integratorsignal processing
Op-amp integrators and some common integrators actually are low-pass filters in practice. So if a signal is DC, then it does not integrate but only multiplies by some amount plus some transient response.
Now of course integration over DC cannot be done for a long period of time, as this would definitely break the circuit. But for a limited time, is there any integrator that does proper integration for DC and some range of frequencies?
A 2nd order low-pass butterworth filter as you have shown will have a similar frequency response as shown below in trace 2: -
So, if 500 Hz is re-normalized to coincide with "1" along the x-axis, you will see that at 5000 Hz, the attenuation is 40 dB down and below 500 Hz it's fairly flat all the way to DC (but with some gain dictated by R3 and R4 in your circuit).
In the specification it says "Signals need to be attenuated by at least 80dB relative to difference frequency of 400Hz" but I'm not sure what this means?
It certainly doesn't apply to a 2nd order filter as you have drawn in your question. It might mean that at 900 Hz (500 Hz plus 400 Hz) the attenuation needs to be at least 80 dB and, if that is the case then a fifth order butterworth filter would be required.
It could even imply that what you want is a pass band filter with 80 dB (or greater) attenuation 400 Hz either side of the centre at 500 Hz.
But, at the end of the day, it's down to you understanding the full specification you are given so, maybe publish a little bit more of it.
The initial condition at the start of the simulation is computed by the simulator by inserting conductances (with value gmin) to ground. It is thanks to this it "knows" that the output of the amplifier will start at 0 Volt at t=0.
To change the initial condition you can either insert a load resistor to the desired voltage, or you can (in most simulators) override the initial condition with an .IC statement.
If you are happy with 0 Volts as the initial condition and the average (DC) value, then you need to change the phase of the square wave so the first edge comes after 1/4 of the period.
Best Answer
This circuit:
simulate this circuit – Schematic created using CircuitLab
will do exactly what you want. Many times a designer will place a resistor in parallel with C1 to limit the gain at DC. If you don't want this behavior you can simply leave it off. Just be prepared to deal with your output hitting the rails in some circumstances.