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.

System Equations

  • $\begingroup$ What are the $V_i$'s ? Are they forces ? $\endgroup$
    – drC1Ron
    Commented Jun 22, 2022 at 11:58

2 Answers 2


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.


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. $$\frac{sX_i(s)}{F_j(s)}=sT_{ij}(s)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.