Transfer function reduction math

Jun 11, 2024 Last reply: 2 years ago 7 Replies

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:



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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).


<snip>

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,

Right, I see that.

Sure, here's the full page in question:

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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.

(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:

formatting link

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,

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.

This one:

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

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

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