Hi all, I am trying to evaluate some expressions consisting of: fg*hx*y where all letters denote function of time t. ie. f is f(t), y is y(t), x is x(t), ... fg denotes function multiplication. g*h denotes function convolution. I want to figure out is the "convolution" the same priority as "multiplication"? ie. can I evaluate the above expression from left to right, or from right to left or in any other orders? I am trying to manipulate the expression and to simplify the expression... thanks a lot
what priority does convolution "*" have? The same as multiplication?(dot?)
Started by ●November 11, 2004
Reply by ●November 11, 20042004-11-11
In article <cn0pn7$56$1@news.Stanford.EDU>, lucy <losemind@yahoo.com> wrote:> Hi all, > > I am trying to evaluate some expressions consisting of: > > fg*hx*y > > where all letters denote function of time t. > > ie. f is f(t), y is y(t), x is x(t), ... > > fg denotes function multiplication. > g*h denotes function convolution. > > I want to figure out is the "convolution" the same priority as > "multiplication"? > > ie. can I evaluate the above expression from left to right, or from right to > left or in any other orders?If you are the one who wrote the expression, then you are the only one who knows what it means. In any case, (fg)*h is in general not equal to f(g*h) so you shouldn't write fg*hx*y .
Reply by ●November 12, 20042004-11-12
"lucy" <losemind@yahoo.com> wrote in message news:<cn0pn7$56$1@news.Stanford.EDU>...> Hi all, > > I am trying to evaluate some expressions consisting of: > > fg*hx*y > > where all letters denote function of time t. > > ie. f is f(t), y is y(t), x is x(t), ... > > fg denotes function multiplication. > g*h denotes function convolution.I'm not sure I understand. Your problem is f(t) [ times ] g(t) [ convolve ] h(t) [ times ] x(t) [ convolve ] y(t)? I have never seen anything like that, that I can remember. I would go back to the derivations of this expression and see if there are some clues to how to order the operations.> I want to figure out is the "convolution" the same priority as > "multiplication"? > > ie. can I evaluate the above expression from left to right, or from right to > left or in any other orders?I first guess (but I may be wrong!) would be that you should do the convolutions first, and the mutliplications last.> I am trying to manipulate the expression and to simplify the expression... > > thanks a lotRune
Reply by ●November 12, 20042004-11-12
"lucy" <losemind@yahoo.com> wrote in message news:<cn0pn7$56$1@news.Stanford.EDU>...> Hi all, > > I am trying to evaluate some expressions consisting of: > > fg*hx*y > > where all letters denote function of time t. > > ie. f is f(t), y is y(t), x is x(t), ... > > fg denotes function multiplication. > g*h denotes function convolution. > > I want to figure out is the "convolution" the same priority as > "multiplication"? > > ie. can I evaluate the above expression from left to right, or from right to > left or in any other orders? > > I am trying to manipulate the expression and to simplify the expression... > > thanks a lotHi Lucy , I always thought that convolution is commutative , see http://www-structmed.cimr.cam.ac.uk/Course/Convolution/convolution.html#commut for example. When I did a quick google on commutative convolution however I found several links claiming that that's not always the case so you might want to look at some of them and see if that's true in your case. Best of Luck - Mike
Reply by ●November 12, 20042004-11-12
In article <bf2a46eb.0411120532.1e2cdc67@posting.google.com>, mpyarwood@hotmail.com (Mike Yarwood) wrote:> "lucy" <losemind@yahoo.com> wrote in message > news:<cn0pn7$56$1@news.Stanford.EDU>... > > Hi all, > > > > I am trying to evaluate some expressions consisting of: > > > > fg*hx*y > > > > where all letters denote function of time t. > > > > ie. f is f(t), y is y(t), x is x(t), ... > > > > fg denotes function multiplication. > > g*h denotes function convolution. > > > > I want to figure out is the "convolution" the same priority as > > "multiplication"? > > > > ie. can I evaluate the above expression from left to right, or from right > > to > > left or in any other orders? > > > > I am trying to manipulate the expression and to simplify the expression... > > > > thanks a lot > > Hi Lucy , > > I always thought that convolution is commutative , see > http://www-structmed.cimr.cam.ac.uk/Course/Convolution/convolution.html#commut > for example. When I did a quick google on commutative convolution > however I found several links claiming that that's not always the case > so you might want to look at some of them and see if that's true in > your case. > > Best of Luck - MikeFor convolution alone, commutative means... f(t) * g(t) = g(t) * f(t) For convolution alone associative means... f(t) * [g(t) * h(t)] = [f(t) * g(t)] * h(t) I believe that the question really being posed is the following: Are convolution and multiplication associative? 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. -- Remove _me_ for e-mail address
Reply by ●November 12, 20042004-11-12
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. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Reply by ●November 12, 20042004-11-12
G. A. Edgar 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.A condition is that the signal is bandlimited. Your signal isn't. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 12, 20042004-11-12
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 -- Remove _me_ for e-mail address
Reply by ●November 12, 20042004-11-12
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. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●November 15, 20042004-11-15
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> Jerry > -- > Engineering is the art of making what you want from things you can get. > �����������������������������������������������������������������������
