1 Controller Optimization according to the Modulus Optimum

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1 Controller Optimization according to the Modulus Optimum w G K (s) F 0 (s) x The goal of applying a control loop usually is to get the control value x equal to the reference value w. x(t) w(t) X(s) W (s) F W (s) = X(s) W (s) for all s = jω ω R. The goal of applying modulus optimum ist to achieve a transfer function of the closed(!) control loop which is constant on value between zero frequency and a limit frequency as high as possible. In technical control systems transfer functions usually are minimal phase transfer functions, i.e. all poles and zeros have a negative real component. In these systems it is sufficient to know the amplification characteristic in case of a minimal phase transfer function the phase characteristic is then defined as well. F W (s) results in F W (s)

2 CounterExample: allpass network filter: G(s) = Ts + Ts G(jωs) = G(jωs) = 2 arctan(t ω) Im{s} T pole T zero Re{s} When dealing with electrical drives or inverters for control issues, the assumption of this system being a minimal phase system usually is ok. 2

3 2 2nd Order Controlled System 2. Modulus Optimum PIcontrol of a st order controlled system fulfills Nyquist criterion without any problems. X* F R F S X V F K. F S F R F S 2nd order controlled system: T F S (s) = ( s + )(T s + ) convention: T > > T 3... X* G K F S X X* F W X F K F R (s) is chosen to get: F K = F R F S F W (jω) = F W (jω) = F S (jω)f R (jω) + F S (jω) F R (jω) 3

4 Example: Controlled System F S (s) has one dominant time constant and several minor time constants Ks F S (s) = ( + T s) ( + T v s) Ks with ( + T s)( + T E s) T E = ν T v Chosen: P I controller F R (s) = K R + T s T s with the requirement of T > T E! Counterexample: Electrical and mechanical time constants of an electrical drive might be close to each other PI control does not work any more. F K (s) = F R (s) F 0 (s) = K R K S st ( + T E s) F W (s) = In the case of F W (s) x would be equal to x! For transfer functions with minimal phase F K(s) + F K (s) = K S K R K S K R + T s + T E T s 2 F W (s) = results from F W (jω) = F W (jω) = K S K R K S K R + jωt + (jω) 2 T e T F W (jω) 2 = K 2 S K2 R (K S K R ω 2 T E T ) 2 + ω2 = K 2 S K2 R K 2 S K2 R + ω4 E + ω2 ( 2T ET K S K R ) 4

5 for small ω : ω 4 E 0 F W (jω) 2 = for ω 2 ( 2T E T K S K R ) = 0 T = 2T E K S K R K R = T 2T E K S In fact the complete closed control loop not a proportional amplifier for all frequencies, but a first order low pass filter in the superposed control loop. Summary: Choose controller F R (s): to compensate the dominant time constant of the controlled system. Requirement: A dominant time constant does exist. to make the transfer function of the closed control loop constant in a wide frequency range: F W (jω) = Analytical solution for PI controllers: F W (jω) = F K = V Tis + T i s F K = T ik s( s + ) ; F W = F W = F K + F K = F 0(jω)G K (jω) + F 0 (jω)g K (jω) ( s + )(T s + ) ; T i = T Z N = Z Z + N + Z N T ik s 2 + T ik s + ( 5 s ω 0 ) 2 + 2D s T ik = T i V ω 0 + is valid in any case.

6 ω 0 = TiK T ik = 2 = T i V = T V 2D ω 0 = T ik T ik D = = 2 TiK 2 TiK! = 2 2 V = T 2 T i = T identical to T E V K R K S 2.2 P controller X* G K F S X X* F W X F K demand: F δ (jω) for ω = 0... ω W, ω W (as far as possible) F w 6

7 standardized form: F δ = ( ) 2 s ω 0 + 2D s ω 0 + F R = V F K = F R F S = F W = F K + F K = V ( s + )(T s + ) = N D = Numerator Denominator N D + N D Eigenvalues: Roots of the characteristic equation: = T s 2 + (T + )s + + V = 0 = N D + N V T s 2 + (T + )s + + V s,2 = (T + ) ± (T + ) 2 4T ( + V ) 2T Discussion of Root Locus: influence of the controller: root locus with V as parameter. crit. D s 45 T 7

8 characteristic points of the root locus: V = 0: s = T s 2 = ( ) (T + ) 2 4 T ( + V ) = 0: s = s 2 = 2 T + else: conjugated complex poles conclusions: (P controller) no stability problems damping decreases with increasing V the more distance there is between T and, the bigger must be parameter V to gain acceptable dynamics of the closed control loop(critical damping) T s K s s W by control s an optimized controller provides good position of the eigenvalues of the closed contol loop. 8

9 controller optimization : There is only a single parameter: V T s 2 + (T + )s + + V = 0 st solution s,2 = (T + ) ± (T + ) 2 4T ( + V ) 2T s,2 = (T + ) ± (T + ) 2 4T ( + V ) 2T Re(s )! = Im(s ) 45 only numerator of s,2 is to be treated: (T + ) 2 = (T + ) 2 + 4T ( + V ) + V = 2(T + ) 2 4T V = + 2 2T 2 nd solution 2 standardized form one term is (see above) F W = V V + T V + s2 + T + V + s + = V W ( ) 2 s ω 0 + 2D s ω 0 + increasing V means: ω 0 D 0 V W usual (aperiodic damping)) 2 modulus optimum V W = V V + V + 2D ω 0 = = T + T ω 0 v + D = T + 2 T (V + ) 9

10 modulus optimum: 2D 2 = D = 2 2 D q 0,5 2 2 in the case of critical damping:. ITAE(integral of time adapted error) 2. modulus optimum } are fulfilled presetting: D = D crit. = T + 2 = 2 2 T (V + ) (T + ) 2 2T = V + for example: V = (T + ) 2 2T = + 2 2T T = 4 V = = V W = V V + = 2 3 0

11 0,67 F H e( ) 33% V I K t dynamic behaviour: ok but not extremely good stationary deviation too big e e 0 is control signal e cannot be 0! P control of a low pass system results in. guaranteed stability 2. decreasing damping factor (D) with increasing controller gain (V ) 3. the more distance between T and the higher is critical gain V crit

12 2.3 Differential Controllers basic idea: control signal derived from de(t), 2 types: PD, PID dt F R F S. PD controller F R V V T d Td F R = V Tds + T d s + F K = V Tds + T d s + ( s + )(T s + ) V F K = (T d s + )(T s + ) 2

13 T pole shift to the left: (a) more dynamic behaviour (b) more control power last example: T d =, T d 0, T d V = T 2 + T d 2 2T T d dynamic behaviour (P D) ω 0 ω (P ) 0 = T = 4 T = 40T d = ; V δ = V V + = 20 2 e( ) = 5% ω 0 = 20 + (2 + ) 0, = V + = 0, 95 T 20 3 = 70 = 8, 4 = 0T d why not factor 0? ( T d = 0, ) phase shift by T did not have any influence so far but now it has 3

14 2. PID control P I D = PD controller with an I channel in parallel (more accurate, but worse in dynamics ( like P PI)) F R = V Tis + T i s low gain at ω ω d, high gain elsewhere Tds + T d s + V T Td d (t) t F R fast leading V T d Td d no phase shift 4

15 (a) pole shift to the left: more accuracy (b) pole shift to the right: more dynamics T' d T controller with D channel: F K = T ik s(t d s + ) (a) fast response is only possible when there is enough control power (small signal behaviour large signal behaviour) (b) increasing harmonics (inverters, digital sensors, e.g position sensors) alternative solution (if possible): make use of (i.e sense) a derivation of the controlled value from the controlled system itself e.g. C i C ( ) L i C = C du dt 5

16 T dx dt x* T x same F K (s) as with PID control, but there is no differentiation process in the controller. 2.4 Symmetrical optimum In the case of T > 7 a pole shift does not change the situation significantly. T s PI controller replaces pole at /T by pole in the origin. s This cannot work, if one of the poles already is in the origin (or close by)! 6

17 T i The most extreme situation is: T s + T s G (s) K F (s) 0 V,T i T m F K = V Tis + T i s ( s + )T m s = V T i T m s Tis + 2 s + = F F 2 T = T impossible i 2 F K,0 not stable 7

18 control loop with double integrator Requirements for stability and sufficient damping (Nyquist criterion) F K (jω d ) = arg(f K (jω d )) = π + Ψ d arg(f K ) = arg(f ) + arg(f 2 ) = π + arg(f 2 ) = π + Ψ d Ψ d = arg(f 2 (jω d )) F K,0 d d If T i, the characteristic of F 2 is: F 2 d s T i 8

19 Demand: T i shoud be as small as possible espacially as possible at ϕ = Ψ d otherwise the response on disturbancesis slow. ϕ = arg(f 2 ) should be max. F 2 (s) = T is + s + F 2 (jω) = + jωt i + jω ϕ = arg(f 2 ) = arctan ωt i arctan ω T + T i for any ω the resuling angle is positiv! dϕ dω ω=ω d = T i + ωd 2T + i 2 + ωd 2 2 T i ( + ω 2 d 2 ) ( + ω 2 d i ) = 0 T i T (T i T )ω 2 d T i = 0 ω 2 d T i =! = 0 (Maximum!) ω d = Ti 9

20 needed: T i mathematical solution: Ψ d = arg(f 2 ) Ψ d = arctan ω d T i arctan ω d T 2 Ti Ψ d = arctan arctan T2 optimized PI controller: calculation of T i based on Ψ d ; then calculation of V How to calculate T i from Ψ d T i a b T i a = ( ) Ti 2 b = a + = ( ) Ti + 2 sin Ψ = a T i b = T i + ( ) Ti sin Ψ + sin Ψ = T i T ( ) 2 Ti (sin Ψ ) = ( + sin Ψ) T i = + sin Ψ sin Ψ T i = + sin Ψ sin Ψ 20

21 now: calculation of V F K d d j F K (jω d ) = at transition frequency F K (jω d ) = V (ωd T i ) ωd 2T 2 + it m (ωd ) 2 +, ω d = Ti (see page 9) = V T i T i T m + T i T i + = V T m ( + T i + T i T i ) = V T 2 Ti Ti = V = T m T m for each T i (Ψ d ) there is a V V = T m Ti 2

22 4 The resulting open loop characteristic Im 3 2 2,25 0,6 0,75 0,5 0,4 2 q= d Re 22

23 The resulting open loop frequency charcteristic 2 F 2 F (lg) T i a d + a F K d 2 d 2 for symmetrical reasons: then = aω d = a 2 T i T i = a 2 ; to calculate a: F K (s) = normalized frequency: = = V T i T m s Tis + 2 s + V a 2 T m s a2 s + 2 s +, V = T m = T m Ti a a 3 T2 2 s a2 s + 2 s + q = s ω d = s T i = a open loop: F K (q) = aq + aq 2 ( a q + ) 23

24 closed loop: F g = F g (q) = F K = N + F K N + D aq + q 3 + aq 2 + aq = N(q) D(q) }{{} Eigenvalues: D(q) = 0; easy to realize: q = is a solution further solutions: q 2,3 = a ( ) 2 a ± j 2 2 (a ) 2 ( ) 2 a q 2,3 = + = 2 2 a a= j q q 2 pole 36 q= a>3 a>3 a zero a=3 q 3 a j 24

25 D 2,3 > 2 2! F g (q) = aq + q 3 + aq 2 + aq = N(q) D(q) F g (q) = aq + (q 2 + (a )q + )(q + )! = aq + (q 2 + 2Dq + )(q + ) 2D = a a = + 2D T i = a 2 T V = T m a a opt. = 2, 6 D, a optimized according to ITAE: D opt. = 0, 7 F g (q) = aq + q 3 + aq 2 + aq + 30% (t) t 25

26 to compensate overshoot: filter in reference channel in general: τ i : normalized time t i optimized according to ITAE: F W = τ 2q + τ q + τ 2 = D opt. = 0, 7 τ = a a opt. = 2, 6 a 26

27 a simple low pass filter would be too slow (acc. to Leonhardt),3 without reference filter w(t),0 F Filter ( q) q aq 0,5 F Filter ( q) aq d t symmetrical optimum step responses 27

28 2.5 Limit between modulus optimum and symmetrical optimum scheme Z 2 Z x* x controller T 3 T PI controller F R = V Tis + T i s s T 3 T 2 a T T opt 2 s a) T s b) 28

29 a) modulus optimum T < a 2 opt. b) symmetrical optimum T > a 2 opt. a opt. = 2, 6 a 2 opt. = = 6, 75 29

MAS107 Control Theory Exam Solutions 2008

MAS107 Control Theory Exam Solutions 2008 MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve