Question 12.7: Find a closed-form expression for the state output q(t) of t......
Find a closed-form expression for the state output q(t) of the system, described by
A={\left[\begin{array}{l}{2\ 1}\\ {3\ 4}\end{array}\right]},\quad B={\left[\begin{array}{l}{1}\\ {0}\end{array}\right]},
using the frequency-domain method, with zero initial conditions and the input u(t), the unit-step function.
(sI − A) = s\left [ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right ] -\left [ \begin{matrix} 2 & 1 \\ 3 & 4 \end{matrix} \right ] =\left [ \begin{matrix} s-2 & -1 \\ -3 & s-4 \end{matrix} \right ]
(s I-A)^{-1}={\frac{1}{s^{2}-6s+5}}\left [ \begin{matrix} s-4 & 1 \\ 3 & s-2 \end{matrix} \right ] =\left [ \begin{matrix} \frac{s-4}{s^{2}-6s+5} & \frac{1}{s^{2}{-}6s{+}5} \\ \frac{3}{s^{2}{-}6s{+}5} & \frac{s-2}{s^{2}{-}6s{+}5} \end{matrix} \right ]
The eigenvalues are {1, 5}.
The zero-state component of the state vector is
q_{z s}(s)=(s I-A)^{-1}B X(s)=\left [ \begin{matrix} \frac{s-4}{s^{2}-6s+5} & \frac{1}{s^{2}{-}6s{+}5} \\ \frac{3}{s^{2}{-}6s{+}5} & \frac{s-2}{s^{2}{-}6s{+}5} \end{matrix} \right ]\left [ \begin{matrix} 1 \\ 0 \end{matrix} \right ] \frac{1}{s}
=\left [ \begin{matrix} \frac{s-4}{s(s^{2}-6s+5)} \\ \frac{3}{s(s^{2}-6s+5)} \end{matrix} \right ]=\left [ \begin{matrix} {\frac{-0.8}{s}}+{\frac{0.75}{s-1}}+{\frac{0.05}{s-5}} \\ {\frac{0.{{6}}}{s}}-{\frac{0.75}{s-1}}+{\frac{0.15}{s-{{5}}}} \end{matrix} \right ]
Taking the inverse Laplace transform, we get
q_{z s}(t)=\left[\begin{array}{c}{{-0.8+0.75e^{t}+0.05e^{5t}}}\\ {{0.6-0.75e^{t}+0.15e^{5t}}}\end{array}\right]u(t)
Using the similarity transformation method also, the state output can be computed. The state transition matrix associated with the diagonal system matrix Λ is
\left[\begin{array}{l l}{e^{t}}&{0}\\ {0}&{e^{5t}}\end{array}\right]u(t)
Pe^{Λt}P ^{−1} =\left [ \begin{matrix} 1 & 1 \\ -1 & 3 \end{matrix} \right ] \left[\begin{array}{l l}{e^{t}}&{0}\\ {0}&{e^{5t}}\end{array}\right]\left [ \begin{matrix} 0.75 & -0.25 \\ 0.25 & 0.25 \end{matrix} \right ]
=e^{A t}=\left[{\begin{array}{c}{{0.75e^{t}+0.25e^{5t}-0.25e^{t}+0.25e^{5t}}}\\ {{-0.75e^{t}+0.75e^{5t}}}\ \ \ {{0.25e^{t}+0.75e^{5t}}}\end{array}}\right]u(t)
which is the same as the inverse Laplace transform of
(sI − A)^{−1} =\left [ \begin{matrix} \frac{s-4}{s^{2}-6s+5} & \frac{1}{s^{2}{-}6s{+}5} \\ \frac{3}{s^{2}{-}6s{+}5} & \frac{s-2}{s^{2}{-}6s{+}5} \end{matrix} \right ]