Systems Analysis and Control

Transcription

1 Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 2: Systems Defined by Differential Equations

2 Introduction In this Lecture, you will learn: A numerical Illustration of Feedback How to use differential equations to define a System. Identify the Inputs and Outputs Model the dynamics Newton s Laws Voltage Laws Put in First-Order (State-Space) Form M. Peet Lecture 2: Control Systems 2 / 30

3 Lets Start with an Example without Dynamics Just for Motivation CRUISE CONTROL Plant: The Automobile (Car) Input: Throttle Position, θ throttle. Output: Real Velocity, v true. Dynamics: No Dynamics! Speed is proportional to throttle (proportional gain). v true = 10 θ throttle The gain factor is 10mph/. M. Peet Lecture 2: Control Systems 3 / 30

4 Cruise Control: Open-Loop vs. Closed-Loop Open Loop Control First lets start with open loop control v desired θ throttle v true Control Engine Desired System Gas Speed Speed Actuator: Throttle Controller: Input: Desired Velocity, v desired. Output: Throttle, θ throttle. We use a simple controller based on our knowledge of the plant. θ throttle = 1 10 v desired M. Peet Lecture 2: Control Systems 4 / 30

5 Cruise Control: Open-Loop vs. Closed-Loop Closed Loop Control: The Error signal Now lets try using closed loop control v true - v desired θ throttle v Control true Engine Desired e velocity System Speed Speed Actuator: Throttle Sensor: Real Velocity Look at the CONTROLLER: Input: Error in Velocity, e velocity = v true v desired. Output: Throttle, θ throttle. The feedback controller simply amplifies the error signal by a positive factor k. θ throttle = k e velocity = k (v true v desired ) M. Peet Lecture 2: Control Systems 5 / 30

6 Closed Loop vs. Open Loop (Solving for v true ) Open Loop: Two relations: v true = 10 θ throttle and θ throttle = 1 10 v desired we have v true = v desired = v desired So there is No Error in the open-loop control Closed Loop: We also have two relations: v true = 10 θ throttle and θ throttle = k (v true v desired ) Combining these, we get v true = 10 k(v true v desired ). Lets Choose k = 10. v true = Closed-loop control has 1% Error. 10 k k v desired = v desired =.99v desired. M. Peet Lecture 2: Control Systems 6 / 30

7 Impact of Error and Disturbances Comparison: Open Loop: No error Closed Loop: Small error Error gets very small if k, since Question: Why use feedback? Answer: Life is Messy. Problems: v true = 10 k k v desired v desired. Modeling Errors: What if our model is wrong by 10%, so v true = 11 θ throttle Disturbances: An Incline of i disturbance degrees will reduce the throttle by.5/. θ throttle =.5 i disturbance M. Peet Lecture 2: Control Systems 7 / 30

8 Impact of Error and Disturbances Open Loop Let v desired = 50mph, i disturbance = 1. Recalculate for the open loop case: Incline i disturbance.5 v desired θ throttle v true Control 11 System v true = 11(θ throttle.5 i disturbance ) θ throttle = 1 10 v desired = 5 we have v true = 11(5 +.5) = 60.5mph Which is NOT ACCEPTABLE!!!. M. Peet Lecture 2: Control Systems 8 / 30

9 Impact of Error and Disturbances Closed Loop Recalculate for the closed loop case: Plant with Disturbance: v true = 11 (θ throttle.5 i disturbance ) Controller: θ throttle = k (v true v desired ) = k(v true 50) Combine expressions and solve for v true. v true = 11( k v true + 50 k +.5) = 11 k v true k Solving for v true yields v true = 11k k = 50 = = 49.6mph 111 Note: Solving for v true is called Closing the Loop. M. Peet Lecture 2: Control Systems 9 / 30

10 A Brief Review of Modeling The previous model of an engine was a static model. In this class, all models will be either static. differential equations. The modeling of physical systems using differential equations was introduced by Newton in You will need to know how to derive Differential Equation models. From physical and electronic systems The first differential equation model: Newton s Second Law: d 2 x(t) = F/m dt2 F(t) m x(t) M. Peet Lecture 2: Control Systems 10 / 30

11 Review: Modeling Differential Equations Nonlinear Differential Equations: dx (t) = f(x(t), u(t)) dt y(t) = g(x(t), u(t)) Where This is a first-order differential equation u(t) is the Input y(t) is the Output x is the state variable. position, heading, velocity, etc. f, g are functions (possibly vector-valued). M. Peet Lecture 2: Control Systems 11 / 30

12 Review: Equations of Motion (EOMs) Linear Equations Usually, our equations of motion will be Linear. e.g. where a is a constant. Linear equations are better because ẍ(t) = bẋ(t) + ax(t) The Laplace Transform exists (Lecture 4) Stability is easy ẋ = ax is stable if a < 0 and unstable if a 0. Nonlinear Equations should be Linearized (Lecture 3)!!! M. Peet Lecture 2: Control Systems 12 / 30

13 Review: Equations of Motion Higher Orders or Multiple Variables Types of EOM: Coupled Differential Equations: ẋ = ax + bz ż = cx + dz The motion of x affects the motion of z and vice-versa. Higher Order Derivatives:... x = aẍ + bẋ + cx Commonly obtained from Newton s Second law. AKA Or anything with inertia... F = mẍ ẍ(t) = 1 m F (t). M. Peet Lecture 2: Control Systems 13 / 30

14 Dynamic Model: Suspension System Mass-Spring Model We wish to study the motion of the vehicle subject to disturbances. Model the car as a solid mass Control the vertical motion of the car (x(t)) Inputs: Force, f(t). Outputs: Displacement, y(t) = x(t). Definition 1. A system with one input and one output is Single-Input, Single-Output (SISO). A system with more than one input or more than one output is Multi-Input Multi-Output (MIMO) M. Peet Lecture 2: Control Systems 14 / 30

15 Dynamic Model: Suspension System Mass-Spring Model Plant Dynamics: Equations of Motion Spring Force: Opposes motion in x with spring constant K. F s (t) = Kx(t) Damper Force: Opposes motion in ẋ with damping coefficient f v Newton s Second Law: System Model: F d (t) = f v ẋ(t) mẍ(t) = F s (t) + F d (t) + f(t) ẍ(t) = K m x(t) f v m ẋ(t) + 1 m f(t) y(t) = x(t) M. Peet Lecture 2: Control Systems 15 / 30

16 Standard Forms Frequency Domain Once we have our dynamic model ẍ(t) = K m x(t) f v m ẋ(t) + 1 m f(t) y(t) = x(t) This model can be expressed in two standard forms Transfer Function State-Space We will discuss these in more depth soon. For now: Differential Equations Output Equation Transfer Function: Apply the Laplace Transform to both equations and solve for the output. s 2 x(s) = K m x(s) f v m sx(s) + 1 m f(s) y(s) = x(s) which yields y(s) = 1 ms 2 + f v s + K u(s) Differential Equations Output Equation M. Peet Lecture 2: Control Systems 16 / 30

17 Suspension System with Wheel Dynamics More Detailed Model x1 Now, we add the dynamics of the wheel. mc There are two states: States: x2 Vehicle Position, x1 mw Wheel Position, x2 u Our Input is the position of the surface of the road. Inputs: Road Surface, u M. Peet Lecture 2: Control Systems 17 / 30

18 Suspension Model: Free Body 1 This time we write the dynamics of both the wheel and the car. x 1 m c x 2 m w u Car Dynamics: Equations of Motion Spring 1 Force on Car: F s1,c (t) = K 1 (x 1 (t) x 2 (t)) Damper Force on Car: F d,c (t) = c(ẋ 1 (t) ẋ 2 (t)) Newton s Second Law: m c ẍ 1 (t) = F s1,c (t) + F d,c (t) = K 1 (x 1 (t) x 2 (t)) c(ẋ 1 (t) ẋ 2 (t)) M. Peet Lecture 2: Control Systems 18 / 30

19 Suspension Model: Free Body 2 x 1 m c x 2 m w u Wheel Dynamics: Equations of Motion Spring 1 Force on Wheel: F s1,w (t) = K 1 (x 1 (t) x 2 (t)) Spring 2 Force on Wheel: F s2,w (t) = K 2 (x 2 (t) u(t)) Damper Force on Wheel: F d,w (t) = c(ẋ 1 (t) ẋ 2 (t)) Newton s Second Law: m w ẍ 2 (t) = F s1,w (t) + F s2,w (t) + F d,w (t) = K 1 (x 1 (t) x 2 (t)) K 2 (x 2 (t) u(t)) + c(ẋ 1 (t) ẋ 2 (t)) M. Peet Lecture 2: Control Systems 19 / 30

20 Equations of Motion Combining the dynamics, we get the coupled system dynamics. x 1 m c x 2 m w u m w ẍ 2 (t) = K 1 (x 1 (t) x 2 (t)) K 2 (x 2 (t) u(t)) + c(ẋ 1 (t) ẋ 2 (t)) m c ẍ 1 (t) = K 1 (x 1 (t) x 2 (t)) c(ẋ 1 (t) ẋ 2 (t)) [ ] x1 (t) y(t) = x 2 (t) This is quite complicated. To simplify, we would like to use a Standard Form. M. Peet Lecture 2: Control Systems 20 / 30

21 Other Sources of Models Angular Momentum Newton s Second Law Applied to Rigid Bodies The rate of change of angular momentum is given by Mi = Iα = I θ α = θ is the angular acceleration in inertial coordinates. I is the moment of inertia, which varies by object. M i are the moments applied to the body. M. Peet Lecture 2: Control Systems 21 / 30

22 Other Sources of Models Voltage Laws Kirchhoff s Current Law (KCL): Current is conserved at each junction ik = 0 Kirchhoff s Voltage Law (KVL): Net Voltage change around any loop is zero. V k = 0 k These are combined with standard voltage laws such as voltage drop across a resister, inductor and capacitor: V r (t) = Ri r (t) d dt i L(t) = 1 L V L(t) d dt V c(t) = 1 C i c(t) M. Peet Lecture 2: Control Systems 22 / 30

23 Review: Equations of Motion State-Space State-Space is a way of writing first order differential equation using matrices. We write x = A x where x is a vector and A R n n is a square matrix. Example: d dt x 1 x 2 x = Is equivalent to writing the three differential equations ẋ 1 = x 1 + x 3 ẋ 2 = 2x 1 ẋ 3 = x 2 + x 3 Writing equations in state-space has many advantages x 1 x 2 x 3 M. Peet Lecture 2: Control Systems 23 / 30

24 Review: Equations of Motion Multiple Variables and State-Space Consider the system ẋ = ax + by ẏ = cx + dy When we have multiple coupled equations, the best option is: Convert to State-Space: [ [ [ ] d x a b x = dt y] c d] y Which is easily expressed as where x is a vector. A is a matrix. ẋ = Ax The equation describes the motion of the vector. M. Peet Lecture 2: Control Systems 24 / 30

25 State-Space Form for Systems Definition 2. State-Space Form is a convenient way of representing multivariate or linear MIMO systems using 4 matrices. u is the vector of Inputs. y is the vector of Outputs. x is the State. ẋ(t) = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) u R m, y R p, and x R n can be vectors of any dimension. However, the matrices must be the right size: A R n n C R p n B R n m D R p m u R n means u is a real vector of length n. C R p n means C is a matrix with p rows and n columns. M. Peet Lecture 2: Control Systems 25 / 30

26 Putting Things in State-Space Form Reducing Higher Order Dynamics When we have higher order dynamics,... x (t) = aẋ(t) + bx(t) + u(t) y(t) = x(t) + u(t) we can still use state-space form by Introducing new variables. Procedure: Define a new variable for every derivative term except for the the highest order one. e.g. Let x1 = x, x 2 = ẋ and x 3 = ẍ. Add a new first order differential equation for each new variable. e.g. ẋ1 = x 2 and ẋ 2 = x 3 Then put in state-space form. So, for example, we would have 3 equations ẋ 1 (t) = x 2 (t) ẋ 2 (t) = x 3 (t) ẋ 3 (t) = ax 2 (t) + bx 1 (t) + u(t) M. Peet Lecture 2: Control Systems 26 / 30

27 Putting Things in State-Space Form Using our first-order equations: ẋ 1 (t) = x 2 (t); ẋ 3 (t) = ax 2 (t) + bx 1 (t) + u(t) ẋ 2 (t) = x 3 (t) y(t) = x 1 (t) + u(t) We construct the matrix representation: ẋ(t) = d x x 1 0 x 2 (t) = x 2 (t) + 0 u(t) dt x 3 b a 0 x 3 1 y(t) = [ ] x 1 (t) + [ 1 ] u(t) So that x 2 x A = B = 0 b a 0 1 C = [ ] D = [ 1 ] M. Peet Lecture 2: Control Systems 27 / 30

28 State-Space Form: Suspension System Recall the dynamics: m w ẍ 2 (t) = K 1 (x 1 (t) x 2 (t)) K 2 (x 2 (t) u(t)) + c(ẋ 1 (t) ẋ 2 (t)) m c ẍ 1 (t) = K 1 (x 1 (t) x 2 (t)) c(ẋ 1 (t) ẋ 2 (t)) [ ] x1 (t) y(t) = x 2 (t) Define the new variables z i z 1 = x 1 z 2 = ẋ 1 z 3 = x 2 z 4 = ẋ 2 [ ] z1 (t) Which yields the following set of equations: y(t) =, z 3 (t) ż 1 (t) = z 2 (t) ż 2 (t) = K 1 (z 1 (t) z 3 (t)) c (z 2 (t) z 4 (t)) m c m c ż 3 (t) = z 4 (t) ż 4 (t) = K 1 (z 1 (t) z 3 (t)) K 2 (z 3 (t) u(t))) + c (z 2 (t) z 4 (t)) m w m w m w M. Peet Lecture 2: Control Systems 28 / 30

29 Constructing State-Space Systems ż 1 (t) = z 2 (t) ż 2 (t) = K 1 z 1 (t) c z 2 (t) + K 1 z 3 (t) + m c m c m c ż 3 (t) = z 4 (t) ż 4 (t) = K 1 z 1 (t) + m w [ ] z1 (t) y(t) = z 3 (t) d dt z 1 z 2 z 3 z 4 (t) = y(t) = c m c z 4 (t) c ( K1 z 2 (t) + K ) 2 z 3 (t) m w m w m w K1 m c K 1 m w [ c m c K 1 m c c m c 0 0 ( 0 ) 1 c K m w 1 m w + K2 m w ] z 1 [ ] z 2 0 z 3 (t) + u(t) 0 z 4 c m w c m w z 4 (t) K 2 m w u(t) z 1 z 2 z 3 z 4 (t) K2 m w u(t) M. Peet Lecture 2: Control Systems 29 / 30

30 Summary What have we learned today? A Static Model of Cruise-Control Simple static model and Control Open Loop Control Closed Loop Control Benefits of Feedback Dynamic Models Including Inputs and Outputs Using Newton s Laws MIMO and SISO systems Other sources of models (Kirchhoff s Laws) State-Space State-Space Form M. Peet Lecture 2: Control Systems 30 / 30