2.7: Chapter References

Solutions to Practice Problems

1. The solutions for P2.1 are
   1. \(F\left(s\right)=\dfrac{3s+8}{(s+2)(s+3)}\)
   2. \(G\left(s\right)=\dfrac{2}{s^2}e^{-3s}\)
   3. \(F\left(s\right)=\dfrac{2}{s+2}e^{-2(s+2)}\)
   4. \(F\left(s\right)=10\dfrac{(s+5)e^{-π(s+5)}}{{\left(s+5\right)}^2+1}\)

1. The solutions for P2.2 are
   1. \(y\left(t\right)=8(1-e^{-2t})\)
   2. \(y\left(t\right)=e^{-t}{\mathrm{c}\mathrm{os} 2t\ }\)
   3. \(f\left(t\right)=te^{-2t}+\dfrac{t^2}{2}e^{-2t}\)
   4. \(f\left(t\right)={\mathrm{c}\mathrm{os2} \left(t-3\right)u\left(t-3\right)\ }\)

1. Three-step solution process in Section 2.2:

Step 1) The quarter car free-body diagrams.

Step 2) Equations of motion

\[m_b{\ddot{x}}_1\left(t\right)+b_s\left({\dot{x}}_1\left(t\right)-{\dot{x}}_2(t)\right)+k_s\left(x_1\left(t\right)-x_2\left(t\right)\right)=0\nonumber \]

\[m_w{\ddot{x}}_2\left(t\right)+b_s\left({\dot{x}}_2\left(t\right)-{\dot{x}}_1\left(t\right)\right)+k_s\left(x_2\left(t\right)-x_1\left(t\right)\right)+k_wx_2(t)=f\left(t\right)\nonumber \]

Step 3) Transfer functions

\[X_1(s)=\dfrac{b_ss+k_s}{m_bs^2+b_ss+k_s}X_2(s)\nonumber \]

\[X_2\left(s\right)=\dfrac{1}{m_ws^2+b_ss+\left(k_s+k_w\right)}F\left(s\right)+\dfrac{b_ss+k_s}{m_ws^2+b_ss+(k_s+k_w)}X_1(s)\nonumber \]

1. \[\dfrac{(J+mr^2)}{r}\ddot{x}\left(t\right)+\dfrac{b}{r}\dot{x}\left(t\right)=-τ(t)\nonumber \]

    

\[G_1\left(s\right)=\dfrac{Θ(s)}{T(s)}=\dfrac{1}{\left(J+mr^2\right)s^2+bs}=\dfrac{1}{s\left[\left(J+mr^2\right)s+b\right]}\nonumber \]

\[G_2\left(s\right)=\dfrac{X(s)}{T(s)}=\dfrac{-r}{\left(J+mr^2\right)s^2+bs}=\dfrac{-r}{s\left[\left(J+mr^2\right)s+b\right]}\nonumber \]

1. The coefficient matrices for the five-element beam model are

\[J=J_e\left[ \begin{array}{cc} \begin{array}{ccc} 52 & 42 & 30 \\ 42 & 40 & 30 \\ 30 & 30 & 28 \end{array} & \begin{array}{cc} 18 & 6 \\ 18 & 6 \\ 18 & 6 \end{array} \\ \begin{array}{ccc} 18 & 18 & 18 \\ 6 & 6 & 6 \end{array} & \begin{array}{cc} 16 & 6 \\ 6 & 4 \end{array} \end{array} \right] \nonumber\nonumber \]

\[B=b_e\left[\begin{array}{ccccc}
2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
0 & -1 & 2 & -1 & 0 \\
0 & 0 & -1 & 2 & -1 \\
0 & 0 & 0 & -1 & 2
\end{array}\right] \nonumber\nonumber \]

\[K=k_e\left[\begin{array}{ccccc}
2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
0 & -1 & 2 & -1 & 0 \\
0 & 0 & -1 & 2 & -1 \\
0 & 0 & 0 & -1 & 2
\end{array}\right] \nonumber\nonumber \]

The fluid flow-induced torque vector \(Γ\) is expressed as

\[\ Γ\mathrm{(}\mathrm{t)}=\dfrac{l}{2}\left[ \begin{array}{c} \begin{array}{c} f^f_1\left(t\right)+2\sum^5_{j=2}{f^f_j\left(t\right)} \\ f^f_2\left(t\right)+2\sum^5_{j=3}{f^f_j\left(t\right)} \end{array} \\ \begin{array}{c} f^f_3\left(t\right)+2\sum^5_{j=4}{f^f_j\left(t\right)} \\ f^f_4\left(t\right)+2f^f_5\left(t\right) \\ f^f_5\left(t\right) \end{array} \end{array} \right] \label{2.210}\nonumber \]

1. The following are the free body diagrams for \(J\) and \(m.\)

The equations of motion for \(J\) and \(m\) are as follows.

\[j\ddot{θ}\left(t\right)+b_m\dot{θ}\left(t\right)=τ\left(t\right)+rf(t)\nonumber \]

\[m\ddot{x}\left(t\right)=-rf\left(t\right), x\left(t\right)=rθ(t)\nonumber \]

b) The Laplace transforms of all of the equations are

\(\begin{array}{c} \begin{array}{ccc} I_a\left(s\right) & = & \dfrac{1}{L_as+R_a}\left(E_a\left(s\right)+E_m\left(s\right)\right) \ \\ T(s) & = & K_τI_a\left(s\right), E_m\left(s\right)\ =K_mΩ\left(s\right) \end{array} \\ \begin{array}{ccc} Ω\left(s\right) & = & \dfrac{1}{Js+b_m}\left(T\left(s\right)-T_l(s)\right), X\left(s\right)=\dfrac{-1}{ms^2}F(s) \ \\ Ω\left(s\right) & = & sΘ\left(s\right), \ X\left(s\right)=rΘ\left(s\right), sX\left(s\right)=rΩ(s) \ \end{array} \end{array}\) &

c) The system block diagram is

d) The equation of motion from \(E_a(s)\) to \(Θ(s)\) is \[G\left(s\right)=\dfrac{Θ(s)}{E_a(s)}=\dfrac{K_τ}{s\left\{(L_as+R_a)((J+ms^2)s+b_m)+K_τK_m\right\}}\nonumber \]

The linearized model at \(v_0=30\ m/sec\) is \[1000δ\dot{v}\left(t\right)+60.1δv\left(t\right)=δf_m\left(t\right) \to G\left(s\right)=\dfrac{ΔV(s)}{ΔF_m(s)}=\dfrac{1}{1000s+60.1}\nonumber \]​​​​​​​

1. Solution for steady state \(x_0=2>0\) with \({\dot{x}}_0={\ddot{x}}_0=0\)
2. The linearized differential equation and transfer function are shown below.

\[2δ\ddot{x}\left(t\right)+6δ\dot{x}\left(t\right)+7δx\left(0\right)=δf_a(t) \to \dfrac{\Delta X(s)}{\Delta F_a(s)}=\dfrac{1}{2s^2+6s+7}\nonumber \]

1. Complete the problem based on Example 2.19.

1. Complete the reading assignment and write a summary by following these steps below.
   1. Reading references [21] and [22] and summarizing the Kriging gray-box modeling method into a step-by-step process.
   2. Creating a MATLAB example of a physical system in your research area and generating the associated Kriging gray-box model following the process developed in Step 1).
   3. Summarizing the Kriging gray-box modeling process and your example into a technical report.

1. Complete the reading assignment and write a summary.
   1. Reading references [10] and [11] and summarizing the \(q\)-Markov Cover system identification method into a step-by-step process.
   2. Creating a MATLAB example of a physical system in your research area and generating the \(q\)-Markov Cover system model following the process developed in Step 1).
   3. Summarizing the \(q\)-Markov Cover modeling process and your identification example into a technical report.