A _really_ basic question

"Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message
news:muNAc.58254$7H1.1960064@news20.bellglobal.com...
>
> "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 applied
> to
> > 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?

I don't think so.  y(t) is a function of time and not stated as being a
function of x,y or z (spatial dimensions) or .....

y(t) is a forcing function or a system response function.  If it's a system
response function then you can't just say what its value is at certain times
and call that a boundary condition.  Let's see if I can come up with an
example....

First of all, OLDEWCCs (ordinary linear differential equations with constant
coefficients) represent lumped systems like filters.  In this case, we don't
talk about boundary conditions because the "boundaries" aren't so evident.
But, the "ports" on a filter could be viewed as simple boundaries.

However, partial differential equations are used to represent "field"
problems or distributed "systems" where there can be boundaries to the
fields or bounds applied to the systems.  So, I gave a simple example of a
field - which was a resistive sheet.  The dependent variables are potential
and flow that vary spatially and with time.  There are initial conditions
and there are boundary conditions of potential and flow.  The boundary
conditions are what exists in terms of either potential or flow at the
boundaries.

The boundary conditions are generally fixed - as in fixed potential or fixed
flow.  A couple of examples would be:
- fixed potential along an edge of a sheet and the flow is arbitrary /
deterministic subject to the system.  This could be the surface of a sphere
just as well as a sheet.
- fixed flow along an edge of a sheet (no potential applied) and the
potential along that edge is arbitrary / deterministic subject to the
system.
- a combination of the two if an edge potential is sourced through a source
impedance ... and source impedance becomes part of the "system".
- a combination of the two if an edge zero flow is actuall sourced through a
high source impedance .. and source impedance.  Neither the potential or the
flow is fixed in either of these cases.  The conditions that we'd otherwise
call "boundary" conditions aren't so useful anymore and vary with time
according to system dynamics and any time variation of the "sources".

In every case, the dependent variables are local flow and potential.  Time
is an independent variable.  So, other than choosing some arbitrary time to
call t=0, it has nothing to do with the field's boundary conditions as
described above.

Now, I suppose that one could map the situation such that potential and time
were the dependent variables and flow were the independent variable but that
hasn't got much physical meaning to *me*.

If you take a vibrating string and fix the ends then you have set the flow
to be zero (i.e the string cannot move).  The potential (tension) at the
boundary varies according to the position of the vibrating string.  It
varies with time without making the zero flow boundary condition
time-variant.  The vibrating string is just a simple version of wave motion
in a box or a cylinder - which is where all this came up isn't it?

Fred

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
news:9fCdnU1bZJ8CIE7dRVn-gg@centurytel.net...
> I don't think so.  y(t) is a function of time and not stated as being a
> function of x,y or z (spatial dimensions) or .....

Yes, y(t) is a function of time...  I could have used y(x), and set y=0 at
x=0 and x=\pi, but then I would have to change mid-argument from talking
about time-invariance to shift-invariance or space-invariance.

> The vibrating string is just a simple version of wave motion
> in a box or a cylinder - which is where all this came up isn't it?

Not quite.  If you use x in place of t, then solving the equation in the
example I gave finds the possible standing wave shapes on a vibrating
string.  Perhaps it was confusing to use t instead of x, but it doesn't
change the math, and I thought it was established in the thread.  All this
came up in response to my claim that LTI systems have all the exponentials
as eigenfunctions, and I thought it was pretty clear that we were talking
about operators on eigenfunctions with time as their argument.

"Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message news:<flCAc.39066$nY.1221300@news20.bellglobal.com>...
> "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.

OK. One function that satisfies the differential equation (without
boundary conditions imposed) is

   y(t) =    A exp(jwt) +    B exp(-jwt)
   y'(t)= jw A exp(jwt) - jw B exp(-jwt)
   y"(t)=-w^2A exp(jwt) + w^2B exp(-jwt).

We find that the eigenvalues are +/-w^2 for this 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.

Here I'm lost. The solution to this problem, with boundary conditions, is 

    y(t)= sin_n(pi*n*t) = 1/j2(exp(jpi*n*t)-exp(-jpi*n*t)).

There are infinately many such solutions, one for every natural 
number n. What do you men by saying that this solution is excluded 
because "it is not Linear Time Invariant"? What, then, is the 
requirement for a system to be LTI? That the solution is a constant?

> The eigenfunctions that meet all the imposed conditions, sin(kt), are linear
> combinations of exponential eigenfunctions that have the same eigenvalue.

Well, that may be one way of saying it. I would prefer to say that 
there are *pairs* of eigenvalues, +/- w^2, as argued above.

> 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.

OK, From your earlier posts, I got the impression that you meant that 
only LTI systems have eigenfunctions, and all systems that have 
eigenfunctions are LTI. I can't make sense of that. In the lines above, 
there appears to be a partial consent that non-LTI systems may have 
some relevance. My point has been, from my first post in this thread, 
that the "LTI" property is not exclusive to linear differential 
operators that can be expanded in series of eigenfunctions, nor 
is every such operator necessarily of the "LTI" type. 

Which inspires the question "what does 'LTI' mean?"

To me, an "LTI" system is one that have constant coefficients, i.e. 
an linear operator L acting on a function f(t) such that 

   Lf(t) = L(a,b,c,d,...)f(t)

where a,b,c,d,... are coefficients of the operator L and are *not* 
functions of the function argument t. There are other linear operators 
Q, e.g. the cylindrical wave equation, where some coefficients of Q are 
functions of the function argument r, i.e.

   Qg(r)=Q(a(r),b(r),c(r),d(r),...)g(r)

These equations are still of Sturm-Liouville type, and their solutions 
can still be expressed in terms of eigenfunctions and all that Hilbert 
space stuff. 

Now, whether those operators are LTI is basically an irrelevant question, 
since the *physical* (not mathematical) property "time" never enters the 
discussion. These equations apply to spatial variables and the coordinate 
systems  do not change with time. Which is why I can't see why your 
argument about "the exponential being an eigenfunction because the system 
is LTI" works.

Rune

"Rune Allnor" <allnor@tele.ntnu.no> wrote in message
news:f56893ae.0406190310.48999e19@posting.google.com...
> > -y''=ay
> >
> OK. One function that satisfies the differential equation (without
> boundary conditions imposed) is
>
>    y(t) =    A exp(jwt) +    B exp(-jwt)

Right, but that is a linear combination of 2 simple exponentials of the
given form.  *All* of the simple exponentials are eigenfunctions too, e.g.,

y(t)=exp(5t)
=> y'(t)=5*exp(5t)
=> y''(t)=25*exp(5t)
=> -y''(t)=(-25)*y(t)

and

y(t)=exp(4jt)
...
=> -y''(t)=16*y(t)

> > 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.
>
> Here I'm lost. The solution to this problem, with boundary conditions, is
>
>     y(t)= sin_n(pi*n*t) = 1/j2(exp(jpi*n*t)-exp(-jpi*n*t)).
>
> There are infinitely many such solutions, one for every natural
> number n. What do you men by saying that this solution is excluded
> [...]

Your solution works (without the pi factor, because I put the upper boundary
condition at t=pi instead of t=1), but all of the simple exponential
eigenfunctions are excluded by the boundary conditions, because they
have y(t) != 0 everywhere.

> because "it is not Linear Time Invariant"?

I only said here that the boundary conditions were not time invariant,
meaning only that they will admit solutions like y(t)=sin(t), but not
y(t)=sin(t-.5).

> > The eigenfunctions that meet all the imposed conditions, sin(kt), are
> > linear combinations of exponential eigenfunctions that have the same
> > eigenvalue.
>
> Well, that may be one way of saying it. I would prefer to say that
> there are *pairs* of eigenvalues, +/- w^2, as argued above.

I appreciate you trying to interpret what I said in a way that makes sense.
We obviously had a misunderstanding above, though, and here I meant simply
what I said.

sin(3t) is a solution, for example, and it is the difference between two
exponential eigenfunctions of the differential operator:
sin(3t)=exp(3jt) - exp(-3jt).

And those exponentials have the same eigenvalue:

y(t)=exp(3jt)
=> y''(t) = -9exp(3jt)
=> -y''(t) = 9y(t)

and

y(t)=exp(-3jt)
=> -y''(t) = 9y(t)

> OK, From your earlier posts, I got the impression that you meant that
> only LTI systems have eigenfunctions, and all systems that have
> eigenfunctions are LTI.

Oh, no.  Many systems have eigenfunctions .  The assertion I made that you
refuted was that LTI systems have *all* the exponentials as eigenfunctions,
which is true, and leads to the great utility of the Laplace and Z
transforms in signal processing.

Between then and now, I Googled around to see what sort of equations you
would be talking about when you mention Sturm-Liouville and curvilinear
coordinate systems.  Let me reorder what you said below:

 > These equations are still of Sturm-Liouville type, and their solutions
> can still be expressed in terms of eigenfunctions and all that Hilbert
> space stuff.
>
> Now, whether those operators are LTI is basically an irrelevant question,
> since the *physical* (not mathematical) property "time" never enters the
> discussion. These equations apply to spatial variables and the coordinate
> systems  do not change with time. Which is why I can't see why your
> argument about "the exponential being an eigenfunction because the system
> is LTI" works.

When we're talking about operators on functions of space instead of time,
then we call them "LSI", or "linear, shift-invariant", or "linear, space
invariant", instead of LTI.  I guess I confused Fred on this point too.  The
only difference is the letter your use for the argument -- t or x.

So, using x, we say that an operator M is LSI iff we always have:

h(x)=f(x)+g(x) => (Mh)(x) = (Mf)(x)+(Mg)(x)
and
h(x)=f(x+d) => (Mh)(x) = (Mf)(x+d)

on functions of t, we say an operator is LTI iff we always have:

h(t)=f(t)+g(t) => (Mh)(t) = (Mf)(t)+(Mg)(t)
and
h(t)=f(t+d) => (Mh)(t) = (Mf)(t+d)

In both cases, LTI or LSI is a property of the operator M, and means the
same thing.

> My point has been, from my first post in this thread,
> that the "LTI" property is not exclusive to linear differential
> operators that can be expanded in series of eigenfunctions,

Agree.

> nor is every such operator necessarily of the "LTI" type.

Agree here, too.  The Laplacian, for example, is LSI in rectangular
coordinates, but in cylindrical coordinates it isn't shift-invariant,
because the derivatives get coefficients that depend on r, so h(r)=f(r+d)
does not imply (Lh)(r) = (Lf)(r+d).

--
Matt

"Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message news:<6M6Bc.49682$nY.1536978@news20.bellglobal.com>...
> "Rune Allnor" <allnor@tele.ntnu.no> wrote in message
> news:f56893ae.0406190310.48999e19@posting.google.com...

> > My point has been, from my first post in this thread,
> > that the "LTI" property is not exclusive to linear differential
> > operators that can be expanded in series of eigenfunctions,
> 
> Agree.
> 
> > nor is every such operator necessarily of the "LTI" type.
> 
> Agree here, too.  The Laplacian, for example, is LSI in rectangular
> coordinates, but in cylindrical coordinates it isn't shift-invariant,
> because the derivatives get coefficients that depend on r, so h(r)=f(r+d)
> does not imply (Lh)(r) = (Lf)(r+d).

I think we agree. It may be that we approached the basic question in the
first post from slightly different angles. Since filters, as we know them 
in DSP, probably are best described as initial value problems, confusion 
got worse.

But it's fun to shake the dust out of the brain every once in a while.

Rune