Se ha denunciado esta presentación.
Se está descargando tu SlideShare. ×

K10692 control theory

Anuncio
Anuncio
Anuncio
Anuncio
Anuncio
Anuncio
Anuncio
Anuncio
Anuncio
Anuncio
Anuncio
Anuncio
Próximo SlideShare
Graph Regularised Hashing
Graph Regularised Hashing
Cargando en…3
×

Eche un vistazo a continuación

1 de 13 Anuncio

Más Contenido Relacionado

Presentaciones para usted (20)

Similares a K10692 control theory (20)

Anuncio

Más reciente (20)

K10692 control theory

  1. 1. Sampled-Data Systems :
  2. 2. State Description of Sampled CT plants • A model of an A/D converter: Samplerf(t) f(k) k t 0 1 2 3
  3. 3. A model of D/A converter ZOH t k0 1 2 3 f(k) f+(t) f+(t)=f(k); kT <= t< (k+1)T
  4. 4. DT systemCTsystem sampler Discrete time system ZOH Interconnection of DT and CT system u(k) u+(t) y(t) y(k) Equivalent Discrete Time system
  5. 5. x(k+1)=Fx(k)+gu(k) y(k)=cx(k)+du(k) bdθe Aθg T 0  F = eAt Find eigen values By equation | λI – A | = 0 Get λ1, λ2 … elt = g(l) = b0 + b1 l el2t = g(l2) = b0 + b1 l2 el1t = g(l1) = b0 + b1 l1 eAt = b0+b1A Thus we find λ1, λ2.
  6. 6. Example : • From the given BD find out state equation. The state variable defined by: x1(t)=q(t), x2(t)=dq(t)/dt State Eqn are given by: dx(t)/dt=Ax(t)+bu+(t) y(t)=cx(t) 12-May-16
  7. 7. . 12-May-16 Gh0(s) 1 s(s+5) u(t) - + T=1s u(k) ZOH plant Q(t) u+(t)
  8. 8. • A= 0 1 b= 0 c=[1 0] 0 5 1 Find eigen values, l1 =0, l2=-5 elt=g(l)=b0+b1l eAt=b0+b1A= 1 1/5(1-e-5T) 0 e-5T 12-May-16 Example :
  9. 9. g= 0.2(T-0.2+0.2e-5T) 0.2(1-e-5T) for T=0.1 sec F = 1 0.0787 0 0.6065 g = 0.043 0.0787 bdT eAg  0
  10. 10. Obtain Z-domain TF using MATLAB • . 12-May-16 ZOH 1 s(s+2) R(s) E(s) E*(s) T=1s C(s) _ +
  11. 11. Matlab Simulation : num=1 den=[1 2 0] T=1 [numz,denz]=c2dm(num,den,T,'ZOH') printsys(numz,denz,'z') sys = tf(numDz,denDz,-1) axis([-1 1 -1 1]) zgrid 12-May-16
  12. 12. num = 1 den = 1 2 0 T = 1 numz = 0 0.2838 0.1485 denz = 1.0000 -1.1353 0.1353 num/den = 0.28383 z + 0.1485 ------------------------ z^2 - 1.1353 z + 0.13534 numDz = 1 denDz = 1.0000 -0.3000 0.5000 Transfer function: 1 ----------------- z^2 - 0.3 z + 0.5 Sampling time: unspecified
  13. 13. P = tf(1,[1, 1e-5,0]); T = 1.0/20; C = ss([-1.5 T/4; -2/T -.5],[ .5 2;1/T 1/T],... [-1/T^2 -1.5/T], [1/T^2 0],T); Pd = c2d(P,T,'zoh'); systemnames = 'Pd C'; inputvar = '[ref]'; outputvar = '[Pd]'; input_to_Pd = '[C]'; input_to_C = '[ref ; Pd]'; sysoutname = 'dclp'; cleanupsysic = 'yes'; sysic; [yd,td] = step(dclp,20*T); M = [0,1;1,0;0,1]*blkdiag(1,P); t = [0:.01:1]'; u = ones(size(t)); y1 = sdlsim(M,C,u,t); plot(td,yd,'r*',y1{:},'b-') axis([0,1,0,1.5]) xlabel('Time: seconds') title('Step response: discrete (*) and continuous') y2 = sdlsim(M,C,u,t,1,0,[0.25;0]); plot(td,yd,'r*',y1{:},'b-',y2{:},'g--') axis([0,1,0,1.5]) xlabel('Time: seconds') title('Step response: nonzero initial condition') M2 = [0,1,1;1,0,0;0,1,1]*blkdiag(1,1,P); t = [0:.001:1]'; dist = 0.1*sin(41*t); u = ones(size(t)); [y3,meas,act] = sdlsim(M2,C,[u dist],t,1); plot(y3{:},'-',t,dist,'b--',t,u,'g-.') xlabel('Time: seconds') title('Step response: disturbance (dashed) and output (solid)')

×