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

Short answer

Perform partial-fraction expansion: (z+1)/((z+0.5)(z+0.8)) = 5/3/(z+0.5) − 2/3/(z+0.8). Define states X1=U/(z+0.5), X2=U/(z+0.8). The diagonal state-space form has G=diag(−0.5,−0.8), H=[1;1], C=[5/3, −2/3].

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 diagonal canonical form.

The diagonal canonical form uses the partial-fraction expansion. Each distinct pole becomes a decoupled first-order state.

Poles

z2+1.3z+0.4=(z+0.5)(z+0.8)z^2 + 1.3z + 0.4 = (z+0.5)(z+0.8)

The poles are p1=−0.5p_1 = -0.5 and p2=−0.8p_2 = -0.8 (distinct). b0=0b_0 = 0.

Partial fractions

Y(z)U(z)=z+1(z+0.5)(z+0.8)=c1z+0.5+c2z+0.8\frac{Y(z)}{U(z)} = \frac{z+1}{(z+0.5)(z+0.8)} = \frac{c_1}{z+0.5} + \frac{c_2}{z+0.8} c1=[z+1z+0.8]z=−0.5=0.50.3=53=1.6667c2=[z+1z+0.5]z=−0.8=0.2−0.3=−23=−0.6667\begin{aligned} c_1 &= \left[\frac{z+1}{z+0.8}\right]_{z=-0.5} = \frac{0.5}{0.3} = \frac{5}{3} = 1.6667\\ c_2 &= \left[\frac{z+1}{z+0.5}\right]_{z=-0.8} = \frac{0.2}{-0.3} = -\frac{2}{3} = -0.6667 \end{aligned}

States

X1(z)=U(z)z+0.5X_1(z) = \dfrac{U(z)}{z+0.5}, X2(z)=U(z)z+0.8X_2(z) = \dfrac{U(z)}{z+0.8}:

x1(k+1)=−0.5x1(k)+u(k)x2(k+1)=−0.8x2(k)+u(k)y(k)=53x1(k)−23x2(k)\begin{aligned} x_1(k+1) &= -0.5x_1(k) + u(k)\\ x_2(k+1) &= -0.8x_2(k) + u(k)\\ y(k) &= \tfrac{5}{3}x_1(k) - \tfrac{2}{3}x_2(k) \end{aligned}

Diagonal canonical form

[x1(k+1)x2(k+1)]=[−0.500−0.8][x1(k)x2(k)]+[11]u(k)\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} -0.5 & 0 \\ 0 & -0.8 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \end{bmatrix}u(k) y(k)=[53−23][x1(k)x2(k)]y(k) = \begin{bmatrix} \tfrac{5}{3} & -\tfrac{2}{3} \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}

Check

5/3z+0.5−2/3z+0.8=53z+43−23z−13(z+0.5)(z+0.8)=z+1z2+1.3z+0.4\frac{5/3}{z+0.5} - \frac{2/3}{z+0.8} = \frac{\tfrac{5}{3}z + \tfrac{4}{3} - \tfrac{2}{3}z - \tfrac{1}{3}}{(z+0.5)(z+0.8)} = \frac{z+1}{z^2+1.3z+0.4} 
        +-->[1/(z+0.5)]--x1-->[ 5/3]--+
 u -----|                            (+)--> y
        +-->[1/(z+0.8)]--x2-->[-2/3]--+

Both eigenvalues lie inside the unit circle, so the system is stable. In this form each mode can be seen directly.

✓ Answer checked by Claude Opus 5.5 (high effort)

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