Question 12.4: Solve the problem of Example 12.2 using the frequency-domain......

The initial state vector is

q(0^{-})=\left[\begin{array}{c}{{-\frac{15}{8}}}\\ \ \ \ \ {{\frac{29}{16}}}\end{array}\right]

as derived in Example 12.2 from the given initial output conditions.

(s I-A)=s\left [ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right ] -\left[\begin{array}{c c}{{-4}}&-4\\ {{1}}&{{0}}\end{array}\right]=\left[\begin{array}{c c}{{s+4\;4}}\\ {{-1\;s}}\end{array}\right]

(s I-A)^{-1}=\frac{1}{s^{2}+4s+4}{\left[\begin{array}{cc}{s} \ \ \ \ {-4}\\ {1} \ \ s+4\end{array}\right]}=\left [ \begin{matrix} \frac{s}{s^{2}+4s+4} & \frac{-4}{s^{2}+4s+4} \\ \frac{1}{s^{2}+4s+4} & \frac{s+4}{s^{2}+4s+4} \end{matrix} \right ]

We used the fact that I = e^{A0}  to check the computation of e^{At} . In the frequency domain, the corresponding check, using the initial value theorem of the Laplace transform, is \mathrm{lim}_{s\rightarrow\infty}s(s{I}-A)^{-1}=I.

The zero-input component of the state vector is

q_{z i}(s)=(s I-A)^{-1}q(0^{-})=\left [ \begin{matrix} \frac{s}{s^{2}+4s+4} & \frac{-4}{s^{2}+4s+4} \\ \frac{1}{s^{2}+4s+4} & \frac{s+4}{s^{2}+4s+4} \end{matrix} \right ]\left [ \begin{matrix} -\frac{15}{8} \\ \frac{29}{16} \end{matrix} \right ]

=\left [ \begin{matrix} \frac{-{\frac{15}{8}}s-{\frac{29}{4}}}{s_{}^{2}+4s+4} \\ \frac{{\frac{29}{16}}s+{\frac{43}{8}}}{s_{}^{2}+4s+4} \end{matrix} \right ] =\left [ \begin{matrix} -\frac{\frac{7}{2}}{(s+2)^{2}}-\frac{\frac{15}{8}}{s+2} \\ {\frac{\frac{7}{4}}{(s+2)^{2}}}+{\frac{\frac{29}{16}}{s+2}} \end{matrix} \right ]

Taking the inverse Laplace transform, we get

q_{z i}(t)=\left[\begin{array}{l l}{{-\frac{15}{8}e^{-2t}-\frac{7}{2}t e^{-2t}}}\\ {{\frac{29}{16}e^{-2t}+\frac{7}{4}t e^{-2t}}}\end{array}\right]

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}{s^{2}+4s+4} & \frac{-4}{s^{2}+4s+4} \\ \frac{1}{s^{2}+4s+4} & \frac{s+4}{s^{2}+4s+4} \end{matrix} \right ]\left [ \begin{matrix} 1 \\ 0 \end{matrix} \right ] \frac{1}{s}

=\left[\begin{array}{l}{{{\frac{1}{s^{2}+4s+4}}}}\\ {{{\frac{1}{s(s^{2}+4s+4)}}\displaystyle}}\end{array}\right]=\left[\begin{array}{l l}{{\frac{1}{(s+2)^{2}}}}\\ {{\frac{\frac{1}{4}}{s}-\frac{\frac{1}{2}}{(s+2)^{2}}-\frac{\frac{1}{4}}{s+2}}}\end{array}\right]

Taking the inverse Laplace transform, we get

q_{z s}(t)=\left [ \begin{matrix}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ te^{-2t} \\ \textstyle{\frac{1}{4}}-{\frac{1}{4}}e^{-2t}-{\frac{1}{2}}t e^{-2t} \end{matrix} \right ]

Using the output equation, the output can be computed as given in Example 12.2.