Linear State-Space Control Systems
Prof. Kamran Iqbal
College of Engineering and Information Technology
University of Arkansas at Little Rock
Course Schedule
Session Topic
1. State space models of linear systems
2. Solution to State equations, canonical forms
3. Controllability and observability
4. Stability and dynamic response
5. Controller design via pole placement
6. Controllers for disturbance and tracking systems
7. Observer based compensator design
8. Linear quadratic optimal control
9. Kalman filters and stochastic control
10. LM in control design
Linear System Stability
Transfer Function
• Consider a linear system
𝑥 = 𝐴𝑥 + 𝐵𝑢
𝑦 = 𝐶𝑥 + 𝐷𝑢
• Transfer function
𝐻 𝑠 = 𝐶 𝑠𝐼 − 𝐴 −1𝐵 + 𝐷 =𝐶 𝐸1𝑠
𝑛−1+𝐸2𝑠𝑛−2+⋯+𝐸𝑘 𝐵
𝑠𝑛+𝑎1𝑠𝑛−1+⋯+𝑎𝑛−1𝑠+𝑎𝑛
+ 𝐷
Let 𝑠𝐼 − 𝐴 = 𝑠𝑛 + 𝑎1𝑠𝑛−1 +⋯+ 𝑎𝑛 = 𝑠 − 𝑠𝑖
𝜈𝑖 … 𝑠 − 𝑠𝑘𝜈𝑘
Then, using partial fractions
𝐻 𝑠 = 𝐻1 𝑠 + 𝐻2 𝑠 + ⋯+ 𝐻𝑘 𝑠 + D
where 𝐻𝑖 𝑠 =𝑅1𝑖
𝑠−𝑆𝑖+
𝑅𝑖2
𝑠−𝑆𝑖2 +⋯+
𝑅𝜈𝑖𝑖
𝑠−𝑆𝑖𝜈𝑖
Asymptotic Stability
• Impulse response
𝐻 𝑡 = 𝐻1 𝑡 + 𝐻2 𝑡 + ⋯+ 𝐻𝑘 𝑡 + 𝐷𝛿(𝑡)
where 𝐻𝑖 𝑡 = 𝑅1𝑖 + 𝑅2𝑖𝑡 + ⋯+𝑅𝜈𝑖𝑡
𝜈𝑖−1
𝜈𝑖−1 !𝑒𝑠𝑖𝑡
Then 𝐶𝑒𝐴𝑡𝐵 = 𝑅1𝑖 + 𝑅2𝑖𝑡 + ⋯+𝑅𝜈𝑖𝑡
𝜈𝑖−1
𝜈𝑖−1 !𝑒𝑠𝑖𝑡𝑘
𝑖=1
Thus,
𝑅𝑒 𝑠𝑖 < 0 for all 𝑖 → asymptotically stable
𝑅𝑒 𝑠𝑖 > 0 for some 𝑖 → unstable
𝑅𝑒 𝑠𝑖 = 0 for some 𝑖
– 𝑠𝑖 is a simple root → stable but not asymptotically stable
– 𝑠𝑖 is a repeated root → unstable
Asymptotic Stability
• Consider unforced system 𝑥 = 𝐴𝑥
Then, for any matrix 𝐴
𝐴 = 𝑇−1𝐽𝑇, 𝑒𝐴𝑡 = 𝑇−1𝑒𝐽𝑡𝑇
where matrix 𝐽 is block diagonal, and for each Jordan block
𝑥 = 𝐽𝜆𝑥, 𝐽𝜆 = 𝜆𝐼 + 𝑁
𝑒𝐽𝜆𝑡 = 𝑒𝜆𝑡𝐼 𝑒𝑁𝑡 = 𝑒𝜆𝑡
1 𝑡 … 𝑡𝑘−1
𝑘−1 !
1 … 𝑡𝑘−2
𝑘−2 !
… …… 1
Thus, 𝑒𝐽𝜆𝑡 includes terms of the form 𝑡𝑖𝑒𝜆𝑡, 𝑖 = 0, … , 𝑘 − 1
If 𝑅𝑒 𝜆𝑖 < 0, then lim𝑡→∞
𝑡𝑖𝑒𝜆𝑡 = 0
BIBO Stability
• Bounded Input Bounded Output Stability
𝑦 𝑡 = ℎ 𝑡 − 𝜏 𝑢 𝜏 𝑑𝜏𝑡
0
𝑦 𝑡 ≤ ℎ 𝑡 − 𝜏 𝑢 𝜏 𝑑𝜏𝑡
0
If 𝑢(𝑡) < 𝑐, then 𝑦 𝑡 ≤ 𝑐 ℎ 𝑡 − 𝜏 𝑑𝜏𝑡
0
• A system is BIBO stable if and only if
ℎ 𝜏 𝑑𝜏∞
0< 𝑀
• Asymptotic stability → BIBO stability
Lyapunov Stability
• The unforced system 𝑥 = 𝐴𝑥 is asymptotically stable if for a positive
definite matrix 𝑄 the Lyapunov equation 𝑃𝐴 + 𝐴𝑇𝑃 = −𝑄 has a
unique positive definite solution
• Proof:
Consider a Lyapunov function: 𝑉 𝑥 = 𝑥𝑇𝑃𝑥
Then 𝑉 𝑥 = −𝑥𝑇𝑄𝑥
𝑉 𝑡 ≤ −𝛼𝑉(𝑡) or 𝑉 𝑡 ≤ −𝛼𝑉(0)
where 𝛼 =𝜆𝑚𝑖𝑛 𝑄 𝑥𝑇𝑥
𝜆𝑚𝑎𝑥 𝑃 𝑥𝑇𝑥
Alternatively, assume that eigenvalues of A have negative real parts
Then 𝑃 = 𝑒𝐴𝑇𝑡𝑄𝑒𝐴𝑡𝑑𝑡
∞
0 is the unique positive definite solution
Hurwitz Stability
• Let 𝐻 𝑠 =𝑁 𝑠
𝐷 𝑠
where 𝐷(𝑠) = 𝑠𝑛 + 𝑎1𝑠𝑛−1 +⋯+ 𝑎𝑛−1𝑠 + 𝑎𝑛
• Let 𝐻 =
𝑎1 𝑎3 … …
1 𝑎2 … …
𝑎1 𝑎3 …
1 𝑎2 …
be the 𝑛 × 𝑛 Hurwitz matrix
• Then zeros of 𝐷 𝑠 are confined to LHP if and only if all leading
minors of 𝐻 are strictly positive:
𝐷1 = 𝑎1, 𝐷2 =𝑎1 𝑎31 𝑎2
, … , 𝐷𝑛 = 𝐻 > 0
System Poles and Zeros
• For a SISO system
𝐻 𝑠 =𝑏0𝑠
𝑛+𝑏1𝑠𝑛−1+⋯+𝑏𝑛
𝑠𝑛+𝑎1𝑠𝑛−1+⋯+𝑎𝑛−1𝑠+𝑎𝑛
=𝑁 𝑠
𝐷 𝑠
– System zeros: zeros of 𝑁 𝑠
– System poles: zeros of 𝐷(𝑠)
• For a square MIMO system
𝐻 𝑠 =𝐸1𝑠
𝑛−1+𝐸2𝑠𝑛−2+⋯+𝐸𝑘
𝑠𝑛+𝑎1𝑠𝑛−1+⋯+𝑎𝑛−1𝑠+𝑎𝑛
+ 𝐷
– System zeros: zeros of 𝐻 𝑠
– System poles: zeros of 𝐻−1 𝑠
Example: Missile Dynamics
Missile dynamics (𝑀𝑞=0)
𝛼 𝑞
=𝑍𝛼
𝑉1
𝑀𝛼 0
𝛼𝑞 +
𝑍𝛿𝑉
𝑀𝛿
𝛿
𝛼𝑁 = 𝑍𝛼𝛼 + 𝑍𝛿𝛿
Add an actuator: 𝛿 =1
𝜏𝑢 − 𝛿 , then
𝐻 𝑠 =1
𝜏𝑠+1 𝑍𝛿𝑠
2+𝑍𝛼𝑀𝛿−𝑍𝛿𝑀𝛼
𝑠2−𝑍𝛼𝑉𝑠−𝑀𝛼
Let 𝑉 = 1253𝑓𝑡
𝑠, 𝑍𝛼 = −4170
𝑓𝑡
𝑠2, 𝑍𝛿 = −1115
𝑓𝑡
𝑠2, 𝑀𝛼 = −248
𝑟𝑎𝑑
𝑠2, 𝑀𝛿 = −662
𝑟𝑎𝑑
𝑠2,
𝜏 = 0.01
𝐻 𝑠 = −1115 𝑠2−2228
0.01𝑠+1 𝑠2+3.33𝑠+248
Poles: 𝑠 = −100,−1.67 ± 𝑗15.65; zeros: 𝑠 = ±47.2
𝐺𝑀 = 11.13
Aircraft Longitudinal Motion
Let 𝑥 = Δ𝑢, 𝛼, 𝑞, 𝜃 ′; 𝑢 = 𝛿𝐸
Δ𝑢 = 𝑋𝑢Δ𝑢 + 𝑋𝛼𝛼 − 𝑔𝜃 + 𝑋𝐸𝛿𝐸
𝛼 =𝑍𝑢
𝑉Δ𝑢 +
𝑍𝛼
𝑉𝛼 + 𝑞 +
𝑍𝐸
𝑉𝛿𝐸
𝑞 = 𝑀𝑢Δ𝑢 +𝑀𝛼𝛼 +𝑀𝑞𝑞 +𝑀𝐸𝛿𝐸
𝜃 = 𝑞
Let 𝑋𝑢, 𝑋𝛼 , 𝑋𝐸 ,𝑍𝑢
𝑉,𝑍𝛼
𝑉,𝑍𝐸
𝑉, 𝑀𝑢, 𝑀𝛼 , 𝑀𝑞, 𝑀𝐸 =
[−0.0507, −3.861, 0, −0.00117, −0.5164, −0.0717, −0.000129, 1.4168,
−0.4932,−1.645]
Then,
𝑠 = −1.705, 0.724, −0.0394 ± 𝑗0.2
Dynamic Response: Stability Margins
• Define the return difference
– For SISO systems, 𝑇 𝑗𝜔 = 1 + 𝐺(𝑗𝜔)
– For MIMO systems, 𝑇 𝑗𝜔 = 𝐼 + 𝐺(𝑗𝜔)
• SISO Stability Margins
– Gain margin: 𝛾 = 𝑇 𝑗𝜔2 , 𝜃𝐺 𝜔2 = 180°
– Phase margin: 𝜙 = 2 sin−1𝑇 𝑗𝜔1
2, 𝐺 𝑗𝜔1 = 1
• MIMO Stability margins:
Let 𝜎 𝑇(𝑗𝜔) ≥ 𝛼, then
– 𝐺𝑀 =1
1±𝛼
– 𝑃𝑀 = ±2 sin−1𝛼
2
Note, these resulting stability margins are extremely conservative
Dynamic Response: Rise Time
• Define system centroidal rise time as:
𝑡 = 𝑡ℎ(𝑡)𝑑𝑡∞0
ℎ(𝑡)𝑑𝑡∞0
= −𝐻′ 0
𝐻 0
Let 𝐻 𝑠 = 𝐶Φ(𝑠)𝐵 where Φ 𝑠 = 𝑠𝐼 − 𝐴 −1
Then 𝐻 0 = −𝐶𝐴−1𝐵, 𝐻′ 0 = −𝐶𝐴−2𝐵, 𝑡 =𝐶𝐴−2𝐵
𝐶𝐴−1𝐵
Alternatively, 𝑡 =𝐴−1
𝐴
1/2
– First order system: 𝑡 =1
𝜔0
– Second order system: 𝑡 =2𝜉
𝜔0, 𝜔𝐵𝑡 ≈ 1
– For 𝐻 𝑠 = 𝐻1 𝑠 𝐻2(𝑠), 𝑡 = 𝑡 1 + 𝑡 2
– For 𝐻 𝑠 =𝐾𝐺 𝑠
1+𝐾𝐺(𝑠), 𝑡 𝐶𝐿 =
1
1+𝐾𝐺 0𝑡 𝐺
