Discrete Time State Equations

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

Standard form

For

$$\frac{Y(z)}{U(z)} = \frac{b_0 + b_1z^{-1} + b_2z^{-2}}{1 + a_1z^{-1} + a_2z^{-2}}$$

the observable canonical form (Ogata) is

$$\mathbf{x}(k+1) = \begin{bmatrix} 0 & -a_2 \\ 1 & -a_1 \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} b_2 - a_2b_0 \\ b_1 - a_1b_0 \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\mathbf{x}(k) + b_0u(k)$$

Derivation

Cross-multiplying gives $Y(z) = -a_1z^{-1}Y - a_2z^{-2}Y + b_1z^{-1}U + b_2z^{-2}U$ (here $b_0 = 0$), so

$$Y(z) = z^{-1}\Big[b_1U - a_1Y + z^{-1}\big(b_2U - a_2Y\big)\Big]$$

Define $X_2(z) = Y(z)$ and $X_1(z) = z^{-1}(b_2U - a_2Y)$. Then

$$\begin{aligned} x_1(k+1) &= -a_2x_2(k) + b_2u(k)\\ x_2(k+1) &= x_1(k) - a_1x_2(k) + b_1u(k)\\ y(k) &= x_2(k) \end{aligned}$$

Coefficients

Result

$$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & -0.368 \\ 1 & 1.368 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0.264 \\ 0.368 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}$$

Check

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

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

The pulse transfer function is $F(z) = \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + D$, obtained by taking the z-transform of the state equations with zero initial state.

Given

$$\mathbf{G} = \begin{bmatrix} -a_1 & 1 & 0 \\ -a_2 & 0 & 1 \\ -a_3 & 0 & 0 \end{bmatrix},\quad \mathbf{H} = \begin{bmatrix} h_1 \\ h_2 \\ h_3 \end{bmatrix},\quad \mathbf{C} = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix},\quad D = b_0$$

Step 1: $z\mathbf{I} - \mathbf{G}$ and its determinant

$$z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z+a_1 & -1 & 0 \\ a_2 & z & -1 \\ a_3 & 0 & z \end{bmatrix}$$

Expanding along the first row:

$$\begin{aligned} |z\mathbf{I}-\mathbf{G}| &= (z+a_1)z^2 + 1\,(a_2z + a_3)\\ &= z^3 + a_1z^2 + a_2z + a_3 \end{aligned}$$

Step 2: First row of the adjoint

Only the first row of $(z\mathbf{I}-\mathbf{G})^{-1}$ is needed, because $\mathbf{C} = [1\ 0\ 0]$. The first row of the adjoint (the cofactors of the first column, transposed) is

$$\begin{bmatrix} z^2 & z & 1 \end{bmatrix}$$

Check: cofactor $C_{11} = z\cdot z - 0 = z^2$, $C_{21} = -(-1\cdot z - 0) = z$, $C_{31} = (-1)(-1) - 0 = 1$.

Step 3: $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H}$

$$\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{\begin{bmatrix} z^2 & z & 1 \end{bmatrix}\begin{bmatrix} h_1 \\ h_2 \\ h_3 \end{bmatrix}}{z^3+a_1z^2+a_2z+a_3} = \frac{h_1z^2 + h_2z + h_3}{z^3+a_1z^2+a_2z+a_3}$$

Step 4: Add D

$$\begin{aligned} F(z) &= b_0 + \frac{h_1z^2 + h_2z + h_3}{z^3+a_1z^2+a_2z+a_3}\\ &= \frac{b_0z^3 + (h_1 + a_1b_0)z^2 + (h_2 + a_2b_0)z + (h_3 + a_3b_0)}{z^3 + a_1z^2 + a_2z + a_3} \end{aligned}$$

Remark

This is the observable canonical form written with the state order reversed. Comparing with $\dfrac{b_0z^3+b_1z^2+b_2z+b_3}{z^3+a_1z^2+a_2z+a_3}$ gives $h_i = b_i - a_ib_0$, which is the standard result for the observable form. Since there is one input and one output, the "matrix" is the 1×1 function above.

Answer: $F(z) = \dfrac{b_0z^3 + (h_1+a_1b_0)z^2 + (h_2+a_2b_0)z + (h_3+a_3b_0)}{z^3+a_1z^2+a_2z+a_3}$

When the denominator has a repeated pole, the diagonal form is not possible, and the Jordan canonical form is used. The repeated pole gives a Jordan block with 1 on the superdiagonal.

Partial fractions

$$\frac{Y(z)}{U(z)} = \frac{5}{(z+1)^2(z+2)} = \frac{c_1}{(z+1)^2} + \frac{c_2}{z+1} + \frac{c_3}{z+2}$$ $$\begin{aligned} c_1 &= \left[\frac{5}{z+2}\right]_{z=-1} = 5\\ c_2 &= \frac{d}{dz}\left[\frac{5}{z+2}\right]_{z=-1} = \left[\frac{-5}{(z+2)^2}\right]_{z=-1} = -5\\ c_3 &= \left[\frac{5}{(z+1)^2}\right]_{z=-2} = 5 \end{aligned}$$

So $\dfrac{Y(z)}{U(z)} = \dfrac{5}{(z+1)^2} - \dfrac{5}{z+1} + \dfrac{5}{z+2}$.

Choice of states

$$X_3(z) = \frac{U(z)}{z+2},\qquad X_2(z) = \frac{U(z)}{z+1},\qquad X_1(z) = \frac{X_2(z)}{z+1} = \frac{U(z)}{(z+1)^2}$$

This gives

$$\begin{aligned} x_1(k+1) &= -x_1(k) + x_2(k)\\ x_2(k+1) &= -x_2(k) + u(k)\\ x_3(k+1) &= -2x_3(k) + u(k)\\ y(k) &= 5x_1(k) - 5x_2(k) + 5x_3(k) \end{aligned}$$

Jordan canonical form

$$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \\ x_3(k+1) \end{bmatrix} = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \\ x_3(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} 5 & -5 & 5 \end{bmatrix}\mathbf{x}(k) + 0\cdot u(k)$$

Check: $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \dfrac{5}{z^3+4z^2+5z+2} = \dfrac{5}{(z+1)^2(z+2)}$.

The pulse transfer function (matrix) relates the z-transform of the output to that of the input, with all initial conditions zero. It is found by taking the z-transform of the state and output equations.

Given system

$$\begin{aligned} \mathbf{x}(k+1) &= \mathbf{G}\mathbf{x}(k) + \mathbf{H}\mathbf{u}(k)\\ \mathbf{y}(k) &= \mathbf{C}\mathbf{x}(k) + \mathbf{D}\mathbf{u}(k) \end{aligned}$$

Here $\mathbf{x}$ is an n-vector, $\mathbf{u}$ an r-vector, $\mathbf{y}$ an m-vector, and $\mathbf{G}$ ($n\times n$), $\mathbf{H}$ ($n\times r$), $\mathbf{C}$ ($m\times n$), $\mathbf{D}$ ($m\times r$) are constant matrices.

Derivation

1. Take the z-transform of the state equation, using $\mathcal{Z}[\mathbf{x}(k+1)] = z\mathbf{X}(z) - z\mathbf{x}(0)$:

$$z\mathbf{X}(z) - z\mathbf{x}(0) = \mathbf{G}\mathbf{X}(z) + \mathbf{H}\mathbf{U}(z)$$

2. Put $\mathbf{x}(0) = \mathbf{0}$ (a transfer function is defined for zero initial state):

$$(z\mathbf{I} - \mathbf{G})\mathbf{X}(z) = \mathbf{H}\mathbf{U}(z)$$

3. Pre-multiply by $(z\mathbf{I}-\mathbf{G})^{-1}$, which exists for all $z$ except the eigenvalues of $\mathbf{G}$:

$$\mathbf{X}(z) = (z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H}\mathbf{U}(z)$$

4. Take the z-transform of the output equation:

$$\mathbf{Y}(z) = \mathbf{C}\mathbf{X}(z) + \mathbf{D}\mathbf{U}(z)$$

5. Substitute $\mathbf{X}(z)$:

$$\mathbf{Y}(z) = \left[\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + \mathbf{D}\right]\mathbf{U}(z)$$

Result

$$\mathbf{F}(z) = \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + \mathbf{D} = \frac{\mathbf{C}\,\operatorname{adj}(z\mathbf{I}-\mathbf{G})\,\mathbf{H}}{|z\mathbf{I}-\mathbf{G}|} + \mathbf{D}$$

$\mathbf{F}(z)$ is an $m\times r$ matrix. Element $F_{ij}(z)$ is the pulse transfer function from input $u_j$ to output $y_i$. For a single-input single-output system it is a scalar.

Properties

• Characteristic equation: $|z\mathbf{I}-\mathbf{G}| = 0$. The poles of $\mathbf{F}(z)$ are eigenvalues of $\mathbf{G}$, unless a pole is cancelled by a zero.
• Invariance: a change of state variables $\mathbf{x} = \mathbf{P}\hat{\mathbf{x}}$ gives $\hat{\mathbf{G}} = \mathbf{P}^{-1}\mathbf{G}\mathbf{P}$, $\hat{\mathbf{H}} = \mathbf{P}^{-1}\mathbf{H}$, $\hat{\mathbf{C}} = \mathbf{C}\mathbf{P}$, and

$$\hat{\mathbf{C}}(z\mathbf{I}-\hat{\mathbf{G}})^{-1}\hat{\mathbf{H}} + \mathbf{D} = \mathbf{C}\mathbf{P}\,\mathbf{P}^{-1}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{P}\,\mathbf{P}^{-1}\mathbf{H} + \mathbf{D} = \mathbf{F}(z)$$

So the pulse transfer function is unique, even though the state-space form is not.

• $\mathbf{D} \neq 0$ means a direct feed-through, and the numerator degree equals the denominator degree.

Example

For $\mathbf{G} = \begin{bmatrix} 0 & 1 \\ -0.16 & -1 \end{bmatrix}$, $\mathbf{H} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$, $\mathbf{C} = [1\ \ 0]$, $D = 0$:

$$F(z) = \frac{[1\ \ 0]\begin{bmatrix} z+1 & 1 \\ -0.16 & z \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix}}{z^2 + z + 0.16} = \frac{1}{z^2 + z + 0.16}$$

The pole $z=-2$ is repeated, so the Jordan canonical form is used. The numerator and denominator have the same degree, so there is a direct term $D = b_0$.

Separate the direct term

Denominator: $(z+2)^2(z+3) = z^3 + 7z^2 + 16z + 12$.

$$\frac{2z^3+6z+5}{z^3+7z^2+16z+12} = 2 + \frac{-14z^2 - 26z - 19}{(z+2)^2(z+3)}$$

So $D = b_0 = 2$.

Partial fractions

$$\frac{Y(z)}{U(z)} = 2 + \frac{c_1}{(z+2)^2} + \frac{c_2}{z+2} + \frac{c_3}{z+3}$$

Let $N(z) = 2z^3 + 6z + 5$, so $N(-2) = -16 - 12 + 5 = -23$ and $N(-3) = -54 - 18 + 5 = -67$.

$$\begin{aligned} c_1 &= \left[\frac{N(z)}{z+3}\right]_{z=-2} = \frac{-23}{1} = -23\\ c_2 &= \frac{d}{dz}\left[\frac{N(z)}{z+3}\right]_{z=-2} = \left[\frac{(6z^2+6)(z+3) - N(z)}{(z+3)^2}\right]_{z=-2} = \frac{30 + 23}{1} = 53\\ c_3 &= \left[\frac{N(z)}{(z+2)^2}\right]_{z=-3} = \frac{-67}{1} = -67 \end{aligned}$$ $$\frac{Y(z)}{U(z)} = 2 - \frac{23}{(z+2)^2} + \frac{53}{z+2} - \frac{67}{z+3}$$

States

$X_2 = \dfrac{U}{z+2}$, $X_1 = \dfrac{X_2}{z+2}$, $X_3 = \dfrac{U}{z+3}$:

$$\begin{aligned} x_1(k+1) &= -2x_1(k) + x_2(k)\\ x_2(k+1) &= -2x_2(k) + u(k)\\ x_3(k+1) &= -3x_3(k) + u(k)\\ y(k) &= -23x_1(k) + 53x_2(k) - 67x_3(k) + 2u(k) \end{aligned}$$

Jordan canonical form

$$\mathbf{x}(k+1) = \begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -3 \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} -23 & 53 & -67 \end{bmatrix}\mathbf{x}(k) + 2u(k)$$

Check: $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + 2$ gives back $\dfrac{2z^3+6z+5}{(z+2)^2(z+3)}$.

The pulse transfer function matrix is $\mathbf{F}(z) = \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + \mathbf{D}$. Here there is one input and two outputs, so $\mathbf{F}(z)$ is 2×1.

Step 1: $(z\mathbf{I}-\mathbf{G})^{-1}$

$$z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z+4 & -1 \\ 2 & z+1 \end{bmatrix}$$ $$|z\mathbf{I}-\mathbf{G}| = (z+4)(z+1) + 2 = z^2 + 5z + 6 = (z+2)(z+3)$$ $$(z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{(z+2)(z+3)}\begin{bmatrix} z+1 & 1 \\ -2 & z+4 \end{bmatrix}$$

Step 2: $(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H}$

$$(z\mathbf{I}-\mathbf{G})^{-1}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \frac{1}{(z+2)(z+3)}\begin{bmatrix} z+2 \\ z+2 \end{bmatrix} = \begin{bmatrix} \dfrac{1}{z+3} \\[2mm] \dfrac{1}{z+3} \end{bmatrix}$$

Step 3: Multiply by C (D = 0)

$$\mathbf{F}(z) = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} \dfrac{1}{z+3} \\[2mm] \dfrac{1}{z+3} \end{bmatrix} = \begin{bmatrix} \dfrac{2}{z+3} \\[2mm] \dfrac{3}{z+3} \end{bmatrix}$$

Remarks

• $\dfrac{Y_1(z)}{U(z)} = \dfrac{2}{z+3}$ and $\dfrac{Y_2(z)}{U(z)} = \dfrac{3}{z+3}$.
• The eigenvalue $z = -2$ is cancelled by a zero: the mode at $-2$ is not excited by $u$ (it is uncontrollable). This is why the transfer functions are first order although the system is second order.
• Both eigenvalues ($-2$, $-3$) lie outside the unit circle, so the system is unstable.

Answer: $\mathbf{F}(z) = \begin{bmatrix} 2/(z+3) \\ 3/(z+3) \end{bmatrix}$

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

Normalise

$$\frac{Y(z)}{X(z)} = \frac{2z^3 + 1.5z^2 + 2.5z + 2}{z^3 + 2.5z^2 + z + 1.5}$$

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

$$\begin{aligned} b_3 - a_3b_0 &= 2 - 3 = -1\\ b_2 - a_2b_0 &= 2.5 - 2 = 0.5\\ b_1 - a_1b_0 &= 1.5 - 5 = -3.5 \end{aligned}$$

Controllable canonical form

With $x(k)$ as the input, let $u(k)$ denote it, and choose states as successive delays of an intermediate variable:

$$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \\ x_3(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -1.5 & -1 & -2.5 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \\ x_3(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} -1 & 0.5 & -3.5 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \\ x_3(k) \end{bmatrix} + 2u(k)$$

Check

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

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]--+

Discretization with a zero-order hold gives $\mathbf{G} = e^{\mathbf{A}T}$ and $\mathbf{H} = \left(\int_0^T e^{\mathbf{A}\lambda}d\lambda\right)\mathbf{B}$, with $\mathbf{C}$ and $\mathbf{D}$ unchanged. The sampling period is not given, so the result is found for a general T and then evaluated for T = 1 s.

Step 1: State transition matrix $e^{\mathbf{A}t}$

$$\mathbf{A} = \begin{bmatrix} 0 & 1 \\ 0 & -2 \end{bmatrix},\quad \mathbf{B} = \begin{bmatrix} 0 \\ 1 \end{bmatrix},\quad \mathbf{C} = \begin{bmatrix} 0 & 1 \end{bmatrix}$$ $$(s\mathbf{I}-\mathbf{A})^{-1} = \begin{bmatrix} s & -1 \\ 0 & s+2 \end{bmatrix}^{-1} = \begin{bmatrix} \dfrac{1}{s} & \dfrac{1}{s(s+2)} \\[2mm] 0 & \dfrac{1}{s+2} \end{bmatrix}$$ $$e^{\mathbf{A}t} = \mathcal{L}^{-1}\left[(s\mathbf{I}-\mathbf{A})^{-1}\right] = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2t}) \\ 0 & e^{-2t} \end{bmatrix}$$

Step 2: G and H

$$\mathbf{G} = e^{\mathbf{A}T} = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2T}) \\ 0 & e^{-2T} \end{bmatrix}$$ $$\mathbf{H} = \int_0^T \begin{bmatrix} \tfrac{1}{2}(1-e^{-2\lambda}) \\ e^{-2\lambda} \end{bmatrix}d\lambda = \begin{bmatrix} \tfrac{1}{4}\left(2T - 1 + e^{-2T}\right) \\ \tfrac{1}{2}(1-e^{-2T}) \end{bmatrix}$$

Discretized model

$$\mathbf{x}(k+1) = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2T}) \\ 0 & e^{-2T} \end{bmatrix}\mathbf{x}(k) + \begin{bmatrix} \tfrac{1}{4}(2T-1+e^{-2T}) \\ \tfrac{1}{2}(1-e^{-2T}) \end{bmatrix}u(k),\qquad y(k) = \begin{bmatrix} 0 & 1 \end{bmatrix}\mathbf{x}(k)$$

For T = 1 s ($e^{-2} = 0.1353$):

$$\mathbf{G} = \begin{bmatrix} 1 & 0.4323 \\ 0 & 0.1353 \end{bmatrix},\qquad \mathbf{H} = \begin{bmatrix} 0.2838 \\ 0.4323 \end{bmatrix}$$

Step 3: Pulse transfer function

$$(z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{(z-1)(z-e^{-2T})}\begin{bmatrix} z-e^{-2T} & \tfrac{1}{2}(1-e^{-2T}) \\ 0 & z-1 \end{bmatrix}$$

With $\mathbf{C} = [0\ \ 1]$ only the second row is needed:

$$\begin{aligned} F(z) &= \frac{[0\ \ z-1]\,\mathbf{H}}{(z-1)(z-e^{-2T})} = \frac{(z-1)\tfrac{1}{2}(1-e^{-2T})}{(z-1)(z-e^{-2T})}\\ &= \frac{\tfrac{1}{2}(1-e^{-2T})}{z-e^{-2T}} \end{aligned}$$

For T = 1 s:

$$F(z) = \frac{0.4323}{z - 0.1353}$$

Remark

The output $y = x_2$ is the velocity, so the integrator pole at $z = 1$ cancels. The result matches the direct ZOH transform of $1/(s+2)$: $(1-z^{-1})\mathcal{Z}\left[\frac{1}{s(s+2)}\right] = \frac{0.5(1-e^{-2T})}{z-e^{-2T}}$.

Answer: $F(z) = \dfrac{0.5(1-e^{-2T})}{z-e^{-2T}}$, which is $\dfrac{0.4323}{z-0.1353}$ for T = 1 s.

The pulse transfer function matrix is $\mathbf{F}(z) = \mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} + \mathbf{D}$. There is one input and two outputs, so $\mathbf{F}(z)$ is 2×1.

Given

$$\mathbf{G} = \begin{bmatrix} 3 & 1 \\ -2 & 1 \end{bmatrix},\quad \mathbf{H} = \begin{bmatrix} 1 \\ 1 \end{bmatrix},\quad \mathbf{C} = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix},\quad \mathbf{D} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

Step 1: Inverse of $z\mathbf{I}-\mathbf{G}$

$$z\mathbf{I}-\mathbf{G} = \begin{bmatrix} z-3 & -1 \\ 2 & z-1 \end{bmatrix},\qquad |z\mathbf{I}-\mathbf{G}| = (z-3)(z-1) + 2 = z^2 - 4z + 5$$ $$(z\mathbf{I}-\mathbf{G})^{-1} = \frac{1}{z^2-4z+5}\begin{bmatrix} z-1 & 1 \\ -2 & z-3 \end{bmatrix}$$

Step 2: Multiply by H

$$(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{1}{z^2-4z+5}\begin{bmatrix} z \\ z-5 \end{bmatrix}$$

Step 3: Multiply by C

$$\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \frac{1}{z^2-4z+5}\begin{bmatrix} z + (z-5) \\ 2z + (z-5) \end{bmatrix} = \frac{1}{z^2-4z+5}\begin{bmatrix} 2z-5 \\ 3z-5 \end{bmatrix}$$

Step 4: Add D

$$\begin{aligned} F_1(z) &= \frac{2z-5}{z^2-4z+5} + 1 = \frac{z^2-2z}{z^2-4z+5}\\ F_2(z) &= \frac{3z-5}{z^2-4z+5} \end{aligned}$$

Result

$$\mathbf{F}(z) = \begin{bmatrix} \dfrac{z(z-2)}{z^2-4z+5} \\[3mm] \dfrac{3z-5}{z^2-4z+5} \end{bmatrix}$$

The poles are $z = 2 \pm j1$ ($|z| = \sqrt5 > 1$), so the system is unstable.

Answer: $Y_1(z)/U(z) = \dfrac{z^2-2z}{z^2-4z+5}$, $Y_2(z)/U(z) = \dfrac{3z-5}{z^2-4z+5}$

In the controllable canonical form the states are successive delays of an intermediate variable. The denominator coefficients form the last row of $\mathbf{G}$.

Write in powers of z

$$\frac{Y(z)}{U(z)} = \frac{z^{-1}+2z^{-2}}{1+4z^{-1}+3z^{-2}} = \frac{z+2}{z^2+4z+3}$$

So $a_1 = 4$, $a_2 = 3$, $b_0 = 0$, $b_1 = 1$, $b_2 = 2$.

Derivation

Introduce $Q(z)$ with $\dfrac{Q(z)}{U(z)} = \dfrac{1}{z^2+4z+3}$, so $Y(z) = (z+2)Q(z)$. Choose $x_1(k) = q(k)$ and $x_2(k) = q(k+1)$:

$$\begin{aligned} x_1(k+1) &= x_2(k)\\ x_2(k+1) &= q(k+2) = -3x_1(k) - 4x_2(k) + u(k)\\ y(k) &= q(k+1) + 2q(k) = 2x_1(k) + x_2(k) \end{aligned}$$

Controllable canonical form

$$\begin{bmatrix} x_1(k+1) \\ x_2(k+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -3 & -4 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix}u(k)$$ $$y(k) = \begin{bmatrix} 2 & 1 \end{bmatrix}\begin{bmatrix} x_1(k) \\ x_2(k) \end{bmatrix}$$

Check: $\mathbf{C}(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H} = \dfrac{[2\ \ 1]\begin{bmatrix} 1 \\ z \end{bmatrix}}{z^2+4z+3} = \dfrac{z+2}{z^2+4z+3}$.