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Controllable canonical form of (z+1)/(z²+1.3z+0.4)

Short answer

With denominator coefficients a1=1.3, a2=0.4 and numerator coefficients b0=0, b1=1, b2=1, introduce an intermediate variable q(k). The controllable canonical form is x(k+1)=[0 1; −0.4 −1.3]x(k)+[0;1]u(k), y=[1 1]x(k).

Consider the following system: Y(z)/U(z)=(z+1)/(z²+1.3z+0.4). Obtain the state space representation of the above pulse transfer function in the controllable canonical form.

In the controllable canonical form the state variables are successive time shifts of an intermediate variable q(k)q(k). G\mathbf{G} is a companion matrix with the denominator coefficients in its last row, and H=[0  1]T\mathbf{H} = [0\ \ 1]^T.

Coefficients

Y(z)U(z)=z+1z2+1.3z+0.4=0⋅z2+1⋅z+1z2+1.3z+0.4\frac{Y(z)}{U(z)} = \frac{z+1}{z^2+1.3z+0.4} = \frac{0\cdot z^2 + 1\cdot z + 1}{z^2 + 1.3z + 0.4}

a1=1.3a_1 = 1.3, a2=0.4a_2 = 0.4, b0=0b_0 = 0, b1=1b_1 = 1, b2=1b_2 = 1.

Derivation

Introduce Q(z)Q(z) such that

Q(z)U(z)=1z2+1.3z+0.4,Y(z)=(z+1)Q(z)\frac{Q(z)}{U(z)} = \frac{1}{z^2+1.3z+0.4},\qquad Y(z) = (z+1)Q(z)

In the time domain, q(k+2)+1.3q(k+1)+0.4q(k)=u(k)q(k+2) + 1.3q(k+1) + 0.4q(k) = u(k) and y(k)=q(k+1)+q(k)y(k) = q(k+1) + q(k). 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)=−0.4x1(k)−1.3x2(k)+u(k)y(k)=x1(k)+x2(k)\begin{aligned} x_1(k+1) &= x_2(k)\\ x_2(k+1) &= -0.4x_1(k) - 1.3x_2(k) + u(k)\\ y(k) &= x_1(k) + x_2(k) \end{aligned}

Controllable canonical form

[x1(k+1)x2(k+1)]=[01−0.4−1.3][x1(k)x2(k)]+[01]u(k)\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -0.4 & -1.3 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k) y(k)=[11][x1(k)x2(k)]y(k) = \begin{bmatrix} 1 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}

Check

C(zI−G)−1H=[11][z+1.31−0.4z][01]z2+1.3z+0.4=1+zz2+1.3z+0.4\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{\begin{bmatrix} 1 & 1 \end{bmatrix}\begin{bmatrix} z+1.3 & 1 \\ -0.4 & z \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix}}{z^2+1.3z+0.4} = \frac{1 + z}{z^2+1.3z+0.4}

✓ Answer checked by Claude Opus 5.5 (high effort)

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