# Low Pass Filter to Isolate DC component

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?

• Change "M" to "Meg".
– jonk
Commented Aug 5, 2020 at 16:59
• In other words, your source is running at 100 millihertz = 0.1 Hz, which is obvious from the 10-second period. Commented Aug 5, 2020 at 17:09
• @Aaron The triangle wave is likely due, instead, to "numerical behaviors" related to truncation/rounding in floating point notation. You are using a very fast sine wave and a long period. If you'd like to see more "normal" plotting, try reducing the frequency by a factor of 100 and reduce the time by a factor of 1000 and see how that plots out.
– jonk
Commented Aug 5, 2020 at 17:36
• @jonk yes a triangle signal is almost equal to a sine signal with very low frequency. Convert DC to AC doesn't produce a perfect sine wave, it is just little changes over time but not continuous discrete. Commented Aug 5, 2020 at 17:36
– jonk
Commented Aug 5, 2020 at 17:37