Jeroen Boschma wrote:> > Jerry Avins wrote: > >>Ken Prager wrote: >> >> >>>In article <121120041011212732%edgar@math.ohio-state.edu.invalid>, >>> "G. A. Edgar" <edgar@math.ohio-state.edu.invalid> wrote: >>> >>> >>> >>>>In article <10p9gpemtk5l47b@corp.supernews.com>, Ken Prager >>>><prager_me_@ieee.org> wrote: >>>> >>>> >>>> >>>>>ie., is the following true... >>>>> >>>>>f(t)[g(t) * h(t)] = [f(t)g(t)] * h(t) >>>>> >>>>>If you write it out in terms of integrals (or sums for the discrete time >>>>>case), you will see that it is true. >>>>> >>>>>Ken P. >>>>> >>>> >>>>??? >>>> >>>>No. Try f = g = h where f(x)=1 on [0,1] and = 0 elsewhere. >>>>Then f[g*h] is zero on (1,2) but [fg]*h isn't. >>> >>> >>>I jumped the gun and yep, you are right. >>> >>>Here's another example. Let g be a unit impulse, u(n). >>> >>>Then >>> >>> f[g*h] = f[u*h] = fh >>> >>>but >>> >>> [fg]*h = [f(0)u]*h = f(0)h >>> >> >>Ken, you were right at the start. Your counterexample would work only >>if an impulse were band limited. Instead, it simply doesn't apply to a >>valid sampling. >> > > > Maybe I don't understand your conclusion because I'm not that good in English, but in general: > > f(t)[g(t) * h(t)] ~= [f(t)g(t)] * h(t) (~= : 'does not equal') > > Enough bandlimited examples can be constructed. > > JeroenA unit impulse is not bandlimited. It cannot be validly sampled. Is that still unclear? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
what priority does convolution "*" have? The same as multiplication?(dot?)
Started by ●November 11, 2004
Reply by ●November 15, 20042004-11-15
Reply by ●November 16, 20042004-11-16
Jerry Avins wrote:> > Jeroen Boschma wrote: > > > > > Jerry Avins wrote: > > > >>Ken Prager wrote: > >> > >> > >>>In article <121120041011212732%edgar@math.ohio-state.edu.invalid>, > >>> "G. A. Edgar" <edgar@math.ohio-state.edu.invalid> wrote: > >>> > >>> > >>> > >>>>In article <10p9gpemtk5l47b@corp.supernews.com>, Ken Prager > >>>><prager_me_@ieee.org> wrote: > >>>> > >>>> > >>>> > >>>>>ie., is the following true... > >>>>> > >>>>>f(t)[g(t) * h(t)] = [f(t)g(t)] * h(t) > >>>>> > >>>>>If you write it out in terms of integrals (or sums for the discrete time > >>>>>case), you will see that it is true. > >>>>> > >>>>>Ken P. > >>>>> > >>>> > >>>>??? > >>>> > >>>>No. Try f = g = h where f(x)=1 on [0,1] and = 0 elsewhere. > >>>>Then f[g*h] is zero on (1,2) but [fg]*h isn't. > >>> > >>> > >>>I jumped the gun and yep, you are right. > >>> > >>>Here's another example. Let g be a unit impulse, u(n). > >>> > >>>Then > >>> > >>> f[g*h] = f[u*h] = fh > >>> > >>>but > >>> > >>> [fg]*h = [f(0)u]*h = f(0)h > >>> > >> > >>Ken, you were right at the start. Your counterexample would work only > >>if an impulse were band limited. Instead, it simply doesn't apply to a > >>valid sampling. > >> > > > > > > Maybe I don't understand your conclusion because I'm not that good in English, but in general: > > > > f(t)[g(t) * h(t)] ~= [f(t)g(t)] * h(t) (~= : 'does not equal') > > > > Enough bandlimited examples can be constructed. > > > > Jeroen > > A unit impulse is not bandlimited. It cannot be validly sampled. Is that > still unclear? >No, that's trivial given the fourier transform of the impulse. But I was under the impression that by your statement ">>Ken, you were right at the start." you support Ken's first statement in his message from Friday 15:05:> >>>>>ie., is the following true... > >>>>> > >>>>>f(t)[g(t) * h(t)] = [f(t)g(t)] * h(t) > >>>>> > >>>>>If you write it out in terms of integrals (or sums for the discrete time > >>>>>case), you will see that it is true.The above statement from Ken is false (counterexamples can be made easily).> Jerry > -- > Engineering is the art of making what you want from things you can get. > �����������������������������������������������������������������������
