3.4 HW 5

Date due and handed in March 18,2010

3.4.1 Problem 3.23 (a)

Figure 3.3: Problem description

3.4.1.1 Part(a)

Labeling the output from the branches as follows

Figure 3.4: Problem description part(a) labeled

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*}

3.4.1.2 Part b

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*}

3.4.1.3 Part c

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*}

3.4.1.4 Part d

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*}

3.4.1.5 Part e

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*}