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Routh-Hurwitz criterion

  1. Apr 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the stability of ##s^3 + 3s^2 + 2(1 + K_p)s + 2K_i = 0##.

    2. Relevant equations
    Routh-Hurwitz criterion

    3. The attempt at a solution
    By the Routh-Hurwitz stability criterion, we have
    $$
    \begin{array}{ccc}
    s^3 & 1 & 2(1 + K_p)\\
    s^2 & 3 & 2K_i\\
    s^1 & \frac{6(1 + K_p) - 2K_i}{3} & 0\\
    s^0 & 2K_i & 0
    \end{array}
    $$
    From the ##s^3## line, we have that ##2(1 + K_p) > 0##; therefore, ##K_p > -1##. From the ##s^2## and ##s^0## line, we have that ##2K_i > 0##; thus, ##K_i > 0##. From the ##s^1## line, we have that ##6K_p - 2K_i + 6 > 0##; therefore,
    $$
    \frac{K_i}{K_p} < \frac{3}{K_p} + 3.
    $$
    I am supposed to conclude
    $$
    0 < \frac{K_i}{K_p} < 13.5.
    $$
    From line ##s^2##, we get greater than zero, but what do I do to go from ##\frac{3}{K_p} + 3## to ##13.5##?
     
    Last edited: Apr 8, 2014
  2. jcsd
  3. Apr 8, 2014 #2

    donpacino

    User Avatar
    Gold Member

    To be honest I do not know. I went through the routh-Hurwitz criterion and got the same results you did

    If you plug in 10 for ki and 1 for kp the system is unstable (i ran it in matlab)
    therefore the conclusion that 0<ki/kp<13.5 is incorrect

    I recommend confirming that all the information you provided us is correct, and all the information provided to you is correct.
     
  4. Apr 8, 2014 #3
    Is the information from the book:
    I had to use the link since the image appears too big on the site.
    http://i.imgur.com/M5Ks5up.jpg?1

    M5Ks5up.jpg
     
    Last edited: Apr 8, 2014
  5. Apr 9, 2014 #4
    Must be an error in your book. Your Routh table is correct, so to avoid any sign changes in the first column for ##K_p > 0, K_i > 0##, we must have:
    [tex]
    \begin{align}
    6(1 + K_p) - 2K_i \geq 0 \Leftrightarrow K_i \leq 3(1 + K_p) \quad (1)
    \end{align}
    [/tex]
    Although (1) being true is necessary and sufficient for stability, the weaker condition:
    [tex]
    K_i \leq 3K_p \Leftrightarrow \frac{K_i}{K_p} \leq 3
    [/tex]
    is sufficient.
     
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