Question 12.6: Derive the state-space model of the system in Example 12.1 w......

P=\left[{\begin{array}{r r}{1\,2}\\ {-3\,4}\end{array}}\right],\quad P^{-1}=\left [ \begin{matrix} \frac{4}{10} & -\frac{2}{10} \\ \frac{3}{10} & \frac{1}{10} \end{matrix} \right ]

{\overline{{A}}}=P A P^{-1}=\left[{\begin{array}{r r}{1\,2}\\ {-3\,4}\end{array}}\right]\left [ \begin{matrix} -\frac{5}{6} & -\frac{1}{2} \\ -\frac{1}{3} & 0 \end{matrix} \right ] \left [ \begin{matrix} \frac{4}{10} & -\frac{2}{10} \\ \frac{3}{10} & \frac{1}{10} \end{matrix} \right ] =\left [ \begin{matrix} -\frac{13}{60} & -\frac{1}{60} \\ \frac{119}{60} & -\frac{37}{60} \end{matrix} \right ]

\overline{{B}}=P B=\left[{\begin{array}{r r}{1\,2}\\ {-3\,4}\end{array}}\right]\left [ \begin{matrix} \frac{1}{2} \\ 0 \end{matrix} \right ] =\left [ \begin{matrix} \frac{1}{2} \\ -\frac{3}{2} \end{matrix} \right ]

\overline{{{C}}}=C P^{-1}={\big[}0\;1{\big]}\left [ \begin{matrix} \frac{4}{10} & -\frac{2}{10} \\ \frac{3}{10} & \frac{1}{10} \end{matrix} \right ] =\left [ \begin{matrix} \frac{3}{10} & \frac{1}{10} \end{matrix} \right ]

The state-space model of a second-order continuous system with the new state vector is shown in Fig. 12.6. This realization requires more components than that shown in Fig. 12.4. However, it must be noted that, while the minimum number of components is of great importance, there are other criteria, such as less coefficient sensitivity, that could decide which of the realizations of a system is suitable for a particular application.

The transfer function, using the new state-space model, is computed as follows.

(s I-{\overline{{A}}})=s\left [ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right ] -\left [ \begin{matrix} -\frac{13}{60} & -\frac{1}{60} \\ \frac{119}{60} & -\frac{37}{60} \end{matrix} \right ] =\left [ \begin{matrix} s+\frac{13}{60} & \frac{1}{60} \\ -\frac{119}{60} & s+\frac{37}{60} \end{matrix} \right ]

(s I-{\overline{{A}}})^{-1}={\frac{1}{s^{2}+{\frac{5}{6}}s+{\frac{1}{6}}}}\left [ \begin{matrix} s+\frac{37}{60} \ \ \ -\frac{1}{60} \\ \ \ \ \ \frac{119}{60}\ \ s+\frac{16}{60} \end{matrix} \right ] =\left [ \begin{matrix} \frac{s+\frac{37}{60}}{s^{2}+\frac{5}{6}s+\frac{1}{6}} & \frac{-{\frac{1}{60}}}{s^{2}+{\frac{5}{6}}s+{\frac{1}{6}}} \\ \frac{\frac{119}{60}}{s^{2}+\frac{5}{6}s+\frac{1}{6}} & \frac{s^{2}+\frac{13}{60}}{s^{2}+\frac{5}{6}s+\frac{1}{6}} \end{matrix} \right ]

H(s)=\left[\frac{3}{10}\quad\frac{1}{10}\right]\left [ \begin{matrix} \frac{s+\frac{37}{60}}{s^{2}+\frac{5}{6}s+\frac{1}{6}} & \frac{-{\frac{1}{60}}}{s^{2}+{\frac{5}{6}}s+{\frac{1}{6}}} \\ \frac{\frac{119}{60}}{s^{2}+\frac{5}{6}s+\frac{1}{6}} & \frac{s^{2}+\frac{13}{60}}{s^{2}+\frac{5}{6}s+\frac{1}{6}} \end{matrix} \right ] \left [ \begin{matrix} \frac{1}{2} \\ -\frac{3}{2} \end{matrix} \right ] =\frac{\frac{1}{6} }{(s^{2}+\frac{5}{6}s+\frac{1}{6})}

The transfer function, using the old state-space model, is computed as follows.

(s I-A)=s\left [ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right ] -\left[\begin{array}{cc}{{-\frac56 \ \ -\frac12}}\\ \ \ \ \ {{\frac13}}\ \ \ \ \ \ \ \ {{0}}\end{array}\right]=\left[\begin{array}{l l}{{s+\frac56~\frac12}}\\ \ \ \ \ \ {{-\frac13}}\ {{s}}\end{array}\right]

(s I-A)^{-1}={\frac{1}{s^{2}+{\frac{5}{6}}s+{\frac{1}{6}}}}\left [ \begin{matrix} s \ \ \ -\frac{1}{2} \\ \frac{1}{3}s \ +\frac{5}{6} \end{matrix} \right ] =\left [ \begin{matrix} \frac{s}{s^{2}+\frac{5}{6}s+\frac{1}{6}} & \frac{-\frac{1}{2}}{s^{2}+\frac{5}{6}s+\frac{1}{6}} \\ \frac{\frac{1}{3}}{s^{2}+\frac{5}{6}s+\frac{1}{6}} & \frac{s+\frac{5}{6}}{s^{2}+\frac{5}{6}s+\frac{1}{6}} \end{matrix} \right ]

H(s)=[0~~~1]=\left [ \begin{matrix} \frac{s}{s^{2}+\frac{5}{6}s+\frac{1}{6}} & \frac{-\frac{1}{2}}{s^{2}+\frac{5}{6}s+\frac{1}{6}} \\ \frac{\frac{1}{3}}{s^{2}+\frac{5}{6}s+\frac{1}{6}} & \frac{s+\frac{5}{6}}{s^{2}+\frac{5}{6}s+\frac{1}{6}} \end{matrix} \right ] \left [ \begin{matrix} \frac{1}{2} \\ 0 \end{matrix} \right ] =\frac{\frac{1}{6}}{(s^{2}+\frac{5}{6}s+\frac{1}{6})},

which is the same as that obtained above.