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LQR Design: Choosing Q & R matrices for specific eigenvalues

  1. Mar 23, 2012 #1
    Assuming we have a closed loop system (A-BK), with stable eigenvalues, how would one choose matrices Q and R such that the eigenvalues of (A-BK) are exactly [-1,-2]?

    LTI System:
    [itex]\dot{x}=\left[ \begin{array}{cc}
    0& 1 \\
    0 & 0 \\
    \end{array} \right]x+\left[ \begin{array}{c}
    0 \\
    1 \\
    \end{array} \right]u [/itex]

    The performance measure is given by:

    [itex]V=\int_0^\infty \left( x'\left[ \begin{array}{cc}
    1 & 0 \\
    0 & 2 \\
    \end{array} \right]x+u^2\right)\mathrm{d}t.
    [/itex]

    So the initial Q is :
    [itex]
    Q=\left[ \begin{array}{cc}
    1 & 0 \\
    0 & 2 \\
    \end{array} \right]
    [/itex]

    and R = 1

    However, we want to choose a new matrix for Q and R so that we have stable eigenvalues:
    [itex]
    \lambda=-1 \\
    \lambda=-2 \\
    [/itex]

    I have gone round and round with this, but I cannot see how to work back through the Hamiltonian Matrix to achieve a desired value for Q & R.

    The Hamiltonian is:
    [itex]
    H=\left[
    \begin{array}{cc}
    A & -BR^{-1}B' \\
    -Q & -A' \\
    \end{array} \right]
    [/itex]

    and the stable Eigenvalues of H (ones with negative real parts) are exactly the eigenvalues of the closed loop system A-BK.

    Please help if you can. Even something to point me in the right direction will help.

    BTW, just iterating by hand the matrix Q, I came up with eig(H) = -1,-2 using:

    [itex]
    Q=\left[ \begin{array}{cc}
    4 & 0 \\
    0 & 5 \\
    \end{array} \right]
    [/itex]

    Yet, I must show how to achieve something like this mathematically.

    Thanks!
     
  2. jcsd
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