Mechanical Engineering - Control and Actuating Devices for Mechanical Systems

Completed notes of the course

Complete course

Politecnico di Milano School of Industrial and InformationEngineering Department of Mechanical EngineeringControl and Actuating Devices for Mechanical SystemsTeacher: Ing. Francesco Braghin Teacher: Ing. Edoardo Sabbioni Trainer: Ing. Davide Enrico QuadrelliLecture Notes Author: Fabio Santoro Academic Year 2021/2022 Pag. 3 Preface This handout contains all the topics covered in theCADMScourse of the academic year 2021-2022. The content was adapted from the lecture notes and slides used during the lectures of Professors Braghin, Quadrelli and Sabbioni. The main topics dealt with in this handout are: 1.Introduction: chapter of introduction to the course with a fast recap of the arguments covered in the preview course"Meccanica del le Vibrazioni"on the third year of the bachelor degree in Mechanical Engineering 2.Stability Analysis: this topic deals all the arguments related to the stability analysis (direct and indirect criteria) of a mechanical system with also the application of the Laplace Transform 3.Control: in this part are dealt all the arguments related to the control of a mechanical system 4.Actuating: this topic deals about the actuators used to control a mechanical system (here it will applied the theory of stability analysis and control) 5.Modern Control Theory: this is the last topic of the course and it deals about the modern techniques of control system compared with the classic theory (covered in the preview parts) Fabio Santoro(Contents link)CADMS - Lecture Notes Pag. 3 Pag. 5 Contents I Introduction11 1 Feedforward and Feedback Control131.1 Feedforward control strategy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13 1.2 Feedback control strategy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13 1.3 Feedforward+Feedback control strategy. . . . . . . . . . . . . . . . . . . . . . . . . . . .14 1.4 FF Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .141.4.1 FF: transient response. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15 1.4.2 FF: steady-state related to the unknown force. . . . . . . . . . . . . . . . . . . . .16 1.4.3 FF: steady-state related to the target. . . . . . . . . . . . . . . . . . . . . . . . .18 1.5 FB Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .201.5.1 FB: transient response. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21 1.5.2 FB: steady-state related to the unknown force. . . . . . . . . . . . . . . . . . . . .22 1.5.3 FB: steady-state related to the target. . . . . . . . . . . . . . . . . . . . . . . . .23 II Stability Analysis25 2 Stability Analysis: 1 dof System272.1 Stability: definition. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .27 2.2 Force Fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .27 2.3 Stability Analysis: 1 dof system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .28[1] Non-linear equation of motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .28 [2] Static equilibrium position. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .28 [3] Linearization of the equation of motion. . . . . . . . . . . . . . . . . . . . . . . . . . .29 [4] Stability of the linearized system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .30Case I. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .30 Case II. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31 Case III. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .32 Case IV. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33 Case V. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33 Case VI. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .34 2.4 Summary of 1 dof Stability Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .35 3 Stability Analysis: 2-n dofs System373.1 Description of the problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .37 3.2 Static equilibrium Position. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .38 3.3 Linearization of the system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .38 3.4 Solution of the system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .39 3.5 Stability analysis: Conservative Force Field (CFF). . . . . . . . . . . . . . . . . . . . . .40Case 1: Global Stiffnes Matrix positive definite. . . . . . . . . . . . . . . . . . . . . . . .41 Case 2: Global Stiffnes Matrix not positive definite. . . . . . . . . . . . . . . . . . . . . .41Determinant less than zero. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .42 Determinant greater than zero. . . . . . . . . . . . . . . . . . . . . . . . . . . . .43 Summary: conservative force field. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .44 3.6 Stability analysis: Non-Conservative Force Field (NCFF). . . . . . . . . . . . . . . . . .45(Contents link)CADMS - Lecture Notes Pag. 5 Case 1: β >0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .45 Determinant less than zero. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .45 Determinant greater than zero. . . . . . . . . . . . . . . . . . . . . . . . . . . . .45 Case 2:β 1overdamped system •Underdamped system:h 1These cases will not be covered in this chapter. This argument is part of the course "Meccanica del le Vibrazioni", please see [1] at section(2.4.1)at page 125 or in the appendix (A) at page213 1.4.2FF: steady-state related to the unknown force Now let’s analysis the particular solutionx p,u( t)related to the unknown force (disturbance)f u( t): m·¨ x+r·˙ x+k·x=f u( t)(1.4.7) As represented in figure (1.6), we can use the Fourier Transform to pass from the time domain to the frequency domain and calculate the response for each single harmonic. In general the Fourier Transform to the unknown force is: fu( t) = 0 Fu,m2 +∞ X j=1| F u,0,j| · cos (Ω j· t+ Ψ j) Usually this force hasn’t a mean value, it contains uncontrolled fluctuations.Ω jand Ψ jare, respec- tively, the frequency and the phase of thejth harmonic. The genericjth frequency can be obtained by the first frequencyΩ 0: Ωj= j·Ω 0Pag. 16 CADMS - Lecture Notes(Contents link) CHAPTER 1. FEEDFORWARD AND FEEDBACK CONTROL Pag. 17Fig 1.6: Introduction - Fourier transform for the unknown force If we analyze a single harmonic, we can use the complex notation to solve the equation (1.4.7): fu( t) =|F u,0| · cos (Ωt+ Ψ) =Re |F u,0| · ei (Ωt+Ψ) =Re |F u,0| · ei Ψ ·ei Ωt
=Re Fu,0· ei Ωt
Fig 1.7: Introduction - Force harmonic in Complex plane m·¨ x+r·˙ x+k·x=f u( t) =Re Fu,0· ei Ωt
The generic solution is:xp,u( t) =Re Xp,u,0· ei Ωt
For simplicity we can consider the complex equation:m·¨ x+r·˙ x+k·x=F u,0· ei Ωt xp,u( t) =X p,u,0· ei Ωt ˙ x p,u( t) =iΩX p,u,0· ei Ωt ¨ x p,u( t) =−Ω2 Xp,u,0· ei Ωt −mΩ2 +irΩ +k
·X p,u,0·  ei Ωt =F u,0·  ei Ωt We can explicit the Frequency Response Function (FRF):(Contents link)CADMS - Lecture Notes Pag. 17 Pag. 18 CHAPTER 1. FEEDFORWARD AND FEEDBACK CONTROLG (iΩ) =X p,u,0F u,0= 1− mΩ2 +irΩ +k= |G(iΩ)| ·eiφ (iΩ) Let’s represent the modulus and the phaseFRF:Fig 1.8: Introduction - Modulus and Phase of FRF The solutionx p,u( t)is: xp,u( t) =Re xp,u,0· ei Ωt
=Re G(iΩ)·F u,0· ei Ωt
=Re |G(iΩ)| ·eiφ (iΩ) · |F u,0| · ei Ψ ·ei Ωt =Re |G(iΩ)| · |F u,0| · ei Ωt ·eiφ (iΩ) ·ei Ψ =Re |G(iΩ)| · |F u,0| · ei [Ωt+φ(iΩ)+Ψ] xp,u( t) =|G(iΩ)| · |F u,0| · cos [Ωt+φ(iΩ) + Ψ] 1.4.3FF: steady-state related to the target Now let’s analysis the particular solutionx p,r( t)related to the control forcef c: m·¨ x+r·˙ x+k·x=f c= b m·¨ x r+ b r·˙ x r+b k·x r(1.4.8) Let’s apply again the Fourier Transform xr=+ ∞ X j=1| X r,0| cos (Ω jt + Ψ j) fc( t) =b m·¨ x r+ b r·˙ x r+b k·x r =+ ∞ X j=1 −b mΩ2 +ib rΩ +b k ·X r,0· ei Ω jt =+ ∞ X j=1F c,0,j· ei Ω jt xp,r=+ ∞ X j=1Re Xp,r,0,j· ei Ω jt
=+ ∞ X j=1Re G(iΩ j) ·F c,0,j· ei Ω jt
Pag. 18 CADMS - Lecture Notes(Contents link) CHAPTER 1. FEEDFORWARD AND FEEDBACK CONTROL Pag. 19Fig 1.9: Introduction - Fourier Transform to the control force xr= X r,0· ei Ωt xp,r= X p,r,0· ei Ωt fc( t) = −b mΩ2 +ib rΩ +b k |{z} R(iΩ)· X r,0 |{z} Fc,0· ei Ωt We can define the functionR(iΩ): R(iΩ) =F c,0X r,0= −b mΩ2 +ib rΩ +b k xp,r= G(iΩ)·F c,0· ei Ω =G(iΩ)·R(iΩ)·ei Ω if and only if we have a very good estimation of the system parameters, we haveb m=m,b r=rand bk=k, soG(iΩ)·R(iΩ) = 1and: xp,r= X r,0· ei ΩFig 1.10: Introduction - Representation of the modulus of R(iΩ)(Contents link)CADMS - Lecture Notes Pag. 19 Pag. 20 CHAPTER 1. FEEDFORWARD AND FEEDBACK CONTROLOmega maxis an important parameter: using an F Fcontrol system we can better represent the response only for the frequency belowOmega max. As represented in the chart ofR(iΩ)if we haveOmega max≪ ω 0because the actuators cannot represent the following situations: •Resonance zone:Omega max∼ ω 0the actuator must provide a force near to zero with big displacement •Seismographic zone:Omega max≫ ω 0the actuator must provide a very big force with small displacement Most of actuators work with a linear relation between force and displacement (for big displacement they apply big forces and for small displacement they apply small forces), and this relationship is in the quasi-static zone. 1.5FB Analysis Let’s consider a 1 dof systemFig 1.11: Introduction - Feedforward Dynamic Model and the block diagram of theF Bcontrol system with the hypothesis of ideal actuator:x rFB ontrollerl 0 fd+ f uSystemx+e ,˙ ef c- Fig 1.12: Introduction - Feedback analysis block scheme eand˙ eare the errors defined as: e=x r− x˙ e= ˙x r− ˙ x The expression of the control forcef cbecomes: fc= k p· e(t) +k d· ˙ e(t) =k p· (x r− x) +k d· ( ˙x r− ˙ x)(1.5.1) wherek pand k dare, respectively, the proportional and the derivative gain, they are both positive and constantLet’s use the d’Alembert principle to write the equation of motion and the expression (1.5.1): m·¨ x+r·˙ x+k·x=f c+ f u= k p· (x r− x) +k d· ( ˙x r− ˙ x) +f u m·¨ x+ (r+k d) |{z} rt· ˙ x+ (k+k p) |{z} kt· x=f u+ k p· x r+ k d· ˙ x r m·¨ x+r t· ˙ x+k t· x=f u+ k p· x r+ k d· ˙ x r(1.5.2)Pag. 20 CADMS - Lecture Notes(Contents link) CHAPTER 1. FEEDFORWARD AND FEEDBACK CONTROL Pag. 21r tand k tare, respectively, the global damping and the global stiffness. The inputs of (1.5.2) are f u andx r. The solution of the motion is: x(t) =x h( t) +x p,xr( t) +x p,u( t)(1.5.3) •x h( t): is the transient response obtained by the solution of the homogeneous solution that depends only by the physical parametersm,randk •x p,xr( t): is the particular solution for the steady-state response related by the targetx r •x p,u( t): is the particular solution for the steady-state response related by the unknown forcef u 1.5.1FB: transient response Let’s consider the homogeneous equation: m·¨ x+r t· ˙ x+k t· x= 0(1.5.4) The trial solution is: xh( t) =x h,0· eλt λ∈C m·λ2 +r t· λ+k t
·x h,0·  eλt = 0(1.5.5) The non-trivial solution of (1.5.5) is the solution of the characteristic equation: m·λ2 +r·λ+k= 0λ 1/2= −r t2 m±r ry2 m 2 −k tm ω0,c=rk tm : natural frequencyh c=r t2 mω0: damping factorα c=r t2 m= h·ω 0,c λ1/2= −α c± ω 0,c·ph 2 c− 1 =              h c< 1underdamped control led system hc= 1 critical control led system hc> 1overdamped control led system •Underdamped controlled system:h c< 1the frequency of the damped system is ω c= ω 0,c·p1 −h2 c. The solution is: x(t) =x 0,1· eλ 1t +x 0,2· eλ 2t =e− α ct ·[A·cos (ω ct ) +B·sin (ω ct )] In the figure (1.13) is represented the transient response of the solution. The motion is oscillating because the roots are complex and conjugateλ 1/2= −α c± iω cand they are represented on the Gauss plane in figure (1.14)(Contents link)CADMS - Lecture Notes Pag. 21 Pag. 22 CHAPTER 1. FEEDFORWARD AND FEEDBACK CONTROLFig 1.13: Introduction - Solution of the underdamped controlled system Fig 1.14: Introduction - Roots of underdamped system on Gauss plane We can analyze how the roots position on the Gauss plane is affected by the constantsk pand k d:(a) variation byk p(b) variation byk pand k d Fig 1.15: Introduction - Variation of the position on the roots If we increase only the proportional gaink p, the oscillations are faster because the frequency is increased. In we increase both the proportional and derivative gaink pand k d, the oscillations are faster and also they last less becauseα cis increased. •Critical system:h= 1and Overdamped system:h >1These cases will not be covered in this chapter. This argument is part of the course "Meccanica del le Vibrazioni", please see [1] at section(2.4.1)at page 125 or in the appendix (A) at page213 1.5.2FB: steady-state related to the unknown force Now let’s analysis the particular solutionx p,u( t)related to the unknown force (disturbance)f u( t). We will apply again the Fourier Transform:Pag. 22 CADMS - Lecture Notes(Contents link) CHAPTER 1. FEEDFORWARD AND FEEDBACK CONTROL Pag. 23m ·¨ x+r t· ˙ x+k t· x=f u( t) =F u,0· ei Ωt (1.5.6) The trial solution is again:x p,u= X p,u,0· ei Ωt . Let’s substitute it: −mΩ2 +ir tΩ + k t
·X p,u,0·  ei Ωt =F u,0·  ei Ωt We can define the functionD(jΩ)as the FRF of the unknown motion of the FB system. It is very similar toG(jΩ): D(jΩ) =X p,u,0F u,0= 1− mΩ2 +ir tΩ + k t= 1− mΩ2 +i(r+k d) Ω + ( k+k p)Fig 1.16: Introduction - Modulus of D(iΩ)transfer function We can analyze how the modulus ofD(iΩ)is affected by the constantsk pand k d:(a) variation byk p(b) variation byk pand k d Fig 1.17: Introduction - Variation of theD(iΩ) Ifk p↑ , we obtainω c,0↑ andh c↓ . Ifk d↑ , we obtainh c↓ . Fork p→ ∞ the FRFD→0, so the effects of the disturbance are totally neglected. A proper selection of the gains in FB control can reduce the effects of disturbance 1.5.3FB: steady-state related to the target Now let’s analysis the particular solutionx p,r( t)related to the control forcef c: m·¨ x+r t· ˙ x+k t· x=f c= k p· x r+ k p· ˙ x r(1.5.7) We can apply again the Fourier Transform to analyze each harmonic: xr= X r,0· ei Ωt xp,r= X p,r,0· ei Ωt(Contents link)CADMS - Lecture Notes Pag. 23 Pag. 24 CHAPTER 1. FEEDFORWARD AND FEEDBACK CONTROL −mΩ2 +ir tΩ + k t
·X r,0·  ei Ωt = (ik dΩ + k p) ·X p,r,0·  ei Ωt We can define the transfer functionL(iΩ)and let’s plot its modulus: L(iΩ) =X p,r,0X r,0= ik dΩ + k p− mΩ2 +ir tΩ + k t The target is to havex r= x, soL= 1Fig 1.18: Introduction - Modulus of L(iΩ)transfer function We can analyze how the modulus ofD(iΩ)is affected by the constantsk pand k d:(a) variation byk p(b) variation byk pand k d Fig 1.19: Introduction - Variation of theL(iΩ) Ifk p↑ , we obtainω c,0↑ andh c↓ . Ifk d↑ , we obtainh c↓ . The FB control system can approximate frequency near and lower than the natural control frequencyPag. 24 CADMS - Lecture Notes(Contents link) Pag. 25 II-Stability Analysis (Contents link)CADMS - Lecture Notes Pag. 25 Pag. 27 2.Stability Analysis: 1 dof System 2.1Stability: definition The stability analysis consists in the study the behaviour of the system in response to a perturbation. If we perturb a system that works around a position, if it returns to that position after the perturbation the system is stable, otherwise it is unstable. The position around that the system work could be: •Static Equilibrium Position (the system doesn’t move)x=x 0and ˙ x= ¨x= 0 •Steady-State position˙ x= ˙x 0and ¨ x The theory of Lyapunov that study the stability analysis is explained in appendix (D) at page229 2.2Force Fields Let’s consider a 1 dof dumped system like represented in figure (2.1)Fig 2.1: Stability Analysis - 1dof - 1 dof damped system This system in moved by the control forcef c, so the equation of the motion is: m¨ x+r˙ x+kk=f c(2.2.1) Now we recap the response of the system by each kind of control:feedforwardcontrol (FF) and feedbackcontrol (FB) F F F B fc= b m¨ x r( t) +b r˙ x r( t) +b kxr( t)f c= k p( x p− x) +k d( x d− x) fc= f c( t)f c= f c( x,˙ x, t) m¨ x+ (r+r d) ˙ x+ (k t+ k p) x=k px r+ k d˙ x r We can see that the control force that acts on the system with FF control is only a time-depend function(f c= f c( t)), so the system is not affected by instability. Instead if consider the FB control, the control force could be also a force field. A force field can affect the stability of the system. A generic force field has the following form: f=f(x,˙ x,¨ x, t)(2.2.2) and we can distinguish two type of force field: •Conservative force field (i.e. gravitational field)(Contents link)CADMS - Lecture Notes Pag. 27 Pag. 28 CHAPTER 2. STABILITY ANALYSIS: 1 DOF SYSTEM• Non-Conservative force field (i.e. control forces, friction forces,...) Usually the control forces are non-conservative force fields and, it will be proved further on, only non-conservative force field can affect the stability of the system. 2.3Stability Analysis: 1 dof system To study the stability of 1 dof system we can follow a procedure:1.Write the non-linear equation of motion of the system (by using the d’Alembert principle or theLagrange equation) 2.Calculate the static equilibrium position (there can be more than one) 3.Linearize the non-linear equation of motion around each static equilibrium position calculated 4.Study the stability of the linearized system 5.Use the Lyapunov Theorem (D) at page229to study the stability of the non-linear system aroundthat specific static equilibrium position. In particular referring to the table (D.1) at page230 Let’s follow this list to study the stability a generic 1 dof system represented in figure (2.2). The system is forced by a force field that depends only by the displacement and the speed.Fig 2.2: Stability Analysis - 1dof - 1 dof damped system forced by a force field and representation of the internal forces using d’Alembert principle [1] Non-linear equation of motion Let’s use the d’Alembert principle like represented in figure2.2to write the non-linear equation of motion: m·¨ x+r·˙ x+k·x |{z} linear= f(x,˙ x) |{z} non−linear(2.3.1) [2] Static equilibrium position To calculate the static equilibrium position we have to impose the following conditions at the equation (2.3.1): x=x 0˙ x= 0 ¨x= 0(2.3.2) Now we have a non-linear algebraic equation to calculate the static equilibrium positionx 0 k·x 0= f(x 0, 0)(2.3.3) Remind thatx 0can be more than one as represented in figure (2.3)Pag. 28 CADMS - Lecture Notes(Contents link) CHAPTER 2. STABILITY ANALYSIS: 1 DOF SYSTEM Pag. 29Fig 2.3: Stability Analysis - 1dof - Example of calculation of the static equilibrium positions The linearization of the system must be done around eachx 0,i [3] Linearization of the equation of motion Now we have to linearize the non-linear equation of motion (2.3.1). In this problem, the left part of the equation is already linear (this cannot be true in every case), but the right part, that is to say the force field, is non linear, so linearize it around the static equilibrium position. To linearize the force field let’s use the Taylor’s expansion at the first order: f(x,˙ x)≃f(x 0, 0) +∂ f∂ x     0 |{z} −k F· (x−x 0) +∂ f∂ ˙ x    0 |{z} −r F·  ˙ x− *0 x0 f(x,˙ x) =f(x 0, 0)−k F· (x−x 0) −r F· ˙ x(2.3.4) kFand r Fare constant. The rate of the straight line in the intersection between the force field and the elastic force as represented in figure (2.4)Fig 2.4: Stability Analysis - 1dof - Linearization of the force field Let’s write the combination between the equation (2.3.1) and (2.3.4) m·¨ x+r·˙ x+k·x≃f(x 0, 0)−k F· (x−x 0) −r F· ˙ x(2.3.5) Now we have to substitute the global variables(x,˙ x,¨ x)with the variables of the perturbed motion  e x,˙ e x,¨ e x     e x=x−x 0 ˙ e x= ˙x ¨ e x= ¨x m·¨ e x+r·˙ e x+k·e x+   k·x 0=   f(x 0, 0)−k F· e x−r F·˙ e x m·¨ e x+r·˙ e x+k·e x=−k F· e x−r F·˙ e x m·¨ e x+ (r+r F) ·˙ e x+ (k+k F) ·e x= 0 m·¨ e x+r t·˙ e x+k t· e x= 0(2.3.6)(Contents link)CADMS - Lecture Notes Pag. 29 Pag. 30 CHAPTER 2. STABILITY ANALYSIS: 1 DOF SYSTEMThe equation (2.3.6) is an homogeneous ODE. The stability is related to the transient response of the system and depends by the sign ofr tand k t rt⋚ 0k t⋚ 0 rF⋚ 0k F⋚ 0 •r t< 0ifr F< 0and|r F| >|r| •k t< 0ifk F< 0and|k F| >|k| [4] Stability of the linearized system The trial solution of the equation (2.3.6) is: e x=e x 0· eλt λ2 ·m+λ·r t+ k t= 0 characteristic equation λ1/2= −r t2 m±r rt2 m 2 −k tm = −α±√∆ α=r t2 m∆ = rt2 m 2 −k tm = ω2 0· h2 −1
(2.3.7) ω0=rk tm h =r t2 mω 0 e x(t) =e x 0,1· eλ 1t +e x 0,2· eλ 2t We have to study the sign ofr tand k tand the behaviour of the stability of the system. All the cases are collected in the table (2.1) CasesIIIIIIIVVVI r t+++--- k t++-++- h =r t2 mω 0≥ 1< 1n.d.≥ 1< 1n.d. Tab 2.1: Stability Analysis - 1dof - 1 dof stability cases Case I In this case we have:r t> 0,k t> 0and|h| ≥1with the definition described in (2.3.7), so: h≥1⇒∆>0k t> 0⇒  √∆   < |α|r t> 0⇒α >0 λ1/2=( −|α 1| −|α 2| both the roots are real and negative e x(t) =e x 0,1· e−| α 1| t +e x 0,2· e−| α 2| t In figure (2.5) there is the representation of the roots on Gauss plane and the transient response of the displacemente xPag. 30 CADMS - Lecture Notes(Contents link) CHAPTER 2. STABILITY ANALYSIS: 1 DOF SYSTEM Pag. 31Fig 2.5: Stability Analysis - 1dof - Roots in Gauss plane and transient response to case I In the case thatr t> 0,k t> 0and|h| ≥1the system isasymptotical ly stable Case II In this case we have:r t> 0,k t> 0and|h| 0,k t> 0,|h| ≥1andh̸ = 0the system isasymptotical ly stable In the same condition, but withh= 0, we have: rt= 0 α= 0 m·¨ e x+k t· e x= 0(2.3.8) λ1/2= − ±iω 0(Contents link)CADMS - Lecture Notes Pag. 31 Pag. 32 CHAPTER 2. STABILITY ANALYSIS: 1 DOF SYSTEMThe roots are complex conjugate with a null real part, so they are pure imaginary. The frequency of the oscillations is the natural frequency of the equation (2.3.8) e x(t) =A·cos (ω 0t ) +B·sin (ω 0t ) In figure (2.7) there is the representation of the roots on Gauss plane and the transient response of the displacemente xFig 2.7: Stability Analysis - 1dof - Roots in Gauss plane and transient response to case II with h= 0 In the case thatr t> 0,k t> 0andh= 0the system isstable, but not asymptotical ly Case III In this case we have the following conditions:                r t> 0⇒α >0 kt< 0⇒k F< 0,|k F| > k⇒∆ = rt2 m 2 −k tm |{z} 0> 0⇒ |√∆ |>|α| λ1/2= −α±√∆ = ( |α 1| −|α 2| Both roots are real, but one is positive and one is negative e x(t) =e x 0,1· e| α 1| t |{z} →+∞+ e x 0,2· e−| α 2| t |{z} →0 In figure (2.8) there is the representation of the roots on Gauss plane and the transient response of the displacemente xPag. 32 CADMS - Lecture Notes(Contents link) CHAPTER 2. STABILITY ANALYSIS: 1 DOF SYSTEM Pag. 33Fig 2.8: Stability Analysis - 1dof - Roots in Gauss plane and transient response to case III In the case thatr t> 0andk t< 0the system is affected byStatic Instability Case IV In this case we haver t< 0,k t> 0andh≥1 h≥1⇒∆>0r t< 0⇒α 0⇒ |√∆ |0± v u u u t rt2 m 2 −k tm |{z} >0= ( |α 1| |α 2| Both the roots are real and positivee x(t) =e x 0,1· e| α 1| t +e x 0,2· e| α 2| t In figure (2.9) there is the representation of the roots on Gauss plane and the transient response of the displacemente xFig 2.9: Stability Analysis - 1dof - Roots in Gauss plane and transient response to case IV In the case thatr t< 0,k t> 0andh≥1the system is affected byStatic Instability Case V In this case we haver t< 0,k t> 0andh 0> 0 βis always positive, soλ2 I ,I I= −γ±√β are real numbers Case 1: Global Stiffnes Matrix positive definite [K t] is positive definite, so ∆ =det([K t]) >0k xx, k yy> 0 γ=k xx+ k yy2 > 0 β=γ2 −∆ |{z} >0        →   pβ   < |γ| The roots are: λ2 I ,I I= −γ±pβ =( − |σ 1| − |σ 2|λ 1/2±p− | σ 1| =±iω 1 λ3/4±p− | σ 2| =±iω 2 The two square-roots are both real and negative, so all the solutions are complex conjugate:(a) Square-Roots(b) Solutions Fig 3.3: Stability Analysis - 2-n dofs - CFF: case 1 ω1and ω 2are the two natural frequencies of the system affected by the force field e z( t) =4 X j=1e z0 ,j· eλ jt =e z0 ,1· eλ 1t +e z0 ,2· eλ 2t +e z0 ,3· eλ 3t +e z0 ,4· eλ 4t =e z0 ,1· eiω 1t +e z0 ,2· e− iω 1t |{z} ∗(t)+ e z0 ,3· eiω 2t +e z0 ,4· e− iω 2t |{z} ∗∗(t) IfFis a CFF and [K t] is positive definite, the system isstable, but not asymptotical ly(Contents link)CADMS - Lecture Notes Pag. 41 Pag. 42 CHAPTER 3. STABILITY ANALYSIS: 2-N DOFS SYSTEMFig 3.4: Stability Analysis - 2-n dofs - CFF: Case 1 - Representation of the solutions Case 2: Global Stiffnes Matrix not positive definite The matrix[K t] is not positive definite, so we have to consider the sign of the determinant of the global stiffness matrix Determinant less than zero det([K t]) |γ| λ2 I ,I I= −γ±pβ =( +|σ 1| − |σ 2|λ 1/2±p+ |σ 1| =±α λ3/4±p− | σ 2| =±iω Both the square-roots are both real, one positive and one negative, so the solutions of the positive square-root are real, one positive and one negative, and the solutions of the negative square-root are complex conjugate(a) Square-Roots(b) Solutions Fig 3.5: Stability Analysis - 2-n dofs - CFF: case 2 with determinant less than zeroPag. 42 CADMS - Lecture Notes(Contents link) CHAPTER 3. STABILITY ANALYSIS: 2-N DOFS SYSTEM Pag. 43e z( t) =4 X j=1e z0 ,j· eλ jt =e z0 ,1· eλ 1t +e z0 ,2· eλ 2t +e z0 ,3· eλ 3t +e z0 ,4· eλ 4t =e z0 ,1· e+ αt |{z} ∗(t)+ e z0 ,2· e− αt |{z} ∗∗(t)+ e z0 ,3· eiωt +e z0 ,4· e− iωt |{z} ∗∗∗(t)Fig 3.6: Stability Analysis - 2-n dofs - CFF: Case 2 - Determinant less than zero - Representation of the solutions In this case, because of∗(t), the system isstatical ly unstable "A system is statical ly unstable if there is at least one positive real root" Determinant greater than zero∆ =det([K t]) = k xx· k yy− k xy· k yx> 0⇒k xx, k yy< 0 γ=k xx+ k yy2 < 0β=γ2 −∆ |{z} >0⇒   pβ   < |γ| λ2 I ,I I= −γ±pβ =( +|σ 1| +|σ 2|λ 1/2±p+ |σ 1| =±α 1 λ3/4±p+ |σ 2| =±α 2 Both the square-roots are real and positive, so all the solutions are real are real coupled positive and negative. The solution is: e z( t) =4 X j=1e z0 ,j· eλ jt =e z0 ,1· eλ 1t +e z0 ,2· eλ 2t +e z0 ,3· eλ 3t +e z0 ,4· eλ 4t =e z0 ,1· e+ α 1t |{z} ∗(t)+ e z0 ,2· e− α 1t |{z} ∗∗(t)+ e z0 ,3· e+ α 2t |{z} ∗∗∗(t)+ e z0 ,4· e− α 2t |{z} ∗∗∗∗(t)(Contents link)CADMS - Lecture Notes Pag. 43 Pag. 44 CHAPTER 3. STABILITY ANALYSIS: 2-N DOFS SYSTEM(a) Square-Roots(b) Solutions Fig 3.7: Stability Analysis - 2-n dofs - CFF: case 2 with determinant greater than zeroFig 3.8: Stability Analysis - 2-n dofs - CFF: Case 2 - Determinant greater than zero - Representation of the solutions Summary: conservative force field•[K t] is positive definite−→stable, but not asymptotic •[K t] is not positive definite−→static instabilityPag. 44 CADMS - Lecture Notes(Contents link) CHAPTER 3. STABILITY ANALYSIS: 2-N DOFS SYSTEM Pag. 453.6Stability analysis: Non-Conservative Force Field (NCFF) IfFis a non-conservative force field, the relation (E.1.8) at page232is not valid, so k xy̸ =k yxand the matrix[K t] is not symmetric. The termβ⋚0 Case 1:β >0 Determinant greater than zero ∆ =det([K t]) >0β=γ2 −∆ |{z} >0  pβ   < |γ| •γ >0λ 2 I ,I I= −γ±pβ =( − |σ 1| − |σ 2|λ 1/2±p− | σ 1| =±iω 1 λ3/4±p− | σ 2| =±iω 2 so, the system isstable, but not asymptotical ly •γ n iii)poles̸ =zeros−→NandDat minimum order iv)f∈R Thesislim t→0f (t) =f(0) = lim s→+∞s ·F(s) important for assessing initial condition Proof g(t) =˙ f(t) L[g(t)] =Lh ˙ f(t)i G(s) =s·F(s)−f(0) lim s→+∞G (s) = lim s→+∞[ s·F(s)−f(0)] 0 = lims→+∞[ s·F(s)]−lim s→+∞f (0) 0 = lims→+∞[ s·F(s)]−f(0) lim t→0f (t) =f(0) = lim s→+∞[ s·F(s)] Q.E.D. 4.4.7Final Value Theorem Hypothesis i)F(s) =N (s)F (s)= a nsn +a n−1sn −1 +· · ·+a 0b msm +b m−1sm −1 +· · ·+b 0 . roots ofN(s)(solution ofN(s) = 0) are calledZEROS roots ofD(s)(solution ofD(s) = 0) are calledPOLES ii)m > n iii)poles̸ =zeros−→NandDat minimum order iv)f∈R v)Re(poles)1⇒ SIMOsystem (single-input multi-output) m>1,p=1⇒ MISOsystem (multi-input single-output) m>1,p>1⇒ MIMOsystem (multi-input multi-output) The system (5.1.4) is in time-domain. Let’s apply what seen in chapter (4) L"( ˙ x( t) = [A]·x( t) + [B]·u( t) y( t) = [C]·x( t)# (s)−→( L[ ˙x( t) = [A]·x( t) + [B]·u( t)] (s) L y( t) = [C]·x( t) (s) −→( s·X( s)−x(0) = [ A]·X( s) + [B]·U( s) Y( s) = [C]·X( s)−→( (s·[I]−[A])·X( s) =x(0) + [ B]·U( s) Y( s) = [C]·X( s) We can expressX( s)andY( s)                     X( s) = (s·[I]−[A])− 1 ·x(0) + ( s·[I]−[A])− 1 ·[B]·U( s)Laplace transform of the system’s response(state variables) Y( s) = [C]·(s·[I]−[A])− 1 ·x(0) |{z} free motion contribution+ [ C]·(s·[I]−[A])− 1 ·[B]·U( s) |{z} forced motion contributionLaplace transform of the system’s output We can define thematrix of transfer function[G(s)] [G(s)] p×m= [ C] p×n· (s·[I]−[A])− 1 n×n· [B] n×m The expression ofY( s)becomes: Y( s) = [C]·(s·[I]−[A])− 1 ·x(0) + [ G(s)]·U( s) If we consider homogeneous initial conditions(x(0) = 0) the expression ofY( s)becomes: Y( s) = [G(s)]·U( s) [G(s)]is the generic transfer function and each elementijcan be calculated as a ratio of a numerator and a denominator. We have to remark that the matrices[B]and[C]are constant matrices and, referring to appendixFat page233, the matrix(s·[I]−[A])− 1 is equal to a matrix of the algebraic component, that each element is divided by the determinant of the same matrix(s·[I]−[A]), so we can sayPag. 58 CADMS - Lecture Notes(Contents link) CHAPTER 5. STATE-SPACE MODEL OF A SYSTEM Pag. 59[ G(s)] =[ G(s)∗ ]det (s·[I]−[A])= [ C]·(s·[I]−[A])∗ ·[B]det (s·[I]−[A]) Each element is calculated: Gij( s) =N G( s)D G( s)=G ∗ ij( s)det (s·[I]−[A])= Y i( s)U j( s) We definepolesas: DG( s) = 0 det(s·[I]−[A]) = 0 Instead we definezerosas: NG( s) = 0 G∗ ij( s) = 0 The poles are properties only of the system, but the zeros depends by the INPUT-OUTPUT combi- nation. For simplicity let’s consider a SISO system, so the relation between the input and the output is Y(s) =G(s)·U(s) we can use also block representation for all the type of systems:G (s) (1×1)U (s)Y (s)(a) SISO[ G(s)] (1×p)U (s)Y 1( s)Y 2( s). . .Y p−1( s)Y p( s)(b) SIMO[ G(s)] (m×1)U 1( s)U 2( s). . .U m−1( s)U m( s)Y (s)(c) MISO[ G(s)] (m×p)U 1( s)U 2( s). . .U m−1( s)U m( s)Y 1( s)Y 2( s). . .Y p−1( s)Y p( s)(d) MIMO Fig 5.1: Stability Analysis - S.S. Model - TF block representation for all systems(Contents link)CADMS - Lecture Notes Pag. 59 Pag. 60 CHAPTER 5. STATE-SPACE MODEL OF A SYSTEM5.2Compositions of TF 5.2.1Series of TF Let’s considern−T Fin seriesT F 1U (s)T F 2· · · T F nY n( s)U (s)T F seriesY n( s)Y 1( s)Y 2( s)Y n−1( s)Y i( s) =T F i· Y i−1( s)Y n( s) =n Y i=1T F i· U(s) =T F series· U(s) 5.2.2Parallel of TF Let’s considern−T Fin parallelT F 1T F 2. . .T F n. U (s)+ Y (s)Y 1( s)Y 2( s)Y n( s)U (s)T F parallelU (s)Y i( s) =T F i· U(s)Y(s) =n X i=1T F i· U(s) =T F parallel· U(s) 5.2.3Feedback loop TF Now let’s consider a simple feedback loop block diagramG (s)H (s)Y (s)U (s)U (s)L (s)U (s)+ - Y (s) =G(s)·[U(s)−H(s)·Y(s)] Y(s) =G(s)·U(s)−G(s)·H(s)·Y(s) [1 +G(s)·H(s)]·Y(s) =G(s)·U(s) L(s) =U (s)F (s)= G (s)1 + G(s)·H(s) L(s)is the close loop transfer function,G(s)is the forward path andH(s)is the feed-back pathPag. 60 CADMS - Lecture Notes(Contents link) Pag. 61 6.Transfer Function There are many ways to represent graphically a Transfer Function. The frequency response function (FRF)as a complex function of the variableΩthat is the frequency of the oscillations. G(iΩ) 6.1Modulus and Phase representation TheFRFfunction can be evaluated as a product of a modulus and a phase, that both depend by the frequency: G(iΩ) =|G(iΩ)| ·eiϕ (Ω) Let’s consider for example a2nd order oscillating system, theFRFis: GH(iΩ) =1− mΩ +irΩ +k The expressions of the modulus and the phase are: |GH(iΩ)|=1q ( −mΩ2 +k)2 + (Ωr)2ϕ (Ω) = arctan −rΩ− mΩ2 +k We can plot these functions:Fig 6.1: Stability Analysis - T.F. - Plot of Modulus and Phase of the FRF (Contents link)CADMS - Lecture Notes Pag. 61 Pag. 62 CHAPTER 6. TRANSFER FUNCTION6.2Bode Diagram A second way to represent the modulus and the phase of theFRFis calledBode diagramor logarithmic plot. It consists again into plot two separated diagrams, one for the modulus and one for the phase. Both of them diagrams have on the horizontal axis the base 10 logarithmic of the frequencyΩ. The modulus of the transfer function is expressed indB (decibels): 20·log 10| G(iΩ)| The Bode diagram is very used to draw approximately the behaviour of the real curve with multiple straight lines that represent theAsymptotic Bode Diagram. Now let’s plot here the Bode diagram, with also the respectively asymptotic one, of the system previously introduced (1 dof2nd order system). In figure (6.2)the blue curve is the real curvewhilethe red curve is the asymptotic curve.Fig 6.2: Stability Analysis - T.F. - Bode Diagram of 2nd order system The asymptotic curve represents the real curve far from the eigenfrequencies, but near them, as shown in the plot, the dynamic amplifications are neglected. We can consider a generic TF (6.2.1) with many poles and many zeros, so we imagine to decompose the numerator and the denominator in products of elementary binomials (which represent the1st order systems) and trinomials (which represent the2nd order systems) T F(s) =µ gs m·Q zk(1 + τ zk· s)·Q zk 1 +2 h zkω 0,z k· s+s 2ω 2 0,z k!Q pk(1 + τ pk· s)·Q pk 1 +2 h pkω 0,p k· s+s 2ω 2 0,p k! (6.2.1) mugis the static gain, mis the number of poles in the origin, the term(1 +τ k· s)represents a1st order system an at last the term 1 +2 h kω 0,k· s+s 2ω 2 0,k! represents a2nd order system. We can substitute s=iΩ: T F(iΩ) =µ g( iΩ)m ·Q zk(1 + iΩτ zk) ·Q zk 1 +2 ih zkω 0,z k· Ω−Ω 2ω 2 0,z k!Q pk(1 + iΩτ pk) ·Q pk 1 +2 ih pkω 0,p k· Ω−Ω 2ω 2 0,p k! (6.2.2) We can calculate the modulus expressed in decibel of (6.2.2), we have to remember the properties of the modulus and the logarithms20·log 10( |T F(iΩ)|) =|T F(iΩ)| dB:Pag. 62 CADMS - Lecture Notes(Contents link) CHAPTER 6. TRANSFER FUNCTION Pag. 63| T F(iΩ)| dB= 20 ·log 10              µ g( iΩ)m ·Q zk(1 + iΩτ zk) ·Q zk 1 +2 ih zkω 0,z k· Ω−Ω 2ω 2 0,z k!Q pk(1 + iΩτ pk) ·Q pk 1 +2 ih pkω 0,p k· Ω−Ω 2ω 2 0,p k!               = 20·log 10     | µ g|| (iΩ)m |·Q zk| (1 +iΩτ zk) | ·Q zk     1 +2 ih zkω 0,z k· Ω−Ω 2ω 2 0,z k!     Q pk| (1 +iΩτ pk) | ·Q pk     1 +2 ih pkω 0,p k· Ω−Ω 2ω 2 0,p k!           = 20·log 10( |µ g| ) + 20·log 10    1( iΩ)m    + +X zk20 ·log 10( |1 +iΩτ zk| ) +X zk20 ·log 10     1 +2 ih zkω 0,z k· Ω +−Ω 2ω 2 0,z k    ! + +X pk20 ·log 10    11 + iΩτ pk    +X pk20 ·log 10            11 + 2 ih pkω 0,p k· Ω−Ω 2ω 2 0,p k             = 20·log 10( |µ g| )−m·20·log 10( |iΩ|) + +X zk20 ·log 10( |1 +iΩτ zk| ) +X zk20 ·log 10     1 +2 ih zkω 0,z k· Ω +−Ω 2ω 2 0,z k    ! + −X pk20 ·log 10( |1 +iΩτ pk| )−X pk20 ·log 10     1 +2 ih pkω 0,p k· Ω− −Ω 2ω 2 0,p k    ! And we can evaluate the phase of (6.2.2),∠(T F(iΩ)): ∠(T F(iΩ)) =∠     µ g( iΩ)m ·Q zk(1 + iΩτ zk) ·Q zk 1 +2 ih zkω 0,z k· Ω−Ω 2ω 2 0,z k!Q pk(1 + iΩτ pk) ·Q pk 1 +2 ih pkω 0,p k· Ω−Ω 2ω 2 0,p k!      =· · · =∠(µ g) −m·∠(iΩ) + +X zk∠ (1 +iΩτ zk) +X zk∠ 1 +2 ih zkω 0,z k· Ω +−Ω 2ω 2 0,z k! + −X pk∠ (1 +iΩτ pk) −X pk∠ 1 +2 ih pkω 0,p k· Ω− −Ω 2ω 2 0,p k! We can analyze each single term and represent the modulus and the phase of the transfer function:•Constant gainT F (iΩ) =µ g |T F(iΩ)| dB= 20 ·log 10( |µ g| )∠(T F(iΩ)) =    0 ifµ g> 0 −πifµ g< 0(Contents link)CADMS - Lecture Notes Pag. 63 Pag. 64 CHAPTER 6. TRANSFER FUNCTION(a) µ g> 0(b) µ g< 0 Fig 6.3: Stability Analysis - T.F. - Bode Diagram for Constant gain •Pole in the originT F (iΩ) =1i Ω |T F(iΩ)| dB= −20·log 10(Ω) ∠(T F(iΩ)) =−π2 Fig 6.4: Stability Analysis - T.F. - Bode diagram for pole in the origin •1st order pole: T F(iΩ) =11 + iτΩ •1st order zero: T F(iΩ) = 1 +iτΩ• 2st order pole: T F(iΩ) =11 + 2 ihω 0Ω −Ω 2ω 0 •2st order zero: T F(iΩ) = 1 +2 ihω 0Ω −Ω 2ω 0Pag. 64 CADMS - Lecture Notes(Contents link) CHAPTER 6. TRANSFER FUNCTION Pag. 65(a) 1st order pole(b) 1st order zero(c) 2nd order pole(d) 2nd order zero Fig 6.5: Stability Analysis - T.F. - Bode Diagram for: .(Contents link)CADMS - Lecture Notes Pag. 65 Pag. 66 CHAPTER 6. TRANSFER FUNCTION6.3Polar Diagram An other graphical representation of theFRFG(iΩ)is the polar diagram. It consists into a continuous line obtained plotting each single point of the modulus and the phase of theFRFas represented in figure (6.6a). The line is obtain by lettingΩvary from0to+∞. In figure (6.6b) is represented the polar diagram of a 1 dof2nd order system.(a) Single point(b) Polar plot Fig 6.6: Stability Analysis - T.F. - Polar diagram It is possible to draw a qualitative shape diagram from the Bode diagram 6.4Nyquist Diagram The Nyquist diagram is the last way to represent graphically theFRFwith the same definition of the polar diagram, but the frequencyΩvaries from−∞to+∞. It is equally to mirror the polar diagram respect to real axis. In figure (6.7) is represented the polar diagram of a 1 dof2nd order system.Fig 6.7: Stability Analysis - T.F. - Nyquist diagram Pag. 66 CADMS - Lecture Notes(Contents link) Pag. 67 7.Nyquist and Bode criteria 7.1Nyquist Criterion 7.1.1Criterion work in progress 7.1.2Special Cases•1 pole at the origin: let’s consider the variable s=δ·eiθ θ∈h −π2 , π2 i counterclockwise TheGH(s)function is: GH(s) =a nsn +a n−1sn −1 +· · ·+a 0s ·(b m−1sm −1 +· · ·+b 0)m > n lim s→0GH (s) = lim s→0a 0s ·b 0= lim s→0ks = lim s→0kδ ·eiθ= lim δ→0kδ |{z} +∞· e− iθ clockwiseFig 7.1: Stability Analysis - NY.&Bode criterion- 1 pole at the origin •r poles at the origin: let’s consider the variable s=δ·eiθ θ∈h −π2 , π2 i counterclockwise TheGH(s)function is: GH(s) =a nsn +a n−1sn −1 +· · ·+a 0s r ·(b m−rsm −r +· · ·+b 0)m > n(Contents link)CADMS - Lecture Notes Pag. 67 Pag. 68 CHAPTER 7. NYQUIST AND BODE CRITERIAlim s→0GH (s) = lim s→0a 0s r ·b 0= lim s→0ks r= lim s→0kδ r ·eirθ= lim δ→0kδ r |{z} +∞· e− irθ clockwise ofr2 rotationsFig 7.2: Stability Analysis - NY.&Bode criterion- r pole at the origin 7.2Bode Criterion 7.2.1Phase and Gain Margins Let’s consider a part of Nyquist diagram curve onGH−plane, we can trace an unitary circle. As shown in figure (7.3) we highlight two particular points: the intersection between the Nyquist curve and the unitary circle(Phase Margin), the intersection between the Nyquist curve and the real axis(Gain Margin)Fig 7.3: Stability Analysis - NY.&Bode criterion- Phase and Gain Margins on Nyquist diagram We must remember that the Bode diagram is logarithmic and the modulus is expressed in decibels:|GH| dB= 20 ·log 10| GH| Let’s define thegain crossover frequencyΩ gc, as represented in figure (7.4a), is the frequency to which the real modulus curve intersects the0dB−axisor: |GH(Ω gc) |= 1⇒ |GH(Ω gc) | dB= 0 Pro jecting this frequency on the phase diagram (7.4a) we intersect the phase curve. We define thephase marginρ mas:Pag. 68 CADMS - Lecture Notes(Contents link) CHAPTER 7. NYQUIST AND BODE CRITERIA Pag. 69ρ m≜ ∠ GH(Ω gc) −π(7.2.1)     ρ m> 0if∠GH(Ω gc) >−π ρm< 0if∠GH(Ω gc) 1 Gm> 0if|GH(Ω pc| 0 ρm> 0(7.2.5) 7.2.2.3Definition of: Robust Stability A system is robust stable if and only ifG m≫ 0andρ m≫ 0. (Practically the these conditions are valid if they assumeG m≥ 6dBandρ m≥ 30◦ )Pag. 70 CADMS - Lecture Notes(Contents link) Pag. 71 8.Root Locus 8.1Description The Root Locus is an indirect method to analyze the stability of a mechanical system. It consists to find the locus of the roots on the Gauss plane (or better thes-plane) of the characteristic equation of the closed-loop transfer function which is affected by a parameter (the gaink p) that varies from zero to infinity. Such a plot clearly shows the contributions of each open loop pole or zero to the locations of the closed loop poles. The root locus indicates in which way the open loop poles and zeros should be modified so that the response of the closed loop system matches stability and performance requirements. By using root locus analysis, it is possible to determine the value of the control gaink p: •that guarantees closed-loop system stability •that will make damping ratio of the dominant closed-loop poles as required Moreover, if the location of an open-loop pole or zero is a system variable, then the root locus method suggests the way to choose the location of an open-loop pole or zero. Let’s now summarize the general rules and procedures for constructing the root loci of a closed-loop system in the case of negative feedback.Y refk pG (s)Y (s)H (s)+E (s)F (s)- Fig 8.1: Stability Analysis - Root Locus - Block representation of closed-loop system in Laplace domain The closed-loop transfer function is (remind thatG(s) =N G/D Gand H(s) =N H/D H): L(s) =k pG (s)1 + k pH (s)G(s) =k pN G( s)D G( s)1 + k pN H( s)D H( s)N G( s)D G( s) =k pN G( s)D G( s)· D H( s)D G( s)D H( s)D G( s) +k pN H( s)N G( s) L(s) =k pN G( s)D H( s)D H( s)D G( s) +k pN H( s)N G( s) The characteristic equation of the feedback control system is:(Contents link)CADMS - Lecture Notes Pag. 71 Pag. 72 CHAPTER 8. ROOT LOCUS1 + k pG (s)H(s) = 0 1 +k pN G( s)D G( s)N H( s)D H( s)= 0 DG( s)D H( s) +k pN G( s)N H( s) = 0 DGH( s) +k pN GH( s) = 0 Let’s consider thatG(s)H(s)hasnpoles (the order ofD GH( s)) andmzeros (the order ofN GH( s)). The Root Locus represents how the poles of the closed-loop system (the poles ofL(s)) vary ask p from0and+∞. So let’s evaluate the characteristic equation at the extremities of the range[0; +∞): •k p= 0k pN GH( s) +D GH( s) =D GH( s) = 0 Fork p= 0 , the poles ofL(s)are equal to the roots ofD GH( s). So, the starting point of the root-locus branches (kp= 0) are the poles of the loop transfer function. •k p→ +∞k pN GH( s) +D GH( s)∼k pN GH( s) = 0 Fork p→ +∞, the poles ofL(s)are equal to the roots ofN GH( s). Thus, the root-locus branches end at the zeros of the loop transfer function. Ifn=m, all root-locus branches end at the zeros of the loop transfer function. Ifn > m, we have the constantq=n−mbranches end at infinity. If the gaink p> 0it is the case of negative-feedback. If the gaink p< 0it is the case of positive- feedback. The rules that we will follow for the negative case must be modified for the positive one. 8.2Negative Feedback We shall now summarize the basic rules for drawing the root locus for a negative feedback systemY refG (s)Y (s)H (s)+E (s)- Fig 8.2: Stability Analysis - Root Locus - Block representation of negative feedback We can write the characteristic equation and we can re-arrange it highlighting the zerosz i, the poles piand the control gain k p: 1 +G(s)H(s) = 0 1 +k p· (s+z 1) ·(s+z 2) · · · · ·(s+z m)( s+p 1) ·(s+p 2) · · · · ·(s+p n)= 0 with(n > m). Based on previous considerations, the following rules can be given for drawing the root locus of a closed-loop system in the case of negative feedback: 1.Place open loop poles and zeros in thes-plane 2.The root locus has as many branches as there are open-loop poles The characteristic equation is:kpN GH( s) +D GH( s) = 0Pag. 72 CADMS - Lecture Notes(Contents link) CHAPTER 8. ROOT LOCUS Pag. 73If the degree nof the polynomialD GH( s)is higher than the degreemof the polynomialN GH( s), the degree of the closed-loop system is equal to the degree of the open-loop system. Thus, the closed-loop system has as many poles as the open-loop one. Since the root locus represents how the poles ofL(s)vary as the control gaink pvaries from zero to infinity, then there arenbranches (a branch is just the trace one pole makes as the gaink pis varied) 3.The root locus is symmetric about the real axis As all roots of the characteristic equation are either real or complex conjugate pairs so that the root locus is symmetrical about the real axis 4.The root locus branches start from open-loop poles and end at open-loop zeros or at infinity See previous considerations 5.Real-axis segments of the root locus are to the left of an odd number of real-axis finite poles/zeros If the total number of poles and zeros to the right of a point on the real axis is odd, this point lies on the locus. Starting at+∞and moving along the real axis toward the left, the root locus lies on the real axis to the left of an odd number of real axis open loop poles or zeros (in any combination). Multiplicities of poles and zeros must be taken into account when counting. The locations of the open loop poles or zeros (right half plane, left half plane, or origin) does not matter; only the cumulative number of poles or zero (odd or even) is important. Poles or zeros off the real axis are not included in determining the real axis portion of the root locus. 6.AsymptotesIfn > m, there will beq=n−mbranches of the root locus going to infinity as kp→ +∞. For largek p, they will follow asymptotes that meet at a common point (centroid) on the real axis and make specified angles with respect to the positive real axis. The angles of asymptotes, θa, and the center of asymptotes, σ a, are given by: θa= ±kq π k = 1,3,5, . . . , q σ a=n P i=1p i−m P i=1z in −m wherep iand z iare the open-loop pole and zero locations, respectively. Complex poles and zeros are included in the calculation ofσ a. The angles of the asymptotes depend only onn−m, not on the actual locations of the poles or zeros. 7.Break-away and break-in points If the root locus on the real axis lies in the interval between two adjacent open-loop poles, there will always be a break-away point between the poles where the root locus leaves the real axis. If the root locus on the real axis lies in the interval between two adjacent open-loop zeros, there will always be a break-in point between the zeros where the root locus enters the real axis. All break-in and break-away points, can be found evaluating the value ofsat whichk ptakes on minimum or maximum values: kpN GH( s) +D GH( s) = 0⇒k p= −D GH( s)N GH( s) dkpds = 0 ⇒− NGH( s)·d (D GH( s))ds − d (N GH( s))ds · D GH( s)N 2 GH( s)= 0 ⇒N GH( s)·d (D GH( s))ds − d (N GH( s))ds · D GH( s) = 0(Contents link)CADMS - Lecture Notes Pag. 73 Pag. 74 CHAPTER 8. ROOT LOCUS8.3Positive Feedback We shall now summarize the basic rules for drawing the root locus for a negative feedback systemY refG (s)Y (s)H (s)+E (s)+ Fig 8.3: Stability Analysis - Root Locus - Block representation of positive feedback We can write the characteristic equation and we can re-arrange it highlighting the zerosz i, the poles piand the control gain k p: 1−G(s)H(s) = 0 1 +k p· (s+z 1) ·(s+z 2) · · · · ·(s+z m)( s+p 1) ·(s+p 2) · · · · ·(s+p n)= 0 1−| k p| · (s+z 1) ·(s+z 2) · · · · ·(s+z m)( s+p 1) ·(s+p 2) · · · · ·(s+p n)= 0 with(n > m). Based on previous considerations, the following rules can be given for drawing the root locus of a closed-loop system in the case of negative feedback: 1.Place open loop poles and zeros in thes-plane 2.The root locus has as many branches as there are open-loop poles 3.The root locus is symmetric about the real axis 4.The root locus branches start from open-loop poles and end at open-loop zeros or at infinity See previous considerations 5.Real-axis segments of the root locus are to therightof an odd number of real-axis finite poles/zeros 6.Asymptotes It changes the angles of asymptotes: θa= ±2 kq π k =±1,±2, . . . , q 7.Break-away and break-in pointsPag. 74 CADMS - Lecture Notes(Contents link) Pag. 75 III-Control (Contents link)CADMS - Lecture Notes Pag. 75 Pag. 77 9.P - Controller 9.1Description TheP-Control leris a proportional controller. Let’s analyze a simple problem.Fig 9.1: Control - P controller - 1 dof damped system The equation of motion can be obtained by using the Lagrange equation or the d’Alembert principlem·¨ x+r·˙ x+k·x=f(9.1.1) ω0=rk m |{z} eigenf requencyh =r2 mω 0 |{z} damping factor< 1Underdamped We consider an underdamped system because most of the systems are like that.The forcingfis a force field depending only by the displacement, or better it is proportional to the error of the position: f=k p· (x ref− x) =k p· e(t)(9.1.2) kpis the proportional gain, we can assume it positive ( k p> 0) and constant. Let’s apply (9.1.2) in (9.1.1) and calculate the new parameters of the system: m·¨ x+r·˙ x+k·x=k p· (x ref− x)(9.1.3) m·¨ x+r·˙ x+ (k+k p) ·x=k p· x ref(9.1.4) (k+k p) is the overall stiffness of the controlled system. ω0,c=rk +k pm ⇒ ω 0,c> ω 0 hc=r2 mω c,0⇒ h c< h ω0,c> ω 0means that the resonance frequency of the controlled system is higher than the non- controlled one.h c< h means that the peak of the modulus of the FRF of the controlled system is higher of the non-controlled one it also means that the oscillations of the transient response last more.(Contents link)CADMS - Lecture Notes Pag. 77 Pag. 78 CHAPTER 9. P - CONTROLLER9.2Stability Analysis 9.2.1Roots of characteristic equation•make reference to homogeneous equation of (9.1.4): m·¨ x+r·˙ x+ (k+k p) ·x= 0 •substitute the general trial solutionx=X 0· eλt m·λ2 +r·λ+ (k+k p) ·X 0·  eλt = 0 (X 0= 0 trivial solution) •solve the characteristic equation m·λ2 +r·λ+ (k+k p) = 0 λ1/2= −r2 m±s r2 m 2 − k+k pm  =−α±iω The system is stable ifRe(λ i) 0. The P-controller does not affect the stability of the system 9.2.2Eigenvalues of state space matrix•rewrite equation of motion in state space form     m ·¨ x+r·˙ x+ (k+k p) ·x=k p· x ref ˙ x= ˙x      ¨ x=−rm · ˙ x−( k+k p)m · x+k pm · x ref ˙ x= ˙x   ¨ x ˙ x   |{z} ˙ z=  − rm −( k+k p)m 1 0  |{z} [A]·   ˙ x x   |{z} z+   k p 0   |{z} [B]· x ref |{z} u˙ z= [ A]·z+ [ B]·u– zis the state variables vector –[A]is the state space matrix –[B]is the input matrix –uis the inputs vector •Determinate the eigenvalues of[A] det(λ·[I]−[A]) = 0 det  λ +rm k +k pm −1λ  = 0 λ+rm  ·λ+ 1·k +k pm = 0 λ2 +rm λ +k +k pm = 0 m·λ2 +r·λ+ (k+k p) = 0 which is the same conclusion of the characteristic equationPag. 78 CADMS - Lecture Notes(Contents link) CHAPTER 9. P - CONTROLLER Pag. 799.2.3Stability criteria: direct and indirect •Convert the equation of motion in Laplace domain with the assumption of homogeneous initial conditions m·¨ x+r·˙ x+ (k+k p) ·x=k p· x ref L[. . .] (s)↓ m·s2 +r·s+ (k+k p) ·X(s) =k p· X ref( s)X refk p1 ms 2 +rs+kX (s)H (s) = 1+E (s)F (s)- Fig 9.2: Control - P controller - Block representation of the system in Laplace domain •calculate the Open-Loop TF and the Closed-Loop TF G(s)H(s) =k pms 2 +rs+k |{z} G(s)· 1 |{z} H(s)L (s) =k pms 2 +rs+ (k+k p) •Apply a stability criterion. We can choose between 9.2.3.1Nyquist criterion TheGHfunction is GH=k pms 2 +rs+k It has two poles: the roots ofms2 +rs+k= 0. The poles are in general complex and conjugate with negative real part.GHhasn’t any zeros. Now let’s draw the Nyquist diagram:Fig 9.3: Control - P controller - Nyquist diagram s= 0GH=k pk static gain s→ ∞GH∼k pms 2→ 0(Contents link)CADMS - Lecture Notes Pag. 79 Pag. 80 CHAPTER 9. P - CONTROLLERunstable poles of GH: P=0 encirclements: N=0P =−N⇒The system is stable Now let’s see how the stability is affected by the constants: The system is stable∀k pFig 9.4: Control - P controller - Variation of the Nyquist diagram by k p 9.2.3.2Bode criterion (only for minimum phase system) Let’s represent theGHon the Bode diagram:Fig 9.5: Control - P controller - GHon Bode diagram Now we can analyze the variation of the bode diagram by the coefficientk p:Fig 9.6: Control - P controller - Variation of the Bode diagram by k pPag. 80 CADMS - Lecture Notes(Contents link) CHAPTER 9. P - CONTROLLER Pag. 819.2.3.3Root locus criterion σa=P pi−P ziq =− 2 r2 m− 02 = −r2 m θa= ±kq · π k= 1,3, . . .=±π2 The system is always stable.Fig 9.7: Control - P controller - Root Locus Criterion: s-plane L(s) =G1 + GH= X (s)X ref( s)= k pms 2 +rs+k+k p 9.3Performance Analysis (only for stable cases) The definitions of the time-domain performance indexes are covered if appendixHat page237 9.3.1Time domain Let’s apply a step input: xref= step(t) =( 1t≥0 0t 0, kd> 0) and constant. Let’s apply (10.1.2) in (10.1.1) and calculate the new parameters of the system: m·¨ x+r·˙ x+k·x=k p· (x ref− x) +k d· ( ˙x ref− ˙ x) m·¨ x+ (r+k d) ·˙ x+ (k+k p) ·x=k p· x ref+ k d· ˙ x ref(10.1.3) (k+k p) is the overall stiffness of the controlled system and(r+k d) is the overall damping of the controlled system. ω0,c=rk +k pm ⇒ ω 0,c> ω 0 hc=r +k d2 mω c,0⇒ h c⋚ h ω0,c> ω 0means that the resonance frequency of the controlled system is higher than the non- controlled one. We see also that the eigenfrequecy of the controlled system is independent by the pro- portional gain. We cannot do any prevision about the relation between the damping factor. Fork d↑ the damping factorh c↑ . Fork p↑ the damping factorh c↓ .(Contents link)CADMS - Lecture Notes Pag. 87 Pag. 88 CHAPTER 10. PD - CONTROLLER10.2Stability Analysis 10.2.1Roots of characteristic equation•make reference to homogeneous equation of (10.1.3): m·¨ x+ (r+k d) ·˙ x+ (k+k p) ·x= 0 •substitute the general trial solutionx=X 0· eλt m·λ2 + (r+k d) ·λ+ (k+k p) ·X 0·  eλt = 0 (X 0= 0 trivial solution) •solve the characteristic equation m·λ2 + (r+k d) ·λ+ (k+k p) = 0 λ1/2= −( r+k d)2 m±s r+k d2 m 2 − k+k pm  =−α±iω The system is stable ifRe(λ i) 0. The PD-controller does not affect the stability of the system 10.2.2Eigenvalues of state space matrix•rewrite equation of motion in state space form     m ·¨ x+ (r+k d) ·˙ x+ (k+k p) ·x=k p· x ref+ k d· ˙ x ref ˙ x= ˙x       ¨ x=−( r+k d)m · ˙ x−( r+k d)m · x+k pm · x ref+k dm · ˙ x ref ˙ x= ˙x   ¨ x ˙ x   |{z} ˙ z=  − ( r+k d)m −( k+k p)m 1 0  |{z} [A]·   ˙ x x   |{z} z+  k pk d 0 0  |{z} [B]·   x ref ˙ x ref   |{z} u˙ z= [ A]·z+ [ B]·u– zis the state variables vector –[A]is the state space matrix –[B]is the input matrix –uis the inputs vector •Determinate the eigenvalues of[A] det(λ·[I]−[A]) = 0 det  λ +( r+k d)m k +k pm −1λ  = 0 λ+( r+k d)m  ·λ+ 1·k +k pm = 0 λ2 +( r+k d)m λ +k +k pm = 0 m·λ2 + (r+k d) ·λ+ (k+k p) = 0 which is the same conclusion of the characteristic equationPag. 88 CADMS - Lecture Notes(Contents link) CHAPTER 10. PD - CONTROLLER Pag. 8910.2.3Stability criteria: direct and indirect •Convert the equation of motion in Laplace domain with the assumption of homogeneous initial conditions m·¨ x+ (r+k d) ·˙ x+ (k+k p) ·x=k p· x ref+ k d· ˙ x ref L[. . .] (s)↓ m·s2 + (r+k d) ·s+ (k+k p) ·X(s) =k p· X ref( s) +k d· s·X ref( s) m·s2 + (r+k d) ·s+ (k+k p) ·X(s) = (k p+ k d· s)·X ref( s) We can use a relation between the gainsk pand k dintroducing a time constant τ d: kd= k p· τ d m·s2 + (r+k d) ·s+ (k+k p) ·X(s) =k p· (1 +s·τ d) ·X ref( s)X refk p· (s·τ d+ 1)1 ms 2 +rs+kX (s)H (s) = 1+E (s)F (s)- Fig 10.2: Control - PD controller - Block representation of the system in Laplace domain •calculate the Oped-Loop TF and the Closed-Loop TF G(s)H(s) =k p· (s·τ d+ 1)ms 2 +rs+k |{z} G(s)· 1 |{z} H(s)L (s) =k p· (s·τ d+ 1)m ·s2 + (r+k p· τ d) ·s+ (k+k p) •Apply a stability criterion. We can choose between 10.2.3.1Nyquist criterion TheGHfunction is GH=k p· (s·τ d+ 1)ms 2 +rs+k It has two poles: the roots ofms2 +rs+k= 0. The poles are in general complex and conjugate with negative real part.GHhas one zero equal toZ 1= −1/τ d. Now let’s draw the Nyquist diagram:Fig 10.3: Control - PD controller - Nyquist diagram s= 0GH=k pk static gain s→ ∞GH∼k pτ dms → 0(Contents link)CADMS - Lecture Notes Pag. 89 Pag. 90 CHAPTER 10. PD - CONTROLLERunstable poles of GH: P=0 encirclements: N=0P =−N⇒The system is stable Now let’s see how the stability is affected by the constants:(a) variation byk p(b) variation byk d( τ d) Fig 10.4: Control - PD controller - Variation of the Nyquist diagram by the constants The system is stable∀k pand ∀k d( ∀τ d) 10.2.3.2Bode criterion (only for minimum phase system) We assume that:kpk > 1 Let’s represent theGHon the Bode diagram:Fig 10.5: Control - PD controller - GHon Bode diagram     G m= ∞(>0) ρm> 0⇒ The system is stable Now let’s analyze the variation of the asymptotic curve by the constants:Pag. 90 CADMS - Lecture Notes(Contents link) CHAPTER 10. PD - CONTROLLER Pag. 91(a) variation byk p(b) variation byk d( �