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From where comes the pole and zero in real worl...??

Discussion in 'Homework Help' started by koolguy, Sep 30, 2011.

  1. koolguy

    koolguy Active Member

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    I know some easy examples, but this are not for me.
    Please tel how it is calculated..?
     
  2. dragonwarrior

    dragonwarrior New Member

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    the poles are found by equating the denominator of the eqn to zero...
    the zeros are found by equating numerator to zero
     
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  3. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,


    I'll do a little example so you can see how it is done, then you can apply it to your actual problem. This is actually very easy to do as you'll see.

    Say we have a function like this:
    3/(s+4)

    (this is our function after we've calculated the transfer function with all gains included)
    and we want to find the value of the output after a very long time has passed (this is part of how we calculate the error). We assume a unit step input has been applied:
    1/s

    so we multiply our function by this and we get:
    (1/s)*3/(s+4)=3/(s*(s+4))

    Now the Final Value Theorem states that the solution for infinite time (or a very long time after the step has been applied) can be found by:
    f(t=+inf)=s*F(s-->0)

    or in words, the infinite time solution can be found by finding the result of the zero frequency solution multiplied by 's'. So for our little example:
    3/(s*(s+4))

    which is the output with a unit step input applied, we first get:
    s*3/(s*(s+4))

    and after simplification we get the original function back:
    3/(s+4)

    and now we take the limit as s approaches zero:
    lim[s-->0] 3/(s+4)

    and that gives us:
    3/4

    or simply 0.75 as the final value output. That's the final output, and since we had an input that was the unit step, we subtract:
    Error=1-0.75

    and we get:
    Error=0.25

    So the error in this case is 25 percent. If this turned out to be too large, we'd have to go back and try more gain in the circuit.

    So the steps are really quite simple:
    1. Multiply the function F(s) by the unit step 1/s.
    2. Multiply that result by 's'.
    3. Take the limit as s approaches zero, that's the final value.
    4. Subtract the final value from '1', that gives the error.

    Sometimes you have to be ready to handle a sinusoidal steady state output that's the only catch, but in many problems it is just a DC value.
     
    Last edited: Oct 12, 2011
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  4. dave

    Dave New Member

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  5. koolguy

    koolguy Active Member

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    Hi again sir,

    The example was interesting and clearing the basic doubt.
    please give more examples.
     
  6. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,


    Here is another example taken from an actual circuit that can be built and tested or simulated to compare calculated results with the simulated results. The circuit is shown in the diagram attached. For simulation you can make the gains shown as '1' as op amp voltage followers. For the other gain G you can make that a non inverting op amp with the gain set to whatever we want G to be.


    In the diagram we see on top a feedback system. The input is a unit step.

    We make R1=R2=R3=1 and C1=C2=C3=1 later for simplicity.

    The transfer function is:
    G/(1+G*H)

    and H is:
    H=1/(s*R*C+1)^3

    and we'll leave G as G for now.

    So the transfer function is:
    G/(1+G/(s*R*C+1)^3)

    which simplifed is:
    G*(s*R*C+1)^3/(G+(s*R*C+1)^3)

    and replacing R with 1 and C with 1 as above we get:
    G*(s+1)^3/(G+(s+1)^3)

    Multiplying top and bottom out we get:
    (s^3+3*s^2+3*s+1)*G/(s^3+3*s^2+3*s+1+G)

    So we have:
    F(s)=(s^3+3*s^2+3*s+1)*G/(s^3+3*s^2+3*s+1+G)

    Now we compute the output with a step input, we get:
    F(s)=(1/s)*(s^3+3*s^2+3*s+1)*G/(s^3+3*s^2+3*s+1+G)

    Now we multiply F(s) above by s and we get:
    (s^3+3*s^2+3*s+1)*G/(s^3+3*s^2+3*s+1+G)

    Now we take the limit of the above as s goes to zero:
    lim[s-->0](s^3+3*s^2+3*s+1)*G/(s^3+3*s^2+3*s+1+G)

    which equals:
    G/(G+1)

    So the final value output is G/(G+1) which is solved for any gain G.
    We could have make it a tiny bit easier by replacing G with the
    actual gain first, but this way we have the solution for any gain.
    Any gain that is not going to cause an oscillation that is !

    Now the steady state error however is the final value subtracted from 1, so
    we have:
    SSError=1-G/(G+1)

    Since we have this formula now we can take a look at the SSE for several gain
    values and see what we end up with here...

    With G=1, we have:
    SSE=1-1/(1+1)=1-1/2=1/2=0.50

    so we have 50 percent error with a gain of only 1.

    With G=2 we have:
    SSE=1-2/(2+1)=1-2/3=1/3=0.3333

    so we have 33 percent error with a gain of 2, and we can
    note that the error went down already.

    With G=5 we have:
    SSE=1-5/(5+1)=1-5/6=1/6=0.1667

    so we have about 17 percent error now, and that error went
    down once again.

    With G=7 we have:
    SSE=1-7/(7+1)=1-7/8=1/8=0.1250

    so now we have 12.5 percent error and again the error went
    down.

    Now with reserves we make G=8 and so we have:
    SSE=1-8/(8+1)=1-8/9=1/9=0.1111

    and it LOOKS like the error went down again,
    but the problem now is that the poles moved either onto the j axis or
    into the right half plane (see lower diagram in the attachement),
    so we can not use this gain in the circuit (unless of course we are
    building an oscillator, which we do not consider to be a valid solution
    when we are looking for the steady state error).


    So in conclusion we see that we can get better and better steady state
    error by increasing the feedforward gain, but at some point the circuit becomes
    unstable so we can not go any higher with the gain.
     

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    Last edited: Oct 12, 2011
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  7. koolguy

    koolguy Active Member

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    What the use of multiply function by's'??

    Please give some hint to calculate this transfer function..
     
  8. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,

    Multiply by 's' because that's part of the Final Value Theorem. If you want to use that theorem, you have to multiply by 's'.

    The transfer function is obtained by looking for what G is and what H is, then calculating:
    G/(1+G*H)

    G is whatever we make the gain, and H is found by analyzing the section with the resistors and capacitors.
    Each individual section is 1/(s*R*C), and there are three sections, and the three convolved come out to
    1/(s*R*C)^3

    so that's what H is. Having now both G and H, we use the above G/(1+G*H) to calculate the response.
     
  9. koolguy

    koolguy Active Member

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    Hi again sir,

    I am confused that num. are zero and den. are poles, but in your fig. how you have calculated or mark that poles and zero on real and imag. axis?
     
  10. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,


    Here's the thing...

    When the denominator of an equation goes to zero, the equation goes to infinity, and we call that a 'pole'.
    For example, in this simple equation:
    V=1/s

    if we allow s to approach zero the equation becomes (informally):
    V=1/0

    and as you know when you divide anything other than zero by zero you get an infinity (pole), so we say we have a pole at s=0.

    Now with a pole at s=0 we note that we have no imaginary part there because s is really to be interpreted as a complex variable:
    s=0+0*j

    so the point to plot would be at the origin of the imaginary plane, which would be (0,0).

    For the equation:
    V=1/(s^2+s+1)

    we find the roots of:
    s^2+s+1=0

    and now we see that we have two complex roots:
    s=-0.5+0.866*j
    s=-0.5-0.866*j

    so we would plot two roots, one at the point (-0.5,+0.866) and the other at (-0.5,-0.866).

    So you see how we get the points to plot now? You know how to find the roots of an equation?

    There's one catch. If the numerator has a zero that cancel a pole, theoretically we assume that it cancels perfectly. In real life this doesnt always work as neatly as that though, but assuming it does for now say we have:
    V=(s+1)*(s+2)/((s+2)*(s+3)*(s+4))

    in the above we see that the numerator has (s+2) and so does the denominator, so we can cancel the pole caused by (s+2). Of course this simplifies then to:
    V=(s+1)/((s+3)*(s+4))

    so here we would not plot the pole caused by (s+2) because the matching zero in the numerator canceled it.

    You know how to find the complex roots of an equation?
     
    Last edited: Oct 14, 2011
  11. koolguy

    koolguy Active Member

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    Yes, i have done it..but if you have an easy method you can tell...
     
  12. koolguy

    koolguy Active Member

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    Yes, i have done it..but if you have an easy method you can tell...
     
  13. koolguy

    koolguy Active Member

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    One thing more:-
    X' = Ax + Bx
    Y = Bx + DU

    what the use of D in actual...?
     
  14. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,


    Wow you seem to be changing subjects a bit in this thread :)

    I think you meant this:

    X'=Ax+Bu
    Y=Cx+Du

    correct?

    D is the feedthrough gain. That is, it is the gain that bypasses everything else in the system and gets summed directly to the output. This is sort of evident by looking at the equation that shows that Du sums directly to Cx and becomes part of the output as just a simple sum of a gain times the input vector.
    In many systems D is not present, but then again often C is just an identity matrix or we dont even really think about it.
     
    Last edited: Oct 14, 2011
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  15. koolguy

    koolguy Active Member

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    Hi,
    we also find something like this:-
    -1
    C(s) = A*(S*I - A)*B ... can you tell more about this..in practical example.
    R(s)
     
    Last edited: Oct 14, 2011
  16. Jugurtha

    Jugurtha Member

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    Ritesh,

    I sent you a message the other day, to which you replied... I thought you were a Normal human being with lots of questions first. I even made you an analogy saying "It's like helping a blind man painting a Gioconda".


    I already answered another thread of yours that you drifted, like this one... It is crystal clear to me now that you absolutely know NOTHING at all about ANYTHING.

    There is a minimum amount of knowledge required to even ask a question about a topic. You don't have that minimum amount.

    I have checked your posting history, and it is impressive. Only questions, very vague questions about things you absolutely ignore the very minimum to entitle you to ask a question.

    I am not used to pointing things out this way, but you are some piece of work.

    Brownout asked you if you had Control Theory and Differential Equations down, you said you have that down. That's the only reason I answered the way I did. Obviously, you know nothing about these topics.

    Each time you get an answer, you pick a key word and ask a question about that (like you did on damping), then someone posts another answer, and you pick another word and ask about it.

    Each time a person asks you if you know how to something, you say "Yes, but can you give another way".. You know nothing, and it is a pattern.

    I invite everyone who posted on this thread to check your posting history. 5 pages threads, lots of messages and everytime, the same pattern.

    You are a waste of bandwidth, a waste of time.

    To whoever is moderating, I am sorry if I crossed the line, although to my defense, I am just stating facts that could easily be checked, I am not difamating with wrong accusations.. Either the pattern of this amusing fellow, his questions, his lack of any basics whatsoever just irritate me.

    I am not going to say much, other than the fact I wonder how you managed to not get kicked out of this forum...

    All my best, "Dear Sir" or whatever expressions you use that make me dumber every time I answer you...
     
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  17. koolguy

    koolguy Active Member

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    I like really..........
    see my interest is to see in practical uses of theory not only understanding without any use of it in real work, so i try to find its application...
    due to which most of people don't like this attitude and take my question in -ve way....
     
  18. koolguy

    koolguy Active Member

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    Reply please Mr al........
     
  19. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again Ritesh,

    I'll try and get back here tomorrow with a simple example to show how this works. I've been a bit busy lately so my time got used up :)

    Friend had a small problem with something i built a long time ago too today so i had to go over and fix that. Something broke so it had to be recalibrated.

    We can start with:




    For the circuit shown in the attachment, we have:

    A=-1/RC, and
    B=1/RC, and
    u=E, and so we have:






    Next, we'll look at the form of the transform to move into the frequency domain directly from the state vector differential equation.

    To get into the frequency domain we can use this form:



    where

    is the identity matrix of the same order as the system,

    are the same matrices as previously described.

    With A equal to [-1/RC] and B equal to [1/RC] the order of the system is 1, so the identity matrix of this system is simply [1]. This gives us:



    Since the inverse of s+1/RC is RC/(s*RC+1), we have:



    So the frequency domain representation of the state equation is:




    Question: How does this result compare with a straight up AC analysis of the circuit in the attachment?
    Answer: Rest assured, it is *exactly* the same.
     

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    Last edited: Oct 20, 2011
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  20. koolguy

    koolguy Active Member

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    Hi again sir,

    The example can be solved with transfer function also but why to use Sate variable equation ? as i know we use Sate variable equation for MIMO
    Can you tell me about matrix D..??
     
  21. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,


    You posted a question about the state equations so that's why i made the example like that.
     
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