An integrator circuit has the same configuration as a filter circuit, but why does it not filter the Vin?

Question

it has 10k ohms 0.1 microF and drove with 50kHz

Answer 1

It does, if it is an integrator.

If something that has a varying response for different frequencies is a filter, then I think it's a filter.

Check out the frequency response, try a few frequencies.

You might find a pattern to the attenuation of frequencies.

If you have Laplace you might look at the Laplace transform for an integrator: Maybe something like 1/s?

This looks something like 1/jOmega for simple stuff. [1/(j2 pi f)]

Looks like a filter.

Answer 2

For the values given the 3dB point is about fc = 160Hz. So the Filter circuit (aka integator) will definitely attenuate (filter out) the 50kHz source.

with the values given the time constant is about 1ms, which means the output will be fully charged after about 3 time constants (some people use 5). So the output has settled after about 3ms. But since the source is changing every 20us the output never gets fully charged, that is since the input signal period, 20us, is well below the circuit time constant, 3msec, this circuit will act as an integrator on this input signal. However, since the 3dB cutoff is 160Hz and the source frequency is 50kHz there will be almost no signal at the output.

To "see" the circuit act as an integrator without too much attenuation use a source frequecy of 1kHz.

Answer 3

If it interferes with the input signal, Vin, then you change the input.  So you have to make sure the input to the integrator, amplifier, multiplier, etc. has a very high input impedance compared to the output impedance of the input stage.  You don't want to change the input!