Timeline for Why is the output of an LTI system the convolution of the input funtion and the impulse response?

7 events

when toggle format	what		by	license	comment
Jun 22, 2011 at 1:02	vote	accept	PFiver
Jun 22, 2011 at 0:08	answer	added	Sridhar Ramesh		timeline score: 1
Jun 21, 2011 at 23:11	answer	added	Steve		timeline score: 1
Jun 21, 2011 at 21:10	history	edited	Yemon Choi		added math-physics tag, since this more a question of physics than of maths
Jun 21, 2011 at 20:48	comment	added	LSpice		thing to do. For that, I rather like Terry Tao's explanation (mathoverflow.net/questions/5892/what-is-convolution-intuitively/…) of convolution as a ‘blurring’ process; but my physics is too weak to know whether one can draw any sort of sensible connection between the optics he describes there and the circuit behaviour in which you're interested.
Jun 21, 2011 at 20:47	comment	added	LSpice		I think that the key difference between the transformations $x \mapsto f \circ x$ and $x \mapsto h * x$ is that the output from the former depends only on the instantaneous behaviour of $x$—one can determine the response right now by knowing only the input right now—whereas the former allows the behaviour of $x$ at all (past) times to have an effect on the present response. The latter behaviour is what one expects out of a real-world circuit. Of course, this doesn't say why convolution (as opposed to any other integral transform—they all exhibit this sort of behaviour) is the ‘right’
Jun 21, 2011 at 19:50	history	asked	PFiver	CC BY-SA 3.0