Automatic control - linearization with two states

Question

I don't understand a part of the solution to c) of this problem, namely linearization at a stationary point using Taylor series and partial derivatives.

I am used to linearization of \$ f(x,u) \$ (two variables) using Taylor series but not \$ f(x_1,x_2,u) \$ (three variables) at a stationary point.

Taylor for two variables is given by (where second row is ignored):

Are we supposed to linearize using Taylor series for three variables or just find the partial derivative of \$f_1\$ and \$ f_2 \$? We get a term \$ f(x^0_1,x^0_2,u^0) \$ (which looks like \$ f(a,b) \$ from Taylor series formula for two variables above) present in the solution which is set to zero that is not present if we were to just find the partial derivative of \$f \$ with respect to all terms and just sum them up?

Sorry if the question is a bit messy since I have trouble connecting the mathematical theory with the control theory.

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Solution:

Answer 1

I believe you are aware of the concept of the Jacobian matrix. This is a case of Linearization of Multistate Models. See, for example, pages 6-7 of Linearization of Nonlinear Models.

Yes, that linearization uses the calculation of the partial derivatives of three functions with three variables (two states \$x_1\$, \$x_2\$ at one input \$u\$) around the equilibrium point \$Q\$. The definition of the Taylor series expansion (neglecting the h.o.t. - high-order terms) already suggests the use of the deviation variables: $$\Delta x_1 = x_1-x_{1_Q}$$ $$\Delta x_2 = x_2-x_{2_Q}$$ $$\Delta u = u - u_Q$$

with which the final linearized model is built. Also, note that, for example: \$\dot{\Delta x}=\dot{x_1}\$, since \$\dot{x_{1_Q}}=0\$.