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Observable canonical form of (0.368z⁻¹+0.264z⁻²)/(1−1.368z⁻¹+0.368z⁻²)

Short answer

Observable canonical form: x(k+1)=[[0, -0.368],[1, 1.368]]x(k)+[0.264, 0.368]^T u(k), y(k)=[0, 1]x(k). This realises the transfer function (0.368z⁻¹+0.264z⁻²)/(1-1.368z⁻¹+0.368z⁻²).

Obtain the state space representation of the following pulse transfer function in observable canonical form: Y(z)/U(z)=(0.368z⁻¹+0.264z⁻²)/(1-1.368z⁻¹+0.368z⁻²).

In the observable canonical form the denominator coefficients appear in the last column of G\mathbf{G}, and the numerator coefficients appear in H\mathbf{H}.

Standard form

For

Y(z)U(z)=b0+b1z−1+b2z−21+a1z−1+a2z−2\frac{Y(z)}{U(z)} = \frac{b_0 + b_1z^{-1} + b_2z^{-2}}{1 + a_1z^{-1} + a_2z^{-2}}

the observable canonical form (Ogata) is

x(k+1)=[0−a21−a1]x(k)+[b2−a2b0b1−a1b0]u(k),y(k)=[01]x(k)+b0u(k)\mathbf{x}(k+1) = \begin{bmatrix} 0 & -a_2 \\ 1 & -a_1 \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} b_2 - a_2b_0 \\ b_1 - a_1b_0 \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\mathbf{x}(k) + b_0u(k)

Derivation

Cross-multiplying gives Y(z)=−a1z−1Y−a2z−2Y+b1z−1U+b2z−2UY(z) = -a_1z^{-1}Y - a_2z^{-2}Y + b_1z^{-1}U + b_2z^{-2}U (here b0=0b_0 = 0), so

Y(z)=z−1[b1U−a1Y+z−1(b2U−a2Y)]Y(z) = z^{-1}\Big[b_1U - a_1Y + z^{-1}\big(b_2U - a_2Y\big)\Big]

Define X2(z)=Y(z)X_2(z) = Y(z) and X1(z)=z−1(b2U−a2Y)X_1(z) = z^{-1}(b_2U - a_2Y). Then

x1(k+1)=−a2x2(k)+b2u(k)x2(k+1)=x1(k)−a1x2(k)+b1u(k)y(k)=x2(k)\begin{aligned} x_1(k+1) &= -a_2x_2(k) + b_2u(k)\\ x_2(k+1) &= x_1(k) - a_1x_2(k) + b_1u(k)\\ y(k) &= x_2(k) \end{aligned}

Coefficients

CoefficientValue
a1a_1−1.368
a2a_20.368
b0b_00
b1b_10.368
b2b_20.264

Result

[x1(k+1)x2(k+1)]=[0−0.36811.368][x1(k)x2(k)]+[0.2640.368]u(k)\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & -0.368 \\ 1 & 1.368 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0.264 \\ 0.368 \end{bmatrix}u(k) y(k)=[01][x1(k)x2(k)]y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}

Check

C(zI−G)−1H=0.368z+0.264z2−1.368z+0.368\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{0.368z + 0.264}{z^2 - 1.368z + 0.368}

This is the given function (multiply numerator and denominator by z−2z^{-2}).

✓ Answer checked by Claude Opus 5.5 (high effort)

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