[sdiy] 90-degree phase displacement network calculations

Ian Fritz ijfritz at comcast.net
Tue Jan 12 02:54:07 CET 2021


Hello Bernie et al --

I just remembered that the old Bell technical papers are free online. 
Here is the link to the Darlington paper:

https://archive.org/details/bstj29-1-94

Ian



On 1/9/2021 9:44 PM, Bernard Arthur Hutchins, Jr wrote:
> David - what you show as the Landen transform (in the box) is EXACTLY 
> called iteration............!   - Bernie
> ------------------------------------------------------------------------
> *From:* David G Dixon <dixon at mail.ubc.ca>
> *Sent:* Saturday, January 9, 2021 8:16 PM
> *To:* Bernard Arthur Hutchins, Jr <bah13 at cornell.edu>; 
> synth-diy at synth-diy.org <synth-diy at synth-diy.org>
> *Subject:* RE: [sdiy] 90-degree phase displacement network calculations
> Hello Bernie,
> No, it is not iterative.  It is closed form.  The step-wise procedure is 
> an implementation of the Landen transformation to enable calculation of 
> elliptic sine from circular sine.  That is all.  If Excel had a function 
> call for elliptic sine, I would use that instead.  Once the elliptic 
> sine is calculated the poles are obtained directly.  Once half the poles 
> are obtained, then you just take the logarithms, set them negative, and 
> take the antilogs to find the other half.
> The reason for the steps is that you must transform the k and k' values 
> using a simple formula until you obtain k = 0 and k' = 1.  At that 
> point, sn = sin, so you can simply find the sines of the angles 
> (2n-1)*(pi/2)/4N and take those as the elliptic sines.  Then, you go 
> back up the chain of k and k' values, calculating the elliptic sines 
> from those below (of which the first one, on the bottom of the table, is 
> a circular sine).  When you get to the top of the table, you have an 
> elliptic sine at the desired modulus.  This is then used to calculate 
> the pole directly.
> To make this more plain, here is a table showing the calculations in Excel.
> 	4N 	32 	1 	3 	5 	7 	9 	11 	13 	15
> i 	k 	k' 	sn(π/64) 	sn(3π/64) 	sn(5π/64) 	sn(7π/64) 	sn(9π/64) 
> sn(11π/64) 	sn(13π/64) 	sn(15π/64)
> 0 	— 	*0.000100* 	*0.319549* 	*0.758823* 	*0.929639* 	*0.980794* 
> *0.994857* 	*0.998630* 	*0.999635* 	*0.999903*
> 1 	0.999800 	0.019998 	0.164092 	0.459577 	0.679394 	0.820798 
> 0.903443 	0.949062 	0.973452 	0.986269
> 2 	0.960788 	0.277283 	0.084258 	0.248264 	0.399666 	0.532763 
> 0.644823 	0.735785 	0.807437 	0.862536
> 3 	0.565823 	0.824527 	0.053899 	0.160873 	0.265418 	0.366039 
> 0.461422 	0.550469 	0.632324 	0.706368
> 4 	0.096175 	0.995364 	0.049181 	0.147064 	0.243512 	0.337584 
> 0.428367 	0.514981 	0.596592 	0.672415
> 5 	0.002323 	0.999997 	0.049068 	0.146731 	0.242980 	0.336890 
> 0.427556 	0.514103 	0.595700 	0.671559
> 6 	0.000001 	1.000000 	0.049068 	0.146730 	0.242980 	0.336890 
> 0.427555 	0.514103 	0.595699 	0.671559
> 7 	0.000000 	1.000000 	0.049068 	0.146730 	0.242980 	0.336890 
> 0.427555 	0.514103 	0.595699 	0.671559
> 8 	0.000000 	1.000000 	*0.049068* 	*0.146730* 	*0.242980* 	*0.336890* 
> *0.427555* 	*0.514103* 	*0.595699* 	*0.671559*
> ** 	** 	** 	** 	** 	** 	** 	** 	** 	** 	**
> ** 	omega 	*sn* 	0.319549 	0.758823 	0.929639 	0.980794 	0.994857 
> 0.998630 	0.999635 	0.999903
> 	100.000000 	*cn* 	0.947570 	0.651297 	0.368472 	0.195047 	0.101294 
> 0.052333 	0.027000 	0.013925
> 		*pole* 	296.533 	85.830 	39.636 	19.887 	10.182 	5.240 	2.701 	1.393
> 	f 	*log pole* 	2.472074 	1.933639 	1.598090 	1.298561 	1.007822 
> 0.719369 	0.431521 	0.143830
> 	628.318531 	*−log pole* 	-2.472074 	-1.933639 	-1.598090 	-1.298561 
> -1.007822 	-0.719369 	-0.431521 	-0.143830
> 		*−pole* 	0.003372 	0.011651 	0.025230 	0.050285 	0.098215 	0.190823 
> 0.370236 	0.718076
> 		*RC* 	*0.471948* 	*0.136603* 	*0.063083* 	*0.031651* 	*0.016205* 
> *0.008340* 	*0.004299* 	*0.002216*
> ** 	** 		*0.000005* 	*0.000019* 	*0.000040* 	*0.000080* 	*0.000156* 
> *0.000304* 	*0.000589* 	*0.001143*
> 
> So, you see, there is no iteration.  You simply have to transform k' 
> until it equals 1.  I have determined that this always occurs within 8 
> steps, as shown.  Here it actually occurred in 6 steps, and the 
> calculated sn values were equal to the sine value until i = 4 to six 
> decimal places.
> The calculation for "p" is closed-form, but there is no more convenient 
> way to calculate elliptic sines.
> ------------------------------------------------------------------------
> *From:* Bernard Arthur Hutchins, Jr [mailto:bah13 at cornell.edu]
> *Sent:* Saturday, January 09, 2021 1:35 PM
> *To:* synth-diy at synth-diy.org
> *Cc:* David G Dixon
> *Subject:* [sdiy] 90-degree phase displacement network calculations
> 
> [*CAUTION:* Non-UBC Email]
> 
> Thanks again David –
> 
> I see. Your function is NOT closed form – it is iterative; 
> self-terminating at 8 “for loops” – absolutely NOTHING wrong with that.
> 
> In equiripple designs we have the powerful “alternation theorem” to tell 
> us WHEN we have “found” an acceptable ER solution (at least - close 
> enough). At the same time, the mathematicians assure us (prove!) that we 
> will never find closed form!(Don’t you hate it when you are shown that 
> they are right?)
> 
> What happens (if you know) if you terminate after just a couple of 
> iterations?  That is: How does convergence progress?
> 
> -Bernie
> 
> 
> 
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-- 
ijfritz.byethost4.com



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