# Transfer function for double cart system

System: Define X2 = Y2;

I've described the system with the following diff equation: $$f_{tot} = m_1\ddot{x_1} + k(x_2-x_1)+m_2\ddot{x_2}+B(\dot{x_2}-\dot{x_1})$$ where m1, m2, k and B are Cart mass 1, Cart mass 2, Spring constant and dampner constant respectively. All derivatives are in respect to time, $t$.

I get the following after Laplace: $$F(s) = m_1s^2X_1(s)+k(X_2(s)-X_1(s)) + m2s^2X_2(s)+B(sX_2(s)+sX_1(s))$$

$$G(s) = \frac{Output}{Input} = \frac{X_2(s)}{F(s)} = ???$$

But the problem is that I can't define $X_1(s)$. Should $X_1(s)$ be interpreted as $1$? Is the above formula correct?

## 2 Answers

The input in your system is the force $f(t)$ and the output is $x_2(t)$. I take it we are assuming zero IC. The free body diagram for the motion of two carts can be modeled as \begin{align} m_1\ddot{x}_1 &= f(t) + k(x_2-x_1) + b(\dot{x}_2-\dot{x}_1)\\ m_2\ddot{x}_2 &= k(x_1-x_2) + b(\dot{x}_1-\dot{x}_2) \end{align} Taking the Laplace transform, we obtain \begin{align} m_1s^2X_1(s) &= F(s) + k(X_2(s)-X_1(s)) + bs(X_2(s)-X_1(s))\tag{1}\\ m_2s^2X_2(s) &= k(X_1(s)-X_2(s)) + bs(X_1(s)-X_2(s))\\ X_1(s)&=X_2(s)\frac{k+b+m_2s^2}{k+bs}\tag{2} \end{align} Substitute equation $(2)$ into equation $(1)$ to determine you transfer function.

\begin{align} m_1s^2X_1(s) &= F(s) + k(X_2(s)-X_1(s)) + bs(X_2(s)-X_1(s))\tag{1}\\ m_2s^2X_2(s) &= k(X_1(s)-X_2(s)) + bs(X_1(s)-X_2(s))\\ X_1(s)&=X_2(s)\frac{k+b\color{red}s+m_2s^2}{k+bs}\tag{2} \end{align}