LTI Systems and Impulse Response

["Followup-To:" header set to sci.math.]
On 2011-01-14, Chris Bore <chris.bore@gmail.com> wrote:
> On Jan 11, 9:17 am, vv <vanam...@netzero.net> wrote:
>> In his book, "A Course in Digital Signal Processing", Porat states
>> that it is a common misconception is that every Linear Time-Invariant
>> system has an impulse response.  He then quotes an example from
>> Kailath's book, "Linear Systems".  The crux of the argument is to
>> define the class of inputs such that the impulse does not belong to it
>> and then claim the system has no impulse response.  This seems like
>> cheating :-).  I seem to be missing the point of the following example
>> from Porat (p. 34):
>>
>> Let x(t) be in the input family iff (1) x(t) is continuous, except at
>> a countable number of points t; (2) the discontinuity at each such
>> point is a finite jump, i.e., the limits at both sides of the
>> discontinuity exist; (3) the sum of absolute values of all
>> discontinuity jumps is finite.  Let y(t) be the sum of all jumps of
>> x(s) at discontinuity points s < t.  This system is linear and time-
>> invariant, but it has no impulse response because delta(t) is not in
>> the input family.  Consequently its response cannot be described by a
>> convolution.
>>
>> I guess the last sentence may hold the key, which I need to ponder
>> about more...
>
> The argument is valid in its own terms. If you do not allow an impulse
> input then you can deny there is an impulse response.

There's a bit more to it than that.  The input family does contain
sequences of functions that converge to delta, and so one could try to
define the impulse response as the limit of y_i(t) as x_i(t) tends to
delta(t).

Unfortunately, it turns out that this limit is not well defined.  For
some obvious choices of x_i(t), like any continuous ones, or
discontinuous ones with even symmetry, the limit is y(t) = 0, which is
problematic enough; a non-trivial linear system certainly should not
have a zero impulse response.  Worse yet, though, if you choose x_i(t)
to be continuous on one side of the origin and discontinuous on the
other, y_i(t) will diverge as x_i(t) tends to delta(t).

So I'd say the real moral of the story is that, on input domains which
don't include impulses, you can have linear time-invariant systems
which really don't have a well-defined impulse response in any sense,
and can't be consistently extended to inputs that include impulses.

-- 
Ilmari Karonen
To reply by e-mail, please replace ".invalid" with ".net" in address.

On Jan 14, 1:30&#4294967295;am, HardySpicer <gyansor...@gmail.com> wrote:
> Never even heard of a Banach space and been using sig processing for a
> very long time!

"Signal Theory" by Lewis Frank is referenced not infrequently in
signal processing literature.

Tim Wescott: "Do you have the page number or chapter number for
Kailath's example?  I've got the book by my elbow, I'd like to look it
up."  No, I don't have Kailath's book and hence unable to quote the
page number in which the example that Porat quotes is present.

Ilmari Karonen: Based on what you have explained, it seems that the
example has more to it than what meets the eye of a less
mathematically sophisticated engineer.  Thanks.  I still need to
digest what you've said.

--vv

On 2011-01-17, VV <vanamali@netzero.net> wrote:
>
> Ilmari Karonen: Based on what you have explained, it seems that the
> example has more to it than what meets the eye of a less
> mathematically sophisticated engineer.  Thanks.  I still need to
> digest what you've said.

It's not really that complicated.  The input space as defined does not
contain the delta function, but we could still approximate it using
nascent delta functions which do belong in the input space.

For example, let x(t) = g(t/a)/a, where g is a bounded piecewise
smooth function with integral 1.  Then x(t) belongs in the input
space, and x(t) -> delta(t) as a -> 0 (in the usual weak sense).

Now, let's say g(t) is the "hat function" g(t) = min(0, 1-abs(t)).
This function is continuous, and so is therefore x(t), which makes
y(t) = 0 regardless of a.

What if we choose a discontinuous g(t), like the "top hat function"
g(t) = 1 for -1/2 < t < 1/2, g(t) = 0 otherwise?  Since x(t) is now
piecewise constant (and 0 for t < -1/2a), y(t) = x(t), and therefore
y(t) -> delta(t) as a -> 0.

Even worse, what if we combine these two into an asymmetric function
like g(t) = 1 if 1/2 < t <= 0, g(t) = 1-t if 0 < t <= 1 and g(t) = 0
otherwise?  Now x(t) has a single discontinuity at -1/2a, where it
jumps up by 1/a, and thus y(t) = 0 for t <= -1/2a and y(t) = 1/a
otherwise.  Therefore, as a -> 0, lim y(t) = 0 for t < 0, but lim y(t)
diverges to infinity for all t >= 0.

The fact that all these limits are different should make it obvious
that this system in fact does not have a well definable impulse
response, and that it's not just a matter of the input space being
arbitrarily constrained.

-- 
Ilmari Karonen
To reply by e-mail, please replace ".invalid" with ".net" in address.