On Wed, Nov 16, 2011 at 7:24 AM, V G wrote: >> It is only continuous for >> non-periodic signals - > > Could you explain this again? > >> periodic ones will be a collection of different >> delta functions with different "weights" (the value you get when >> integrating around the immediate neighborhood of the delta function). > > (According to wikipedia) - I thought that the integral of the delta > function =A0=3D 1. > I only have time to answer this one question right now, although you are asking good questions and getting good answers from others, too. The integral of the delta function is 1, although you can multiply the delta function by a "weight" constant and then its integral will be equal to that weight constant. If you have a continuous time signal which is exactly periodic for all time, then it can be expressed as a discrete sum of up to a countably-infinite number of sinusoids, each with a frequency, phase, and amplitude. This means that the same amount of information can be conveyed by a countably-infinite number of discrete dirac delta functions (which give you frequency information by their position on the horizontal axis) and their weights (which, if allowed to be complex, give amplitude and phase information). If you have a continuous time signal which is not periodic, it can convey an uncountably-infinite quantity of information, even in a finite time (assuming no noise). Therefore, in general, it will require a continuous function in the frequency domain to convey the same amount of information, because no discrete set of (even a countably-infinite number of) delta functions can convey the same amount of information. If you have a strictly-periodic function which abruptly terminates to zero outside of a finite time window, then you can represent this by a Fourier series plus start and stop times. However, if you process this through the Fourier Transform, the result is that the dirac delta functions become finite in amplitude and spread-out in width (frequency), yielding a continuous function, even though strictly-speaking, you do not actually need a continuous function to represent this amount of information. The Fourier Transform will not actually converge as a normal integral if you try it on periodic signals. If you allow delta functions as the result of the integral then you can consider it to converge in this special function space which allows delta functions as if they were finite (because they are integrable themselves to a finite number). Another neat trick: if y(t)=3Da(t)*b(t), then Y(f) =3D convolution of A(f) and B(f), where Y(f) is the FT of Y, etc. And the converse is also true: if Y(f)=3DA(f)*B(f) then y(t)=3Dconvolution of a(t) and b(t). One more incredibly neat factoid: the various pairs of quantities in Quantum mechanics which satisfy the Heisenberg Uncertainty Principle (like position and velocity, momentum and energy, etc.) are Fourier Transform pairs, that is that position wavefunction is the inverse FT of the velocity wavefunction and velocity wavefunction is the FT of position wavefunction. This is another way to think of QM and uncertainty - a wavefunction which is well-defined (delta-function-like) in one domain (like position) will be widely-spread-out in the other domain (like velocity), just like a constant DC signal has a tiny bandwidth and a quick impulse has a very wide bandwidth, Sean --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .