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How can I determine the damping coefficient that minimizes the amplitude of vibrations? This is an extension of Coupled ODEs that model a quad rotor \begin{align} m_{eq}\ddot{y}_1 &= k_{eq}\Bigl[y_1 + \frac{m_{eq}}{m_b}\sum_jy_j\Bigr] + c\Bigl[\dot{y}_1 + \frac{m_{eq}}{m_b}\sum_j\dot{y}_j\Bigr] - m_{eq}gy_1 + F_1(t)\\ m_{eq}\ddot{y}_2 &= k_{eq}\Bigl[y_2 + \frac{m_{eq}}{m_b}\sum_jy_j\Bigr] + c\Bigl[\dot{y}_2 + \frac{m_{eq}}{m_b}\sum_j\dot{y}_j\Bigr] - m_{eq}gy_2 + F_2(t)\\ m_{eq}\ddot{y}_3 &= k_{eq}\Bigl[y_3 + \frac{m_{eq}}{m_b}\sum_jy_j\Bigr] + c\Bigl[\dot{y}_3 + \frac{m_{eq}}{m_b}\sum_j\dot{y}_j\Bigr] - m_{eq}gy_3 + F_3(t)\\ m_{eq}\ddot{y}_4 &= k_{eq}\Bigl[y_4 + \frac{m_{eq}}{m_b}\sum_jy_j\Bigr] + c\Bigl[\dot{y}_4 + \frac{m_{eq}}{m_b}\sum_j\dot{y}_j\Bigr] - m_{eq}gy_4 + F_4(t) \end{align} where $F_i(t) = F_i\cos(\omega t + \phi)$


I tried converting to the state space equation but this doesn't make the system any easier to deal with:

By letting $q_1 = y_1$, $q_2 = \dot{y}_1$, $q_3 = y_2$, $q_4 = \dot{y}_2$, $q_5 = y_3$, $q_6 = \dot{y}_3$, $q_7 = y_4$, and $q_8 = \dot{y}_4$, we obtain the following state equations. \begin{align} \dot{\mathbf{q}} &= \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ a & b & C & d & C & d & C & d\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ C & d & a & b & C & d & C & d\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ C & d & C & d & a & b & C & d\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ C & d & C & d & C & d & a & b \end{bmatrix}\mathbf{q} + \frac{1}{m_{eq}} \begin{bmatrix} F_1(t)\\ F_2(t)\\ F_3(t)\\ F_4(t) \end{bmatrix}^{\intercal} \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}\\ \mathbf{y} &= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}\mathbf{q} \end{align} where $a = \frac{k_{eq}}{m_{eq}} + \frac{k_{eq}}{m_b} - g$, $b = \frac{c}{m_{eq}} + \frac{c}{m_b}$, $C = \frac{k_{eq}}{m_b}$, and $d = \frac{c}{m_b}$.

If I take $\mathcal{L}^{-1}\{(s\mathbb{I} - \mathbf{A})^{-1}\}$ with Mathematica, the solution is ridiculously long.

I have also tried setting up $[m]\ddot{y} + [c]\dot{y} + [k]y = [F]$ but I am not sure how to solve this equation either.

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You should use complex numbers. Assume the solution is of the form $y_i(t)=\tilde Y_i\mathrm e^{\mathrm i\omega t}$ where $\tilde Y_i$ is complex. Write also $\tilde F_i=F_i\mathrm e^{\mathrm i\phi_i}$ such that $F_i(t)=\mathrm{Re}(\tilde F_i\mathrm e^{\mathrm i\omega t})$. The system of equations becomes "simply" (it's "only" of the fourth order) $$ (\omega^2+b\mathrm i\omega+a)\tilde Y_i+(C+d\mathrm i\omega) \sum_j\tilde Y_j=-\tilde F_i.$$ Let us write $$\mathbb U=\begin{pmatrix}1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{pmatrix}.$$ $\mathbb U$ has characteristic polynomial equal to $\Delta=X^4-4X^3$ has $\Delta$ has only two roots : 0 and 4. The determinant of the system is the determinant of $A=(\omega^2+b\mathrm i\omega+a)\mathbb I+(C+d\mathrm i\omega)\mathbb U$, so is equal to $$\delta=(C+d\mathrm i\omega)^4 \,\Delta\left(-\frac{\omega^2+b\mathrm i\omega+a}{C+d\mathrm i\omega}\right)=(C+d\mathrm i\omega)^4+4(\omega^2+b\mathrm i\omega+a)(C+d\mathrm i\omega)^3.$$ The solutions of the system are given by Cramér's rules: $\tilde Y_i=\mathrm{det}B_i/\delta,$ where $B_i$ is the matrix made of the original matrix $A$ in which one replace the $i^{\rm th}$ column by the right hand side $(\tilde F_i)_{i}^T$. The damping corresponds to the imaginary part of the solutions $\tilde Y_i$. From this point, you should be able to take and find the optimal damping coefficient.

Note. There seems to be some mistakes in your starting equations. The sign in front of $k_{\rm eq}$ is wrong and leads to exponential explosion of $y_i$. Moreover, the term $-mgy$ is dimensionally inconsistent.

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  • $\begingroup$ I noticed that yesterday as well. It should read -k_{eq}( + ) and -c( + ). I forgot the negative when plugging in z. $\endgroup$ – dustin Dec 9 '14 at 17:36
  • $\begingroup$ How should the gravity term be written then? $\endgroup$ – dustin Dec 9 '14 at 17:55
  • $\begingroup$ If you have a, b, C, d defined as I did, your U should be $$\begin{bmatrix}0 & 1 & 1 & 1\\1 & 0 & 1 & 1\\1 & 1 & 0 & 1\\1 & 1 & 1 & 0\end{bmatrix}$$ $\endgroup$ – dustin Dec 9 '14 at 19:49
  • $\begingroup$ With the U I supplied, $\Delta = (x - 3)(x + 1)^3$. Also, making the negative correction leads to $$(a + bi\omega - \omega^2)\mathbb{I} + (C + di\omega)\mathbb{U} = A$$ $\endgroup$ – dustin Dec 10 '14 at 0:28

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