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Forma canรณnica observable en tiempo discreto

Considere un sistema con funciรณn de transferencia,

Y(z)U(z)=b1zโˆ’1+b2zโˆ’2+โ€ฆ+bnzโˆ’n1+a1zโˆ’1+a2zโˆ’2+โ€ฆ+anzโˆ’n\frac{Y(z)}{U(z)} = \frac{b_1z^{-1} + b_2z^{-2} + \ldots + b_nz^{-n}}{1 + a_1z^{-1} + a_2z^{-2} + \ldots + a_nz^{-n}} Y(z)+a1zโˆ’1Y(z)+a2zโˆ’2Y(z)+โ€ฆ+anzโˆ’nY(z)=b1zโˆ’1U(z)+โ€ฆ+bnzโˆ’nU(z)Y(z) + a_1z^{-1}Y(z) + a_2z^{-2}Y(z)+\ldots+a_nz^{-n}Y(z) = b_1z^{-1}U(z) + \ldots + b_nz^{-n}U(z) Y(z)=zโˆ’1[b1U(z)โˆ’a1Y(z)]+zโˆ’2[b2U(z)โˆ’a2Y(z)]+โ€ฆ+zโˆ’n+1[bnโˆ’1U(z)โˆ’anโˆ’1Y(z)]+zโˆ’n[bnU(z)โˆ’anY(z)]Y(z) = z^{-1}\left[b_1 U(z) - a_1Y(z)\right] + z^{-2}\left[b_2U(z) - a_2Y(z)\right] + \ldots + z^{-n+1} \left[b_{n-1}U(z) - a_{n-1}Y(z)\right] + z^{-n}\left[b_nU(z) - a_{n}Y(z)\right] Y(z)=zโˆ’1[b1U(z)โˆ’a1Y(z)+zโˆ’1{b2U(z)โˆ’a2Y(z)+โ€ฆ+zโˆ’1(bnโˆ’1U(z)โˆ’anโˆ’1Y(z)+zโˆ’1[bnU(z)โˆ’anY(z)])}]Y(z) = z^{-1}\left[b_1U(z) - a_1Y(z) + z^{-1}\left\{b_2 U(z) - a_2Y(z) + \ldots + z^{-1}\left(b_{n-1} U(z) - a_{n-1}Y(z) + z^{-1}\left[b_n U(z) - a_n Y(z)\right]\right)\right\}\right]

Se definen las variables de estado,

{X1(z)=zโˆ’1[bnU(z)โˆ’anXn(z)]X2(z)=zโˆ’1[bnโˆ’1U(z)โˆ’anโˆ’1Xn(z)+X1(z)]โ‹ฎXnโˆ’1(z)=zโˆ’1[b2U(z)โˆ’a2Xn(z)+Xnโˆ’2(z)]Xn(z)=zโˆ’1[b1U(z)โˆ’a1Xn(z)+Xnโˆ’1(z)]=Y(z)(a)\tag{a}\begin{cases} \begin{aligned} X_1(z) &= z^{-1}\left[b_nU(z) - a_nX_n(z)\right]\\ X_2(z) &= z^{-1}\left[b_{n-1}U(z) - a_{n-1}X_n(z) + X_1(z)\right]\\ &\vdots\\ X_{n-1}(z) &= z^{-1}\left[b_2U(z) - a_2X_n(z) + X_{n-2}(z)\right]\\ X_n(z) &= z^{-1}\left[b_1U(z) - a_1X_n(z) + X_{n-1}(z)\right] = Y(z)\\ \end{aligned} \end{cases}

Multiplicando ambos lados del sistema aa por zz:

{zX1(z)=[bnU(z)โˆ’anXn(z)]zX2(z)=[bnโˆ’1U(z)โˆ’anโˆ’1Xn(z)+X1(z)]โ‹ฎzXnโˆ’1(z)=[b2U(z)โˆ’a2Xn(z)+Xnโˆ’2(z)]zXn(z)=[b1U(z)โˆ’a1Xn(z)+Xnโˆ’1(z)]Y(z)=Xn(z)(b)\tag{b}\begin{cases} \begin{aligned} zX_1(z) &= \left[b_nU(z) - a_nX_n(z)\right]\\ zX_2(z) &= \left[b_{n-1}U(z) - a_{n-1}X_n(z) + X_1(z)\right]\\ &\vdots\\ zX_{n-1}(z) &= \left[b_2U(z) - a_2X_n(z) + X_{n-2}(z)\right]\\ zX_n(z) &= \left[b_1U(z) - a_1X_n(z) + X_{n-1}(z)\right]\\ Y(z) &= X_n(z) \end{aligned} \end{cases}

Aplicando la transformada inversa al sistema bb:

{x1(k+1)=โˆ’anxn(k)+bnu(k)x2(k+1)=x1(k)โˆ’anโˆ’1xn(k)+bnโˆ’1u(k)โ‹ฎxnโˆ’1(k+1)=xnโˆ’2(k)โˆ’a2xn(k)+b2u(k)xn(k+1)=xnโˆ’1(k)โˆ’a1xn(k)+b1u(k)y(k)=xn(k)(c)\tag{c}\begin{cases} \begin{aligned} x_1(k+1) &= - a_nx_n(k) + b_nu(k) \\ x_2(k+1) &= x_1(k) - a_{n-1}x_n(k) + b_{n-1}u(k) \\ &\vdots\\ x_{n-1}(k+1) &= x_{n-2}(k) - a_2x_n(k) + b_2u(k)\\ x_n(k+1) &= x_{n-1}(k) - a_1x_n(k) + b_1u(k)\\ y(k) &= x_n(k) \end{aligned} \end{cases}

En forma matricial:

x(k+1)=[000โ€ฆ0โˆ’an100โ€ฆ0โˆ’anโˆ’1010โ€ฆ0โˆ’anโˆ’2โ‹ฎโ‹ฑโ‹ฎโ‹ฎ00โ€ฆ10โˆ’a200โ€ฆ01โˆ’a1]x(k)+[bnbnโˆ’1โ‹ฎb2b1]u(k)y(k)=[00โ€ฆ01]x(k)โŸForma canoหŠnica observable\underbrace{\begin{aligned} x(k+1) &= \begin{bmatrix} 0 & 0 & 0 & \ldots & 0 & -a_n\\ 1 & 0 & 0 & \ldots & 0 & -a_{n-1}\\ 0 & 1 & 0 & \ldots & 0 & -a_{n-2}\\ \vdots & & \ddots & & \vdots & \vdots\\ 0 & 0 & \ldots & 1 & 0 & -a_{2}\\ 0 & 0 & \ldots & 0 & 1 & -a_{1}\\ \end{bmatrix}x(k) + \begin{bmatrix} b_n\\b_{n-1}\\\vdots\\b_2\\b_1 \end{bmatrix}u(k)\\ \\ y(k) &= \begin{bmatrix} 0 & 0 & \ldots & 0 & 1 \end{bmatrix} x(k) \end{aligned}}_\text{Forma canรณnica observable}

Ejercicio

Sea el sistema,

y(k)+5y(kโˆ’1)+6y(kโˆ’2)=u(kโˆ’1)+u(kโˆ’2)y(k) + 5y(k-1) + 6y(k-2) = u(k-1) + u(k-2)
  1. Simular el sistema, y(0)=2y(0) = 2, y(1)=โˆ’3y(1) = -3, u(k)=cosโก(k)u(k) = \cos(k). Graficar y(k)y(k)

    y(k+2)+5y(k+1)+6y(k)=u(k+1)+u(k)y(k+2) + 5y(k+1) + 6y(k) = u(k+1) + u(k) y(k+2)=โˆ’5y(k+1)โˆ’6y(k)+u(k+1)+u(k)y(k+2) = - 5y(k+1) - 6y(k) + u(k+1) + u(k)
     function yk2 = fcn(yk,yk1,k)
     uk1 = cos(k+1);
     uk = cos(k+2);
    
     yk2 = -5*yk1 - 6*yk + uk1 + uk;
    
  2. Obtener la forma canรณnica observable. Graficar y(k)y(k) y comparar con el punto 1.

    x(k+1)=[0โˆ’61โˆ’5]x(k)+[11]u(k)y(k)=[01]x(k)\begin{aligned} x(k+1) &= \begin{bmatrix} 0 & -6\\ 1 & -5\\ \end{bmatrix}x(k) + \begin{bmatrix} 1\\1 \end{bmatrix}u(k)\\ y(k) &= \begin{bmatrix} 0 & 1 \end{bmatrix}x(k) \end{aligned}

    Para k=0k = 0:

    {x2(0)=y(0)=2x2(1)=x1(0)โˆ’5x2(0)+u(0)=x1(0)โˆ’10+1=y(1)=โˆ’3x1(0)=9โˆ’3=6\begin{cases} x_2(0) = y(0) = 2\\ x_2(1) = x_1(0) -5 x_2(0) + u(0) = x_1(0) -10 + 1 = y(1) = -3\\ x_1(0) = 9-3 = 6 \end{cases}
     function [xk1,yk] = fcn(xk,k)
     A = [0 -6
          1 -5];
     B = [1 1]';
     C = [0 1];
    
     uk = cos(k);
    
     xk1 = A*xk + B*uk;
     yk = C*xk;
    

    Simulando ambos sistemas:

    Simulaciรณn

    Se realiza la comparaciรณn de las salidas y(k)y(k):

    Comparaciรณn

    โ–ก\square