Question 11.6: Find a closed-form expression for the output y(n) of the sys......

The initial state vector

q(0)= \left [ \begin{matrix} \frac{17}{3} \\ \frac{16}{3} \end{matrix} \right ]

is derived in Example 11.2 from the given initial output conditions.

(zI − A) = \left [ \begin{matrix} z & -2 \ \ 3 \\ & -1 \ \ z \end{matrix} \right ] and (zI − A) ^{-1}= \left [ \begin{matrix} \frac{z}{z^{2}-2z+3}  -{\frac{3}{z^{2}-2z+3}} \\ {\frac{1}{z^{2}-2z+3}} \ \ \ {\frac{z-2}{z^{2}-2z+3}} \end{matrix} \right ]

As a check on (zI − A) ^{-1} , we use the initial value theorem of the z-transform to verify that

\operatorname*{lim}_{z\rightarrow\infty}z(z I-A)^{-1}=I=A^{0}

The transform of the zero-input component of the state vector is

Q_{z i}(z)=z(z I-A)^{-1}q(0)

= z \left [ \begin{matrix} \frac{z}{z^{2}-2z+3} -{\frac{3}{z^{2}-2z+3}} \\ {\frac{1}{z^{2}-2z+3}} \ \ \ {\frac{z-2}{z^{2}-2z+3}} \end{matrix} \right ]\left [ \begin{matrix} \frac{17}{3} \\ \frac{16}{3} \end{matrix} \right ] =\frac{z}{3}\left [ \begin{matrix} \frac{17z-48}{z^{2}-2z+3} \\ \frac{16z-15}{z^{2}-2z+3} \end{matrix} \right ]

= \left [ \begin{matrix} \frac{({\frac{17}{6}}+j{\frac{31}{6{\sqrt{2}}}})z}{z-1-j{\sqrt{2}}} + \frac{({\frac{17}{6}}-j{\frac{31}{6{\sqrt{2}}}})z}{z-1+j{\sqrt{2}}} \\ \frac{(\frac{8}{3}-j\frac{1}{6\sqrt{2}})z}{z-1-j\sqrt{2}} +\frac{(\frac{8}{3}+j\frac{1}{6\sqrt{2}})z}{z-1+j\sqrt{2}} \end{matrix} \right ]

Finding the inverse z-transform and simplifying, we get the zero-input component of the state vector as

q_{z i}(n)=\left [ \begin{matrix} {\textstyle\frac{17}{3}}({\sqrt{3}})^{n}\cos(\tan^{-1}({\sqrt{2}})n)-{\textstyle\frac{31}{3{\sqrt{2}}}}({\sqrt{3}})^{n}\sin(\tan^{-1}({\sqrt{2}})n)) \\ {\textstyle\frac{16}{3}}({\sqrt{3}})^{n}\cos(\tan^{-1}({\sqrt{2}})n)+{\textstyle\frac{1}{3{\sqrt{2}}}}({\sqrt{3}})^{n}\sin(\tan^{-1}({\sqrt{2}})n)) \end{matrix} \right ] u(n)

The transform of the zero-state component of the state vector is

= \left [ \begin{matrix} \frac{z}{z^{2}-2z+3} & -{\frac{3}{z^{2}-2z+3}} \\ {\frac{1}{z^{2}-2z+3}} & {\frac{z-2}{z^{2}-2z+3}} \end{matrix} \right ]\left [ \begin{matrix} \frac{z}{z-1} \\ 0 \end{matrix} \right ] =z\left [ \begin{matrix} \frac{{{{z}}}}{({z-1})({{z}}^{2}-2z+3)} \\ \frac{{{{1}}}}{({z-1})({{z}}^{2}-2z+3)} \end{matrix} \right ]

=\left [ \begin{matrix} {\frac{({\frac{1}{2}})z}{z-1}}-\frac{{\frac{1}{4}}(1+j\sqrt{2})z}{z-1-j\sqrt{2}}-\frac{{\frac{1}{4}}(1-j\sqrt{2})z}{z-1+j\sqrt{2}} \\ \frac{(\frac{1}{2})z}{z-1}\,-\,\frac{\frac{1}{4}z}{z-1-j\sqrt{2}}\,-\,\frac{\frac{1}{4}z}{z-1+j\sqrt{2}} \end{matrix} \right ]

Finding the inverse z-transform and simplifying, we get the zero-state component of the state vector as

q_{z s}(n)=\left [ \begin{matrix} {\frac{1}{2}}-{\frac{1}{2}}({\sqrt{3}})^{n}\cos(\tan^{-1}(\sqrt{2})n)+{\frac{1}{\sqrt{2}}}({\sqrt{3}})^{n}\sin(\tan^{-1}(\sqrt{2})n) \\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad {\textstyle\frac{1}{2}}\stackrel{}{-}{\textstyle\frac{1}{2}}(\sqrt{3})^{n}\cos(\tan^{-1}(\sqrt{2})) \end{matrix} \right ]u(n)

Using the output equation, the output can be computed as given in Example 11.4.