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Electrical Engineering - Control Systems - Design Via Frequency Respon

  1. Mar 22, 2014 #1
    done editing

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

    I'm given the following transfer function that represents a physical apparatus

    [itex]G(s) = \frac{X(s)}{F(s)} = \frac{3.333*10^{4}}{s^{2}}[/itex]

    I'm asked to design a lead compensator to achieve closed-loop stability. The requirements are that it must have a transient response of [itex]16%[/itex] overshoot, a settling time of [itex]2 mS[/itex] for a step input.

    2. Relevant equations

    [itex]ζ = -\frac{ln(\frac{OS}{100})}{\sqrt{∏^{2} + ln^{2}(\frac{OS}{100})}}[/itex]
    [itex]ω_{BW} = \frac{4}{T_{s}ζ}\sqrt{(1-2ζ) + \sqrt{4ζ^{4} - 4ζ^{2} + 2}}[/itex]
    [itex]Θ_{m} = arctan(\frac{2ζ}{\sqrt{-2ζ^{2} + \sqrt{1 + 4ζ^{4}}}})[/itex]

    3. The attempt at a solution

    I've spent some time trying to solve this problem. I've looked through my notes, my textbook, the internet and still am lost. I've looked at the solutions manual and it doesn't seem to help either. So I was hoping that someone could point me in the right direction.

    I start off by finding the damping ratio since

    [itex]ζ = -\frac{ln(\frac{OS}{100})}{\sqrt{∏^{2} + ln^{2}(\frac{OS}{100})}} = -\frac{ln(\frac{16}{100})}{\sqrt{∏^{2}+ln^{2}(\frac{16}{100})}} ≈ 0.503[/itex]

    I know find the bandwidth frequency [itex]ω_{BW}[/itex]

    [itex]ω_{BW} = \frac{4}{T_{s}ζ}\sqrt{(1-2ζ) + \sqrt{4ζ^{4} - 4ζ^{2} + 2}} = \frac{4(10^{3} mS)}{(2 mS)(0.503) S}\sqrt{(1-2(0.503)^{2}) + \sqrt{4(0.503)^{4} - 4(0.503)^{2} +2}} ≈ 5,044.122 \frac{1}{S}[/itex]

    Next I find the gain evaluated at [itex]jω_{BW}[/itex] [itex]G(jω_{BW})[/itex]

    [itex]G(jω_{BW}) = \frac{3.333*10^{4}}{(j5,044.122 \frac{1}{S}} ≈ -0.131*10^{-2}[/itex]

    Next I find the phase margin [itex]Θ_{m}[/itex]

    [itex]Θ_{m} = arctan(\frac{2ζ}{\sqrt{-2ζ^{2} + \sqrt{1 + 4ζ^{4}}}}) = arctan(\frac{2(0.503)}{\sqrt{-2(0.503)^{2} + \sqrt{1 + 4(0.503)^{4}}}}) ≈ 52.068°[/itex]

    I'm not really sure were to go from here but all of the above work seems right and are all formulas that I can find in my book. The solutions manual does this next

    [itex]0.013K = \frac{1}{\sqrt{2}}[/itex]

    and solves for [itex]K[/itex]. I'm not sure where this formula comes from. The solutions manual says it's the compensators gain. But that doesn't really help me. The equation is clearly

    [itex]G(jω_{BW})K = ?[/itex]

    I'm not really sure what the right side of the equation is.

    Thanks for any help.
     
    Last edited: Mar 22, 2014
  2. jcsd
  3. Mar 23, 2014 #2
    Whats the representation for lead compensator? Closed loop stability implies you're using feedback, and the stability condition is:

    1 + G(s)*C(s) = 0, where C(s) is your compensator.

    Edit: You'll also want to add a gain K in front of the compensator...so K doesn't just appear out of nowhere. Also for stability, the solution for K usually has a given range of possible values (like K between 0 and 5), so you solutions manual exact solution for K is odd to me.

    Edit2: Should've been more clear, for the characteristic equation I gave you above, the roots of that equation are equivalent to the poles of the system and therefore must be in the left half plane for stability.
     
    Last edited: Mar 23, 2014
  4. Mar 23, 2014 #3
    For a lead network [itex]G_{c}(s) = \frac{1}{β}\frac{s + \frac{1}{T}}{s + \frac{1}{βT}}[/itex] I'm still unsure where [itex]\frac{1}{\sqrt{2}}[/itex] came from though. There's a standard representation for the compensator [itex]C(s)[/itex]?

    Edit 1:

    From the characteristic equation

    [itex]1 + kG(s)C(s) = 0[/itex]
    [itex]kG(s)C(s) = -1[/itex]
    [itex]kG(s) = - \frac{1}{C(s)}[/itex]

    This is apparently what the solutions manual is doing? And apparently [itex]C(s) = -\sqrt{2}[/itex] But I'm not sure why it's [itex]-\sqrt{2}[/itex] had not had the solutions manual and not sure were that came from
     
    Last edited: Mar 23, 2014
  5. Mar 23, 2014 #4
    its something like what you wrote: C(s) = (s + a)/(s + b), where conditions on a,b determine if its lead or lag. Your lead network form is more useful because there exists equations for T and beta.

    Have you tried to solve for T,beta? Then its likely the 1/sqrt(2) falls out of the stability condition 1 + KGC = 0
     
  6. Mar 23, 2014 #5
    My text says that we can find the frequency, [itex]ω_{max}[/itex] at which the maximum phase angle, [itex]Θ_{max}[/itex], occurs

    [itex]ω_{max} = \frac{1}{T\sqrt{β}}[/itex]

    But this doesn't really seem to help since I don't know the frequency [itex]ω_{max}[/itex], unless it's the bandwidth frequency [itex]ω_{BW}[/itex], which I don't think it is.

    My text also states that the maximum phase shift of the compensator [itex]θ_{max}[/itex] is

    [itex]θ_{max} = arctan(\frac{1 - β}{2\sqrt{β}}) = arcsin(\frac{1 - β}{1 + β})[/itex]

    Which doesn't really help since I don't know what [itex]θ_{max}[/itex].

    My text also states that the compensator's magnitude at [itex]ω_{max}[/itex] is

    [itex]|G_{c}(jω_{max})| = \frac{1}{\sqrt{β}}[/itex]

    This equation looks strongly similar to where I got stuck. If this correct than [itex]β = 2[/itex]. Which I'm not sure how was obtained exactly.

    Thanks for your help

    My gut says use the bandwidth frequency in this equation

    [itex]ω_{max} = \frac{1}{T\sqrt{β}}[/itex]
    [itex]Tω_{BW} = \frac{1}{\sqrt{B}} = T(5,044.122 \frac{1}{s}) = \frac{1}{\sqrt{β}}[/itex]

    But I'm unsure what I'm supposed to use for [itex]T[/itex]. Apparently if I'm not mistaken [itex]\frac{1}{T}[/itex] and [itex]\frac{1}{βT}[/itex] are the break frequencies which I don't know how to obtain without the Bode Plot
     
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