Products
Rewards 
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY 
BUSINESS MANAGER

Advertise your business, and reach millions of students around the world.

HOLOOLY 
TABLES

All the data tables that you may search for.

HOLOOLY 
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY 
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY 
HELP DESK

Need Help? We got you covered.

Chapter 7

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)

Step-by-Step

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

MATLAB Verified Solution

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)].