filterfourier
I have a question based on low pass filtering. If I input a signal to this low pass filter, why does the output file have an AC (of very low frequency).
The input signal is$$ v_i(t) = 2 + sin(2*pi*100Mhz*t)$$ According to the filter response, the $$ sin(100Mhz*t) $$ should be attenuated by a very large number, thus resulting in a coefficient << 0 and leaving the DC component of 2 left. The output however is $$ v_o = 2 + sin(2*pi*0.1t) $$
Where does that AC signal come from?
The values in the circuit produce a natural resonance at about 89Hz based on this formula: -
Natural resonance = \$\dfrac{1}{2\pi \sqrt{R_1R_2C_1C_2}}\$
This circuit can also produce what to the uninitiated might think of as strange behaviour - it can actually resonate i.e. produce a significnatly higher amplitude at a range of frequencies just when you might have expected it to be attenuating those frequencies. This is component value dependent.
I believe this is what you are seeing. I think you are seeing this because you have one of the capacitors at the wrong value. For instance, if the 47n were in fact a 1nF capacitor the circuit would resonate at about 600 Hz with a significant peak: -
I used an online sallen-key calculator and plugged in the R and C numbers to get this. Here is the link.
I would encourage you to check the capacitor and resistor values you used.
You have not thought this through. You have not specified exactly what sort of signal you expect to see, or how accurately you need to know it. But let's say it's a step function, and since you're using a 12-bit A/D, you need to know its' amplitude to 1 lsb. So, for a full-scale input step, your filter must settle to one part in 4096 in 16 msec. For a simple RC lowpass, the response to a step is an inverse exponential with a response proportional to e^-t/RC. For this to settle to 1/4096 will take about 9 time constants, so the 1/RC = 16 usec/9. In other words, you need C at about 560 pF rather than 5 nF.
But all of this assumes that the signal behaves as a step, AND that you need perfect accuracy from the A/D. Note, for instance, that if the step occurs 8 usec before an A/D sample, your filter will not permit a perfect sample for 16 usec, but that sample will not occur until 24 usec after the step (8 usec to the first, inaccurate sample, then another 16 usec until the correct value is available). So you need to figure out exactly what accuracy you need, and when you need it. Only then can you design your filter.
Best Answer
The AC signal comes from the source, a passive linear filter can only change the amplitude and the phase, it cannot change the frequency of the waveform that it is filtering.
So the AC diagram shows the amplitude for all frequencies, and the chart below (in the OP) shows the output for one frequency.