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I have an exercise where the solution says

For the given system $$A_{fb}= A+bf = \begin{bmatrix} 0 & 1 \\ -1+f_1 & -0.1+f_2 \end{bmatrix} $$

The closed loop system ($A_{fb},~b,~c$) is again in controller canoncial form, therefore we can directly observe that the coefficients of the closed-loop characteristic polynomial are $$\bar a_0 =1.0 -f \\ \bar a_1 = 0.1 -f_2$$

These can be compared with those of a standard 2nd ord system $$s^2+2 \zeta \omega_n s + \omega_n^2 = 0 $$ $$ \bar a_1 = 2 \zeta \omega_n ~~~~~~~~~ \bar a_0 = \omega_n^2 $$

Substituting in the design conditions ($\zeta = 0.7,~t_s=5.0$) gives $$ t_s = 5.0 = \frac{4.6}{\zeta \omega_n},~~~ \omega_n = 1.314 $$

So the solution is $$ \bar a_1 = 2 \times 0.7 \times 1.314 \\ \bar a_0 = 1.314^2 $$

and finally $$ f_1 = \frac{1.0}{1.0} - (1.314)^2 = -0.73 \\ f_2 =\frac{0.1}{1.0} - 2 \times 0.7 \times 1.314 = -1.74 $$

Question

Where do $ \bar a_0 $ and $ \bar a_1 $ come from? I can find them nowhere in my textbook. I know the concept of $s^2+2 \zeta \omega_n s + \omega_n^2$ and I know that $ \bar a_0 $ and $ \bar a_1 $ are taken from the matrix but what do theses coefficients mean? I look on the Internet and can find no answer. Other explaination like on https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-30-feedback-control-systems-fall-2010/lecture-notes/MIT16_30F10_lec11.pdf , I understand but there is also no usage of $ \bar a_0 $ and $ \bar a_1 $ or similar.

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Those are the coefficients of the characteristic polynomial, whose roots are the eigenvalues/poles of the system. Namely when your system is in the controllable canonical form, this Wikipedia page has a slight different form similar to your 6th slide, but it comes down to the same thing, it is easy to find the characteristic polynomial.

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