Controller and Observer Design (Design of Control System in State Space) (Design of Control System in State Space by Pole placement)

References Dr. Radhakant Padhi, Asstt. Prof, IISC, Bangalore, through NPTEL Modern Control Engineering by Katsuhiko Ogata, PHI Pvt. Ltd New Delhi

Pole placement Observer Design

Outline Philosophy of observer design Full order observer Reduced (minimum) order observer

Philosophy of Observer Design In practice all the sate variable are not available for feedback. Possible reason include: Non availability of sensors Expensive sensors Available sensors are not acceptable (due to high noise, high power consumption etc) A device (or a computer program) that estimate or observes state variables is called a state observer or simply an observer.

Philosophy of Observer Design… If the observer observes all state variables of the system, regardless of whether some state variables are available for direct measurement, it is called a full order observer. If the observer observes only the unmeasurable state variables (n-m, n = no of sates, m= measurable state= no of o/p), it is called a minimum order state observer or reduced order state observer.

State Observer Block Diagram y Plant X 𝑋 + u Observer (full order) Ke Observed Sate Vector 𝑋 C - 𝑦 y- 𝑦

State Observer Block Diagram.. Plant: let plat dynamics is (plant state space equation) 𝑋 =𝐴𝑋+𝐵𝑢 (Single i/p) 𝑦=𝐶𝑋 (Single o/p) Let the observed state is 𝑋 and let the observer dynamic (observer state space equation) be 𝑋 = 𝐴 𝑋 + 𝐵 𝑢+ 𝐾 𝑒 𝑦 Error 𝐸𝑟𝑟𝑜𝑟 𝐸=𝑋− 𝑋 Note: We are taking sate matrix of plant and observer (𝐴 & 𝐴 ) different but finally we will show that both are same

Observer design concepts Error dynamics 𝐸 = 𝑋 − 𝑋 put the values of 𝑋 and 𝑋 ⇒ 𝐸 = 𝐴𝑋+𝐵𝑢 − 𝐴 𝑋 + 𝐵 𝑢+ 𝐾 𝑒 𝑦 Add and subtract 𝐴 𝑋 and put 𝑦=𝐶𝑋 ⇒ 𝐸 = 𝐴 𝑋− 𝐴 𝑋+ 𝐴𝑋+𝐵𝑢 − 𝐴 𝑋 + 𝐵 𝑢+ 𝐾 𝑒 𝐶𝑋 ⇒ 𝐸 = 𝐴− 𝐴 𝑋+ 𝐴 (𝑋− 𝑋 )+ 𝐵− 𝐵 𝑢− 𝐾 𝑒 𝐶𝑋 ⇒ 𝐸 = 𝐴− 𝐴 𝑋+ 𝐴 𝐸+ 𝐵− 𝐵 𝑢− 𝐾 𝑒 𝐶𝑋 ⇒ 𝐸 = 𝐴 𝐸+ 𝐴− 𝐴 − 𝐾 𝑒 𝐶 𝑋+ 𝐵− 𝐵 𝑢 Strategy Make the above error dynamics independent of X ⇒ 𝐴− 𝐴 − 𝐾 𝑒 𝐶 =0 Make the above error dynamics independent of u ⇒ 𝐵− 𝐵 =0

Observer design concepts … This leads to 𝐴 =𝐴− 𝐾 𝑒 𝐶 𝐵 =𝐵 So error dynamics 𝐸 = 𝐴 𝐸 ⇒ 𝐸 = 𝐴− 𝐾 𝑒 𝐶 𝑋 Observer Dynamics (Observer State space equation) 𝑋 = 𝐴 𝑋 + 𝐵 𝑢+ 𝐾 𝑒 𝑦 ⇒ 𝑋 =(𝐴− 𝐾 𝑒 𝐶) 𝑋 +𝐵𝑢+ 𝐾 𝑒 𝑦 ⇒ 𝑋 =𝐴 𝑋 +𝐵𝑢+ 𝐾 𝑒 (𝑦−𝐶 𝑋 ) ⇒ 𝑋 =𝐴 𝑋 +𝐵𝑢+ 𝐾 𝑒 (𝑦− 𝑦 ) Note: Finally we have shown that sate matrix of plant & observer are same (A= 𝐴 )

Observer Design (Full Order) Goal: As we know error 𝐸 = 𝐴− 𝐾 𝑒 𝐶 𝑋 Dynamic behavior of above error vector is determined by the eigenvalues of matrix 𝐴− 𝐾 𝑒 𝐶. If matrix 𝐴− 𝐾 𝑒 𝐶 is stable matrix, the error vector will converge to zero for any initial error vector e(0). E will become zero => 𝑋 =𝑋 (Observed sates = Actual sates) So obtain gain matrix Ke such that the error dynamics are asymptotically stable with sufficient speed of response. Hence the design of full order observer becomes that of determining an appropriate Ke such that 𝐴− 𝐾 𝑒 𝐶 has desired eigenvalues. Thus, the problem here becomes the same as the pole placement problem in controller design. Necessary and sufficient condition for existence of Ke: The system should be completely observable.

Comparison of controller & observer design Controller Design Observer Design Dynamics 𝑋 = 𝐴−𝐵𝐾 𝑋 Dynamics 𝐸 = 𝐴− 𝐾 𝑒 𝐶 𝐸 Objective: 𝐴𝑠 𝑡→∞, 𝑋→0 Objective: 𝐴𝑠 𝑡→∞, 𝐸→0 Note: 1. Note the position of Matrix K in controller design and matrix Ke in observer design. Position can be made same if we take transpose of 𝐴− 𝐾 𝑒 𝐶 . 2. As we know Eigen value of A = Eigen value of AT ⇒𝜆 𝐴 =𝜆( 𝐴 𝑇 ) 3. So 𝜆(𝐴− 𝐾 𝑒 𝐶 )=𝜆 𝐴− 𝐾 𝑒 𝐶 𝑇 =𝜆 𝐴 𝑇 − 𝐶 𝑇 𝐾 𝑒 𝑇 4. Observer Dynamics may be written as 𝐸 = 𝐴 𝑇 − 𝐶 𝑇 𝐾 𝑒 𝑇 𝐸 This is similar to controller design of a following system know as dual system

Observer Design as a dual problem Consider the following system 𝑍 = 𝐴 𝑇 𝑍+ 𝐶 𝑇 𝑣 (v=i/p) 𝑤= 𝐵 𝑇 𝑍 (w=o/p) If we do controller design of above system and find out value of state feedback matrix K. This will be same as the observer design of following system 𝑋 =𝐴𝑋+𝐵𝑢 𝑦=𝐶𝑋 Above two systems given by equ(1) & (2) are know as dual systems. The value of matrix Ke is obtained by 𝐾 𝑒 = 𝐾 𝑇 ---(1) ---(2)

Observer Design as a dual problem… Hence observer design of a system is nothing but simply doing the controller design of its dual system. So the methods will be same as for controller design. Method 1: Direct substitution method (when order of system n≤3) Method 2: Bass-Gura Approach Method 3: Ackermann’s formula

Pole Placement Technique The closed loop poles should lie 𝜇 1 , 𝜇 2 ,… 𝜇 𝑛 . Which are their “desired locations”.

Observer Design by method 1: (n≤3) Let the system is 𝑋 =𝐴𝑋+𝐵𝑢 steps are Step 1: Check observability of the system Step 2: Put 𝐾 𝑒 = 𝑘 𝑒1 𝑘 𝑒2 𝑘 𝑒3 𝑇 So 𝐸 = 𝐴− 𝐾 𝑒 𝐶 𝐸 Step 3: Write characteristic equations of above system 𝑠𝐼−(𝐴− 𝐾 𝑒 𝐶) =0 Step 4: Write Desired characteristic equation 𝑠− 𝜇 1 𝑠− 𝜇 2 𝑠− 𝜇 3 =0 Step 5: Compare above two characteristic equations and solve for k1, k2, k3 by equating the power of s on both sides

Controller Design using Method 2: Bass-Gura Approach Step 1: Check controllability of the system Step 2: Form a characteristic equation using matrix A. i.e 𝑠𝐼−𝐴 = 𝑠 𝑛 + 𝑎 1 𝑠 𝑛−1 + 𝑎 2 𝑠 𝑛−2 … 𝑎 𝑛−1 𝑠 1 + 𝑎 𝑛 find ai’s Step 3: find the transformation matrix T if system is not in first companion Q=WNT Q=I (Identity Matrix) If system is in controllable canonical form Step 4: Write the desired characteristic equation 𝑠− 𝜇 1 … 𝑠− 𝜇 𝑛 =𝑠 𝑛 + 𝛼 1 𝑠 𝑛−1 + 𝛼 2 𝑠 𝑛−2 … 𝛼 𝑛−1 𝑠 1 + 𝛼 𝑛 find 𝛼i’s

Controller Design using Method 2: Bass-Gura Approach… Step 5: The required matrix is

Controller Design using Method 3:Ackermann’s Formula For an arbitrary positive integer n ( number of states) Ackermann’s formula for the state feedback gain matrix K is given by

Example Example 1: Consider the system defined by 𝑋 =𝐴𝑋+𝐵𝑢 & 𝑦=𝐶𝑋 where 𝐴= 0 20.6 1 0 𝐵= 0 1 𝐶= 0 1 Design a full order state observer, Assume that desisered eigenvalues of the observer matrix are 𝑠=−1.8+𝑗2.4 and 𝑠=−1.8−𝑗2.4. Solution: First check the observability of above system

Example .. Observability matrix 𝑁= 𝐶 𝑇 𝐴 𝑇 𝐶 𝑇 = 0 1 1 0 𝑁 =−1 so rank of N =2. Hence system is completely observable. Now we will solve this problem with previous three methods

Example .. Method 1: Direct substitution method Put 𝐾 𝑒 = 𝑘 𝑒1 𝑘 𝑒2 𝑇 So 𝐸 = 𝐴− 𝐾 𝑒 𝐶 𝐸 Write characteristic equations of above system 𝑠𝐼−(𝐴− 𝐾 𝑒 𝐶) =0 𝑠 0 0 𝑠 − 0 20.6 1 0 + 𝑘 𝑒1 𝑘 𝑒2 0 1 =0 ⇒ 𝑠 2 + 𝑘 𝑒2 𝑠+ −20.6+ 𝑘 𝑒1 =0

Example… Write Desired characteristic equation 𝑠− 𝜇 1 𝑠− 𝜇 2 =0 𝑠− 𝜇 1 𝑠− 𝜇 2 =0 𝑠+1.8−𝑗2.4 𝑠+1.8+𝑗2.4 =0 ⇒ 𝑠 2 +3.6𝑠+9=0 comparing above two characteristic equations ke1 = 29.6, ke2 = 3.6 So 𝐾 𝑒 = 29.6 3.6

Example Method 2: Characteristic equation of the given system 𝑠𝐼−𝐴 = 𝑠 −20.6 −1 𝑠 ⇒Φ 𝐴 = 𝑠 2 −20.6 Comparing with Φ 𝐴 = 𝑠 2 + 𝑎 1 𝑠+ 𝑎 2 a1 = 0, a2 = -20.6

Example… Desired Characteristic equation 𝑠− 𝜇 1 𝑠− 𝜇 2 =0 𝑠+1.8−𝑗2.4 𝑠+1.8+𝑗2.4 =0 ⇒ 𝑠 2 +3.6𝑠+9=0 Comparing with 𝑠 2 + 𝛼 1 𝑠+ 𝛼 2 ⇒ 𝛼 1 =3.6, 𝛼 2 =9 Sate feedback gain matrix Ke is 𝐾= 𝑄 −1 𝛼 2 − 𝑎 2 𝛼 1 − 𝑎 1 Where Q = WNT = I (identity matrix as system is in controllable canonical form) 𝐾= 29.6 3.6

Example… Method 3: Ackermann’s Formula 𝐾 𝑒 =Φ 𝐴 𝐶 𝐶𝐴 −1 0 1 𝐾 𝑒 =Φ 𝐴 𝐶 𝐶𝐴 −1 0 1 Φ 𝐴 = 𝐴 2 +3.6𝐴+9𝐼 Φ 𝐴 = 0 20.6 1 0 2 +60 0 20.6 1 0 +9 0 20.6 1 0 ⇒Φ 𝐴 = 29.6 74.16 3.6 29.6 𝑁= 𝐶 𝐶𝐴 = 0 1 1 0

Example… So 𝐾 𝑒 = 29.6 74.16 3.6 29.6 0 1 1 0 −1 0 1 ⇒ 𝐾 𝑒 = 29.6 3.6

Reduced order design?

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