cmd_sys_1.py

Created on March 05, 2024
1.47 KB
Exercice 3

         2
H(p)= --------
       p(p+2)
       
           2
H(jw)= ----------
        jw(jw+2)
        
                    
Module :
                  2
|H(jw)| = ------------------
           w(2^2 + w^2)^1/2
           
Argument :

phi(H)=-90-arctg(w/2)


Marge phase initiale:delta phi i

|H|=1 -> w(2^2 + w^2)^1/2 = 2

pour w = 1 -> ...=sqrt(5)>2
pour w = 0.5 -> ...= ...<2
pour w = 0.9 -> ...= 2

=> wc = 0.9 rad/s

phi(H)=-90 * tg^-1(0.9/2)= -114deg

delta phi i = -180+114 = -66deg

/////////

Reglage :
pour w = 3 et deltaf = -60deg

phi(H)=-90-tg^-1(3/2) = -146deg

-180 = phi(H)+deltaf+phi(cor)
-180 = -146 - 60 + phi(cor)
phi(cor) = 26deg

        1+sin(phimax)
=> a = ---------------
        1-sin(phimax)
        
        1+sin(30)
     = ----------- = 3
        1-sin(30)

wx = 3 = 1/T*sqrt(a)

=> T = 1/3*sqrt(3) = 0.2

|H|w=3 = 2/3 * 1/(2^2 + 3^2)
       = 2/3 * 1/sqrt(13)
       = 0.18

            (1+jawt)
C(jw)= K * ----------
            (1+jwt)
            
              sqrt(1+(3*0.2*3)^2)
|C|w=3 = K * --------------------
              sqrt(1+(3*0.2)^2)
              
       = K * 1.76
       
puisque |H|w=3 * |C|w=3 = 1

on peut poser 0.18*1.76*K=1
K=3.16

///////////////////////////////

E(p)= Ref(p) / 1+Bo(p)

E(t)t->00 = (p*ε(p))p=0 =0.15

Bo(p)= C(p)*H(p) 

      (1+aTp)     2
= K * ------- * ------
      (1+Tp)    p(p+2)

E(t)t->00 = 2/2K = 1/K=0.15
=> K=6

Remarque :
On a trouve K=3 par avance de 
phase et il faut k=6 pour la 
précision