why there is a sag at the output of high pass filter?! I am checking transient response under pulse stimulus. There are theories available but nobody explains it without the mathematical forms. Is there any physical reason behind it?!
Sure. Its a basic thing but I could not find any proper physical reasoning.
Since the high frequency components are missing thus the sagging comes in. But I am trying to understand the behavior of voltage across res and cap individually. Also for a longer sequence the amplitude is fluctuating.
The HPF is simply rolling off the low (not high) frequency component of the square-wave, thus the sag in the voltage.
You would need response down to DC for the square-wave to have no sag.
The physical reason for this is that the capacitor in series with the signal that generates the HP function cannot conduct DC.
Think of the change in the square-wave voltage as a step function between two DC levels. Since the capacitor cannot carry DC, the DC level of the step will sag with time. With enough time the voltage will decay back to zero.
You can see this if you lower the frequency of the square-wave sufficiently.
O.K., you are talking about first order high-pass. That's really simple.
You can understand the behavior better if you start with a single voltage step. Pulse waveform is just a superposition of multiple rising and falling step functions.
Voltage step response clarifies that the voltage across the capacitor can only change continuously. It holds the initial zero voltage for the first moment and is then charged through the resistor.
bit_an, in post #4, the waveform is from a circuit simulator or an actual oscilloscope? If it is form an oscilloscope, which one? It looks to me like the VirtualBench from NI.
Nice clean waveforms :-D.