+emcgon Thanks! Be sure to check out the videos Matt Duff (from Analog Devices) did; there's a bunch of them, all very good.

I second that. Very well explained and thought out. Will check out Matt Duff as well. Thank you!

I suppose you had to filter through a lot of tutorials to get here eh?

@@magerehenk7579 i see wht u did there :)

If you just want the DC value of a signal, you'll want to remove all signal components whose frequencies are above zero hertz, so you'd want a low-pass filter. What cut-off frequency you choose for your low-pass filter will depend on the frequency components of your input signal, as you want to attenuate all those components as much as possible. Note also that a low-pass filter is effectively an integrator in the time domain (think of a simple RC filter), so generally the lower your cut-off frequency, the longer it will take for the filter to settle on the DC value of the signal.

My understanding is that in a low-pass filter, when dc and lower frequencies are passed from input to output, the capacitor facilitates this by slowly getting charged. In order to meet a very low cut-off frequency requirement, let me assume I use a very large time constant (RC) which, let's say, is larger than the period T of the input signal. During each positive excursion, the capacitor will try to charge towards the peak value, though slowly, but will lose very little charge before the next positive excursion arrives. Does this not mean, the capacitor will not average out over time but instead keep riding on the peak . If on the other hand, I am restricted to use a RC time constant which is lesser than T, I will have no freedom to choose the theoretical maximum RC time constant. Kindly let me know if my understanding is correct. Thanks.

You state that the capacitor "will lose very little charge before the next positive excursion". This suggests that the capacitor discharges more slowly than it charged. If you apply a signal, say a sine wave, to an RC low pass filter, why would the capacitor charge at a different rate than it discharges?

The charge and discharge time of the capacitor is assumed to be the same. I consider a saw-tooth wave where the sweep time from 0v - peak (say 10v) takes 7T/8 secs and the fly-back period lasts for T/8 secs. If the charge and discharge time constants (RC) is comparable to 7T/8 secs, the capacitor will have sufficient time to charge to the peak value while getting only T/8 secs to discharge, before the next cycle starts. So, while RC will give the rate at which the capacitor will charge or discharge, the actual level to which it has come down on discharging will be determined by the fly-back time, viz. T/8 secs. Assuming it has lost only 2v before the next cycle started, the average value will be (10v+8v)/2 = 9v. On the other hand if I had allowed the fly-back period period of T/4 secs for the capacitor to discharge, the average would have decreased further. My question is, without caring for such variable average values dictated by the time periods of the impressed signal, how do I recover the true Fourier dc component present in the impressed signal. To cite another case, I want to know how I can recover the Fourier average of a full wave rectified signal whose average is some finite quantity and not zero. The charge and discharge time constants are assumed same and equal to T/4 secs. Thanks.

An RC low pass filter is an integrator, thus with the correctly selected R and C values it will give you the average of your input signal. In your example of a 0-10v sawtooth wave, the center of that sawtooth is 5v. Being a symmetrical, periodic waveform, the sawtooth spends just as much time above 5v as it does below 5v, hence the average of that signal is 5v, which is the DC component of that waveform. If you generate a 0-10v sawtooth signal with a function generator and feed it into an appropriate RC low pass filter, that's exactly what you'll see. Change the signal to a square, triangle, or sine wave, the average remains the same. When you say "without caring for such variable average values dictated by the time periods of the impressed signal, how do I recover the true Fourier dc component", I assume you mean that you want to get the average voltage of the signal without caring about the frequency of the signal. The answer is, I don't think you can. Integrating a function over some period of time gives you the average, but different frequency signals necessarily complete their cycles over different time periods. For example, the period of a 10MHz signal is 100nS, so 1mS would be more than enough time to see many periods of the signal and determine its average value very accurately. A 1Hz signal on the other hand wouldn't complete even one cycle in that same 1mS time frame, so you wouldn't be able to accurately determine its average value. So, if you know that your input signal is of a higher frequency, then you can make your R and C values smaller, and thus obtain your average DC voltage faster. But if you must accommodate very low frequency signals, then your R and C values will have to be larger, and it will necessarily take more time to get the average DC value that you're looking for.