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I'm trying to model a basic ball and beam system using Euler-Lagrange Equation. My system looks something like this Source: Modelling and Control of Ball and Beam System using Coefficient Diagram
Method (CDM) based PID controller

I have come up with this final Euler-Lagrange Equation:

enter image description here

Where JB is the ball's moment of inertia, r is the radius of the ball, m is the mass of the ball, g is the acceleration constant, β is the ratio of d to L, rB is the position of the ball along the beam, and finally θ is the gear angle.

The Euler-Lagrange Equation was acquired after finding the partial derivatives with respect to rB.

My question is: How do I proceed with finding the transfer function? I have seen two research papers straight away cancelling the term with $\dot{θ}$ from the equation, changing the equation to Laplace domain, and finding rB to $θ$. I'm assuming that this was done due to an assumption, but I'm unable to figure out what this assumption is. Is this a correct way to do it?

Alternatively, can I proceed with leaving the $\dot{θ}$ term and changing it to the Laplace domain, and again find the transfer function from there?

Also, the research paper has proceeded with finding the transfer function from the Euler-Lagrange equation taken by finding the partial derivatives with respect to rB. What about finding it with respect to $θ$?

I'm a bit confused, so I'd appreciate some clarification. Thanks!

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How do I proceed with finding the transfer function?

The concept of (Laplace transform) transfer functions are applicable for linear systems. So, you need to linearise your equations of motion. Linearisation uses the concept of Taylor series.

In the example given in your question, that reduces to removing the term $\dot{\theta}^2$. This comes from the assumption used for linearization.

but I'm unable to figure out what this assumption is.but I'm unable to figure out what this assumption is.

The assumption is that values of $\dot{r},\ \ddot{r},\ \theta,\ \dot\theta,\ \ddot\theta$ are small such that $\dot{r}^2,\ \ddot{r}^2,\ \theta^2,\ \dot\theta^2,\ \ddot\theta^2, \ \dot{\theta}\dot{r},\ \theta\dot\theta$ etc. are negligibly small compared to $\dot{r},\ \ddot{r},\ \theta,\ \dot\theta,\ \ddot\theta$.

Alternatively, can I proceed with leaving the θdot term and changing it to the Laplace domain, and again find the transfer function from there?

Just to clarify, the papers (I think) are not cancelling $\dot{\theta}$ terms, they are cancelling terms which are the product of two or more "small-valued" variables; e.g. $\theta\dot{\theta}$ or $\dot{\theta}^2$ or $\theta^2$.

What about finding it with respect to θ

I assume that the input to your system is defined as $\theta$. If the input to the system was something else, like torque to the disc, then, you are right. The partial derivative with respect to θ does need to be taken.

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what if "θ˙^2" has a considerable value, such that the equation cannot be linearized. How would it be approached?

It is better to follow a text book for knowing the proper way to do this. I will attempt to illustrate the specific case below. I follow the Taylor series expansion as mentioned above.

Let the steady value of $\dot{\theta}$ be $a$. Then, $$ \dot{\theta}^2_{(a+\delta_\dot{\theta})} = a^2 + \frac{2\dot{\theta}|_{a}}{1!} \delta_{\dot{\theta}} + \frac{2}{2!} \delta^{2}_{\dot{\theta}} = \color{blue}{a^2} + \color{red}{ \frac{2 a}{1!} \delta_{\dot{\theta}} } + \frac{2}{2!} \delta^{2}_{\dot{\theta}} $$

The term in blue is part of the steady state and is usually not accounted for deviations from steady state which we want to analyse. The term in red is the linear part which we will use for further analysis of deviations from steady state. The variable $\dot\theta$ is now replaced by its variation from steady state value $a$. The new variable which will appear in the linear analysis is denoted $\delta_{\dot{\theta}}$. However, a lot of textbooks will not introduce a new symbol for this purpose. They will reuse the original symbol $\dot{\theta}$ and the reader needs to understand from context, if they are referring to the original variable or the variation from its steady state value. Note that the term is red is linear in the new variable $\delta_{\dot{\theta}}$

The variation $\delta_{\dot{\theta}}$ is assumed to be a small value. If the variation from steady state value is also not small, then linear analysis is not suitable and one needs to do non linear analysis and transfer functions cannot be used (IMO).

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  • $\begingroup$ Thanks a lot for the clarification. It is much more clear now. To make sure I got it correct: In order to find the transfer function of a system, I need to have a linear equation, and that is done by cancelling "θ˙^2". This term can only be cancelled by assuming that "θ˙^2" is of a negligible value, thus having little effect on the equation. Am I correct here? If so, what if "θ˙^2" has a considerable value, such that the equation cannot be linearized. How would it be approached? $\endgroup$
    – Zelreedy
    Jul 23 at 12:14

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