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## Q. 7.EX.10

Transformation of Thermal System from Control to Modal Form
Find the matrix to transform the control form matrices in Eq. (7.15) into the modal form of Eq. (7.17).

$\pmb A_c = \begin{bmatrix} -7 & -12 \\ 1 & 0 \end{bmatrix}, \ \ \ \ \pmb B_c = \begin{bmatrix}1 \\0 \end{bmatrix} ,$           (7.15a)

$\pmb C_c = \begin{bmatrix} 1 & 2 \end{bmatrix} , \ \ \ \ D_c =0,$             (7.15b)

$\pmb A_m = \begin{bmatrix} -4 & 0 \\ 0 & -3 \end{bmatrix}, \ \ \ \ \pmb B_m = \begin{bmatrix} 1 \\ 1 \end{bmatrix} ,$        (7.17a)

$\pmb C_m = \begin{bmatrix} 2 & -1 \end{bmatrix} , \ \ \ \ D_m=0,$                      (7.17b)

## Verified Solution

According to Eqs. (7.37) and (7.38), we need first to find the eigenvectors and eigenvalues of the $\pmb A_c$ matrix. We take the eigenvectors to be

$\begin{bmatrix} t_{11} \\ t_{21} \end{bmatrix}$           and            $\begin{bmatrix} t_{12} \\ t_{22} \end{bmatrix}$

$\pmb {TA} = \pmb{FT} \\ \begin{bmatrix} \pmb t_1 & \pmb t_2 & \pmb t_3 \end{bmatrix} \begin{bmatrix} p_1 & 0 & 0 \\ 0 & p_2 & 0 \\ 0 & 0 & p_3 \end{bmatrix} = \pmb F \begin{bmatrix} \pmb t_1 & \pmb t_2 & \pmb t_3 \end{bmatrix} .$                 (7.37)

$p_i \pmb t_i = \pmb {Ft}_i, \ \ \ \ i = 1, 2, 3.$                 (7.38)

The equations using the eigenvector on the left are

$\begin{bmatrix} -7 & -12 \\ 1 & 0 \end{bmatrix} \begin{bmatrix}t_{11} \\ t_{21} \end{bmatrix} = p \begin{bmatrix} t_{11} \\ t_{21} \end{bmatrix} ,$                 (7.39a)

$−7t_{11} − 12t_{21} = pt_{11},$                         (7.39b)

$t_{11} = pt_{21}.$                              (7.39c)

Substituting Eq. (7.39c) into Eq. (7.39b) results in

$−7pt_{21} − 12t_{21} = p^2t_{21},$                        (7.40a)

$p^2t_{21} + 7pt_{21} + 12t_{21} = 0,$                      (7.40b)

$p^2 + 7p + 12 = 0,$                           (7.40c)

$p = −3, −4.$                              (7.40d)

We have found (again!) that the eigenvalues (poles) are $−3$ and $−4$ ; furthermore, Eq. (7.39c) tells us that the two eigenvectors are

$\begin{bmatrix} -4t_{21} \\ t_{21} \end{bmatrix}$                and                 $\begin{bmatrix} -3t_{22} \\ t_{22} \end{bmatrix} ,$

where $t_{21}$ and $t_{22}$ are arbitrary nonzero scale factors. We want to select the two scale factors such that both elements of $\pmb B_m$ in Eq. (7.17a) are unity. The equation for $\pmb B_m$ in terms of $\pmb B_c$ is $\pmb {TB}_m = \pmb B_c,$ and its solution is $t_{21} = −1$ and $t_{22} = 1.$ Therefore, the transformation matrix and its inverse$^5$ are

$\pmb T = \begin{bmatrix} 4 & -3 \\ -1 & 1 \end{bmatrix}, \ \ \ \ \pmb T^{-1} = \begin{bmatrix} 1 & 3 \\ 1 & 4 \end{bmatrix}.$                     (7.41)

Elementary matrix multiplication shows that, using $T$ as defined by Eq. (7.41), the matrices of Eqs. (7.15) and (7.17) are related as follows:

$\pmb A_c = \begin{bmatrix} -7 & -12 \\ 1 & 0 \end{bmatrix}, \ \ \ \ \pmb B_c = \begin{bmatrix}1 \\0 \end{bmatrix} ,$           (7.15a)

$\pmb C_c = \begin{bmatrix} 1 & 2 \end{bmatrix} , \ \ \ \ D_c =0,$             (7.15b)

$\pmb A_m = \begin{bmatrix} -4 & 0 \\ 0 & -3 \end{bmatrix}, \ \ \ \ \pmb B_m = \begin{bmatrix} 1 \\ 1 \end{bmatrix} ,$        (7.17a)

$\pmb C_m = \begin{bmatrix} 2 & -1 \end{bmatrix} , \ \ \ \ D_m=0,$                      (7.17b)

$\pmb A_m = \pmb T^{−1} \pmb A_c \pmb T, \ \ \ \ \ \pmb B_m = \pmb T^{−1} \pmb B_c, \\ \pmb C_m = \pmb C_c \pmb T, \ \ \ \ \ \ D_m = D_c.$                    (7.42)

These computations can be carried out by using the following MATLAB statements

## Script Files

T = [4 −3; −1 1];
Am = inv(T)*Ac*T;
Bm = inv(T)*Bc;
Cm = Cc*T;
Dm = Dc;

$^5$ To find the inverse of a $2 × 2$ matrix, you need only interchange the elements subscripted “$11$ ” and “$22$ ,” change the signs of the “$12$ ” and the “$21$ ” elements, and divide by the determinant [$= 1$ in Eq. (7.41)].