Lecture 13-Linearization

Chap 2: Linearizatin

$$y-y_0 = y_0'(x-x_0)$$

example 1

nonlinear r: $i_r = e^{V_r}$
so

$$C \frac{dV}{dt} + i_r -i(t) - 2 = 0$$

replace

$$V = V_0 + \delta V\\ i_r = e^{V_r} = e^{V_0 + \delta V}$$

So we obtain, here C = 1

$$\frac{d\delta V}{dt} + e^{V_0 + \delta V} - 2 = i(t)$$

replace $e^{V_0 + \delta V} \approx e^{V_0} + e^{V_0} \delta V$\
So we havee

$$\frac{d\delta V}{dt} + e^{V_0} + e^{V_0} \delta V - 2 = i(t)$$

Thus means

$$\frac{d\delta V}{dt} + e^{V_0} \delta V = i(t) + 2 - e^{V_0}$$

Because $i_r|_{V_r = V_0} = e^{V_0} = 2, V_0 = ln(2) = 0.693$
so, simplyfied equation:

$$\frac{d\delta V}{dt} + 2 \delta V = i(t)$$

Then do laplace transform $\mathbb{L}$

$$(s+2)\delta V(s) - \delta V(0_-) = I(s)$$

because

$$\int_{0_-}^{\infty} \frac{df(t)}{dt}e^{-st}dt = e^{-st} f(t)|_{0_-}^{\infty} + s \int_{0_-}^{\infty} f(t)e^{-st}dt\\ \mathbb{L}[\frac{df(t)}{dt}] = s\mathbb{L}[f(t)] - f(0_-)$$

example 2

$$M \frac{d^2 y(t)}{dt^2} = Mg - \frac{i^2(t)}{y(t)}\\ V(t) = R i(t) + L \frac{d i(t)}{dt}$$

V(t) is input, now select varaible

$$X_1 = y(t)\\ X_2 = \frac{dy(t)}{dt}\\ X_3 = i(t)$$

So we obtain

$$\dot{X_1} = X_2\\ \dot{X_2} = -\frac{X_3^2}{MX_1} + g\\ \dot{X_3} = - \frac{R}{L} X_3 + V(t)$$

Find balanced point, suppose input $V(t) = V_0 + \Delta V(t), i(t) = i_0 + \Delta i(t), y(t) = y_0 + \Delta y(t)$\
Then we have, by set $\dot{X} = 0$

$$0 = \frac{dy(t)}{dt} |_{y(t) = y_0}\\ 0 = Mg- \frac{i_0^2}{y_0}\\ V_0 = R i_0$$

now express Phase variable balanced poit with input balanced point $V_0$

$$i_0 = \frac{V_0}{R}\\ y_0 = \frac{i_0^2}{Mg} = \frac{V_0^2}{R^2 Mg}$$

Thus means

$$X_{10} = y_0 = \frac{V_0^2}{R^2 Mg}\\ X_{20} = \frac{dy(t)}{dt} |_{y(t) = y_0} = 0\\ X_{30} = i_0 = \frac{V_0}{R}$$

Now define $\delta X \equiv X - X_0, \delta V = V - V_0$

$$\dot{\delta} X_1= \delta X_2\\ \dot{\delta} X_2= \frac{\partial[-\frac{X_3^2}{MX_1}]}{\partial X_1}|_{X=X_0} \delta X_1 + \frac{\partial[-\frac{X_3^2}{MX_1}]}{\partial X_3}|_{X=X_0} \delta X_3\\ = [\frac{X_3^2}{MX_1^2}]|_{X=X_0} \delta X_1 + [-\frac{2X_3}{MX_1}]|_{X=X_0} \delta X_3\\ = \frac{R^2Mg^2}{V_0^2}\delta X_1 + [\frac{-2Rg}{V_0}] \delta X_3\\ \dot{\delta} X_3 = - \frac{R}{L} \delta X_3 + \delta V(t)$$

Thus means

$$\dot{\delta} X = \begin{bmatrix} 0 & 1 & 0\\ \frac{R^2Mg^2}{V_0^2} & 0 & \frac{-2Rg}{V_0}\\ 0 & 0 & - \frac{R}{L} \end{bmatrix} \delta X + \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \delta V$$

Then calculate the poles of linearized model:\
if poles are all in left plane <=> the model is stable!!!

Then verify it is Controllable or Not?

$$|C_M| =[B \quad AB \quad A^2B]\ =\begin{bmatrix} 0 & 0 & \frac{-2Rg}{V_0}\\ 0 & \frac{-2Rg}{V0} & \frac{2R^2g}{LV0}\\ 1 & - \frac{R}{L} & \frac{R^2}{L^2} \end{bmatrix} \neq 0$$

If so, we could control it, to make the model stable