Digital and Non-Linear Control Design of Control Systems in State Space

Lecture Outline Introduction Pole Placement Topology of Pole Placement Pole Placement Design Techniques Using Transformation Matrix P Direct Substitution Method Ackermann’s Formula

Introduction One of the drawbacks of frequency domain methods of design is that after designing the location of the dominant second-order pair of poles, we keep our fingers crossed, hoping that the higher-order poles do not affect the second-order approximation. What we would like to be able to do is specify all closed-loop poles of the higher-order system.

Introduction Frequency domain methods of design do not allow us to specify all poles in systems of order higher than 2 because they do not allow for a sufficient number of unknown parameters to place all of the closed-loop poles uniquely. One gain to adjust, or compensator pole and zero to select, does not yield a sufficient number of parameters to place all the closed-loop poles at desired locations.

Introduction Remember, to place n unknown quantities, you need n adjustable parameters. State-space methods solve this problem by introducing into the system Other adjustable parameters and The technique for finding these parameter values On the other hand, state-space methods do not allow the specification of closed-loop zero locations, which frequency domain methods do allow through placement of the lead compensator zero.

Introduction Finally, there is a wide range of computational support for state-space methods; many software packages support the matrix algebra required by the design process. However, as mentioned before, the advantages of computer support are balanced by the loss of graphic insight into a design problem that the frequency domain methods yield.

Pole Placement In this lecture we will discuss a design method commonly called the pole-placement or pole-assignment technique. We assume that all state variables are measurable and are available for feedback. If the system considered is completely state controllable, then poles of the closed-loop system may be placed at any desired locations by means of state feedback through an appropriate state feedback gain matrix.

Pole Placement The present design technique begins with a determination of the desired closed-loop poles based on the transient-response and/or frequency-response requirements, such as speed, damping ratio, or bandwidth, as well as steady-state requirements. By choosing an appropriate gain matrix for state feedback, it is possible to force the system to have closed-loop poles at the desired locations, provided that the original system is completely state controllable.

Topology of Pole Placement Consider a plant represented in state space by 𝒙 =𝑨𝒙+𝑩𝑢 𝑦=𝑪𝒙

Topology of Pole Placement In a typical feedback control system, the output, y, is fed back to the summing junction. It is now that the topology of the design changes. Instead of feeding back y, we feed back all of the state variables. If each state variable is fed back to the control, u, through a gain, ki, there would be n gains, ki, that could be adjusted to yield the required closed-loop pole values.

Topology of Pole Placement The feedback through the gains, ki, is represented in following figure by the feedback vector K. 𝒙 =𝑨𝒙+𝑩(𝑟−𝑲𝒙) 𝒙 =𝑨𝒙+𝑩𝑟−𝑩𝑲𝒙 𝑦=𝑪𝒙 𝒙 =(𝑨−𝑩𝑲)𝒙+𝑩𝑟

Topology of Pole Placement For example consider a plant signal-flow graph in phase-variable form

Topology of Pole Placement Each state variable is then fed back to the plant’s input, u, through a gain, ki, as shown in Figure

Pole Placement We will limit our discussions to single-input, single-output systems (i.e. we will assume that the control signal u(t) and output signal y(t) to be scalars). We will also assume that the reference input r(t) is zero. 𝑦 𝒙 =(𝑨−𝑩𝑲)𝒙+𝑩𝑟 𝒙 =(𝑨−𝑩𝑲)𝒙 𝑢=−𝑲𝒙

Pole Placement 𝒙 =(𝑨−𝑩𝑲)𝒙 The stability and transient response characteristics are determined by the eigenvalues of matrix A-BK. If matrix K is chosen properly Eigenvalues of the system can be placed at desired location. And the problem of placing the regulator poles (closed-loop poles) at the desired location is called a pole-placement problem. 𝒙 =(𝑨−𝑩𝑲)𝒙

Pole Placement There are three approaches that can be used to determine the gain matrix K to place the poles at desired location. Using Transformation Matrix P Direct Substitution Method Ackermann’s formula All those methods yield the same result.

Pole Placement (Using Transformation Matrix P) Following are the steps to be followed in this particular method. Check the state controllability of the system 𝐶𝑀= 𝐵 𝐴𝐵 𝐴 2 𝐵 ⋯ 𝐴 𝑛−1 𝐵

Pole Placement (Using Transformation Matrix P) Following are the steps to be followed in this particular method. Transform the given system in CCF. 𝑠𝐼−𝐴 = 𝑠 𝑛 + 𝑎 1 𝑠 𝑛−1 + 𝑎 2 𝑠 𝑛−2 +⋯+ 𝑎 𝑛−1 s+ 𝑎 𝑛

Pole Placement (Using Transformation Matrix P) Following are the steps to be followed in this particular method. Obtain the desired characteristic equation from desired Eigenvalues. If the desired Eigenvalues are 𝜇 1 , 𝜇 2 , ⋯ , 𝜇 𝑛 (𝑠− 𝜇 1 )( 𝑠− 𝜇 2 ) ⋯ 𝑠− 𝜇 𝑛 = 𝑠 𝑛 + 𝛼 1 𝑠 𝑛−1 + 𝛼 2 𝑠 𝑛−2 +⋯+ 𝛼 𝑛−1 𝑠+ 𝛼 𝑛

Pole Placement (Using Transformation Matrix P) Following are the steps to be followed in this particular method. Compute the gain matrix K. 𝑲= 𝛼 𝑛 − 𝑎 𝑛 𝛼 𝑛−1 − 𝑎 𝑛−1 ⋯ 𝛼 2 − 𝑎 2 𝛼 1 − 𝑎 1

Pole Placement (Using Transformation Matrix P) Example-1: Consider the regulator system shown in following figure. The plant is given by 𝑥 1 𝑥 2 𝑥 3 = 0 1 0 0 0 1 −1 −5 −6 𝑥 1 𝑥 2 𝑥 3 + 0 0 1 𝑢(𝑡) The system uses the state feedback control u=-Kx. The desired eigenvalues are 𝜇 1 =−2+𝑗4, 𝜇 2 =−2−𝑗4 , 𝜇 3 =−1. Determine the state feedback gain matrix K.

Pole Placement (Using Transformation Matrix P) Example-1: Step-1 First, we need to check the controllability matrix of the system. Since the controllability matrix CM is given by We find that rank(CM)=3. Thus, the system is completely state controllable and arbitrary pole placement is possible. 𝑥 1 𝑥 2 𝑥 3 = 0 1 0 0 0 1 −1 −5 −6 𝑥 1 𝑥 2 𝑥 3 + 0 0 1 𝑢(𝑡) 𝐶𝑀= 𝐵 𝐴𝐵 𝐴 2 𝐵 = 0 0 1 0 1 −6 1 −6 31

Pole Placement (Using Transformation Matrix P) Example-1: Step-2 (Transformation to CCF) The given system is already in CCF 𝑥 1 𝑥 2 𝑥 3 = 0 1 0 0 0 1 −1 −5 −6 𝑥 1 𝑥 2 𝑥 3 + 0 0 1 𝑢(𝑡)

Pole Placement (Using Transformation Matrix P) Example-1: Step-3 Determine the characteristic equation Hence 𝑥 1 𝑥 2 𝑥 3 = 0 1 0 0 0 1 −1 −5 −6 𝑥 1 𝑥 2 𝑥 3 + 0 0 1 𝑢(𝑡) 𝑠𝐼−𝐴 = 𝑠 3 +6 𝑠 2 +5𝑠+1=0 𝑠𝐼−𝐴 = 𝑠 3 + 𝑎 1 𝑠 2 + 𝑎 2 𝑠+ 𝑎 3 𝑎 1 =6, 𝑎 2 =5, 𝑎 3 =1

Pole Placement (Using Transformation Matrix P) Example-1: Step-4 The desired characteristics polynomial can be computed using desired eigenvalues 𝜇 1 =−2+𝑗4 𝜇 2 =−2−𝑗4 𝜇 3 =−1 Hence (𝑠− 𝜇 1 )( 𝑠− 𝜇 2 ) ⋯ 𝑠− 𝜇 𝑛 =(𝑠+2−4𝑗)( 𝑠+2+4𝑗) 𝑠+10 𝑠+2−4𝑗 𝑠+2+4𝑗 𝑠+10 = 𝑠 3 +14 𝑠 2 +60𝑠+200 = 𝑠 3 + 𝛼 1 𝑠 2 + 𝛼 2 𝑠+ 𝛼 3 𝛼 1 =14, 𝛼 2 =60, 𝛼 3 =200

Pole Placement (Using Transformation Matrix P) Example-1: Step-4 State feedback gain matric K is then calculated as 𝑎 1 =6, 𝑎 2 =5, 𝑎 3 =1 𝛼 1 =14, 𝛼 2 =60, 𝛼 3 =200 𝑲= 𝛼 3 − 𝑎 3 𝛼 2 − 𝑎 2 𝛼 1 − 𝑎 1 𝑲= 199 55 8

Pole Placement (Using Transformation Matrix P) 𝑥 1 𝑥 2 𝑥 3 = 0 1 0 0 0 1 −1 −5 −6 𝑥 1 𝑥 2 𝑥 3 + 0 0 1 𝑢(𝑡) State diagram of the given system ∫ 𝑥 3 𝑥 2 𝑥 3 𝑥 2 𝑥 1 𝑥 1 -6 -5 + 𝑢(𝑡)

𝑢=− 199 55 8 𝑥 1 𝑥 2 𝑥 3 𝑢=−𝑲𝑥 𝑲= 199 55 8 199 55 8 + -1 𝑥 3 ∫ 𝑥 3 𝑥 2 ∫ 𝑥 2 𝑥 1 ∫ 𝑥 1 𝑢(𝑡) + + + -6 + + -5 + + -5

Pole Placement (Direct Substitution Method) Following are the steps to be followed in this particular method. Check the state controllability of the system 𝐶𝑀= 𝐵 𝐴𝐵 𝐴 2 𝐵 ⋯ 𝐴 𝑛−1 𝐵

Pole Placement (Direct Substitution Method) Following are the steps to be followed in this particular method. Define the state feedback gain matrix as And equating 𝑠𝐼−𝐴+𝐵𝐾 with desired characteristic equation. 𝑲= 𝑘 1 𝑘 2 𝑘 3 ⋯ 𝑘 𝑛 (𝑠− 𝜇 1 )( 𝑠− 𝜇 2 ) ⋯ 𝑠− 𝜇 𝑛 = 𝑠 𝑛 + 𝛼 1 𝑠 𝑛−1 + 𝛼 2 𝑠 𝑛−2 +⋯+ 𝛼 𝑛−1 s+ 𝛼 𝑛

Pole Placement (Using Direct Substitution) Example-1: Consider the regulator system shown in following figure. The plant is given by 𝑥 1 𝑥 2 𝑥 3 = 0 1 0 0 0 1 −1 −5 −6 𝑥 1 𝑥 2 𝑥 3 + 0 0 1 𝑢(𝑡) The system uses the state feedback control u=-Kx. The desired eigenvalues are 𝜇 1 =−2+𝑗4, 𝜇 2 =−2−𝑗4 , 𝜇 3 =−1. Determine the state feedback gain matrix K.

Pole Placement (Using Transformation Matrix P) Example-1: Step-1 First, we need to check the controllability matrix of the system. Since the controllability matrix CM is given by We find that rank(CM)=3. Thus, the system is completely state controllable and arbitrary pole placement is possible. 𝑥 1 𝑥 2 𝑥 3 = 0 1 0 0 0 1 −1 −5 −6 𝑥 1 𝑥 2 𝑥 3 + 0 0 1 𝑢(𝑡) 𝐶𝑀= 𝐵 𝐴𝐵 𝐴 2 𝐵 = 0 0 1 0 1 −6 1 −6 31

Pole Placement (Using Transformation Matrix P) Example-1: Step-2 Let K be Desired characteristic polynomial is obtained as Comparing the coefficients of powers of s 𝑲= 𝑘 1 𝑘 2 𝑘 3 𝑠𝐼−𝐴+𝐵𝐾 = 𝑠 0 0 0 𝑠 0 0 0 𝑠 − 0 1 0 0 0 1 −1 −5 −6 + 0 0 1 𝑘 1 𝑘 2 𝑘 3 = 𝑠 3 + 6+ 𝑘 3 𝑠 2 + 5+ 𝑘 2 𝑠+1+ 𝑘 1 𝑠+2−4𝑗 𝑠+2+4𝑗 𝑠+10 = 𝑠 3 +14 𝑠 2 +60𝑠+200 14= 6+ 𝑘 3 𝑘 3 =8 60= 5+ 𝑘 2 𝑘 2 =55 200=1+ 𝑘 1 𝑘 1 =199

Pole Placement (Ackermann’s Formula) Following are the steps to be followed in this particular method. Check the state controllability of the system 𝐶𝑀= 𝐵 𝐴𝐵 𝐴 2 𝐵 ⋯ 𝐴 𝑛−1 𝐵

Pole Placement (Ackermann’s Formula) Following are the steps to be followed in this particular method. Use Ackermann’s formula to calculate K 𝐾= 0 0 ⋯ 0 1 𝐵 𝐴𝐵 𝐴 2 𝐵 ⋯ 𝐴 𝑛−1 𝐵 −1 ∅(𝐴) ∅ 𝐴 = 𝐴 𝑛 + 𝛼 1 𝐴 𝑛−1 +⋯+ 𝛼 𝑛−1 𝐴+ 𝛼 𝑛 𝐼

Pole Placement (Ackermann’s Formula) Example-1: Consider the regulator system shown in following figure. The plant is given by 𝑥 1 𝑥 2 𝑥 3 = 0 1 0 0 0 1 −1 −5 −6 𝑥 1 𝑥 2 𝑥 3 + 0 0 1 𝑢(𝑡) The system uses the state feedback control u=-Kx. The desired eigenvalues are 𝜇 1 =−2+𝑗4, 𝜇 2 =−2−𝑗4 , 𝜇 3 =−1. Determine the state feedback gain matrix K.

Pole Placement (Using Transformation Matrix P) Example-1: Step-1 First, we need to check the controllability matrix of the system. Since the controllability matrix CM is given by We find that rank(CM)=3. Thus, the system is completely state controllable and arbitrary pole placement is possible. 𝑥 1 𝑥 2 𝑥 3 = 0 1 0 0 0 1 −1 −5 −6 𝑥 1 𝑥 2 𝑥 3 + 0 0 1 𝑢(𝑡) 𝐶𝑀= 𝐵 𝐴𝐵 𝐴 2 𝐵 = 0 0 1 0 1 −6 1 −6 31

Pole Placement (Ackermann’s Formula) Following are the steps to be followed in this particular method. Use Ackermann’s formula to calculate K 𝐾= 0 0 1 𝐵 𝐴𝐵 𝐴 2 −1 ∅(𝐴) ∅ 𝐴 = 𝐴 3 + 𝛼 1 𝐴 2 + 𝛼 2 𝐴+ 𝛼 3 𝐼 𝛼 𝑖 are the coefficients of the desired characteristic polynomial. 𝑠+2−4𝑗 𝑠+2+4𝑗 𝑠+10 = 𝑠 3 +14 𝑠 2 +60𝑠+200 𝛼 1 =14, 𝛼 2 =60, 𝛼 3 =200

Pole Placement (Ackermann’s Formula) 𝑥 1 𝑥 2 𝑥 3 = 0 1 0 0 0 1 −1 −5 −6 𝑥 1 𝑥 2 𝑥 3 + 0 0 1 𝑢(𝑡) ∅ 𝐴 = 𝐴 3 +14 𝐴 2 +60𝐴+200𝐼 ∅ 𝐴 = 0 1 0 0 0 1 −1 −5 −6 3 +14 0 1 0 0 0 1 −1 −5 −6 2 +60 0 1 0 0 0 1 −1 −5 −6 +200 1 0 0 0 1 0 0 0 1 ∅ 𝐴 = 199 55 8 −8 159 7 −7 −34 117

Pole Placement (Ackermann’s Formula) 𝐵 𝐴𝐵 𝐴 2 𝐵 = 0 0 1 0 1 −6 1 −6 31 ∅ 𝐴 = 199 55 8 −8 159 7 −7 −34 117 𝐾= 0 0 1 𝐵 𝐴𝐵 𝐴 2 −1 ∅(𝐴) 𝐾= 0 0 1 0 0 1 0 1 −6 1 −6 31 −1 199 55 8 −8 159 7 −7 −34 117 𝐾= 199 55 8

Pole Placement Example-2: Consider the regulator system shown in following figure. The plant is given by Determine the state feedback gain for each state variable to place the poles at -1+j, -1-j,-3. (Apply all methods) 𝑥 1 𝑥 2 𝑥 3 = 1 2 1 0 1 3 1 1 1 𝑥 1 𝑥 2 𝑥 3 + 1 0 1 𝑢(𝑡)

State Controllability A system is completely controllable if there exists an unconstrained control u(t) that can transfer any initial state x(to) to any other desired location x(t) in a finite time, to ≤ t ≤ T. uncontrollable controllable

State Controllability Controllability Matrix CM System is said to be state controllable if

State Controllability (Example) Consider the system given below State diagram of the system is

State Controllability (Example) Controllability matrix CM is obtained as Thus Since 𝑟𝑎𝑛𝑘(𝐶𝑀)≠𝑛 therefore system is not completely state controllable.

State Observability A system is completely observable if and only if there exists a finite time T such that the initial state x(0) can be determined from the observation history y(t) given the control u(t), 0≤ t ≤ T. observable unobservable

State Observability Observable Matrix (OM) The system is said to be completely state observable if

State Observability (Example) Consider the system given below OM is obtained as Where

State Observability (Example) Therefore OM is given as Since 𝑟𝑎𝑛𝑘(𝑂𝑀)≠𝑛therefore system is not completely state observable.

Output Controllability Output controllability describes the ability of an external input to move the output from any initial condition to any final condition in a finite time interval. Output controllability matrix (OCM) is given as

Home Work Check the state controllability, state observability and output controllability of the following system