MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.F ( t )FdpF ( t )mm a s sB ( v i s c o u s f r i c t i o n )v ( t )v e l o c i t yF ( t )dpv ( t )e n g i n eKem + B v = Fd vd t++F ( t )F ( t )dpqg a s p e d a lc a r m o d e lVc a r s p e e d� Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 21 Reading: • Nise: Chapter 1 1 Cruise Control Example (continued from Lecture 1) From Lecture 1 the model for the car is: where the propulsion force Fp(t) is proportional to the gas pedal depression θ: Fp(t) = Kθ(t) so that dv m + Bv = Keθ(t) + Fd(t)dt The “Open-Loop”Dynamic Response: Let’s examine how the car will respond to commands at the gas-pedal. Assume Fd(t) = 0, so that we can write the differential equation as m Ke v˙ + v = θ B B and compare this to the standard form for a first-order ODE: τy˙ + y = f(t) where the time-constant τ = m/B, and the forcing function f(t) = KB e θ(t). 1copyright cD.Rowell 2008 2–100t2 t3 t 4 tt ( s e c s )vo0 . 6 vo0 . 8 vo0 . 4 vo0 . 2 vov e l o c i t y v ( t )e = 0 . 3 6 7 9e = 0 . 1 3 5 3e = 0 . 0 4 9 8e = 0 . 0 1 8 3- 1- 2- 3- 4Consider 1) The “coast-down” response from an initial speed v(0) = v0 with θ = 0. The homogeneous response is v(t) = v0e−= v0e− tB m t τwhich has the form and after a period 4τ we have v < 0.02 v0. 2) The response to a “step” in the command θ(t). Assume that the car is at rest, that is v(0) = 0, and we ”step on the gas” so that θ(t) = θ. The differential equation is then m Ke v˙ + v = θ B B with v(0) = 0. (a) The steady-state (final) speed is Ke vss = θ B (found by letting all derivatives to be zero), and (b) the differential equation may be solved to give the dynamic response: B m tKeθ v(t) = vss(1 − e−τ ) = (1 − e− t)B which is shown below: 2–2e n g i n eKem + B v = Fd vd t++F ( t )F ( t )dpqc a r m o d e lVc a r s p e e dKce ( t )+f e e d b a c k p a t hV ( t )d-e r r o rc o n t r o l l e r0tv s sv ( t )4 tA f t e r 4t t h e r e s p o n s e i s w i t h i n2 % o f t h e f i n a l v a l u e .s t e a d y - s t a t e r e g i o nt r a n s i e n t r e g i o n2 Closed-Loop Control Now design the controller. Assume we will use error-based control. In other words, given a desired speed for the car vd, and the measured car speed v(t), we define the error e(t) as the difference e(t) = vd(t) − v(t), and choose a control law that is based on e(t) θ(t) = Kce(t) = Kc(vd(t) − v(t)) where Kc is the controller gain. Thus the control effort is proportional to the error, and the block diagram becomes: Note that the control law states: • if v < vd then e > 0 - depress gas pedal. • if v = vd then e = 0 - set θ = 0 (do nothing). • if v > vd then e < 0 - set θ < 0 (apply brakes). The new differential equation is dv m + Bv = KcKe(vd(t) − v) + Fd(t)dt and rearranging dv m + (B + KcKe)v = KcKevd(t) + Fd(t)dt 2–3s t e a d y - s t a t et r a n s i e n tty = Ks sy ( t )4ta t t = 4t , t h e r e s p o n s e i s a p p r o x .0 . 9 8 y .s s0which is the closed-loop differential equation. We can write this as m dv KcKe 1 + v = vd(t) + Fd(t)B + KcKe dt B + KcKe B + KcKe and compare with the standard first-order form τy˙ + y = u(t) where τ = m/(B + KcKe) is the closed-loop time-constant. The important thing to note is that feedback has modified the ODE. 3 Questions: Assume there are no external disturbance forces, that is Fd(t) ≡ 0 for now, (a) If we command the car to travel at a steady speed vd, what speed will it actually reach? To find the steady-state speed vss, set dv = 0 and solve for vss, giving dt KcKe vss = vdB + KcKe or vss < vd, for B > 0. Note that as we increase the controller gain so that KcKe � B then vss vd. → (b) How has the dynamic response affected by the feedback? Consider the first-order ODE dyτ + y = Ku(t). dt The response to a steady input u(t) = 1 with initial condition y(0) = 0 is t y(t) = K(1 − e− τ ) 2–4K i n c r e a s i n gs t e a d y - s t a t e s p e e da p p r o a c h e s v .tcdvvdt d e c r e a s e s0F ( t )pv ( t )F ( t)F = m g s i n (q)dpv(t)q( a ) o n l e v e l g r o u n d( b ) o n a n i n c l i n eFor the car under feedback control, this translates to a steady-state speed KcKe vss = vdB + KcKe which is a function of the controller gain Kc, and a closed-loop time constant τ that is m τ = B + KcKe which is also a function of Kc, and as Kc increases, τ decreases, meaning that the car responds more quickly to changes in the gas pedal. 4 The Effect of an External Disturbance Force Now consider the car on an incline under closed-loop control: (a) On horizontal ground Fd = 0, and we showed: KcKe vss = vdB + KcKe (b) On the incline there is a constant disturbance force Fd = −mg sin φ acting down …
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