DC-Servomechanism. Alberto Bemporad. Corso di Complementi di Controllo Digitale a.a. 2001/2002

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Transcription:

DC-Servomechanism Alberto Bemporad Corso di Complementi di Controllo Digitale a.a. 2001/2002 Figure 1: Servomechanism model (modello del servomeccanismo). 1

Corso di Complementi di Controllo Digitale, Università di Siena, a.a. 2001/2002 2 1 Servomechanism Model The position servomechanism is schematically described in Fig. 1. It consists of a DC-motor (motore in corrente continua), gearbox (scatola di riduzione), elastic shaft (albero di trasmissione elastico) and load (carico meccanico). We derive the equations of the system, in order to obtain a model suitable for control tasks. To this aim, let us consider each physical component of the system. 1.1 DC Motor - Armature Control (Controllo in Tensione 1. di Armatura) V = Ri + e where V = applied armature voltage (tensione di armatura), i =armature current (corrente di armatura), R =resistance, e =back e.m.f. (forza contro-elettromotrice)

Corso di Complementi di Controllo Digitale, Università di Siena, a.a. 2001/2002 3 2. e = k b ω M 3. where ω M = θ M =angular velocity of motor-shaft, k b =constant φ = k f i f where φ =air gap flux, i f =field current (corrente di campo), 4. k f =constant T M = k 1 φi where T M =torque (coppia ) developed by the motor, k 1 =constant 5. By steady-state power balance (equilibrio della potenza in condizioni di regime) T M ω M = ei

Corso di Complementi di Controllo Digitale, Università di Siena, a.a. 2001/2002 4 and therefore k 1 k f i f = k b = k T, k T =motor constant 6. (costante del motore) J M ω M = T M β M ω M T r where J M =equivalent moment of inertia of motor, β M =equivalent viscous friction coefficient (coefficiente di attrito viscoso) of motor, T r =other torques. In summary: V = Ri + k T ω (1) J M ω M = k T i β M ω M T r (2) 1.2 Gear Box (Scatola di Riduzione) Consider the gear box depicted in Fig. 2. 1. Same distance traveled by the two wheels θ 1 r 1 = θ 2 r 2

Corso di Complementi di Controllo Digitale, Università di Siena, a.a. 2001/2002 5 Figure 2: Gear box. where r 1,r 2 = wheel radii (raggi delle ruote dentate) 2. and therefore θ 1 = r 2 = N 2 = ρ θ 2 r 1 N 1 where N 1,N 2 =number of teeth (numero di denti), ρ =gear ratio (rapporto di riduzione). 3. By differentiation, ω 1 = ρω 2 4. Power transmission (no loss) (potenza trasmessa, trascu-

Corso di Complementi di Controllo Digitale, Università di Siena, a.a. 2001/2002 6 rando le perdite) and therefore T 1 ω 1 = T 2 ω 2 T 1 T 2 = 1 ρ By referring again to Fig. 1, we have for θ 1 = θ M, θ 2 = θ s, T 1 = T r, T 2 = T θ s = 1 ρ θ M (3) T r = 1 ρ T (4) where θ M =angular displacement of motor shaft (posizione angolare dell angolo motore), θ s =angular displacement of load-side gear (posizione angolare a valle della scatola di riduzione), T =torque acting on the load (coppia agente sul carico).

Corso di Complementi di Controllo Digitale, Università di Siena, a.a. 2001/2002 7 1.3 Elastic Shaft (Albero di trasmissione elastico) The shaft has finite torsional rigidity (rigidità torsionale) k θ : T = k θ (θ L θ s ) (5) where θ L =angular displacement of load (posizione angolare del carico). 1.4 Load Dynamics (Dinamica del Carico Meccanico) J L ω L = β L ω L T (6) where ω L = θ L =angular load velocity (velocità angolare del carico), J L =equivalent moment of inertia of load, β L =equivalent viscous friction coefficient of load.

Corso di Complementi di Controllo Digitale, Università di Siena, a.a. 2001/2002 8 1.5 Differential Equations of the System By collecting the previous relation, we can write the ω L = k ( θ θ L θ ) M β L ω L (7) J L ρ J L ω M = k ( ) T V kt ω M β Mω M + k ( θ θ L θ ) M (8) J M R J M ρj M ρ 1.6 State-Space Model (Modello in Forma Spazio di Stato) By setting x p [θ L θl θ M θm ], the model can be described by the following state-space form 0 1 0 0 k θ β L k θ ẋ p = J L JL 0 ρj L 0 0 0 1 k θ ρj M 0 k θ [ ] θ L = 1000 x p [ ] θ L = 0100 x p [ ] T = k θ 0 k θ ρ 0 x p J M ρ 2 J M β M+k 2 T /R x p + 0 0 0 k T RJ M V

Corso di Complementi di Controllo Digitale, Università di Siena, a.a. 2001/2002 9 Since the steel shaft (albero in acciaio) has finite shear strength (resistenza al taglio), determined by a maximum admissible τ adm =50N/mm 2, the torsional torque T must satisfy the constraint T 78.5398 Nm (9) Moreover, the input DC voltage V has to be constrained within the range V 220 V (10) The model is transformed in discrete time by sampling (campionando) every T s =0.1sand using a zero-order holder on the input voltage. 1.7 Parameters The parameters of the system are reported in Table 1.

Corso di Complementi di Controllo Digitale, Università di Siena, a.a. 2001/2002 10 2 Esercizio 1. Descrivere con uno script Matlab il modello del sistema (tempo continuo) 2. Calcolare la risposta di θ L, θ L e T dovuta ad un gradino di tensione V = 120 V 3. Calcolare la risposta all impulso 4. Calcolare l evoluzione libera da condizione iniziale θ L (0) = θ M (0) = 0, θ L = 1 rad/s, θ M =20 rad/s 5. Ottenere un modello a tempo discreto del sistema (tempo di campionamento T s =0.1 s) 6. Confrontare la risposta al gradino del sistema discretizzato e del sistema tempo continuo.

Corso di Complementi di Controllo Digitale, Università di Siena, a.a. 2001/2002 11 Table 1: Model parameters Symbol Value (MKS) Meaning L S 1.0 shaft length d S 0.02 shaft diameter J S negligible shaft inertia J M 0.5 motor inertia β M 0.1 motor viscous friction coefficient R 20 resistance of armature k T 10 motor constant ρ 20 gear ratio k θ 1280.2 torsional rigidity J L 50J M nominal load inertia β L 25 load viscous friction coefficient T s 0.1 sampling time