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<DIV dir=ltr align=left><FONT color=#0000ff size=2 face=Arial><SPAN
class=017472418-12012021>I think that the key concept here is that the pole
calculation is an exact, closed-form expression based on elliptic sine and
cosine. Also, the period of the elliptic sine and cosine is given by
the modulus.</SPAN></FONT></DIV>
<DIV dir=ltr align=left><FONT color=#0000ff size=2 face=Arial><SPAN
class=017472418-12012021></SPAN></FONT> </DIV>
<DIV dir=ltr align=left><FONT color=#0000ff size=2 face=Arial><SPAN
class=017472418-12012021>The only "iterative" thing here is that mathematicians
have so far failed to give us a convenient way to calculate the elliptic
functions. However, I have demonstrated that a simple function call in
Excel is possible.</SPAN></FONT></DIV>
<DIV dir=ltr align=left><FONT color=#0000ff size=2 face=Arial><SPAN
class=017472418-12012021></SPAN></FONT> </DIV>
<DIV dir=ltr align=left><FONT color=#0000ff size=2 face=Arial><SPAN
class=017472418-12012021>So, ask yourself this question: What if
there had been built-in Excel functions for sn(u,k) and cn(u,k), and
another function call for the period, such that the functions could have been
calculated by, say, SN(3*SNPERIOD(k)/64,k)? If such functions were
available in Excel already, would there have been any argument that this wasn't
a closed-form solution to the filter pole problem? I would suggest
that there would not.</SPAN></FONT></DIV>
<DIV dir=ltr align=left><FONT color=#0000ff size=2
face=Arial></FONT> </DIV><FONT color=#0000ff size=2 face=Arial></FONT><BR>
<DIV dir=ltr lang=en-us class=OutlookMessageHeader align=left>
<HR tabIndex=-1>
<FONT size=2 face=Tahoma><B>From:</B> Liam Wall [mailto:liam.wall@gmail.com]
<BR><B>Sent:</B> Tuesday, January 12, 2021 2:40 AM<BR><B>To:</B> Guy
McCusker<BR><B>Cc:</B> David G Dixon; synth-diy@synth-diy org; Bernard Arthur
Hutchins, Jr; Brian Willoughby<BR><B>Subject:</B> Re: [sdiy] 90-degree phase
displacement network calculations<BR></FONT><BR></DIV>
<DIV></DIV><SPAN
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Non-UBC Email]</SPAN></SPAN></SPAN>
<DIV dir=ltr>
<DIV>Thanks David for sharing this stuff.</DIV>
<DIV><BR></DIV>
<DIV>Agree with what Guy has said below, and would add to that all the same
applies to standard sine, exp, ln, and sqrt --- these are functions that we
can't compute exactly for all input values, but we know how to compute
sufficiently good approximations in a known (short) time. All considered "closed
form". As far as I can tell "closed form" isn't really a precisely defined
mathematical term. In fact, quoting the opening line of the Wikipedia page "The
set of operations and functions admitted in a closed-form expression may vary
with author and context". So are sn and K in or out? It's pretty much
arbitrary.<BR></DIV>
<DIV><BR></DIV>
<DIV>I've no idea how much accuracy Excel can manage. I think floating point
doubles are 64 bits in most languages? In the interests of science I calculated
the k'(i) series starting with k'(0)=0.0001 using 2048 bit floating point values
(the most I have handy). Excel gave David k'(6)=1.0; all the extra bits gave
k'(6)=0.9999999999990897, and k'(7) == 1.0.<BR></DIV></DIV><BR>
<DIV class=gmail_quote>
<DIV dir=ltr class=gmail_attr>On Tue, 12 Jan 2021 at 09:33, Guy McCusker <<A
href="mailto:guy.mccusker@gmail.com">guy.mccusker@gmail.com</A>>
wrote:<BR></DIV>
<BLOCKQUOTE
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class=gmail_quote>
<DIV dir=ltr>While I agree with David that it is essentially fruitless to
argue over whether something is "iterative" or "recursive", trying to figure
out how I would describe this algorithm gave me the impetus I needed to
study it a bit, so it helped me at least.
<DIV><BR></DIV>
<DIV>From a computational perspective, iteration and recursion are
equivalent, and they both describe processes that may in principle be
non-terminating. But there is a difference in the spirit of how things
are expressed. Iteration somehow means "do these steps again and again" while
recursion means "compute this function in terms of other invocations of
itself". (The computational equivalence means you can always hack one to look
like the other so the distinction is not at all sharp really.) A very common
use of "iteration" is "do these steps again and again until some calculated
error is small enough." With that in mind I looked at the theory and practice
of what David is doing here. </DIV>
<DIV><BR></DIV>
<DIV>The expressions for the poles in terms of elliptic functions are exact.
However they rely on the elliptic integral which requires some work to
compute. David uses the Landen transformation to do this. Again, the Landen
transformation is exact. No approximations are in play... yet.</DIV>
<DIV><BR></DIV>
<DIV>The equations David wrote for computing sn via the Landen transformation
are recursive in spirit: the equations show how to calculate sn( , ) using an
expression ** which also contains calls to sn( , ) **. That is
what gives it a recursive nature in my eyes. The equations also show when
this recursion can terminate: when the "k" term is zero, sn reduces to sin,
which presumably we know how to compute, so we do not need to perform further
recursive calls.</DIV>
<DIV><BR></DIV>
<DIV>However, if computing *exactly*, the recursion will never
terminate, as Ian Fritz pointed out -- the k term in the recursive calls will
never become exactly zero. </DIV>
<DIV><BR></DIV>
<DIV>For practical purposes we don't need it to be exactly zero: it just needs
to be close enough that sin is a good enough approximation to sn. What David's
code does is to compute the sequence of k values until they get close enough
to zero. This computation feels like an iterative approximation, though I
could live with it being called recursive</DIV>
<DIV><BR></DIV>
<DIV>David has determined that 8 steps are enough to get close enough for his
purposes, which may simply mean close enough that Excel thinks the answer
is 0. Once that is accepted, we have a fixed way to calculate good
approximations to the values we are looking for. </DIV>
<DIV><BR></DIV>
<DIV>Well, that's my take on all this anyway. I'm finding this interesting
from both a mathematical and a practical point of view. This list is
great!</DIV>
<DIV><BR></DIV>
<DIV>Guy.</DIV>
<DIV><BR></DIV>
<DIV><BR></DIV>
<DIV><BR></DIV>
<DIV><BR></DIV>
<DIV><BR></DIV>
<DIV><BR></DIV>
<DIV>
<DIV><BR></DIV>
<DIV><BR></DIV></DIV></DIV><BR>
<DIV class=gmail_quote>
<DIV dir=ltr class=gmail_attr>On Mon, Jan 11, 2021 at 11:57 PM David G Dixon
<<A href="mailto:dixon@mail.ubc.ca" target=_blank>dixon@mail.ubc.ca</A>>
wrote:<BR></DIV>
<BLOCKQUOTE
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class=gmail_quote><U></U>
<DIV>
<P><FONT size=2 face="Courier New">Well, Michael, we're basically arguing
about the meaning of the words "iteration" and "recursion" at this point,
and I find this argument to be utterly fruitless.<BR><BR>To me, iteration is
something that is required when approximate solutions are sought and the
criterion is convergence. The solution to the phase displacement
problem is exact. The poles are given by the following closed-form
equation:<BR><BR><IMG border=0 hspace=0 alt="" align=bottom
src="cid:017472418@12012021-113C"></FONT></P>
<P><FONT size=2 face="Courier New">The only recursive part of this whole
problem is that 2K is a function of k' which must be determined by a
recursive process (as far as I know). This is a feature of elliptic
sines and has nothing to do with the filter pole calculation, which is
completely closed form. Elliptic functions differ from circular
functions in that the latter have fixed periods (2*pi), but the former have
periods which are functions of their modulus k. However, the period is
a unique function of the modulus. Hence, when you specify the modulus
you are also specifying the period. However, to actually calculate the
value of the period from the modulus, you need to engage in a recursive
calculation. THIS IS NOT ITERATION. We are not approximating or
approaching some idealized solution by iterating to the correct
period. The period is determined by the modulus -- there is nothing
approximate about it. The solution to this problem is
exact.</FONT></P>
<P><FONT size=2 face="Courier New">How's that for an analytical and
thought-provoking response?</FONT></P>
<P><BR><BR><BR><BR><FONT size=2 face="Courier New">-----Original
Message-----<BR>From: Michael E Caloroso [</FONT><A
href="mailto:mec.forumreader@gmail.com" target=_blank><FONT size=2
face="Courier New">mailto:mec.forumreader@gmail.com</FONT></A><FONT size=2
face="Courier New">]<BR>Sent: Monday, January 11, 2021 3:29 PM<BR>To: David
G Dixon<BR>Cc: Ian Fritz; Bernard Arthur Hutchins, Jr; <A
href="mailto:synth-diy@synth-diy.org"
target=_blank>synth-diy@synth-diy.org</A>; Brian Willoughby<BR>Subject: Re:
[sdiy] 90-degree phase displacement network calculations<BR><BR>[CAUTION:
Non-UBC Email]<BR><BR>Well that was a highly analytical and thought
provoking conclusion<BR><BR>MC<BR><BR>On 1/11/21, David G Dixon <<A
href="mailto:dixon@mail.ubc.ca" target=_blank>dixon@mail.ubc.ca</A>>
wrote:<BR>> Hello Ian,<BR>><BR>> Well, I'm getting a bit tired
about arguing about this, so my official<BR>> answer is...
whatever.<BR>><BR>> Cheers<BR>> Dave<BR>><BR>> -----Original
Message-----<BR>> From: Ian Fritz [</FONT><A
href="mailto:ijfritz@comcast.net" target=_blank><FONT size=2
face="Courier New">mailto:ijfritz@comcast.net</FONT></A><FONT size=2
face="Courier New">]<BR>> Sent: Monday, January 11, 2021 6:44 AM<BR>>
To: David G Dixon; 'Bernard Arthur Hutchins, Jr';<BR>> <A
href="mailto:synth-diy@synth-diy.org"
target=_blank>synth-diy@synth-diy.org</A><BR>> Cc: 'Brian
Willoughby'<BR>> Subject: Re: [sdiy] 90-degree phase displacement network
calculations<BR>><BR>> [CAUTION: Non-UBC Email]<BR>><BR>> That
looks not to be true. The difference between two successive k'(i)<BR>>
values clearly can not be zero. The process is a (rapidly)
converging<BR>> iterative one.<BR>><BR>> In case you can't see
this, the proof is trivial:<BR>> Suppose k'(i) = k'(i-1)<BR>> Then
from the second equation, k(i) = k(i-1) Now the first equation<BR>>
yields 0 = k(i)-k(i-1) = [1-k'(i-1)]/[1+k'(i-1)] -<BR>>
[1-k'(i-2)]/[1+k'(i-2)]<BR>> This can be generally true only if k'(i-2) =
k'(i-1) So by induction,<BR>> all the k'(i) values are the
same.<BR>><BR>> A sequence either iterates or it doesn't -- it can't
just drop dead in<BR>> the middle of the street.<BR>><BR>>
Ian<BR>> (math minor, including some pretty tough analysis
courses)<BR>><BR>><BR>> On 1/11/2021 2:23 AM, David G Dixon
wrote:<BR>><BR>>> ........ There are no "approximate"
answers, and this problem is not<BR>>> one where one gets closer and
closer to the true solution with each<BR>>> step. That would be
an iterative solution, and as I've said ad<BR>>> nauseum, this is not
that problem.<BR>><BR>><BR>><BR>>
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