Date due and handed in March 18,2010
Labeling the output from the branches as follows
Then the differential equation becomes\[ y_{1}^{\prime \prime }=u-2y_{1}^{\prime }-26y_{1}\] While the output equation become\[ y=29y_{1}\] Let \(x_{1}=y_{1}\)\[ \left . \begin{array} [c]{c}x_{1}=y_{1}\\ x_{2}=y_{1}^{\prime }\end{array} \right \} \rightarrow \left . \begin{array} [c]{c}x_{1}^{\prime }=y_{1}^{\prime }=x_{2}\\ x_{2}^{\prime }=y_{1}^{^{\prime \prime }}=u-2y_{1}^{\prime }-26y_{1}=u-2x_{2}-26x_{1}\end{array} \right \} \] Hence\begin{align*} \begin{pmatrix} x_{1}^{\prime }\\ x_{2}^{\prime }\end{pmatrix} & =\overset{A}{\overbrace{\begin{pmatrix} 0 & 1\\ -26 & -2 \end{pmatrix} }}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} +\overset{B}{\overbrace{\begin{pmatrix} 0\\ 1 \end{pmatrix} }}u\left ( t\right ) \\ y\left ( t\right ) & =\overset{C}{\overbrace{\begin{pmatrix} 29 & 0 \end{pmatrix} }}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} +\overset{D}{\overbrace{\left ( 0\right ) }}u\left ( t\right ) \end{align*}
To find \(e^{At}\) use the eigenvalue approach. Find find \(\left \vert A-\lambda I\right \vert \)\[ \left \vert A-\lambda I\right \vert =\left \vert \begin{pmatrix} 0 & 1\\ -26 & -2 \end{pmatrix} -\lambda \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \right \vert =\left \vert \begin{pmatrix} -\lambda & 1\\ -26 & -2-\lambda \end{pmatrix} \right \vert =-\lambda \left ( -2-\lambda \right ) +26 \] Now solve \(-\lambda \left ( -2-\lambda \right ) +26=0\) or \(\lambda ^{2}+2\lambda +26=0\), which has solutions \begin{align*} \lambda _{1} & =-1+5j\\ \lambda _{2} & =-1-5j\ \end{align*}
Hence we have the following 2 equations to solve for \(\beta _{0}\) and \(\beta _{1}\)\begin{align*} e^{\lambda _{1}t} & =\beta _{0}+\lambda _{1}\beta _{1}\\ e^{\lambda _{2}t} & =\beta _{0}+\lambda _{2}\beta _{1} \end{align*}
Solving we find \begin{align*} \beta _{0} & =e^{-t}\left ( \cos 5t+\frac{1}{5}\sin 5t\right ) \\ \beta _{1} & =\frac{1}{5}e^{-t}\sin 5t \end{align*}
Hence \begin{align*} e^{At} & =\beta _{0}+\beta _{1}A\\ & =e^{-t}\left ( \cos 5t+\frac{1}{5}\sin 5t\right ) \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} +\frac{1}{5}e^{-t}\sin 5t\begin{pmatrix} 0 & 1\\ -26 & -2 \end{pmatrix} \\ & =e^{-t}\begin{pmatrix} \cos 5t+\frac{1}{5}\sin 5t & \frac{1}{5}\sin 5t\\ \frac{-26}{5}\sin 5t & \cos 5t-\frac{1}{5}\sin 5t \end{pmatrix} \end{align*}
To find matrix \(\left ( j\omega I-A\right ) ^{-1}\)\begin{align*} j\omega I-A & =j\omega \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} -\begin{pmatrix} 0 & 1\\ -26 & -2 \end{pmatrix} \\ & =\begin{pmatrix} j\omega & 0\\ 0 & j\omega \end{pmatrix} -\begin{pmatrix} 0 & 1\\ -26 & -2 \end{pmatrix} \\ & =\begin{pmatrix} j\omega & -1\\ 26 & j\omega +2 \end{pmatrix} \end{align*}
Hence\begin{align*} \begin{pmatrix} j\omega & -1\\ 26 & j\omega +2 \end{pmatrix} ^{-1} & =\frac{\begin{pmatrix} j\omega +2 & 1\\ -26 & j\omega \end{pmatrix} }{\left ( j\omega \right ) \left ( j\omega +2\right ) +26}=\frac{\begin{pmatrix} j\omega +2 & 1\\ -26 & j\omega \end{pmatrix} }{-\omega ^{2}+2j\omega +26}\\ & =\frac{1}{-\omega ^{2}+2j\omega +26}\begin{pmatrix} j\omega +2 & 1\\ -26 & j\omega \end{pmatrix} \end{align*}
To find the frequency response function. Assuming zero initial conditions, from equation 3.10.4 in the book\begin{align*} H\left ( j\omega \right ) & =C\left ( j\omega I-A\right ) ^{-1}B\\ & =\begin{pmatrix} 29 & 0 \end{pmatrix} \frac{1}{-\omega ^{2}+2j\omega +26}\begin{pmatrix} j\omega +2 & 1\\ -26 & j\omega \end{pmatrix}\begin{pmatrix} 0\\ 1 \end{pmatrix} \\ & =\frac{1}{-\omega ^{2}+2j\omega +26}\begin{pmatrix} 29 & 0 \end{pmatrix}\begin{pmatrix} 1\\ j\omega \end{pmatrix} \\ & =\frac{29}{-\omega ^{2}+2j\omega +26} \end{align*}
Hence \[ \left \vert H\left ( j\omega \right ) \right \vert =\frac{29}{\left \vert -\omega ^{2}+2j\omega +26\right \vert }=\frac{29}{\sqrt{\left ( 26-\omega ^{2}\right ) ^{2}+4\omega ^{2}}}\] And phase is\begin{align*} \arg \left ( H\left ( j\omega \right ) \right ) & =\arg \left ( 29\right ) -\arg \left ( -\omega ^{2}+2j\omega +26\right ) \\ & =-\tan ^{-1}\frac{2\omega }{26-\omega ^{2}} \end{align*}
The state solution is\[ x\left ( t\right ) ={\displaystyle \int \limits _{0}^{t}} e^{A\tau }Bu\left ( \tau \right ) d\tau \] and \[ y\left ( t\right ) =Cx\left ( t\right ) ={\displaystyle \int \limits _{0}^{t}} Ce^{A\tau }Bu\left ( \tau \right ) d\tau \] Hence, let \(u\left ( \tau \right ) =\delta \left ( t\right ) \), then\begin{align*} h\left ( t\right ) & =Ce^{At}B\\ & =\begin{pmatrix} 29 & 0 \end{pmatrix} e^{-t}\begin{pmatrix} \cos 5t+\frac{1}{5}\sin 5t & \frac{1}{5}\sin 5t\\ \frac{-26}{5}\sin 5t & \cos 5t-\frac{1}{5}\sin 5t \end{pmatrix}\begin{pmatrix} 0\\ 1 \end{pmatrix} \\ & =e^{-t}\begin{pmatrix} 29 & 0 \end{pmatrix}\begin{pmatrix} \frac{1}{5}\sin 5t\\ \cos 5t-\frac{1}{5}\sin 5t \end{pmatrix} \\ & =e^{-t}\left ( \frac{29}{5}\sin 5t\right ) \xi \left ( t\right ) \end{align*}
