linearizing a non linear state space equation

Question

I have this 1-DOF helicopter simulator.

The motor's armature resistance is $R$, inductance $L$ and motor constant $K$ and reaction torque on the motor shaft is $τ$. $T$ is equal to $k_1w_2^2$ and $\tau$ is equal to $k_2w_2^2$.

$$w_1 = \dot{\varphi}_1$$

$$w_2=\dot{\theta}$$

Τhe differential equation for the current is:

$$\frac{\partial I}{\partial t}=\frac{E - RI - K_1w_2}{L}$$

in which $E$ is the input armature voltage and the equation for the torque is

$$\frac{\partial w_2}{\partial t}=\frac{KI - K_2w_2^2}{J} $$

in which $k_2w_2^2$ is the load torque and finally for the last equation I used lagrange method to derive equation of motion of mass $m$ and $M$.

$$ T = \frac{1}{2}mb^2\dot{\varphi}^2 + \frac{1}{2}M\alpha^2\dot{\varphi}^2 + \frac{1}{2}Jw_2^2 $$

$$ U = mgb\sin\varphi - Μg\alpha\sin\varphi $$

which finally gives

$$ mb^2\dot{w} + M\alpha^2\dot{w}_1 + bmg\cos\varphi - Mg\alpha\cos\varphi = -Tb = K_1w_2^2b $$

$$ \frac{\partial w_1}{\partial t} = \frac{-k_1w_1^2 + Mg\alpha\cos\varphi - mgb\cos\varphi}{M\alpha^2 + mb^2} $$

I wanna know whether everything here is right and how to linearize it around $\varphi = 0$ cause when I try to do that $\dfrac{\partial w_1}{\partial t}$ becomes a constant which is:

$$ \frac{\partial w_1}{\partial t} = \frac{Mg\alpha - mgb}{M\alpha^2 + mb^2} $$

Answer 1

The Lagrangian for your 1D pitch system is $\mathcal{L} := T - U$, which is

$$ \mathcal{L} = \frac{1}{2}(mb^2 + Ma^2)\dot{\phi}^2 + \frac{1}{2}J\dot{\theta}^2 + (Ma - mb)g\sin\phi $$

The Euler-Lagrange Equations are

$$ \frac{d}{dt}\bigg(\frac{\partial\mathcal{L}}{\partial \dot{q}_k}\bigg) + \frac{\partial\mathcal{L}}{\partial q_k} = Q_k $$ This gives the EOM

$$ (mb^2 + Ma^2)\ddot{\phi} + (Ma - mb)g\cos\phi = -Tb = -k_1\dot{\theta}^2b $$ Under a small angle approximation, $\cos\phi \approx 1$, so this leaves

$$ (mb^2 + Ma^2)\ddot{\phi} + (Ma - mb)g = -k_1b\dot{\theta}^2, $$ or in another form, $$ \ddot{\phi} = \frac{mb-Ma}{mb^2 + Ma^2}g - \frac{k_1b}{mb^2 + Ma^2}\dot{\theta}^2 = \beta_1 - \beta_2\dot{\theta}^2. $$ Now, I don't know if these are the correct equations for your system, but based on what you outlined in your question, this is what the EOM are for $\phi(t)$.

You also have a coupled system of ODE's for $I(t)$ and $\theta(t)$ since from $\dot{\theta} = \omega_2$ it follows that

$$ \begin{align} \ddot{\theta} &= \frac{K}{J}I - \frac{k_2}{J}\dot{\theta}^2\\ \dot{I} &= \frac{E}{L} - \frac{R}{L}I - \frac{k_1}{L}\dot{\theta} \end{align} $$

Again, this is all based off your model that you wrote in the problem statement. So, in total, you have three ODE's for the three variables $\theta,\phi,I$. Given that this system of 3 ODE's is nonlinear, there is no analytical solution. What you can do to simulate this is solve the system in MATLAB for example using ODE45 or just any differential equaton solver.

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If you want to linearize this system, you can do the following. First, put the system in state space form, i.e. let $x = [\phi,\dot{\phi},\theta,\dot{\theta},I]^\intercal$ so that the full dynamics are

$$ \begin{align} \dot{x}_1 &= x_2\\ \dot{x}_2 &= \beta_1\cos x_1 - \beta_2 x_4^2\\ \dot{x}_3 &= x_4\\ \dot{x}_4 &= \frac{K}{J}x_5 - \frac{k_2}{J}x_4^2\\ \dot{x}_6 &= \frac{E}{L} - \frac{R}{L}x_5 - \frac{k_1}{L}x_4. \end{align} $$ Notice that the nonlinear equations are the second one and the fourth one. So right now you system like $\dot{x} = f(x)$, where $f(x)$ is a nonlinear function (5 by 1 vector). To linearize this sytem into the form $\dot{x} = Ax$, you can set

$$ A = \bigg[\frac{\partial f}{\partial x}\bigg]_0, $$ evaluated at some nominal conditions, which will give a 5 by 5 matrix. In your case this gives

$$ A = \begin{bmatrix} 0 & 1 & 0 & 0 & 0\\ -\beta_1\sin x_1 & 0 & 0 & -2\beta_2 x_4 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -2\frac{k_2}{J}x_4 & \frac{K}{J}\\ 0 & 0 & 0 & -\frac{k_1}{L} & -\frac{R}{L} \end{bmatrix}_0 $$ So you need to come up with some reference point for linearization for $\phi$ and $\dot{\theta}$, and then you can simulate the linearized system.