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Controllable canonical form of (4z³+3z²+5z+4)/(2z³+5z²+2z+3)

Short answer

Normalize denominator to monic: divide by 2. Controllable canonical form: G=[[0,1,0],[0,0,1],[-1.5,-1,-2.5]], H=[0;0;1], C=[-1,0.5,-3.5], D=2. Coefficients from bᵢ-aᵢb₀: b₃-a₃b₀=-1, b₂-a₂b₀=0.5, b₁-a₁b₀=-3.5. Input renamed u(k) to avoid confusion with state.

Obtain the state space representation of the following pulse transfer function in controllable canonical form: Y(z)/X(z)=(4z³+3z²+5z+4)/(2z³+5z²+2z+3).

The controllable canonical form needs a monic denominator, so first divide the numerator and denominator by 2.

Normalise

Y(z)X(z)=2z3+1.5z2+2.5z+2z3+2.5z2+z+1.5\frac{Y(z)}{X(z)} = \frac{2z^3 + 1.5z^2 + 2.5z + 2}{z^3 + 2.5z^2 + z + 1.5}
b0b_0b1b_1b2b_2b3b_3a1a_1a2a_2a3a_3
21.52.522.511.5

Since the degrees are equal, D=b0=2D = b_0 = 2 and

b3−a3b0=2−3=−1b2−a2b0=2.5−2=0.5b1−a1b0=1.5−5=−3.5\begin{aligned} b_3 - a_3b_0 &= 2 - 3 = -1\\ b_2 - a_2b_0 &= 2.5 - 2 = 0.5\\ b_1 - a_1b_0 &= 1.5 - 5 = -3.5 \end{aligned}

Controllable canonical form

The input here is called x(k)x(k), so rename it u(k)u(k) to avoid confusion with the state vector. Choosing the states as successive delays of an intermediate variable gives:

[x1(k+1)x2(k+1)x3(k+1)]=[010001−1.5−1−2.5][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 \\ -1.5 & -1 & -2.5 \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)=[−10.5−3.5][x1(k)x2(k)x3(k)]+2u(k)y(k) = \begin{bmatrix} -1 & 0.5 & -3.5 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \\ x_3(k) \end{bmatrix} + 2u(k)

Check

C(zI−G)−1H=−3.5z2+0.5z−1z3+2.5z2+z+1.5\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \dfrac{-3.5z^2 + 0.5z - 1}{z^3 + 2.5z^2 + z + 1.5}, and adding D=2D = 2 gives 2z3+1.5z2+2.5z+2z3+2.5z2+z+1.5\dfrac{2z^3 + 1.5z^2 + 2.5z + 2}{z^3 + 2.5z^2 + z + 1.5}, which is the given function.

✓ Answer checked by Claude Opus 5.5 (high effort)

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