# Coefficients of a closed-loop characteristic equation

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.