Re: Transfer function reduction math

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Sujet : Re: Transfer function reduction math
De : user (at) *nospam* example.net (bitrex)
Groupes : sci.electronics.design
Date : 12. Jun 2024, 05:05:52
Autres entêtes
Message-ID : <66691ea0$0$954$882e4bbb@reader.netnews.com>
References : 1 2 3 4
User-Agent : Mozilla Thunderbird
On 6/11/2024 11:17 PM, Don wrote:
bitrex wrote:
Don wrote:
bitrex wrote:
>
<snip>
>
Seems like they're doing some kind of tanh interpolation but it's not
entirely obvious to me how they get from equation (3) to the expression
in (5).
>
Unless the fuzzy form of your scan deceives my eyes, it appears the
numerator and denominator are multiplied by the conjugate to obtain
(4) from (3).
>
A clearer scan may enable me to continue.
>
Danke,
>
>
(not sure if my response posted as I don't see it on my newsreader,
apologies if this reply appears twice)
>
Sure, here's the full page in question:
>
<https://imgur.com/a/as3jfNo>
>
I have a hardcopy from an academic library which is a relatively massive
(800+) page tome so difficult to get a good scan of...the only full-text
online I can find is on Springerlink (blech) and despite my having an
"institutional login" that should grant access to it. it never seems to
work with them.
 Your first followup was indeed posted.
 The first three steps from (4) to (5) are easy-peasy:
      tanh(s) = (e^s - e^-s) / (e^s + e^-s)
     H(s) = Q(s) / D(s) = (e^s - e^-s) / (e^s + e^-s)
     (e^s + e^-s)Q(s) = (e^s - e^-s)D(s)
 Control Theory must now be reviewed by me in order to continue.
Thanks, I think I see sorta see how (5) is derived now. I believe they mean by their notation tanh(phi(s)) and phi(s) = arctan(Q(s)/D(s)).
The denominator of (4) will be real, and the portions of the numerator that are an even function times an even function will be real and the parts that are anything else will be imaginary, cuz in e^ix = cos(x) + i sin(x) the sin is imaginary and sin is an odd function, when each term in the expanded fraction is expressed as a magnitude and phase angle.
Then there's a logarithmic form of the arctangent, arctan(z) = -i/2*ln[(1 + iz)/(1 - iz)] and for z(s) = Q(s)/D(s) as decomposed into even and odd parts in (4), I think plugging that form into tanh(arctan(z)) = (e^z - e^-z) / (e^z + e^-z) should then give (5), though I haven't grunged it all out to check.

# # #
 "Lecture Notes in Control and Information Sciences" seems to be a series
of books, each about three hundred pages long. Where do you find page
808?
 Danke,
 
This one:
<https://link.springer.com/book/10.1007/BFb0006119?page=6>

Date Sujet#  Auteur
11 Jun 24 * Transfer function reduction math6bitrex
11 Jun 24 `* Re: Transfer function reduction math5Don
12 Jun 24  `* Re: Transfer function reduction math4bitrex
12 Jun 24   `* Re: Transfer function reduction math3Don
12 Jun 24    `* Re: Transfer function reduction math2bitrex
12 Jun 24     `- Re: Transfer function reduction math1bitrex

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