# 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 – Mahendra Gunawardena 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? – John Smith 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? – John Smith 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. – John Smith 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.