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Pulse TF matrix of a continuous state model

Short answer

Discretize with a zero-order hold: G=e^{AT}, H=∫e^{Aλ}dλ B. For T=1 s, the transfer functions from u to the state are X1(z)/U(z)=(0.2838z+0.1485)/((z−1)(z−0.1353)) and X2(z)/U(z)=0.4323/(z−0.1353).

State space representation of a system is given by [ẋ₁; ẋ₂]=[0 1; 0 -2][x₁; x₂]+[0; 1]u. Obtain the pulse transfer function matrix.

The continuous model is first discretized with a zero-order hold (G=eAT\mathbf{G} = e^{\mathbf{A}T}, H=∫0TeAλdλ B\mathbf{H} = \int_0^T e^{\mathbf{A}\lambda}d\lambda\,\mathbf{B}). The pulse transfer function matrix is then (zI−G)−1H(z\mathbf{I}-\mathbf{G})^{-1}\mathbf{H}. No output equation is given, so the matrix is found from u to the state vector x\mathbf{x}, for general T and for T = 1 s.

Step 1: eAte^{\mathbf{A}t}

A=[010−2],(sI−A)−1=[1s1s(s+2)01s+2]\mathbf{A} = \begin{bmatrix} 0 & 1 \\ 0 & -2 \end{bmatrix},\qquad (s\mathbf{I}-\mathbf{A})^{-1} = \begin{bmatrix} \dfrac{1}{s} & \dfrac{1}{s(s+2)} \\[2mm] 0 & \dfrac{1}{s+2} \end{bmatrix} eAt=[112(1−e−2t)0e−2t]e^{\mathbf{A}t} = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2t}) \\ 0 & e^{-2t} \end{bmatrix}

Step 2: G and H

G=[112(1−e−2T)0e−2T],H=∫0T[12(1−e−2λ)e−2λ]dλ=[14(2T−1+e−2T)12(1−e−2T)]\mathbf{G} = \begin{bmatrix} 1 & \tfrac{1}{2}(1-e^{-2T}) \\ 0 & e^{-2T} \end{bmatrix},\qquad \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}(2T-1+e^{-2T}) \\ \tfrac{1}{2}(1-e^{-2T}) \end{bmatrix}

For T = 1 s: G=[10.432300.1353]\mathbf{G} = \begin{bmatrix} 1 & 0.4323 \\ 0 & 0.1353 \end{bmatrix}, H=[0.28380.4323]\mathbf{H} = \begin{bmatrix} 0.2838 \\ 0.4323 \end{bmatrix}.

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

(zI−G)−1=1(z−1)(z−e−2T)[z−e−2T12(1−e−2T)0z−1](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} X(z)U(z)=[(2T−1+e−2T)z+(1−e−2T−2Te−2T)4(z−1)(z−e−2T)12(1−e−2T)z−e−2T]\frac{\mathbf{X}(z)}{U(z)} = \begin{bmatrix} \dfrac{(2T-1+e^{-2T})z + (1-e^{-2T}-2Te^{-2T})}{4(z-1)(z-e^{-2T})} \\[3mm] \dfrac{\tfrac{1}{2}(1-e^{-2T})}{z-e^{-2T}} \end{bmatrix}

For T = 1 s

X(z)U(z)=[0.2838z+0.1485(z−1)(z−0.1353)0.4323z−0.1353]\frac{\mathbf{X}(z)}{U(z)} = \begin{bmatrix} \dfrac{0.2838z + 0.1485}{(z-1)(z-0.1353)} \\[3mm] \dfrac{0.4323}{z-0.1353} \end{bmatrix}

Interpretation

  • If the output is y=x1y = x_1 (position), the pulse transfer function is the first element. This is the ZOH equivalent of 1/[s(s+2)]1/[s(s+2)].
  • If the output is y=x2y = x_2 (velocity), it is the second element, the ZOH equivalent of 1/(s+2)1/(s+2).
  • For any output y=Cxy = \mathbf{C}\mathbf{x}, the pulse transfer function is C\mathbf{C} times this vector.

✓ Answer checked by Claude Opus 5.5 (high effort)

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