Ch 14: Feedback Control systems


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1 Ch 4: Feedback Control systems Part IV A is concerned with sinle loop control The followin topics are covered in chapter 4: The concept of feedback control Block diaram development Classical feedback controllers Closedloop transfer functions Closedloop transient response Closed loop stability Root locus diarams
2 4.,4. Feedback control systems  example Stirred heatin tank system under feedback control
3 4.3 Classical feedback controllers P PI PID PD G(s) K c G(s) K c (TIs) G(s) K s c D ( Is) G(s) K c s D
4 4.4 Closed loop transfer functions (/3) d h y h y s c v d d s c v c v d d hy y y hy y u u d d c v d c v c v )) ( ( ) (
5 Closed loop transfer functions (/3) This can be eneralized: CLTF f L Where: f L Product of all the transfer functions in the direct path between the output (y) and input (y d or d) Product of all the transfer functions in the entire loop
6 Closed loop transfer functions (3/3) Example 4. d h h d h h y h h y c v c v d d c v c v y
7 4.5. Closed loop transient response Let s simplify the process: Transfer function: y c c h y d d c h d
8 4.5. Transient response of st order system under Pcontrol Stirred heated tank Valve dynamics neliible Measurements perfect Transfer function of the controller: (s) K c c d ( s) s ( s) s where pvc p residence time K =
9 Transient response of st order system under Pcontrol Unit Step in Setpoint (/) The transfer function simplifies to: y( s) K * * s s Where: Closed loop ain: closed loop time constant KKC K* KK * KKC C Transient response of the system: y( t) K *( e t /* )
10 Transient response of st order system under Pcontrol Unit Step in Setpoint (/) Notifications: As t, y K* y approaches K* exponentially K* will always be less than for all finite values of K C offset = y d y( ) Offset= KK c KKc As K C the offset tends to zero
11 Transient response of st order system under Pcontrol Unit Step in disturbance c d c d c c d KK K KK s s K s y s s KK s s y s d s y * * * ) ( ) ( ) ( ) ( * ) *( ) ( /* t K d e t y c d KK offset y y offset 0 ) (
12 Transient response of st order system under P control Unit Step in disturbance Similarly unit step in disturbance causes offset, which is KK C
13 Effect of Pcontrol to st order systems (/) Open loop: K y( s) u( s) s Closed loop: ( v =, h=) Where: K * y( s) yd ( s) * s KKC K* KK C * KKC
14 Effect of Pcontrol to st order systems (/) Modify the dynamic behavior Retain the order Effect to the time constant: KK C > 0 * < (speeds up system)
15 Closedloop transient response example 4.3 (Pcontrol) Three tanks K c = ( s) (s 6 )(4s )(6s ) 6Kc (s )(4s )(6s ) y( s) yd ( s) 6Kc (s )(4s )(6s ) y( s) 6 (s )(4s )(6s ) 6 s
16 4.5.3 Effect of PIcontrol on st order systems Can exhibit overshoot (zero involved) Dependin on the poles the response can approach its ultimate value exponentially (real and neative roots), or it can damp oscillatory (complex conjuates) With positive root unstable y( s) s I KK ( I s ) ( KK ) s KK I C C C s
17 4.5.3 Effect of PIcontrol on st order systems (IMPORTANT!) The closed loop transfer function has a firstorder numerator polynomial and a second order denominator i.e. It has one zero and two poles Under P control, the closed loop transfer function was first order, with one pole and no zero The effect of PI control is to add a pole and a zero to the process transfer function.
18 4.6 Stability analysis Characteristic equation and it s roots If (and only if) all roots of the Characteristic equation (denominator polynomial of the closed loop transfer function) have neative real parts, the system is stable c y c h y d s For low order polynomial characteristic equations, the roots are easy to find and the stability analysis is therefore easy to carry out. For hiher order polynomials, however, findin the roots can be quite tedious
19 4.6.3 Routh s stability test How many roots of a polynomial have positive real parts? Only for polynomial equations (no dead times) Actual values of the roots doesn t have to be calculated
20 Routh s Stability test procedure (/3) Write the characteristic equation in the standard form: a 0 s n + a s n + +a n s +a n s+a n =0 where a 0 is positive (multiply with  if necessary) Inspect the coefficients a 0 a n: If just one coefficient is neative, at least one root lies in the RHP unstable system If all the coefficients a 0 a n are positive, proceed
21 Routh s Stability test procedure (/3) Generate the Routh array: Row a 0 a a 4... Row a a 3 a 5 Row 3 b b b 3 Row 4 c c c 3 Row (n+) φ b b c c aa a0a a aa4 a0a a ba 3 ab b ba 5 ab 3 b
22 Routh s Stability test procedure (3/3) Analyze array: If all the elements in the first column are positive and nonzero, all roots lie in LHP If any element is neative, there is at least one root in RHP. Number of sin chanes number of RHP roots If all elements vanish in the nth row, there is a pair of roots in the imainary axis. It can be located by solvin equation: p s +p =0, where p and p are elements in row (n)
23 Routh  example 4.5 s 4 +s 3 +5s +6s+=0 Routh array: Row 5 Row 6 Row 3 Row 4 5 Row 5 b b a a a a 56 0 a Stable
24 Routh  example s 3 44s s 0 Row 48 Row 44 Row Row 4? The elements in the 3rd row vanish. Let s solve the imainary roots from row : p s +p =0 44s 0 s j
25 4.6.4 Direct substitution Stability limits can be solved by substitutin s j to the characteristic equation and so findin the stability limits for the unknown controller parameters. ( when a closed loop system is on the vere of instability, at least one pair of roots of the characteristic equation must lie on the imainary axis of the complex plane)
26 Direct substitution  example 48s 3 48 j 3 ( s 44 ) j Now both the imainary and real part must be zero resultin in two solutions: =0, K c = /6 or = ±/, K c = 5/3 /6 < K c < 5/3 is the stable rane of controller ain s ( 6K j ( 6K C C ) ) ( 6K ) 44 0 C 0 0
27 4.7 The Root Locus diaram A Plot in the complex plane of all the roots r,r,,r n as the control ain K c is varied over some rane 0 < K c < K c,max Visualization of stability When we want to know more than just a stability limit The entire history of how chanes in Kc values affect the roots of the characteristic equation.
28 Root locus example (/) Three thank system plus Pcontroller: Characteristic equation: 48s 3 44s s ( 6 KC numerical polynomial root finder, when 0 K c 3.0 ) 0
29 KC r r r3 00,670,50,5 0,0040,050,050,506 0,050,8j+0,6j 0,80,6j 0,553 0, 0,40+0,9j 0,400,9j 0,637 0,50,093+0,35j 0,0930,35j 0,730,045+0,47j 0,0450,47j 0,88,67 0,5j 0,5j 0,97 3 0,06+0,63j 0,060,63j ,04
30 Observations For Kc=0, there are three real roots= open loops poles of (x): 0,670,500,5 For 0<=Kc<0,004, all the roots are real and there is no oscillatory response For 0,004<Kc<,67, two of the roots are complex with neative real parts so that the response is damped oscillatory
31 Observations For Kc>,67, two of the roots are complex with positive real parts so that the response is oscillatory with increasin amplitude and unstable. Since the frequency of the sinusoidal portion of a system s oscillatory response is determined by the imainary part of the characteristic equation roots, the oscillatory response takes on hiher and hiher frequency as Kc increases.
32 Observations Since the third root r3 stays real and becomes more neative as Kc increases, the transient response arisin from the term becomes faster as Kc increases e r 3 A 3 t
33 Root locus example (/)
34 Root locus eneral features The diaram will show as branches of the poles of c h Symmetrical about the real axis The loci has n branches, r of which terminate at the zeros of c h, while the remainin (nr) branches o to infinity as Kc increases.
35 Root locus eneral features The closed loop system response is nonoscillatory for points on the loci that lie completely on the real axis. When the loci branch away from the real axis, the closed loop system response becomes oscillatory. The closed loop system is on the vere of instability at the point where one of the branches first intersects the imainary axis in oin from the left to the riht half of the complex plane.
36 Root locus example 4. c h Kc( s )( s ) (0.5s )(s )(4s )
37 Root locus example 4. (time delays) c h Kc (0s ) e 5s
38 Ch. 5: Conventional Feedback Controller Desin Controller desin principles Controller tunin: with fundamental process models CT with Approximate process model CT with Frequency response models CT without a model
39 5. Controller desin principles Controller type P, PI, PID, PD Parameters for the controller Performance criteria: Stability Steady state Dynamic response
40 Performance criteria Minimum rise time t r Min. settlin time t s Specified max. Overshoot a < a,spec Specified max. Decay ratio a /a < d r
41 Performance criteria example 5. Three water tanks, PIcontrol: 6 ( s) (s )(4s )(6s ) 6Kc( I s ) y( s) y( d) 6K ( s ) s(s )(4s )(6s ) c I I
42 Time interal performance criteria Interal absolute error Int. Squared error (lare errors) Int. Timeweihted absolute error (lon times) Int. Timeweihted squared error (lare errors at lon times) IAE ISE ITAE ITSE 0 0 ( t) dt 0 0 ( t) dt t ( t) dt t ( t) dt
43 5.. Controller types: P Accelerate the control system response Leaves nonzero steadystate offset Use when offsets are unimportant or when there is a natural interator Liquid level controls, for example (depends on the case!)
44 Controller types: PI Eliminates offset Response becomes oscillatory As K c increases, the interal action increases the propensity towards instability Use when offsets cannot be tolerated
45 Controller types: PD Anticipatory and stabilizin effect of derivative action A nonzero steadystate offset Seldom used can tolerate hiher controller ains
46 Controller types: PID No offset Derivative action curbs oscillation Derivative action amplifies noise components in noisy sinals Use when important to compensate natural sluishness and the system is noise free
47 5.3 Controller tunin with fundamental process models Optimization of Timeinteral Criteria: The three tank system PIcontrol CONSYDEXproram:
48 5.3. Model Followin PID Controller Desins Findin the PID controller parameters that cause the actual closedloop system to behave in a precribed fashion.
49 5.3. Model Followin PID Controller Desins; Direct synthesis The closed loop transfer function for a process under feedback control: A desired trajectory represented by the transfer function q(s) We obtain c y( s) yd ( s) h y( s) q( s) y ( s) c c d ( q q )
50 5.3. Model Followin PID Controller Desins; Direct synthesis Fully explained later Reference trajectory: = 3. Some mathematics p.56 equations for control parameters (PID): K c I D r r 4 r r r K K Some help parameters: ) ( ) ( ) ( s s s y s y s q r r r d ) )( )( ( ) ( 3 s s s K s ) ( ) )( )( ( ) ( ) ( ) )( )( ( ) ( 3 3 s s s s s s s K s s s s c r r r c
51 Direct synthesis  example Three tanks Reference trajectory: τ r =4.0, ζ r =0.5 K c ( s) (s )(4s )(6s ) y( s) q( s) y ( s) s q d ( s) 6s r 4s s r r 0.5 I D 0.50
52 5.3. Model Followin PID Controller Desins; Pole Placement Desins Controller desin based on pole placement By adjustin K c, I, D it is possible to locate the closedloop poles wherever we want in the complex plane (p. 530) The closed loop poles of a feedback control system are the roots of the characteristic equation h 0 c
53 Stability marins Zieler Nichols With Pcontrol et the controller ain in the vere of instability (K cu ), The perioid of the sustained oscillation at this ain: P u Get the parameters from table:
54 ZielerNichols  example K cu =.667 P u =.56 P: K c = And so on
55 5.4 Controller tunin with approximate process model Simple models for controller tunin Process characterization: Reaction curve can be approximated by step response of a first order system with time delay.
56 Process characterization Parameters: K, and a are enouh to characterize the process: ym () K A y () m
57 Process characterization  example Three tanks
58 5.4. Tunin rules For example Zieler Nichols CohenCoon, ITAE, Direct Synthesis, IMC (p )
59 Tunin rules example (three tanks) PI PID
60 5.5 Controller tunin with Frequency response models; stability marin methods The basic principle involves determinin the controller parameters that will brin the closedloop system to the vere of instability( the closed loop system is at border between stability and instability) and back off on the tunin parameters to provide some stability marin. Determinin the point of closed loop instability from frequency response models
61 5.5 Controller tunin with Frequency response models; stability marin methods y d c v y y m h Openin the closed loop y m v c hy d y v c y d L 0 vch
62 5.5 Controller tunin with Frequency response models Nyquist diaram for the loop transfer function= a plot of the transfer function in the complex plane L L = C ( j) Re( ) j Im( ) L ( j) AR( ) e Kartesian form j ( ) AR L ( j) Polar form ( ) L ( j) tan Im( ) Re( )
63 Nyquist Stability Criterion (/3) The imainary axis in the Cartesian coordinates corresponds to the curve in the Nyquist plot (and its mirror imae) RHP (unstable roots) corresponds to the interior of the curve
64 Nyquist Stability Criterion (/3) If N is the net number of times that the curve L (s) encircles the point (,0) in the clockwise direction, and P is the number of unstable openloop poles of L (s) (if P>0 unstable open loop process) The number of unstable roots of the closedloop characteristic equation is U=N+P ( + G(s)=0
65 Nyquist Stability Criterion (3/3) Stable openloop process P=0 closed loop process is unstable only if there are net clockwise encirclements of (,0) Unstable openloop process ( P> 0) There must be at least P counterclockwise encirclements of (,0) for the controller to stabilize the process
66 Nyquist Stability Criterion Since the controller ain Kc control the size of AR, the Nyquist plot may therefore be used in conjunction with the Nquist stability criterion to determine stability limits on the controller parameter values.
67 Determinin stability limits. Get the L (s) for the process. Obtain the Nyquist plot usin value K c = 3. Determine the critical point at which the ain neutral Nyquist plot intersects the real axis ( γ, 0) 4. Determine the value of the ultimate ain ( the value of K required to cause the Nyquist plot to intersect the real axis at (,0) K cu and stability limits from the critical point K cu K c
68 Nyquist  example Three tanks (surprise ) Determine rane of stable controller ains a) Pcontrol Intersection with real axis at (0.6,0) system is unstable, when K c > /0.6 =.667 b) PIcontrol Intersection with real axis at (0.73,0) system is unstable, when K c > /0.73 =.37 c) PIDcontrol Unstable, when 0.05 < K c <0.674
69 Bode stability criterion (/) An alternate form of frequency domain representation is the Bode plot. Critical point AR= for the phase anle  80 deree. If the AR of the openloop transfer function is reater than when φ =  80 the closedloop system is unstable
70 Bode stability criteria (/) Specify all parameters, but K c, et bode diaram(kc=) Determine the crossover frequency ω c the value of ω for which φ =  80 Determine the critical value AR c at this crossover frequency The ultimate value K cu is obtained as: K cu AR c
71 Bode stability criteria  example Gain neutral bodediaram: ω c = 0.5 AR c =0.60 stable, when K c <.667
72 5.5.3 stability marin tunin methods in the frequency domain; Gain and Phase marins Determine the critical points ain and phase marins. GM AR c PM ( 80) These are quantitative measures, how far system is from bein On the vere of instability Stable GM> and PM> 0 GM about,7 and PM about 30 deree
73 ZielerNichols stability marins from Bode diaram Get the bodediaram for Pcontrolled K c = system (ain neutral) Get the critical value of AR K cu AR c P u Get the parameters for desired controller from table: c
74 5.6 Controller tunin without a model Experimental determination of closed loop stability criteria ZielerNichols AutoTunin Tunin with relay controller control amplitude h resultin output amplitude A K cu 4h A
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