# [sdiy] 90-degree phase displacement network calculations

Brian Willoughby brianw at audiobanshee.com
Sun Jan 10 03:08:57 CET 2021

```I'm confused. (or at least I was)

Iteration seems to be tautologically equivalent to repeating specific steps until a desired result is found, without knowing the specific number of steps that will be required in advance. Perhaps I am thinking only of the computer science sense of the term, iterate.

David, are you distinguishing between iterative processes that are infinite approximations as opposed to iterative processes that converge on an exact answer? Searching my question for myself, I see that "closed form" is basically defined as a finite iteration, although there is some disagreement as to which operations and functions are considered valid for that definition. I can see why limits, differentials, and integrals are excluded, because they're basically infinite even though precise.

In any case, I'm seeing how computer science terms and mathematics terms are not always consistent across both fields. And here I was already thinking that it's tough enough for us electrical engineers because mathematics wants 'i' to stand for the square root of -1, while physics and EE wants 'i' to stand for current, and thus choose 'j' to stand for the square root of -1.

Brian

On Jan 9, 2021, at 17:16, David G Dixon <dixon at mail.ubc.ca> wrote:
> 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.
>
> <LandenTrans.png>    <PoleCalc.png>
>
> 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.

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