Ok genius iam also approaching the problem same way like you I don't use matheMatical way my question is so simple because LTI depends on convulution here's my question below Convulution is nothing but stacks and scales the input that's why the input to an amplifier is stacked and scaled or amplified but in filter design it attenuate frequency so I don't know how it regret certain frequency by stacking and scaling the input if possible some one explain to me
The impulse response of a simple high pass filter includes a delta function + an opposite sign decaying exponential. The impulse response of the simple low pass filter is just the decaying exponential. If you think about what happens when high and low frequency sinusoids are convolved with these two you'll see that the delta function produces a qualitatively different result depending on whether then sinusoid period is much longer or shorter than the exponential decay constant. In the first case you get nothing if the sinusoidal period is long compared to the time constant since the delta function has an area of opposite sign to the exponential. In the second case (just the exponential) you get a big output since the sign of the sinusoid doesn't change in the decay time. The situation is reversed when the sinusoid period is short compared to the exponential decay time. Then in the first case (high pass) the exponential convolved with the sinusoid gives zero, but the delta function produces an unattenuated copy of the sinusoid, which is all that survives. The simple exponential (low pass filter) just has the exponential (no delta function) which gives zero when convolved with a high frequency sinusoid. Hope that helps!
@@sspickle still could not able to understand but iam going to case study it you are awesome made my day I published this question in more than 4 channel s now I recieved an reply thank you sir you are my teacher
