"Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message news:_LsAc.36154$nY.1146137@news20.bellglobal.com...> H(s)... eigenvalue...eigenfunction e^s.Sorry, that's e^(st)>H(z)...eigenfunction e^z.and e^(zn)
A _really_ basic question
Started by ●June 17, 2004
Reply by ●June 18, 20042004-06-18
Reply by ●June 18, 20042004-06-18
Matt Timmermans wrote:> "Bob Cain" <arcane@arcanemethods.com> wrote in message > news:caslgn02uq1@enews2.newsguy.com... > >>Thanks Matt. That's just what I'm looking for. Eigenstuff >>stands outside my education and I'm always a bit mystified >>when people start speaking it. > > > Just remember that the term is as simple as I'm saying it is -- the result > of an operator applied to one of its eigen-whatevers is just that > eigen-whatever multiplied by a scalar eigenvalue.I wish someone had told me that a long time ago. I'm sure it was there to see but somehow the simplicity of the way you state it did not jump off the page. :-) Come to think of it, until fairly recently, the signifigance would have been totally lost on me. So the big distinction the Fourier decomposition has over all the others is that the basis is the eigenfunction of LTI systems which is what we usually deal with. Got it. That then gives me a more formal reason for thinking that basis has little relevance to hearing, especially in the use of sinusoids to measure its capabilities. Hearing is neither linear nor time invarient, and probably could care less about orthonormality, yet we measure it in terms of the magnitude of this supposed eigenvalue. Hrmph! :-)>>>Other LTI >>>systems can have additional eigenfunctions, but exponentials like this > > are > >>>eigenfunctions of *every* LTI system. >> >>Got an example of such a system and its eigenfunction? > > > Every continuous filter is an LTI system. H(s) gives the eigenvalue for the > eigenfunction e^s. When H(s_i) is the same for multiple s_i abscissas, any > linear combination of the corresponding e^(s_i) eigenfunctions will be > another eigenfunction.You lost me here. I don't know what s_i means nor "is the same for multilple s_i abcissas". If you could reduce that further I'd appreciate it.> Zero-phase filters have sines and cosines as > eigenfunctions, for example, because they have H(jw)=H(-jw).Or linear-phase if we subtract out the delay. Got it.> > > Ah, then he will be using a formal definition of the word "vector" that > includes continuous functions, discrete functions, finite vectors, and > anything else you can make a Hilbert space out of.That's the problem. He uses too much of the really correct (almost Bourbaki) formalism which makes it tough sledding if you don't have all that stuff under your belt.> > Note also that wavelet analysis is an explicit departure from the > time-invariance of Fourier analysis, so all this LTI stuff may not apply.Hopefully I can get far enough to see how he steps away from that, 'cause LTI is where he starts. Thanks Again, Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein
Reply by ●June 18, 20042004-06-18
"Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message news:<pwrAc.34979$nY.1119228@news20.bellglobal.com>...> "Rune Allnor" <allnor@tele.ntnu.no> wrote in message > news:f56893ae.0406170228.2fe9b8d9@posting.google.com... > > "Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message > news:<Xf9Ac.24496$nY.949518@news20.bellglobal.com>... > > > For LTI systems, it actually works for any x(t) = (a+jb)*exp(t*(x+jy)), > > > i.e., any single value at any single point on the S or Z plane. Other > LTI > > > systems can have additional eigenfunctions, but exponentials like this > are > > > eigenfunctions of *every* LTI system. > > > > Eh... I don't think the exponentials are a caused by the system being > > LTI, rather, I think exponentials are eigen vectors of particular > > differential equations. > > I'm not sure what you mean by "caused", but many different kinds of > operators have eigenfunctions or eigen-something-elses. In all cases, the > meaning of the term is simple -- the result of the operator applied to an > eigen-whatever-the-domain-is is that eigen-whatever multiplied by a scalar > eigenvalue. > > All linear time-invariant systems have all of the exponentials of the above > form as eigenfunctions. > > > If you express the physical problem in a cylindrical or spherical > > coordinate system, the systems are LTI but some of the eigen vectors > > might be Bessel functions or Legendre polynomials instead of exponentials. > > Those systems aren't linear and time-invariant.They are linear and, when present in partial differential equations describing physical systems, also time-invariant. The physical coordinate system does not change with time. The Bessel and other functions do, however, scale with their argument. These types of differential equations and their eigenfunctions are studied in what I know as Sturm-Liouville systems. Apparently, just about every linear differential equation that appears in mathemathical physics is of this type. Rune
Reply by ●June 18, 20042004-06-18
"Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message news:<_LsAc.36154$nY.1146137@news20.bellglobal.com>...> "Bob Cain" <arcane@arcanemethods.com> wrote in message > news:caslgn02uq1@enews2.newsguy.com... > > Thanks Matt. That's just what I'm looking for. Eigenstuff > > stands outside my education and I'm always a bit mystified > > when people start speaking it. > > Just remember that the term is as simple as I'm saying it is -- the result > of an operator applied to one of its eigen-whatevers is just that > eigen-whatever multiplied by a scalar eigenvalue.Agreed.> > > Other LTI > > > systems can have additional eigenfunctions, but exponentials like this > are > > > eigenfunctions of *every* LTI system. > > > > Got an example of such a system and its eigenfunction? > > Every continuous filter is an LTI system.Are you sure "continous" implies "LTI"? Could you elaborate on that? Preferably with references to the literature.> H(s) gives the eigenvalue for the > eigenfunction e^s. When H(s_i) is the same for multiple s_i abscissas, any > linear combination of the corresponding e^(s_i) eigenfunctions will be > another eigenfunction. Zero-phase filters have sines and cosines as > eigenfunctions, for example, because they have H(jw)=H(-jw). > > A discrete filter is linear and time-invariant too, with the understanding > that time shifts are integral multiples of the sample period. Similar to the > continuous case, every H(z) gives the eigenvalue for the eigenfunction e^z.What about LMS and RMS adaptive filters? Kalman filters? I think they are discrete, but I would hardly call them "Time Invariant". If one did, all that fuzz about convergence rates would be obsolete.> > My question came from starting to read Mallat's "A > > Wavelet Tour Of Signal Processing" and he thinks in terms of > > vectors and Hilbert spaces. I doubt I'll get far with this, > > he presumes too formal a math education, but I thought I'd > > give it a go. > > Ah, then he will be using a formal definition of the word "vector" that > includes continuous functions, discrete functions, finite vectors, and > anything else you can make a Hilbert space out of. > > Note also that wavelet analysis is an explicit departure from the > time-invariance of Fourier analysis, so all this LTI stuff may not apply. > Eigenthings are used all over the place, but I don't remember the concept > being fundamental to wavelet analysis, so I can't guess at what you would > actually be seeing in Mallat's book.Agreed. Wavelet analysis is easily understood in terms of Hilbert space theory, but it's not obvious how or why wavelets should be eigenfunctions of any useful linear operator. Rune
Reply by ●June 18, 20042004-06-18
"Rune Allnor" <allnor@tele.ntnu.no> wrote in message news:f56893ae.0406180203.ead534b@posting.google.com...> > Every continuous filter is an LTI system. > > Are you sure "continous" implies "LTI"? Could you elaborate on that? > Preferably with references to the literature.I was using "filter" in the colloquial sense that I thought Bob would interpret it in, i.e. a filter that convolves an impulse response with a signal.> What about LMS and RMS adaptive filters? Kalman filters? I think they > are discrete, but I would hardly call them "Time Invariant". If one did, > all that fuzz about convergence rates would be obsolete.But perhaps I should have spoken more precisely. -- Matt
Reply by ●June 18, 20042004-06-18
"Rune Allnor" <allnor@tele.ntnu.no> wrote in message news:f56893ae.0406172209.7a73dbc1@posting.google.com...> > Those systems aren't linear and time-invariant. > > They are linear and, when present in partial differential equations > describing physical systems, also time-invariant. The physical coordinate > system does not change with time. The Bessel and other functions do, > however, scale with their argument. > > These types of differential equations and their eigenfunctions are > studied in what I know as Sturm-Liouville systems. Apparently, just > about every linear differential equation that appears in mathemathical > physics is of this type.In rectangular coordinate systems, the Sturm-Liouville boundary conditions disallow the arbitrary exponentials. A simple example: y'' + ay=0, y(0)=0, y(pi)=0 a is the unspecified eigenvalue, which is written into the problem. A simple rewrite: -y''=ay shows that we are looking for eigenfunctions of the "negated second derivative" operator. That operator is LTI, and any exponential solves that equation, but the boundary conditions, which are not time-invariant, along with the requirement for a real solution, exclude them all. The eigenfunctions that meet all the imposed conditions, sin(kt), are linear combinations of exponential eigenfunctions that have the same eigenvalue. In curvilinear coordinate systems, the linear differential operators themselves will describe non-LTI systems. I'm not sufficiently familiar with the physics to provide a simple realistic example, but if you can, I will try to show how the system fails to be LTI. -- Matt
Reply by ●June 18, 20042004-06-18
Matt Timmermans wrote:> "Rune Allnor" <allnor@tele.ntnu.no> wrote in message > news:f56893ae.0406172209.7a73dbc1@posting.google.com... > >>>Those systems aren't linear and time-invariant. >> >>They are linear and, when present in partial differential equations >>describing physical systems, also time-invariant. The physical coordinate >>system does not change with time. The Bessel and other functions do, >>however, scale with their argument. >> >>These types of differential equations and their eigenfunctions are >>studied in what I know as Sturm-Liouville systems. Apparently, just >>about every linear differential equation that appears in mathemathical >>physics is of this type. > > > In rectangular coordinate systems, the Sturm-Liouville boundary conditions > disallow the arbitrary exponentials. A simple example: > > y'' + ay=0, y(0)=0, y(pi)=0 > > a is the unspecified eigenvalue, which is written into the problem. > > A simple rewrite: > > -y''=ay > > shows that we are looking for eigenfunctions of the "negated second > derivative" operator. > > That operator is LTI, and any exponential solves that equation, but the > boundary conditions, which are not time-invariant, along with the > requirement for a real solution, exclude them all. > > The eigenfunctions that meet all the imposed conditions, sin(kt), are linear > combinations of exponential eigenfunctions that have the same eigenvalue. > > In curvilinear coordinate systems, the linear differential operators > themselves will describe non-LTI systems. I'm not sufficiently familiar > with the physics to provide a simple realistic example, but if you can, I > will try to show how the system fails to be LTI. > > -- > MattI don't get it. The boundary conditions are constant. How is that not LTI? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 18, 20042004-06-18
"Jerry Avins" <jya@ieee.org> wrote in message news:40d31094$0$2988$61fed72c@news.rcn.com...> I don't get it. The boundary conditions are constant. How is that not > LTI?They not time invariant, because they are defined at particular instants t=0 and t=\pi.
Reply by ●June 18, 20042004-06-18
"Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message news:wBMAc.58077$7H1.1938159@news20.bellglobal.com...> > "Jerry Avins" <jya@ieee.org> wrote in message > news:40d31094$0$2988$61fed72c@news.rcn.com... > > I don't get it. The boundary conditions are constant. How is that not > > LTI? > > They not time invariant, because they are defined at particular instantst=0> and t=\pi.If they aren't time invariant, then what is the forcing function applied to them? Do the cylinder walls wiggle? Maybe I don't know what you mean by boundary conditions. Normally I think of boundary conditions as being "zero flow" or "zero potential" or "fixed" of either one. That sort of condition doesn't change with time. Or, there can be a forcing function that varies with time but that isn't part of the LTI. For example, one can simulate flow over an airfoil using a resistive sheet with constant potential at opposite edges and zero flow along the edges parallel to the direction of flow - where the width is great enough that the lines of flow are parallel enough - so that extension in the cross-flow direction would not be productive. Then cut out the airfoil shape out of the center of the sheet pointing the the desired direction and measure the potential or flow at points on the sheet. How is this different? A drum has constant boundary conditions on the cylinder doesn't it? Of course the heads move but they can be analyzed as having zero movement on their periphery. etc. etc.... What am I missing? Fred
Reply by ●June 18, 20042004-06-18
"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:AOSdnd0waurdAU7d4p2dnA@centurytel.net...> If they aren't time invariant, then what is the forcing function appliedto> them? Do the cylinder walls wiggle? > What am I missing?What?? Why do I need a forcing function, or a cylinder, or walls? The boundary conditions, which I gave, are: y(0)=0 and y(\pi) = 0, i.e, y is 0 at time zero and y=0 at time \pi. Those conditions admit solutions like y=sin(t), but if you shift that solution in time to make, say y=sin(t-1), it now violates those conditions, so the conditions are not time invariant. Simple, yes?
