When can the diagonal canonical form be used?

When all zeros are at the origin

Asked 13/22
2081 Chaitra · 4 marks
2080 Chaitra · 8 marks

Diagonal canonical form of (z⁻¹+2z⁻²)/(1+0.7z⁻¹+0.12z⁻²)

Short answer

Convert to positive powers: (z+2)/[(z+0.3)(z+0.4)]. Partial fractions: 17/(z+0.3) - 16/(z+0.4). Diagonal canonical form: G=diag(-0.3,-0.4), H=[1;1], C=[17,-16], D=0. Distinct poles yield decoupled states; each state corresponds to one pole.

Obtain the state space representation of the following pulse transfer function in the diagonal canonical form: Y(z)/U(z)=(z⁻¹+2z⁻²)/(1+0.7z⁻¹+0.12z⁻²).

The diagonal canonical form is obtained from the partial-fraction expansion when all poles are distinct. Each pole becomes one decoupled state.

Write in powers of z

\frac{Y(z)}{U(z)} = \frac{z^{-1}+2z^{-2}}{1+0.7z^{-1}+0.12z^{-2}} = \frac{z+2}{z^2+0.7z+0.12} = \frac{z+2}{(z+0.3)(z+0.4)}

Poles: $p_1 = -0.3$ , $p_2 = -0.4$ (distinct). $b_0 = 0$ .

Partial fractions

\begin{aligned} c_1 &= \left[\frac{z+2}{z+0.4}\right]_{z=-0.3} = \frac{1.7}{0.1} = 17\\ c_2 &= \left[\frac{z+2}{z+0.3}\right]_{z=-0.4} = \frac{1.6}{-0.1} = -16 \end{aligned}

\frac{Y(z)}{U(z)} = \frac{17}{z+0.3} - \frac{16}{z+0.4}

States

$X_1(z) = \dfrac{U(z)}{z+0.3}$ , $X_2(z) = \dfrac{U(z)}{z+0.4}$ :

\begin{aligned} x_1(k+1) &= -0.3x_1(k) + u(k)\\ x_2(k+1) &= -0.4x_2(k) + u(k)\\ y(k) &= 17x_1(k) - 16x_2(k) \end{aligned}

Diagonal canonical form

\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} -0.3 & 0 \\ 0 & -0.4 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \end{bmatrix}u(k)

y(k) = \begin{bmatrix} 17 & -16 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}

Check

\frac{17}{z+0.3} - \frac{16}{z+0.4} = \frac{17z + 6.8 - 16z - 4.8}{(z+0.3)(z+0.4)} = \frac{z+2}{z^2+0.7z+0.12}

        +-->[1/(z+0.3)]--x1-->[17]---+
 u -----|                            (+)--> y
        +-->[1/(z+0.4)]--x2-->[-16]--+

✓ Answer checked by Claude Opus 5.5 (high effort)

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