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Input Pole and Forced Response

  1. Sep 10, 2011 #1
    I am reading a book on Control Theory by Norman Nise.

    I don't get his description of forced and natural response based on poles and zeroes.

    I have posted a screenshot of the page http://www.flickr.com/photos/66943862@N06/6134272368/sizes/l/in/photostream/" [Broken].

    I understand upto Eq 4.2.

    I do not understand how he arrives at Points 1,2,3 & 4 after Eq 4.2

    I guess "Pole of the Input function" means when s = 0.
    How does this generate the form of a forced response?

    Likewise for points 2, 3 & 4?

    If anyone can help, it would be great - I spent a lot of time trying to figure this out. No other text books have this kind of description.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 10, 2011 #2

    jim hardy

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    maybe you'd show the paragraphs where he defines "Forced" and "Natural" response.
     
  4. Sep 11, 2011 #3

    jim hardy

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    Philby

    see if this is any help

    www.dspguide.com/CH32.PDF

    the Laplace Transform and S-plane are for me just tools that i do not understand but have learned to use..
    I can use them to solve problems, just as a monkey uses a stone to open a nut without understanding the physics behind momentum and impulse.

    So for me to attempt to explain them to you would be less than honest, i'd be faking it.

    I think your author is trying to demonstrate the "WHY it works" before the "HOW" - and often it's easier to learn the other way round.

    Thanks for the question - this is an area i have wanted to re-visit for thirty years. You got me going again. That Chapter 32 link looks intriguing.
    Maybe one of our more mathematically-enabled denizens has some words of wisdom.

    old jim
     
  5. Sep 11, 2011 #4
    I did quite a few control system designs. I did study Laplace transform and all. But the main tool I use is still Bode Plot. I simply draw the graph out, work on the poles and zeros to get a one pole crossover at 0dB loop gain. I simply treat taming a close loop system like taming an opamp.

    Actually finding the characteristic of poles is the hard part. Particular if you mix mechanical devices, filament heating delay in temperature control etc., that they are not clear cut like a pole created by a cap or an amp roll off. You have to put in as constant dead time and complicates the whole thing. Once you characterizing all the poles, a Bode plot works just fine.....at least for me in over 5 different projects.
     
  6. Sep 11, 2011 #5

    jim hardy

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    I have to tie everything to something physical.

    Those terms in your transfer function will each do something, that is perform a math operation, to an input signal.

    If there is as "s" in the numerator it will attempt to differentiate the input.
    If there is a number added to the "s" in numerator it will pass the DC component of a signal.
    If there is an "s" in denominator it will attempt to integrate the input.
    If there is a number added to the "s" in denominator the output will approach final value exponentially.


    If there is a "s+n", in a term, then there is a time constant associated with the operation it is going to perform.
    The time constant is usually called "tau", greek for t.

    I am accustomed to the terms having form :
    tau * s + 1, usually written ts+1

    but often as in your case they are written s + 1/tau instead - author's choice.

    Your transfer function

    s+2
    -----
    s+5

    i would have written as

    .5s+1
    ------ X (2/5)
    .2s+1

    The 1 in numerator tells me that for any nonzero input there will be a nonzero final value, ie it passes the DC component.
    and the .5 there says there'll be a transient component related to how much the input moves in .5 second


    and the (.2s + 1) denominator tells me the output will approach final value with a .2 second time constant.

    For a step input,
    the ratio of [Tau numerator/Tau denominator] tells how large the initial transient will be, 2.5 X the step. After getting multiplied by that (2/5) in parentheses, 2.5 X (2/5) = 1 (how convenient.)

    i say the output will step to 1 and exponentially approach final value with .2 sec time constant, and final value will be 1X (2/5) = .4

    so i say: output will step to 1 and decay to .4 with .2 sec time constant.

    That's just how i have learned to use these things. Like a stone ax.

    ----------

    Let's see what he did:
    He writes for output eq 4.2 :
    2/5 + 3/5e-5t

    2/5 is the final value

    and e-5t is a transient with 0.2 sec time constant, .2 being tau from denominator

    the e-5t term starts out at 1 and decays to zero, so when multiplied by his 3/5
    starts at 3/5 and decays to zero

    so output is the sum of:
    DC gain component of 2/5
    plus a transient starting at 3/5 and decaying to zero

    so output steps to 5/5 and decays to 2/5 on a .2 sec time constant
    same as:::: steps to 1 and decay to .4

    _----------------------------

    point being not a mathematical tap-dance

    but

    do not despair
    this technique is handy for shortcutting around some of that algebra

    and it's real handy for building opamp circuits
    a resistor in parallel with a cap makes ts + 1 , and t= r X c
    so your transfer function could have been realized with a single op-amp, two resistors and two caps
    and it's what is called in the trade a "Lead-Lag" and is used for aligning phase

    and soon you'll be evaluating stability from just looking at the graphs

    and if math challlenged me can learn to use this, you will doubtless go me one better and actually understand it.

    I hope somebody else chimes in, i'm actually embarassed at my lack of knowledge here. .



    old jim
     
    Last edited: Sep 11, 2011
  7. Sep 11, 2011 #6

    jim hardy

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    Yungman -

    my friends all preferred Bode plots.
    But the youngsters coming out of school today can actually use this stuff.

    "Math is a young man's game " g h hardy, "a mathematician's apology"
     
  8. Sep 12, 2011 #7
    I have been using Bode Plot for a long time, I studied Laplace Transform only 6 years ago. I understand it. But I just never really try it out. Bode plot serve me very well so far and the important thing is to get the job done.

    As I said, the hard part is not the math, it's to determine how to model some of the poles in the system.

    One thing I always do that is different from a lot of the circuits I saw. I always use two to three opamps cascade to form the forward gain of the loop so I can have total control of all the forward poles and zeros. For example say I use gain of 10 on each amp and cascade three to get forward gain ( so called open loop gain in conventional sense) of 60dB. The difference of using it this way is because the frequency response of the forward chain is very high and predictable. Then I roll off the forward gain by injecting my own poles and where the zeros. So I can have absolute control in how I want to roll off the forward gain. A lot of the circuits use the opamp open loop gain and add onto it. It is much harder to control and also any other pole you add will further decrease the frequency response of the circuit.

    One thing, I still refuse to surrender to the young yet!!! I took the the first class in the last 28 years, the ODE class three years ago and I still competed to get the first in the class. I got second only because the professor allow student to drop one out of four exams and the guy that got the first blew out on one and got only 80. He got 100 on the other three. I got 97 average on all four only due to stupid mistake on everyone of them!!!! I should have got the first if they count all four. No I refuse to give in to the young!!!! I don't go to school because The two good college near me are Stanford U and Santa Clara U, both are super expensive and I don't want to pay that kind of money to go there at my age. San Jose State is pretty much a joke. I communicated with the professors in SJSU and study the EM and did the homework the first go around. It was so easy that I don't think student really understand EM with that book. Then I went through the second book and then the third book, so I repeat 3 times just to get a feel of EM. No I don't surrender to the young....even with half my memory!!!!:rofl: I finished PDE and I still plan on studying Complex analysis and Numerical analysis in the future after antenna before I tackle the J D Jackson Classical Electrodynamics.....the holy grail for EM.
     
    Last edited: Sep 12, 2011
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