Question 11.9: Derive the state-space model of the system in Example 11.1 w......
Derive the state-space model of the system in Example 11.1 with the new state vector that is related to old state vector as
\begin{array}{c}{{\overline{{{q}}}_{1}(n)=2q_{1}(n)+q_{2}(n)}}\\ {{\overline{{{q}}}_{2}(n)=q_{1}(n)-q_{2}(n)}}\end{array}
Verify that the transfer function remains the same using either state-space model.
P={\left[\begin{array}{cc}{2}&{1}\\ {1}&{-1}\end{array}\right]},\quad P^{-1}={\left[\begin{array}{l l}{{\frac{1}{3}}} \ \ \ \ \ \ {{\frac{1}{3}}}\\ {{\frac{1}{3}} \ -{\frac{2}{3}}}\end{array}\right]}
\overline{{{A}}}=P A P^{-1}=\left [ \begin{matrix} 2 & 1 \\ 1 & -1 \end{matrix} \right ] \left [ \begin{matrix} 2 & -3 \\ 1 & 0 \end{matrix} \right ] \left [ \begin{matrix} \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{matrix} \right ] =\left [ \begin{matrix} -\frac{1}{3} & \frac{17}{3} \\ -\frac{2}{3} & \frac{7}{3} \end{matrix} \right ]
\overline{{B}}=P B={\left[\begin{array}{cc}{2}&{1}\\ {1}&-1\end{array}\right]}{\left[\begin{array}{l}{1}\\ {0}\end{array}\right]}={\left[\begin{array}{l}{2}\\ {1}\end{array}\right]}
\overline{{{C}}}=C P^{-1}=\left[\,1\,-2\,\right]\left [ \begin{matrix} \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{matrix} \right ] =\left [ \begin{matrix} -\frac{1}{3} & \frac{5}{3} \end{matrix} \right ]
The state-space model of a second-order discrete system with the new state vector is shown in Fig. 11.9. The transfer function, computed using the new state-space model, is
H(z)=\left[-\frac{1}{3}\quad\frac{5}{3}\right]\left [ \begin{matrix} z+\frac{1}{3} \ \ -\frac{17}{3} \\ \ \ \ \ \ \frac{2}{3}z-\frac{7}{3} \end{matrix} \right ] ^{-1}\left[2\atop1\right]+2=\frac{z-2}{(z^{2}-2z+3)}+2,
which is the same as that obtained in Example 11.7.
