Reviewer: prof. Ing. Miroslav Olehla, CSc. Osvald Modrlák, Lukáš Hubka Technical University of Liberec, 2014 ISBN

Transcription

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2 Bibliographic reference to this document: MODRLÁK, O. a L. HUBKA. Automatic Control in Mechatronics. 1st edition. Liberec: Technical University of Liberec, Faculty of Mechatronics, ISBN Reviewer: prof. Ing. Miroslav Olehla, CSc. Osvald Modrlák, Lukáš Hubka Technical University of Liberec, 2014 ISBN

3 ACKNOWLEDGEMENT This textbook preparation was supported by: Project ESF CZ.1.07/2.2.00/ Modernization of Didactic Methods and Innovation of Teaching Technical Subjects.

4 ABSTRACT The publication Automatic Control in Mechatronics summarize basic theoretical knowledge from the automatic control area. This knowledge background should create a starting point for successful design of mechatronics process automatic control. There is served the overview and the description of the theoretical approaches and applied methods used in automatic control including the software add on support in Matlab in the form of examples. The analysis and the synthesis of systems in the time domain and frequency domain is discussed. The publication includes description from the crucial mathematical apparatus over several PID controller design methods to robust control application. This paper is appropriate study source for self-studies, examination preparation or for the real application support. Keywords Automatic control, Mechatronic, PID controller, Identification, State control. ABSTRAKT Publikace Automatic Control in Mechatronics shrnuje základní poznatky z automatického řízení, které tvoří základ znalostí pro úspěšný návrh automatického řízení mechatronického systému. Je poskytnut přehled a popis teoretických postupů a aplikovaných metod automatického řízení, včetně ukázek softwarové podpory v Matlabu a příkladů. Je tu rozebrána analýza a syntéza systémů v časové i frekvenční oblasti od základního matematického aparátu pro popis systému, přes několik návrhových metod PID regulátorů až po vybrané postupy aplikace robustního řízení. Tento text je vhodným studijním materiálem při samostudiu a přípravě na zkoušky i podporou při aplikacích. Klíčová slova Automatické řízení, mechatronika, PID regulátor, identifikace, stavové řízení.

5 PREFACE Dear colleague, this electronic text was prepared for the purpose of realize basic overview in the automatic control theory. The text should making your examination preparation easier. This text ought to supplement the lectures of this semester. We hope that with the recommended literature this text creates a good basis for your study of control in Mechatronics. We wish you good understanding of the matter and excellent examination results!

6 CONTENTS 1 ANALYSIS OF DYNAMIC SYSTEMS Linear System Input-Output relation Transfer function, definition, poles, zeros Poles, Zeros and Time Responses Poles, partial fraction expansion Time domain specifications Mathematical modelling with state-space representation State-space model of a DC motor Dynamic response of the state equations Transfer function Transfer function with real poles and zeros Transfer function with dead time Scaling consideration Terms describing in control Questions Linear systems The Block Diagram Elementary block diagrams Block diagram simplifications Signal-Flow Graphs Mason s Rule Questions Block Diagram Control system configuration Open and closed loop control Models of control systems configuration Question Control System Configuration Root Locus Techniques Introduction Model of a feedback system Description of PID Controllers with gain, zeros and poles What is the Root Loci Basic properties of the Root Loci Guidelines for Sketching a Root Locus Properties and Construction of the Root Loci Questions about the Root Locus Technique Frequency domain analysis Frequency responses Nyquist plots Bode plots Bode plot in systems with pure time delay Questions Frequency domain analysis Nyquist Stability Criterion Zeros and poles of open-and closed-loop function The Argument Principle Nyquist condition of stability

7 Analysis of Dynamic Systems Nyquist criterion Checking the number of encirclements Gain and phase margin Question Nyquist Stability Criterion FEEDBACK CONTROL Basic properties of feedback Basic structure of control Closed-loop systems The basic equations in control Control of Dynamic Error P controller PI controller PID controller PD controller Time domain specifications Control steady state error Feedback system sensitivity Sensitivity studies of feedback control PID Controller tuning Procedure for ON-LINE Tuning Design Rules for PID Controller by a known unit response PID Controller tuning for dynamic performance Integral error measures cost functions Controller tuning based on numerical optimization Numerical optimization in Matlab Questions Basic properties of feedback The root- locus design method Questions The Root Locus design The Ackermann frequency design method of PID controllers Introduction to loop shaping Bode plots of PD controller Bode plots of PI controller Bode plots of PID controller P compensator compensation procedure PD Compensator design procedure PI compensation design procedure PID compensation design procedure Questions The frequency response method ENHANCEMENTS TO SINGLE-LOOP FEEDBACK CONTROL Feed forward control Control systems with an auxiliary measured and manipulated variable Control system with an auxiliary measured variable Control system with an auxiliary manipulated variable The Principle and strategy of cascade control Internal model control

8 3.4.1 Dynamic control law inverse response dynamic Factorization techniques Internal model control structure Questions Enhancements to single loop PID feedback control INTRODUCTION TO STATE SPACE CONTROL State-space representation of SISO systems Control (reachable) canonical form Observable canonical form Similarity transformation Complete controllability Completely observability Kalman form Minimal realization Model reduction Time scale decomposition Balanced reduction State estimation State feedback control Control law and control feedback structures State feedback with integral control Questions State space control MATHEMATICAL DESCRIPTION OF MIMO SYSTEMS A Set of differential equation Transfer function matrix Steady-state of output variable BASICS PRINCIPLES OF ROBUST CONTROL Introduction Norms H2, H Model uncertainty in dynamic systems Multiplicative uncertainty Additive uncertainty Robust stability of a feedback system with unstructured uncertainty Robust performance and control with multiplicative uncertainty Robust performance Mixed-sensitivity H control Rules for parameter setting of the filters WP = W1 and W Questions Principle of robust control REFERENCES 157 8

9 Analysis of Dynamic Systems 1 ANALYSIS OF DYNAMIC SYSTEMS 1.1 LINEAR SYSTEM Linear time-invariant systems have two important attributes: 1. A linear systems response complies with the principle of superposition. 2. The response of a linear time invariant system can be expressed as the convolution of the input with the unit impulse response of the system. The Superposition principle states that if the system has an input that can be expressed as a sum of signals, then the response of the system can be expressed as the sum of individual responses to the respective signals. The equation (1-1) is called the convolution integral. y(t) = u(t τ) g(τ)dτ (1-1) where y(t) is the response, g(t) is the unit pulse function (weighting function), u(t) is the input (manipulated variable). The Final Value Theorem. If all poles of the transform s Y(s) are in the left half of the s-plane then the steady-state of the variable y(t) is possible to find by (1-2). Similarly, the correspondence for y(0+) is the equation (1-3). y( ) = lim t y(t) = lim s 0 s Y(s) (1-2) y(0 +) = lim t 0+ y(t) = lim s s Y(s) (1-3) 1.2 INPUT-OUTPUT RELATION Transfer function, definition, poles, zeros Suppose there are linear, continuous-time-invariant systems with lumped parameters that are described by the differential equation y (n) + a n 1 y (n 1) + + a 1 y (1) + a 0 y = b m u (m) + b 1 u (1) + b 0 u, for m n (1-4) 9

10 If all initial conditions are set to zero and the Laplace transformation is applied to both sides of this equation, one obtains the following algebraic equation: Y(s)(s n + a n 1 s n a 1 s + a 0 ) = (b m s m + b 1 s + b 0 )U(s) (1-5) The transfer function is F(s) = B(s) A(s) = b m s m + b 1 s + b 0 s n + a n 1 s n a 1 s + a 0 = Y(s) U(s) Y(s) = F(s) U(s) (1-6) where A(s) and B(s) describe the numerator and denominator polynomials, respectively. The roots of A(s) = 0 are called the system poles, the roots of B(s) = 0 are called the system zeros. Let consider the transfer function of the system F(s) = B(s) A(s). A system F(s) is strictly proper if F(jω) 0 as ω. A system F(s) is semi-proper if F(jω) D 0 as ω. A system F(s) which is strictly proper or semi-proper is proper. A system F(s) is improper if F(jω) as ω. A proper system is physically realizable and can be described by a state-space description. If m < n we say the model is strictly proper. If m = n we say the model is semi-proper. If m n we say that the model is proper. The transfer function of a continuous time system is a rational transfer function and can have several definition. The transfer function is the ratio of the Laplace transform of the output to the Laplace transform of input with zero as its initial condition. OR The transfer function is the Laplace transform of the unit pulse function Poles, Zeros and Time Responses We now examine a number of fundamental properties of poles and zeros of transfer functions. It is known that a transfer function (rational function) can be written in the following form F(s) = B(s) A(s) = b m s m + b 1 s + b 0 s n + a n 1 s n 1 = + + a 1 s + a 0 = K m (s s m i=1 bi) i=1(s s bi ) n = b k=1(s s k ) m n k=1(s s k ) Recall, the polynomial in denominator (1-7) 10

11 Analysis of Dynamic Systems A(s) = s n + a n 1 s n a 1 s + a 0 = (s s k ) is called the characteristic polynomial, s k are the poles, and the equation n k=1 (1-8) A(s) = s n + a n 1 s n a 1 s + a 0 = 0 (1-9) is called the characteristic equation. A special class of transfer functions arises when all poles and zeros lie on the left half of the complex plane s. These transfer functions are called minimum-phase transfer functions. If the transfer functions have a zero in the right half plane (RHP) than that is called non-minimum phase. If a transfer function is referred to as a stable transfer function, the implication is that all poles are in the open left half plane (LHP). If it is said to be unstable, it has at least one pole in the closed RHP. The poles are called stable or unstable, depending on whether they lie in the open LHP or closed RHP Poles, partial fraction expansion It is known that any scalar rational transfer function can be expanded into partial fractions. Each partial fraction contains a single real pole, a complex conjugate pair, or multiple combinations with repeated poles. The form of the partial fraction depends on the kind of the factors in the product form of A(s). Case formula (s a) (s a) m (s a)(s a ) [(s a)(s a )] 2 Case 1. Unrepeated factors Case description Tab. 1-1: Factors types Unrepeated factors for unrepeated roots a Repeated factors for m th repeated root a Complex factors, complex roots a = α ± iβ Repeated complex factors repeated complex roots a 1,2 = α ± iβ The poles are s 1 = 0, s 2 = a, s 3 = b. Unrepeated factors require partial fractions Case 2. Repeated factors F(s) = B(s) A(s) = 2s + 1 s(s + a)(s + b) = A s + B s + a + C s + b. The poles are s 1 = 0, s 2 = s 3 = a, s 4 = s 5 = s 6 = b. Repeated factors require partial fractions F(s) = B(s) A(s) = 2s + 1 s(s + a) 2 (s + b) 3 = A s + B 2 (s + a) 2 + B 1 s + a + C 3 (s + b) 3 + C 2 (s + b) 2 + C s + b. 11

12 Case 3. Unrepeated complex factors The poles are s 1 = 0, s 2 = b, s 3 = a, a = a + iβ, a = α iβ, (s a)(s a ) = (s + α) 2 + β 2. Unrepeated complex factors require partial fractions or in the form F(s) = B(s) A(s) = 2s + 1 s(s b)(s a)(s a ) = A s + F(s) = B(s) A(s) = 2s + 1 s(s b)[(s + α) 2 + β 2 ] = A s + Case 4. Repeated complex factors In this case the partial fractions are of the form B s b + B s a + Ps + Q [(s a)(s a )] 2 + Ms + N (s a)(s a ). Example 1-1 Response to the unit step the case of two single real poles. a 1 y (1) + a 0 y = b 0 u Y(s) = Y(s) = b 0 a 1 1 s + a 0 s = a 1 b 0 a 1 s + a 0 a 1 U(s), F(s) = b 0 a 0 s b 0 a 0 K s + a = 0 s a 1 Ms + N (s a)(s a ). Ms + N (s + α) 2 + β 2. b 0 a 1 s + a 0 = K s + 1 τ K τs + 1 where K = b 0 is gain, τ = a 1 is the time constant, the term K for s a 0 a 0 s 0 = 0 is forced mode and the term K for s 1 = 1 is natural mode. τ s+ 1 τ The response (transient function) of the system is y(t) = h(t) = b 0 (1 e a 1 t a 0 ) = K (1 e 1 τ t ) a 0 Example 1-2 Response to the unit step the case two single real poles and one zero s B = 1. where the term A s ( s + 1) F(s) = 2 (s + 1)(s + 2), U(s) = 1 s ( s + 1) A = 2 (s + 1)(s + 2) s=0 Y(s) = A s + B s C s Y(s) y(t), s + 2 3e 2t, 4 s + 1 4e t, 1 s 1(t) B is forced mode and terms, C are natural modes. s+1 s+2 ( s + 1) = 1, B = 2 s(s + 2) ( s + 1) = 4, C = 2 s(s + 1) = 3 s= 1 s= 2 12

13 Analysis of Dynamic Systems Y(s) = 1 s 4 s s Y(s) y(t), s + 2 3e 2t, 4 s + 1 4e t, 1 s 1(t) Forced pole -flat System pole System pole System zero Conclusions: Fig. 1-1: Transfer characteristic (left), zero-pole placement in s-flat (right) 1. A pole of the input function generates the form of the forced response. 2. A pole of the transfer function generates the form of the natural response. 3. A pole on the real axis generates an exponential response e αt where the α is the pole location on the real axis. The further to the left a pole is on the negative axis, the faster the exponential transient response will decrease to zero. 4. Together, the zeros and poles generate the amplitudes for both the forced and natural responses. Example 1-3 Response to the unit step the case of twice repeated pole. Find the unit step response of the transfer function F(s) = 1 (2s+1) 2. Solution. H(s) = F(s) 1 s = 1 (2s + 1) 2 1 s = 1 4 s(s + 0.5) 2 Laplace transform of the unit step response has one root equal to zero and two multiply roots s 1 = 0, s 2 = s 3 = 0,5. Break up H(s) by partial fraction expansion where terms H(s) = 1 4 s(s + 0,5) 2 = A s + B 1 s + 0,5 + B 2 (s + 0.5) 2 B 1 s+0,5, B 2 (s+0,5) 2 represents repeated poles s 2 = s 3 = 0.5. Coefficients are calculated 1 A = s 4 s(s + 0.5) 2 s=0 1 = 1, B 2 = (s + 0.5) 2 4 s(s + 0.5) 2 = 0.5, s=

14 B 1 = d 1 ds [(s )2 s(s + 0.5) 2 ] = d 1 4 ds s = s= 0.5 s= 0.5 Using the inverse Laplace transform for partial fraction expansion H(s) = 1 4 s 2 = 1 s= s(s + 0.5) 2 = 1 s 1 s (s + 0.5) 2 1 e 0.5t 0.5t e 0.5t = h(t) Example 1-4 Response to the unit step the case of a pair of complex-conjugate poles Calculate the response to the unit step for the transfer function ω n 2 F(s) = K s 2 + 2ξω n + ω2 n If a system has two energy accumulators then the energy can go from one accumulator to another. This is the case if water flows through two tanks that are connected by a pipe. This is sketched in the Fig In this case we cannot model the system with two connected first-order systems in the series form. Q I1 h ( y 1 t) 1 Q I 2 h ( y 2 t) 2 Q O2 Fig. 1-2: Second order system We must use the second-order transfer function in the following form F(s) = K 2 ω n s 2 + 2ξω n + ω2 = K n (Ts) 2 + 2ξTs + 1 where ω n ξ K T = 1/ω n is natural or undamped natural frequency is known as the damping factor is the gain of the transfer function/plant is a multiple time constant ω d = ω n 1 ξ 2 is damped natural frequency for ξ (0,1) The Laplace transform of the transient function can be expressed as follows H(s) = K ω n 2 s(s 2 + 2ξω n + ω n2 ) The system has two complex-conjugate poles s 1 and s 2, s 3 = 0. 14

15 Analysis of Dynamic Systems s 1,2 = 2ξω n± 4(ξω n ) 2 4ω2 n = ξω 2 n ± ω n ξ 2 1 = ξω n ± iω n 1 ξ 2 = ξω n ± iω d Pole location is in the Fig Expanding in partial fractions H(s) = K ω n 2 s(s 2 + 2ξω n + ω n2 ) = = K [ A s + C s + ξω n iω n 1 ξ 2 C + s + ξω n + iω n 1 ξ 2] we obtain coefficients A (for s 3 = 0) by using ω n 2 A = s 2 + 2ξω n + ω2 = 1 n s=0 Complex constant C is calculated for s 1 = ξω n + iω n 1 ξ 2 from the equation s 1 i d s 2 n s 1 n Im n Re Fig. 1-3: Pole position in s-flat C = ω n 2 s(s + ξω n + iω n 1 ξ 2 ) s=s 1 ω n 2 = ( ξω n + iω n 1 ξ 2 )( ξω n + iω n 1 ξ 2 + ξω n + iω n 1 ξ 2 ) ξ = i0.5 1 ξ 2 Complex conjugate C is equal C = 0.5 i0.5 ξ 1 ξ 2 The absolute value is obtained from C = ξ2 = ξ 2 1 ξ 2 and the phase is equal φ = arctan I{C} = arctan ξ R{C} 1 ξ2. We can obtain the following unit step response by applying the inverse Laplace transform 1 h(t) = K [1 1 ξ 2 e ξωnt ξ cos [(ω n 1 ξ 2 ) t arctan 1 ξ 2]]. 15

16 Tab. 1-2: Classification by damping factor ξ Damping factor ξ Poles Classification 0 < ξ < 1 s 1,2 = ξω n ± iω n 1 ξ 2 Underdamped ξ = 1 s 1,2 = ω n Critically damped ξ = 0 s 1,2 = ±iω n Undamped (constant amplitude) ξ > 1 s 1,2 = ξω n ± iω n 1 ξ 2 Over damped 1 < ξ < 0 s 1,2 = ξω n ± iω n 1 ξ 2 Undamped (raising amplitude) We have defined two parameters associated with the second-order system ξ and ω n. Other parameters associated with the underdamped response are: percent overshoot, peak time, settling time, rise time Time domain specifications Specifications for a control system design often involve certain requirements associated with the time response of the system. The requirements for a step response are expressed in terms of the standard quantities illustrated in Fig The rise time, t r is the time it takes the system to reach the vicinity of its new set point. The settling time, t s is the time it takes to reach and stay within ±2 % of the steady state. The overshoot, M P (%) is the largest overshoot from the system of all overshoots. The largest overshoot is divided by its final value (and is often expressed as a percentage). The peak time, t p is the time it takes the system to reach the maximum overshoot point. Conclusions poles Every pole generates a special component called natural mode in the system response to a pulse input. This mode is present in the system response to any given input. Dominant or slow poles are poles that are closer to the stability boundary than the rest of the system poles. We can say that the dominant poles decrease more slowly than the rest. Conclusions zeros The effect that zeros have on the response of a transfer function is a little more subtle than the effect of the poles. Zeros do not influence the exponents, but they influence the constants in partial-fraction expansion. This causes the zeros and poles to generate together the amplitudes for both the forced and natural responses. 16

17 Analysis of Dynamic Systems M P +1 % t P t s t r Fig. 1-4: Time domain specification illustration 1.3 MATHEMATICAL MODELLING WITH STATE- SPACE REPRESENTATION State-space model of a DC motor A modern method of writing electrical network equations is the state-variable method. It will be demonstrated on the modelling of a DC motor. Consider a separately excited DC motor that has the equivalent circuit diagram in Fig. 1-5 and the rotational system. The armature is modelled as a circuit with resistance R M, connected in series with inductance L M, and an applied voltage u M. The back emf is represented by the voltage source u i when the rotor rotates. Armature current is i M, magnetic flux in the air gap is Φ M that is supposed to be constant. The voltage rotational speed equation is u i (t) = k i ω M (t) (1-10) where k i is the back-emf constant and ω M (t) is rotor angular velocity. The rotational system is represented by its rotor inertia J M, viscous-friction coefficient B M, and φ(t) is rotor displacement. The torque M d represents the load torque. The torque M M developed by the motor is proportional to the air-gap flux and the armature current i M Φ Fig. 1-5: DC motor M M 17

18 M M (t) = k M i M (t) (1-11) where k M is the torque constant. The current equation can be written as di M (t) dt = 1 L M u M (t) R M L M i M (t) 1 L M u i (t) (1-12) The torque (moment) equation of the motor rotor has the form d 2 φ M (t) dt 2 = 1 M J M (t) 1 M M J d (t) B M dφ M (t) M J M dt We know that the rotor equation velocity is equal ω M (t) = dφ M(t) dt Substituting back-emf Eq. (1-11) into Eq. (1-13) yield (1-13) (1-14) di M (t) dt = R M L M i M (t) K i L M ω M (t) + 1 L M u M (t) (1-15) In a similar way we can substitute the rotor velocity given by Eq. (1-15) and motor torque equation (1-13) into Eq. (1-15) that yields dω M (t) dt = k M i J M (t) B M dφ M (t) 1 M M J M dt J d (t) (1-16) M Equations (1-14), (1-15) and (1-16) are modelling the DC motor. The same equation can be represented in the state-variable representation in the matrix form x (t) = A x(t) + B u(t) y(t) = C x(t) (1-17) x 1 (t) i M (t) where x(t) = [ x 2 (t)] = [ ω M (t)] is the state vector, u(t) = [ u M(t) ] is the input vector, M x 3 (t) φ M (t) d (t) y(t) = ω M (t) is the output-the rotor velocity, A = [ R M L M k M J M k i L M 0 B M 0 J M 0 1 0] 1 0 L M, B = [ 0 1 ], C = [0 1 0]. J M 0 0 Generally, the state-equation for a SISO (Simple Input-Simple Output) system can be written in the form x (t) = A x(t) + b u(t) (1-18) y(t) = C x(t) + D u(t) (1-19) The matrices and vector are named as follows x(t) state vector [n 1] A system matrix [n n] 18

19 Analysis of Dynamic Systems x (t) derivative of state vector b input vector [n 1] y(t) output C output matrix [1 n] u(t) input D feed-forward matrix/scalar [1] The block diagram of the SISO state-space equation is depicted in Fig (t) (t) t (t) (t) Fig. 1-6: Block diagram of general state space equation The state equations of dynamic systems are called state-space representation. (n first-order differential equations) Dynamic response of the state equations Our aim is now to find the dynamic response from the state equations and to discover the relationship between the state equation and the transfer function description. Taking the Laplace transform Eq. (1-18) we obtain x (t) = A x(t) + b u(t) s X(s) x(0) = A X(s) + b U(s) If we collect the terms involving X(s) on the left side, keeping in mind that order is very important in matrix multiplications, we find that [s I A] X(s) = x(0) + b U(s) If we multiply both sides by the inverse of si A, then X(s) = [s I A] 1 x(0) + [s I A] 1 b U(s) (1-20) The output of the system is Y(s) = C X(s) + D U(s) = C [s I A] 1 x(0) + C [s I A] 1 b U(s) + D (1-21) U(s) This equation expresses the output response to an initial condition vector x(0) and an external forcing input u(t). The transfer function is defined for zero initial conditions and is given by G(s) = Y(s) U(s) = C [s I A] 1 b + D (1-22) The inversion can be calculated 19

20 [s I A] 1 adj[s I A] adj[s I A] = = det[s I A] Δ(s) G(s) = Y(s) adj[s I A] = C b + D = B(s) U(s) Δ(s) A(s) The characteristic polynomial is Δ(s) = A(s) = det[si A]. If the determinant is set to zero det[s I A] = (s) = A(s) = a 0 + a 1 s + + a n 1 s n 1 + s n = 0, we obtain characteristic equation and the roots of the characteristic equation are eigenvalues of the state matrix A. Example 1-5 Given the system represented in state space by equation x (t) = [ ] x(t) + [0 1 ] u(t), y(t) = [ 1 0] x(t), x(t = 0) = 0 Find the transfer function and the eigenvalues and the poles and zeros. Solution. Applying the Laplace transform of both sides of the state equation yields sx(s) = [ ] X(s) + [0 0 ] U(s) [s [ ] [ 0 1 ]] X(s) 2 3 Determinant = [ 0 1 ] U(s) X(s) = [ s s + 3 ] [ 0 1 ] U(s) det [ s 1 2 s + 3 ] = s(s + 3) + 2 = s2 + 3s adj [ s 1 s 1 [ 2 s + 3 ] = 2 s + 3 ] s s 2 + 3s + 2 = [ s 2 + 3s + 2 s 2 + 3s + 2 ] 2 s s 2 + 3s + 2 s 2 + 3s + 2 s Y(s) = C X(s) = [1 0] [ s 2 + 3s + 2 s 2 + 3s + 2 ] [ 0 2 s 1 ] U(s) s 2 + 3s + 2 s 2 + 3s + 2 s Y(s) = [ s 2 + 3s + 2 s 2 + 3s + 2 ] [0 1 ] U(s) = 1 s 2 + 3s + 2 U(s) The transfer function is F(s) = Y(s) U(s) = 1 s 2 +3s+2. (1-23) (1-24) Eigenvalues and poles of the system are s 1 = 1, s 2 = 2. The system does not have zeros! 20

21 Analysis of Dynamic Systems 1.4 TRANSFER FUNCTION Transfer function with real poles and zeros A rational transfer function can be described as a ratio in factored zero-pole form F(s) = b m(s s B1 )(s s B2 ) (s s Bm ) s r a n (s s 1 )(s s 2 ) (s s n ) Transfer function with real poles in the Bode Form is F(s) = K (1 + st B1)(1 + st B2 ) (1 + st Bm ) s r (1 + st 1 )(1 + st 2 ) (1 + st n ) where T i = 1 is time constant of system, K = b 0 is gain. s i a 0 Transfer function with complex pole and zero. (1-25) (1-26) If we consider a dynamic system with complex pole and zero, then the transfer function in Bode form is F(s) = K (1 + st B1)(1 + st B2 ) (1 + st Bm )[(st B ) 2 + 2ξsT B + 1] s r (1 + st 1 )(1 + st 2 ) (1 + st n )[(sτ) 2 + 2ξsτ + 1] Transfer function with dead time (1-27) In a real system the signal velocity is bounded. In practice, pure time delay may be encountered in various types of systems, especially systems with hydraulic, pneumatic or mechanical transmissions (conveyor belt). The system responds to the input changing after a given time interval is called time delay, which is denoted by the symbol T d. Fig. 1-7: Systems with time delay Fig. 1-7 illustrates systems in which transportation lag or pure time delays are observed. The Fig. 1-7a depicts the classical transport of solid materials. Fig. 1-7b illustrates the control thickness of rolled steel plates. The transport delay between the thickness at the rollers and the measuring point is given by the equation T d = d v, where d is distance [m] and v is velocity [m/s]. Fig. 1-7c outlines an arrangement in which two differential fluids are to be mixed in appropriate proportions. To assure that a homogenous solution is measured, the monitoring point is located some distance from the mixing point. The time delay T d [s], proves for all systems as a time shift at T d seconds (see Fig ). 21

22 Fig. 1-8: Transient response for a system with time delay L{y(t T d ) 1(t T d )} = Y(s) e st d (1-28) where T d is a shift to the right. Suppose, the Laplace transform of g 0 (t) is the transfer function F 0 (t). Then the time shifted function g(t T d ) (with time delay T d ) has the transfer function G(s) = F(s) = L{g(t T d ) 1(t T d )} = F 0 (s) e st d (1-29) where F 0 (s) is the transfer function without a time delay and T d is the time delay. If a time delay or dead time T d is introduced in the input signal u(t), one obtains instead of the differential equation A n y(t) (n) + + a 1 y(t) (1) + a 0 y(t) = b 0 u(t T d ) + + b 0 u(t T d ) (m) Transfer function of a plant with time delay T d and with real poles and zero has the form F(s) = K (1 + st B1)(1 + st B2 ) (1 + st Bm ) s r (1 + st 1 )(1 + st 2 ) (1 + st n ) e st d (1-30) Scaling consideration The analysis of models (minimal realization, order reduction) for real objects requires scaling of the transfer function. It means that the outputs and inputs are approximately the same. Let (*) represent the scale variables. The scaled output is Since the input-output relation is Similarly, the scaled input is Also, so the scaled input-output relation is y (t) = S o y(t) Y (s) = S o Y(s) Y(s) = F(s)U(s) Y (s) = S o F(s)U(s) u (t) = S i u(t) U(s) = S i U(s) U(s) = S i 1 U (s) Y (s) = S o F(s)S i 1 U (s) = F (s)u (s) The scaled transfer function is equal 22

23 Analysis of Dynamic Systems F (s) = S o F(s)S i 1 (1-31) Terms describing in control Equation Transfer function Transient function P-block y = Ku K K Blocks t I-block y = K I udτ 0 K I s K I t D-block y = K D u K D s K D δ(t) TDblock y = Ku(t T D ) Ke st D K 1(t T D ) PT1- block τy (1) + y = Ku K τs + 1 K (1 e t τ) DT1- block τy (1) + y = K D u K D s τs + 1 K t τ e τ K [1 Ae αt cos (ωt T2- block T 2 y + 2ξTy + y = Ku K (Ts) 2 +2ξTs+1 ξ arctan 1 ξ 2)], 1 A = 1 ξ 2, ω = 1 1 ξ2 T Questions Linear systems 1) What is the definition of a transfer function? 2) What are the properties of systems whose responses can be described by transfer functions? 3) What is the Laplace transform of f(t x) 1(t x) if the transform of f(t) is F(s)? 4) What is a time delay, a time constant, gain? Write a transfer function with those parameters. 5) State the Final Value Theorem. 6) What is the most noticeable effect of a zero in the right half-plane on the step response of the second order system? 7) What is scaling? 8) What causes poles in the response to a unit step input? 23

24 9) What causes zeros in the response to a unit step input? 10) What is the rise time, settling time, the overshoot and the peak time? 11) What is damping factor, natural frequency, frequency and multiple time constant? 12) What is a non-minimum phase transfer function? 1.5 THE BLOCK DIAGRAM The components of the block diagrams are: signals, blocks/systems (transfer functions), summing junctions and pickoff points. Block has only one input and output. See Fig Fig. 1-9: Block symbols The characteristic of the summing junction is that the output signal is the algebraic sum of the inputs signal (Fig. 1-10a-b). A pickoff point distributes the input signal U(s) undiminished to several output branches. For attribute and equivalent outputs of the pickoff points see (Fig. 1-10d-e). Fig. 1-10: Summing junction (a-c), pickoff points (d-e) We will examine elementary topologies for interconnecting subsystems and derive the single transfer representation for each of them. These common topologies will form the basis for reducing more complicated systems to a single block Elementary block diagrams Cascade (Series) form The signals Y 1 (s), and Y(s) are equal: Parallel form Y 1 (s) = F 1 (s)u(s) Y(s) = F 2 (s)y 1 (s) = F 2 (s)f 1 (s)u(s) F yu (s) = Y(s) U(s) = F 2(s)F 1 (s) Y(s) = Y 1 (s) + Y 2 (s) = F 1 (s)u(s) + F 2 (s)u(s) = [F 1 (s) + F 2 (s)]u(s) Overall transfer function is Feedback form The input signal Y 1 (s) is equal F yu (s) = Y(s) U(s) = F 1(s) + F 2 (s) 24

25 Analysis of Dynamic Systems Overall transfer function is Y 1 (s) = U(s) + Y 2 (s) Y 2 (s) = F 2 (s)y(s) Y(s) = F 1 (s)y 1 (s) = F 1 (s)[u(s) + F 2 (s)y(s)] F 1 (s)u(s) + F 1 (s)f 2 (s)y(s) = Y(s) F yu (s) = Y(s) U(s) = F 1 (s) 1 F 1 (s)f 2 (s) Block diagram simplifications The three elementary cases given in Fig Fig can be used in combination to solve, by repeated reduction, any transfer function defined by a block diagram. The manipulation can be tedious and subject to error when the topology of the diagram is complicated. Next table shows the equivalent forms for block diagram algebra. Fig. 1-11: Series form Fig. 1-12: Parallel form Fig. 1-13: Feedback form 25

26 Tab. 1-3: Selected equivalent forms for block diagram algebra Schema Transfer function Description Y = FU + Y 1 Y = FU + 1 F Y 2F = = FU + Y 2 Moving a block to the right past a summing junction. Y = F(Y 1 + Y 2 ) Moving a block to the left past a summing junction. Y = Y 1 F + Y 2 F Y = FU Y = FU Y = FU Moving a block to the right past a pickoff point. Y = FY 1 Y 1 = Y 1 Y = FY 1 Y 1 = 1 F Y Moving a block to the left past a pickoff point. Y 1 = 1 F Y = 1 F FY 1 = Y 1 26

27 Analysis of Dynamic Systems Example 1-6 Find the overall transfer function of the system shown in Fig Solution Fig. 1-14: Three systems plant 1. Moving the block F 1 to the right past the pickoff point. 2. Form equivalent cascade system in the feedback path F 1 F 2. Change the pickoff points. 3. Feedback is equal F 12 = 1 1+F 1 F 2 and parallel form has transfer function F 13 = F 1 + F 3. The overall transfer function is F = F 12 F 13 = F 1 F 2 (F 1 + F 3 ) = F 1 + F F 1 F Signal-Flow Graphs A signal-flow graph (SFG) may be regarded as a simplified version of a block diagram. The signal flow graph components are: systems (Fig. 1-15a), interconnection of systems and signals (circles in Fig. 1-15b). 27

28 Fig. 1-15: System (a), signals (b) Junction points node represents a variable. The nodes are connected by line segments called branches. The branches have associated branch gain and direction. A signal can transmit through a branch only in the direction of the arrow. The Input Node (Source) is a node that has only outgoing branches. The Output Node (Sink) is a node that has only incoming branches. Interconnections. A block diagram and the corresponding signal flow graph. Fig. 1-16: Block diagram and corresponding signal-flow graph The following example will show the use of a Signal flow equation for calculating an overall transfer function. Example 1-7 With the help of signal flow equation determine the overall transfer function for system on the Fig. 1-14! Solution 28

29 Analysis of Dynamic Systems We chose a node-signal behind the first summing junction X and behind the second summing junction Y. Signal flow equations are X = U F 1 F 2 X Y = XF 1 + XF 3 We have only two equations and three variables X, U, Y and so we have two possibilities: 1) One variable can be chosen that give us any good result! 2) We can choose the ratio of two variables F = Y U. We will use option 2. Both equations will be divided by U and it yields X(1 + F 1 F 2 ) = U X U (1 + F 1F 2 ) = 1 X(F 1 + F 3 ) Y = 0 X U (F 1 + F 3 ) Y U = 0 X [ 1 + F 1F 2 0 F 1 + F 3 1 ] [ U ] = [ 1 Y 0 ] U The equations are rewritten into the matrix equation. Crammer s rule will be used to find a solution Y = D Y U, where D is the determinant of system matrix, D U D Y U is the determinant of the adapted system matrix. D = det [ 1 + F 1F 2 0 F 1 + F 3 1 ] = (1 + F 1F 2 ), D Y U = det [ 1 + F 1F 2 1 F 1 + F 3 0 ] = (F 1 + F 3 ) Overall transfer function is Mason s Rule F(s) = Y U = F 1 + F F 1 F 2 The Example 1-7 has shown us that the use of the block diagram simplification method is not generally easy. It is necessary to have some experiences with this method. So we show you a method that is based on signal-flow graph theory and it is known as Mason s rule. Let us introduce some definitions. Paths. A path is a collection of a continuous succession of branches traversed in the same direction. Loops. A loop is a path that originates and terminates on the same node along which no other node is encountered more than once. Mason s formula generally has four components that must be evaluated. 29

30 Loop gain: The product of branch gains found by traversing a path that starts at a node and ends at the same node without passing through any other node more than once and following the direction of the signal flow. Forward-path gain. The product of gains found by traversing a path from the input node to the output node of the signal-flow graph in the direction of the signal flow. Non-touching loops. Loops that do not have any nodes in common. Non-touching loops gain. The product of loop gains from non-touching loops taken two, three, four, etc., at a time. Non-touching forward-paths and loops. Forward-paths that do not have any nodes in common with a loop. The transfer function of a system represented by a signal flow graph or block diagram is G(s) = Y(s) U(s) = k T kδ k Δ where k is the number of forward paths, T k is the k th forward-path gain Δ = 1 loop gains (1-32) + products of gains of all possible combinations of two nontouching loops products of gains of all possible combinations of three nontouching loops + Δ k = 1 loop gains that does not touch the k th forward path products of gains of all possible combinations + of two nontouching loops that do not touch the k th forward path products of gains of all possible combinations of three nontouching loops that do not touch the k th forward path + Example 1-8 Find the overall transfer function for the block diagram in the previous example. Solution. Forward Path Gains: F 1, F 3 Loop Path Gain: F 1 F 2 There are not any non-touching loops and forward path! The overall transfer function is F = F 1+F 3 1+F 1 F 2. Mason s rule is useful for solving complicated block diagrams by hand! It yields the solution in the sense that it provides an explicit input/output relationship for the system 30

31 Analysis of Dynamic Systems represented by the diagram. The advantage compared with repeated block reduction is that it is systematic and algorithmic rather than problem-dependent. Example 1-9 In the Fig there is a given block diagram. There are two inputs U, D and two outputs Y, X. Fig. 1-17: Four systems plant Tasks: 1) Find the transfer function F yu, U = 1(t), D = 0. 2) Find the transfer function F yd, D = 1(t), U = 0. 3) Find the transfer function F xu, U = 1(t), D = 0. Solution. 1) U = 1(t), D = 0; transfer function F yu Forward Path Gains: F 1, F 3 Loop Path Gains: F 1 F 2, F 1 F 4, F 4, F 3 F 4 Non-touching forward paths: F.P. F 1 is not touching the loop path F 4 Non-touching loops: F 1 F 2, F 4, F 3 F 4 Determinant is D = 1 [F 1 F 2 F 1 F 4 F 4 F 3 F 4 ] S i 1 F.P. F 3 is not touching the loops paths F 4, F 1 F 2 + [F 1 F 2 ( F 4 ) + F 1 F 2 ( F 3 F 4 )] S i 2 Determinant of the first forward path k = 1 marked as D 1 is equal D 1 = 1 [ F 4 ] S i 1 = 1 + F 4 Determinant of the second first forward path k = 2 marked as D 2 is equal D 2 = 1 [F 1 F 2 F 4 ] S i 1 + [F 1 F 2 ( F 4 )] S i 2 = 1 F 1 F 2 + F 4 F 1 F 2 F 4 31

32 The overall transfer function is F yu is equal F yu (s) = Y(s) U(s) = F 1 (1 + F 4 ) + F 3 (1 F 1 F 2 + F 4 F 1 F 2 F 4 ) 1 (F 1 F 2 F 1 F 4 F 4 F 3 F 4 ) + (F 1 F 2 ( F 4 ) + F 1 F 2 ( F 3 F 4 )) 2) D = 1(t), U = 0; transfer function F yd Forward Path Gain: F 3 Loop Path Gains: The Loops Gains and their locations are not changed! Non-touching forward paths: F.P. F 3 is not touching the loops paths F 4, F 1 F 2 The overall transfer function F yd is equal F yd (s) = Y(s) D(s) = F 3 (1 F 1 F 2 + F 4 F 1 F 2 F 4 ) 1 (F 1 F 2 F 1 F 4 F 4 F 3 F 4 ) + (F 1 F 2 ( F 4 ) + F 1 F 2 ( F 3 F 4 )) 3) U = 1(t), D = 0; transfer function F xu Forward Path Gain: F 1 Loop Path Gains: The Loops Gains and their locations are not changed! Non-touching forward paths: F.P. F 1 is not touching the loops paths F 4 The overall transfer function is F xu is equal F xu (s) = X(s) U(s) = F 1 (1 + F 4 ) 1 (F 1 F 2 F 1 F 4 F 4 F 3 F 4 ) + (F 1 F 2 ( F 4 ) + F 1 F 2 ( F 3 F 4 )) Questions Block Diagram 1) Write an elementary block diagram with its transfer functions! 2) Draw the block diagram algebra! 3) Draw the equivalent forms for moving a block to the left past a summing junction! 4) Draw the equivalent forms for moving a block to the right past a pickoff point! 5) What is a signal flow graph and its rules? 6) What is the open and closed loop response and what are its characteristic equations? 7) By applying the Final Value Theorem determine the final values of the error for a step set point change for the transfer function F = 1 under P, PI. s+1 8) What is Mason s rule, a path, a loop and a loop gain? 9) Explain the non-touching loops and non-touching forward path! 10) Write Mason s rule and explain it! 1.6 CONTROL SYSTEM CONFIGURATION Open and closed loop control Now we describe two control system configurations: open-loop and closed-loop. 32

33 Analysis of Dynamic Systems Open-loop Systems A generic open-loop is shown in Fig It consists of a subsystem called an input transducer that coverts the form of the input to that used by the controller. The controller drives the process or plant. The input w = w (t) is sometimes called the reference while the output can be called the controlled variable y = y(t). The input d = d(t) is called the disturbance. The controller output u = u(t) is called the manipulated variable. Fig. 1-18: Open loop control system Closed-loop (Feedback Control) Systems The generic architecture of a closed-loop system is shown in Fig The input transducer converts the form of the input w to the form used by the controller. An output transducer, or sensor, measures the output response and converts it into the form used by the controller. The summing junction algebraically adds the signal from the input w to the signal from the output, which arrives via the feedback path, the return path from the output to the summing junction. Fig. 1-19: Closed loop (Feedback Control) Systems 33

34 1.6.2 Models of control systems configuration This chapter begins with the creation of the basic equations for open-loop, feedback, feedback and feed forward control and the comparison of feedback structures with open loop control. The open loop control The output is given Y(s) = F U (s)r W (s)w(s) + F D (s)d(s) (1-33) where R W (s) is a transfer function of an open-loop controller (Feed-forward controller). The error between the reference input and system output is E(s) = W(s) Y(s) = W(s) F U (s)r W (s)w(s) F D (s)d(s) = [1 F U (s)r W (s)]w(s) F D (s)d(s) (1-34) Fig. 1-20: Open-loop control model system The feedback control The model of the feedback structure in Fig is shown in Fig. 1-21a. Imagine a system with a disturbance modelled by the transfer function G D (s). The feedback control of the system is depicted in the Fig. 1-21b. The output is equal Y(s) = R(s)G U(s) 1 + R(s)G U (s) W(s) + G D (s) D(s) (1-35) 1 + R(s)G U (s) The following notation and terminology will be used. L(s) = R(s)G U (s) is loop transfer function, S(s) = T(s) = 1 = 1 is sensitivity function, 1+R(s)G U (s) 1+L(s) R(s)G U (s) 1+R(s)G U (s) = L(s) 1+L(s) is complementary sensitivity function. The complementary sensitivity function and sensitivity function meet the equation T(s) + S(s) = 1 Using the defined terminology the resulting output is equal Y(s) = T(s)W(s) + G D (s)s(s)d(s) (1-36) 34

35 Analysis of Dynamic Systems Fig. 1-21:a) Feedback/closed-loop control with transfer function of sensor b) Feedback/closed-loop control model system Two degrees-of-freedom feedback control (Fig. 1-22a) The input U(s) is equal U(s) = (W(s) Y(s))R(s) + R W (s)w(s) The output is Y(s) = R(s)G U(s) + R W (s)g U (s) G D (s) W(s) + D(s) (1-37) 1 + R(s)G U (s) 1 + R(s)G U (s) where R W (s) is the transfer function of the feed-forward controller. The feedback and feed-forward control (Fig. 1-22b) Y(s) = R(s)G U(s) 1 + R(s)G U (s) W(s) + G D(s) R M (s)g U (s) D(s) (1-38) 1 + R(s)G U (s) where R M (s) is the transfer function of the feed- forward controller. 35

36 Fig. 1-22: Two degrees-of-freedom feedback control (a), Feedback and feed-forward control (b) Question Control System Configuration 1) Draw the generic open and closed-loop system! 2) Draw the model of an open and closed loop system, describe the blocks and write the open and closed loop transfer function! 3) What is the loop transfer function, sensitivity function and complementary sensitivity function? 4) Draw the diagram of feedback and feed-forward control! What is feed-forward controller? What is feedback and feed-forward control model? 1.7 ROOT LOCUS TECHNIQUES Introduction The linear control theory uses LTI models for analysis and synthesis. But real industrial systems can change their parameters over time. Very often engineers face the problem that the dynamic of the closed loop plant is changing over time. The change of the dynamic of the plant can be proved as a change of the closed loop gain over time. The question is: How are the poles of the closed loop transfer function changed when the gain in the closed loop changes? W. R. Evans developed a specific technique that shows how changes in one of the system s parameter modify the roots of the characteristic equation. The method gives rules for plotting the path of the roots and the plot is called the Root Locus. The Root Locus technique is used for analysis and synthesis of the control systems Model of a feedback system Fig. 1-23: Feedback system with timechanging parameters Root locus, is a graphical representation of the closed-loop poles as a system parameters are varied. The root locus technique is commonly used to study the effect of loop gain variations. We will start our study with the basic feedback system shown in Fig. 1-21a) that will be simplified by choosing G D (s) = F(s) = 0 (see Fig. 1-24). 36

37 Analysis of Dynamic Systems Fig. 1-24: The basic closed-loop block diagram The measured output Y(s) of the block diagram can be changed as shown in Fig Fig. 1-25: Closed-loop block diagram The closed-loop block diagram in Fig is the unity feedback, which has in the forward path the transfer function L(s) that is called the open-loop transfer function (see Fig. 1-26a). The block diagrams of the open-loop transfer function and the closed-loop are shown in Fig Fig. 1-26: Open-loop block diagram (a), and closed-loop block diagram (b) The open loop transfer function is equal L(s) = R(s)G(s)H(s) = R(s)G U (s) (1-39) Where R(s) is a transfer function of a controller (mostly PID), G U (s) is a transfer function of extended plant model Description of PID Controllers with gain, zeros and poles We suppose that the changing dynamic of the plant can be approximated by modelling with a feedback gain. The easiest way is to separate the Gain from the controller. That means that the transfer function of the controller shall be written in the Bode Form! PI Controller Transfer function of a PI controller can be written as R(s) = r 0s + r r 0 (s + r 1 1 r0 ) = = r 0 (s s BR1 ) s s s = Gain R 0 (s) = Gain (1 + st BR1) s (1-40) 37

38 where Gain is the gain of the controller, T BR1 = 1 s BR1 is the zero of the controller. PID controller with real zeros The transfer function of the PID controller can be written in the following form R(s) = r 0s + r 1 + r 2 s 2 = r 2 (s s BR1 )(s s BR2 ) s s = Gain (1 + st (1-41) BR1)(1 + st BR2 ) = Gain R s 0 (s) where the real zeros of the controller are s BR1, s BR2 ; Gain is the gain of the PID controller. PID controller with complex zeros: s RB12,= α ±i ω. R(s) = r 0s + r 1 + r 2 s 2 s = r 2 (s s BR1 )(s s BR2 ) s = Gain (1 + 2ξT ns + (T n s) 2 ) = Gain (1 + βs + (γs)2 ) s s = Gain R 0 (s) where (1 + βs + (γs) 2 ) is complex factor and s BR1,2 = α ± iω are complex zeros. (1-42) The transfer function R 0 (s) represents a transfer function that is determined by the pole and zeros of the controller (1 + st BR1 )(1 + st BR2 ), for real zeros R(s) = s (1 + βs + (γs) 2 ) {, for complex zeros s In general, the transfer function of a PID controller can be written as follows (1-43) R(s) = P + I s + Ds = Gain R 0(s) = K R 0 (s) (1-44) where Gain, K is the gain of a PID controller. 0 Gain, R 0 (s) is the structure of the controller determined the pole and by zeros of the controller. Now we must introduce a variation of the feedback gain to our model. Using (1-44) that yields L(s) = R(s)G(s)H(s) = K R 0 (s)g(s)h(s) = K B 0(s) (1-45) A 0 (s) where R(s) = K R 0 (s) is transfer function of the controller, K is the gain of the feedback 0 K <, R 0 (s) is transfer function of controller defined by its zeros and poles What is the Root Loci The open loop transfer function is the product of the controller, the system gain and the gain of the sensors. For the closed-loop block diagrams in the Fig. 1-23, Fig. 1-24, Fig the closed-loop transfer functions are Y S (s) W(s) = R(s)G(s) 1 + R(s)G(s)H(s), Y(s) W(s) = R(s)G(s)H(s) 1 + R(s)G(s)H(s) (1-46) 38

39 Analysis of Dynamic Systems You can see the characteristic equation whose roots are the poles of this transfer function is only one and can be expressed in the form This equation can be rewritten in the following form 1 + R(s)G(s)H(s) = 0 (1-47) 1 + L(s) = 1 + K B 0(s) A 0 (s) = 1 + K L 0(s) = 0, L 0 (s) = B 0(s) A 0 (s) (1-48) 1 + L(s) = 1 + K B 0(s) A 0 (s) = A 0(s) + K B 0 (s) = 0 A 0 (s) (1-49) where 0 K < is the parameter/gain of the loop or the gain of the controller. The closed-loop characteristic equation is equal to the zeros of Eq. (1-49), make sure of the following equation Y(s) W(s) = B R(s)G(s)H(s) K 0 (s) 1 + R(s)G(s)H(s) = A 0 (s) 1 + K B = 0(s) A 0 (s) K B 0 (s) A 0 (s) + K B 0 (s) (1-50) Now we will study the roots of the equation (1-49) when the parameter K is changed!!! Root Loci The trajectories of roots of the characteristic equation (1-45) when the parameter K is changed in the interval 0 K <, is called Root Loci Basic properties of the Root Loci What condition must every point of the trajectories of roots satisfy? Every point must comply with the characteristic equation! Now we are ready to investigate the equation (1-49) and (1-48). The equation (1-49) can be rewritten 1 + K B 0(s) A 0 (s) = 0 B 0(s) A 0 (s) = 1 K (1-51) Notice in Fig that to satisfy the equation (1-45) in the form (1-51) the following equations must be satisfied simultaneously: 1. Condition to Magnitude 2. Condition to Angles B 0(s) A 0 (s) = 1 K, 0 K < (1-52) φ = arg { B 0(s) A 0 (s) } = (2i + 1)π, i = ±0, ±1,, K 0 (1-53) From the equation (1.7-6) we define the root locus this way: 39

40 The root locus is the set of values of s for which hold the equation (1-51). That must be satisfied as the real parameter K varies from 0 to +. The characteristic equation of the closed-loop system has the form as in Eq. (1-49). Fig. 1-27: Pictured equation (1-51) Guidelines for Sketching a Root Locus The construction of the root loci is basically a graphical problem but some of the properties are derived analytically. The graphical construction of the root loci is based on knowledge of the poles and zeros of the function L(s). L(s) = K B 0(s) A 0 (s) = K b ms m + b m 1 s m b 1 s + b 0 s n + a n 1 s n a 1 s + a 0 = K L 0 (s) (1-54) The open loop transfer function can be expressed in terms of the product of factors L(s) = F 0 (s) = K b m(s s b1 )(s s b2 ) (s s bm ) (s s a1 )(s s a2 ) (s s an ) = K b m m j=1(s s bj ) n (s s ak ) where the zeros and poles of L(s) are real or complex-conjugate pairs. k=1 (1-55) Applying the condition in Eqs. (1-49), (1-51), (1-52), (1-53), (1-55) we have the condition for a point s that is lying on the root loci L(s) = K b m m j=1 (s s bj ) n (s s ak ) k=1 = 1 K (1-56) Where s s bj is the absolute value of the vector, that originates in the point s bj and terminates in the point s, s s ak is the absolute value of the vector, that originates in the point s k and terminates in the point s. 40

41 Analysis of Dynamic Systems We can define the root locus in terms of the phase condition. Applying the condition (1-52), (1-53) we get the condition for a point s that is lying on the root loci m L(s) = (s s bj ) (s s ak ) = (2i + 1) 180 (1-57) j=1 n k=1 where L(s) = arg{l(s)}, (s s bj ) = arg{(s s bj )} = ψ j 0, (s s k ) = arg{(s s ak )} = φ k 0, i = ±1, ±2, The graphical interpretation of Eq. (1-56) is that any point s on the root loci (for 0 K < ) must satisfy the condition (1-57). The difference between the sums of the angles of the vectors drawn from the zeros and those from the poles of L(s) to s, which is lying on the root loci, is an odd multiple of 180. An illustration of the use of the previous equations for checking the conditions of the root loci will be show in the following example. Example 1-10 Let us consider the transfer function L(s) = Solution i if it is lying on root loci! The location of the poles and zero is in Fig For the vectors can be written as v = v e iφ = v φ. 1) s B1 = 1, v 2 = v 2 ψ 1 = 45 arctan 6 3 = ) s 1 = 0, v 1 = v 1 φ 1 = 52 arctan 6 4 = ) s 2 = 2, v 3 = v 3 φ 2 = 40 arctan 6 2 = ) s 2 = 3, v 4 = v 4 φ 3 = 37 arctan 6 1 = s+1 s(s+2)(s+3). Check the trial point s = 45 A Φ = [ ( )] = This point is not a point from the root loci!!! Fig. 1-28: Poles and zero configuration 41

42 Example 1-11 Let us consider the transfer function L(s) = 1. Check the point z = 1 + i. s(s+1) Solution. The poles are s 1 = 0, s 2 = 2. The location of the poles is shown in Fig ) s 1 = 0, v 1 = v 1 φ 1 = 2e 3 4 π = ) s 2 = 2, v 2 = v 2 φ 2 = 2e 1 4 π = 2 45 arg z = φ = φ 1 + φ 2 = = 180 = π This point is lying on the root loci. The gain K is equal K = b m m j=1 (s s bj ) n (s s ak ) k=1 = Properties and Construction of the Root Loci Number of branches on the Root Loci The number of branches of the root loci is equal to the order of the polynomial A 0 (s). Example 1-12 Open loop transfer function L(s) = B 0 (s) = 1. How many branches has the root A 0 (s) s(s+2) loci? Solution. Poles: s 1 = 0, s 2 = 2, P = 2, zeros Z = 0, see Fig Fig. 1-29: Poles and zero configuration = 2 42

43 Analysis of Dynamic Systems Fig. 1-30: Graphical solution of Root Loci for L(s) = Starting and ending points of the Root Loci For closed-loop gain yields Eq. (1-51). 1 s(s+2) For K = 0 the equation (1-51) has the form 0 = A 0 (s), that means for K 0 the points on the root loci are the poles of the open loop. For K the equation (1-51) has the form B 0 (s) = 0, that means for K A 0 (s) points on the root loci are at the zeros of the open loop. Example 1-13 s+1 The transfer function is L(s) =. Determine: s(s+2)(s+3) 1) The number of branches and its starting and ending points. 2) How many branches end in zeros of open loop. 3) Plot the root loci with the function. Solution. 1) Number of poles and zeros of the open loop is P = 3, Z = 1. Root loci has 3 branches, which begin in the poles s 1 = 0, s 2 = 2, s 3 = 3. 2) One branch ends in the zero s B1 = 1. 3) Root loci is depicted in Fig

44 Fig. 1-31: Graphical solution of Root Loci for L(s) = s+1 s(s+2)(s+3) Symmetry of the Root Loci The root loci are symmetrical with respect to the real axis of the s-plane. Root Loci on the Real Axis On a given section of the real axis, the root loci are found in the section only if the total number of poles and zeros of the open-loop transfer function to the right of the section is odd. Number of Root Loci branches ending in infinity If P is the number of poles and N is the number of zeros of the open-loop function, then (P N) branches of root loci are ending in infinity. Example 1-14 Consider this open-loop transfer function L(s) = number of branches that end in infinity? Solution. s+1 s(s+4)(s 2 +2s+2). Determine the Number of poles of the open-loop P = 4, number of zeros Z = 1. Number of branches ending in infinity is 3. The symmetry is clearly shown in Fig

45 Analysis of Dynamic Systems Fig. 1-32: Graphical solution of Root Loci for L(s) = Angles of departure and Angles of arrival of the Root Loci s+1 s(s+4)(s 2 +2s+2) The angle of departure or arrival of a root locus at a pole zero, respectively, of L(s) denotes the angle of the tangent to the locus near the point. From an r-multiple pole of an open-loop r-branches leave with the angles of departures φ k = 1 r [(n P n L 1 ) π + k 2π], where n P is the number of poles and zeros lying to right of the multiple pole, n L is the number of poles and zeros lying to left of the multiple pole, k = 1,2,, r. Example 1-15 Consider the open-loop transfer function L(s) = 1 departure from the pole s 1 = s 2 = s 3 = 1. Solution. s(s+1) In accordance with the definition n P = 1, n L = 0 is the angle φ equal to 3. Determine the angles of φ k = 1 r [(n P n L 1 ) π + k 2π] = 1 [(1 0 1) π + k 2π] 3 k = 1 φ 1 = 1 3 [ 0π + 1 2π] = 2 3 π k = 2 φ 2 = 1 3 [ 0π + 2 2π] = 4 3 π k = 3 φ 3 = 1 3 [ 0π + 3 2π] = 6 π = 2π 3 45

46 The angles φ 1, φ 2, φ 3 are drawn in Fig Fig. 1-33: Graphical solution of Root Loci for L(s) = Intersection of the Asymptotes 1 s(s+1) 3 The intersect of the 2 n m asymptotes of the root loci lies on the real axis of the s-plane, at x 0 = n s m k=1 k j=1 s Bj n, where there are s P Z k=1 k the sum of the m poles of the open-loop, j=1 s Bj the sum of the zeros of the open-loop, P number of poles of the open-loop, Z number of zeros of the open-loop. The angle of the asymptotes with real axis is equal α 0 = ± π P Z.. Intersection of the Root Loci with the Imaginary Axis The point where the root loci intersect the imaginary axis of the s-plane, if any, exist and the corresponding values of K may be determined by means of the Routh Hurwitz criterion. Asymptotes of the Root Loci as s For large values of s, the root loci are asymptotic to asymptotes with angles given by Θ i = 2i+1 180, where i = 0,1,, n m 1, n is number of poles of n m L(s), m is number of zeros of L(s). 46

47 Analysis of Dynamic Systems Breakaway Points of the Root Loci The breakaway points on the root loci of 1 + K L(s) = 0 must satisfy d ds [B 0 (s) ] = 0. A 0 (s) s=x vet Example 1-16 Consider the transfer function L(s) = Determine: 1 s(s+4)(s 2 +4s+30). 1) The angles of asymptotes and their intersection with the real axis. 2) Breakaway points of the root loci. Solution of 1): The angles of asymptotes are given by the formula α 0 = ± π P Z = ± π 4 0 = ±π 4. The intersection of root loci with the real axis by x 0 = k=1 s k j=1 s Bj = P Z Solution of 2): n m i i = 2. Breakaway points of the Root Loci are given by the formula in definition where the point must satisfy the equation d ds [B 0(s) A 0 (s) ] = 0 s=x vet d 1 ds s(s + 4)(s 2 + 4s + 20) = 4s3 + 24s s + 80 (s(s + 4)(s 2 + 4s + 20)) 2 = 0. This is equal to zero for x vet = 2. The Root loci are shown in the Fig Fig. 1-34: Graphical solution of Root Loci for L(s) = 1 s(s+4)(s 2 +4s+30) 47

48 Example 1-17 Consider the transfer function of the open-loop L(s) = s+4. Determine: s(s+2) 1) The breakaway points of the root loci. 2) The root loci. Solution of 1): According to the definition the condition must be satisfied. This yields d ds [B 0(s) A 0 (s) ] = 0 d ds s=x vet s + 4 s(s + 2) = s2 + 8s + 8 (s(s + 2)) 2 = 0 Breakaway points such that x 1vet = 1.17, x 2vet = Solution of 2): Root loci are shown in Fig Fig. 1-35: Graphical solution of Root Loci for L(s) = Questions about the Root Locus Technique s+4 s(s+2) 1) Give a definition for the root locus. 2) Write the basic conditions of the root locus. 3) Explain the conditions of magnitude and angles for a point on the root loci. 4) What properties of root loci construction do you know? 5) Explain starting and ending points of the root loci. 6) Explain the numbers of branches on the root loci. 1.8 FREQUENCY DOMAIN ANALYSIS Frequency responses Frequency response is very important in communication and control systems. Most of the signals to be processed are either sinusoidal or composed of sinusoidal components. The starting point for frequency domain analysis of a linear system is the transfer function. 48

49 Analysis of Dynamic Systems Fig. 1-36: Output for LTI system with sinusoidal input It is known from linear system theory that when the input to a linear time-invariant system is sinusoidal u(t) = u 0 sin ωt with amplitude u 0 and frequency ω. The steady-state output signal of the system, will be a sinusoid with the same frequency, but with a different amplitude and phase y(t) = y 0 sin(ωt + φ), where y 0 is the amplitude of the output sine wave and φ is the phase shift in degrees or radians. A linear system s response to sinusoidal input is called frequency response. The Bode form of a transfer function F(s) is obtained by the substitution s = iω in F(s). F(iω) = R{F(iω)} + ii{f(iω)} = F(iω) e iφ(ω) (1-58) F(iω) = R{F(iω)} 2 + I{F(iω)} 2 (1-59) φ(ω) = arctan I{F(iω)} R{F(iω)} = F(iω) (1-60) Nyquist plot depicts a graphical frequency response F(iω) in the complex plain for frequency 0 ω Nyquist plots The plot is created using the formulas (1-58) that calculates the absolute value F(iω) and the phase φ(ω). The phase is calculated according to (1-60). Fig. 1-37: An example of Nyquist plot 49

50 A sketch of a frequency response in the complex plane can be drawn following steps: 1) The transfer function F(s) is rewritten to the Bode form of the transfer function F(iω). 2) Separate real and imaginary parts by multiplying the numerator and denominator complex by the conjugate number. 3) Find the start point of the Nyquist plot for lim ω 0 R{F(iω)}, lim ω 0 I{F(iω)}. 4) Find the end of the Nyquist plot for ω : lim ω R{F(iω)}, lim ω I{F(iω)}. 5) Find the intersections with the real axis for ω PI that satisfies the condition I{F(iω PI )} = 0. 6) Find the intersections with the imaginary axis for ω PR that satisfies the condition R{F(iω PR )} = 0. 7) Calculate a few points for frequencies between ω PR and ω PI. The procedure is demonstrated in the following examples. Example 1-18 Transfer function F(s) = 2 = 2. Draw the Nyquist plot for the (2s+1) 3 8s 3 +12s 2 +6s+1 transfer function F(s). Solution. 1) The Bode form of the transfer function is F(iω) = 2 8(iω) 3 12ω 2 + i6ω + 1 = 2 (1 12ω 2 ) + iω(6 8ω 2 ) = 2[(1 12ω2 ) iω(6 8ω 2 )] (1 12ω 2 ) 2 + ω 2 (6 8ω 2 ) 2 2) Separation of real and imaginary parts R{F(iω)} = 3) Start point of the Nyquist plot 4) End point of the Nyquist plot 5) Intersections with the real axis 2(1 12ω 2 ) (1 12ω 2 ) 2 + ω 2 (6 8ω 2 ) 2 2ω(6 8ω 2 ) I{F(iω)} = (1 12ω 2 ) 2 + ω 2 (6 8ω 2 ) 2 lim R{F(iω)} = 2, lim I{F(iω)} = 0 ω 0+ ω 0+ lim R{F(iω)} = ω 0, lim I{F(iω)} = 0 + ω 2ω(6 8ω 2 ) I{F(iω)} = (1 12ω 2 ) 2 + ω 2 (6 8ω 2 ) 2 = 0 ω PR 2 = 3 4, ω PR = R{F(iω PR = 0.866)} = 6) Intersections with the imaginary axis 2 (1 12ω 2 ) 2 = 2 8 =

51 Analysis of Dynamic Systems R{F(iω)} = 2(1 12ω 2 ) (1 12ω 2 ) 2 + ω 2 (6 8ω 2 ) 2 = 0 ω PI 2 = 1 12, ω PI = ω(6 8ω 2 ) I{F(iω = 0.289)} = (1 12ω 2 ) 2 + ω 2 (6 8ω 2 ) 2 2 ω(6 8ω 2 ) = ) Calculating a few points between frequencies ω PR and ω PI. For chosen ω = 0.5 is R{F(i0.5)} = 0.5 and I{F(i0.5)} = 0.5. The Nyquist plot is in Fig Fig. 1-38: Nyquist plot for F(s) = 2 (2s+1) 3 Example Transfer function is F(s) = = 2. Draw the Nyquist plot for (s 2 +2s+2)(s 0.5) s s 2 +s 1 the transfer function F(s). Solution. 1) The Bode form of the transfer function is equal F(iω) = 2 (iω) 3 1.5ω 2 + iω 1 = (3ω2 + 2) iω(1 ω 2 ) ω ω 4 + 4ω ) Separation of real and imaginary parts (3ω 2 + 2) R{F(iω)} = ω ω 4 + 4ω 2 + 1, I{F(iω)} = iω(1 ω 2 ) ω ω 4 + 4ω ) Start point of the Nyquist plot 4) End point of the Nyquist plot lim R{F(iω)} = 2, lim I{F(iω)} = 0 ω 0+ ω 0+ 51

52 5) Intersections with the real axis I{F(iω)} = lim R{F(iω)} = ω 0, lim I{F(iω)} = 0 + ω iω(1 ω 2 ) ω ω 4 + 4ω = 0 ω PR = 1, R{F(iω PR = 1)} = 0.8 6) Intersections with the imaginary axis do not exists for 0 ω <. 7) Calculating a few points between frequencies ω PR and ω PI. For chosen ω = 0.5 is R{F(i0.5)} = and I{F(i0.5)} = ) The Nyquist plot is in Fig Fig. 1-39: Nyquist plot for F(s) = 2 (s 2 +2s+2)(s 0.5) Example 1-20 A system is approximated by the transfer function F(s) = 2s+1 plot. Solution. 1) The Bode form of the transfer function is s(s 1) F(iω) = 2iω + 1 iω(iω 1) = 3ω2 + iω( 2ω 2 + 1) ω 2 (ω 2 + 1) 2) Separation of real and imaginary parts R{F(iω)} = 3) Start point of the Nyquist plot 3 ω 2 + 1, I{F(iω)} = 2ω2 + 1 ω(ω 2 + 1) lim R{F(iω)} = 3, lim I{F(iω)} = + ω 0+ ω 0+. Draw the Nyquist 52

53 Analysis of Dynamic Systems 4) End point of the Nyquist plot 5) Intersections with the real axis lim R{F(iω)} = ω 0, lim I{F(iω)} = 0 ω I{F(iω)} = 2ω2 + 1 ω(ω 2 + 1) = 0 ω PR = 0.707, R{F(iω PR = 0.707)} = 2 6) Intersections with the imaginary axis do not exist for 0 ω <. 7) The Nyquist plot is in Fig Fig. 1-40: Nyquist plot for F(s) = 2 (s 2 +2s+2)(s 0.5) Bode plots Working with frequency response, it is convenient to replace s with iω. F(iω) = F(iω) e iφ(ω) (1-61) log{f(iω)} = log F(iω) + iφ(ω) log e (1-62) Every frequency characteristic can be divided into two parts The logarithm of magnitude plot log F(iω) versus log ω Phase plot φ(ω) versus log ω Magnitude plot in decibels F(iω) db = 20 log F(iω). 53

54 The Bode plot can be presented as the magnitude in decibels versus log ω and the phase in degrees versus log ω. Open-loop transfer function in the form Bode form F(iω) F(s) = K (1 + st B1)(1 + st B2 ) (1 + st Bm )[(st B ) 2 + 2ξsT B + 1] s r (1 + sτ 1 )(1 + sτ 2 ) (1 + sτ n )[(sτ) 2 + 2ξsτ + 1] F(iω) = K (1 + iωt B1)(1 + iωt B2 ) (1 + iωt Bm )[(iωt B ) 2 + 2ξiωT B + 1] (iω) r (1 + iωτ 1 )(1 + iωτ 2 ) (1 + iωτ n )[(iωτ) 2 + 2ξiωτ + 1] Magnitude in decibels db Phase φ(ω) A(iω) db = 20 log K + 20 log 1 + iωt B log 1 + iωt Bn + 20 log (iωt B ) 2 + 2ξiωT B + 1 r 20 log iω 20 log 1 + iωt 1 20 log 1 + iωt n 20 log (iωt) 2 + 2ξiωT + 1 φ(ω) = arctan{k} + arctan{1 + iωt B1 } + + arctan{1 + iωt Bn } + arctan{(iωt B ) 2 + 2ξiωT B + 1} r arctan{iω} arctan{1 + iωt 1 } arctan{1 + iωt n } arctan{(iωt) 2 + 2ξiωT + 1} All transfer functions are composed of four factors (terms): Gain factor K Factors match the r-multiply poles or zero at origin (iω) r Factors match real p-multiply poles or zeros (1 + iωt) ±p Factors match complex conjugate poles and zeros [(iωt) 2 + 2ξiωT + 1] ±1 Plotting the term (factor) Gain K Magnitude of K in decibels db This magnitude does not depend on the frequency ω! Phase The phase does not depend on the frequency ω! (1-63) (1-64) (1-65) (1-66) K db = 20 log K (1-67) K > 0 φ = 0 K < 0 φ = π (1-68) Plotting the term (factor) matching the r-multiply poles or zero at origin 54

55 Analysis of Dynamic Systems The magnitude in db is given (iω) ±r db = 20 log (iω) ±r = ±r 20 log iω (1-69) The magnitude plot of this term is a straight line in semi logarithmic coordinates with a slope ±r 20dB/decade. This line passes through the 0 db axis at ω = 1. The phase is equal φ(ω) = ±r arctan{iω} = ±r arctan ( ω 0 ) = ±r 90 (1-70) The procedure of making a sketch of term s ±r is as follows: Magnitude in db Determine the multiplicity r. Draw the straight line ±r 20dB/dec that passes through the 0 db axis at ω = 1. Phase in degree The phase is equal φ(ω) = ±r 90. Example 1-21 Transfer functions F 1 (s) = 1 s, F 2(s) = 1 s 2, F 3(s) = s, F 4 (s) = s 2. Draw the Bode plots (magnitude and phase)! Solution. Fig. 1-41: Bode plot for F 1 (s) = 1 s, F 2(s) = 1 s 2, F 3(s) = s, F 4 (s) = s 2 55

56 Plotting the term (factor) match real or p-multiply poles or zeros The magnitude in db is (1 + iωt) ±p db = 20 log (1 + iωt) ±p = ±p 20 log 1 + iωt = ±p 20 log 1 + (Tω) 2 (1-71) The magnitude of this term approaches one asymptote at very low frequencies and another asymptote at very high frequencies: The frequency ω B = 1 is called the break point (frequency) / corner frequency. T a) For ωt 1 the magnitude is lim (1 + ω 0 iωt)±p = 20 log 1 ±p = 0 db b) For ωt 1 the magnitude is lim (1 + ω iωt)±p = lim ±p 20 log 1 + (Tω) 2 = ±p 20 log (iωt) 2 ω = ±p 20 log iωt At very high frequencies, ωt we can approximate (Tω) by (Tω) 2. The actual Bode plot is a smooth curve, and deviates only slightly from the straight-line approximation. c) For ωt = 1 the magnitude is 1. Phase: (1 + iωt) ±p ωt=1 = 20 log 2 ±p = ±p 20 log 2 = ±p 3.01 db The actual magnitude curve lies above that break point by factor Δ = ±p. 3 db. For phase asymptote is: φ(ω) = arctan{±p (1 + iωt)} = ±p arctan(ωt) (1-72) a) For ωt 1 the phase is equal to 0 b) For ωt 1 the phase is going to lim φ(ω) = ±p arctan( ) = ±p π = ±p 90 ωt 2 c) For ωt = 1 is the phase = ±p. 45 φ(ω) ωt=1 = ±p arctan(1) = ±p π = ±p

57 Analysis of Dynamic Systems The procedure of making a sketch of (1 + iωt) ±p db is as follows. The magnitude curve (db) can also be easily drawn by using the following low and high asymptotes. Determine the multiplicity p. Locate the break/corner frequency ω B = 1 on the frequency axis. T The low-frequency asymptote originates from the break point frequency ω B = 1 T and is zero for ω < ω B. The high-frequency asymptote originates from the break point frequency ω B = 1 T and has the slope ±p 20dB/dec for ω > ω B. The maximum deviation (error) of approximation by asymptotes is at the break frequency ω B = 1 and is equal Δ = ±p. 3 db. T The phase curve can be approximated by using three straight segments. For low frequencies from ω B 5 to 0 by the low asymptote φ = 0. For high frequencies from 5 ω B to by the high asymptote φ = ±p 90. For high frequencies from ω B 5, 5 ω B by the abscissa, which is going from 0 at ω B 5 to 90 at 5 ω B. The centre of the abscissa has the phase ±p 45. (The abscissa is tangent to the phase curve at the break point ω B = 1 T ) Example 1-22 Consider a transfer function F 1 (s) = 10 (2s+1) 2. Sketch the Bode plot. Solution. Magnitude in db 1) The magnitude of the gain 20 log(10) = 20 db. 2) The pole has multiplicity of two, p = 2. 3) Break/corner frequency ω B = ) The high asymptote has the slope 40 db/dec and both high and low originate at frequency ω B = 0.5. Phase 1) The abscissa goes from 0 at ω B 5 = 0.1 to 90 at 5 ω B = 2.5. The centre of the abscissa has the phase ±p 45 = 90. 2) The low frequency asymptote is φ = 0 originates at ω B 5 = 0.1 and goes to 0. 3) The high frequency asymptote is φ = 90 originates at 5 ω B = 2.5 and goes to. See the Bode plot in the Fig

58 Example 1-23 Fig. 1-42: Bode plot for F 1 (s) = 10 (2s+1) 2 Consider a transfer function F 1 (s) = 10(10s+1). Sketch the Bode plot. s(2s+1) 3 Solution numerical pre-processing: The transfer function consists of three terms (factors) that matches: 1) The gain K = 10. 2) The real zero for the term (10s + 1) is s B = 0.1, p = +1. 3) The zero pole for term 1 s at origin with r = 1. 4) The real pole for term (2s + 1) 3 is s 1 = s 2 = s 3 = 0.5, the multiplicity p = 3. Solution preparation to graphical result: 1) The magnitude of the gain 20 log(10) = 20 db. The phase is zero. Lines (1). 2) The real zero s B =

59 Analysis of Dynamic Systems Magnitude: The break/corner frequency ω BZ = 0.1. The low asymptote originates at ω BZ = 0.1 and is going to 0. The high asymptote has the slope of 20 db/dec and originates at ω BZ = 0.1 and is going to. Phase: The abscissa originates at ω = ω BZ = 0.02 and at the centre is the phase The low asymptote φ = 0 originates at ω = 0.02 and goes 0. The high asymptote φ = +90 originates at frequency ω = 5 ω BZ = 0.5 and goes to. Lines (3). 3) The zero pole at origin 1 s. The magnitude in db is a straight line 20 db/dec that passes through 0 db axis at ω = 1. The phase is φ = 90. Lines (2). 4) The real pole s 1 = s 2 = s 3 = 0.5. Magnitude in db: The pole has the multiplicity three, p = 3. Break/corner frequency ω BP = 0.5. The high asymptote has the slope 60 db/dec and both high and low originate at frequency ω BP = 0.5. Phase: The abscissa goes from 0 at ω BP 5 = 0.1 to 90 at 5 ω BP = 2.5. The centre of the abscissa has the phase 3 45 = 135. The low asymptote φ = 0 originates at ω BP 5 = 0.1 and goes to 0. The high asymptote φ = +90 originates at ω = 5 ω BP = 2.5 and goes to. Lines (4). Solution graph: By graphically adding all Bode plots of the terms (factors) we obtain the Bode plot of the given transfer function as a result. See the Fig. 1-43! 59

60 Fig. 1-43: Bode plot for F 1 (s) = 10(10s+1) s(2s+1) 3 Plotting the term (factor) match the complex conjugate poles and zeros Call ω n a natural frequency, and let ξ is the damping ratio. These parameters satisfy the conditions T = 1, 0 ξ 1. The term can be rewritten in [(iωt) 2 + 2ξiωT + 1] ±1 = ω n [(i ω ω n ) 2 + 2ξi ω ω n T + 1] ±1. The transfer function has the forms F(s) = K s 2 + 2ξω n s + ω2 = K n ( s ω ) 2 + 2ξ ( s n ω ) + 1 n = ω n 2 K (Ts) 2 + 2ξTs + 1 (1-73) 60

61 Analysis of Dynamic Systems Fig. 1-44: Example of Bode plot for the term with complex conjugates poles Bode plot in systems with pure time delay Transfer function in systems with pure time delay can be written as: F(s) = K (1 + st B1)(1 + st B2 ) (1 + st Bm )[(st B ) 2 + 2ξsT B + 1] s r (1 + sτ 1 )(1 + sτ 2 ) (1 + sτ n )[(sτ) 2 e + 2ξsτ + 1] st d Bode form F(s) with pure time delay is: (1-74) F(iω) = F 0 (iω) e st d = F 0 (iω) e iφ 0 (ω) e iωt d = F 0 (iω) e i[φ 0 (ω) ωt D ] (1-75) You can see that the magnitude in a system with time delay is equal to the magnitude without time delay. The phase is F(iω) = F 0 (iω) (1-76) φ(ω) = φ 0 (ω) ωt D (1-77) The phase with time delay is equal to the phase without the time delay from that is subtracted for each ω the amount of angle ωt D. 61

62 Example 1-24 Draw the Nyquist plot for F(s) = 2 (2s+1) 2 e 2s. Solution. Fig. 1-45: Nyquist plot for system with time delay Questions Frequency domain analysis 1) What is a frequency response? 2) For given transfer function F(s) calculate the magnitude and phase! 3) What is Nyquist plot? 4) What is Bode plot? 5) What is the transfer function for Bode plot? F(s) = 200(s+0.1) and draw the s(s+5)(s+10) asymptotes of Bode plot! 6) Write the phase of Bode plot for the transfer function F(s) = 200(s+0.1). s(s+5)(s+10) 7) Draw the Bode plot for the transfer function F(s) = 100(s+0.1) s 2 (s+10)! 8) Plot the terms K, (iω) 3, (1 + i10ω) 2. 9) How will you draw the asymptote of the phase? Give an example! 1.9 NYQUIST STABILITY CRITERION Zeros and poles of open-and closed-loop function Consider again the closed-loop diagram with the measured output Y(s). The closed-loop block diagram in the Fig is the unity feedback that has in the forward path the transfer function L(s) that is called the open-loop transfer function (loop transfer function). 62

63 Analysis of Dynamic Systems L(s) = K R 0 (s)g(s)h(s) = K B 0(s) A 0 (s) The closed loop transfer function of a SISO system is F YW (s) = Setting the equation (1-78) to (1-79) we obtain L(s) 1 + L(s) B F YW (s) = L(s) K 0 (s) 1 + L(s) = A 0 (s) K B 0 (s) 1 + K B = 0(s) A 0 (s) + K B 0 (s) A 0 (s) and the characteristic equation of the closed-loop transfer function is equal (1-78) (1-79) (1-80) Δ(s) = A 0 (s) + K B 0 (s) = 0 (1-81) Now we will investigate the denominator of the equation (1-79) 1 + L(s) = A 0(s) + K B 0 (s) A 0 (s) = 0 (1-82) Since the characteristic equation is obtained by setting the denominator polynomial to zero, the roots of the characteristic equation are the zeros of (1-82). Now, we can summarize the pole-zero relationship of the various transfer functions. Loop transfer function zeros: Loop transfer function poles: Closed-loop transfer function poles: Poles of poles1 + L(s): The Argument Principle zeros of L(s) poles of L(s) zeros of 1 + L(s) = roots of characteristic equation poles of L(s) The Nyquist stability criterion is based on a result from complex variable theory known as the argument principle. The argument principle can be stated as follows: Let F 0 (s) be a single-valued function that has a finite number of poles in the splane. Suppose that an arbitrary closed path c, is chosen in the s-plane so that the path does not go through any one of the poles or zeros of F 0 (s). The corresponding c F locus mapped in the F 0 (s)-plane (marked c) will encircle the origin as many times as the difference between the number of zeros and poles of F 0 (s) is. In equation form, the principle of argument is stated as ΔΦ = arg{f 0 (s)} = 2π(Z P) where there are Z number of zeros of F 0 (s) encircled by the s-plane locus c, P is number of poles of F 0 (s) encircled by the s-plane locus c. The application of the argument principle is demonstrated on the following example. 63

64 Example 1-25 Consider the transform function F(s) = 2(s+2) (s+1) 2 (s+3). The function F(s) has tree poles s 1 = s 2 = 1, s 3 = 3 and one zero s B = 2 (see Fig. 1-46a). Choose an arbitrary closed path c in the counterclockwise direction so that all poles and zero of F(s) are encircled. Then by mapping this path c to the F(s)-plain will be determined the path c F, which will encircle the origin of F(s)-plain. arg{f(s)} = 2π(Z P) = 2π(1 3) = 2 2π The origin is encircled twice but the direction of c F is opposite (see the Fig. 1-46b). Fig. 1-46: a) s-plain: poles, zero encircled by path c, b) The corresponding path c F mapped in the F(s) -plane Nyquist condition of stability Nyquist discovered that the argument principle could be applied to solve the stability problem if the s-plane locus c is taken to be one that encircles the entire right-half of the s-plane. Figure 1.9-4a illustrates a c locus with a counter-clockwise direction, which encircles the entire right- half of the s-plane. Of course, the c locus must not go through any poles and zeros of the term {1 + L(s)}. For closed-loop stability the number of zeros Z P which are inside the right - half of the s-plane of the term {1 + L(s)} must be equal to zero. Thus the Nyquist criterion follows from the argument principle that has the following form Fig. 1-47: Path c encircling the entire right-half of the s-plane ΔΦ = arg{1 + L(s)} = 2π(Z P P P ) = Z P = 0 = 2π P P (1-83) Where P P is the number of poles {1 + L(s)} inside the right-half s-plane. Z P is the number of zeros {1 + L(s)} inside the right-half s-plane Nyquist criterion Generally, the mapping by complex function F 0 (s) is not easy. For engineers it is important, that Nyquist created the path c (see Fig. 1-48) so that the branches of the path c F 64

65 Analysis of Dynamic Systems in the F(s)-plane would have a physical interpretation provided that there is one pole equal to zero. This path is with the clockwise direction and is called the Nyquist path. The Nyquist path is generally composed of three segments (portions): c I, c II and c IV. i. Segment c I runs along the iω axis from s 0 + to +i, excluding all the indentations. ii. Segment c II runs from s +i to s i along the semicircle with infinitive radius R. iii. Segment c III (similar to c I ) runs along the iω axis from s iω to s 0, excluding all the indentations. iv. Segment c IV runs with all the small indentation on the real axis with radius r 0 0. The Nyquist condition of stability for the Nyquist path (running clockwise direction) satisfies the following equation Now we can state Fig. 1-48: Nyquist path arg{1 + L(s)} = 2π ( P P ) = 2π P P (1-84) A closed-loop system is stable, if the corresponding c F locus mapped in the {1 + L(s)}-plane will encircle the origin as many times as the number of poles P P that are in the right-half s-plane. The encirclement, if any, must be made in the counter clockwise direction. You see that if the Nyquist path is specified, the stability of a closed/loop system can be determined by plotting the locus of c F in the complex plane {1 + L(s)}. Variable s takes on values along the Nyquist path c. We can easily draw trajectories in the L(s)-plane, but we shall investigate the trajectory of c F in the complex plane {1 + L(s)}. The plane {1 + L(s)} is created by moving the plane{l(s)} to left by 1. Since the function L(s) is generally known, it would be simpler to construct the L(s) plot that corresponds to the Nyquist path, and the same conclusion on the stability of the closed loop system can be obtained by observing the behaviour of the L(s) plot with respect to the [ 1,0i] point in the L(s)-plane. Investigation of the Nyquist criterion (1-84) and the mapping of Nyquist path results in the following four conclusions: i. Segment c I runs along the iω axis from s 0 + to +i corresponds to the single branch of the Nyquist plot of L(iω). ii. Segment c II runs from s +i to s i along the semicircle with infinitive radius R corresponds to lim s L(s). iii. Segment c III (similar to c I ) runs along the iω axis from s iω to s 0 corresponds to the single branch of the Nyquist plot of L( iω). 65

66 iv. The origin of the Δ(s) = {1 + L(s)}-plane corresponds to the [ 1,0i] point in the L(s)-plane. Thus the point [ 1,0i] in the L(s)-plane becomes the critical point for the determination of closed-loop stability The Nyquist criterion can be stated Fig. 1-49: Nyquist path (a), Nyquist plot (b) A closed-loop system with loop transfer function L(s) is stable, if the plot L(s) that corresponds to the Nyquist path, encircles the critical point [ 1,0i] in the counter-clockwise direction as many as times as the number of poles P P that lie in the right-half of the s-plane. This satisfies the following equation Nyquist condition ΔΦ = arg{1 + L(s)} = arg{[1 + L(s)] 1} = 2π P P Checking the number of encirclements The Nyquist criterion can be applied to a minimum as well as a non-minimum phase transfer function. We can create two Nyquist paths shown in Fig The Nyquist path in Fig. 1-50a is the original, whereas the path in Fig. 1-50b encircles not only the entire right-half s-plane, but also all the poles and zeros of L(s) on the iω-axis. The poles and zeros that are included by the left half s-plane are considered to be stable, and those encircled to the entire right half s-plane are considered to be unstable. The mapped Nyquist path in the L(s)-plane must be closed therefore it is necessary to find the connection between branches. It is the pole s = 0 on the iω-axis. If the pole s = 0 with multiplicity r is in the left half s-plane, then the segment c IV of the Nyquist path can be expressed as s = lim ρ 0 ρ e iφ for π 2 < φ < π 2, where the angle φ varies in the counter clock direction. The poles and zeros are considered to be stable Fig. 1-50a) or to be unstable Fig. 1-50b)! Suppose the transfer function L(s) has the form 66

67 Analysis of Dynamic Systems L(s) = β m s m + β 1 s + β 0 s r (α n s n + + α 1 s + α 0 ) If we set s = ρ e iφ to (1-85) and s 0 then we obtain (1-85) The magnitude is equal The change of the angle ΔΦ is equal lim L(s) = lim L(ρ e iφ β 0 ) = lim s 0 ρ 0 ρ 0 α 0 (ρ) r e irφ (1-86) A = lim ρ 0 L(ρ e iφ ) = (1-87) ΔΦ = arg {lim L(ρ e iφ 2 )} = rφ π = r π ρ 0 (1-88) π 2 Fig. 1-50: The left half s-plane includes poles on the iω axis (a), poles on the iω axis are encircled in the right half of the s-plane The equation (1-88) can be stated as follows. If the pole s = 0 with multiplicity r is in the left half of the s-plane, then the connection of one end of the Nyquist plots given by the point lim ω 0 L(iω) to the other end at the given point lim ω 0+ L(iω) will be described by an arc with an infinity radius R = lim ρ 0 L(ρ e iφ ) = and with the changing angle ΔΦ = r π in the clockwise direction. If the pole s = 0 with multiplicity r is encircled in the right half s-plane, than the poles and zeros are considered to be unstable!!! It is possible to write ΔΦ = r π. 67

68 The result can be stated as follows. If the pole s = 0 with multiplicity r is in the right half of the s-plane, then the connection of one end of the Nyquist plots given by the point lim ω 0 L(iω) to the other end at the given point lim ω 0+ L(iω) will be described by an arc with an infinity radius R = lim ρ 0 L(ρ e iφ ) = and with the changing angle ΔΦ = r π in the counter-clockwise direction. The application of the Nyquist criterion may have the following steps The loop transfer function L(s) = R(s)F(s) must be determined. The number of poles n R that lie in the right half of the s-plane, can be determined. The number of poles n I lying on the iω axis, must be specified. If the poles that are lying on the iω axis are considered: to be stable then they must be included in the left s-plane, the number of poles lying in the right half of the s-plane is not changed to be unstable these poles are encircled in the entire right s-plane, then the number of poles lying in the right half of s-plane is changed to n R + n I. The Nyquist path is defined in the s-plane as in step 2) as shown in Fig The Nyquist condition ΔΦ = arg{[1 + L(s)] 1} = 2π P P has to be calculated. The L(s) = R(s)F(s) plot corresponding to the Nyquist path is constructed in the L(s)-plane. For the pole s = 0 with the multiplicity r the connection must be made. The connection arc has the angle ΔΦ = r π in the clockwise direction or ΔΦ = r π in the counter-clock wise direction. The number of encirclement of the critical point [ 1,0i] can be calculated. If the real encirclement is equal to the Nyquist condition with the counterclockwise direction we can say the closed-loop is stable!!! The number of encirclement N E of the critical point can be easily determined by applying the Nyquist half-line p. It is a half line that must be drawn from the critical point in the auxiliary direction so that the half-line intersects L(s) plots. You can see that the curve c intersects the halfline on the points (+ar 1, ar 1, +ar 2 ) in order. The number of encirclement is equal N E = i(+ar i ) j ( ar j ). 68

69 Analysis of Dynamic Systems Fig. 1-51: Nyquist half-line The following examples serve to illustrate the application of the Nyquist criterion. Example 1-26 Consider the transfer function of the open loop L(s) = loop stable? Solution. 1) The transfer function is given. 2 (s 2 +2s+2)(s 0.5). Is the closed 2) One pole s 1 = 0.5 is in the right s-plane, pole s 2,3 = 1 ± i that is in the left s- plane. There are no poles on the iω-axis. 3) The Nyquist path is in Fig on the left side. The number of poles lying in the right s-plane is P p = 1. 4) Nyquist stability condition is arg{1 + L(s)} = 2π P P = 2π. 5) Nyquist plot of L(s) is depicted in Fig on the right side. 6) Nyquist straight line p 1 has only one arrow in the counter-clockwise, so the number of encirclement is +1. 7) The Nyquist half-line p 1 is drawn in Fig (right). You can see the real encirclement is equal +1, ΔΦ = arg{l(s)} = 2π P P = 2π in the counterclockwise direction. The closed-loop will be stable! 69

70 Fig. 1-52: Nyquist path (left), Nyquist plot (right) Example 1-27 Consider the transfer function L(s) = 2s+1. Is the closed-loop stable? s(s 1) Solution. 1) The transfer function is given. 2) One pole s 1 = 1 is in the right s-plane, pole s 2 = 0 is on the iω-axis. We assume the pole s = 0 with multiplicity r = 1 is in the left half s-plane. 3) The Nyquist path is in Fig. 1-53a. The number of poles lying in the right s-plane is P P = 1. 4) Nyquist stability condition is arg{1 + L(s)} = 2π P P = 2π. 5) Nyquist plot of L(s) is depicted in Fig. 1-53b. The end lim ω 0 L(iω) of the Nyquist plot is connected to the other end lim ω 0+ L(iω) by an arc with the infinite radius and with angle ΔΦ = r π = π. 6) The Nyquist straight line p 2 has only one arrow in the counter-clockwise direction, so the number of encirclement is +1. You can see that the real encirclement is equal to +1, so that the phase is ΔΦ = arg{l(s)} = 2π P P = 2π in the counterclockwise direction. 7) The closed-loop will be stable! 70

71 Analysis of Dynamic Systems Gain and phase margin Fig. 1-53: Nyquist path (a), Nyquist plot (b) In general, we are interested not only in the absolute stability of a system, but also how stable the system is. In that case we are speaking about relative stability. Relative stability in the frequency domain is measured by how close the Nyquist plot of L(s) is to the critical point [ 1,0i]. Gain margin (GM). In the frequency domain, gain margin is used to indicate the closeness of the intersection of the negative real axis made by the Nyquist plot of L(s) to the [ 1,0i] point. If the Nyquist plot of L(s) crosses the negative real axis between 1 and 0 then 1 the gain margin is defined as GM =. Phase crossover frequency ω L(iω 180 ) 180 is where the Nyquist curve L(iω) crosses the negative real axis between 1 and 0. The gain margin in Bode plots is defined as 1 GM = 20 log L(iω 180 ) = 20 log L(iω 180) (1-89) The graphical interpretation of definition of the gain margin is shown in Fig. 1-54a. Phase margin (PM). The gain margin is only a one-dimensional representation of the relative stability of a closed loop system. To include the effect of phase shift on stability, we introduce the phase margin. First, we give some definitions. The phase margin is defined as PM = Φ M = L(iω c ) Gain crossover frequency ω c is the frequency where L(iω c ) crosses the circle with diameter 1. Another definition of phase margin can be stated as. 71

72 Phase margin (PM) is defined as the angle in the degrees through which the L(iω) plot must be rotated around the origin so that the gain crossover passes through the [ 1,0i] point. Fig. 1-54b shows the graphical interpretation of definition of phase margin. Fig. 1-54: Definition of gain margin (a), definition of phase margin (b) The interpretation of gain and phase margin in Bode plots is illustrated in Fig You can see that: 1) The gain margin is positive and the system is stable if the magnitude of L(iω) at the phase crossover is negative in db. That is, the gain margin is measured below the 0 db axis. If the margin is measured above the 0 db axis, the gain margin is negative and the system is unstable. 2) The phase margin is positive and the system is stable if the phase of L(iω) is greater than 180 at the gain crossover. That is, the phase margin is measured above the 180 axis. If the phase margin is measured below the 180 axis, the phase margin is negative, and the system is unstable. Fig. 1-55: Gain and phase margin in Bode plot 72

73 Analysis of Dynamic Systems Question Nyquist Stability Criterion 1) What is the simplified Nyquist criterion? 2) Explain relation between zeros and poles of an open and closed-loop transfer function! 3) The Argument Principle. 4) Formulate the Nyquist condition of stability, the Nyquist path and check the number of encirclements! 5) What should you to do when there are poles on the imaginary axis? 6) Determine the gain and phase margin in a complex plane and in Bode plot! 73

74 2 FEEDBACK CONTROL 2.1 BASIC PROPERTIES OF FEEDBACK Basic structure of control Generally, two generic control configurations are used: open-loop and closed-loop systems. Open-loop systems A generic open-loop system is shown in Fig. 2-1 and consists of a subsystem called Input transducer that converts the form of the input to that used by the controller. The controller drives the process or plant with controller output u. The input is w sometimes called the reference while the output y S can be called the controlled variable. The disturbance is an undesirable input to the plant. Open-loop systems do not correct disturbances and are simply commanded by the reference signal. Fig. 2-1: Open loop system 2.2 CLOSED-LOOP SYSTEMS The generic architecture of a closed-loop system is shown in Fig The input transducer converts the form of the input w to the form of the controller w. The summing junction algebraically subtracts the feedback signal y from the desired value/set point signal w, which arrives via the feedback path. This signal is generally called the error or the actuating signal. The controller generates signal u(t) known as the manipulating signal, which is the input to the process (plant). 74

75 Feedback control The closed-loop system compensates for disturbances by measuring the output response y(t), feeding that measurement back through a feedback path, and comparing that response to that input w(t) at the summing junction. If there is any difference between the two responses, the system drives the plant, via error (actuating signal) e(t), to make a correction. If there is no difference, the system does not drive the plant, since the plant response is already the desired response. Fig. 2-2: Closed loop control The basic equations in control A block scheme of a technological process is shown in Fig We can denote (identify) as an extended/generalized plant- that part of the scheme that consists of the plant/process and the measuring devices (see Fig. 2-4). We recall that the plant contains amplifiers with transducers, transmitters and a technological process with an actuator. Fig. 2-3: Scheme of a controlled plant/process with measuring devices Fig. 2-4: Extended/generalized controlled plant 75

76 The measuring device consists of sensors with transducers or transmitters. The dynamic of the extended plant is generally approximated by three transfer functions F V (s), F S (s) and H(s). The transfer function F V (s) approximates the dynamic of the amplifiers, transmitters and the actuator, the transfer function F P (s) approximates the dynamic of the process and the transfer function H(s) approximates the dynamic of measuring devices (see the Fig. 2-5). Fig. 2-5: Transfer function of extended controlled plant Finally, the transfer function notation is often simplified by lumping all instrumentation and process dynamic into one term F U. This is equivalent to the following expression F U (s) = Y(s) U(s) = F V(s) F S (s) H(s) = F(s) (2-1) The classical control system uses models that are described by transfer functions. The block diagram of open-loop control with disturbance is in Fig. 2-6.The block diagram of closed-loop control with disturbance model is in Fig Model of an open-loop control system. The Laplace transform of output is equal Fig. 2-6: Model of an open-loop control system Y(s) = F(s) R W (s) W(s) + D(s) (2-2) b where F(s) = m s m + +b 1 s+b 0, R s n +a n 1 s n 1 + +a 1 s+a W (s) is transfer function of a feed forward 0 controller, W(s) is Laplace transform of set point/desired value, D(s) is Laplace transform of the disturbance. From Eq. (2-2) it is obvious that the disturbance cannot be compensated. Model of a closed-loop control For closed-loop control/feedback control, Fig. 2-7 gives the basic structure of the area of interest but with disturbances (incoming of sensor noise is disregard). 76

77 Feedback control Fig. 2-7: Model of a feedback control system Transfer Function Tab. 2-1: Description of symbols in Fig. 2-7 Variables/signals R(s) Controller Y S (s) F V (s) Transmission, transducer and valve-actuator Y(s) F P (s) Process transfer function D(s) Disturbance H(s) Sensor, transducer, and transmission W(s) Final controlled variable/ unmeasured Measured value of control variable Set point/ Reference value/ Desired value F D (s) Disturbance transfer function U(s) Manipulated variable The Laplace transform of output Y(s) is Y(s) = F YW (s) W(s) + F YD (s) D(s) = = R(s)F V(s)F P (s)h(s) 1 + R(s)F V (s)f P (s)h(s) W(s) + F D (s)h(s) 1 + R(s)F V (s)f P (s)h(s) D(s) (2-3) where F YW (s) = Y W (s) = R(s)F V (s)f P (s)h(s) W(s) 1+R(s)F V (s)f P (s)h(s) Y D (s) = F D (s)h(s) is closed loop disturbance. D(s) 1+R(s)F V (s)f P (s)h(s) Set point response is Disturbance response is Y W (s) = F YW (s) W(s) = Y D (s) = F YD (s) D(s) = is set point transfer function, F YD (s) = R(s)F V(s)F P (s)h(s) 1 + R(s)F V (s)f P (s)h(s) W(s) (2-4) F D (s)h(s) 1 + R(s)F V (s)f P (s)h(s) D(s) (2-5) Using the general transfer function (2-1) the block diagram can be simplified in the form shown in the Fig

78 Fig. 2-8: Block diagram of a general closed-loop feedback control system Tab. 2-2: Description of symbols in Fig. 2-8 y(t) Controlled variable w(t) Set point/desired input variable d(t) Disturbance u(t) Manipulated variable e(t) = w(t) y(t) Error/Correcting variable Simplified closed loop disturbance transfer function Simplified set point transfer function F YD (s) = Y D(s) D(s) = F D (s) 1 + R(s)F U (s) F YW (s) = Y W(s) W(s) = R(s)F U(s) 1 + R(s)F U (s) The general closed-loop transfer function model can be applied to any specific system by substituting the transfer function models for the loop elements. 2.3 CONTROL OF DYNAMIC ERROR P controller P controller = proportional only controller. The output from a P controller is proportional to the error and is equal to (2-6) (2-7) u P (t) = P e(t) (2-8) where P represents the proportional gain, e(t) represents the error. Transfer function of P controller is R P (s) = U P(s) E(s) = P (2-9) 78

79 Feedback control The proportional mode is simple, provides an adjustment of the manipulated variable, does not provide zero offset although it reduces the error, speeds the dynamic response, and can cause instability if tuned improperly PI controller PI controller = proportional-integral controller. A PI controller has two factors ( terms) and two tuning parameters, one proportional to the error and the other proportional to the integral of the error. The PI controller output is equal to t u PI (t) = P e(t) + I e(τ)dτ = K R [e(t) + 1 e(τ)dτ] T (2-10) I 0 where P represents proportional gain, I represents integral gain, K R represents all controller gain, T I represents the integral time, e(t) is the error. The PI controller transfer function is Integral mode R PI (s) = U PI(s) E(s) = P + I Ps + I = = K s s R T Is + 1 T I s where there is e(0) an initial condition. u I (t) = K R T I e(τ)dτ 0 t 0 t (2-11) + e(0) (2-12) The integral mode is simple; achieves zero offset; adjusts the manipulated variable in a slower manner than the proportional mode, thus giving poor dynamic performance; and can cause instability if tuned improperly PID controller PID controller = proportional-integral-derivative controller. A PID controller has three terms(factors) and three tuning parameters, one proportional to the error the other proportional to the integral of the error and uses knowledge of the derivative of the error. The derivative mode allows the controller to predict where the error is heading and compensate for it. The PID controller output in regards tom the time-domain is u PID (t) = P e(t) + I e(τ)dτ 0 t + D de(t) dt where P is the proportional gain, I is the integral gain, D represents the derivative gain. Often the output is written in the follow form u PID (t) = K R [e(t) + 1 T I e(τ)dτ 0 t (2-13) + T D de(t) dt ] (2-14) where K R is all controller gain, T I is the integral time, T D is the derivative time, e(t) is the error. 79

80 A transfer function of a PID controller R PID (s) = U PID(s) = P + I E(s) s + Ds = Ds2 + Ps + I s = K R T DT I s 2 + T I s + 1 T I s (2-15) The ideal PID algorithm is not physically realizable, that means no technical instrument can take perfect derivative. In this case the real PID controller has the following transfer function R PID (s) = U PID(s) E(s) = P + I s + Ds T I s + 1 = K 1 R ( + T Ds N s + 1 T I s T F s + 1 ) (2-16) The derivative mode is important if the error is zero or very small but in spite of that the disturbances already begin to affect the controlled variable (the derivative is already present). In this situation, the error and integral error are nearly zero, but a substantial change in the manipulated variable would seem to be appropriate because the rate of change of the controlled variable is large. This situation is addressed by the derivative mode. u D (t) = K R T D de(t) dt = D de(t) dt = D dy(t) dt = D d(w(t) y(t)) dt = w = const. The derivative mode is simple; does not influence the final steady state-value of error; provides rapid correction based on the rate of change of the controlled variable; but can cause undesirable high-frequency variation in the manipulated variable PD controller (2-17) PD controller = proportional - derivative controller. The PD controller output in the timedomain is u PD (t) = P e(t) + D de(t) dt Transfer function of a PD controller Time domain specifications (2-18) R(s) = U(s)/E(s) = P + Ds (2-19) Control performance is the ability of a control system to achieve the desired dynamic responses, as indicated by the control performance measures, over an expected range of operating conditions. Next, we carefully define control performance by specifying several goals to be balanced concurrently. The goals are 80

81 Feedback control 1) Controlled variable performance. The well-tuned controller should provide satisfactory performance for one or more measures of the behaviour of the controlled variable. 2) Model error. Linear dynamic models always have errors, because the plant is nonlinear and its operations change. The ability of a control system to provide good performance, if the plant dynamic changes, is often termed robustness. 3) Manipulated-variable behaviour. The most important variable, other than the controlled variable, is the manipulated variable. We shall choose the common goal of preventing excessive variation in the manipulated variable by defining limits on its allowed variation, as explained shortly. Specifications for a control system design often involve certain requirements associated with the time response of the system. The requirements for a step response are expressed in terms of the standard quantities is shown in Fig Fig. 2-9: Time domain specifications Consider the second order and the transfer function F(s) = ω n s 2 +2ω n ξs+ω n 2. The roots are s 1,2 = 2ξω n±2ω n ξ 2 1 = ξω 2 n ± iω n 1 ξ 2 = σ ± iω, where σ = ξω n ξ = σ, ω n and ω = ω d = ω n 1 ξ 2, ξ 0,1). The rise time T R is the time it takes the system to reach the vicinity of its new set point. It can be calculated approximately from the equation T R = 1.8 ω n ω n = 1.8 T R (2-20) 81

82 Rise time T R is inversely proportional to natural frequency ω n. Since the circle with radius ω n1 on the s-plane is locus of constant natural frequency ω n1, it is also the locus for constant rise time T R (see Fig. 2-10b). The settling time T S is the time it takes the system transients to reach and stay within ±2 % of the steady state. T S = 4.6 σ σ = 4.6 T S (2-21) Settling time is inversely proportional to the real part of the pole. Since vertical lines on the s-plane are lines of constant real value, they are also lines for constant settling time (see Fig. 2-10b). The overshoot M P is the maximum amount the system overshoots its final value divided by its final value (and is often expressed as a percentage). πξ M P (%) = e 1 ξ % ξ = ln ( M P(%) 100 ) π 2 + ln ( M P(%) 100 ) (2-22) We know that φ = sin 1 ξ, radial lines are lines of constant ξ. Since percent overshoot is only a function of ξ, radial lines are thus lines of constant overshoot. This is depicted in Fig. 2-10b. Fig. 2-10: Image of time domain specification in s-plane The peak time T P is the time it takes the system to reach the maximum overshoot point T P = π ω ω = π T P (2-23) 82

83 Feedback control The equation (2-23) shows that T P is inversely proportional to the imaginary part of the pole. Since horizontal lines on the s-plane are lines of the constant imaginary value, they are also lines of constant peak time (see Fig. 2-10b). In design synthesis we wish to specify T R, M P, and T S and to ask where the poles need to be so that the actual responses are less than or equal to these specifications Control steady state error Steady-state error is the difference between the input and the output for a prescribed test input as t. Test inputs used for steady-state error are in Tab Tab. 2-3: Test inputs for steady state error Signal type Function in time domain Laplace image Step w(t) = 1(t) W(s) = 1 s Ramp w(t) = t 1(t) W(s) = 1 s 2 Parabola w(t) = t 2 1(t) W(s) = 1 s 3 In the unity feedback system drawn in the Fig. 2-5, the error E W (s) (D(s) = 0) is E W (s) = R(s)F U (s) W(s) = L(s) W(s) (2-24) Suppose the input w(t) is equal w(t) = t k 1(t), k = 0,1,2, W(s) = 1 sk+1. For k = 0 it is step input, for k = 1 it is a ramp input and for k = 2 it is a parabola input, known as an acceleration input. The open-loop transfer function can be described in the following form: R(s)F U (s) = L(s) = L 0(s) s n (2-25) where n is the number of integrators (integrals) in the open loop. Of course, the integrators (integral) can be incorporated both by the control system and the plant. The structure of open-loop must be created by the control engineers by choosing an appropriate structure of a control system. The application of the Final Value Theorem to the error results in the following formula lim t e(t) = lim s E(s) = lim s 0 From this equation we can see at once that s 1 1 s L = lim 0(s) sk+1 s 0 s n 1) If n > k then e(t) = 0. 2) If n < k then e(t). 3) If n = k = 0 then e( ) = 1 1+L(0) = 1 1+K 0. 4) If n = k 0 then e( ) = sn+1 s n +L(0) 1 s n+1 = 1 L 0 (0). s n+1 s n + L 0 (s) 1 s k+1 (2-26) 83

84 Looking back at the expression (2-25) we can see that the error constants depend on the number n of integrators in the loop with the unity feedback. K P = lim s 0 L(s), n = 0 K V = lim s 0 sl(s), n = 1 K A = lim s 0 s 2 L(s), n = 2 Steady-state error as a function of system type and reference inputs is shown in Tab Type Step (position)/e( ) Ramp(velocity)/e( ) Parabola/e( ) (2-27) (2-28) (2-29) Type K P Type 1 0 Type K V 1 K A 2.4 FEEDBACK SYSTEM SENSITIVITY Sensitivity consideration can be important in the design of control systems. Since all physical elements have properties that change with the environment and time, the parameters of a controlled plant are not constant stationary over the entire operating life of the control system. Technically, a good control system should be very insensitive to parameter variations of the controlled plant. Using sensitivity functions we can quantify the effects of (adverse) changes in the parameters of the functions. Consider a function where P is a vector of parameters. Y = Y(P) (2-30) Based on the assumption that a proportional change in parameters dp/p (infinitesimal) of the function Y = Y(P) corresponds to the proportional change in the function dy/y (infinitesimal). Then it is possible to define the sensitivity function as a fraction S P Y = dy Y dp P = percentage change in Y percentage change in P where S P Y is the sensitivity function of the complex variable s. (2-31) 2.5 SENSITIVITY STUDIES OF FEEDBACK CONTROL Sensitivity studies in the frequency domain of linear control systems are based on the sensitivity function (2.1-26). Consider a single feedback-control structure in Fig The controlled plant, which can change its parameters, is described by approximation of the transfer function F U (s) = F(s). The controller has the transfer function R = R(s) and it does not change its parameters! For the single-loop control system configuration in Fig the closed-loop transfer function can be written 84

85 Feedback control F YW (s) = Y(s) W(s) = R(s)F(s) 1 + R(s)F(s) = M(s) = M(F(s)) (2-32) Fig. 2-11: Feedback control structure The sensitivity of M(s) with respect to the transfer function of the controlled plant F(s) is S F M = dm(s) M(s) df(s) F(s) = F(s) M(s) dm df (2-33) where M(s) is the transfer function of the closed/loop, F(s) is the transfer function of the controlled plant. Substituting Eq. (2-32) into Eq. (2-33) yields = S F M = F(s)(1 + R(s)F(s)) R(s)F(s) dm(s) M(s) df(s) F(s) = = F(s) M(s) dm(f) df = R(s)(1 + R(s)F(s)) R2 (s)f(s) (1 + R(s)F(s)) 2 = R(s)F(s) = S (2-34) The sensitivity function of the closed-loop transfer function M(s) is a function of the complex variable s. The magnitude of S F M (iω) = S(iω) is equal S F M (iω) = where L(iω) = R(iω)F(iω) R(iω)F(iω) = L(iω) = S(iω) (2-35) The plot of magnitude of S F M (iω) = S(iω) versus the frequency gives an indication of the sensitivity of the system as a function of frequency. The ideal robust situation is for S(iω) to assume a small value S(iω) 1 over a wide range of frequencies. As an example, a magnitude of a sensitivity function is plotted and shown in Fig Example 2-1 Consider an approximation of the controlled plant as the transfer function F(s) = 10 (2s+1) 3 with the PI controllers R 1(s) = s and R 2 (s) = s. The 85

86 Amplitude Singular Values (db) magnitude of S 1 (iω) for R 1 and S 2 (iω) for R 2 is plotted as shown in Fig and step responses for the set point are in Fig Sigma S1, S2, Pr S1 S Frequency (rad/sec) Fig. 2-12: Magnitudes of S 1, S Step T1, T2, Pr Time (sec) Fig. 2-13: Step T 1, T 2 86

87 Feedback control In general, it is desirable to formulate a design criterion in the following manner S = S F M (iω) = R(iω)F(iω) k (2-36) where k is a positive real number. The maximum peak of the sensitivity function is defined as M S = max ω S(jω) (2-37) The general closed-loop feedback control system in Fig. 2-8 is the basis for analysis of dynamic behaviour using sensitivity function S. The set point transfer function F YW (s) = Y(s) for D(s) = 0 is equal W(s) F YW (s) = Y(s) W(s) = F U(s)R(s) 1 + F U (s)r(s) (2-38) F EW (s) = E W(s) W(s) = F U (s)r(s) = S(s) (2-39) The closed loop disturbance transfer function F YD (s) = Y(s) for W(s) = 0 is equal The output Y(s) is equal D(s) F YD (s) = Y(s) D(s) = F D (s) 1 + F U (s)r(s) = F D(s) S(s) (2-40) F ED (s) = E D(s) D(s) = F D (s) 1 + F U (s)r(s) = F D(s) S(s) (2-41) Y(s) = F U(s)R(s) 1 + F U (s)r(s) W(s) + F D (s) 1 + F U (s)r(s) D(s) = S(s)F U (s)r(s)w(s) + S(s)F D (s)d(s) The closed-loop response can be rewritten in the form (2-42) Y(s) = T(s)W(s) + S(s)F D (s)d(s) (2-43) The following notation and terminology are used Loop transfer function L(s) Eq. (2-44) Sensitivity function S(s) Eq. (2-45) Complementary sensitivity function T(s) Eq. (2-46) We see that S is the closed loop transfer function from the output disturbances to the output, while T is the closed loop transfer function from the reference signal to the output. For the sensitivity and complementary sensitivity function S, T follows the identity in Eq. (2-47). L(s) = R(s) F(s) (2-44) S(s) = (1 + R(s) F(s)) 1 = (1 + L(s)) 1 (2-45) 87

88 Singular Values (db) T(s) = (1 + R(s) F(s)) 1 R(s) F(s) = (1 + L(s)) 1 L(s) (2-46) Example 2-2 Consider transfer function F(s) = S(s) + T(s) = 1 (2-47) 2 s 3 +3s 2 +2s+1 G(iω) using the function sigma in Matlab. Solution. clear all B = [2]; A = [ ]; G = tf(b, A); H2 = norm(g, 2) Hinf = norm(g, inf) [MaxSig, frek] = norm(g, inf) sigma(g) title('sigma Pr1') grid Program result: H2 = Hinf = MaxSig = frek = Plot the singular value responses 20 Sigma Pr Frequency (rad/sec) Fig. 2-14: Singular value of F(iω) 2.6 PID CONTROLLER TUNING The proportional-integral-derivative PID controller has three adjustable tuning constants that enable the engineer, through judicious selection of their values, to adjust the algorithm to a wide range of process applications. Many methods can be used to determine the tuning constant values. 88

89 Feedback control The tuning constants must be derived using the same algorithm that is applied in the control system Procedure for ON-LINE Tuning Hand tuning We suppose that: 1) there is a real closed loop system with a PID controller or a model closed-loop system, 2) the model must be able to model the maximal movement constraints in the form of saturation limits (see Fig. 2-15). Tuning procedure Fig. 2-15: Closed-loop feedback model with saturation of the actuator 1) Set gains P, I, D to zero. 2) Gradually increase the proportional gain P until the transient response of closedloop system is fast enough (speeds up the dynamic response). 89

90 Fig. 2-16: Effect of changing the proportional coefficient on closed loop with PID 3) Choose the appropriate proportional gain and begin to increase gradually the integral gain. The focus must be the steady error and its behaviour. Fig. 2-17: Effect of changing the integration coefficient on closed loop with PID 4) Finally increase the derivative gain to damp oscillations of the closed-loop system. 90

91 Feedback control Fig. 2-18: Effect of changing the derivative coefficient on closed loop with PID Ziegler-Nichols Procedure for ON-LINE Tuning This method was designed by J.G.Ziegler and N.B.Nichols in 1942 and is widely known. We suppose that: 1) there is a real closed loop system with a PID controller or a model of closed-loop system. The structure of the PID controller is chosen in the form R(s) = K R [1 + 1 T i s + T Ds] (2-48) where K R is the controller proportional gain, T I is the integral time constant, T D is the derivative time constant. 2) the model must be able to model the maximal movement constraints in the form of saturation limit (see Fig. 2-15). Tuning procedure 1) Set the gain K R to low value, T I should be set to maximum (infinitive) and T D is switched OFF (zero). Now you can start to operate the closed loop. The controller is in automatic mode. 2) Gradually increase the proportional gain K R until the transient response of closedloop system oscillates continuously at a constant amplitude. 3) The gain that causes the sustained oscillation is called ultimate gain K RU and the stability boundary of the closed loop has been reached. The period of this oscillation is called the ultimate period T RU (see Fig. 2-19). 91

92 Fig. 2-19: Ziegler-Nichols parameters setting 4) Parameter settings for the Ziegler-Nichols method are in the Tab Tab. 2-4: Parameters settings for Ziegler-Nichols method Controller type K R T I T D P 0.5 K RU - - PI PID 0.45 K RU 0.65 K RU T RU 1.2 T RU Design Rules for PID Controller by a known unit response Ziegler-Nichols Controller Parameters Settings This empirical method was developed by Ziegler-Nichols and was published in The transfer function of the controller is supposed in the form as in Eq. (2-48). The relation for tuning P, PI, PID controller is based on the data obtained by the process. It is supposed that the unit step response of the plant can be approximated by a response of a first order system with a time delay F(s) = - T RU 8 K Ts + 1 e T ds (2-49) where K is the gain of the unit step response, T is time constant and T d is time delay. Parameter settings of PID controller can be calculated according to the formulas given in the Tab

93 Feedback control Fig. 2-20: Unit step response approximated by 1 st order response with time delay Tab. 2-5: Ziegler-Nichols controller parameters for known input response tuning method Controller type K R T I T D P PI PID T KT d T 10 T d - KT d T 2 T KT d 0.5 T d d Cohen-Coon parameters The method was developed by Cohen and Coon and is based on a first order +time delay process model as in Eq. (2-49). A set of tuning parameters were empirically developed to yield a closed-loop response with a decay ratio ¼. The tuning parameters are shown in Tab

94 Tab. 2-6: Cohen Coon tuning parameters Controller type K R T I T D P T K T d (1 + T d 3 T ) - - I T K T d (0.9 + T d 12 T ) T d ( T d T 3 T d T ) - D T ( 4 K T d 3 + T 6 T d 4 T ) T d (32 + d T ) T d T T d T 2.7 PID CONTROLLER TUNING FOR DYNAMIC PERFORMANCE By quantifying time domain performance we apply integral performances. We obtain a cost function. The tuning parameter of the PID controller are based on the optimization of a chosen cost function Integral error measures cost functions There exist several standard forms of coast functions. The usual rule is the operand is in quadrate or in absolute value to reach only positive values of operand and the operand includes signals from the closed loop such as error signal, manipulated variable and/or their derivatives, sometimes in multiplication with time. Among criterions belongs: The integral of the Absolute value of the Error (IAE) J(P, I, D) = e(t) e( ) dt = e (t) dt 0 where e(t) is the error signal and e( ) = lim t e(t) is the steady state error. The Integral of the product of Time and the Absolute Error value (ITAE) The Integral of square error (ISE) J(P, I, D) = t e(t) e( ) dt = t e (t) dt The Generalized integral square error 0 J(P, I, D) = [e(t) e( )] 2 dt = e (t) 2 dt (2-50) (2-51) (2-52) 94

95 Feedback control J(P, I, D) = {[e(t) e( )] 2 + κ[u(t) u( )] 2 }dt 0 = (e (t) 2 + κu (t) 2 )dt 0 J(P, I, D) = {[e(t) e( )] 2 + κ [ de(t) 2 dt ] } dt 0 = (e (t) 2 + κ [ de(t) 2 dt ] ) dt 0 J(P, I, D) = {[e(t) e( )] 2 + κ [ du(t) 2 dt ] } dt 0 = (e (t) 2 + κ [ du(t) 2 dt ] ) dt 0 (2-53) (2-54) (2-55) where u(t) is the manipulated variable, u( ) = lim t u(t) is the steady state of manipulated variable and κ is the coefficient, which causes damping. The bigger is κ the greater is the damping effect! Model Controller tuning based on numerical optimization Suppose that the structure of a control system is known as modelled in Fig (left). The dynamic of the manipulated variable u(t) is approximated by the transfer function F U (s) and the dynamic of the not measured disturbance d(t) is approximated by the transfer function F d (s). Criterion-cost function Suppose that parameter tuning can achieve a minimization of one of the cost functions Eq. (2-50) to Eq. (2-55). Minimization J(P, I, D) min (2-56) Scalar cost functions J(P, I, D) which includes three control parameters P, I, D is given or selected. The task is to find the minimum of a chosen cost function specified by the transfer functions F d (s), F U (s), set points w(t), supposed disturbance d(t) and chosen simulation time T SIM. 95

96 Fig. 2-21: Model of closed loop system with the criterion Eq. (2-53) Numerical optimization in Matlab The criterion in Eq. (2-53) is used to show the basic program structure in Matlab. The optimization structure description: The kernel of the optimization is the Matlab program 'PIDopt.m' that specifies: 1) Polynomials of the transfer functions F U (s) = B(s) and F A(s) d(s) = C(s). A(s) 2) Sample period Ts and simulation time Tsim. 3) Initial parameters P0, I0, D0. 4) Weight coefficient Kappa. 5) Disturbances du, d and set point w. The Matlab program is performing the minimization. The criterion is calculated in the Simulink program 'PIDkr1.m' (see Fig on the right side) and the Simulink program is called from the Matlab program by the minimization function 'x = fminsearch('fpidkr1',x,options);'. The optimization function 'fminsearch' evaluated the function that is to be minimized by calling the function 'fpidkr1.m'. This function uses the command 'sim('pidkr1',tsim)' which executes the Simulink model 'PIDkr1.mdl'. After the optimization the simulation program 'PIDsim.mdl' is called (the same structure as 'PIDkr1.mdl'). The logical structure of the optimization is shown in the Fig on the left side. %PIDopt1 integral{e^2 + kappa * [u-u(nek)^2)]} clear all; close all; clc; global P I D Tsim 96

97 Feedback control A = [ ]; B = [ ]; C = [ ]; Ts = 0.05; %Sample period Tsim = 10; %Simulation time dt = 0.01; %Simulation step P0 = 1; I0 = 0.5; D0 = 0.5; N = 20; %Setting of initial parameter N = length(a); Kappa = 1; %Weight coefficient w = 1; du = 0; d = 0; US = -A(n)/B(n)*w; Kw = A(n)/B(n); %Gain coefficient disp('optimalization OF PID PARAMETERS') disp('criterion: J = integral {e^2 + kappa * [u-u(inf)^2)},'),kappa P = P0; I = I0; D = D0; sim('pidkr1', Tsim); disp('criterion initial value:'); krit1 x = [P I D]; PIDpoc = x X = [P I D]; OPTIONS = optimset('tolfun', 1e-10, 'MaxFunEvals', 100); x = fminsearch('fpidkr1', x, OPTIONS); disp('optimalized parameters of PID-Controller:') PID = x sim('pidkr1', Tsim); disp('criterion for optimal parameters:');krit1 sim('pidsim',tsim); PIDsim; %function f=fpidkr1(x) function f = fpidkr1(x) global P I D Tsim P = x(1); I = x(2); D = x(3); sim('pidkr1',tsim); f = krit1; Fig. 2-22: Graphical results of the optimization program Questions Basic properties of feedback 1) What is extended controlled plant, how is its model? Write transfer function of closedloop system set point transfer function, closed loop disturbance. 2) Draw and explain the scheme of a controlled plant with measuring devices. Give advantages of feedback control! 97

98 3) Describe P and PI controller, integral and proportional mode! 4) Describe ideal and real PID controller, derivative mode, its transfer function and unit step response! 5) What is control performance and goals? What are time domain specifications? 6) What is steady error and what type of system do you know? 7) Describe the procedure of hand tuning! 8) What is Cohen Coon tuning? 9) What is the sensitivity function? 10) Write the sensitivity of the closed-loop transfer function! Explain its implementation! 11) Explain the significance of the maximum peak of the sensitivity function! 12) Write the response of the closed-loop system using sensitivity function! What type of sensitivity function do you know? 13) Formulate a design criterion using the magnitude of S! 14) Explain integral performance and its use by tuning? 15) Controller tuning as an optimization problem! 2.8 THE ROOT- LOCUS DESIGN METHOD A program packet in Matlab called SISO Design Tool is used for the Root-Locus Design Method. The tool is a graphical-user interface (GUI) that allows you to design compensators. The SISO Design Tool is made up of the following: 1) The SISO Design Task Node in the Control and Estimation Tools Manager, a user interface (UI) that facilitates the design of compensators for single-input, singleoutput feedback loops through a series of interactive panes. 2) The Graphical Tuning Window, a graphical user interfaces (GUI) for displaying and manipulating the Bode, root locus, and Nichols plots for the controller currently being designed. This window is titled SISO Design for Design Name. 3) The Graphical Tuning Window by default displays the root locus and Bode diagrams for your imported systems. The two are dynamically linked; for example, if you change the gain in the root locus, it immediately affects the Bode diagrams as well. 4) The SISO Design Task-associated LTI Viewer (For instructions on how to operate the LTI Viewer, see LTI Viewer). 5) A tool that automatically generates compensators using PID, internal model control (IMC), or linear-quadratic-gaussian (LQG) methods. 6) A response optimization tool that automatically tunes the system to satisfy design requirements (available if you have Simulink Response Optimization installed). Controller tuning 1) In the synthesis, the pole and zeros of a transfer function R(s) of the controller define the structure of the controller. R(s) = U(s) E(s) = P + I s + Ds = Ds2 + Ps + I = D (s s BR1)(s s BR2 ) s s = Gain (s s BR1)(s s BR2 ) s 2) To create a PID controller, a second zero point must be connected. (2-57) 98

99 Feedback control 3) If it is necessary to control an oscillating plant, both the zero points of the controller should be situated in the vicinity of the complex poles of the transfer function of the plant. The synthesis is demonstrated in the example. Example 2-3 The transfer function of the plant is F U (s) = satisfy the condition a) I controller %P < 20%, b) PI controller %P < 20% and T s < 6 s. Solution of a) 1 s 2 +2s+1. Try to find controller which For required occasional overshoot %P, the = 0.6 is chosen. Transfer function of controller R o (s) = Gain 1 s, that is the controller pole lies at s R = 0. Where the curve of the geometric locus of roots intersects the straight line of the damping, the value Gain creates the required occasional overshoot = 0.6. The desired overall gain of the controller (see Fig. 2-23) is Gain = Solution is presented on the Fig Solution of b) Fig. 2-23: Root loci of the closed loop (left), step response to w(t) = 1 (right) For the PI controller, the oscillation time is T s < 6 s. PI controller has form R o (s) = r 0 s+r 1 = Gain s s BR. Zero point s s s BR of the transfer function R 0 (s) has been set to s BR = The desired overall gain of the controller Gain lies in the point of intersection of the straight line for = 0.6 and of the straight line for T s = 6 s with the curve of the geometric locus of the roots, see Fig The amplification is Gain = The PI controller has been adjusted as follows R(s) = 1.17 s+0.73 s = s. 99

100 Fig. 2-24: Curve of geometric locus of roots of PI controller (left), step response to w(t) = 1 (right) Questions The Root Locus design 1) How can you use the root locus method for design? 2) Describe the transfer function of controller by root locus design! 2.9 THE ACKERMANN FREQUENCY DESIGN METHOD OF PID CONTROLLERS Introduction to loop shaping By loop shaping we mean a design procedure that involves explicitly shaping the magnitude of the loop transfer function L(iω) = R(iω) F(iω). Consider the most commonly used control system configuration with the controller R(s) placed in series with the controlled process as shown in Fig The controller task is to improve stability and control performance of the controlled system. This means, the design of the controller must change-shape the open loop transfer function so, that the closed-loop transfer function will satisfy the given specifications and conditions. Fig. 2-25: Block diagram of a closed-loop feedback control system The open-loop transfer function (loop transfer function) is L(s) = R(s) F(s). To create this we use the properties of Bode magnitude and phase plots. Remember that L(iω) db = R(iω) db + F(iω) db φ 0 (ω) = arg{l(iω)} = arg{r(iω)} + arg{f(iω)} (2-58) 100

101 F [db] Feedback control You see the following results: 1) Bode plots of the plant F(s) and the controller R(s) can be draw separately. 2) The resulting Bode plot of the open-loop transfer function L(s) = R(s)F(s) is given by the graphical addition of the Bode plots of the plant and the controller. 3) The controller-compensator changes the undesired shape of the Bode plot. 4) The structure-transfer function of the controller must be created so, that the stability and given specifications of the closed-loop are satisfied Bode plots of PD controller For the frequency design method according to Ackermann, it is useful to draw the asymptotic approximation of the controllers. The used form of the transfer function of the PD compensator is R(s) = r 0 + r 2 s = r 0 ( r 2 r 0 s + 1) = K R (T D s + 1) (2-59) where K R = r 0, T D = r 2 r 0 r 0 = K R, r 2 = T D r 0. R(iω) db = 20 log K R + 20 log it D ω + 1 (2-60) where ω DZ = r 0 is the break frequency (because it is the break between low-and high r 2 frequency asymptotes. For ω 0 is R(iω) db = 20 log K R = 20 log r 0, so the low-frequency asymptote is equal 20 log r 0. The high-frequency asymptote has the gradient (slope) of +20 db/dec see Fig The phase φ(ω) = arg{r(iω)} can be written in the form φ(ω) = arg{k R } + arg{it D ω + 1} = arctan(ωt D ) = arctan (ω r 2 r 0 ) (2-61) [ ] log [rad/sec] Fig. 2-26: Bode plot of the PD controller 101

102 At the break frequency ω DZ the equation (2-61) shows the phase to be 45. At the low frequencies equation (2-61) shows that the phase is 0. At high-frequencies equation (2-61) shows that the phase is 90. It is possible to draw the asymptotes of the Bode of a PD controller as shown in Fig using: a) Break frequency ω DZ = r 0 ; r 2 b) Proportional gain r 0 in decibels 20 log r 0 ; c) The asymptotes of the phase plot create three primes. The low-frequency asymptote for ω ω 1 = ω DZ creates the zero axis. 5 The high-frequency asymptote for ω ω 2 = 5ω DZ creates the prime φ = π. 2 To draw the third line, start at ω 1 = ω DZ below the break point frequency 5 [ω 1 = ω DZ ; φ 5 1 = 0] with 0 phase, and draw a line with gradient (slope) +45 /dec passing through +45 at the break frequency and continuing to 90 one decade above the break frequency [ω 2 = 5ω DZ ; φ infl = 90 ]. The line segment goes through the point [ω DZ ; φ DZ = 45 ] Bode plots of PI controller PI compensator asymptotic approximation works with the following formula of PI controller R(s) = r 0 + r ( r 0 1 s = r r1 s + 1) 1 = K s R (1 + 1 T I s ) = K R T Is + 1 T I s (2-62) where K R = r 0 ; r 1 = K R T I is overall gain and T I is integral time. The frequency-response magnitude of R(iω) db in decibel is R(iω) db = 20 log K R T I + 20 log it I ω log iω and the break-frequency is placed on ω IZ = 1 T I = r 1 r 0. = 20 log K R T I + 20 log (T I ω) log ω Frequency asymptotes are derive by the following way. For ω 0 the magnitude is (2-63) lim R(iω) db = 20 log K R 20 lim log ω, ω 0 T I ω 0 (2-64) so the low frequency asymptote has the gradient (slope) 20 db/dec. For ω is the magnitude 102

103 F [db] Feedback control lim ω R(iω) db = 20 log K R T I + 20 log T I + lim ω 0 log ω T I 20 lim ω 0 log ω = 20 log K R 20 log T I + 20 log T I = 20 log K R, so the high-frequency asymptote is constant line on the level of 20 log K R. The phase φ(ω) = arg{r(iω)} can be written in the form φ(ω) = arg { K R T I } + arg{it I ω + 1} arg{iω} = 0 + arctan(ωt I ) π 2 = arctan(ω T I ) π 2, K R T I > 0 (2-65) (2-66) It is possible to draw the asymptotes of the Bode of PI controller as shown in Fig using: a) Break frequency ω IZ = r 1. r 0 b) Gain r 0 in decibels 20 log r 0. c) The asymptotes of the phase plot create three primes. The low-frequency asymptote for ω ω 1 = ω IZ creates the prime φ = π. 5 2 The high-frequency asymptote for ω ω 2 = 5ω IZ creates the zero axis. To draw the third line, start at ω 1 = ω IZ below the break point frequency [ω 5 1 = ω DZ ; φ 5 1 = π ] with 90 phase, and draw a line with gradient (slope) +45 /dec 2 passing through 45 at the break frequency and continuing to 0 one decade above the break frequency [ω 2 = 5ω IZ ; φ infl = 0 ]. The line segment goes through the point [ω IZ ; φ IZ = 45 ] [ ] log [rad/sec] Fig. 2-27: Bode plot of the PI controller 103

104 2.9.4 Bode plots of PID controller PID compensator asymptotic approximation works with the following formula of PID controller, where there are assumed real zeros in numerator of the fraction. R(s) = r 0 + r ( r 2 1 s + r r1 s 2 + r 0 r1 s + 1) 2s = r 1 s ( K R (2-67) TI s + 1) = T I (T s D s + 1) The relations between r 0, r 1, r 2 and K R, T I, T D are determined by comparing the power of s in the equation (2-67). R(s) = T I ( K R TI s + 1) s (T D s + 1) = T I ( K RT D T I s 2 + ( K R TI + T D ) s + 1) ( r 2 r1 s 2 + r 0 r1 s + 1) = r 1 s By comparing coefficients yields r 0 = K R + T D T I, r 1 = T I, r 2 = K R T D. The frequency response magnitude of R(iω) db in decibel is s R(iω) db = 20 log T I + 20 log i K R T I ω log iω + 20 log it D ω + 1 = 20 log T I + 20 log ( K 2 R ω) log ω T I + 20 log (T D ω) The phase φ(ω) = arg{r(iω)} can be expressed in the form (2-68) φ(ω) = arg{t I } + arg {i K R T I ω + 1} arg{iω} + arg{it D ω + 1} = arg{t I } + arctan (ω K R T I ) π 2 + arctan(ωt D) = arctan (ω K R T I ) π 2 + arctan(ωt D), T I > 0 The break frequencies are ω IZ = T I K R and ω DZ = 1 T D. The asymptotes of Bode of PID controller can be drawn as shown in Fig using: a) Break frequencies ω IZ = T I K R and ω DZ = 1 T D. b) Gain of K R in decibels 20 log K R. (2-69) c) The asymptotes of the phase plot create complicated shape in dependence of the distance between break frequencies and are combination of PI and PD controller. 104

105 F [db] Feedback control [ ] log [rad/sec] Fig. 2-28: Bode plot of the PID controller P compensator compensation procedure Classical compensation refers to shaping the loop transfer function with the aim of fulfilling the stability conditions using easy rules. The design procedure: 1) The Bode plot of the transfer function of the uncompensated system is produced. 2) The phase and gain margin of the uncompensated system are determined from the Bode plot. 3) The transfer function of the P controller is written R(s) = r 0. 4) For the specified (desired) phase-margin requirement Φ P, the new gain-crossover frequency ω P corresponding to this phase margin is found on the Bode plot. The magnitude plot of the compensated transfer function must pass the 0 db axis at this new gain-crossover frequency in order to get the desired phase margin. 5) To bring the magnitude curve of the uncompensated transfer function down to 0 db at the new gain-crossover frequency ω P, the P controller must provide the amount of attenuation equal to the gain of the magnitude curve F(iω P ) db at new gain crossover frequency. F(iω P ) db = 20 log r 0 r 0 = 10 F(iω P ) db 20 (2-70) Example 2-4 Consider the transfer function F(s) = 10 (2s+1) 3. Stabilize and improve control performance using the P controller. Solution. 105

106 1) The Bode asymptotes plot of the uncompensated system are plotted. 2) The uncompensated system is unstable. 3) P controller. 4) We set the desired phase margin Φ P = 30. The new gain crossover frequency ω P corresponding to this phase margin is found on the Bode plot and is ω P = 0.6 rad, and the magnitude attenuation F(iω P ) db = F(i0.6) db = 8.6 db. 5) The P controller must provide the amount of attenuation F(iω P ) db = 20 log r 0 r 0 = = Fig. 2-29: P compensation PD Compensator design procedure 1. The Bode plot of the transfer function of the uncompensated system is made. 2. The phase and gain margin of uncompensated system are determined from the Bode plot. 3. The PD controller transfer function is in form R(s) = r 0 + r 2 s = r 0 ( r 2 r 0 s + 1). 4. The design principle of the PD controller involves the placing of the corner frequency of the controller ω DZ = r 0 r 2, such that an effective improvement of the phase margin is realized at the new gain crossover frequency. 106

107 Feedback control Example 2-5 Consider the transfer function F(s) = 10 (2s+1) 3. Stabilize and improve control performance using a PD controller. Solution 1) The Bode asymptotes plot of the uncompensated system are plotted. 2) You can see the uncompensated system is unstable. 3) PD controller. 4) Choose phase margin, select break frequency, compute PD controller parameters. a) Var.1: We have set ω DZ = r 0 r 2 = 0.5 rad and r 0 = 1, r 2 = 2. We draw the compensated system and we see, the system is stable and the phase margin is Φ 22. b) Var.2: We can improve the stability by setting the desired phase margin to Φ P = 45. The new gain crossover frequency ω P corresponding to this phase margin is found on the Bode plot is ω P = 0.5 rad, and the magnitude attenuation is F C (iω P ) db = F C (0.5i) db = 11 db. The P part of the PD controller provides the amount of attenuation F(iω P ) db = 20 log r 0 r 0 = = For the new gain r 0 = the gain r 2 is ω DZ = r 0 = 0.5 rad r r 2 = r 0 = = ω DZ 0.5 Fig. 2-30: PD compensation 107

108 2.9.7 PI compensation design procedure 1. The Bode plot of the transfer function of the uncompensated system is produced. 2. The phase and gain margin of the uncompensated system are determined from the Bode plot. 3. The transfer function of the PI controller is written as R(s) = r 0 + r 1 r0 r1 s+1 = (r 0s+r 1 ) s s r 1. s 4. For the specified phase margin requirement Φ P, the new gain crossover frequency ω P corresponding to this phase margin is found on the Bode plot. The magnitude plot of the compensated transfer function must pass the 0 db axis at this new gain crossover frequency in order to get the desired phase margin. 5. To bring the magnitude curve of the uncompensated transfer function down to 0 db at the new gain crossover frequency ω P, the PI controller must provide the amount of attenuation equal to the gain of the magnitude curve at new gain crossover frequency. 6. We have assumed that although the gain crossover frequency is altered by attenuating the magnitude of the controller, the original phase is not affected by the PI controller. To ensure the desired phase margin (not be affected by the phase of the PI controller), it is necessary to move the crossover frequency ω ZI of the controller to left. That we set ω ZI = ω P 10 = r 1 r 0 r 1 = r 0 ω ZI = r 0 ω P 10 (2-71) Example 2-6 Consider the transfer function F(s) = 10 (2s+1) 3. Stabilize and improve control performance using the PI controller. Solution. 1) The Bode asymptotes plot of the uncompensated system are plotted. 2) The uncompensated system is unstable. 3) PI Controller. 4) We can improve the stability by setting the desired phase margin to Φ P = 20. The new gain crossover frequency ω P corresponding to this phase margin is found on the Bode plot is ω P = 0.67 rad, and the magnitude of attenuation is F(iω P ) db = F(0.67i) db = 6.48 db. 5) The P part of the PI controller provides the amount of attenuation F(iω P ) db = 20 log r 0 r 0 = = ) We set ω ZI = ω P = r 1 r 10 r 1 = r 0 ω ZI = r 0 ω P = = = 108

109 Feedback control Fig. 2-31: PI compensation PID compensation design procedure 1. Consider a PID controller consisting of a PI portion connected in cascade with a PD portion. Its transfer function is written as R(s) = r 0 + r 1 s + r 2s = (1 + T D s) (K R + T I s ). 2. Draw the Bode plot of the plant. 3. Suppose the PD part can be tuned. Select the frequency ω D so that the stability will increase. Calculate the value T D from the equation T D = 1/ω D. 4. Calculation specification of the PI controller. a. Select the desired phase margin Φ P and read from the Bode plot the new desired crossover frequency ω P. From the Bode plot we can now read or calculate F(iω P ) db. This is the part of the magnitude which needs to be compensated by the gain K R of the PID controller. b. The gain K R can be calculate from the equation (2-70). c. The controller will be cautious if the crossover frequency of the PI part ω ZI is shifted to left. The frequency can be calculated from equation (2-71). Example 2-7 Consider the transfer function F(s) = 10 (2s+1) 3. Stabilize and improve control performance using a PID controller. Solution. 1) The Bode asymptotes plot of the uncompensated system are plotted. 2) The uncompensated system is unstable. 109

110 3) Set the frequency to ω D = 0.9 rad. Calculate the value T D from the equation T D = = ) Calculation of the PI part a) The desired phase margin Φ p = 20. The new desired cross frequency is ω P = 0.7. The part of the magnitude that needs to be compensated by the gain K R of the PID controller is F(iω P ) db = F(i0.7) db = 6.75 db. b) The gain K R can be calculate from the equation K R = 10 F(iω P ) db 20 = = c) The gain T I can be calculated as T I = K R ω ZI = K R ω P 10 = = The parameters of PID controller are r 1 = T I = 0.032, r 2 = T D K R = 0.51, r 0 = K R + T D T I = Fig. 2-32: PID compensation Questions The frequency response method 1) Why is a Bode plot used in the design of a controller design? 2) Draw the Bode plot asymptotic approximation of PD controller! 3) Draw the Bode plot asymptotic approximation of PI controller! 4) Draw the Bode plot asymptotic approximation of PID controller! 5) Describe P and PD compensation! 6) Describe the PI compensation! 7) Describe PID compensation! 110

111 Enhancements to single-loop feedback control 3 ENHANCEMENTS TO SINGLE-LOOP FEEDBACK CONTROL According to our experience and knowledge, a single loop feedback control with a PID controller gives good performance and is easy to use. On the other hand, with special tasks such as disturbance rejection, the results are not always optimal. To improve the feedback performance we must take advantage of additional knowledge about the process dynamic and control objectives. As additional information about the process we can use: 1. Additional measurements of process input. 2. Additional measurements of process output. 3. Explicit modelling in the control calculation. 4. Modification of the PID algorithm and tuning to match the control objective. 3.1 FEED FORWARD CONTROL The primary disadvantage to feedback only- control is that a disturbance must pass through the controlled system. After that the influence of the disturbance can be measured on the output variable and only then can the controller output changed. However enhancements can be achieved by additional measuring of the disturbance (see Fig ) of the process. Therefore it is preferable to have a sensor that measures the disturbance and adjust the manipulated input before the process output changes. Consider a block diagram of the control system in Fig It is a feed forward/feedback system. The output from the sensor S 1, which measures the disturbance d m (t), is the input to the forward controller R m (s). S 1 Fig. 3-1: Block diagram of the feed-forward and feedback control, the measured disturbance is d m 111

112 The transfer function F y dm is equal F ydm = F m + R m F U 1 + RF U (3-1) We can see that the only way to ensure that y(t) = 0 is to satisfy the following equation F m (s) + R m (s)f U (s) = 0 (3-2) Solving equation (3-2), we find the transfer function of the forward controller R m (s) = F m(s) C(s) A(s) = F U (s) D(s) B(s) (3-3) Notice that the equation (3-3) requires the inverse of the process model. This controller might be physically not-realizable. Therefore in some cases only the static feed-forward can be used R mstat (s) = lim s 0 R m (s) = c 0 a 0 d 0 b 0 (3-4) Another structure for a feed forward/feedback system is on Fig Fig. 3-2: Block diagram of the feed-forward and feedback control, the measured disturbance is d 2 The transfer function of the forward controller can be written R m F 1 F 2 + F 1 = 0 R m = 1 F 2 (3-5) 3.2 CONTROL SYSTEMS WITH AN AUXILIARY MEASURED AND MANIPULATED VARIABLE Control system with an auxiliary measured variable Control system with an auxiliary measured variable represents an additional measure of process output. Consider a plant with an unmeasured disturbance in the Fig. 3-3a). The core idea of a control system with auxiliary measured variable is to make use of the opportunity as follows: 1. To divide the plant into two parts, Plant 1 ; and Plant 2 (see Fig. 3-3b) so, that the disturbance injection point is placed between Plant 1 and Plant The measured variable lies between the disturbance injection point and the output. 112

113 Enhancements to single-loop feedback control Fig. 3-3: Plant, structure of divided plant, model of divided plant with a disturbance A controlled plant model with a disturbance dynamic approximated by the transfer function F d (s) is shown in Fig. 3-3c). The model of the plant is described by the partial transform function F 1, F 2. A control system with an auxiliary measured variable is depicted in Fig You can see that a set point w P for the auxiliary measured variable y P must be implemented. Fig. 3-4: Structure of a feedback control with auxiliary measured variable y P It is easy to check that this control structure improves the disturbance rejection well, but there is a difference between the steady-state values of the set point w(t) and the controlled variable y(t). This disadvantage can be eliminated by using an additional (auxiliary) P controller (R 2 (s) = r 02 ) as shown in Fig

114 Fig. 3-5: Control with auxiliary measured variable y P and two controllers R 1, R 2 Consider the usual transfer functions of controllers R 1 (s) = r 01 + r 11, R s 2(s) = r 02 and plant transfer functions F 1 (s) = B 1 (s), F A 1 (s) 2(s) = B 2 (s), then the set point transfer is A 2 (s) F YW (s) = R 1 F 1 F 2 (r 01 s + r 11 )B 1 B 2 = 1 + R 1 F 1 F 2 + R 2 F 2 A 1 A 2 s + (r 01 s + r 11 )B 1 B 2 + r 02 B 2 A (3-6) 1 For the unit set point is applied final value theorem and the steady state equals lim y(t) = lim sy W(s) = t s 0 (r 01 s + r 11 )B 1 B 2 = lim s 1 s 0 A 1 A 2 s + (r 01 s + r 11 )B 1 B 2 + r 02 B 2 A 1 s = 1 (3-7) Control system with an auxiliary manipulated variable Consider the controlled process in Fig with unmeasured disturbances d U (t) and d(t). The condition for applying the control system with an auxiliary manipulated variable is that the controlled system can be divided into two parts and an auxiliary manipulated variable u P can be found using an auxiliary manipulated injection point as shown in Fig. 3-6, Fig Fig. 3-6: Auxiliary manipulated variable control scheme 1 st option 114

115 Enhancements to single-loop feedback control Fig. 3-7: Auxiliary manipulated variable control scheme 2 nd option The two parts are described by two transfer functions F 1 (s) and F 2 (s). Disturbances are counteracted by two manipulated variables u and an auxiliary manipulated variable u P. The auxiliary manipulated variable u P is generated by an additional auxiliary controller R P. The tuning strategy of R P is to make the auxiliary closed loop as fast as possible due to the rapid cancelling of disturbances. In the steady state the u P (t) must be zero and the P or PD controller can be used. 3.3 THE PRINCIPLE AND STRATEGY OF CASCADE CONTROL Generally we can say that a cascade control uses multiple measurements and a single manipulated input. The condition for applying cascade control is once again that the controlled system can be divided into two sections (see Fig. 3-8). The controlled variable y(t) = y 1 (t) is measured as an auxiliary controlled variable y 2 (t) behind the section 2 (R 2 F 2 ), for which the dynamic is approximated by the transfer function F 2 (s). As shown in Fig. 3-8 the main controller R 1 does not directly affects the actuator, but provides the reference value (set point) for the underlying auxiliary controller R 2. The auxiliary controller together with the second plant section F 2 form the auxiliary control loop, which is inside the main control loop. Fig. 3-8: Block diagram of cascade control Disturbance d 2 (t) in the second plant section F 2 will be already controlled by the auxiliary controller R 2. The disturbance then has less influence on the first section (F 1 ) and the main controller R 1 then has only to act slightly for disturbance d 2 (t) compensation. The task of the main controller is to secure the set point w(t) and compensate the disturbance d 1 (t) via the feedback. The controlled variable Y(s) is given by Y(s) = F Y W (s) W(s) + F Y D1 (s) D 1 (s) + F Y D2 (s) D 2 (s) (3-8) 115

116 where F Y W (s) = F Y D2 (s) = R 1 (s)r 2 (s)f 1 (s)f 2 (s), F 1+R 1 (s)r 2 (s)f 1 (s)f 2 (s)+r 2 (s)f 2 (s) Y D 1 (s) = 1+R 2 (s)f 2 (s) 1+R 1 (s)r 2 (s)f 1 (s)f 2 (s)+r 2 (s)f 2 (s). R 2 (s)f 2 (s) If we denote F A (s) = and F 1+R 2 (s)f 2 (s) AD(s) = 3-8 can be redrawn see Fig R 2 (s)f 2 (s) F 1 (s) 1+R 1 (s)r 2 (s)f 1 (s)f 2 (s)+r 2 (s)f 2 (s), then the block diagram in Fig. Design of a cascade control system Fig. 3-9: Rearranged block diagram of cascade control 1) First starts the parameters setting of the auxiliary control system, tuning the R 2 controller for a given section F 2 of the plant for disturbances. The auxiliary controller system must be fast (high bandwidth) and therefore a P or PD controller is to choose for R 2. 2) After the parameters tuning of the main controller R 1 is performed for the plant transfer function F 2 F 1. The controller R 1 has the task of removing steady-state error in the controlled variable. Therefore, it is expedient to use a PI controller as the plant has real poles. Fig and Fig show two examples of industrial cascade control systems. Fig. 3-10: Cascade control in heating water production 116

117 Enhancements to single-loop feedback control Requirements for cascade control. Fig. 3-11: Cascade control of superheated steam The secondary loop process dynamics must be at least four times as fast as primary loop process dynamic. The secondary loop must have influence over the primary process. The secondary loop must be measured and controllable. Reasons not to use cascade the cost of measuring the secondary variable, the additional costs of setting the auxiliary input. 3.4 INTERNAL MODEL CONTROL Dynamic control law inverse response dynamic As an example of explicit modelling in the control calculation we introduce the inverse response dynamic. Consider the model of a feed forward SISO control system shown in Fig Fig. 3-12: Model of a feed forward control system First, we suppose that the unmeasured disturbance d(t) = 0. Then the Laplace output is equal Y(s) = F(s) R W (s) W(s) (3-9) where are F(s) = B(s) A(s), R W(s) is the transfer function of a feed forward controller, W(s) is the Laplace transform of set point/desired value. We find that the transfer function of the feed forward controller is F(s) R(s) = 1 R ideal (s) = 1 F(s) = A(s) B(s) (3-10) 117

118 This controller is not physical realizable! An appropriate filter must be used. R real (s) = R(s) = A(s) A(s) Λ(s) = B(s) B(s) 1 (λs + 1) n (3-11) where Λ(s) = 1 (λs+1) n is the transfer function of the filter and n A B. Notice that the zeros of the process model become the poles of the controller when the inverse of the model is used for control system design. This creates an unstable controller if the process model has zeros in the right half plane! If this is the case then the zeros must be factored out before using the model inverse to design the controller. The order of the controller is equal to the order of the model system. The design of a controller by using the model inverse yields a controller that has the order of the controller equal to the order of the model system. The order of the filter must be chosen so that the controller is proper. We cannot use standard equipment and algorithms to implement an advanced control concept. The controlled output is Y(s) = F(s) R(s) W(s) = B(s) A(s) A(s) Λ(s) W(s) = Λ(s) W(s) B(s) 1 (3-12) = (λs + 1) n W(s) Example 3-1 Consider the following transfer function F(s) = controller. Solution. 2 (15s+1)(3s+1). Design the open loop According to Eq. (3-11) the transfer function of the controller is R(s) = A(s) Λ(s) = B(s) (15s+1)(3s+1) 1 2 (λs+1) 2. The controlled variable is Y(s) = F(s) R(s) W(s) = 2 (15s+1)(3s+1) 1 W(s) = 1 W(s). (15s+1)(3s+1) 2 (λs+1) 2 (λs+1) Factorization techniques In general, we factor the process model in the following fashion F(s) = F + (s) F (s) (3-13) where F + (s) contains the noninvertible/unstable elements (zeros and poles), F (s) contains the invertible/stable elements (zeros and poles). The following is an explanation of simple and all-pass factorization 1) Simple factorization The simple factorization approach is to simply place the RHP zeros in the noninvertible part of the process model as demonstrated in the following example. 118

119 Enhancements to single-loop feedback control Example 3-2 Consider the following transfer function F(s) = 5 ( 2s+1) (15s+1)(3s+1) unstable element! Find controller response in IMC control structure. Solution.. Find the stable and 5 The factorization results in F(s) = F + (s) F (s) = ( 2s + 1). The (15s+1)(3s+1) 5 stable element is F (s) =, the unstable element is F (15s+1)(3s+1) +(s) = ( 2s + 1). The controller response is Y(s) = F(s) R(s) W(s) = (F + (s) F (s)) ( 1 = F + (s) Λ(s) W(s) = 2s+1 (λs+1) 2 W(s). 2) All-pass factorization F (s) Λ(s)) W(s) = The all-pass factorization approach places the RHP zero in the noninvertible part of the process model, but it also places a pole at the reflection of the RHP zero Generally the unstable roots are denoted as s k = (+ α k ± iω k ). The non-stable polynomial can be expressed as A + = (s s 1 )(s s 2 ) = (s (+ α 1 ± iω 1 ))(s (+ α 2 ± iω 2 )) Factorized polynomial can be written in the form A = A A + (3-14) where A + is a polynomial with stable roots, the polynomial A + = (s ( α 1 ± iω 1 ))(s ( α 2 ± iω 2 )) and it has replaced the non-stable roots by stable roots s k = ( α k ± iω k ). This method replaces the unstable factors with stable ones. The stable factor is moved to the stable part. In the unstable part the unstable factor remains in nominator that is divided by its stable term. The overall transfer function does not changed, but the stable part is minimized in the quadratic function. Example 3-3 Consider the following transfer function F(s) = 5 ( 2s+1) (15s+1)(3s+1) unstable element! Find controller response in IMC control structure. Solution. The factorization results in F(s) = F + (s) F (s) = 2s+1 2s+1 5 (2s+1) element is F (s) =. Find the stable and (15s+1)(3s+1) 5 (2s+1), the unstable element is F (15s+1)(3s+1) +(s) = 2s+1. 2s+1 The controller response is Y(s) = F(s) R(s) W(s) = 2s+1 (2s+1)(λs+1) 2 W(s).. The stable 119

120 It is obvious that some form of feedback is needed to correct for model uncertainty as well as any disturbances entering the process. The method that we develop to account for model uncertainty and disturbances is known as internal model control (IMC) Internal model control structure The internal model control structure is shown in Fig The distinguishing characteristic of this structure is the process model, which is in parallel with the actual process. Fig illustrates that both the controller and model exist as computer computations. It is convenient to treat them separately for design and analysis. Fig. 3-13: The internal model control structure Blocks marks and signals on Fig are: F(s) process d(t) disturbance F M (s) process model u(t) manipulated input (controller output) R W (s) internal model controller w(t) setpoint y(t) measured process output ε(t) modified setpoint The structure of an IMC controller is given in Fig The signal to the controller is Consider some limiting cases 1) Perfect model with no disturbance. Fig. 3-14: IMC controller Δ(s) = (F(s) F M (s)) U(s) + D(s) (3-15) Conditions F(s) = F M (s), D(s) = 0. The relationship between W(s) and Y(s) is Y(s) = F(s) R W (s) W(s). 120

121 Enhancements to single-loop feedback control Result: If the controller R W (s) is stable and the process F(s) is stable, then the closed-loop system is stable. Recall that a standard feedback controller could actually destabilize a process if we did not correctly choose the tuning parameters. 2) Perfect model with a disturbance effect. Conditions F(s) = F M (s), D(s) 0 and the feedback signal is Δ(s) = D(s). 3) Model uncertainty Conditions F(s) F M (s), D(s) = 0. The feedback signal is Δ(s) = (F(s) F M (s)) U(s). 4) Model uncertainty and disturbance effect. Conditions F(s) F M (s), D(s) 0. Y(s) = F(s)R W (s) W(s) + 1 F M (s)r W (s) D(s). 1+R W (s)(f(s) F M (s)) 1+R W (s)(f(s) F M (s)) To recap. The reasons for feedback control include the following: Unmeasured disturbances Model uncertainty Faster response than the open-loop system Closed-loop stability of an open-loop unstable system Questions Enhancements to single loop PID feedback control 1) What is feed forward control? What is the problem of physical realization? 2) Explain the design using the inverse response dynamic, what is the filter? 3) Why is the factorization technique used? 4) Explain the internal model control structure! 5) Describe control systems with an auxiliary manipulated variable! 6) Describe control systems with an auxiliary controlled variable! 7) What is cascade control, what is its structure and how is it tuned? 8) Describe an example of industrial cascade control! 121

122 4 INTRODUCTION TO STATE SPACE CONTROL 4.1 STATE-SPACE REPRESENTATION OF SISO SYSTEMS Control (reachable) canonical form Let us consider a system with the transfer function F(s) = B(s) A(s) = b m s m + b 1 s + b 0 s n + a n 1 s n a 1 s + a 0 = Y(s) U(s) (4-1) In order to develop a state description of this system, we construct a block diagram that corresponds to the transfer function and the differential equation using only isolated integrators as the dynamic elements. a) Introducing an auxiliary (help) variable x(t) that has no derivative on input. The Laplace transform of output Y(s) can be expressed from the equation A(s)Y(s) = B(s)U(s) Y(s) = B(s) U(s) A(s) = B(s)X(s) (4-2) where X(s) is the auxiliary (help) variable. With help variable X(s) we obtain the equation X(s) = U(s) A(s)X(s) = U(s) A(s) (s n + a n 1 s n a 1 s + a 0 )X(s) = U(s) (4-3) Applying an inverse Laplace transform to the equation (4-3) then the differential equation for the auxiliary variable x(t) is x (n) (t) + a n 1 x (n 1) (t) + + a 1 x (1) (t) + a 0 x(t) = u(t) (4-4) b) The output signal y(t) is found as the weighted sum of derivatives of help variable x(t). The Laplace transform of output Y(s) can be obtained from the equation (4-2) Y(s) = B(s)X(s) = (b n 1 s n 1 + b n 2 s n b 1 s + b 0 )X(s) (4-5) The output of the time domain y(t) is equal to the weight sum 122

123 Introduction to state space control y(t) = b n 1 x (n 1) (t) + b n 2 x (n 2) (t) + + b 1 x (1) (t) + b 0 x(t) (4-6) The state variables are taken as the derivative of the auxiliary function x(t) x 1 (t) = x(t) x 2 (t) = x (t) x 3 (t) = x (t) x n (t) = x (n 1) (t) x 1(t) = x 2 (t) x 2(t) = x 3 (t) x 3(t) = x 4 (t) x n(t) = a n 1 x n (t) a 1 x 2 (t) a 0 x 1 (t) + u(t) Substituting the derivative of the help function x(t) in the equation (4-6) for the state variable in equation (4-7), we obtain for the output variable y(t) equation (4-7) y(t) = b n 1 x n (t) + b n 2 x n 1 (t) + + b 1 x 2 (t) + b 0 x 1 (t) (4-8) The structure of the state vector x, the system matrix A, input matrix B and the output matrix C (D = 0) is x 1 (t) x 2 (t) x x(t) = 3 (t) 0 0 0, A = 0 0, B =, x n 1 (t) [ x n (t) ] [ a 0 a 1 a 2 a n 1 ] [ 1] C = [b 0 b 1 b 2 b n 1 ] (4-9) Fig. 4-1: Block diagram for control canonical form Example 4-1 Transfer function is F(s) = 1.5. Calculate the dynamic model in a control 2s 2 +3s+1 canonical form. Draw a block diagram for Simulink. Solution. 123

124 Transfer function will be modified in the form of F(s) = 0.75 s s+0.5. This clearly shows that a 0 = 0.5, a 1 = 1.5, b 0 = Applying equation (4-7) and (4-9) we obtain x (t) = [ x 1(t) x 2(t) ] = [ ] [x 1(t) x 2 (t) ] + [0 1 ] u(t), y(t) = [ ] [ x 1(t) x 2 (t) ]. The block diagram is in Fig The step response and time responses of x 1 (t), x 2 (t) are depicted in Fig Fig. 4-2: Block diagram (left), state and output trajectories (right) Observable canonical form The transfer function (4-1) can be rewritten as a differential equation y (n) (t) + a n 1 y (n 1) (t) + + a 1 y (1) (t) + a 0 y(t) = b n 1 u (n 1) (t) + + b 1 u (1) (t) + b 0 u(t) After n-times integration of the equation (4-10) we obtain an integral equation t y(t) = a n 1 y(τ)dτ a n 2 y(τ)dτ 2 a 0 y(τ)dτ n 0 0 t + b n 1 u(τ)dτ + + b 0 u(τ)dτ n 0 t 0 t 0 t (4-10) (4-11) The block diagram of the system to solve the equation (4-11) is shown in Fig If we denoted the output from the first integrator as x 1 (t) = y(t), then it is possible to denote the output from the second output as x 2 (t) and so on. From this block diagram it is clear that the state variables are equal to x 1(t) = a n 1 x 1 (t) + x 2 (t) + b n 1 u(t) x 2(t) = a n 2 x 1 (t) + x 3 (t) + b n 2 u(t) x n 1(t) = a 1 x 1 (t) + x n (t) + b 1 u(t) x n(t) = a 0 x n (t) + b 0 u(t) (4-12) 124

125 Introduction to state space control Fig. 4-3: Block diagram for observable canonical form The structure of the state vector x(t), the system matrix A, the input matrix B and the output matrix C (D = 0) for the observable canonical form are x 1 (t) a n x 2 (t) a n x x(t) = 3 (t) a, A = n x n 1 (t) a [ x n (t) ] [ a Similarity transformation Consider the state space equation C = [1 0 0] x (t) = A x(t) + B u(t) y(t) = C x(t) + D u(t) 0, B = 1 0] [ b n 1 b n 2 b n 3, b 1 b 0 ] (4-13) (4-14) Let s assume that the dynamic state system equations of (4-14) are transformed into another set of the same dimension by the following transformation where P is an [n n] nonsingular matrix, so that The transformed dynamic equation are written dx (t) dt x(t) = P x (t) (4-15) x (t) = P 1 x(t) (4-16) = A x (t) + B u(t) y (t) = C x (t) + D u(t) (4-17) where A = P 1 A P and B = P 1 B, C = C P, D = D. The transformation above is called a similarity transformation. 125

126 4.1.4 Complete controllability The concepts of controllability and observability were introduced by Kalman and are important for modern control system theory. The process is said to be completely controllable if every state variable of the process can be controlled to reach a certain objective in finite time by some unconstrained control u(t). You can understand if any of the state variables is independent of the control u(t), there would be no way to drive the particular state variable to a desired state. We can define the condition when the system is controllable. The state x(t) is said to be controllable at t = t 0 if there exists a piecewise continuous input u(t) that will drive the state to any final state x(t f ) for a finite time (t f t 0 ) > 0. The system described by (4-1) and/or (4-14) is controllable if the controllability matrix Co [n n] = [B A B A 2 B A n 1 B] has a rank of n Completely observability Consider a given state variable model for which the matrices A, B, C and D are known and the input u(t) and output y(t) are measured. The question remains is it possible to find the vector x(t)? If any of the states cannot be observed from the measurements of the output, the state is said to be unobservable. In the following frame we will define observability and its conditions. The state x(t 0 ) is said to be observable if for any given input u(t), there exists a finite time t f t 0 such that the knowledge u(t) for t 0 t t f, matrices A, B, C and D; and the output y(t) for t 0 t t f are sufficient to determine x(t f ). The system described by (4-1) and/or (4-14) is observable if the observability matrix C C A Ob = C A 2 has a rank of n. [ C A n 1 ] Kalman form Controllability and observability are closely related to the properties of transfer functions. If the input-output transfer function of a linear system has pole-zero cancellation, the system will be uncontrollable and unobservable. If the input-output transfer function does not have pole-zero cancellation, the system is always controllable and observable. In general the state vector can be split into sub-vectors involving the four possible combinations of controllability and observability. This structure is known as the Kalman form and is composed of blocks as depicted in Fig. 4-4: uncontrollable and unobservable controllable and unobservable (nc & no) (c & no) uncontrollable and observable (nc & o) 126

127 Introduction to state space control controllable and observable (c & o) Minimal realization Fig. 4-4: Block decomposition Minimal realization minimal order of transfer function is a transfer function where practicable poles-zeros cancel each other out. It may not look difficult but you will see some problems during execution. 4.2 MODEL REDUCTION When determining a dynamic control model it is common that the order of the model is higher than required to fulfil the control purpose. In this case order reduction is recommended. The basis for the reduction is to find modes that are unachievable and unobservable that can be excluded. You can see there are some modes with little influence. These modes are cancelled out by reducing the order. This forces the approximating dynamic behaviour to keep the static gain or magnitude in a given range of frequencies. Two common methods of model reduction are used: time scale decomposition and balanced reduction Time scale decomposition If a dynamic system has two different time scales in behaviour, it can be decomposed into two systems (one is slow x 1 and very fast x 2 ) such as x (t) = A x(t) + B u(t) [ x 1(t) x 2(t) ] = [A 11 A 12 ] [ x 1(t) A 22 x 2 (t) ] + [B 1 ] u(t) B 2 A 21 y = C x(t) = [C 1 C 2 ] [ x 1(t) x 2 (t) ] (4-18) If the behaviour of x 1 (t) is slow, then the second state sub-vector is diminishing x 2(t) 0, and we can calculated x 2 (t). The reduced order system will be x 1(t) = [A 11 A 12 ] [ x 2(t) = 0 A 21 x 1 (t) + A 22 x 2 (t) + B 2 u(t) = 0 x 2 (t) = A 1 22 (A 21 x 1 (t) + B 2 u(t)) A 22 x 1 (t) 1 (A 21 x 1 (t) + B 2 u(t)) ] + [B 1] u(t) 0 (4-19) 127

128 After some modification we obtain x 1(t) = A 11 x 1 (t) A 12 A 1 22 (A 21 x 1 (t) + B 2 u(t)) + B 1 u(t) The reduced order system is x 1(t) = (A 11 A 12 A 1 22 A 21 ) x 1 (t) + (B 1 A 12 A 1 22 B 2 ) u(t) (4-20) This approximation is called residualisation Balanced reduction By reducing the model order the input/output relevance of each mode may be directly considered. In the balance reduction, the less achievable and observable modes are candidates to be deleted. Let us explain it in an example. Example 4-2 Given a transfer function in factorized form F(s) = minimal realization. Solution. Factorization: Rewriting to the form F(s) = 1 (s+0.01)(s+0.2)(s+5). Find the 1 = (s+0.01)(s+0.2)(s+5) s+0.01 s+0.2 s+5 5 From the factorization we see that the factor represents a mode that has its s+5 response many times faster as the other modes, so that it s dynamic can be disregarded. 5 Therefore the mode can be approximate by 1. The reduced transfer function after pole-zero cancellation (residualisation) is F R (s) = 0.1 s s+0.2. Partial-fraction expansion: Rewriting to the form F(s) = s+0.01 s+0.2 s+5 s+5 1 (s+0.01)(s+0.2)(s+5) =. From the partial fraction you can see that the gain of the third member is similar to other members, that is, very small. By deleting (truncating) the term 0, the reduced transfer function is F s+5 R(s) = = s+0.01 s (s+0.01)(s+0.2). 4.3 STATE ESTIMATION The state variable cannot usually be directly measured in real systems. But for control, it is necessary to know all the states. The states from measuring the dynamic system and from the state/space model have to be estimated. Here we explain the estimating method for a full-order model of the plant dynamics. Consider a state/space model in the form Eq. (4-14). We can use the state model directly in the first approach for estimating. See Fig

129 Introduction to state space control Fig. 4-5: Open-loop estimator The estimator can be described by these equations dx (t) dt = A E x (t) + B E u(t) y (t) = C x (t) (4-21) where A E, B E are matrices of estimator, y (t) is the estimated output and x (t) is the state vector. To study the dynamic of the estimator, we define the error of the estimator Δx(t) = x(t) x (t) Δx (t) = x (t) dx (t) dt Δx (t) = A x(t) + B u(t) [A E x(t) + B E u(t)] = A x(t) A E x(t) + (B B E ) u(t) (4-22) The error of the estimator does not have to depend on input variable u(k), so that must be satisfied the equation B = B E. When we set A = A E, then the error of the estimator will not depend on the estimated vector nor the error of the estimator. Then the dynamic of this error is given by equations Δx (t) = A Δx(t), Δx (0) = x(0) x (0) (4-23) The error converges to zero in a stable system, but we have no way to influence the rate at which the state estimate converges to true state. To correct the estimator error, we can use information that is present in the difference between the measured and estimated output Δy(t) = y(t) y (t) (4-24) This signal is used as a feedback signal that is added to the estimator s derivative of the state. The correcting signal is scalar and the state derivative is a vector, therefore we use a transform matrix estimator matrix L. The estimator equation is dx (t) dt and the estimator matrix L is defined as = A E x(t) + B E u(t) + L (y(t) y (t)) (4-25) L = [l 1 l 2 l n ] T (4-26) By setting A E = A, B E = B then the scheme for the state estimator equation is in Fig

130 Substituting (4-21) to (4-25), we obtain Fig. 4-6: Block diagram of closed-loop estimator dx (t) = A dt E x (t) + B E u(t) + L (y(t) C x (t)) (4-27) The error of the estimator is then Δx (t) = x (t) dx (t) dt = A x(t) + B u(t) [A E x (t) + B E u(t) + L (y(t) C x (t))] Rewriting this equation Δx (t) = A x(t) A E x (t) + (B B E ) u(t) L y(t) + L C x (t) = A x(t) + (A x (t) A x (t)) A E x (t) + (B B E ) u(t) L C (x(t) x (t)) = A (x(t) x (t)) + (A A E ) (x(t) x (t)) + (B B E ) u(t) L C (x(t) x (t)) When we set A = A E, B = B E, then the error of the estimator has following equation Δx (t) = (A L C) Δx(t) (4-28) where L is the estimator matrix that is chosen to achieve satisfactory error characteristic. The characteristic equation of the error (4-28) is given det[s I (A L C)] = 0 (4-29) For the observer canonical form, the characteristic equation has the form det[s I (A L C)] = s n + (l 1 + α n 1 )s n (l n 1 + α 1 )s + (l n + α 0 ) = 0 (4-30) If we can choose L so that (A L C) has stable and reasonably fast eigenvalues, then Δx 0 and remain there independent of the forcing function u(t) or initial condition x(0). 130

131 Introduction to state space control 4.4 STATE FEEDBACK CONTROL Control law and control feedback structures Consider a dynamic system described by the following state equation: x (t) = A x(t) + B u(t) (4-31) where x(t) is an [n 1] state vector, and u(t) is the scalar input (manipulated variable). Consider that all state variables are measured or estimated then the control law is a linear combination of the state variables and is equal u(t) = K x(t) = [k 1 k 2 k n ] [ ] (4-32) x n where K is the feedback matrix with constant-gain elements. The feedback state control structures for measured and estimated variable are shown in Fig x 1 x 2 Fig. 4-7: Feedback control with measured and estimated state variables By substituting Eq. (4-32) into Eq. (4-31), the closed loop system will be represented by the state equation x (t) = A x(t) B K x(t) = (A B K) x(t) (4-33) The characteristic equation of this closed-loop system is det[s I (A B K)] = 0 (4-34) 131

132 When this determinant is evaluated, this yields an n th order polynomial in s containing the gains k 1,, k n. It will be shown that if the pair [A, B] is completely controllable, a matrix K exists that can give an arbitrary set of eigenvalues of (A B K). That means the n roots can be arbitrary placed. Consider a dynamic system that is represented in the controllable canonical representation, in Eq. (4-9). If the feedback matrix K has the form (4-32) then the matrix (A B K) is equal (A B K) = [ (a 0 + k 1 ) (a 1 + k 2 ) (a 2 + k 3 ) (a n 1 + k n )] It is possible to check that the characteristic equation is det[s I (A B K)] = s n + (a n 1 + k n )s n (a 1 + k 2 )s + (a 0 + k 1 ) = (s s 1 )(s s 2 ) (s s n ) (4-35) where s i are the poles-roots of the closed loop. If we know the poles then the Eq. (4-35) yields the constant k i of the feedback matrix. The equation shows that all coefficients of the state feedback matrix can be influenced. Denoting α i as coefficients of the desired characteristic equation of the closed-loop system, then k i = α i a i, i = 0,1,, n 1 (4-36) By computing of a feedback matrix with constant-gain elements in Matlab support, we can use the function acker. Example 4-3 Consider a dynamic system with state space representation x (t) = [ x 1(t) x 2(t) ] = [ ] [x 1(t) x 2 (t) ] + [0 ] u(t), y(t) = [ ] Determine: [x 1(t) x 2 (t) ] 1) the gain vector of the state controller with the given characteristic equation of closed loop system s 2 + 4s + 4 = (s + 2) 2. 2) Time responses by controlling the initial condition x(0) = [ 1 1 ]. Solution for 1) The system is in the controllable canonical form an the coefficients of the gain vector can be calculating via (4-36). i = 0: k 1 = α 0 a 0 = = 3.5 i = 1: k 2 = α 1 a 1 = = 2.5 The controller has the form K = [ ]. Solution for 2) 132

133 Introduction to state space control The program diagram for Simulink is in Fig. 4-8 on the left side. Time responses of x 1 (t), x 2 (t) are in Fig. 4-8 on the right side. Fig. 4-8: Simulink model (left) and initial condition response (right) A = [0 1; ]; B = [0; 1]; C = [0.75 0]; D = 0; C2 = [1 0;0 1]; % state vector observable K = [ ]; F2 = ss(a,b,c2,d); x0 = [1;1]; % initial condition T = feedback(f2,k); %feedback realization initial(ss(t),x0); %response to initial condition The state feedback control structure shown in Fig is deficient in that it does not track or compensate for the desired value or the disturbance of the type 1 (step function). This s structure can only determine the system from a given initial condition x(0) to the origin of the state space State feedback with integral control In general the control system must track desired outputs and compensate disturbances. The easy solution is to introduce an integral control, just as with a PI controller together with the constant gain state feedback. The block diagram of the structure is shown in Fig The input to the controlled plant is Fig. 4-9: Block diagram of state feedback with integral control u(t) = u R (t) + d U (t) (4-37) 133

134 The state equation is then x (t) = A x(t) + B u(t) = A x(t) + B [u R (t) + d U (t)] (4-38) t Introducing the integral control u n+1 (t) = k n+1 x n+1 (t) = k n+1 x n+1(τ)dτ adds one 0 integrator to the system. From Fig you can see that the output from the n + 1 th integrator is designated as the state variable x n+1 (t). The n + 1 th variable has the following equation x n+1(t) = w(t) y(t) = w(t) C x(t) (4-39) The dynamic equation of the dynamic system in Fig is x n+1(t) = [ x (t) x n+1(t) ] A 0 = [ C 0 ] [ x(t) x n+1 (t) ] + [0 1 ] w(t) + [B 0 ] u R(t) + [ B 0 ] d U (t) (4-40) The actuating signal u R (t) is related to the state variables through the matrix K and the integral feedback k n+1. u R (t) = K x(t) + k n+ x n+1 (t) = [K k n+1 ] [ x(t) x n+1 (t) ] = K n+1 x n+1 (t) (4-41) where is x n+1 (t) = [ x(t) x n+1 (t) ], K n+1 = [K k n+1 ] and x n+1 (t) is a broadened state vector, K n+1 is a broad matrix of state control. Substituting (4-41) into (4-40) we obtain the overall state equation x n+1(t) = [ x (t) x n+1(t) ] A 0 = [ C 0 ] [ x(t) x n+1 (t) ] + [0 1 ] w(t) [B 0 ] [K + [ B 0 ] d U(t) k n+1] Introducing matrix A n+1, B n+1, C n+1 the closed loop can be written in the form x n+1(t) = (A n+1 B n+1 K n+1 ) x n+1 + [ 0 1 ] w(t) + B n+1 d U (t) (4-42) (4-43) y(t) = C n+1 x n+1 where A n+1 = [ A 0 C 0 ] is a matrix [(n + 1) (n + 1)], B n+1 = [ B ] is a matrix [(n + 0 1) 1] and C n+1 = [C 0] is a matrix [1 (n + 1)]. The characteristic polynomial of the closed-loop system with integral control is det[s I (A n+1 B n+1 K n+1 )] (4-44) 134

135 Introduction to state space control The design objective are: Fig. 4-10: Structure of the feedback with integral The steady-state value of the output y(t) follows a step function input with zero error. The (n + 1) eigenvalues of (A n+1 B n+1 K n+1 ) are placed at the desirable location Questions State space control 1) Describe a state space representation of a DC motor. 2) What is the general state space representation? What is dynamic response? 3) What canonical forms do you know? Pleas, describe the forms. 4) Explain: similarity transformation, controllability, observability and its conditions. 5) What is Kalman form? 6) Explain time scale decomposition! 7) What do you understand under balanced reduction? 8) Describe the structure, estimation error and characteristic equation of an estimator. 9) Write the state variable control law and draw the control structure. 10) Explain the pole-placement method. 11) Describe the state feedback with integral control. 135

136 5 MATHEMATICAL DESCRIPTION OF MIMO SYSTEMS 5.1 A SET OF DIFFERENTIAL EQUATION Systems with multiple inputs and - outputs (MIMO) can be described by a set of equations, state equations or by matrix transfer functions. In the following example we will demonstrate the use of differential equations to describe the MIMO systems. Example 5-1 Consider a dynamic system with two inputs u 1, u 2 and two outputs y 1, y 2 that is described by differential equations set y 1(t) + 2y 1(t) + y 2(t) + y 2 (t) = u 1 (t) + u 2(t) + 3u 2 (t) y 1(t) + 2y 1(t) + 4y 2(t) + 4y 2 (t) = 3u 1(t) 4u 1 (t) + u 2(t) + 6u 2 (t) The block diagram is shown in Fig. 5-1a. Putting u 2 (t) = 0, then the outputs y 1 (t), y 2 (t) are described by the set of equations y 1(t) + 2y 1(t) + y 2(t) + y 2 (t) = u 1 (t) y 1(t) + 2y 1(t) + 4y 2(t) + 4y 2 (t) = 3u 1(t) 4u 1 (t) The interactions between u 1 (t) and the output variables y 1 (t), y 2 (t) are plotted in Fig. 5-1b. In a similar way if we put u 1 (t) = 0 then for the outputs y 1 (t), y 2 (t) we get a set of differential equations y 1(t) + 2y 1(t) + y 2(t) + y 2 (t) = u 2(t) + 3u 2 (t) y 1(t) + 2y 1(t) + 4y 2(t) + 4y 2 (t) = u 2(t) + 6u 2 (t) In Fig. 5-1c are plotted the interactions between u 2 (t) and the output variables y 1 (t), y 2 (t). Fig. 5-1: MIMO system interactions 136

137 Mathematical description of MIMO systems 5.2 TRANSFER FUNCTION MATRIX The transfer function matrix can be used as a linear time invariant mathematical model that approximate the dynamic behaviours of MIMO system. It is demonstrated in the following example where the structure of the transfer matrix is shown. Example 5-2 Consider again the plant described by the previous example. y 1(t) + 2y 1(t) + y 2(t) + y 2 (t) = u 1 (t) + u 2(t) + 3u 2 (t) y 1(t) + 2y 1(t) + 4y 2(t) + 4y 2 (t) = 3u 1(t) 4u 1 (t) + u 2(t) + 6u 2 (t) If we take the Laplace transform by zero initial conditions, we obtain the transformed equations as Y 1 (s)s(s + 2) + Y 2 (s)(s + 1) = U 1 (s) + U 2 (s)(s + 3) Y 1 (s)s(s + 2) + Y 2 (s)4(s + 1) = U 1 (s)(3s + 4) + U 2 (s)(s + 6) Writing the equation in a matrix form yields s(s + 2) s + 1 [ s(s + 2) 4(s + 1) ] [Y 1(s) Y 2 (s) ] = [ 1 s + 3 (3s + 4) s + 6 ] [U 1(s) U 2 (s) ] The Laplace transform of the output vector is than 1 1 Y(s) = [ Y 1(s) Y 2 (s) ] = [ s + 2 s ] [ U 1(s) 1 U 1 2 (s) ] s + 1 = [ F 11(s) F 12 (s) F 21 (s) F 22 (s) ] [U 1(s) U 2 (s) ] = F(s) U(s) where F(s) is the transfer function matrix [2 2], Y(s) = [ Y 1(s) ] is the Laplace transformation Y 2 (s) of the output vector [2 1], U(s) = [ U 1(s) U 2 (s) ] is the Laplace transformation of the input vector [2 1]. L-transform of output variables Y 1 (s), Y 2 (s) is than Fig. 5-2: MIMO system structure Y 1 (s) = F 11 (s)u 1 (s) + F 12 (s)u 2 (s) = 1 s + 2 U 1(s) + 1 s U 2(s) Y 2 (s) = F 21 (s)u 1 (s) + F 22 (s)u 2 (s) = U 1 (s) + 1 s + 1 U 2(s) The internal structure diagram of the transfer function matrix is shown in Fig

138 In general the transformed response of the system with p inputs and q outputs in matrix vector form is equal Y(s) = F(s) U(s) (5-1) F 11 (s) F 12 (s) F 1p (s) F where F(s) = 21 (s) F 22 (s) F 2p (s), is the [q p] transfer function matrix, [ F q1 (s) F q2 (s) F qp (s)] Y 1 (s) U 1 (s) Y Y(s) = 2 (s) U is [q 1] transformed output vector, U(s) = 2 (s) is the [p 1] [ Y q (s)] [ U p (s)] transformed input vector. The transfer function between j th input and i th output is defined as Y i (s) = F ij (s)u j (s) F ij (s) = Y i(s) U j (s) (5-2) When all the p inputs are in action, the i th output transform is written Y i (s) = F i1 (s)u 1 (s) + F i2 (s)u 2 (s) + + F ip (s)u p (s) = F ij (s)u j (s) The block diagram representations of multivariable system with p inputs and q outputs is shown in Fig In Fig. 5-3a), the individual input and output are designed, whereas in the block diagram of Fig. 5-3b), the multiplicity of the inputs and outputs is denoted by vectors. p j=1 (5-3) Fig. 5-3: Block diagrams of a multivariable system with p inputs and q outputs Work with the transfer matrix is demonstrated in the following example. Example 5-3 Consider a part of a steam cooler described by the scheme diagram in Fig. 5-4a. The F 11 (s) F 12 (s) system was approximated by the matrix transfer function F(s) = [ F 21 (s) F 22 (s)]. F 31 (s) F 32 (s) The steam temperature T CH and pressure P CH is adjusted by a pressure reducing valve (PRV) and by the flow regulator valve (FRV) ( flow of cooling water). The manipulated variable of the pressure reducing valve is u R1 and the flow regulator valve 138

139 Mathematical description of MIMO systems u R2. Disturbance to the process is the changes in the flow of the steam. Calculate the output transform vector and draw the internal structure diagram for the transfer function matrix F(s)! Solution. The system approximated by the matrix transfer function has p = 2 inputs manipulated variables u R1, u R2, and q = 3 output variables T CH steam temperatures, P CH steam pressure, m CH steam flow. The interactions are shown in Fig. 5-4b). Fig. 5-4: Technological diagram (a), Interactions (b), Internal block diagram structure (c) Using (5-1) the output vector transform is Y 1 (s) F 11 (s) F 12 (s) Y(s) = [ Y 2 (s)] = F(s) U(s) = [ F 21 (s) F 22 (s)] [ U 1(s) U Y 3 (s) F 31 (s) F 32 (s) 2 (s) ] F 11 (s)u 1 (s) F 12 (s)u 2 (s) = [ F 21 (s)u 1 (s) F 22 (s)u 2 (s)] F 31 (s)u 1 (s) F 32 (s)u 2 (s) The Internal block diagram structure is shown in Fig. 5-4c). 5.3 STEADY-STATE OF OUTPUT VARIABLE According to the definition of matrix transfer function F(s) it is possible to define the unit step function h ij (t) and the impulse response g ij (t). Unit step function h ij (t) is the response function of the i th output variable to the unit step 1(t) on the j th input variable by zero initial conditions. Impulse response g ij (t) is the response of the i th output variable when the input on j th is a unit impulse function δ(t) by zero initial conditions. 139

140 What is important are the steady-states that can be evaluated from a) The transformed output vector Y(s) using the final-value theorem lim s Y 1 (s) s 0 lim s Y 2 (s) y( ) = lim s Y(s) = s 0 s 0 [ lim s Y q (s) s 0 ] b) The steady states of the unit step functions and the step inputs y i ( ) = h i1 ( )u 1 ( ) + h i2 ( )u 2 ( ) + + h ip ( )u p ( ) p = h ik ( )u k ( ) k=1 c) The set of differential equation describing the multivariable system We set for steady state (5-4) (5-5) where k 1 and i = 1,2,, q, j = 1,2,, p. Example 5-4 lim y (k) t i (t) = 0, lim u (k) t j (t) = 0 (5-6) 2 1 s+1 s (2s+1) 2 Consider transfer the function F U (s) = [ ]. Calculate the steady-state (2s+1) 2 output vector for given inputs u 1 (t) = 2 1(t), u 2 (t) = 1(t). Solution. The transformed vector output vector is s + 1 s + 1 Y(s) = F U (s) U(s) = [ s s(s + 1) ] = [(2s + 1) 2 (2s + 1) 2 ] s [ s(2s + 1) 2 ] According to (5-4), the steady state vector output is equal 3 y( ) = [ y lim s Y 1( ) 1 (s) lim s ( y 2 ( ) ] = [ s 0 lim s Y 2 (s) ] = s 0 s(s + 1) ) 0.5 s 0 lim s ( [ s 0 s(2s + 1) 2) ] = [ ]. 140

141 Basics principles of robust control 6 BASICS PRINCIPLES OF ROBUST CONTROL 6.1 INTRODUCTION The final part of the course consists of basic principles of robust control of SISO systems. This is a very demanding theoretically but practically very important and a revolutionary part of the field of system control. Why is there no significant installation of robust control in industry? I believe that the root causes are: 1) This part of the field of control is built on mathematically very difficult theory. 2) The resulting algorithm requires an extensive software support. 3) The resulting controllers are of higher orders and are, therefore it is not possible to make simple use of existing control systems and components. 4) Measurement and control technicians have no opportunity to make changes to settings without the necessary knowledge and appropriate software. Usually, they have neither sufficient experience nor theoretical background. Therefore they rightly fear the risks associated with the operation of these control systems. Our task is only to introduce students to the basic principles and ideas of robust control. Students then apply the knowledge gained by means of an appropriate software application to find robust algorithms. Students can verify the basic characteristics of a robust control system and this basis should motivate further study of robust management. 6.2 NORMS H 2, H In the Hilbert space it is possible to define a standard H 2 norm for SISO system G 2 = 1 2π G( iω) G(iω) dω = lim t g 2 (t)dt (6-1) where G(iω) is the transfer frequency function, g(t) is a weight function. It is clear that the norm is finite, if the G( ) = 0. (stable, physically realizable). For the calculation, it is simpler to use the square of the norm that can be written G 2 2 = 1 2π G( iω) G(iω) dω = lim t g 2 (t)dt = g(t) 2 2 (6-2) 141

142 Singular Values (db) Norm H is determined G = sup Re(s)>0 σ (G(s)) = sup σ (G(iω)) ω R (6-3) σ (G(iω)) is the largest singular number for the ω. For calculations of norms H 2 and H in Matlab the function norm is used. Singular numbers are calculated by using the function sigma. This text is only a basic description of these functions and options for your quick orientation and to facilitate work with Matlab in this field. The largest singular number is equal Example Consider the transfer function F(s) =. Calculate H s 3 +3s 2 +2s+1 2, H norms. Solution. Let s show the exact picture of the states. Norms H 2, H is calculated by the function norm, plot G(iω) by the function sigma, see the program. clear all display('h2norma, Hinf, min, peak:') B = [2]; A = [ ]; G = tf(b, A); H2 = norm(g, 2) Hinf = norm(g, inf) [MaxSig, frek] = norm(g, inf) sigma(g) title('sigma Pr1') grid Result: H2norma, Hinf, min, peak: H2 = Hinf = MaxSig = frek = Sigma Pr Frequency (rad/sec) Fig. 6-1: Singular values 142

143 Basics principles of robust control 6.3 MODEL UNCERTAINTY IN DYNAMIC SYSTEMS Uncertainty in the mathematical modelling can be divided into two groups: a) Parametric uncertainty, which represent changes in model parameters in certain intervals, in which the actual parameter may change. b) Dynamic uncertainty, which can be thought of as signals for entry or exit. These signals exhibit dynamic properties that can be modelled using a transfer function, which will place certain restrictions. We will concentrate only on the dynamic uncertainty. Acquiring technical information about the system requires the measuring of inputs and outputs Multiplicative uncertainty If the measured input signal is different from the nominal input signal, this deviation represents an uncertainty. How should we model it? Let the transfer function Δ m (s) approximates the effect of dynamic uncertainty. Index m indicates that this is the input (manipulated) variable. Because a member of Δ m (s) must not affect the stability of the system it must satisfy the condition max ω Δ m(iω) 1 If this condition is expressed as standard norm H, then it can be written as the inequality (6-4) Δ m 1 (6-5) It is known from technical experience that the uncertainty model depends on the frequency and is greater for higher frequencies. It is possible to express this in the frequency transfer W m (iω). The total uncertainty is then given as the product W m (s)δ m (s). The transfer function W m (s) is generally low order and in terms of the mathematical analysis it can be regarded as a weighting function. We assume that the nominal input signal is u = u(t) and the model system is given by the nominal transfer function G nom (s). In addition, we require a model in the absence of uncertainty correspond to the nominal values. These requirements correspond to the input model of uncertainty (see Fig. 6-2), which is called the model of multiplicative uncertainty. Fig. 6-2: Input model uncertainty multiplicative uncertainty The transfer function of the model with multiplicative uncertainty is G(s) = G nom (s) + W m (s)δ m (s)g nom (s) = [1 + W m (s)δ m (s)]g nom (s) (6-6) The frequency transfer function G(iω) is G(iω) = [1 + W m (iω)δ m (iω)]g nom (iω) (6-7) Due to the condition (6-5) it is possible in the extreme case of the equation (6-7) to write 143

144 Substituting Eq. (6-5) to Eq. (6-7) yields Δ m = 1 (6-8) G(iω) G nom (iω) + W m (iω) G nom (iω) (6-9) How can we find the weighting transfer function? Suppose, the uncertainty is represented by a set of transfer functions G k (s) for k = 1,, N. For every function G k (s) the parameters of the transfer function are known. From the inequalities (6-9) it is possible to calculate the relative uncertainty for given k W k (iω) = G k(iω) G nom (iω) G nom (iω) = G k(iω) G nom (iω) 1 (6-10) We know that the weighting transfer functions W m (iω) modulates the uncertainty depending on the frequency ω. From the inequalities (6-9) and equation (6-10) it is possible to estimate the uncertainty tolerance zone upper bound using weighting functions W m (iω) (Multiplicative relative uncertainty) W m (iω) = max k W k(iω) = max for k = 1,, N where G k (iω) is the k th uncertainty). k G k (iω) G nom (iω) G nom (iω) (6-11) characteristic uncertainty (parametric The following example illustrates the determination of a weighting function. Example 6-2 Consider a system approximated by the transfer function G(s) = K. The coefficients Ts+1 lie in the parameter intervals K 1,5, T 0.5,2. The parametric uncertainty is given by the gain and time constant changing. The nominal transfer function is G nom (s) = 2.5. The task is to find the transfer function of W 1.5s+1 m(s), which defines the tolerance of the dynamic uncertainty in the frequency domain for all ω. Solution. We have a set of transfer function that represents the parametric uncertainties of K, T for k = 1,2,3,4. G 1 (s) = 1 0.5s + 1, G 2(s) = 1 2s + 1, G 5 3(s) = 0.5s + 1, G 4(s) = 5 2s + 1 For this set we need to evaluate a set of magnitudes W k (iω) = G k (iω) G nom (iω) 1 for k = 1,2,3,4 (see Fig. 6-3). The condition (6-11) satisfies directly the frequency transfer function W 3 (iω). So the weighting function W m (iω) = W 3 (iω). The weight W m (s) for complex uncertainty according to (6-11) is 5 W m (s) = W 3 (s) = 0.5s = 2.5s s + 1 W m(iω) = 2.5iω iω s + 1 We can check that the condition is fulfilled in the interval of K and T too. 144

145 Basics principles of robust control Fig. 6-3: Set of filters W i (left), checking the condition (right) Additive uncertainty Another measured signal is the output. Regarding the structure of the model of uncertainty of the output with the nominal model then the structure of output uncertainty is defined in Figure The output uncertainty is called the additive uncertainty. Fig. 6-4: Additive uncertainty The transfer function of a system with disturbance modelled with additive uncertainty is G(s) = G nom (s) + W a (s) Δ a (s) (6-12) 6.4 ROBUST STABILITY OF A FEEDBACK SYSTEM WITH UNSTRUCTURED UNCERTAINTY To simplify the task of robust stability we consider an open circuit with the uncertainty on the input according to Fig Fig. 6-5: Model of an open loop with multiplicity uncertainty Consider the nominal model G nom (s) = G 0 (s) and multiplicative uncertainty. We are asking how an uncertainty effects on the stability of the closed loop systems? The stability 145

146 analysis is performed using Nyquist stability criterion. The transfer function of an open loop with a minimal and non-minimal transfer function is L 0 (s) = G 0 (s)r(s) L(s) = G(s)R(s); W(s) = W m (s) (6-13) The non-minimal open loop transfer function L(s) as a function of the nominal transfer function G 0 (s) is given as L(s) = G(s)R(s) = G 0 (s)r(s)[1 + W(s)Δ(s)] (6-14) The equation (6-14) for (s) = 1 can be rewritten in the form The magnitude of (6-15) is L(s) = L 0 (s) + W(s)L 0 (s) (6-15) L(iω) = L 0 (iω) + W(iω)L 0 (iω) (6-16) Comment: The transfer function can be rewritten in the form L(iω) = L 0 (iω) + W(iω) L 0 (iω) = G 0 (iω)r(iω) [1 + W(iω) ] G k (iω) G 0 (iω) = G 0 (iω) R(iω) + G 0 (iω) R(iω) max k G 0 (iω) = G 0 (iω) R(iω) + R(iω) max [ G k(iω) G 0 (iω) ] k = L 0 (iω) + R(iω) Δ max G(iω) You can see the analysis of robust stability of feedback control is based on nominal open loop transfer function L 0 (s) and G(s) that represents the upper band of the plant uncertainty. Consider a nominal open loop transfer function L 0 (s) whose Nyquist plot is depicted in the Fig Applying Nyquist s criterion, the stability condition is explored in the Fig You can see that the system with the uncertainty that is described by the term W m1 (iω) L 0 (iω) (circle 1), is stable because of the stability condition is fulfilled (id does not intersect the real axis). The system with uncertainty that is described by the term W m2 (iω) L 0 (iω) (circle 2) is stable too, because of the line segment 1,0 is intersected by the circle 2 (The critical point [ 1,0i] lies on the left hand side of the Nyquist plot or encirclements are avoided if none of the circle cover the point [ 1,0i]). The system with uncertainty that is described by the expression W m3 (iω) L 0 (iω) (circle 3) is unstable, because the critical point [ 1,0i] lies on the right side of the Nyquist plot. 146

147 Basics principles of robust control Fig. 6-6: Implementation Nyquist stability criterions to a system You can see that the condition stability can be measured by the length of the vector v. The length of the vector v must fulfil The length of the vector v is v > W(iω) L 0 (iω) (6-17) v = v L0 v 1 = [ R{L 0(iω)} I{L 0 (iω)} ] [ 1 0 ] = [R{L 0(iω)} + 1 ] I{L 0 (iω)} (6-18) v = (R{L 0 (iω)} + 1) 2 + I{L 0 (iω)} 2 = 1 + L 0 (iω) (6-19) The condition (6-17) can be rewritten as 1 + L 0 (iω) > W(iω)L 0 (iω) (6-20) A system with uncertainty will be stable if the condition (6-20) is satisfied for ω R. Dividing the inequality (6-20) by 1 + L 0 (iω) we get 1 > W(iω)L 0(iω) 1 + L 0 (iω) (6-21) Putting the sensitivity and complementary sensitivity function into the inequality (6-21) we can modify this inequality in the form of 1 > W(iω) L 0 (iω) 1 + L 0 (iω) = W(iω) T 0(iω) (6-22) Applying the standard norm H than the stability condition is in the form W(iω) T 0 (iω) < 1 (6-23) 147

148 A system with uncertainty will be stable if the condition (6-23) is satisfied for ω R. From the conditions (6-22) the upper bound on complementary sensitive T 0 (iω) in the frequency domain can be obtained in the form T 0 (iω) < 1 W(iω) = 1 W m (iω) (6-24) 6.5 ROBUST PERFORMANCE AND CONTROL WITH MULTIPLICATIVE UNCERTAINTY It is known that robust control design must guarantee: 1. Stability of closed loop by a given uncertainty that is characterized by the transfer function W m. 2. Control performance of controlled variable (control error). 3. Constraints on manipulated variables, which generates the real actuator. In the previous chapter, the robust stability condition was explained and defined. We are going control performance in the next chapter Robust performance Consider the structure of a feedback control system according to Fig Fig. 6-7: Block diagram of a general closed-loop feedback control system We are seeking an appropriate criterion of performance (quality) control to ensure compliance with the requirements for control. From classical control, we know that in the frequency domain design, the controller can be shaped to achieve the desired magnitude (amplitude) L 0 (iω) and the phase characteristics arg{l 0 (iω)} of an open loop. According to the results of the analysis of stability, the control system with multiplicative uncertainty must meet the condition (6-23). That means that the magnitude (absolute value) of the complementary sensitivity function must have the required shape. We know that a nominal complementary sensitivity function T 0 (s) = L 0 (s) 1+L 0 (s) = S 0(s)L 0 (s). For the nominal system the control sensitivity function can be directly used as a performance (quality) criterion directly. Using the weighting filter W P (iω) we define the requirement for the searched sensitivity function S 0 (iω). The sensitivity function contains the transfer function of the controller 148

149 Basics principles of robust control R(s), so by shaping the S 0 (iω) the controller R(s) will be found. The sensitivity transfer function must therefore fulfil the condition S 0 (iω) < Eq. (6-25) can be rewritten in the following form 1 W P (iω) (6-25) W P (iω) S 0 (iω) < 1 W P (iω) < 1 + L 0 (iω) (6-26) Weighting function (filter) W P (iω) = W(iω) can be chosen s W(s) = W P (s) = M + ω B s + A ω (6-27) B where A is the gain for low frequency, M the gain for a high frequency, ω B the break frequency with zero axis. Example 6-3 The following example shows the parameterization of the weighting filter (function) according to (6-27). Chose: 1) Gain for high frequency M = 10, break frequency with zero axis ω B = 2, gain for 1 low frequency A = 0. Adequate magnitudes W P1 (iω) and are in Fig a. W P1 (iω) 2) We leave M = 10, ω B = 2, we chose the gain for low frequency A = Adequate magnitudes W P2 (iω) and are depicted in Fig W P2 (iω) Fig. 6-8: Magnitude of weighting filters W P1 and W P2 Remembering the condition (6-25) for a system with multiplicative uncertainty. Substituting Eq. (6-15) to Eq. (6-25) yields W P (iω) < 1 + L 0 (iω) + W m (iω)l 0 (iω), ω R (6-28) The condition (6-28) represents a robust performance condition and is graphically illustrated in the Nyquist plot in Fig

150 Fig. 6-9: Nyquist plot of robust performance condition The condition (6-28) can be interpreted in the L 0 plain as a circle centred in L 0 (iω) with diameter W m (iω)l 0 (iω) and another circle centred in [ 1,0i] with diameter W P (iω) for given ω. The centre lies on the connecting line 1 + L 0 (iω). Since the circles do not overlap, the condition is satisfied. The modified condition can be written 1 + L 0 (iω) > W P (iω) + W m (iω)l 0 (iω), ω R (6-29) Using sensitivity function S, T then the condition (6-29) can be rewritten in the following form 1 > W P(iω) 1 + L 0 (iω) + W m(iω)l 0 (iω) 1 + L 0 (iω) = W P (iω)s(iω) + W m (iω)t(iω) (6-30) This condition can be written in the following expression max { W P(iω)S(iω) + W m (iω)t(iω) } < 1, ω R ω (6-31) The condition (6-31) can t be directly implemented, therefore it is approximated by the norm H W P(iω)S(iω) W m (iω)t(iω) = max W P(iω)S(iω) 2 + W m (iω)t(iω) 2 ω (6-32) Mixed-sensitivity H control Consider the structure of a feedback control system in Fig

151 Basics principles of robust control Fig. 6-10: Closed loop control The transfer functions F EW (s), F UW (s), F YW (s) a F ED (s), F YD (s), F UD (s) are F EW (s) = G(s)R(s) = S(s), F R(s) UW(s) = 1 + G(s)R(s) = R(s)S(s), F YW (s) = G(s)R(s) 1 + G(s)R(s) = T(s), F 1 ED(s) = 1 + G(s)R(s) = S(s), R(s) F UD (s) = 1 + G(s)R(s) = R(s)S(s), F 1 YW(s) = 1 + G(s)R(s) = S(s). We can simply say that mix-sensitivity control allows us to make demands on the chosen closed-loop transfer function or signals from the control loop (control error e, controlled variable y, and manipulated variable u). For instance in Fig a closed-loop control system with weight function and signals is depicted. Fig. 6-11: Closed-loop system with signals z 1, z 2, z 3 and weight transfer functions W 1, W 2, W 3 You can see the signal z 1 is the impulse function of the error, e = w y that is weighted by the transfer function W 1. The signal z 2 is the impulse function of the manipulated variable u that is weighted by the transfer function W 2 and the signal z 3 is the impulse function of the controlled variable y that is weighted by the transfer function W 3. Using the norm H for optimization, it yields min Rstab W 1 (iω)s(iω) W 2 (iω)r(iω)s(iω) W 3 (iω)t(iω) max ω W 1(iω)S(iω) 2 + W 2 (iω)r(iω)s(iω) 2 + W 3 (iω)t(iω) 2 (6-33) 151

152 From a practical point of view, the optimal H controller is replaced by the sub-optimal H controller, which satisfies the condition W P (iω)s 0 (iω) F ZW = F l (P, R) = W 2 (iω)r(iω)s 0 (iω) W m (iω)t 0 (iω) σ (S(iω)) γ σ(w P 1 (iω)) σ (R(iω)S(iω)) γ σ(w 1 2 (iω)), ω R σ (T(iω)) γ σ(w 3 1 (iω)) < γ (6-34) The transfer function W 1, W 2, W 3 must be chosen. The evolution of the norm (6-33) is solved in Matlab by the toolbox function mixsyn, which is a part of the Robust Control Toolbox Rules for parameter setting of the filters W P = W 1 and W 2 During the synthesis of the controller according to the standard H mixed sensitivity function it is necessary to specify transfer function for the filter W 1, W 2 and W 3 when the function mixsyn in Matlab is used. Weighting filters: W 1 represents the requirements of the control quality W 2 represents the requirements of the manipulated variable W 3 defines the upper limit of the system uncertainty. If it is not known, it enters into the algorithm as an empty parameter, if it is known, so the uncertainty is substituted into the function mixsyn. W 1 (iω) The practical design of the controller uses a method that searches for a sub-optimal controller, which satisfies the condition (6-34). Our task is to parameterize the weighting filters W 1 and W 2. The weighting function must be designed using control techniques. We are going to concentrate ourselves on only one the basic shapes of the transfer function - weighting function W P = W 1 (6-27). Using the inverse weighting 1 function (see the Figure 6.5-6) we can define the upper limit of the amplitudesensitivity function S in which the amplitude of S(iω) cannot lie. An asymptotic 1 approximation of the is depicted in the Fig You can see that the W 1 (iω) approximation has three corresponding parameters A, M and ω B. The selection of a weight function W 1 (s) parameters according to the inverse function 1 can be considered as complicated, because the weighting (shaping) filter in W 1 (iω) frequency domain will determine the control behaviour (performance) in the time domain and shall meet many other requirements. In our course, we do not want to deal with a detailed analysis of the frequency transfer 1 function, but we would like to convey some recommendations to students, allowing W 1 (iω) them to use Matlab function mixsyn for controller design. These rules have a purely engineering background and interpretation without claim to generality and mathematical precision. We consider the starting platform that an integrator, which ensures steady zero control 152

153 Basics principles of robust control deviation, is not treated in a special way! The transfer function of a controller R(s) that we search for, therefore, must contain at least an approximation of an integration component, which will find expression by the real pole, which have to be close to zero. Filter W 1 (s) Fig. 6-12: Approximation of the inverse weight function Because there is no unique prescription how to choose the parameters of the transfer function W 1 (s) we give you the following recommendations: It should be noted that the search filter parameters of W 1 (s) is a method by trial / error. Selection of suboptimal control will also be affected by the penalty function W 2 (s) and of course the uncertainty of the regulated system, which represents a penalty function W 3 (s). The gain of the filter at low frequencies A L should be set low because the transfer function W 1 (s) may then represent an integration component. Assume that A L 0, then the transfer function W 1 (s) will be approaching to the transfer function lim W 1(s) = A L 0 s M + ω B s + ω B A L = s M + ω B r 0s + r 1 s s The filter gain at high frequencies M shoud be M > 1 (M = 2!). It may be noted that under the classical PID control in the transfer function W 1 (s) the parameter r 0 = 1 represents the proportional component. This is evident from M the previous equation. You can verify that the greater the gain M at high frequencies in the determined area, the faster the control processes! In the first step, we choose the frequency of intersections with the zero axes ω B as close to breaking point of the controlled plant as possible! By increasing this frequency a significant acceleration of the control responses can be achieved. Significant acceleration of the control responses can be achieved by increasing the frequency ω B. 153

154 Filter W 2 (s) Filter W 2 (s) can significantly influence the time behavior of the manipulated variable. Generally it recommended that the filter W 2 (s) is chosen independently of the frequency. Recommendation. Reduction of the initial peak of the manipulated variable can be achieved by increasing the absolute value of W 2. Example 6-4 Consider a nominal transfer function G(s) = 2.5. The weight transfer functions are 1.5s+1 chosen as W 1 (s) = 0.1 s+100, W 100s+1 2(s) = 0.5, W m (s) = W 3 (s) = 2.5s+1. Write a 0.5s+1 Matlab program to synthesize H controller with the function mixsyn! Solution. s = zpk('s') G = [2.5/(1.5*s+1)] W1 = [0.1*(s+100)/(100*s+1)]; W2 = [0.5] ; W3 = [(2.5*s+1)/(0.5*s+1)]; [K, CL, GAM] = mixsyn(g, W1, W2, W3) R = tf(k) Program results: GAM = s^ s R = s^ s^ s Example 6-5 Mix-sensitive H synthesis for rotation speed-control of a DC-motor with an elastic clutch. This lab task consists of a PC with I/O Card, DC++-motor M, elastic clutch EC and dynamo T. The system has approximately linear characteristics in the operating range with an oscillating movement in its output. See Fig Fig. 6-13:Rotation speed control of DC motor with elastic clutch process scheme (left), identification measurement (right) The identification measuring is shown in Fig (right). The transfer function is equal G(s) =. Write a Matlab program to synthesize H s s s+1 154

155 Singular Values (db) Basics principles of robust control controller with the function mixsyn! The weight transfer functions are chosen W 1 (s) = 0.01s+1 s+0.01, W 2(s) = 0.1, W 3 (s) = [] is empty. Solution. clc; clear all; close all; %% System setting Bs = [0.2964]; As = [ ]; s = zpk('s') G = /(0.0257*s^ *s^ *s+1) %% Filter setting W1 = (0.01*s+1)/(s+0.01); W2 = 0.1; W3 = []; %% Controller calculation [K, CL, GAM] = mixsyn(g, W1, W2, W3) R = tf(k) [BR, AR] = tfdata(r, 'v'); L = G*K; S = inv(1+l); T = 1-S; sigma(s, 'g', T,'r', GAM/W1, 'g-.', GAM*G/ss(W2), 'r-.') grid %% PID controller to compare P = 0.5; I = 6; D = 1; Program results: GAM = s^ s^ s e04 R = s^ s^ s^ s Singular Values S T /W 1 *G/W Frequency (rad/s) Fig. 6-14: Singular values (sigma function) 155

156 y u Robust PID Robust PID See the initial peak on the manipulated variable for PID control t [s] Fig. 6-15: Output values and input actions for robust and PID controller Questions Principle of robust control t [s] 1) What is the norm H? 2) Draw and describe the model of multiplicity uncertainty! 3) Uncertainty tolerance zone and its calculation (Weight function W m = W 3 ). 4) Robust stability of feedback systems with the multiplicative uncertainty. 5) Draw the condition of robust stability in Bode plot! 6) Robust performance, weighting filter W P, performance condition in Bode plot. 7) Mixed sensitivity, weighting filters W 1, W 2, W 3. 8) Rules for parameter setting of weights function W 1, W

157 References 7 REFERENCES [1] Kuo, B.C.: Automatic control systems. John Wiley & Sons, Inc. New York, ISBN [2] Marlin, T. E.: Process control. Designing processes and control systems for dynamic performance. Mc Graw Hill, 2 nd ed, ISBN [3] Bolton, W.: Mechatronics. Electronic control systems in mechanical and electrical engineering. Pearson Education Limited, 3-rd ed ISBN 10: [4] Bequette, B., W.: Process Control. Modeling, Design and Simulation. Prentice Hall, [5] Goodwin, G., C., Graebe, S., F., Salgado, M.,S.: Control System Design. Prentice Hall [6] Marlin, T. E.:Process control. Designing processes and control systems for dynamic performance. 2 nd ed. Mc Graw Hill, [7] 157

158 Name of Document Automatic Control in Mechatronics Authors doc. Ing. Osvald Modrlák, CSc. Ing. Lukáš Hubka, Ph.D. Target Readership TUL students Publisher Technical University of Liberec Approved by TUL Rector s Office as of 18th of December 2014, ref. number RE 146/14 Published in December 2014 Number of Pages 157 Edition 1st Publication Number This publication has been edited. ISBN