filter
simulate this circuit – Schematic created using CircuitLab
I have faced a problem in understanding, the following circuit. But after deep search, I expect to be Differential Low Pass Filter. If so, why we have make the first Capacitor common with Vin+ and Vin-, and the next two capacitors, we have put a 'ground' between them ?
I'll paraphrase what Dave Tweed says.
If you input a sinewave then there is no misshaping of the signal just an attenuation that is dependent on how high the input frequency is. If you input a square wave, the output will be misshaped because the harmonics (that comprise the square wave) will be attenuated at significantly different levels and you'll likely start to see an output that more resembles a sine wave or triangular wave.
Here's what a 1kHz square wave looks like when it is filtered with the same circuit as the OP has in his question: -
The input is 1Vp-p centred on 3V dc and as you can see the square wave is nearly triangular.
Both capacitors have accumulated a certain amount of charge (and it does not matter if the corresponding voltages are different). If both capacitors act in parallel (one side grounded and the other side connected to the inverting opamp terminal which is on virtual ground) the charge is transferred completely (opamp ideal) to the feedback capacitor.
Best Answer
The filter is both a differential filter and a common mode filter. That is why the caps are configured the way they are. Think about both inputs being joined together - the two caps to ground service the common mode filter function.