Transfer function reduction math

[bitrex]

The "Lecture Notes in Control and Information Sciences" can be a
fascinating book to page through, particularly if you can get your hands
on a hard copy.

The particular one I'm looking through is from the early 1980s and
there's a lot of interesting material related to optimization problems
here, like "Optimal turning strategy for a supercruiser" (aircraft) and
"Optimal maintenance policy and sale date for a machine with random
deterioration and subject to random catastrophic failure"...

A fair bit of the mathematics assumes a certain baseline knowledge of
the field of systems optimization/linear programming/etc and I don't
easily follow most papers, but there are some papers of interest to
electrical engineering, e.g. one about reduction of order of transfer
functions using a minimum-phase approximation, higher order transfer
functions sometimes contain more information than you need for a
restricted bandwidth.

Unfortunately partly due to the pre-Latex typesetting e I'm unclear what
this one is saying exactly:

<https://imgur.com/a/9HIKOEN>

H(s) is just a regular s-domain transfer function with polynomials top
and bottom, so they decompose it into odd and even parts and set it
equal to...what's tanh phi(s) supposed to mean? Tanh(s)phi(s)?
Tanh(phi(s))?

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).

[Don]

bitrex wrote:

<snip>

Post by bitrex
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,

[bitrex]

Post by Don
<snip>

Post by bitrex
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).

Right, I see that.

Post by Don
A clearer scan may enable me to continue.
Danke,

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.

[bitrex]

Post by Don
<snip>

Post by bitrex
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.

[Don]

Post by bitrex

Post by Don
<snip>

Post by bitrex
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)
<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.

# # #

"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,

[bitrex]

Post by Don

Post by bitrex

Post by Don
<snip>

Post by bitrex
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)
<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.
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.

Post by Don
# # #
"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>

[bitrex]

Post by bitrex

Post by Don

Post by bitrex

Post by Don
<snip>

Post by bitrex
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)
<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.
     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

oops, odd*odd will also give an even function.

[bitrex]

Post by bitrex
even and odd parts in (4), I think plugging that form into
tanh(arctan(z)) = (e^z - e^-z) / (e^z + e^-z)

tanh(arctan(z)) = [e^(arctan(z)) - e^(-arctan(z))]/[e^(arctan(z)) +
e^(-arctan(z))], rather.