# Transfer Function of Spring-Damper System

Given the spring-damper system below where $x_i$, $x_i$, and $y_i$ are position values, how do I find the transfer function $\frac{X_o(s)}{X_i(s)}$? Wouldn't it require mass? I tried neglecting the masses in the equation but it zeros out the acceleration term and does not work. Any diea? • Looks like a HW problem. In order to seek help from this community you need to show your work. So what have you done so far, show work Mar 14 '17 at 11:08
• That's a very odd looking naming convention... I'm not sure why it goes $x_i$, $x_0$ and then $y$ when they all seem to be different masses with the same direction of displacement. The $y$, $x_o$ and $x_i$ should all be masses though (at least the diagram implies that).
– JMac
Mar 14 '17 at 11:45
• @MahendraGunawardena I'm simply asking about the masses, I'm not asking for a solution because I'll solve it on my own. Should I assign mass numbers to the squares in between the spring or damper branches? Are they supposed to be masses? Can the problem be even solved if there are no masses? Mar 14 '17 at 12:23
• @JMac I'm sure the naming y is random and bears no significance with the system's orientation relative to displacements $x_o$ and $x_i$. So the squares in between the spring/damper branches are indeed implied to be masses right? The masses can't be zero right? Mar 14 '17 at 12:26
• @MahendraGunawardena Besides, I have tried solving the problem assuming no masses and I couldn't get a transfer function. So I'm asking if there really should be masses or if it's possible to solve that system assuming no masses. Mar 14 '17 at 12:28

First, create the free body diagram for this system. If you cut through the spring $k_1$ and the damper $b_1$ you will get two forces $F_{k_1}=k_1(x_i-x_0)$ and $F_{b_1}=b_1(\dot{x}_i-\dot{x}_0)$ opposing the direction of $x_i$. Writing down Newton's second law of motion for the mass $m_i$ will result in:

$$m_i\ddot{x}_i=-b_1(\dot{x}_i-\dot{x}_0)-k_1(x_i-x_0) \implies m_i\ddot{x}_i+b_1\dot{x}_i+k_1x_i=b_1\dot{x}_0+k_1x_0.$$

Assuming zero initial conditions we can transform this into the Laplace domain:

$$\left[m_is^2+b_1s+k_1\right]x_i(s)=\left[b_1s+k_1\right]x_0 \implies \frac{x_0(s)}{x_i(s)}=\frac{m_is^2+b_1s+k_1}{b_1s+k_1}.$$

If the point $x_i$ is attached to a massless position, then we simply set $m_i=0$ in the previous expression.

I think this problem has the following structure:

• $$y$$ corresponds to the disturbance from the road
• $$x_o$$ corresponds to the travel of the unsprung mass
• $$x_i$$ corresponds to the travel of the sprung mass

This is in accordance to this image. For this problem, I think masses should be assigned to displacement $$x_o$$ and $$x_i$$. Otherwise there is not much point.