1. Consider the 1-DOF system described by the equation of motion, 4ẍ+20ẋ+25x = f.

Transcription

1 Introduction to Robotics (CS3A) Homework #6 Solution (Winter 7/8). Consider the -DOF system described by the equation of motion, ẍ+ẋ+5x = f. (a) Find the natural frequency ω n and the natural damping ratio ξ n of the natural (passive) system (f = ). What type of system is this (oscillatory, overdamped, etc.)? Using Section 7.. from the course reader, we can compare this sytem with mẍ + bẋ + kx = like in Equation 7.9. Thus: k 5 ω n = m = =.5 (b) ξ n = b km = 5 = Since ξ n =, this system is critically damped. Design a PD controller that achieves critical damping with a closed-loop stiffness k CL = 36. In other words, let f = k v ẋ k p x, and determine the gains k v and k p. Assume that the desired position is x d =. The original system is: ẍ + ẋ + 5x = f with input force f. The controller provides this input, using the formula: So, the closed loop equation is: f = k v ẋ k p x ẍ + ẋ + 5x = k v ẋ k p x ẍ + ( + k v )ẋ + (5 + k p )x = This closed loop system behaves just like the natural dissipative system in Section 7.. of the course reader. So, we first compare to Equation 7.9: ẍ + ( + k v )ẋ + (5 + k p )x = mẍ + bẋ + kx = The closed loop stiffness is given by k, the coefficient of the positional term, so: k = 5 + k p = k CL = 36 For the damping requirement, we need to first figure out how the control gains k p and k v affect the damping ratio ξ; we do this by applying Equation 7. to our system: ξ = b km

2 For critical damping: so the coefficient b of ẋ must satisfy ξ = b = km = 36 = Based on our closed loop equation, we have: b = + k v so the gains that we need are and the PD controller is k p =, k v = f = ẋ x (c) Assume that the friction model changes from linear (ẋ) to Coulomb friction, 3sign(ẋ). Design a control system which uses a non-linear model-based portion with trajectory following to critically damp the system at all times and maintain a closed-loop stiffness of k CL = 36. In other words, let f = αf +β and f = ẍ d k v(ẋ ẋ d ) k p(x x d ). Then, find f,α,β,f,k p and k v. Note that f is an m-mass control, and f is a unit-mass control. Use the definition of error, e = x x d. The differential equation for the system is now ẍ + 3sign(ẋ) + 5x = f In order to linearize it, we apply a force f of the form where f = αf + β α =, β = 3sign(ẋ) + 5x For purposes of control, this makes the system look like the unit-mass system: to which we apply the control ẍ = f f = ẍ d k v(ẋ ẋ d ) k p(x x d ) Substituting into our unit-mass system yields the equation ë + k vė + k pe = where e is the position error, e = x x d. Now, we want to choose our gains k p and k v so that we achieve critical damping and the desired closed-loop stiffness. To look at the closed loop stiffness, we need to consider the controlled system before factoring out the mass: ẍ + 3sign(ẋ) + 5x = αf + β ẍ = αf

3 (d) ẍ + αf = (ẍ ẍ d ) + k v(ẋ ẋ d ) + k p(x x d ) = ë + k vė + k pe = The coefficient of the e term is the closed loop stiffness, so: k p = 9. In order to have critical damping, we need to have ξ =. Using Equation 7. we see that the coefficient of ė must be: k v = k p = 6 Thus, the control is f = αf + β α = β = 3sign(ẋ) + 5x f = ẍ d 6(ẋ ẋ d ) 9(x x d ) Given a disturbance force f dist =, what is the steady-state (ë = ė = ) error of the system in part (c)? We can analyze the error by observing the error in the unit-mass system. With a disturbance force added, the system s equation of motion becomes ẍ + 3sign(ẋ) + 5x = f + f dist To linearize the system, we apply a force of the same form as before: f + f dist = f + 3sign(ẋ) + 5x + f dist ( = f + f ) dist + 3sign(ẋ) + 5x This yields a unit-mass system as before, but now it has a disturbance force of f dist /, so the unit-mass system now looks like ẍ = f + f dist With the control from before, we get a unit-mass closed-loop system of ë + 6ė + 9e = f dist For the steady state, when ẍ = ẋ =, we get So, the steady state error is given by 9e = f dist e = f dist 9 = 36.. For a certain RR manipulator, the equations of motion are given by + c + c θ s ( + θ + θ θ ) τ = + c θ s θ τ

4 (a) Assume that joint is locked at some value θ using brakes and joint is controlled with a PD controller, τ = θ (θ θ d ). What is the minimum and maximum inertia perceived at joint as we vary θ? What are the corresponding closed-loop frequencies? For joint locked ( θ = θ = ), the equation of motion for joint is: ( + c ) θ = τ The inertia seen at joint is the coefficient of the θ term, ( + c ). So, this inertia achieves its maximum and minimum values at θ = and θ = 8 : The closed-loop equation for joint is m max = 5, m min = 3 ( + c ) θ + θ + (θ θ d ) = To get an expression for closed loop frequency, we compare our closed loop equation with the generic system of Equation 7.9 (mẍ + bẋ + kx = ). The closed loop frequency is then given by: So, we have ω = k m = ( + c ) m = m max ω min = 5 m = m min ω max = 3 (b) Still assuming that joint is locked, at what values of θ do the minimum and maximum damping ratios occur? What are the minimum and maximum damping ratios? To get an expression for damping ratio, we once again compare our closed loop equation with the generic system of Equation 7.9 (mẍ+bẋ+kx = ). In this case, using Equation 7. the closed-loop damping ratio is given by: ξ = b km = m = = m + c So, the minimum and maximum values of ξ occur at θ = and θ = 8 : ξ min = ξ max = = mmax 5 = mmin 3 (c) Now assume that both joints are free to move, and that this system is controlled by a partitioned PD controller, τ = ατ + β. Design a partitioned, trajectory-following controller (one that tracks a desired position, velocity and acceleration) which will provide a closed-loop frequency of rad/sec

5 on joint and rad/sec on joint and be critically damped over the entire workspace. That is, let τ = θ k d v k v ( θ θ k d ) p k p (θ θ d ), then find the matrices α and β and the vector τ, along with the necessary gains k v i and k p i. The equations of motion are of the form M(θ) θ + V ( θ, θ) = τ to which we apply a vector of torques τ of the form τ = ατ + β To make this look like a unit-mass system, we let α = M(θ), β = V ( θ, θ) which gives the unit-mass system θ = τ To this system, we apply the control τ = θ k d v k v ( θ θ k d ) p k p (θ θ d ), This yields two closed-loop equations ë + k v ė + k p e = ë + k v ė + k p e = where e i is the error at joint i, e i = (θ i θ id ). Now, we need to choose k v i and k p i to achieve critical damping, and to achieve our desired closed-loop frequencies. For a unit-mass system, we choose k p i = ω i k v i = ξ i ω i So, we get k p =, k v = k p =, k v = (d) If θ = 8, what is the steady-state error vector for a given disturbance torque, τ dist = T? The controlled system, with a disturbance torque τ dist is M(θ) θ + V ( θ, θ) = τ + τ dist Substituting in our form for τ = ατ + β yields M(θ) θ M(θ)τ = τ dist

6 This has the form M(θ) ë + K vė + K pe = τ dist where e is the error vector e = θ θ d, and K v and K p are the matrices given by K v = k v k v, K p = In the steady state (ë = ė = ), the equation is which means that the steady state error is For our values, this is: e = ( 3 M(θ)K pe = τ dist e = (M(θ)K p) τ dist ) = k p k p 3 = 5