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Homework Help: Closed Loop transfer function

  1. Oct 20, 2015 #1


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    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    I don't think you need to see the problem to figure this one out, if needed I will post it (takes up a lot of space in the thread).

    I know for a PI controller, the controller transfer function is given

    $$ g_{c} = k_{c} \frac {\tau_{I}s+1}{\tau_{I}s} $$

    and the closed loop transfer function is given

    $$ g_{CL} = \frac {g_{p} {g_{c}}}{1+g_{p}g_{c}} $$

    So I multiply,

    $$ g_{CL} = \frac { (\frac {-0.2735}{s^{2}+6.035s+4.146}) (k_{c} \frac {\tau_{I}s+1}{\tau_{I}s}) }{1 + (\frac {-0.2735}{s^{2}+6.035s+4.146}) (k_{c} \frac {\tau_{I}s+1}{\tau_{I}s})} $$

    However, I don't know what ##k_{c}## or ##\tau_{I}## are, so I am wondering if there is some what I should know these? Then I can have a numerical answer for part (a).
  2. jcsd
  3. Oct 21, 2015 #2


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    Your expression as for gc is not a PI-controller but an ID-controller.

    Generally you could write the transfer function for a PID-controller:

    H(s) = k * ( s + a ) / ( s + b ) , where

    k*a/b = the proportional amplification.
    a = 1 / τd.
    b = 1 / τi.

    So if b=0 you cannot speak of a "proportional amplification".

    But the rewriting of H(s) doesn't solve the problem, because: What are k, a, b ?

    Well, you are the one to choose the values, so that the closed loop will become stable: Calculate the characteristic equation with k, a, b algebraically included, plot a root locus varying k, a, b and determine the ranges wherein the system will be stable.

    Remember that if no steady state error is accepted, b must be zero. ( "Clean" integration of an error ).
  4. Oct 21, 2015 #3
    Please post the full problem so we know the feedback configuration.

    The task in (b) is to find suitable values for those controller parameters using the Routh–Hurwitz stability criterion.

    Just leave them as variables for now and reduce the closed-loop transfer function down to a form where the numerator and denominator polynomials have simple coefficients.
  5. Oct 21, 2015 #4


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  6. Oct 21, 2015 #5


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    That is the expression for a PI controller as given in the textbook.
  7. Oct 21, 2015 #6
    It's a common form of a PI, i.e:
    k_c\frac{\tau_I s + 1}{\tau_I s} = k_c + \frac{k_c}{\tau_I s}
  8. Oct 21, 2015 #7


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    I'm not the one, that has written your textbook. :smile:

    Divide numerator a denominator by τI, then at least the zero of the numerator will become: s = -1 / τd

    P = k*a / b = k*a / 0 doesn't make sense ( when b= 0 ).
  9. Oct 21, 2015 #8


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    I really don't know how to comment that form ( a sum of elements ).

    I don't "feel" that it is "fit" for calculations.

    Writing a PID-controller in the form: H(s) = k * ( s + a ) / ( s + b ), I can immediate plot a zero and a pole in my root locus. How will I plot kc + . . . ?
  10. Oct 21, 2015 #9
    I don't know what you're on about. This transfer function:
    g_{c} = k_{c} \frac {\tau_{I}s+1}{\tau_{I}s}
    represents the transfer function of a PI and it's perfectly fine.

    And this:
    doesn't make any sense. You can plot the root locus for a single parameter, and the set of root loci produced by varying one parameter and holding the others constant have no meaningful relationship to the necessary and sufficient condition for the stability of the OP's system.

    Also, the problem statement specifically asks to apply the Routh–Hurwitz stability criterion, and not root-locus analysis.
  11. Oct 21, 2015 #10
    I wrote it in that form to make it easy for you to recognize the proportional and integral terms, i.e, if you have:
    C(s) = H(s)E(s) = \left(k_c + \frac{k_c}{\tau_I s}\right) E(s) = k_c E(s) + \frac{k_c}{\tau_I s} E(s)
    where ##C(s)## and ##E(s)## is the controller output and system error, respectively.

    So you have a term which is proportional to the error ##k_c E(s)##, and you have a term which is proportional to the integral of the error ##\frac{k_c}{\tau_I s} E(s)##. That's a PI controller.

    It is, however, also a valid transfer function.
  12. Oct 21, 2015 #11


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    Can't help you.

    The expression: H(s) = ( k * ( s + 1/τd ) / s ) suggests a zero at ( -1/τd ; 0 ) and a pole at ( 0 ; 0 ).
    That's what I'm "on about".

    I can plot a lot of root loci, varying k and τd, not simultanious but in steps as for the latter ( maybe changing plot-color as for the latter).
    I can do it in 5 minutes ( I have a computer ), and that's how I do it.

    It's absolutely meaningful.
  13. Oct 21, 2015 #12
    Is this:
    H(s) = k\left(s + \frac{1}{\tau_d s}\right)
    is what you mean, then this:
    isn't right.

    Maybe you'll show me what you have in mind when the OP has solved the assignment using the Routh–Hurwitz stability criterion.
  14. Oct 21, 2015 #13


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    I learned about Routh-Hurwitz many years ago (1976?), but when a PC became available, I forgot everything about it. So now I use root-locus.

    It seems to me, that Routh-Hurwitz is a stable/not stable test, but when using root locus you can see how stable it is, what is the minimum damping ratio at which frequency, and so on, you can add a zero-pole pair near some location, preventing a local problem. That's why I suggest root locus to be used.
  15. Oct 21, 2015 #14
    I still find a lot of uses for the Routh-Hurwitz stability criterion for high-order systems. A modern computer algebra system is a great help in producing Routh arrays.

    I agree, for one parameter.

    If you have some transfer function with parameters A and B, then you can iterate over values of A, and for each value:
    1. Plot the root locus.
    2. From the root locus, record the set of values of B for which the system is stable.

    From that table, you might be able to guess a necessary and sufficient condition for the stability of the system. I'd wager, though, that it's more complicated than you make it out to be.

    Then you can try it with three parameters and feel the swift hand of dimensionality!
  16. Oct 21, 2015 #15


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    I'm glad everyone is learning from my homework problem. So I simplified my expression in my OP to

    $$ g_{CL} = \frac {-0.2735k_{c}(\tau_{I}s+1)}{\tau_{I}s^{3}+6.035 \tau_{I}s^{2} + (4.146 \tau_{I} - 0.2735k_{c} \tau_{I})s - 0.2735 k_{c}} $$

    So my characteristic equation is

    $$ \tau_{I}s^{3}+6.035 \tau_{I}s^{2} + (4.146 \tau_{I} - 0.2735k_{c} \tau_{I})s - 0.2735 k_{c} $$
    The first necessary criteria is that all coefficients are positive, therefore ##\tau_{I} > 0##, and ##4.146 \tau_{I} - 0.2735k_{c}\tau_{I} > 0##, and ##-0.2735k_{c} > 0##.

    This is weird, so it means ##k_{c} < 0##, and from the second coefficient, ##k_{c} > 15.159##, so it doesn't give a proper range of values for ##k_{c}##, I was expecting something like ## a < k_{c} < b##
    Last edited: Oct 21, 2015
  17. Oct 21, 2015 #16
    The first rule of the test is that all coefficients must be non-zero. The second rule is that they must all be positive or they must all be negative.

    I think you forgot to flip the inequality sign (divide by a negative number).
  18. Oct 21, 2015 #17


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    Yes, I see, so ##k_{c} < 15.159##. Well I just thought of something weird.

    ##-0.2735k_{c} > 0 ## and divide through, ## k_{c} < 0 ##

    However, I start with the same and add to the other side, ##0 > 0.2735k_{c}##, then divide through, ##0 > k_{c}##. This is strange
  19. Oct 21, 2015 #18
    Why? In both cases you have ##k_c < 0##.
  20. Oct 21, 2015 #19


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    Woops, my mind is in other places right now :rolleyes:

    So ##k_{c} < 0##, and ##k_{c} < 15.159##, so once again I don't have a range like ##a < k_{c} < b##
  21. Oct 21, 2015 #20
    You have:
    a - b k_c > 0 \Leftrightarrow -b k_c > -a \Leftrightarrow k_c < \frac{a}{b},\quad a >0, b > 0
    If ##k_c < 0## then ##k_{c} < 15.159## is also satisfied.
  22. Oct 21, 2015 #21


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    I get that, but I don't have a range for ##k_{c}##, so some regime where it is stable
  23. Oct 21, 2015 #22
    You're not done yet. You still need to compute the Routh array and apply the third rule of the test.

    To clarify:
    Rule 1 and 2 are necessary conditions that you can check first before you go to the trouble of calculating the Routh array.

    If you're, for instance, calculating the coefficients of the characteristic polynomial by hand, and find one that's zero, then you can stop there - the system is unstable.

    When you have all the coefficients, then you can apply Rule 2 to see if you're able to stop there.

    If not, then procede to Rule 3 of the test. The rules are ordered according to the effort it takes to apply them.
    Last edited: Oct 21, 2015
  24. Oct 21, 2015 #23


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    Okay, here is my routh array

    $$ \begin{bmatrix} \tau_{I} & 4.146\tau_{I}-0.2735k_{c} \tau_{I} & 0 \\ 6.035 \tau_{I} & -0.2735 k_{c} & 0 \\ 4.146 \tau_{I} + (0.045 - 0.274 \tau_{I})k_{c} & 0 \\ -0.2735k_{c} \end{bmatrix} $$

    since I know ##k_{c}## is negative, then ##0.274\tau_{I}-0.045 < 0##, therefore ##\tau_{I} < 0.164##

    So I know the range for the time constant is ##0 < \tau_{I} < 0.164##

    I suppose ##0.274 \tau_{I} - 0.045## is negative, so I divide and get

    ##k_{c} > \frac {4.146 \tau_{I}}{0.274 \tau_{I} - 0.045} ##
    so my gain range is ##\frac {4.146 \tau_{I}}{0.274 \tau_{I} - 0.045} < k_{c} < 0##
    Last edited: Oct 21, 2015
  25. Oct 22, 2015 #24


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    That's not how you do it, using root locus:

    1. Plot the root locus.
    Now, look at the curve. If there are problems problems, you can move/add poles and zeroes that twist the path of the curve. A pole will repel the path, a zero will attrack the path. It's like planning a slalom track by placing red and blue marks:
    Having moved/added poles/zeroes, just redraw the plot: Is it satisfactory? Maybe this zero should be moved a little more to the left ( e.g. dragging it graphically with your mouse )?
    With a little training, you can systematically drag 8 poles and 8 zeroes in turns and redraw/inspect the result.
    As I said: You can do it in 5 minutes ( well, maybe in 10 minutes as for 8 poles/zeroes ).

    2. I don't have to record tables. It's more convenient just to look at the root locus plot.
    The stable area is to the left of the imaginary axis as for the s-plane.

    Attached Files:

    Last edited: Oct 22, 2015
  26. Oct 22, 2015 #25
    That's not necessarily true for ##\tau_I##.

    You have the condition for stability (there's another for ##k_c > 0, \tau_I < 0##):
    k_c < 0, \tau_I > 0, 4.15\tau_I + (0.0453 - 0.274 \tau_I) k_c > 0
    You could pick a very large value for ##\tau_I## that makes ##0.0453 - 0.274 \tau_I## negative, so ##k_c## must be less than some positive number, but you already know that ##k_c < 0##.

    So for a value of ##\tau_I## that makes ##0.0453 - 0.274 \tau_I## negative, the requirement for stability is just that ##k_c < 0##.

    Your result looks good for the case when ##0.0453 - 0.274 \tau_I## is positive.
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