[sdiy] 90-degree phase displacement network calculations

[David G Dixon]

Yes, exactly.  It's RECURSION, not ITERATION.
 
Iteration implies that you have to converge to an answer by a successive
approximation.  This process is exact and you know the answer.
 
I did a bit more research into the Landen transformation and discovered
that, for each successive step, the argument of the function is changing by
a very simple formula.  If the sn on the next step down the table is sn(v),
and the one above is sn(u), then u = (1+k)v.  Hence, all one needs to do is
to calculate the argument for the top row of sn functions, and then one can
use a formula to find the actual function values.  I'll demonstrate this
with a table:  Note that u = (2n-1)K/32 or (2n-1)(2K)/64.  The sn(u) values
at the bottom were calculated directly from these arguments by the
approximation formula given.
 
			32	 1	 3	 5	 7	 9	 11
13	 15	
2K	 i	 k	 k'	 sn(K/32)	 sn(3K/32)	 sn(5K/32)
sn(7K/32)	 sn(9K/32)	 sn(11K/32)	 sn(13K/32)	 sn(15K/32)

21.1933	 0	 1.000000	 0.000100	 0.319549	 0.758823
0.929639	 0.980794	 0.994857	 0.998630	 0.999635
0.999903	
10.5966	 1	 0.999800	 0.019998	 0.164092	 0.459577
0.679394	 0.820798	 0.903443	 0.949062	 0.973452
0.986269	
5.2988	 2	 0.960788	 0.277283	 0.084258	 0.248264
0.399666	 0.532763	 0.644823	 0.735785	 0.807437
0.862536	
2.7024	 3	 0.565823	 0.824527	 0.053899	 0.160873
0.265418	 0.366039	 0.461422	 0.550469	 0.632324
0.706368	
1.7259	 4	 0.096175	 0.995364	 0.049181	 0.147064
0.243512	 0.337584	 0.428367	 0.514981	 0.596592
0.672415	
1.5744	 5	 0.002323	 0.999997	 0.049068	 0.146731
0.242980	 0.336890	 0.427556	 0.514103	 0.595700
0.671559	
1.5708	 6	 0.000001	 1.000000	 0.049068	 0.146730
0.242980	 0.336890	 0.427555	 0.514103	 0.595699
0.671559	
1.5708	 7	 0.000000	 1.000000	 0.049068	 0.146730
0.242980	 0.336890	 0.427555	 0.514103	 0.595699
0.671559	
1.5708	 8	 0.000000	 1.000000	 0.049068	 0.146730
0.242980	 0.336890	 0.427555	 0.514103	 0.595699
0.671559	
	

			u	 0.331145	 0.993435	 1.655724
2.318014	 2.980304	 3.642593	 4.304883	 4.967173

			sn(u)	 0.319549	 0.758823	 0.929639
0.980794	 0.994857	 0.998630	 0.999635	 0.999903

			tanh(u)	 0.319549	 0.758823	 0.929639
0.980794	 0.994857	 0.998630	 0.999635	 0.999903

 
Since k' is so small, the approximation formula for the elliptic sine may be
used:
 

 
Of course, since (k')^2 is only 1E-8, sn(u) ~ tanh(u).  In fact, this is
true to six decimal places until k' is larger than about 0.001.  Above k' =
about 0.005, the approximation starts to become inaccurate at the sixth
decimal place.  That means that, for all of our PDN networks (where the
bandwidth is always at least 3 decades), the elliptic sine can be taken as
the hyperbolic tangent.  One only needs the Landen transformation to find
the actual quarter-period of the tanh.  That's not iterative.  That's
recursive.  In any case, the Landen transformation technique is nearly as
easy and more accurate.

  _____  

From: Ian Fritz [mailto:ijfritz at comcast.net] 
Sent: Saturday, January 09, 2021 11:42 PM
To: Bernard Arthur Hutchins, Jr
Cc: synth-diy at synth-diy.org; David G Dixon; Brian Willoughby
Subject: Re: [sdiy] 90-degree phase displacement network calculations


[CAUTION: Non-UBC Email] 
Hmmm... If you substitute the first equation into the second, then you have
a recursive definition of the components of the vector k'. I would vote for
calling this closed form, since there is an explicit formula for calculating
each component.

As another example, consider, say, the function e^x.  This may be defined
recursively by its Taylor expansion, but I think most folks would consider
e^x a closed form expression.

Ian
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