"Jerry Avins" <jya@ieee.org> wrote in message
news:NJidnbbgud7NNu7ZnZ2dnUVZ_sCdnZ2d@rcn.net...
> Fred Marshall wrote:
>
>> "Jerry Avins" <jya@ieee.org> wrote in message
>> news:r6udnYWUtNaZ_O7ZnZ2dnUVZ_sGdnZ2d@rcn.net...
>>
>>>I know as well as anyone that we couldn't do this stuff without a lot of
>>>math. We certainly wouldn't understand what we did even if we bumbled
>>>into it. But when the math which enables understanding becomes the thing
>>>understood instead of its description, reality is obscured.
>>>
>>>Jerry
>>>--
>>>Engineering is the art of making what you want from things you can get.
>>>�����������������������������������������������������������������������
>>
>>
>> Jerry,
>>
>> I liked your comment so much that I decided to hone it a bit.
>>
>> "Math enables a description of something for better understanding. One
>> might call the math description a model. When the math mechanics are the
>> things understood, to the exclusion of what the math is intended to
>> describe
>> or model, reality can be obscured. Indeed, such narrow focus can lead to
>> mistakes or pathological cases."
>
> Your statement is better than mine. (I bet you spent longer composing it.)
>
> Jerry
At least it appears we agree..
Fred
Reply by Jerry Avins●May 23, 20062006-05-23
Fred Marshall wrote:
> "Jerry Avins" <jya@ieee.org> wrote in message
> news:r6udnYWUtNaZ_O7ZnZ2dnUVZ_sGdnZ2d@rcn.net...
>
>>I know as well as anyone that we couldn't do this stuff without a lot of
>>math. We certainly wouldn't understand what we did even if we bumbled
>>into it. But when the math which enables understanding becomes the thing
>>understood instead of its description, reality is obscured.
>>
>>Jerry
>>--
>>Engineering is the art of making what you want from things you can get.
>>�����������������������������������������������������������������������
>
>
> Jerry,
>
> I liked your comment so much that I decided to hone it a bit.
>
> "Math enables a description of something for better understanding. One
> might call the math description a model. When the math mechanics are the
> things understood, to the exclusion of what the math is intended to describe
> or model, reality can be obscured. Indeed, such narrow focus can lead to
> mistakes or pathological cases."
Your statement is better than mine. (I bet you spent longer composing it.)
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Fred Marshall●May 23, 20062006-05-23
"Jerry Avins" <jya@ieee.org> wrote in message
news:r6udnYWUtNaZ_O7ZnZ2dnUVZ_sGdnZ2d@rcn.net...
>
> I know as well as anyone that we couldn't do this stuff without a lot of
> math. We certainly wouldn't understand what we did even if we bumbled
> into it. But when the math which enables understanding becomes the thing
> understood instead of its description, reality is obscured.
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> �����������������������������������������������������������������������
Jerry,
I liked your comment so much that I decided to hone it a bit.
"Math enables a description of something for better understanding. One
might call the math description a model. When the math mechanics are the
things understood, to the exclusion of what the math is intended to describe
or model, reality can be obscured. Indeed, such narrow focus can lead to
mistakes or pathological cases."
Fred
Reply by Andor●May 23, 20062006-05-23
robert bristow-johnson wrote:
...
> > An instable filter excited with zero should still yield
> > zero output, no?
>
> the answer is no. only in the case of a digital filter with no
> dithering or other external noise source inside and with the initial
> states all zero and an exactly zero input, is what you asks is true.
I implicitely assumed a discrete system (this being comp.dsp) - it is
very hard (as I know from experience) posing riddles in a well-defined
manner. Hidden assumptions almost always creep in.
It was a theoretical game - does there exist a system with the proposed
frequency response? I showed it does. The practical relevance is a
completely other pair of shoes. One of the very few drawbacks of this
group is the consistent mix-up of theory and practice. An effect, I
guess, from having so much working experience and theoretical knwoledge
crowded into one forum.
Reply by Jerry Avins●May 23, 20062006-05-23
Andor wrote:
> Jerry Avins wrote:
> ...
>
>>>Any filter is an oscillator - the kind where there are poles on the
>>>unit circle are a special subset.
>>
>>A very special subset indeed! The only kind that actually oscillate.
>
>
> I also call a sequence y[n] = exp(s n) p(n), where s is a complex
> frequency and p a polynomial, a (discrete) oscillation. Harmonic and
> damped (discrete) oscillators are special cases. The reason I call them
> "oscillations" is that such y_n are the general solutions to linear
> recursive equations - of which impulse responses of linear filters are
> also a special subset :-).
"Calling a tail a leg doesn't make it one." It may be an oscillator
mathematically, but it doesn't actually oscillate. Do equations actually
/do/ anything? Maybe we should call it an "oscillator descriptor".
I know as well as anyone that we couldn't do this stuff without a lot of
math. We certainly wouldn't understand what we did even if we bumbled
into it. But when the math which enables understanding becomes the thing
understood instead of its description, reality is obscured.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by robert bristow-johnson●May 23, 20062006-05-23
Andor wrote:
> Now I'm completely confused :-). My initial idea was the following:
> First filter the signal with an FIR which has a zero at w0.
that's doable.
> Then filter
> the output with the inverse filter - for all sinusoids with frequency
> not equal to w0, the filter would return the input.
that is not precisely doable. but, if you're willing to put in the
resources (number of FIR taps) and consequential delay, you can get
arbitrarily close. so the end result is that you can get arbitrarily
close to an ideal notch filter that passes everything unchanged but
zeros w0, but you can't get it exacty.
> For a sinusoid with
> frequency w0, the output of the FIR would be equal to zero, and so
> would therefore also be the output of the inverse IIR. Here is a block
> diagramm:
>
> x[n] ---> H1(z) ----> u[n] ----> 1/H2(z) ----> y[n]
>
> A real notch filter, like r b-j's, has
>
> H1(z) =3D 1 - 2 cos(w0) z^-1 + z^-2
i do not, nor never made any claim to owning or originating the design
of this particular nor any notch filter. the Cookbook was done to tie
together a bunch of possibly disparate 2nd order IIR designs with as
much common language and common intermediate parameters to sorta
"unify" them all. perhaps those particular shelf filters originated
with me, but none of the others have. if you take other designs from
other sources of any of the other filters, make the substitutions i did
to get rid of tan(w0/2), and possibly dink around with the definition
of Q or bandwidth, you'll get the same equations as in the cookbook.
(so there is no notch filter that is "r b-j's" notch filter.)
> An instable filter excited with zero should still yield
> zero output, no?
the answer is no. only in the case of a digital filter with no
dithering or other external noise source inside and with the initial
states all zero and an exactly zero input, is what you asks is true.
in the analog world, an oscillator often is a simply an unstable filter
with the dominant poles alpha =B1 j*w0, where 0 < alpha << w0. it has
zero input and non-zero output. a digital filter modeling that works
just fine as long as you set the initial states to be something that is
non-zero.
Reply by Andor●May 23, 20062006-05-23
Jerry Avins wrote:
...
> > Any filter is an oscillator - the kind where there are poles on the
> > unit circle are a special subset.
>
> A very special subset indeed! The only kind that actually oscillate.
I also call a sequence y[n] = exp(s n) p(n), where s is a complex
frequency and p a polynomial, a (discrete) oscillation. Harmonic and
damped (discrete) oscillators are special cases. The reason I call them
"oscillations" is that such y_n are the general solutions to linear
recursive equations - of which impulse responses of linear filters are
also a special subset :-).
>
> > Setting the oscillator states by hand
> > (initializing the states) to non-zero or applying a filter input which
> > is non-zero for a duration of length equal to the order of the
> > oscillator is equivalent. One can be computed from the other.
>
> Sure. Every instant's results are the next instant's boundary conditions.
Yup.
Reply by Jerry Avins●May 23, 20062006-05-23
Andor wrote:
> Jerry Avins wrote:
> ...
>
>>Once again, math obscures underlying simplicity.
>
>
> What you all got against maths? Maths _shows_ the underlying
> simplicity.
>
>
>>It's evident that the
>>inverse of a filter with a real zero can't be stable: it has a real
>>pole. You don't need an input to excite such a filter. It's called an
>>oscillator.
>
>
> A pendulum is also on oscillator. There won't be much pendulation (:-)
> if you never give it a first shove (a non-zero input).
A free pendulum is a high-Q resonator A clock's pendulum must swing far
enough to activate the escapement. That's a hard discontinuity and
difficult to model. An analog oscillator can behave similarly -- some
crystals need a jolt to start -- but most start just fine with noise.
Digital filters are another case entirely. Since they necessarily
operate in discrete time, reasoning about passing to limits don't yield
useful insights.
> Any filter is an oscillator - the kind where there are poles on the
> unit circle are a special subset.
A very special subset indeed! The only kind that actually oscillate.
> Setting the oscillator states by hand
> (initializing the states) to non-zero or applying a filter input which
> is non-zero for a duration of length equal to the order of the
> oscillator is equivalent. One can be computed from the other.
Sure. Every instant's results are the next instant's boundary conditions.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Andor●May 23, 20062006-05-23
Jerry Avins wrote:
...
> Once again, math obscures underlying simplicity.
What you all got against maths? Maths _shows_ the underlying
simplicity.
> It's evident that the
> inverse of a filter with a real zero can't be stable: it has a real
> pole. You don't need an input to excite such a filter. It's called an
> oscillator.
A pendulum is also on oscillator. There won't be much pendulation (:-)
if you never give it a first shove (a non-zero input).
Any filter is an oscillator - the kind where there are poles on the
unit circle are a special subset. Setting the oscillator states by hand
(initializing the states) to non-zero or applying a filter input which
is non-zero for a duration of length equal to the order of the
oscillator is equivalent. One can be computed from the other.
Regards,
Andor
Reply by Jerry Avins●May 23, 20062006-05-23
Martin Eisenberg wrote:
> Andor wrote:
>
>>Martin Eisenberg wrote:
>>
>>>Andor wrote:
>>>
>>>
>>>>>If u[n] = 0, then by equation (1.b) we have
>>>>>
>>>>>y[n] - 2 cow(w0) y[n-1] + y[n-2] = 0,
>>>>>
>>>>>which does _not_ imply y[n] = 0 for all n.
>>>>
>>>>Now I'm completely confused :-). My initial idea was the
>>>>following: First filter the signal with an FIR which has a
>>>>zero at w0. Then filter the output with the inverse filter -
>>>>for all sinusoids with frequency not equal to w0, the filter
>>>>would return the input. For a sinusoid with frequency w0, the
>>>>output of the FIR would be equal to zero, and so would
>>>>therefore also be the output of the inverse IIR. Here is a
>>>>block diagramm:
>>>>
>>>>x[n] ---> H1(z) ----> u[n] ----> 1/H2(z) ----> y[n]
>>>>
>>>>A real notch filter, like r b-j's, has
>>>>
>>>>H1(z) = 1 - 2 cos(w0) z^-1 + z^-2
>>>>
>>>>and
>>>>
>>>>H2(z) = 1 - 2 a cos(w0) z^-1 + a^2 z^-2
>>>>
>>>>with 0 < a < 1. If the filter is excited with a sinusoid at
>>>>w0, ie. x[n] = exp(i w0 n), then u[n] and therefore (why?)
>>>>y[n] are equal to zero. However, if a = 1, clearly u[n] is
>>>>still equal to zero. What about y[n]? An instable filter
>>>>excited with zero should still yield zero output, no?
>>>
>>>As you observed before, the output is *indeterminate* as
>>>long as no initial conditions are imposed.
>>
>>But this is also the case for 0 < a < 1.
>
>
> Note that I did not refer to any particular coefficient value.
>
>
>>u[n] is equal to zero, and therefore y[n] is not determined.
>
>
> y[n] is not determined without IC's whatever u[n] is.
>
>
>>However, everybody agrees that y[n] is equal to zero for the
>>standard notch filter, if excited by a sinusoid at the notch
>>frequency.
>
>
> The standard notch filter output decays to zero independent of the
> IC's, but it's not generally identically zero. True, the steady-state
> response to an actual infinite sinusoid is identically zero, but zero
> IC's at some finite starting time may be taken just to summarize the
> infinite history in that state of affairs. In other words, upon
> observing that a sine goes in and zeros come out we're unable to
> distinguish whether the filter has been running forever or has been
> kicked off with zero states a second ago. Not sure if that's germane
> to your concern, though.
Once again, math obscures underlying simplicity. It's evident that the
inverse of a filter with a real zero can't be stable: it has a real
pole. You don't need an input to excite such a filter. It's called an
oscillator.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������