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Consider the system:
a) Design a state-feedback control in the form of u(t) = -k x(t) + Fr(t) to place the
eigenvalues of the closed-loop system at 1,2=-0.5±j and 3 =-0.7 and track a given
constant reference and use MATLAB to plot the response of the system to see whether
y(t) asymptotically tracks the reference signal. (This is a repeat from Mini-project 1)
Answer:
State-feedback control design:
𝑥 = [
0 1
1 2
] 𝑢 + [
0
1
] 𝑢 𝑦 = [0 1]𝑢 𝜆 = 1,2 = −0.5 ± 𝑗
(𝐴 − 𝐵𝑢) = [
0 1
1 2
] − [
0
1
] [𝑢1 𝑢2] = [
0 1
1 − 𝑢1 2 − 𝑢2
]
Characteristics equation, |𝑆𝐼 − (𝐴 − 𝐵𝑢)| = 0
=> |(
𝑆 0
0 𝑆
) − (
0 1
1 − 𝑢1 2 − 𝑢2
)| = 0
=> |
𝑆 −1
𝑢1 − 1 𝑢2 − 2
| = 0
=> (𝑆 + 0.5 ± 𝑗)(𝑆 + 0.5 − 𝑗) = 0
=> 𝑆2
+ 𝑆 + 0.25 + 1 = 0 => 𝑆2
+ 𝑆 + 1.25 = 0
𝑢2 − 2 = 1 𝑢1 − 1 = 1.25
𝑢2 = 3 𝑢1 = 2.25
𝑲 = [𝟐. 𝟐𝟓 𝟑]
Matlab:
k =
Column 1
2.2500
Column 2
3.0000
sys =
A =
x1
x1 0
x2 -1.25
x2
x1 1
x2 -1
B =
u1
x1 0
x2 1
C =
x1
y1 0
x2
y1 1
D =
u1
y1 0
Continuous-time state-space model.
y1 =
-s/(- s^2 + 2*s + 1)
y2 =
(4*s)/(4*s^2 + 4*s + 5)
b) Design an observer to estimate the state of the system.
Answer:
c) For part (a), imagine you only have access to the output y(t). Use the observer that you
designed in part (b), to feedback the estimated state and control the system. Find the state
space representation of the closed loop system, which includes both the controller and the
observer.
Answer:
d) In part (c), use MATLAB to plot the response of the system to see whether y(t)
asymptotically tracks the reference signal r(t).
Answer:
e) Compare the results in part (d) and (a).
The stability of a linear system may be determined directly from its transfer function. An nth
order linear system is asymptotically stable only if all of the components in the homogeneous
response from a finite set of initial conditions decay to zero as time increases. As we can see,
part a is stable compared to the part d.

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Consider the system.docx

  • 1. Consider the system: a) Design a state-feedback control in the form of u(t) = -k x(t) + Fr(t) to place the eigenvalues of the closed-loop system at 1,2=-0.5±j and 3 =-0.7 and track a given constant reference and use MATLAB to plot the response of the system to see whether y(t) asymptotically tracks the reference signal. (This is a repeat from Mini-project 1) Answer: State-feedback control design: 𝑥 = [ 0 1 1 2 ] 𝑢 + [ 0 1 ] 𝑢 𝑦 = [0 1]𝑢 𝜆 = 1,2 = −0.5 ± 𝑗 (𝐴 − 𝐵𝑢) = [ 0 1 1 2 ] − [ 0 1 ] [𝑢1 𝑢2] = [ 0 1 1 − 𝑢1 2 − 𝑢2 ] Characteristics equation, |𝑆𝐼 − (𝐴 − 𝐵𝑢)| = 0 => |( 𝑆 0 0 𝑆 ) − ( 0 1 1 − 𝑢1 2 − 𝑢2 )| = 0 => | 𝑆 −1 𝑢1 − 1 𝑢2 − 2 | = 0 => (𝑆 + 0.5 ± 𝑗)(𝑆 + 0.5 − 𝑗) = 0 => 𝑆2 + 𝑆 + 0.25 + 1 = 0 => 𝑆2 + 𝑆 + 1.25 = 0 𝑢2 − 2 = 1 𝑢1 − 1 = 1.25 𝑢2 = 3 𝑢1 = 2.25
  • 2. 𝑲 = [𝟐. 𝟐𝟓 𝟑] Matlab: k = Column 1 2.2500 Column 2 3.0000 sys = A = x1 x1 0 x2 -1.25 x2 x1 1 x2 -1 B = u1 x1 0 x2 1
  • 3. C = x1 y1 0 x2 y1 1 D = u1 y1 0 Continuous-time state-space model. y1 = -s/(- s^2 + 2*s + 1) y2 = (4*s)/(4*s^2 + 4*s + 5) b) Design an observer to estimate the state of the system. Answer:
  • 4.
  • 5.
  • 6. c) For part (a), imagine you only have access to the output y(t). Use the observer that you designed in part (b), to feedback the estimated state and control the system. Find the state space representation of the closed loop system, which includes both the controller and the observer. Answer:
  • 7. d) In part (c), use MATLAB to plot the response of the system to see whether y(t) asymptotically tracks the reference signal r(t). Answer:
  • 8. e) Compare the results in part (d) and (a). The stability of a linear system may be determined directly from its transfer function. An nth order linear system is asymptotically stable only if all of the components in the homogeneous response from a finite set of initial conditions decay to zero as time increases. As we can see, part a is stable compared to the part d.