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Control calculations for a control horizon of M=1 can be performed either analytically or numerically. For the process model in Exercise 20.1, derive K _{c 1} for \Delta t=1, N=50, and P=5, Q = I and R = 0, using Eq. 20-65. Compare your answer with the analytical result reported by Maurath et al. (1988).
K _{c 1}=\frac{1}{\sum\limits_{i=1}^{p} S_{i}^{2}}\left[S_{1} S_{2} S_{3} \ldots S_{P}\right]
K _{c} \triangleq\left( S ^{T} Q S + R \right)^{-1} S ^{T} Q (20-65)
From the definition of matrix S, given in Eq. 20-28, for P=5, M=1, with S_{i} obtained from Exercise 20.1,
S \triangleq\left[\begin{array}{cccc}S_{1} & 0 & \cdots & 0 \\S_{2} & S_{1} & 0 & \vdots \\\vdots & \vdots & \ddots & 0 \\S_{M} & S_{M-1} & \cdots & S_{1} \\S_{M+1} & S_{M} & \cdots & S_{2} \\\vdots & \vdots & \ddots & \vdots \\S_{P} & S_{P-1} & \cdots & S_{P-M+1}\end{array}\right] (20-28)
S =\left[\begin{array}{l}S_{1} \\S_{2} \\S_{3} \\S_{4} \\S_{5}\end{array}\right]=\left[\begin{array}{c}0 \\0.01811 \\0.06572 \\0.1344 \\0.2174\end{array}\right]
From Eq. 20-65:
K _{c} \triangleq\left( S ^{T} Q S + R \right)^{-1} S ^{T} Q (20-65)
\begin{aligned}K _{c} &=\left( S ^{T} S \right)^{-1} S ^{T} \\K _{c} &=\left[\begin{array}{lllll}0 & 0.2589 & 0.9395 & 1.9206 & 3.1076\end{array}\right]= K _{c 1}{ }^{T}\end{aligned}
Because K _{c 1}{ }^{ T } is defined as the first row of K _{c}, Using the given analytical result,
\begin{aligned}K _{c 1}^{T} &=\frac{1}{\sum\limits_{i=1}^{5}\left(S_{i}^{2}\right)}\left[\begin{array}{lllll}S_{1} & S_{2} & S_{3} & S_{4} & S_{5}\end{array}\right] \\K _{c 1}^{T} &=\frac{1}{0.06995}\left[\begin{array}{lllll}0 & 0.01811 & 0.06572 & 0.1344 & 0.2174\end{array}\right] \\K _{c 1}^{T} &=\left[\begin{array}{lllll}0 & 0.2589 & 0.9395 & 1.9206 & 3.1076\end{array}\right]\end{aligned}
which is the same as the answer that was obtained above using (20-65).