User:Egm6321.f10.team3.Sudheesh/Homeworks/Homework3

= Problem 1 - Motion of a Projectile =

Given
Figure shows the Trajectory of a projectile (ex:Rocket):

Find

 * $$\left( 1 \right)$$ Drive Equation of Motion (EOM)
 * $$\left( 2 \right)$$ Particular case $$\displaystyle k = 0 $$ : Verify $$y\left( x \right)$$ is parabolla
 * $$\left( 3 \right)$$ Consider $$k\ne 0$$, $$\displaystyle {{v}_{x}}=0$$,


 * $$ (3.1) $$ Find $${{v}_{y}}\left( t \right)$$, $$\displaystyle y(t)$$ for $$ \displaystyle m$$ = constant
 * $$ (3.2) $$ Find $${{v}_{y}}\left( t \right)$$, $$\displaystyle y(t)$$ if $$m=m\left( t \right)$$

Solution
Part 1 Consider the trajectory of a projectile (ex. Rocket)


 * Various forces acting on the projectile at time 't' are:


 * 1) Weight of the projectile


 * {| style="width:100%" border="0"


 * $$\displaystyle W=mg$$
 * }
 * }
 * }


 * 2) Inertia force


 * {| style="width:100%" border="0"


 * $${{F}_{a}}=ma=m\cdot \frac{dv}{dt}$$ for particle with constant mass
 * $${{F}_{a}}=ma=m\cdot \frac{dv}{dt}$$ for particle with constant mass


 * }
 * }


 * 3) Air resistance which is proportional to the velocity of particle


 * {| style="width:100%" border="0"


 * $$\displaystyle {{F}_{D}}=k{{v}^{n}}$$
 * }
 * }
 * }

Now consider the force equilibrium in both horizontal and vertical direction


 * a) Force Equilibrium in horizontal direction:


 * {| style="width:100%" border="0"


 * $$\sum{{{F}_{H}}=0}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$m\cdot \frac{d{{v}_{x}}}{dt}+k{{v}^{n}}\cos \alpha =0$$,
 * $$m\cdot \frac{d{{v}_{x}}}{dt}+k{{v}^{n}}\cos \alpha =0$$,


 * where $$\displaystyle {{v}_{x}}\to $$ horizontal component of velocity


 * }
 * }


 * {| style="width:100%" border="0"

$$m\cdot \frac{d{{v}_{x}}}{dt}=-k{{v}^{n}}\cos \alpha $$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.1)
 * }
 * }


 * b) Force Equilibrium in the vertical direction:


 * {| style="width:100%" border="0"


 * $$\sum{{{F}_{V}}=0}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"

$$m\cdot \frac{d{{v}_{y}}}{dt}+k{{v}^{n}}\sin \alpha +mg=0$$,


 * where $$\displaystyle {{v}_{y}}\to $$ vertical component of velocity


 * }
 * }


 * {| style="width:100%" border="0"

$$m\cdot \frac{d{{v}_{y}}}{dt}=-k{{v}^{n}}\sin \alpha -mg$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.2)
 * }
 * }

Part 2 


 * Particular case: When $$\displaystyle k=0$$

Eq.1.1 reduces to


 * {| style="width:100%" border="0"


 * $$m\frac{d{{v}_{x}}}{dt}=0\Rightarrow \frac{d{{v}_{x}}}{dt}=0$$
 * }
 * }
 * }

Integrating the above equation gives:


 * {| style="width:100%" border="0"

$$
 * $${{v}_{x}}\left( t \right)={{c}_{1}}$$
 * $$\displaystyle (Eq. 1.3)
 * $$\displaystyle (Eq. 1.3)
 * }
 * }

Apply 'initial condition' to determine integration constant,$$\displaystyle{{c}_{1}}$$


 * {| style="width:100%" border="0"


 * $${{v}_{x}}\left( t=0 \right)={{v}_}={{c}_{1}}$$
 * }
 * }
 * }

Now Eq.1.3 becomes:
 * {| style="width:100%" border="0"


 * $${{v}_{x}}\left( t \right)={{v}_}$$
 * }
 * }
 * }

Integrate the above equation to obtain ,$$\displaystyle x$$


 * {| style="width:100%" border="0"


 * $$x\left( t \right)={{v}_}t+{{c}_{2}}$$
 * }
 * }
 * }

Then 'Initial condition' is applied to determine ,$$\displaystyle{{c}_{2}}$$


 * {| style="width:100%" border="0"


 * $$x\left( t=0 \right)={{v}_}\times 0+{{c}_{2}}\Rightarrow {{c}_{2}}={{x}_{0}}$$
 * }
 * }
 * }

Therefore,


 * {| style="width:100%" border="0"


 * $$x\left( t \right)={{v}_}t+{{x}_{0}}$$
 * }
 * }
 * }

Now $$\displaystyle t$$, can be expressed in terms of $$\displaystyle x$$,


 * {| style="width:100%" border="0"

$$
 * $$t=\frac{x-{{x}_{0}}}$$
 * $$\displaystyle (Eq. 1.4)
 * $$\displaystyle (Eq. 1.4)
 * }
 * }

Similarly When $$\displaystyle k=0$$, Eq.1.2 ruduces to


 * {| style="width:100%" border="0"


 * $$m\frac{d{{v}_{y}}}{dt}=-mg\Rightarrow \frac{d{{v}_{y}}}{dt}=-g$$
 * }
 * }
 * }

Integrate the above equation to evaluate, $$\displaystyle {{v}_{y}}$$


 * {| style="width:100%" border="0"

$$
 * $${{v}_{y}}\left( t \right)=-gt+{{c}_{3}}$$
 * $$\displaystyle (Eq. 1.5)
 * $$\displaystyle (Eq. 1.5)
 * }
 * }

Apply 'initial condition' to obtain $$\displaystyle {{c}_{3}}$$


 * {| style="width:100%" border="0"


 * $${{v}_{y}}\left( t=0 \right)={{v}_}={{c}_{3}}$$
 * }
 * }
 * }

Now Eq.1.5 becomes,


 * {| style="width:100%" border="0"


 * $${{v}_{t}}\left( t \right)=-gt+{{v}_}$$
 * }
 * }
 * }

Then integrate the above equation to determine, $$\displaystyle y $$


 * {| style="width:100%" border="0"


 * $$y\left( t \right)=\frac{-g{{t}^{2}}}{2}+{{v}_}t+{{c}_{4}}$$
 * }
 * }
 * }


 * $${{c}_{4}}$$ is determined using 'initial condition' as:


 * {| style="width:100%" border="0"


 * $$y\left( t=0 \right)=\frac{-g\times 0}{2}+{{v}_}\times 0+{{c}_{4}}\Rightarrow {{c}_{4}}={{y}_{0}}$$
 * }
 * }
 * }

Therefore,


 * {| style="width:100%" border="0"


 * $$y\left( t \right)=\frac{-g{{t}^{2}}}{2}+{{v}_}t+{{y}_{0}}$$
 * }
 * }
 * }

Now Substitute Eq.1.4 for $$\displaystyle t$$ in the above equation;


 * {| style="width:100%" border="0"

$$y\left( x \right)=-\frac{g}{2}{{\left( \frac{x-{{x}_{0}}} \right)}^{2}}+{{v}_}\left( \frac{x-{{x}_{0}}} \right)+{{y}_{0}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.6)
 * }
 * }

Eq.1.6 is in the form of a parabollic equation. Therefore $$\displaystyle y\left( x \right)$$ is parabolla.

Part 3

= Problem 2- Equation of Motion of Pendulums Connected by a Spring =

Given
Shown in figure is the two Pendulums connected by a spring:

Find

 * $$\left( 1 \right)$$ Derive equation of motion:


 * {| style="width:100%" border="0"

$$
 * $${{m}_{1}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=-k{{a}^{2}}\left( {{\theta }_{1}}-{{\theta }_{2}} \right)-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l$$
 * $$\displaystyle (Eq. 2.1)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.1)


 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $${{m}_{2}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=-k{{a}^{2}}\left( {{\theta }_{2}}-\theta 1 \right)-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.2)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.2)
 * }
 * }


 * $$\left( 2 \right)$$ Write Eq.2.1 and Eq.2.2 in the form of [[media:2010_09_21_14_56_48.djvu|Mtg 13 (c),page2 ]], of:


 * {| style="width:100%" border="0"

$$
 * $$\underset{-}{\overset{\centerdot }{\mathop{x}}}\,=\underset{-}{\mathop{A}}\,\left( t \right)\underset{-}{\mathop{x}}\,\left( t \right)+\underset{-}{\mathop{B}}\,\left( t \right)\underset{-}{\mathop{u}}\,\left( t \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.3)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.3)
 * }
 * }


 * {| style="width:100%" border="0"

$$\underset{-}{\mathop{x}}\,={{\left[ \begin{matrix} {{\theta }_{1}} & \overset{\centerdot }{\mathop{{{\theta }_{1}}}}\, & {{\theta }_{2}} & \overset{\centerdot }{\mathop{{{\theta }_{2}}}}\, \\ \end{matrix} \right]}^{T}}$$ and $$\underset{-}{\mathop{u}}\,={{\left[ \begin{matrix} {{u}_{1}}l & {{u}_{2}}l \\ \end{matrix} \right]}^{T}}$$
 * Given
 * Given
 * }
 * }

Solution

 * $$\left( 1 \right)$$ Derive equation of motion:
 * (a) Consider Free Body Diagram of left pendulum:


 * For small angle:
 * {| style="width:100%" border="0"


 * $$\displaystyle l\sin {{\theta }_{1}}=l{{\theta }_{1}}$$ and $$\displaystyle l\cos {{\theta }_{1}}=l$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$acceleration, {{a}_{1}}=\frac{{{d}^{2}}\left( l{{\theta }_{1}} \right)}{d{{t}^{2}}}=l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$ Inertia force ={{m}_{1}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$\displaystyle Spring force =ka{{\theta }_{1}}-ka{{\theta }_{2}}=ka({{\theta }_{1}}-{{\theta }_{1}})$$
 * }
 * }
 * }

Now using D'Alembert's_principle, sum of the moments about pivot(A)is equal to zero


 * {| style="width:100%" border="0"


 * $$\sum{{{M}_{A}}=0\Rightarrow }\left( {{m}_{1}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}} \right)\cdot l+\left( ka({{\theta }_{1}}-{{\theta }_{2}}) \right)\cdot a+\left( {{m}_{1}}g \right)\cdot l{{\theta }_{1}}-\left( {{u}_{1}} \right)\cdot l=0$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"

$$\Rightarrow {{m}_{1}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=-k{{a}^{2}}({{\theta }_{1}}-{{\theta }_{2}})-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.1)
 * }
 * }


 * (b) Consider Free Body Diagram of right pendulum:




 * For small angle:
 * {| style="width:100%" border="0"


 * $$\displaystyle l\sin {{\theta }_{2}}=l{{\theta }_{2}}$$ and $$\displaystyle l\cos {{\theta }_{2}}=l$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$acceleration, {{a}_{2}}=\frac{{{d}^{2}}\left( l{{\theta }_{2}} \right)}{d{{t}^{2}}}=l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$ Inertia force ={{m}_{2}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$\displaystyle Spring force =ka{{\theta }_{2}}-ka{{\theta }_{1}}=ka({{\theta }_{2}}-{{\theta }_{1}})$$
 * }
 * }
 * }

Using D'Alembert's_principle, sum of the moments about pivot(B)is equal to zero


 * {| style="width:100%" border="0"


 * $$\sum{{{M}_{B}}=0\Rightarrow }\left( {{m}_{2}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}} \right)\cdot l+\left( ka({{\theta }_{2}}-{{\theta }_{1}}) \right)\cdot a+\left( {{m}_{2}}g \right)\cdot l{{\theta }_{2}}-\left( {{u}_{2}} \right)\cdot l=0$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"

$$\Rightarrow {{m}_{2}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=-k{{a}^{2}}({{\theta }_{2}}-{{\theta }_{1}})-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.2)
 * }
 * }


 * $$\left( 2 \right)$$ Write Eq.2.1 and Eq.2.2 in the form of Eq.2.3(system of coupled equation):


 * Eq.2.1 can be rearranged as,


 * {| style="width:100%" border="0"

$$
 * $${{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=\frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}}{{\theta }_{1}}+\frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}}{{\theta }_{2}}+\frac{{{u}_{1}}l}{m{}_{1}{{l}^{2}}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.4)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.4)
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $${{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=\frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}}{{\theta }_{1}}+\frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}}{{\theta }_{2}}+\frac{{{u}_{2}}l}{m{}_{2}{{l}^{2}}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.5)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.5)
 * }
 * Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:
 * Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:


 * {| style="width:100%" border="0"


 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$\left[ \begin{matrix} {{\overset{\centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{2}} \\ {{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}} \\ \end{matrix} \right]=\underbrace{\left[ \begin{matrix} 0 & 1 & 0 & 0 \\   \frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}} & 0 & \frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 & 0 & 1 \\   \frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}} & 0 & \frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}} & 0  \\ \end{matrix} \right]}_{\underset{-}{\mathop{A}}\,}\left[ \begin{matrix} {{\theta }_{1}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\theta }_{2}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{2}} \\ \end{matrix} \right]+\underbrace{\left[ \begin{matrix} 0 & 0 \\   \frac{1}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 \\   0 & \frac{1}{m{}_{2}{{l}^{2}}}  \\ \end{matrix} \right]}_{\underset{-}{\mathop{B}}\,}\left[ \begin{matrix} {{u}_{1}}l \\ {{u}_{2}}l \\ \end{matrix} \right]$$

$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.6)
 * }
 * }


 * Where:


 * {| style="width:100%" border="0"

0 & 1 & 0 & 0 \\   \frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}} & 0 & \frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 & 0 & 1 \\   \frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}} & 0 & \frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}} & 0  \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}


 * }
 * }


 * {| style="width:100%" border="0"

0 & 0 \\   \frac{1}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 \\   0 & \frac{1}{m{}_{2}{{l}^{2}}}  \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * }
 * }

= Problem 7 - Roll Control of Rocket By Actuating Ailerons =

Given



 * {| style="width:100%" border="0"

$$
 * $$\overset{\centerdot }{\mathop{\phi }}\,=\omega $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.1)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.1)
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }\delta $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.2)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.2)
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\overset{\centerdot }{\mathop{\delta }}\,=u$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.3)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.3)
 * }
 * }

Find
Control behavior in form of


 * {| style="width:100%" border="0"

$$
 * $$\overset{\centerdot }{\mathop{x\left( t \right)=\underset{-}{\mathop{A}}\,}}\,\underset{-}{\mathop{x}}\,\left( t \right)+\underset{-}{\mathop{B}}\,\underset{-}{\mathop{u}}\,\left( t \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.4)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.4)


 * }
 * }

Solution

 * Integrating Eq.7.3 gives,


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle \delta =ut$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.5)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.5)
 * }
 * Substitute Eq.7.5 for $$\displaystyle \delta $$, in Eq.7.2:
 * Substitute Eq.7.5 for $$\displaystyle \delta $$, in Eq.7.2:


 * {| style="width:100%" border="0"

$$
 * $$\overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }ut$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.6)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.6)
 * }
 * }


 * Now integrate Eq.7.6:


 * {| style="width:100%" border="0"

$$
 * $$\omega =\frac{-1}{\tau }\omega t+\frac{Q}{2\tau }u{{t}^{2}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.7)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.7)
 * }
 * }


 * Similarly integrating Eq.7.1 gives,


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle \phi =\omega t$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.8)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.8)
 * }
 * }


 * Substitute Eq.7.8 in Eq.7.7 implies,


 * {| style="width:100%" border="0"

$$
 * $$\omega =\frac{-1}{\tau }\phi +\frac{Q{{t}^{2}}}{2\tau }u$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.9)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.9)
 * }
 * }


 * From Eq.7.1 and Eq.7.9,


 * {| style="width:100%" border="0"

$$
 * $$\overset{\centerdot }{\mathop{\phi }}\,=\frac{-1}{\tau }\phi +\frac{Q{{t}^{2}}}{2\tau }u$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.10)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.10)
 * }
 * }


 * Now, Eq.7.6 and Eq.7.10 can be written in the required form as,


 * {| style="width:100%" border="0"

$$\left[ \begin{matrix} \overset{\centerdot }{\mathop{\phi }}\, \\ \overset{\centerdot }{\mathop{\omega }}\, \\ \end{matrix} \right]=\left[ \begin{matrix} \frac{-1}{\tau } & 0 \\ 0 & \frac{-1}{\tau } \\ \end{matrix} \right]\left[ \begin{matrix} \phi  \\ \omega  \\ \end{matrix} \right]+\left[ \begin{matrix} \frac{Q{{t}^{2}}}{2\tau } \\ \frac{Qt}{\tau } \\ \end{matrix} \right]\left[ u \right]$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.11)
 * }
 * }


 * where,


 * {| style="width:100%" border="0"

\frac{-1}{\tau } & 0 \\ 0 & \frac{-1}{\tau } \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}
 * }
 * }


 * {| style="width:100%" border="0"

\frac{Q{{t}^{2}}}{2\tau } \\ \frac{Qt}{\tau } \\ \end{matrix} \right]$$ |-
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * }


 * {| style="width:100%" border="0"

\phi  \\ \omega  \\ \end{matrix} \right]$$ |-
 * $$\underset{-}{\mathop{x}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{x}}\,=\left[ \begin{matrix}
 * }


 * {| style="width:100%" border="0"

|-
 * $$\underset{-}{\mathop{u}}\,=\left[ u \right]$$
 * $$\underset{-}{\mathop{u}}\,=\left[ u \right]$$
 * }