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Controllable canonical form of a third-order system

Short answer

Rewrite as (z³+3z²+5z+4)/(z³+2z²+7z+5). The controllable canonical form matrices are G=[0 1 0; 0 0 1; −5 −7 −2], H=[0;0;1], C=[−1 −2 1], D=1.

Obtain the state space representation of following pulse transfer function in controllable canonical form: Y(z)/U(z)=(1+3z⁻¹+5z⁻²+4z⁻³)/(1+2z⁻¹+7z⁻²+5z⁻³).

Since the numerator and denominator have the same degree, there is a direct term D=b0D = b_0, and the output row uses bi−aib0b_i - a_ib_0.

Standard result

For

Y(z)U(z)=b0zn+b1zn−1+⋯+bnzn+a1zn−1+⋯+an\frac{Y(z)}{U(z)} = \frac{b_0z^n + b_1z^{n-1} + \cdots + b_n}{z^n + a_1z^{n-1} + \cdots + a_n}

the controllable canonical form (Ogata) is

G=[01⋯0⋮⋱⋮00⋯1−an−an−1⋯−a1],H=[0⋮01]\mathbf{G} = \begin{bmatrix} 0 & 1 & \cdots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ -a_n & -a_{n-1} & \cdots & -a_1 \end{bmatrix},\quad \mathbf{H} = \begin{bmatrix} 0 \\ \vdots \\ 0 \\ 1 \end{bmatrix} C=[bn−anb0⋯b1−a1b0],D=b0\mathbf{C} = \begin{bmatrix} b_n - a_nb_0 & \cdots & b_1 - a_1b_0 \end{bmatrix},\quad D = b_0

Coefficients

Y(z)U(z)=1+3z−1+5z−2+4z−31+2z−1+7z−2+5z−3=z3+3z2+5z+4z3+2z2+7z+5\frac{Y(z)}{U(z)} = \frac{1+3z^{-1}+5z^{-2}+4z^{-3}}{1+2z^{-1}+7z^{-2}+5z^{-3}} = \frac{z^3+3z^2+5z+4}{z^3+2z^2+7z+5}
b0b_0b1b_1b2b_2b3b_3a1a_1a2a_2a3a_3
1354275
b3−a3b0=4−5=−1b2−a2b0=5−7=−2b1−a1b0=3−2=1\begin{aligned} b_3 - a_3b_0 &= 4 - 5 = -1\\ b_2 - a_2b_0 &= 5 - 7 = -2\\ b_1 - a_1b_0 &= 3 - 2 = 1 \end{aligned}

Controllable canonical form

[x1(k+1)x2(k+1)x3(k+1)]=[010001−5−7−2][x1(k)x2(k)x3(k)]+[001]u(k)\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \\ x_3(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -5 & -7 & -2 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \\ x_3(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}u(k) y(k)=[−1−21][x1(k)x2(k)x3(k)]+u(k)y(k) = \begin{bmatrix} -1 & -2 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \\ x_3(k) \end{bmatrix} + u(k)

Check

C(zI−G)−1H+D=z2−2z−1z3+2z2+7z+5+1=z3+3z2+5z+4z3+2z2+7z+5\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + D = \frac{z^2 - 2z - 1}{z^3+2z^2+7z+5} + 1 = \frac{z^3+3z^2+5z+4}{z^3+2z^2+7z+5}

✓ Answer checked by Claude Opus 5.5 (high effort)

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