# Control for. Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e

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1 Control for Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e

2 Motion Systems m F

3 Introduction Timedomain tuning Frequency domain & stability Filters Feedforward Servo-oriented design of mechanical systems 3

4 . Time Domain Tuning 4

5 Servo force? F s Mass M Disturbance F d 5

6 Mechanical solution: k d F Disturbance F d Mass M Forcespring-damper: F = k d & Eigenfrequ ency: f = π k M 6

7 Servo analogon: F servo Mass M ServoForce: F s = k p k v & Eigenfrequ ency: f = π k p M k k p v : : servo stiffness servo damping 7

8 Eample: Disturbance F d F s Mass M Slide: mass = 5 kg Required accuracy 0 µm at all times Disturbance (f.e. friction) = 3 N. Required servo stiffness?. Eigenfrequency? 8

9 9 h or s ) ( ) ( : -damper Spring h d h k F & & + = How to move to / follow a setpoint: ) ( ) ( : Controller k k F s v s p s & & + =

10 s F d F disturbance controller process s + - e k p + k v d dt + + F servo F Mass K p /k v -controller or PD-controller 0

11 s+ - e controller k k p v =? =? F servo process k k p v error Trade off stability

12 Concluding remarks time domain tuning A control system, consisting of only a single mass m and a k p /k v controller (as depicted below), is always stable. k p will act as a spring; k v will act as a damper As a result of this: when a control system is unstable, it cannot be a pure single mass + k p /k v controller (With positive parameters m, k p and k v ) s k p M k v

13 Setpoints: s What should s look like as a function of time, when moving the mass? (first order, second order, third order,.?) 3

14 Apply a force F (step profile): F ( t) = M & ( t) F M (t) is second order, when F constant Second order profile requires following information: - maimum acceleration - maimum velocity - travel distance 4

15 Eample Pos Vel Acc ma ma = = = 0π 500π π 6.3rad 63rad.6e3rad / sec / sec 5

16 3 Frequency domain Time domain: Monday and Thursday at :0 Frequency domain: twice a week 6

17 F M amplitude (log) H = F - phase H -80 o frequency (log) 7

18 going from Time-domain to the Frequency-domain weak spring (f =.5 Hz) M M = 5 kg F ecitation force (offset 0. &scaling e-4) Ft ( ) = 400sin( π 7t) response H( 7Hz) / 400= e 4 m/ N H( 7Hz) time in sec 8

19 finding a solution of the equation of motion: choose input: then: solution: F = M F = F sin( ωt) = sin( ω t+ ϕ ) =?; ϕ =? t ( ) = F sin( ωt) + ct+ c Mω H = = F Mω log( H) = log logω M F M H = = F Mω H = ϕ = 80 9

20 measurement mechanics stage amplitude in db frequency in Hz phase in deg

21 Derivation of transfer function - make a model of the dynamics: differential equations - substitute s=d./dt - rearrange the equations and get the transfer function e.g. H(s) - for sinusoids make a Bode plot using s=jω

22 Transfer function: H ( s) ( s) = = F( s) Ms + ds + k F Ms + ds + k

23 consider sinusoidal signals ('Euler notation'): ( t) = ˆ(cosωt + j sin ωt) = e ˆ jωt & ( t) = ωˆ( sin ωt + j cosωt) = jωe ˆ jωt apparently: s = jω for sinusoidal signals Frequency Response Function: s jω H( jω) = Mω + jdω + k 3

24 e k p + k v d dt F F = k p e + k v e& F() s = ( k + k s)() e s transfer function: F Cs () = () e s = ( k k s p + v ) frequency response: p C = k + jk ω p v v 4

25 Amplitude: C = + k p k v ω ω 0 C k p C 0 ω C k v ω C 90 ω ( C ) = log k logω log + v break point: log k = log k + log ω p ω= k k p v v 5

26 Bode plot of the PD-controller: amplitude in db 00 kp = 500 N/m; kv = 0 Ns/m frequency in Hz phase in deg

27 Block manipulation e C(s) H(s) s () es () = CsHs () () C + H - k M p k p k v frequency CH - - frequency 7

28 s + e - C(s) H(s) H c = s = CH + CH 8

29 Four important transfer functions. open loop: Ho( s) = C() s H() s s + F d e + C(s) H(s) - F s +. closed loop: H c ( s) = s ( s) = C( s) H ( s) + C( s) H ( s) 3. sensitivity: 4. process sensitivity: e S() s = () s = + CsHs () () H ps s s F s Hs () () = () = + CsHs () () d 9

30 s + F d e + C(s) H(s) - F s + Derivation of closed-loop transfer functions: start with the output variable of interest go back in the loop, against the signal flow write down the relations, using intermediate variables stop when arrived at the relevant input variable eliminate the intermediate variables 30

31 amplitude in db 0 closed loop phase in deg amplitude in db phase in deg open loop frequency in Hz Eperimental results: stage servo

32 bandwidth: 0 db crossing open loop (cross-over frequency) 3

33 The Nyquist curve litude in db 50 Bode plot.5 Nyquist plot hase in deg 0 frequency amplitude (appr 0.7) phase (appr. -75 deg) frequency 33

34 Stability: The open-loop FRF CH(jω) should have the (-,0) point at left side.5 Nyquist plot Im Re -0.5 ω ω

35 4. Filters Mimo Integral action Differential action Low-pass High-pass Band-pass Notch ( sper ) filter PeeDee PeeEye 35

36 Integral action : X(t) τ i s Y(t) - τ I integral time constant τ I =/k i 0 ω=πf

37 Differential action + u u H = ks = ; s = jω; = ε ε kω +90 ω ε ks u 0 u ε tamme differentiator : = τ d ks s τ d = ω d = πf d ω 0 37

38 38 s s s s u H d d γ τ τ τ τ ε + + = + + = = lead filter γ> ω τ ω= τ ω = 5 τ τ γ τ τ ω ω ω = = c

39 ω i i τ ω = d d τ ω = - + P+I+D = = γ τ τ τ ε s s s k u d d i

40 ε(t) ω - s nd order filter ω s u(t) H = u ε = s ω + k β s ω + β β Top:. ω o = ω β 0 - ε ω=ω 40

41 = = ω β ω ω β ω ε s s s s u H General nd order filters ω ω General: ω ω

42 ω ω + ω ω - ω 0-80 ω ω 4

43 Notch -filter :ω = ω ampl. β β fase

44 W.B.E. 44

45 Loop shaping procedure. stabilize the plant: add lead/lag with zero = bandwidth/3 and pole = bandwidth*3, adjust gain to get stability; or add a pure PD with break point at the bandwidth. add low-pass filter: choose poles = bandwidth*6 3. add notch if necessary, or apply any other kind of first or second order filter and shape the loop 4. add integral action: choose zero = bandwidth/5 5. increase bandwidth: increase gain and zero/poles of integral action, lead/lag and other filters during steps -5: check all relevant transfer functions, and relate to disturbance spectrum 45

46 Implementation issues. sampling = delay: linear phase lag for eample: sampling at 4 khz gives phase lag due to Zero-Order-Hold of: 4 khz 400 Hz 00 Hz. Delay due to calculations 3. Quantization (sensors, digital representation) 46

47 5 Feedforward design 47

48 Why feedforward? Consider the simple motion system m F Control problem: track setpoint 0 0 s Setpoint t [s] s.8 Is this possible with a PD-controller? 48

49 Analysis (IV) -3 0 error [m] K p = [N/m] K p = 6500 [N/m] m = 5 [kg] K v = 60 [Ns/m] t [s] 49

50 Feedforward based on inverse model ms s K p + K v s ms 50

51 Eample: m=5 [kg], b= [Ns/m], nd degree setpoint s [m] vs [ms-] as [ms-] t [s] 5

52 Eample: tracking error, no feedforward viscous damping effect error [m] t [s] 5

53 Eample: tracking error, with feedforward K fv = 0.9, K fa = 0 error [m] = 0.9, = 4.5 K fa K fv t [s] 53

54 feedforward structure sign( & s ) K fc & & s K fa & s K fv s C(s) H(s) 54

55 3rd degree setpoint trajectory.5 s [m] vs [ms-] as [ms-] t [s] 55

56 6. Servo-oriented design of mechanical systems 56

57 Eample of measurement: mechanical system (force to position) modelling understanding the dynamical behaviour 57

58 Three Types of Dynamic Effects - Actuator fleibility - Guidance fleibility - Limited mass and stiffness of frame 58

59 . Actuator fleibility F s Motor k Sensor d 59

60 . Guidance fleibility F s M, J k 60

61 3. Limited mass and stiffness of frame F s Motor Frame 6

62 M M Positioning the load M (while using for feedback): Rule of thumb: Optimal bandwidth with 0 db crossing of open loop between the antiresonance and resonance frequency of the mechanical system. 6

63 Concluding Remarks bit of control into mechanical design bit of mechanics into control design same language ( mechatronics ) 63

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