Mechanical System Transfer Function - Systems Modelling

Question

First of all my english writing is not so good, but I can read it without problems. I've got this mechanical system and I need to get some transfer functions. We made all the force diagrams and got the respective equations, they're eight. Now we need to get the transfer function of $\dot{x}_1\over v_1$, $x_2\over v_3$, $\ddot{x}_3\over v_4$ and $x_4\over v_7$.

Now, assuming that the equations are correct, what we were trying to do was for the first T.F clear all $X_2(s)$ from the second equation and replace it on the first equation, then we clear all $sB_1$, $B_2X_1(s)$ from the new equation that we just got from replacing $X_2(s)$, then we're trying to make $sB_1$, $B_2X_1(s)\over V_1(s)$ but there's no way to clear $V_1(s)$.

It's very difficult to understand, and I don't know if I explained myself, but if someone can tell us what we're doing wrong I'll be really grateful.

Answer 1

First off, the equations you have are not correct. For example, if equations 1 and 5 are added it results in $V_1+V_5=0$. The l.h.s of the first four equations should be $V_1-V_5$, $V_2-V_6$, $V_3-V_7$, and $V_4-V_8$. Then you are have four equations in the four unknowns $X_1$, $X_2$, $X_3$, and $X_4$ after taking Laplace transforms with zero initial conditions.

The general method is to solve for these four unknowns and express them in terms of the $V_i$ and then you get the transfer functions for each input-output pair. The derivatives can then be obtained by multiplying with powers of the Laplace variable $s$.

Since you need just 4 transfer functions, I think it can be done a bit more easily.

For the first transfer function, you are interested only in $V_1$. Set all the other $V_i$ to zero and solve for $\frac{X_1}{V_1}$. Since you want the derivative just multiply the result by $s$.

For the second transfer function, set all the $V_i$ except $V_3$ to zero and solve for $\frac{X_2}{V_3}$. Multiply the result by $s^2$ to obtain the desired transfer function.

I hope you get the idea.

Answer 2

Your system of equations can be written in matrix form $$\mathbf{M}\ddot{\mathbf{x}}=\mathbf{K}\mathbf{x}+\mathbf{B}\dot{\mathbf{x}}+\mathbf{F}$$ Assuming the direction of positive state variables are to the right, we get matrices $$\mathbf{M}=\begin{bmatrix}m_1&0&0&0\\0&m_2&0&0\\0&0&m_3&0\\0&0&0&m_4\end{bmatrix},\hspace{0.5cm}\mathbf{F}=\begin{bmatrix}V_1-V_5\\V_2-V_6\\V_3-V_7\\V_4-V_8\end{bmatrix}$$ $$\mathbf{K}=\begin{bmatrix}-(k_1+k_2)&k_2&0&0\\k_2&-(k_2+k_3)&k_3&0\\0&k_3&-(k_3+k_4)&k_4\\0&0&k_4&-(k_4+k_5)\end{bmatrix}$$ $$\mathbf{B}=\begin{bmatrix}-(b_1+b_2)&b_2&0&0\\b_2&-(b_2+b_3)&b_3&0\\0&b_3&-(b_3+b_4)&b_4\\0&0&b_4&-(b_4+b_5)\end{bmatrix}$$ Assuming zero initial conditions, the Laplace transformed variable $\mathbf{X}(s)$ can be written $$\mathbf{X}(s)=\left(s^2\mathbf{M}-s\mathbf{B}-\mathbf{K}\right)^{-1}\mathbf{F}(s)$$ The matrix transfer function is $$\mathbf{T}(s)=\left(s^2\mathbf{M}-s\mathbf{B}-\mathbf{K}\right)^{-1}=\begin{bmatrix}T_{11}(s)&\cdots&T_{14}(s)\\\vdots&\ddots&\vdots\\T_{41}(s)&\cdots&T_{44}(s)\end{bmatrix}\hspace{1cm}\text{where}\hspace{1cm}T_{ij}(s)=\frac{X_i(s)}{F_j(s)}$$ The force inputs are of the form $F_i = V_i-V_{i+4}$.
The transfer function for $x_4/V_7$ is then $T_{43}(s)$ where $V_3=0$. or equivalently, $F_3<0$.
To get transfer functions that correspond to $\dot{x}$, e.g. $\dot{x}_i/F_j$, you multiply by $s$, i.e. $$\frac{sX_i(s)}{F_j(s)}=sT_{ij}(s)$$