# Lecture 25: Tue Nov 27, 2018

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1 Lecture 25: Tue Nov 27, 2018 Reminder: Lab 3 moved to Tuesday Dec 4 Lecture: review time-domain characteristics of 2nd-order systems intro to control: feedback open-loop vs closed-loop control intro to PID control 0

2 Quiz 2 Results min = 66 5 mean = 91 median =

3 Q2 vs Q QUIZ QUIZ 1 2

4 Q1 + Q min = 92 mean = 166 max = 197 median =

5 Time Domain Properties of 2nd Order Systems 4

6 Underdamped LPF Step Response Can t measure n and directly, but can derive them from measurements of: steady-state value (determines H 0 ) percentage overshoot (determines ) peak time (given, determines n ) settling time (1 + )H 0 y( t ) STEP RESPONSE 1.02H 0 H H 0 0 t pk t settle t 5

7 Start with LPF: 2 H( s ) = H 0 n s n s + n Start with Impulse Response: h( t ) 8 = = = / n t -2 1/ n -4 (Later Find Step Response) 6

8 Underdamped Impulse Response? = d n 1 2 n If poles were on j axis, h( t ) = Ksin( d t)u( t ). By mod property, shifting poles to left introduces an exponential envelope: h( t ) = Ke nt sin( d t)u( t ): h( t ) ENVELOPE t 7

9 To Find the Constant K H( s ) = 2 H n s n s + n 2 H n s n = (complete the square) H 0 n n d = (use table 2 and modulation property) 1 2 s d H 0 n 1 2 n h( t ) = e nt sin( d t)u( t ). K = H n 1 2 oscillates with frequency d t 8

10 Three Different Frequencies natural frequency n damped frequency d resonant frequency r 9

11 Three Different Frequencies natural frequency n : were there no damping (no friction or resistance) + L C n = LC m y k n = k m ---- my + ky = 0 damped frequency d = n (accounts for friction or resistance) 1 2 this is the frequency of oscillations in the time domain resonant frequency r = n (where frequency response peaks) 10

12 Underdamped LPF Step Response Write Laplace transform of step response as: 2 1 Y( s ) = -- 1.H( s ) =. H n s s s n H s As + a + B d s + a d n = (where a = n ) PFE: s 2 : H 0 A = 0 A = H 0 s 1 : 2H 0 a Aa B d = 0 B = H y( t ) = H 0 u( t ) e n t u( t )[Acos( d t) + Bsin( d t)] = H 0 u( t )(1 e n t [cos( d t) sin( d t)])

13 Phasor Addition, or Trig Identity sin 1 2 cos cos( d t) sin( d t) = (sin cos( d t) + cos sin( d t)) = sin( d t + )

14 3 Equiv ways to Write Step Response Underdamped step response for a 2nd-order LPF system can be written as: y( t ) = H 0 u( t )(1 e n t [cos( d t) sin( d t)]) = H 0 u( t )( e n t sin( d t + )) 1 2 = H 0 u( t )( e n t cos( d t )): 1 2 sin 1 2 cos 13

15 y( t ) Time of Peak STEP RESPONSE 0 t t pk To find t p : Set ---- d y( t ) to zero and solve for t. dt IOW set h( t ) = 0! H 0 n 1 2 But h( t ) = e nt sin( d t)u( t ). h( t ) IMPULSE RESPONSE t pk t Zero when sin(. ) argument is equal to the time of the first step-response peak is t pk = = d n

16 Overshoot (1 + )H 0 STEP RESPONSE H 0 0 t pk t Peak overshoot can be found by evaluating step response at t p = / d : y( t p ) = H 0 (1 + e / 1 2 ) overshoot is = e / 1 2. Example: =0.707 = 0.043, or 4.3% of overshoot. 1 From measured overshoot, we can estimate via = log

17 Step response: y( t ) = H 0 u( t )(1 LPF Settling Time When does the envelope decay to 0.02? e n t = e n t sin( d t + )) 1 2 t s t s = ln n n 20 n 15 n 10 n (EXACT) (APPROX) 5 n

18 LPF Step Response (1 + )H 0 = e / = log H 0 0 t t pk = n 1 2 t s n 17

19 Relationship to Pole Locations What stays the same: d, = e / 1 2, or t s n 4 18

20 Relationship to Pole Locations What stays the same: d, = e / 1 2, or t s n 4 SAME SETTLING TIME SAME ENVELOPE SAME OSCILLATION FREQ SAME OVERSHOOT 19

21 Impact of Pole Movement 20

22 Circuit Example R L + v in ( t ) F v out ( t ) Design R and L so that its step response settles in 2 seconds (to ±2% of its steady-state value), with 20% overshoot. 21

23 Solution 1/LC 4/L H( s ) = = s 2 R s + 1/LC s 2 R s + 4/L L L % overshoot = = log L H 2-sec settling time t s = L = 2 = n R --- L = 2 n = = 4 R = 4L = t s

24 Pop Quiz A 2nd-order LPF system has the following step response: (a) Find H 0 (b) (c) Find Find n 23

25 Pop Quiz A 2nd-order LPF system has the following step response: (a) Find H 0 (b) (c) Find Find n 24

26 HPF Step Response? 1 Y( s ) = -- 1 H( s ) = s 2 s s s n s + n = = s s 2 2 n s n 2 s + n n + n s 2 2 d 1 y( t ) = e nt sin( d t + )u( t ) 1 2 s s n s + n R C + L v v in out 3 2 r = n / = sin 1 2 cos (Except for phase change, this is proportional to LPF impulse response!) n 2 n 25

27 Coming Next: Controls 26

29 Connecting Systems Series x( t ) y( t ) H 1 ( s ) H 2 ( s ) Parallel H 1 ( s ) x( t ) y( t ) + H 2 ( s ) Feedback x( t ) y( t ) H 1 ( s ) + H 2 ( s ) 28

30 Overall Transfer Functions Easy: Series H 1 ( s )H 2 ( s ) Parallel H 1 ( s ) + H 2 ( s ) Feedback: Not as easy! Chicken & egg. x( t ) e( t ) y( t ) H 1 ( s ) + H 2 ( s ) E( s ) = X( s ) Y( s )H 2 ( s ) Y( s ) = E( s )H 1 ( s ) E( s ) = Y( s )/H 1 ( s ) Ys H Equate H overall ( s ) = = ( s ) Xs 1+H 1 ( s )H 2 ( s ) 29

31 Control Components REFERENCE r( t ) CONTROLLER G c ( s ) x( t ) y( t ) ACTUATOR G p ( s ) PLANT Reference Desired output y( t ). Plant system we re trying to control. We re stuck with it. Controller Produces input to plant in attempt to produce desired output Examples: cruise control temperature control steering control robotic control rocket guidance 30

32 Two Types of Control OPEN LOOP G p ( s ) REFERENCE r( t ) CONTROLLER ACTUATOR PLANT SENSOR y( t ) G c ( s ) CLOSED LOOP G p ( s ) REFERENCE ERROR CONTROLLER x( t ) ACTUATOR PLANT SENSOR y( t ) 31

33 Two Types of Control OPEN LOOP G p ( s ) REFERENCE r( t ) CONTROLLER ACTUATOR PLANT SENSOR y( t ) G c ( s ) CLOSED LOOP G p ( s ) REFERENCE + ERROR CONTROLLER x( t ) ACTUATOR PLANT SENSOR y( t ) 32

34 Control Components: Heating a House REFERENCE 68 + ERROR CONTROLLER ACTUATOR PLANT SENSOR 33

35 Control Components: Cruise Control REFERENCE 55 MPH + ERROR CONTROLLER AMOUNT OF FUEL ACTUATOR PLANT SENSOR 34

36 Segway REFERENCE + ERROR CONTROLLER ACTUATOR PLANT SENSOR 35

37 Tracking Control Set a speed, temperature, position, angle, etc. and have plant track it. We are given or specify a reference signal r( t ), to be tracked. We want y( t ) = r( t ) How to choose x( t )? 36

38 Option 1: Open Loop Control R( s ) G c ( s ) G p ( s ) Y( s ) Y( s ) = R( s )G c ( s )G p ( s ) Example: G p ( s ) = b, how to choose open-loop controller? s + a 1 G c ( s ) = = G p s s + a b Two Problems: For a unit-step reference, differentiator yields impulse In practice we cannot know G p ( s ) exactly. 37

39 Steady-State Tracking When reference is r( t ) = r 0 u( t ): instantaneous tracking: y( t ) = r( t ) for all t, often impractical steady state tracking: y( ) = lim t y( t ) = r 0 Example: r 0 REFERENCE r( t ) y( t ) STEADY-STATE TRACKS DOESN T TRACK 0 t 38

40 Continue example... Proportional Control Does proportional controller G c ( s ) = K perfectly track in steady state? Y( s ) = R( s )G c ( s )G p ( s ) = solve for y( t ): Kbr 0 ss + a 0 t Is y( ) = r 0? What choice for controller gain K yields perfect steady-state tracking? 39

41 Pop Quiz Can an open-loop controller stabilize an unstable plant? 1 R( s ) G c ( s ) s 2 Y( s ) What open-loop controller G c ( s ) results in perfect tracking of a unit step? How robust is this controller to plant changes, e.g. if G p = 1/(s 2.1)? 40

42 Open Loop Drawbacks requires perfect knowledge of plant model not robust 41

43 Feedback: When output impacts input Advantages: robust to external disturbances robust to model knowledge can linearize system made of nonlinear components Disadvantages: can lead to oscillations can lead to instability 42

44 Closed Loop Control R( s ) E( s ) Y( s ) + G c ( s ) G p ( s ) Derive the closed-loop transfer function (relating r( t ) to y( t )): H( s ) = = Y( s ) R( s ) 43

45 Closed Loop Control R( s ) E( s ) Y( s ) + G c ( s ) G p ( s ) Derive the closed-loop transfer function (relating r( t ) to y( t )): Y( s ) H( s ) = = R( s ) G c ( s )G p ( s ) 1 + G c ( s )G p ( s ) 44

46 Examples of Controllers P PI PD PID K p G c ( s ) K p + K i /s K p + K d s K p +K i /s + K d s R( s ) E( s ) Y( s ) + G c ( s ) G p ( s ) 45

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