Given the controllable system described by
𝑥( 𝑘 + 1) = [
0 0 2
2 1 −2
1 0 1
] x(k) + [
0
1
0.5
] 𝑢( 𝑘), 𝑦( 𝑘) = [1 − 1 2]
a)Find the state feedback gain matrix K so as to locate the closed loop poles at
𝑧1,2 = 0.4 ± 0.6 𝑗 𝑎𝑛𝑑 𝑧3 = −0.2
Use your results of part a) to determine the closed loop pulse transfer function. Note that the
numerators of the open and closed loop (after state feedback) pulse transfer functions are the
same.
Solution:-
CON = [ H G*H G*G*H] = [
0 1 1
1 0 1
0.5 0.5 1.5
] CON-1 = [
1 2 −2
2 1 −2
−1 −1 2
]
CON-1 * G * CON = [
1 2 −2
2 1 −2
−1 −1 2
] [
0 0 2
2 1 −2
1 0 1
] [
0 1 1
1 0 1
0.5 0.5 1.5
] = [
0 0 −2
1 0 1
0 1 2
]
−𝑎3
−𝑎2
−𝑎1
The transformation matrix M = [
𝑎2 𝑎1 1
𝑎1 1 0
1 0 0
] = [
−1 −2 1
−2 1 0
1 0 0
], M-1 = [
0 0 1
0 1 2
1 2 3
]
Therefore, P = CON * M = [
−1 1 0
0 −2 1
0 −0.5 0.5
] and P-1 = M-1 * CON-1 = [
−1 −1 2
0 −1 2
0 −1 4
]
Thus, we have
𝑥̃( 𝑘 + 1) = P-1 G P 𝑥̃( 𝑘) + P-1 h u(k) = [
0 1 0
0 0 1
−2 1 2
] 𝑥̃( 𝑘) + [
0
0
1
] u(k)
The closed loop characteristic equation is given by
Z
3 + α1 z2 + α2 z + α3 = ( z – 0.4 + 0.6 j ) ( z – 0.4 - 0.6 j ) ( z + 0.2)
= z3 -0.6 z2 + 0.36 z + 0.104
Therefore,
𝐾̃ = [ (∝3- 𝑎3 ) (∝2- 𝑎2 ) (∝1- 𝑎1 ) ] = [ -1.896 1.36 1.4]
u(k) == r(k) - 𝐾̃ 𝑥̃( 𝑘) = 𝑟( 𝑘) − 𝐾̃ 𝑃−1
x(k) ==.> or K= 𝐾̃ P-1
K = [ 1.696 -0.864 4.528]
b) The standard controllable form of the system before feedback is given by
𝑥̃ ( 𝑘 + 1) = 𝑃−1
𝐺 𝑃 𝑥̃( 𝑘)+ 𝑃−1
ℎ 𝑢( 𝑘) , 𝑦( 𝑘) = 𝐶 𝑃 𝑥̃(𝑘)
𝑥̃ ( 𝑘 + 1) = [
0 1 0
0 0 1
−2 1 2
] 𝑥̃( 𝑘) + [
0
0
1
] u(k), y(k) = [1 − 1 2] [
−1 1 0
0 −2 1
0 −0.5 0.5
] 𝑥̃( 𝑘)
y(k) = [-1 2 0] 𝑥̃( 𝑘)
Thus
𝑌(𝑧)
𝑈(𝑧)
=
2 𝑧 −1
𝑧3 −2 𝑧2− 𝑧 +1
And the closed loop transfer function will be
𝑌(𝑧)
𝑅 (𝑧)
=
2 𝑧 −1
𝑧3 −0.6 𝑧2+0.36 𝑧 +0.104

State feedback example

  • 1.
    Given the controllablesystem described by 𝑥( 𝑘 + 1) = [ 0 0 2 2 1 −2 1 0 1 ] x(k) + [ 0 1 0.5 ] 𝑢( 𝑘), 𝑦( 𝑘) = [1 − 1 2] a)Find the state feedback gain matrix K so as to locate the closed loop poles at 𝑧1,2 = 0.4 ± 0.6 𝑗 𝑎𝑛𝑑 𝑧3 = −0.2 Use your results of part a) to determine the closed loop pulse transfer function. Note that the numerators of the open and closed loop (after state feedback) pulse transfer functions are the same. Solution:- CON = [ H G*H G*G*H] = [ 0 1 1 1 0 1 0.5 0.5 1.5 ] CON-1 = [ 1 2 −2 2 1 −2 −1 −1 2 ] CON-1 * G * CON = [ 1 2 −2 2 1 −2 −1 −1 2 ] [ 0 0 2 2 1 −2 1 0 1 ] [ 0 1 1 1 0 1 0.5 0.5 1.5 ] = [ 0 0 −2 1 0 1 0 1 2 ] −𝑎3 −𝑎2 −𝑎1 The transformation matrix M = [ 𝑎2 𝑎1 1 𝑎1 1 0 1 0 0 ] = [ −1 −2 1 −2 1 0 1 0 0 ], M-1 = [ 0 0 1 0 1 2 1 2 3 ] Therefore, P = CON * M = [ −1 1 0 0 −2 1 0 −0.5 0.5 ] and P-1 = M-1 * CON-1 = [ −1 −1 2 0 −1 2 0 −1 4 ] Thus, we have 𝑥̃( 𝑘 + 1) = P-1 G P 𝑥̃( 𝑘) + P-1 h u(k) = [ 0 1 0 0 0 1 −2 1 2 ] 𝑥̃( 𝑘) + [ 0 0 1 ] u(k) The closed loop characteristic equation is given by Z 3 + α1 z2 + α2 z + α3 = ( z – 0.4 + 0.6 j ) ( z – 0.4 - 0.6 j ) ( z + 0.2) = z3 -0.6 z2 + 0.36 z + 0.104 Therefore, 𝐾̃ = [ (∝3- 𝑎3 ) (∝2- 𝑎2 ) (∝1- 𝑎1 ) ] = [ -1.896 1.36 1.4] u(k) == r(k) - 𝐾̃ 𝑥̃( 𝑘) = 𝑟( 𝑘) − 𝐾̃ 𝑃−1 x(k) ==.> or K= 𝐾̃ P-1 K = [ 1.696 -0.864 4.528]
  • 2.
    b) The standardcontrollable form of the system before feedback is given by 𝑥̃ ( 𝑘 + 1) = 𝑃−1 𝐺 𝑃 𝑥̃( 𝑘)+ 𝑃−1 ℎ 𝑢( 𝑘) , 𝑦( 𝑘) = 𝐶 𝑃 𝑥̃(𝑘) 𝑥̃ ( 𝑘 + 1) = [ 0 1 0 0 0 1 −2 1 2 ] 𝑥̃( 𝑘) + [ 0 0 1 ] u(k), y(k) = [1 − 1 2] [ −1 1 0 0 −2 1 0 −0.5 0.5 ] 𝑥̃( 𝑘) y(k) = [-1 2 0] 𝑥̃( 𝑘) Thus 𝑌(𝑧) 𝑈(𝑧) = 2 𝑧 −1 𝑧3 −2 𝑧2− 𝑧 +1 And the closed loop transfer function will be 𝑌(𝑧) 𝑅 (𝑧) = 2 𝑧 −1 𝑧3 −0.6 𝑧2+0.36 𝑧 +0.104