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## Q. 20.3

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)

## Verified Solution

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 $(20-65)$.