Mechanical Engineering - Control and Actuating Devices for Mechanical Systems

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1 Chiara Moreschini NOTES OF CONTROL AND AC TUATING DEVICES FOR MECHANICAL SYSTEMS Prof. F. Bucca – AA 2021/22 INDEX 0. Summary on mechanical systems analysis 1. Feedforeward and feedback control a. Feedback control b. Feedforeward control c. Summary and comparison between FF and FB 2. Stability analysis a. Lyapounov theorem b. Stability in 1 DoF system c. Stability in 2 DoFs system (conservative and non - cons . force fields) 3. Classical control a. Laplace domain b. Representations: Bode diagram, polar and Nyquist diagram, root locus c. Nyquist and Bode stability criterion d. Performances parameters e. PD controller f. PI controller g. Control of a multi - DoFs system: co - located and non - co - located control h. Ziegler - Nichols methods i. Disturbances and noises attenuation j. Non - ideal sensors 4. Actuators a. Electric al actuator: DC and AC motor s b. Hydraulic actuator 2 0 . SUMMARY ON MECHANICAL SYSTEM ANALYSIS Equazione di moto dei sistemi Equazione di Lagrange L’ equazione di Lagrange è un modo semplice per scrivere le equazioni di moto di un qualsiasi sistema , che si ottengono con i seguenti passaggi: a. Definire i GdL del sistema : !"# !"!#$%& = !"# #'# − !"& '() !"# #'# = + (')*" ∙ !"# *"&+' / !*&-"' Il numero delle variabili indipendenti usate in seguito del sistema dovrà essere uguale al numero di GdL del sistema. b. Definire la variabile o le variabili indipendenti e la relativa posizione all’istante zero : - '((."/)010 /)"/23)"3)13 → - ( 1 ) = 0 890)"( … c. Definire un sistema di riferimento e le convenzioni di segno per il sistema : 1. Scrivere le forme di energia in funzione delle variabili fisiche : ;)3. ? 1 2 B . C . / + 1 2 E . F . / G . ;)3. ? 1 2 J . ∆ I . / + B . < ℎ . G . ;)3. ? ; 2 < @ " − A < = < 0 B − ℎ ± C ℎ 2 − 1 D = − M ± 3 < → " ( , ) = " ! . & / . " * $ + " % . & / . & " * $ Con % 0 = 2 < < ) smorzamento critico del sistema ℎ = 1 1 # = * 2 , * ! smorzamento adimensionale del sistema M = ℎ < ) = * 2 , coefficiente di smorzamento del sistema < = < ) √ ℎ % − 1 pulsazione smorzata del sistema A seconda del valore di h (ovvero del rapporto tra lo smorzamento effettivo del sistema e lo smor zamento critico), possono presentarsi diverse situazioni: 1. Sistema sovra - smorzato e stabile 2. Sistema in smorzamento critico e stabile 3. Sistema sotto - smorzato e instabile ´ Lo smorzamento adimensionale h è molto importante perché permette di valutare lo smorzamento del sistema in modo immediato, poiché mette in relazione lo smorzamento r con la massa m del sistema e generalizza lo smorzamento ´ Fare molta attenzion e che F ≠ F < : la prima è la pulsazione smorzata del sistema, che diminuisce con l’avanzare del tempo, mentre la seconda è la pulsazione propria del sistema, che caratterizza il sistema libero non smorzato ´ Generalmente lo smorzamento dei sistemi è piccolo quindi < ≅ < ) 8 1. Sistema sovra - smorzato E > E - In un sistema sovra - smorzato lo smorzamento adimensionale è maggiore di 1: ; > ; . → ℎ > 1 Dunque √ ℎ / − 1 > 0 e le radici dell’equazione caratteristica associata all’equazione di moto sono reali e distinte, entrambe negative perché ℎ > √ ℎ % − 1 : : ( ," = < 0 B − ℎ ± C ℎ 2 − 1 D = − M 1 , − M 2 → " ( , ) = " ! . & / $ $ + " % . & / % $ I due esponenziali negativi portano la soluzione immediatamente a zero senza oscillazioni (al massimo con una piccola oscillazione negativa), dunque la soluzione è STABILE 2. Sistema in smorzam. critico E = E - In un sistema in smorzamento critico lo smorzamento adimensionale è esattamente uguale a 1: ; = ; . → ℎ = 1 Dunque √ ℎ / − 1 = 0 e le radici dell’equazione caratteristica associata all’equazione di moto sono reali e coincidenti, inoltre sono negative perché : : ( ," = < 0 B − ℎ ± C ℎ 2 − 1 D = − < 0 ℎ = − M 1 → " ( , ) = " ! . & / $ $ + " % , . & / % $ I due esponenziali negativi portano la soluzione immediatamente a zero senza oscillazioni , dunque la soluzione è STABILE ( condizione limite di stabilità ) 9 3. Sistema sotto - smorzato E < E - In un sistema sotto - smorzato lo smorzamento adimensionale è minore di 1: ; < ; . → ℎ < 1 Dunque √ ℎ / − 1 < 0 e le radici dell’equazione caratteristica associata all’equazione di moto sono complesse coniugate, quindi la soluzione sarà data da un’esponenziale complesso, che può essere espresso in forma trigonometrica : : ( ," = < 0 B − ℎ ± C ℎ 2 − 1 D = − M ± 3 < → " ( , ) = " ! . & / . " * $ + " % . & / . & " * $ = = 9 . & / cos ( < , + i ) Con < = < ) √ ℎ % − 1 I due esponenziali complessi portano la soluzione ad oscillare con un’ampiezza che decresce seguendo l’andamento di . & / , dunque la soluzione è IN STABILE C) Moto forzato L’equazione di moto forzato corrisponde all’equazione di moto completa, ovvero: ! " ̈ + % " ̇ + '" = R ( , ) Con R ( , ) = ) ) + ) ( , ) Questa equazione si risolve sommando l’ integrale generale (ovvero la soluzione dell’equazione di moto libero smorzato o non smorzato trovata precedentemente) all’ integrale particolare dato dalla forzante g ( 1 ) : → " ( , ) = " ,4$4 5"6714 + " 891$"045917 Per trovare la soluzio ne dell’integrale particolare si usa il criterio di similitudine, ovvero la soluzione sarà dello stesso tipo della forzante (costante, sinusoidale, esponenziale, ecc..). Poiché in generale la forzante è data dalla somma di una componente costante e una dipendente dal tempo g ( 1 ) = Q < + Q ( 1 ) si può applicare il principio di sovrapposizione degli effetti (possibile poiché stiamo trattando sistemi lineari) e studiare separatament e le componenti, per poi sommarne le soluzioni. 10 A. Forzante costante S : Analizziamo quindi separatamente l’effetto della componente costante della forzante: ! " ̈ + % " ̇ + '" = ) ) Per il principio di similitudine ipotizziamo che la soluzione sia costante R = R < quindi sostituiamola all’interno dell’equazione e troviamo la soluzione: ' R 0 = ) ) → R 0 = Q 0 J La soluzione trovata rappresenta la posizione di equilibrio statico del sistema , poiché è la soluzione dell’equazione in cui vengono rimossi tutti i termini dinamici e rimangono solo quelli statici (passaggio 5 di Lagrange). B . Forzante variabile S ( T ) B1 . Forzante periodica J / ( K ) Analizziamo poi separatamente l’effetto della componente variabile della forzante: ! " ̈ + % " ̇ + '" = ) ( , ) Q ( 1 ) può essere una funzione di vari tipi (sinusoidale, periodica, aperiodica, random), ma analizziamo solo il caso di forzante periodica . Analizziamo per semplicità il sistema caratterizzato da una forzante con una singola armonica e troviamone la soluzione: ! " ̈ + % " ̇ + '" = ) ) cos ( Ω ) , ) L’integrale particolare per il principio di similitudine sarà del tipo " ! ( $ ) = " ! " 3 ?Ω 0 A quindi sost ituiamo nell’equazione e troviamo la soluzione: " 8 ( , ) = " 8 ) . " ; ! $ = V " 8 ) V cos ( Ω ) , + 2 ) Con " 8 ) = < ! & ; ! % , = " ; ! 1 = + Dove w è il ritardo con cui il sistema meccanico reagisce alla sollecitazione e deriva dalla componente immaginaria di R * < w è diverso dalla fase w + dell’anti - trasformata di Fourier, che rappresenta invece lo sfasamento con cui ogni armonica viene sommata alle altre per ricostruire la forzante Quando abbiamo una forzante periodica definiamo le seguenti quantità: Risposta in frequenza : " 8 ) = ) ) − W ) % ! + 3 W ) % + ' Funzione armonica di trasferimento : 9 ( Ω ) = " 8 ) ) ) = 1 − Ω % ! + 3 Ω % + ' 11 Funzione armonica di trasferimento della forza n te periodica La funzione armonica di trasferimento può anche essere riscritta utilizzando lo smorzamento critico e un nuovo parametro a: Definiamo X = Ω 1 0 → 9 ( Ω ) = 1 ' 1 ( 1 − X % ) − 3 2ℎ X La funzione armonica di trasferimento H è quindi un numero complesso, di cui si possono calcolare modulo e fase , i cui grafici sono molto utili per studiare la risposta del sistema quando sottoposto ad una forzante armonica: | M | = 1 A > 1 ( 1 − N " ) " + ( 2ℎ N ) " O = arctan U − 2ℎ N 1 − N " V In assenza di smorzamento ( ℎ = 0 ) si distinguono 3 diverse condizioni di risposta: • 0 < 1 : Zona quasi statica | M | ≅ ( 2 , O ≅ 0° Il sistema risponde in fase con la forzante e con un’intensità controllata • 0 ≅ 1 : Zona di risonanza | M | → ∞ , O ≅ − 90° Il sistema risponde con uno sfasamento di circa - 90° e tende ad aumentare tantissimo le ampiezze, diventando fortemente instabile • 0 > 1 : Zona sismografica | M | → 0 , O ≅ − 180° Il sistema tende a tende a portare a zero le ampiezze, dunque è stabile, e risponde con uno sfasamento di - 180° 12 In presenza di smorzamento ( ℎ ≠ 0 ) invece si distinguono ancora le 3 diverse zone, ma lo smorzamento tende ad attenuare il fenomeno di risonanza, portando il sistema a variare in modo più dolce e soprattutto a non aumentare ad infinito le ampiezze in corrispondenza della risonanza. La derivata di H è un indice dell’entità dello smorzamento: minore è la derivata (derivata piatta) maggiore è lo smorzamento. Quando ℎ = 0 ,7 ÷ 0 ,8 il grafico tende a non mostrare più il picco di risonanza, ovvero si ottiene un sistema ipercritico in cui scompare l’amplificazione dinamica, ovvero c’è solo smorzamen to senza oscillazioni. Soluzione completa del moto forzato La soluzione completa del moto forzato è quindi data per sovrapposizione degli effetti dalla somma della soluzione dell’equazione di moto libero smorzato o non smorzato R B ( 1 ) ( integrale generale ), dalla soluzione dell’equazione con forzante costante R < e dalla soluzione dell’equazione con costante variabile R * ( 1 ) ( integrale particolare ): " ( , ) = " > ( , ) + " ) + " 8 ( , ) Equazione di moto generalizzata In un sistema generico si imposta l’equazione differenziale e si ricavano massa, smorzamento e costante elas tica generalizzati guardando il coefficiente dei vari termini : Y ∗ " ̈ + Z ∗ " ̇ + [ ∗ " = \ ´ La massa generalizzata include oltre alle masse vere e proprie anche tutti i momenti inerziali dei corpi, per questo spesso viene anche indicata con ] ∗ ´ Le forze generalizzate si indicano con la lettera Q ´ Tutti gli altri termini (come la frequenza propria) si ricavano utilizzando ] ∗ , ^ ∗ , _ ∗ 13 Sistemi ad 1 G d L non lineari Linearizzazione Per risolvere analiticamente l’equazione di moto di un sistema non lineare essa va linearizzata nell’intorno della posizione di equilibrio per poi usare le formule risolutive note per i sistemi lineari (viste nel capitolo precedente). Per fare ciò calcoli amo quindi la posizione di equilibrio statico e poi procediamo con la linearizzazione (ultimi step del metodo di Lagrange): 5. Trovare la posizione di equilibrio statico : *- *# 6 `a = / 45657.8 → # 9:;778 = # ' Dove ^ = ^ < + ^ ( 1 ) = ^ !#&#"(' + ^ !#&#"(' 6. Linearizzare nell’intorno della posizione di equilibrio statico - < trovato. Per linearizzare è necessario seguire i seguenti passaggi: 6a) Trasformazione delle coordinare come oscillazione attorno alla posizione di equilibrio # ' : # = # ' + #b -| è detto moto perturbato e rappresenta la vibrazione nell’intorno della posizione di equilibrio - < , si assume quindi che -| sia piccolo, dunque di t rovarsi in una condizione di moto in piccolo. 6b) Trasformazione delle energie in forma quadratica sfruttando l’espansione in serie di Taylor arrestata al secondo ordine (l’ipotesi di moto in piccolo permette di trascurare gli infinitesimi di ordine superiore): g ( - ) = g ( - < ) + Wg ( - ) W- _ " 0 ∙ -| + 1 2 ' 2 ( ( " ) ' " 2 _ " 0 ∙ -| 2 + Wg ( - ) W - ̇ _ " 0 ∙ -| ̇ + + 1 2 ' 2 ( ( " ) ' " ̇ 2 _ " 0 ∙ -| ̇2 + W / g ( - ) W- W - ̇ _ " 0 ∙ -| -| ̇ 14 I termini nell’espansione di Taylor vanno moltiplicati per la differenza del generico punto y e del punto attorno al quale si sta linearizzando y 0 , che per definiz ione precedente è pari al moto perturbato: y − y < = y| . Energia cinetica : c . = 1 2 ) 0 #b ̇ " Energia potenziale : - = - ' + *- *# 6 # ' ∙ #b + 1 2 d ' ∙ #b " L a rigidezza generalizzata a < è data dalla somma di tre contributi: un contributo legato alla derivata prima dell’energia potenziale elastica a $8 ,D , un contributo legato alla derivata seconda dell’energia potenziale elastica a $8 ,DD e un contributo legato alla derivata seconda dell’energia potenziale gravitazionale a E d ' = d ∗ = * " - * # " 6 # ' = d 9< ,? + d 9< ,?? + d @ '() ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a $8 ,D = J ~ W ∆ I F ( - ) W -  / _ - < a $8 ,DD = J ∆ I < ~ W / ∆ I F ( - ) W - /  _ - < a E = B< W / ℎ . ( - ) W - / _ - < La derivata seconda di V è l’unica compo nente di V a sopravvivere nell’equazione di moto (a partire dalla forma quadratica di Taylor), infatti & < si annulla derivando e la derivata prima 3G 35 � - < si annulla con la componente statica delle forze generalizzate per la condizione di equilibrio . & = > ? 1 2 J . � ∆ I < + ∆ I F ( - ) � / + B . < ℎ . ( - ) G . Energia dissipativa: e = 1 2 * 0 #b ̇ " Lavoro virtuale : Per il lavoro virtuale sviluppiamo in serie le forze espresse nella notazione generalizzata, sepa rando la componente costante Q . < dalla componente variabile Q . ( 1 ) ; in questo caso è sufficiente arrestare la serie al primo ordine : + = - ( ) ' . ∗ / ( ' . ∗ " 0 " ) + - ( ) ' + . ∗ / ( ' . ∗ " + 0 " ) ∙ "1 + - ( ( 3 ) ' . ∗ / ( ' . ∗ " 0 " ) + - ( ( 3 ) ' + . ∗ / ( ' . ∗ " + 0 " ) ∙ "1 → / A = / 45657.8 ( " 0 ) + A B ' ∙ #b + / C7D6E7.8 ( " 0 ) + A B ( 0 ) ∙ #b ^ H = ^ !#&# � " 0 � + J I < -| + ^ F"+ � " 0 � + J I ( 1 ) -| ≅ ≅ ^ !#&# ( " 0 ) + J I < -| + ^ F"+ ( " 0 ) 15 J I ( 1 ) si trascurano perché vogliamo ottenere un’equazione linearizzata a coefficienti costanti, non perché sono piccoli! 6 c ) S ostituzione delle energie in forma quadratica nell’equazione di Lagrange rispetto al moto perturbato (punti 6b + 6c ) : S < -| ̈ + ` < -| ̇ + a < -| + W& W- _ - < ∙ -| = ^ !#&# ( " 0 ) + ^ F"+ ( " 0 ) + J I < -| + J I ( 1 ) -| T rascura t i → Y ) ]^ ̈ + Z ) ]^ ̇ + [ ) ]^ = \ @"A9,"04 Equazione di moto linearizzata 16 1 . FEEDFOREWARD AND FEEDBACK CONTROL Feedback and feedforeward control Feedback and feedforeward control are t wo opposite strategies of control: 1. Feedforeward : the input of the system is controlled by the foreward co n troller so t hat the output is equal to the reference - Very fast - Does NOT affect stability - Does not measure the actual output of the system - If the disturbance is high does not attenuate it - Needs an analytical description of the controlled system 2. Feedback : the input of the system is the difference between the actual output of the system and the reference output, which is put to zero - Slower response - It i s a force field so it CAN affect stability - Doesn’t need an analytical description of the controlled system ´ We can also have controller which combine bot FF and FB controllers to combine their advantages. Feed foreward control (FF) Feedforeward control scheme The FF control works with inverse dynamics , so it takes as input the reference output we want and create a force g ( which grant x to tend to the reference R ) . ! " ̈ + % " ̇ + '" = R 0 + R B + R @ g ( controller force, g F known disturbance , g 7 unknown forces acting on the system By giving R = R ) as input we can compute the force g ( the controller has to apply to obtain the reference as output. Notice that in the mathematical model ( B� , . ̂, J� ) we do not know the value of g 7 so we don’t take it into account when we compute g ( : R 0 = !_ " 1̈ + % ̂" 1̇ + ' a " 1 − R @ However due to the approximations of the coefficients in the mathematical model ( B� , . ̂, J� ) and the presence of an unknown force the res ponse of the system won’t be exactly the reference one but there will be some errors: ! " ̈ + % " ̇ + '" = R B + !_ " 1̈ + % ̂" 1̇ + ' a " 1 17 Effects of FF control If we compute the solutions of the previous equation we can understand how the FF control affects the behavior of the response of the system: ! " ̈ + % " ̇ + '" = R B + !_ " 1̈ + % ̂ " 1̇ + ' a " 1 = R B + R 0 + R @ The solution will be the sum of the homogeneous solution and the 2 particular solutions corresponding to the 2 inputs of the system ( R ) and g 7 ): " ( , ) = " > ( , ) + " 8 ,B ( , ) + " 8 ,1 ( , ) • R B ( 1 ) : The homogeneous solution, which is connected to the transient response of the system, is NOT affected at all by the controller (in fact the controller doesn’t modify the parameter m, r, k), therefore it doesn’t affect stability • R * ,7 ( 1 ) : the first particular solution connected to the unknown forces g 7 could be studi ed by using the Fourier transformation of the forces, finding the response of the system for each armonic present in the input and then sum up all the responses (since the system is linear). By comparing the frequencies present in the input g 7 and the FRF of the system it is possible to see if the inputs are not modified by the system ( quasi - static zone ), if they are attenuated ( seismographic zone ) or if they are amplified ( resonance zone ). f ( g Ω ) = 5 F ,; ' 4 ; ' = 1 − < Ω " + g; Ω + k ⟵ 2,2 We have only passive damping of the forces, which cannot be modified by the FF controller (because it cannot modify the stability through m, r, k) but can only be modified increasing r in the system or changing the characteristics of the system • R * ,) ( 1 ) : the second particular solution connected to the reference output R ) is associated with the capability of the system to follow the reference and achieve the output target , ( g Ω ) = 4 . ,' 5 > ,' = − ' 5 > ' = g A C s + A F − < s " + g ; 5 s + A 5 ⟵ 2,2 The target is achieved when r ( g Ω ) → 1 so the output is equal to the reference Also this time the function can be modified acting on the proportional and derivative gains J * and J F since at the denominator appear the equivalent stiffness and damping of the system J # and . # If we increase only k J and keep constant k K then L ( i Ω ) → 1 in the quasi static zone but there will be a high resonance, if instead we increase both k J and k K we obtain a much better response and we will have L ( i Ω ) → 1 also in the resonance zone Block diagram of FB control 21 Summary and comparison between FF and FB FF control FB control Disturbances CANNOT attuenuate disturbances C AN attenuate disturbances both in the quasi stat ic zone and in the resonance zone Transient response CANNOT modify the transient response of the system CAN modify the transient response of the system Reference CAN achieve the exact reference in theory CANNOT achieve the exact reference in theory but only get close to it Control force 4 . = ̈ + ; ̂ 5 >̇ + Am 5 > → 4 . = 4 . ( 0 ) NOT a f orce field 4 . = A F ( 5 >9I − 5 ) + A C ( 5 >9I ̇ − 5 ̇) → 4 . = 4 . ( 0 , 5 , 5 ̇) Force field ( conservative or non - conservative ) Stability Does NOT affect stability CAN affect stability because it modifies the equivalent damping and stiffness 22 2 . STABILITY ANALYSIS Stability analysis A system is stable if after a perturbation it comes back to its initial state or at least it has a bounded reaction , otherwise it is unstable . Moreover the stability is a property of a point in which the system is, it is NOT a general property of the system! The stability can be studied in 2 main states: 1. Static equilibrium : the system does NOT move 5 = 5 ' , 5 ̇ = 5 ̈ = 0 2. Steady state : the system moves at constant speed 5 ̇ = 5 ̇' , 5 ̈ = 0 The stability is related to the transient response of the linearized system therefore to study stability we have to consider a homogeneous equation of motion (we did not neglect the forces applied but we linearized the system and obtained a homogeneous equation) : ! " ̈ + ( % + A J ) " ̇ + b ' + A K c " = 0 ´ In general for a non - linear system we can have more than one equilibrium position , for each of these position we can have different stability properties so we have to perform different stability analysis for each equilibrium position. Stable, asymptotically stable and unstable systems Depending on the stability properties of a system we can define 3 types of systems: 1. Stable systems : the motion of the system remains bounded into a certain range ∃ P : � | H | � = � ( R / + R ̇ / ) < P ∀ 1 > 0 2. Asymptotically stable systems : the motion of the system not only remains bounded into a certain range P but it also tends to zero when time approaches infinite ∃ P : � | H | � = � ( R / + R ̇ / ) < P ∀ 1 > 0 and lim # L → N | | H | | = 0 3. Unstable systems : the system motion becomes unstable for a certain instant ∃ 1 ̅: � | H | � = � ( R / + R ̇ / ) > P g(. 1 = 1 ̅ Where � | H | � = � ( R / + R ̇ / ) is a perturbation of the system Static and dynamic stability The stability or instability of a system can be dynamic or static depending on how the solution tends to the steady state value: a) Static stability/instability : the solution reaches the steady state value without oscillations ; this behavior is due to the presence of real eigenvalues b) Dynamic stability/instability : the solution reaches the steady state value with oscillations ; this behavior is due to the presence of complex conjugated eigenvalues 23 Lyapounov theorem The Lyapounov theorem allow us to infer if a non - linear system is stable by studying the stability of the associated linearized system : Linearized system Original non - linear system Asymptotically stable ⟹ Stable Stable ⟹ ? Cannot be said Unstable ⟹ Unstable Then by studying when a linear system is stable or unstable we can also define the stability for non - linear system by using the Lyapounov theorem. 24 Stability in 1 DoF system : 6 possible cases Stability cases for a linear 1 DoF system To study the stability of a linear 1 Do F system we have to study the following equation and solution: ! " ̈ + ( % + A J ) " ̇ + b ' + A K c " = 0 ; + A C = ; 5 A + A F = A 5 : ( ," = − ; 5 2 < ± > ? ; 5 2 < @ " − A 5 < = n ' o − ℎ ± C ℎ " − 1 p = − M ± √ ∆ → " ( , ) = " ! . & / . " * $ + " % . & / . & " * $ with % 0 = 2 < < ) critic damping n ' = C 2 % E natural frequency ℎ = 1 1 # = * 2 , * ! adi m. damping M = ℎ < ) = * & 2 , damping coeff . < = < ) √ ℎ % − 1 damped frequency Depending on the value of the coefficients we have 6 different cases: Cases 1. 2. 3. 4. 5. 6. ; 5 + + + - - - A 5 + + - + + - | ℎ | ≥ 1 < 1 NA ≥ 1 < 1 NA Case 1. ; 5 > 0 , A 5 > 0 , | ℎ | ≥ 1 w | ℎ | ≥ 1 ⟹ ∆ > 0 ; 5 > 0 ⟹ q > 0 ⟹ y M ,N = − z M , − z N { √ ∆ { < | q | Both the exponents of the solution are real and negative , therefore we obtain a statically stable system and asymptotically since it tends to zero. STATIC ASYMPTOTIC STABILITY 25 Case 2. ; 5 > 0 , A 5 > 0 , | ℎ | < 1 w | ℎ | < 1 ⟹ ∆ < 0 ; 5 > 0 ⟹ q > 0 ⟹ y M ,N = − z ± | } O C ~ − d N T he exponents of the solution are complex conjugated with negative real part , therefore we obtain a dynamically stable system and asymptotically since it tends to zero. DYNAMIC ASYMPTOTIC STABILITY Particular case 2. | ℎ | = 0 ⟹ y M ,N = ± | } O The exponents of the solution are purely imaginary , therefore we obtain a dynamically stable system with oscillations not damped which therefore do not tend to zero (just stable, not asymptotically). DYNAMIC STABILITY (not asymptotic) 26 Case 3. ; 5 > 0 , A 5 < 0 , | ℎ | � w A 5 < 0 ⟹ ∆ > 0 ; 5 > 0 ⟹ q > 0 ⟹ y M ,N = − z M , + z N { √ ∆ { > | q | T he exponents of the solution are real , one positive and negative , therefore we obtain a statically un stable system due to the contribution of the positive exponent . STATIC IN STABILITY Case 4. ; 5 < 0 , A 5 > 0 , | ℎ | ≥ 1 w | ℎ | > 1 ⟹ ∆ > 0 ; 5 < 0 ⟹ q < 0 ⟹ y M ,N = + z M , + z N { √ ∆ { < | q | Both the exponents of the solution are real and positive , therefore we obtain a statically unstable system. STATIC IN STABILITY 27 Case 5. ; 5 < 0 , A 5 > 0 , | ℎ | < 1 w | ℎ | < 1 ⟹ ∆ < 0 ; 5 < 0 ⟹ q < 0 ⟹ y M ,N = + z ± | } O C ~ − d N T he exponents of the solution are complex conjugated numbers with positive real part , therefore we obtain a dynamically un stable system. DYNAMICAL IN STABILITY Case 6. ; 5 < 0 , A 5 < 0 , | ℎ | � w A 5 < 0 ⟹ ∆ > 0 ; 5 < 0 ⟹ q < 0 ⟹ y M ,N = − z M , + z N { √ ∆ { > | q | The exponents of the solution are real , one positive and negative , therefore we obtain a statically unstable system due to the contribution of the positive exponent. STATIC IN STABILITY N.B. g4 w ; 5 < 0 ; 5 > 0 N�� A 5 < 0 ⟹ ����������� In those case we have instability because ether the real part of the solution is negative or the square root has a higher modulus than alpha ( ± ), therefore their sum gives a negative number. 28 Stability in 2 DoFs system : Conservative and non - conservative force fields 2 DoFs systems The equation of motion of 2 DoFs system can be written in matrix - vector form: [ ! ] 5 ̈ + [ , ] 5 ̇ + [ d ] 5 = 2 If the matrices [M], [R], [K] are diagonal the 2 equations of motion are totally independent ( sistema disaccoppiato ) To analyze the stability of 2 DoFs system we procede following the same steps we used for the 1 DoF system: 1. Write the equation of motion [ ! ] 5 ̈ + [ , ] 5 ̇ + [ d ] 5 = 2 2. Find the static equilibrium position 5 ' 3. If the equations are non - linear, linearize them around the equilibrium position ; the linearized forcing terms can be seen as equivalent stiffness and dampin g therefore we obtain the linearized homogeneous equation with the new equivalent stiffness and damping matrices [ ! ] 5 ̈ + [ , P ] 5 ̇ + [ d P ] 5 = 0 With [ , P ] = [ , B ] + [ , ] , [ d P ] = [ d B ] + [ d ] The initial matrices [M], [R], [K] were diagonal if the system was uncoupled, but the new equivalent matrices are not diagonal (full matrices) , because the forcing term has a coupling effect , so the 2 equations are not more independent 4. From the linearized equation we can now study of the equation of free motion from which we can find the natural frequencies of the system [ ! ] 5 ̈ + [ d P ] 5 = 0 → det ( : " [ � ] + [ d P ∗ ] ) = 0 [ d P ∗ ] = [ ! ] Q ( [ d P ] = � A RR A RS A SR A SS � det ( … ) = 0 → : " ? ,?? = − � ± � � Where � = 2 '' T 2 (( " and � = ? 2 '' T 2 (( " @ " − � A RR A SS − A RS A SR � = � " − ∆ Or also � = ? 2 '' Q 2 (( " @ " + A RS A SR 5. Finally solving the complete linearized equation we can find the solution of the system as small oscillations around the static equilibrium position 29 Types of stability For the 2 DoFs systems we can have all the ty pes of stability we already seen for 1 DoF system plus an additional type: 1. Stability (not asymptotical) 2. Asymptotical stability à static/dynamic 3. Instability à static/dynamic 4. Flutter instability We will see more in detail these types of instability by studying the cases in which a conservative or a non - conservative force field are applied . Conservative and non - conservative force fields Equation of free motion – no damping The force fields applied on the system can be either conservative or non - conservative ; depending on the force applied the response of the system will be different: Consider the system: 5 ̈ + [ d P ] 5 = 0 : " ? ,?? = − � ± � � → : ( ," ,U ,V � ℎ `;` � = A RR + A SS 2 � = � A RR + A SS 2 � " − � A RR A SS − A RS A SR � = � " − ∆ �; � = � A RR − A SS 2 � " − A RS A SR 1. Stability of conservative force fields Equation of free motion – no damping Conservative force fields are fields in which the work of the force does not depend on the trajectory chosen but only on the initial and final points . In this case the response of the system has the following characteristics: : " ? ,?? = − � ± � � �g 0 ℎ � > 0 → : " ? ,?? N��N#� ;`N� In this case we can prove that the stiffness introduced by the force field [ d B ] is a symmetric matrix therefore also the final equivalent matrix will be symmetric: [ d P ] = [ d B ] + [ d ] = � A RR A RS A SR A SS � N��N#� �# 0 � = � " − ∆ → | � | > { � � { → ,` � : " ? ,?? � < 0 ∀ � , � → : ( ," ,U ,V = C : " ? ,?? = ± g n ( ," ∀ � , � So in this case the response is always a rootsquare of a real positive number therefore the response will be always a purely imaginary number , so the system is stable but not asymptotically , because the response will keep oscillating in a bounded region 30 2) [ _ / ] NON symmetric positive definite det [ d P ] < 0 6DC 8> A RR , A SS < 0 In both cases det [ a O ] < 0 and/or J 44 , J 55 < 0 one or both the roots h / D ,DD are positive and real, therefore h P ,/ ,Q ,R will have a positive real part that causes static instability : ,` � : " ? ,?? � > 0 4�; �� 0 we CANNOT have dynamic instability or asymptotic stability because the h / D ,DD are always real numbers and therefore the solutions h P ,/ ,Q ,R can only be purely imaginary or real positive numbers. ´ A matrix is symmetric positive definite (SPD) if it is symmetric, its determinant and the determinants of its main minors are positive; for a 2x2 matrix it means that the elements on the main diagonal must be positive and its determinant must be positive 2. Stability of NON - conservative force fields Equation of free motion – no damping Non - c onservative force fields are fields in which the work of the force does depend s on the trajectory chosen and NOT only on the final and initial points . In this case the response of the system has the following characteristics: : ( ," = − � ± � � �g0 ℎ � ≷ 0 → : ( ," ;`N� �; �� In this case we CANNOT prove that the stiffness introduced by the force field [ d B ] is always symmetric like for conservative force fields therefore also the final equivalent matrix will be NOT symmetric in general: [ d P ] = [ d B ] + [ d ] = � A RR A RS A SR A SS � �� �# 0 & � > 0 → : ( ," ,U ,V = � : " ? ,?? = ± g n ( ," Stable but not asymptotically b) �4 ∆ < 0 �; � < 0 → : ( ," ,U ,V = � : " ? ,?? = ± q Static instability 2) � = � N − ∆ < � Possible only with non - conservative force fields and it can happen if: � = ? 2 '' Q 2 (( " @ " + A RS A SR < 0 → � A RS A SR < 0 � ? 2 '' Q 2 (( " @ " � < { A RS A SR { This is the only case in which the solutions h / D ,DD can be complex conjugated numbers and therefore also the solutions h P ,/ ,Q ,R can be complex conjugated: : " ? ,?? �� 0 N�� � = � + gn ´ We could also write the inverse Laplace transform but we never use it to go back to the time domain because it is too complicated; we instead decompose the functions in many simpler functions of which w e know the anti - transformation ´ We do NOT lose any information passing from the time domain to the Laplace domain and viceversa P roperties in Laplace domain The Laplace transform has many important properties , that can all be demonstrated by applying the definition of Laplace transform : 1. Linearity ℒ [ q4 ( 0 ) + �� ( 0 ) ] = q ℒ [ 4 ( 0 ) ] + � ℒ [ � ( 0 ) ] 2. Time derivation ℒ � 4 ̇( 0 ) � = � 2 ( � ) − 4 ( 0 ) The time derivation consists in multiplying the function by the complex number s for each derivation In general: ℒ � 4 ̈( 0 ) � = � " 2 ( � ) − �4 ( 0 ) − 4 ̇( 0 ) ℒ � 4 ( D ) ( 0 ) � = � D 2 ( � ) − � � D Q 2 4 ( 2 Q ( ) ( 0 ) D 2 Z ( 3. Time integration ℒ o � 4 ( 0 ) �0 5 ' p = 1 � 2 ( � ) The time integration consists in dividing the function by the complex number s for each integration 34 4. Delay in time domain ℒ [ 4 ( 0 − � ) ] = ` Q 4[ 2 ( � ) The delay in time consists in multiplying by a negative exponential in the Laplace domain with the delay as exponent Note that the Laplace transform of g ( 1 ) is only valid for 1 > 0 , but if the signal is delayed the transformation will be valid not for 1 > 0 but for 5 > � 5. Delay in Laplace domain 2 ( � − N ) = ℒ [ ` 65 4 ( 0 ) ] The delay in Laplace consists in multiplying by a positive exponential in time with the delay as exponent Laplace transform of particular functions To compute the Laplace transform of a function we rarely apply the definition but we usually use the transformation of particular functions: �0`� ( 0 ) = w 1 4�; 0 ≥ 0 0 4�; 0 < 0 → ℒ [ �0`� ( 0 ) ] = 1 � g 0 → �0N��` �� ���`� → − � ℎ < 0 → ���0N��` �� ���`� → + � Contribution of complex conj . zeros f = U 1 + " g d 1 ,* � + 4 + d 1 ,* + V The contribution of complex conj. zeros to the Bode diagram of the phase depend on � , that means it depends on the stability of the zeros: � > 0 → �0N��` �� �`;�� → + � � < 0 → ���0N��` �� �`;�� → − � SUMMARY : Stable (sx) Unstable (dx) Poles Real CC − � / 2 − � + � / 2 + � Zeros Real CC + � / 2 + � − � / 2 − � 41 Minimum phase systems Minimum phase systems are systems that have positive gain and only stable poles and stable zeros , therefore the drawing of their Bode diagram is simpler because we have less cases: !g�g 0 ,` ( � 7 ) < 0 ,` ( � 7 ) < 0 47 To apply the Bode criterion we define 2 new stability parameters defined on the open loop TF GH: ´ Gain margin = value of the modulus of GH when the phase is equal to − � , which is defined as phase crossover frequency ( Ω *( ) so it is the frequency in which the Bode diagram of the phase intercepts the line at − � Ω F. ��� ℎ 0 ℎ N0 ∡ � f � g Ω F. � M � g Ω F. � � = − � Definition of the gain margin : � 1 f E � = { f � g Ω F. � M � g Ω F. � { = − | f E | Cb Because � ( _ 4 � Cb = 20 log ( _ 4 = − 20 log f E = − | f E | Cb ´ Phase margin = value of the phase of GH when the modulus is equal to 0 "p , which frequency is defined as gain crossover frequency ( Ω E( ) so it is the frequency in which the Bode diagram of the modulus in dB intercepts the horizontal axis, which is equivalent to the phase in which the simple modulus is equal to 1 Ω @. ��� ℎ 0 ℎ N0 { f � g Ω @. � M � g Ω @. � { Cb = 0 �� �; { f � g Ω @. � M � g Ω @. � { = 1 In particular the phase margin is defined as the phase of GH in correspondence to Ω E( plus � : � E = ∡ { f � g Ω @. � M � g Ω @. � { + � Since Ω *( is the phase in which the Bode diagram intercepts the axis at 0 dB, that is | !l | = 1 it is an index of the distance of the Nyquist diagram from the point - 1 so an index of the number of encirclements around - 1. Similarly the phase margin is an index of the number of encirclements around - 1 because it is the angle between the negative horizontal axis (Re axis) and the phase of GH in Ω @. 48 � 1 f E � > 0 → | f E | Cb < 0 � E > 0 → ∡ ( fM ) > − � If there is no interception of the phase with the line − � it means that the gain margin is infinite, therefore it is surely positive: | f E | Cb → ∞ 0 ℎ `;`4�;` | f E | Cb > 0 ⟹ �0N�g�g0# ´ The Bode criterion can only be applied when we have minimum phase systems because the gain and the phase margin are index of the distance of the Nyquist diagram from - 1. If instead we have unstable poles of GH (so NOT a minimum phase system) the system to be stable must have N= - p encirclements around - 1 so we cannot use the gain and the phase margin ( à Nyquist criterion) ´ If GH has one unstable pole or one unstable zero the Bode criterion cannot be applied; however we always draw the Bode diagrams, even if w e cannot apply this criterion, because they are needed to draw the Nyquist diagram, which instead can always be applied. Bode diagram of GH and of L(s) Starting from the asymptotic Bode diagram of GH we can also find the asymptotic diagrams of L(s) using the following observations: r ( � ) = f 1 + 7 M → r ( � ) ≅ w 1 g4 fM ≫ 1 f g4 fM ≪ 1 | r ( � ) | Cb ≅ w 0 �� g4 | fM | Cb ≫ 0 �� | fM | Cb g4 | fM | Cb ≪ 0 �� ∡ r ( � ) ≅ w 0° g4 | fM | Cb ≫ 0 �� ∡ ( fM ) g4 | fM | Cb ≪ 0 �� So we have to find the point in which GH intercepts the axis at 0 dB and then the Bode diagrams of L(s) will be equal to 0 until that point and equal to the diagrams of GH after that point. To be stable the phase of GH in Ω @. must be above − � To be stable the modulus of GH in Ω F. must be lower than the 0 dB axis 49 Dominant poles The dominant poles are the poles closest to the imaginary axis and they are more important than the others because their contribution on the dynamics of the system is higher than the one of the other poles: if we have a N - DoFs system we can approximate its response with the response of a 1 DoF or 2 DoFs system that has poles similar to the dominant poles of the original system. However if there are unstable (positive) poles that cause instability they will always have the highest effect wrt the dominant stable poles. ´ If the dominant pole is real the response of the system will show approximately a static response. ´ If the dominant poles are complex conjugates the response of the system will show oscillations . ´ The reason why the dominant poles have the highest contribution is that they are associated with the exponentials with the highest exponent in modulus ´ For example when we apply the Ziegler - Nichols methods we apply the first method if the dominant pole is real, because the system will have an “s - shape” response, while we apply the second method if the dominant poles are complex conjugated, because the response will show oscillations (not “s - shape”) 50 Performance parameters Performances parameters To evaluate the response of a system in the time domain we can compare it with a step function (reference output) and study the value of some parameters which are indexes of how much the response of the system is similar to the step function. Rise time K m The rise time is the time needed to reach 90 % of the reference output 0 > → 5 ( 0 ) = 90% 5 >9I T o have a good control we want: 0 > → 0 Percentage of overshoot � . � . The percentage of overshoot is the maximum overshoot (first oscillation) of the response wrt the reference output � . � . → � . � . = 5 E6R − 5 >9I 5 >9I �; � . � . = 5 E6R − 5 W 5 W T o have a good control we want: � . � . → 0 ´ T here are 2 possible definitions of the PO: when we are interested in the actual behavior of the response we use the definition with 5 W while if we are interested to the ideal output we use the definition with 5 >9I ´ For the PD controller the steady state error is different from zero so the 2 definitions are different, while for PI controllers the error is zero ( 5 W = 5 >9I ) so the 2 definitions coincide Settling time K n The settling time is the time needed to reach the interval between ± 5% of the final output (which can be different from 5 >9I ), so it is the interval in which the oscillation s of the output become negligible 0 4 → 5 ( 0 ) = 5 W ± 5% 5 W ( 5 W ≠ 5 >9I ) T o have a g ood control we want: 0 4 ��� Steady state error � W The steady state error is the difference between the steady state response and the reference output at infinity ` W → ` W = 5 >9I − 5 W T o have a good control we want: ` W → 0 51 TYPE OF CONTROLLERS a ) PD controller Equation of PD controller The force introduced by a PD controller to control the system has a proportional part (displacement) and a derivative part (velocity): 4 . = A F � 5 >9I − 5 � + A C ( 5 ̇>9I − 5 ̇) Analysis in time domain The force introduced by the controller is a force field since it depends on the displacement and on the velocity so it is able to modify the stability of the system; if it was only a time dependent force it could not modify the stability of the system. Stability analysis in time The system in the time domain with the application of a PD controller becomes: < 5 ̈ + ; 5 ̇ + A5 = A F � 5 >9I − 5 � + A C ( 5 ̇>9I − 5 ̇) < 5 ̈ + ( ; + A C ) 5 ̇ + � A + A F � 5 = A F 5 >9I + A C 5 ̇>9I Homogeneous equation (steady state) : < 5 ̈ + ( ; + A C ) 5 ̇ + � A + A F � 5 = 0 : ( ," = − ; + A C 2 < ± > U ; + A C 2 < V " − A + A F < = − q ± √ ∆ → 5 ( 0 ) = 5 ( ` Q o ` 7 d 5 + 5 " ` Q o ` Q 7 d 5 n . = > A + A F < ℎ . = ; + A C 2 < n . ,' q . = ℎ C n . = ; 5 2 < So the stability of the system in affected in the following way: ´ �g j J * > 0 J F > 0 ∶ the system is always stable because stiffness and damping in crease and are always positive ´ I g j J * ↑ J F = 0 ∶ the natural frequency of the controlled system F ( increases so the rise time decreases but the dumping decreases ℎ ( so the overshoot increases and also the settling time j J * ↑ → n p ↑ → 0 * ↓ J F = 0 → ℎ p ↓ → � . � . ↑ 0 q ↑ ´ I g j J * ↑ J F ↑ ∶ if instead we increase also J F we can avoid the decrease of the dumping and therefore also reduce the PO and the settling time j J * ↑ → n p ↑ → 0 * ↓ J F ↑ → ℎ p ↑ → � . � . ↓ 0 q ↓ 52 State - space form We can see it also in the state space form in time domain: 5 ̇ = [ � . ] 5 + [ � ] � [ � . ] = � − ; + A C < − A + A F < 1 0 � 5 = � 5 ̇ 5 � � = w 5 ̇>9I 5 >9I � To study the stability from the state space form we compute the eigenvalues of [ k ( ] and we obtain the same characteristic equation we found before for : ( ," det ( : [ � ] − [ � . ] ) = 0 → : → ,` ( : ) < 0 Response in time The response in time varies depen d in g on the values of J * and J F as we saw before: Analysis in Laplace domain We can also study the stability of the system in the Laplace domain , by applying the indirect methods like Nyquist criterion and the Bode criterion (if it is a minimum phase system) or by using the root locus . We do not use direct methods since the computation of the poles of L(s) can be very difficult. Stability analysis in Laplace domain First of all we obtain the equation of motion in the Laplace domain: < 5 ̈ + ; 5 ̇ + A 5 = A F � 5 >9I − 5 � + A C ( 5 ̇>9I − 5 ̇) → [ < � " + ; � + A ] � ( � ) = � A F + A C � � � � >9I ( � ) − � ( � ) � We typically write J F as a function of J * : J F = J * � F [ < � " + ; � + A ] � ( � ) = A F ( 1 + � C � ) � � >9I ( � ) − � ( � ) � � ( � ) � >9I ( � ) = r ( � ) = A F ( 1 + � C � ) 9I ( � ) = lim 4 → ' � r ( � ) 1 � = = lim 4 → ' A F ( 1 + � C � ) 9I − 5 � + A 7 � � 5 >9I − 5 � �0 5 ' → [ < � " + ; � + A ] � ( � ) = U A F + A 7 1 � V � � >9I ( � ) − � ( � ) � We typically write J " as a function of J * : J " = . $ O ! [ < � " + ; � + A ] � ( � ) = A F U 1 + 1 � 7 � V � � >9I ( � ) − � ( � ) � � ( � ) � >9I ( � ) = r ( � ) = A F ? � + 1 � 7 @ � 9I ( � ) = lim 4 → ' � r ( � ) 1 � = = lim 4 → ' A F ? � + 1 � 7 @ � 9I − 5 " ) So in the matrix - vector form we obtain: 2 = w 4 ( 4 " � = w 0 A F ( 5 " ,>9I − 5 " ) � = � 0 0 0 A F � � 5 >9I − 5 � → 2 = [ d . ] � 5 >9I − 5 � Where 5 >9I = � 5 ( ,>9I 5 " ,>9I � 5 = � 5 ( 5 " � And the equation becomes: [ ! ] 5 ̈ + [ , ] 5 ̇ + ( [ d ] + [ d . ] ) 5 = [ d . ] 5 >9I Where [ d ] + [ d . ] = [ d 585 ] Since the matrix of the control [ d . ] only add a positive term on the diagonal, if the initial matrix is stable (symmetric and positive definite), also the final matrix [ d 585 ] will be symmetric and positive definite, therefore the controller does never introduce instability , in particular the P controller will have the same effects of the 1 - DoF system: [ d 585 ] = [ d ] + � 0 0 0 A F � → N��N#� �0N��` ANALYSIS IN LAPLACE DOMAIN In Laplace domain the co - located control can be represented in the block diagram by neglecting the force g P and studying the remaining feedback control: fM = A F f "" = A F < (( � " + ; (( � + A (( det [ Q ( s ) ] 67 2 �� �`;�� : � ( ," 4 �� �`;�� : � ( ," � U ,V { � ( ," { = C " 2 E { � ( ," { = C 2 E { � U ,V { = C U 2 E Now we can study as always the stability in the Laplace domain by considering the Bode and the Nyquist criterion of GH or also the root locus : Bode diagrams The position of the zero will be always between the two couples of poles. { � ( ," { < { � ( ," { < { � U ,V { C 2 E < C " 2 E < C U 2 E By increasing kp the diagram moves upwards and the band width decreases, but the system remains always stable since the phase margin and the gain margin are always positive � E > 0 ∀ A F f E > 0 ∀ A F ���N#� �0N��` ∀ A F By drawing the Nyquist diagram we obtain the same results. Root locus ���N#� �0N��` ∀ A F 68 2) Non co - located control ANALYSIS IN TIME DOMAIN In the non - co - located control if we want to control the variable � N we act indirectly on the other DoF applying a forc e that still depends on � N but on the mass � M : 4 ( = A F ( 5 " ,>9I − 5 " ) , 4 " = 0 So in the matrix - vector form we obtain: 2 = w 4 ( 4 " � = w A F ( 5 " ,>9I − 5 ( ) 0 � = � 0 A F 0 0 � � 5 >9I − 5 � → 2 = [ d . ] � 5 >9I − 5 � Where 5 >9I = � 5 ( ,>9I 5 " ,>9I � 5 = � 5 ( 5 " � And the equation becomes: [ ! ] 5 ̈ + [ , ] 5 ̇ + ( [ d ] + [ d . ] ) 5 = [ d . ] 5 >9I Where [ d ] + [ d . ] = [ d 585 ] This time the matrix of the control [ d . ] add a positive term on the extra - diagonal, so even if the initial matrix is stable (symmetric and positive definite), the final matrix [ d 585 ] could be unstable because is not more symmetric, therefore the controller can introduce instability : [ d 585 ] = [ d ] + � 0 A F 0 0 � → �N� �`�� +5 { � ( ," { = C 2 E { � U ,V { = C U 2 E Now we can study as always the stability in the Laplace domain by considering the Bode and the Nyquist criterion of GH or also the root locus : Bode diagrams The position of the zero usually is after the poles because generally A "( ≫ ; "( { � ( ," { < { � U ,V { < | � ( | By increasing kp the diagram moves upwards and the band width decreases, but this time for some kp the phase margin (or the gain margin) can become negative so we can have instability � E < 0 4�; �� � � ̇̇ 6 = − , 6 r 6 g 6 − A ~ r 6 n + � 6 r 6 5 = � n g 6 � � = w � > � 6 � 5 ̇ = [ � ] 5 + [ � ] � [ � ] = ⎣ ⎢ ⎢ ⎢ ⎡ − � � A ~ � − A ~ r 6 − , 6 r 6 ⎦ ⎥ ⎥ ⎥ ⎤ [ � ] = ⎣ ⎢ ⎢ ⎡ − 1 � 1 r 6 ⎦ ⎥ ⎥ ⎤ Stability : det ( : [ � ] − [ � ] ) = 0 → ,` ( : ) < 0 We can also transform the equations in Laplace domain, study the block diagram of the motor and analyze the stability by using Bode and Nyquist criterion: � ( �� + � ) Ω ( s ) = A ~ � 6 ( � ) − � > ( � ) ( r 6 � + , 6 ) � 6 ( � ) = - 6 ( � ) − A ~ Ω ( s ) NATURAL FEEDBACK Even before introducing a controller the actuators (every type, not just the electrical ones) interact with the mechanical systems and create a natural feedback 79 Open loop TF : f ( � ) M ( � ) = A ~ " ( r 6 � + , 6 ) ( �� + � ) Closed loop TF : r ( � ) = A ~ ( r 6 � + , 6 ) ( �� + � ) + A ~ " 2. Separately excited DC motor The separately excited DC motor is a motor in which the circuits are surrounded by a coil in which flows current , which induce a magnetic flow B on the circuits and makes them rotating by generating a force on the circuits; differently from the permanent magnet DC motor, the magnetic field produced by the coil can be modulated by changing the current flowing in the coil, while with a permanent magnet we cannot modify the magnetic field produced. NOT REQUESTED IN THE EXERCICES ONLY FOR THEORETICAL PART Speed control on permanent magnet DC motor We now consider a permanent magnet DC motor and we apply a speed controller (which can be a P, PD, PI, …, controller). The speed controller can be a P, PD, PI, …, controller: � ���0;���`; : - 6 = A F � Ω >9I − Ω � �e ���0;���`; : - 6 = � A F + � C � � � Ω >9I − Ω � �� ���0;���`; : - 6 = U A F + 1 � 7 � V � Ω >9I − Ω � Usually we do NOT use this type of scheme with only a speed controller because it is important to control also the current for safet y reason. 80 Current control on permanent magnet DC motor Generally when we use a permanent magnet DC motor we introduce both a speed and a current control because it is important to control the current flowing in the circuit, to avoid excessive Joule effect for safety reasons (more complex scheme but safer). Both the speed and the current controllers can be a P, PD, PI, …, controller: Speed controller : speed as input , current as output � ���0;���`; : � 6 ,>9I = A F � Ω >9I − Ω � �e ���0;���`; : � 6 ,>9I = � A F + � C � � � Ω >9I − Ω � �� ���0;���`; : � 6 ,>9I = U A F + 1 � 7 � V � Ω >9I − Ω � Current controller : current as input , voltage as output � ���0;���`; : - 6 = A F � � 6 ,>9I − � 6 � �e ���0;���`; : - 6 = � A F + � C � � � � 6 ,>9I − � 6 � �� ���0;���`; : - 6 = U A F + 1 � 7 � V � � 6 ,>9I − � 6 � Torque control on permanent magnet DC motor If we want to apply a torque control to a DC motor we can use a current controller , since the torque generated by the motor depends on the current: � E ,>9I = A ~ � 6 ,>9I → � 6 ,>9I = � E ,>9I A ~ So the block diagram and the controller will be exactly the same as the p