Linearize $\tau -mgl cos(\theta)$ for $\tau = \tau_0 + \delta \tau, \theta = \theta_0 + \delta \theta$

Question

I'm trying to solve a control problem involving a pendulum, in which the equation of motion is:

$ml^2\frac{d^2 \theta}{d\theta^2} = \tau -mgl cos(\theta)$

I need to linearize $\tau -mgl cos(\theta)$ for $\tau = \tau_0 + \delta \tau, \theta = \theta_0 + \delta \theta$.

The answer to this question is supposed to be $mgl sin(\theta_0)\delta \theta + \delta \tau$, but I just don't see how to get to that answer. I know it probably assumes that $\delta \tau = -mglcos(\theta_0)$, as it is the minimal torque required ot counteract the force of gravity. But the other part, probably some mathematical trick involving the $\delta \theta$, eludes me.

Answer 1

I think you forgot to mention that $(\tau_0, \theta_0)$ is a fixed point, i.e. $\tau_0 - mgl \cos(\theta_0) = 0$. Now, we can use Taylor series expansion around $(\tau_0, \theta_0)$ to obtain

$$ml^2 \frac{d^2 (\theta_0 + \delta \theta)}{dt^2} = \tau_0 + \delta \tau - mgl [\cos(\theta_0) - \sin(\theta_0) \delta \theta + O((\delta \theta) ^2)]$$

Ignoring the higher order terms for small $\delta \theta$, we can obtain

$$ml^2 \frac{d^2 (\delta \theta)}{dt^2} = \delta \tau + mgl \sin(\theta_0) \delta \theta$$

Answer 2

First: $\tau$ is not a state variable but a forcing term (torque in the equation), so the perturbation $\tau = \tau_0 + \delta\tau$ is unnecessary.

Second: In principle, you need not necessarily choose a fixed point equilibrium to linearize about (though it certainly helps for any practical implementation of a controller).

First let's derive the general solution, then we'll see how it works for the specific problem.

Let $x$ be an arbitrary state vector in the Hilbert space $X$ and suppose the time evolution of $x$ is given via $\dot{x} = f(x,t)$ for a smooth mapping $f:X\rightarrow X$. As obareey states, $f$ enjoys a Taylor expansion about the function $y$ given by $$ f(x) = f(y) + \partial_xf(y)(x-y) + \frac{1}{2}\partial^2_xf(y)(x-y)(x-y)^T+\cdots, $$ where $$ \partial_xf(y) = \frac{\partial f}{\partial x}|_{x = y},\ \ \partial^2_xf(y) = \frac{\partial^2 f}{\partial x\partial x^T}|_{x = y}, $$ etc. The linearization is of $f$ is defined to be the truncation of this series to first (that is, linear) order. This produces the linearized equation $$ \dot{x} \approx f(y) + \partial_xf(y)(x-y). $$ Since this approximation is based on the assumption $x = y + \delta$ where $|\delta|/|y| \ll 1$, it follows that $\dot{x} = \dot{y} + \dot{\delta} = f(y) +\dot{\delta}$, and so we have $$ \dot{\delta} = \partial_xf(y)\delta. $$ Since $y$ is presumed known then $x = y + \delta$ is solved when $\delta$ is known and $\partial_xf(y)$ is just a (possibly time-dependent) matrix $A(t)$. By this method we have reduced the nonlinear equation to the equation $$ \dot{\delta} = A(t)\delta. $$

Now on to your problem:

The state vector is $x = [\theta, \dot{\theta}]^T$ and the nonlinear model is $$ \frac{d\theta}{dt} = f_\theta = \dot{\theta},\ \ \frac{d\dot{\theta}}{dt} = f_{\dot{\theta}} = \frac{\tau}{m\ell^2}- \frac{g}{\ell}\cos\theta. $$ Choosing the constant reference $y = [\theta_0,\dot{\theta}_0]^T$, the linearization is equivalent to finding the partials $$\begin{align} \frac{\partial f_\theta}{\partial \theta} = 0, &\ \frac{\partial f_\theta}{\partial \dot{\theta}} = 1\\ \frac{\partial f_\dot{\theta}}{\partial \theta} = \frac{g}{\ell}\sin\theta_0, &\ \frac{\partial f_\dot{\theta}}{\partial \dot{\theta}}=0\\ \end{align}$$ and therefore the linearized system is of the form $$ \dot{\delta} = \begin{bmatrix} 0 & 1\\ \frac{g}{\ell}\sin\theta_0 & 0\\ \end{bmatrix}\delta $$ where $\delta = [\theta-\theta_0,\dot{\theta}-\dot{\theta}_0]^T$. Choosing the fixed point $\theta_0 = \cos^{-1}(\tau/mg\ell)$ and $\dot{\theta}_0 = 0$ we have $$ \begin{align} \frac{d\theta}{dt} &= \dot{\theta}\\ \frac{d\dot{\theta}}{dt} &= \frac{g}{\ell}\sin(\cos^{-1}(\tau/mg\ell))\theta - \frac{g}{\ell}\sin(\cos^{-1}(\tau/mg\ell))\cos^{-1}(\tau/mg\ell) \end{align} $$