Reply by Jeroen Boschma●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.
> �����������������������������������������������������������������������
Reply by Jerry Avins●November 15, 20042004-11-15
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?
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Jeroen Boschma●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.
> �����������������������������������������������������������������������
Reply by Jerry Avins●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 Ken Prager●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 Jerry Avins●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 <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 Ken Prager●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 - Mike
For 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 Mike Yarwood●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 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 - Mike
Reply by Rune Allnor●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 lot