 1kHz, and 10kHz.
The idea now is to find the frequency that caused an output that is 0.707 times the DC output level. That is the cutoff frequency.
Lets look at an example...
Say when you input 1 DC you get an output of 1v DC (you actually should see this with this filter).
Then at 1Hz you get an output of around 1v peak.
Then at 10Hz you get an output of around 1v peak.
Then at 100Hz you still get an output of around 1v peak, maybe a tiny tiny bit less.
Then at 1000Hz you notice you get an output close to 0.7v peak.
Then at 10kHz you get an output around 25mv peak or close to zero.
Now you examine all the values above and notice that when you multiply the reading at 1kHz by the DC level (which was 1) you get 0.7, so that means 1kHz is very close to what the cutoff frequency really is.
So now you set the input side to 900Hz and test again, you get an output of 0.75v peak.
Then you set the input side to 1100Hz and test again, you get an output of 0.68v peak.
Now you can see that 0.707 (the number we are looking for here) is between 900Hz and 1100Hz, so you adjust the frequency generator until you see an output of very close to 0.707v peak and that is the cutoff frequency.
Keep in mind the above was for a filter with an output of 1v DC for an input of 1v DC. If the filter put out 2v DC for an input of 1v DC you would be looking for a number that is twice 0.707 which is 1.41v instead.
Also, this procedure is for a low pass filter only which can pass DC as well as AC. If it only could pass AC we'd have to use a very low frequency like 1Hz as the reference instead of 1v DC.
If this wasnt a low pass filter we'd have to use a different procedure to find the reference output.
Just to note, your filter here should have a cutoff of just over 1kHz.