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Controllable canonical form of (z⁻¹+2z⁻²)/(1+4z⁻¹+3z⁻²)

Short answer

Controllable canonical form: G = [[0,1],[-3,-4]], H = [[0],[1]], C = [2,1]. Obtained from Y/U = (z+2)/(z^2+4z+3) with states as successive delays of intermediate variable.

Obtain a state-space representation of the following pulse-transfer-function system in the controllable canonical form: Y(z)/U(z)=(z⁻¹+2z⁻²)/(1+4z⁻¹+3z⁻²).

In the controllable canonical form the states are successive delays of an intermediate variable. The denominator coefficients form the last row of G\mathbf{G}.

Write in powers of z

Y(z)U(z)=z−1+2z−21+4z−1+3z−2=z+2z2+4z+3\frac{Y(z)}{U(z)} = \frac{z^{-1}+2z^{-2}}{1+4z^{-1}+3z^{-2}} = \frac{z+2}{z^2+4z+3}

So a1=4a_1 = 4, a2=3a_2 = 3, b0=0b_0 = 0, b1=1b_1 = 1, b2=2b_2 = 2.

Derivation

Introduce Q(z)Q(z) with Q(z)U(z)=1z2+4z+3\dfrac{Q(z)}{U(z)} = \dfrac{1}{z^2+4z+3}, so Y(z)=(z+2)Q(z)Y(z) = (z+2)Q(z). Choose x1(k)=q(k)x_1(k) = q(k) and x2(k)=q(k+1)x_2(k) = q(k+1):

x1(k+1)=x2(k)x2(k+1)=q(k+2)=−3x1(k)−4x2(k)+u(k)y(k)=q(k+1)+2q(k)=2x1(k)+x2(k)\begin{aligned} x_1(k+1) &= x_2(k)\\ x_2(k+1) &= q(k+2) = -3x_1(k) - 4x_2(k) + u(k)\\ y(k) &= q(k+1) + 2q(k) = 2x_1(k) + x_2(k) \end{aligned}

Controllable canonical form

[x1(k+1)x2(k+1)]=[01−3−4][x1(k)x2(k)]+[01]u(k)\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -3 & -4 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k) y(k)=[21][x1(k)x2(k)]y(k) = \begin{bmatrix} 2 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}

Check: C(zI−G)−1H=[2  1][1z]z2+4z+3=z+2z2+4z+3\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \dfrac{[2\ \ 1]\begin{bmatrix} 1 \\ z \end{bmatrix}}{z^2+4z+3} = \dfrac{z+2}{z^2+4z+3}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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