More Examples of Linear Systems (LS 3/40)
Author's note: This linear systems series is a collection of notes I took from one of my professors. It's my attempt to get practice with latex and an excuse to have something to blog about. You'll notice some anomalies within the syntax. This is mostly due to the binder library not supporting full latex functionality. I'm looking at you "\substack"! There are some notation shortcuts my professor wrote on the board which I'm unable to reproduce with latex, so I'll default to more conventional ways to represent them. Most of it should be understandable to the reader though. In the future, some of it is expected to change. If you're curious on what failures and where work arounds might've been attempted you can view here for more background.'Linearization
Basically every real world physical system is by definition nonlinear. Some of these systems can be modeled by nonlinear equations, and some of these nonlinear equations can be "approximated" by linear equations under certain conditions. This isn't always true however. Some nonlinear equations, with very small differences in initial states will generate radical seemingly unrelated solutions. This phenomenon is called chaos. This post will assume the nonexistence of chaos. There are several techniques one might use to linearize a system. Below we go through a few examples including a simple mechanical SISO (single-input single-output systems, electrical SISO systems (RLC circuit), and a more complex multi variable system MISO (multi-input single-output) system.begin
Suppose you had a system defined by two state variables
\begin{aligned} \dot x_1& = x_1 x_2 - x_1^3 \\ \dot x_2& = -5x_2 + 9e^{x_1+5x_2}-9+5 \end{aligned}
The minus 9 and plus 5 terms at the end of the second equation were changed to make the solution more user friendly. Note that this represents motion around an equilibrium point. Depending on where you choose to linearize, different solutions will appear.
Let's choose the following;
\begin{aligned} x_1 = 0 = x_{10} \\ x_2 = 0 = x_{20} \end{aligned}
figure 1 - XY Plot with origin
Now Imagine if figure had a disturbance leading to perturbed poles - represented by the area surrounded the origin. Think of the perturbed poles as
where;
\begin{aligned} x_1 &= x_{10} + \tilde x_1 \\ x_2 &= x_{20} + \tilde x_2 \end{aligned}
are small deviations from the origin found at x_{10}, x_{20}
The way we lineaerize this is by taking the Taylor series expansion, thus we have
\begin{aligned} f(x_{1},x_{2}) &= f(x_{10} + \tilde x_1,x_{20} + \tilde x_2) \\ &= f(x_{10},x_{20}) + \frac{\partial f(x_{10},x_{20})}{\partial x_1} \tilde x_1 + \frac{\partial f(x_{10},x_{20})}{\partial x_2} \tilde x_2+ H.O.T \end{aligned}
No need to reach the 2nd derivatives before calling it quits on the higher order terms, unless of course, you want accuracy.
Now, in order to actually solve for a specific partial derivative as stated in the 2nd and 3rd term of the 2nd line, we need invoke an operator known as the Jacobian. Once we invoke the almighty Jacobian, we should have something that looks like...
\begin{aligned} \frac{\partial}{\partial x} f_1(x)= \begin{bmatrix} {\frac{\partial f_1}{\partial x_1}} & {\frac{\partial f_1}{\partial x_2}} \\ {\frac{\partial f_2}{\partial x_1}} & {\frac{\partial f_2}{\partial x_2}} \end{bmatrix} &= {\begin{bmatrix} {x_2 - 3x_1^2} & {x_1} \\ {0+9e^{x_1}e^{5x_2}} & {9e^x_1 5^{5x_2}-5} \end{bmatrix}} \Big|_{x_1 = 0, x_2 = 0} \\ &= \begin{bmatrix} {0} & {0} \\ {9e^0} & {45e^0-5} \end{bmatrix} = \begin{bmatrix} {0} & {0} \\ {9} & {40} \end{bmatrix} = A \end{aligned}
With this task accomplished, we now know the answer for the Jacobian Matrix A and can use it to linearized the equation mentioned above.
\begin{aligned} \dot {\tilde x}(t) = A(t)\tilde x(t) \end{aligned}
Note that we did not define an input for this system. Meaning, the output is only defined by the state vector x.