"kiki" <lunaliu3@yahoo.com> wrote in message news:<cljh80$btq$1@news.Stanford.EDU>...> Dear all, > > Could you please help me understand better by providing some deep thoughts > on why linear systems are commutative? > > If T1 and T2 are linear systems, then y1=T2(T1(x)) and y2=T1(T2(x)) > > y1 and y2 are the same. > > After seeing some examples of non-linear system and linear systems. I got > convinced that for non-linear system this property does not hold. > > But any deeper thinking? Proof? > > Thanks a lot!Given: T1(x) = m1*x + b1 T2(x) = m2*x + b2 Then: T2(T1(x)) = m2 * (m1*x+b1) + b2 = m2*m1*x + m2*b1 + b2 T1(T2(x)) = m1 * (m2*x+b2) + b1 = m1*m2*x + m1*b2 + b1 Thus T2(T1(x)) not equal T1(T2(x)) for linear systems
any deep thinking on why linear systems are commutative?
Started by ●October 25, 2004
Reply by ●October 25, 20042004-10-25
Reply by ●October 25, 20042004-10-25
David C. Ullrich wrote:> and two such commute with each other. (Not that there's any way a > person could tell that that's what was intended by the phrase > "linear system".)Well, there is, actually -- the fact that the message is posted on comp.dsp as well as sci.math. In DSP and electrical engineering circles the term "linear system" has a very specific meaning (see my reply to the original post). I'm assuming that the OP had no idea that the term looses its specific meaning when it's mentioned outside of DSP circles. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●October 25, 20042004-10-25
"steve" <bungalow_steve@yahoo.com> wrote in message news:15da8106.0410251434.5952a95e@posting.google.com...> "kiki" <lunaliu3@yahoo.com> wrote in message > news:<cljh80$btq$1@news.Stanford.EDU>... >> Dear all, >> >> Could you please help me understand better by providing some deep >> thoughts >> on why linear systems are commutative? >> >> If T1 and T2 are linear systems, then y1=T2(T1(x)) and y2=T1(T2(x)) >> >> y1 and y2 are the same. >> >> After seeing some examples of non-linear system and linear systems. I got >> convinced that for non-linear system this property does not hold. >> >> But any deeper thinking? Proof? >> >> Thanks a lot! > > Given: > T1(x) = m1*x + b1 > T2(x) = m2*x + b2 > > Then: > T2(T1(x)) = m2 * (m1*x+b1) + b2 = m2*m1*x + m2*b1 + b2 > T1(T2(x)) = m1 * (m2*x+b2) + b1 = m1*m2*x + m1*b2 + b1 > > Thus T2(T1(x)) not equal T1(T2(x)) for linear systemsT1(x) = m1*x + b1 T2(x) = m2*x + b2 are affine systems but not linear systems...
Reply by ●October 25, 20042004-10-25
"Stephen J. Herschkorn" <herschko@rutcor.rutgers.edu> wrote in message news:fHcfd.14194$bz4.1975817@news4.srv.hcvlny.cv.net...> kiki wrote: > >>Could you please help me understand better by providing some deep thoughts >>on why linear systems are commutative? >> >>If T1 and T2 are linear systems, then y1=T2(T1(x)) and y2=T1(T2(x)) >> >>y1 and y2 are the same. >> >>After seeing some examples of non-linear system and linear systems. I got >>convinced that for non-linear system this property does not hold. >> >>But any deeper thinking? Proof? >> > > I thought this would cause confusion amongst the mathematicians. The OP > is referring to systems theory as covered in electrical engineering, I > believe. If I recall correctly from over twenty years ago when I had to > work with these things, let I be either the nonnegative reals or the > natural numbers. A linear system is a linear map from R^I to R^I. As > another poster has pointed out, you may need time-invariance for > commutativity. That is, letting L_t be the map such that [L_t (x)] (s) > = x(t+s) for all s in I, T and L_t commute for all nonnegative t. > (I may be messing this definition up a little.) > > Signal processing always struck me as applied functional analysis. > > > -- > Stephen J. Herschkorn herschko@rutcor.rutgers.eduHi all, I did not want to confuse engineers and mathematicians... I meant the signal processing system, in the Oppenheim's sense. By linearity I meant the properties of superpostion and scaling... Thank you all!
Reply by ●October 25, 20042004-10-25
"kiki" <lunaliu3@yahoo.com> wrote in message news:<cljh80$btq$1@news.Stanford.EDU>...> Dear all, > > Could you please help me understand better by providing some deep thoughts > on why linear systems are commutative? > > If T1 and T2 are linear systems, then y1=T2(T1(x)) and y2=T1(T2(x)) > > y1 and y2 are the same. > > After seeing some examples of non-linear system and linear systems. I got > convinced that for non-linear system this property does not hold. > > But any deeper thinking? Proof?Nothing is commutative, that is not time invariant. In linear systems it's almost never true that T1T2 = T2T1. And if you had a mailbox and mail, both of which ' are linear systems. would you put the mail in the mailbox or the mailbox in the mail? And non-linear systems that do commute are special systems, rather than merely non-linear systems. Einstein called some of them elevators, and some of them trains.> > Thanks a lot!
Reply by ●October 25, 20042004-10-25
Stephen J. Herschkorn wrote:> Signal processing always struck me as applied functional analysis.And so it should be taught! F.A. should be the basis of all further signal processing course work. Wavelets made that apparent and my bet is that, where it isn't F.A. based, teaching will move in that direction. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein
Reply by ●October 25, 20042004-10-25
"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message news:ajvqn0d9966l0v8ivt08vhs44gmq5d4h5p@4ax.com...> > > >A couple of thoughts: I think you also need to add "time-invariant" to your > >requirement for being commutative. Maybe you were assuming a static system,but> >it's good to be explicit about that. Sometimes shift-invariant is usedinstead> >in a sampled data/digital system. > > > >I don't know about the general case, but for something simple such as T1 andT2> >being different (linear) filters, it certainly holds. A really simpleexample> >if making a band-pass filter out of a low-pass and high-pass filter. Itdoesn't> >matter if you run the signal through the LP first and then run the result > >through the HP or vice versa. (I'm talking the world pure mathematics here,so> >things such as round-off error/quantization noise are ignored.) Are there > >examples where two linear time-invariant systems are not commutative? > > You have any idea why your reply is appearing in a totally different > thread, btw?Nope. Appears correctly for me here in Outlook Express.> Anyway, yes, the sort of linear operators you're talking about here > do commute. Those shift-invariant things are convolution operators, > and two such commute with each other. (Not that there's any way a > person could tell that that's what was intended by the phrase > "linear system".)I assumed the DSP definition of linear system because I was reading the message in comp.dsp. I later noticed it was also posted to sci.math, hence the confusion.> >In the DSP world, the general definition for a linear system is where the > >principle of super-position holds. In layman's terms, put in twice the input > >and you get twice the output (T(g*a) = g*T(a). Or put in a + b and theanswer> >is the same as if you put in a, then b, and added the results (T(a+b) = T(a)+> >T(b)). I noticed this is posted to sci.math as well, and perhaps a different > >definition is used there? > > Assuming that g is a constant and * means multiplication then no, > that's the definition of "linear" here, and two linear operators > most certainly need not commute. You have to add the translation > invariance for that - translation/shift/time-invariance is _not_ > part of the definition of "linear".Right, g is a constant and * means plain old multiplication. (Originally I left out the *, but later added it because I thought it would clarify things--guess I was wrong! :-) Someone else posted a much better formal definition of a linear system in the DSP-sense in this thread, so please refer to that if there is still any confusion. And yes I did mention the "time/shift-invariant" requirement in my post.
Reply by ●October 26, 20042004-10-26
On 2004-10-26 01:45:21 +0200, zzbunker@netscape.net (ZZBunker) said:> And if you had a mailbox and mail, both of which ' > are linear systems. > would you put the mail in the mailbox > or the mailbox in the mail?But my mailbox is highly non-linear! -- Stephan M. Bernsee http://www.dspdimension.com
Reply by ●October 26, 20042004-10-26
Bob Cain <arcane@arcanemethods.com> wrote in message news:<clk60j030rg@enews2.newsguy.com>...> Stephen J. Herschkorn wrote: > > > Signal processing always struck me as applied functional analysis. > > And so it should be taught! F.A. should be the basis of all > further signal processing course work. Wavelets made that > apparent and my bet is that, where it isn't F.A. based, > teaching will move in that direction.I agree with you and I hope you are right. Still, don't be too optimistic about such steps being taken in the near future. The problem is that DSP is an applied dicipline, with lots of the business being dominated by "hands on problem-solving" Electrical Engineers. I've been trying to attempt the FA-based problem solving strategy with practical problems, and obtained certain encouraging results. However, the reactions from the old-timers with 40+ years experience in EE and who "knew the proper way" of doing things, made me abandon the whole thing. And almost DSP as well. Rune
Reply by ●October 26, 20042004-10-26
Stephan M. Bernsee <spam@dspdimension.com> wrote in message news:<2u674mF26ndeqU2@uni-berlin.de>...> On 2004-10-26 01:45:21 +0200, zzbunker@netscape.net (ZZBunker) said: > > > And if you had a mailbox and mail, both of which ' > > are linear systems. > > would you put the mail in the mailbox > > or the mailbox in the mail? > > But my mailbox is highly non-linear!That's only because you use dsp.net. Which doesn't really use mailboxes. It uses the twilight zone, it uses a cookbook, it uses internet packets. Which are by design, both highly nonlinear and highly overpriced.
