# Controllability and Observability of a Circuit through State Equation Inspection

In Kalman's 1963 paper "Mathematical description of Linear Dynamical Systems" he introduces the concepts of controllabilty and observability. In the first example he builds a constant resistance network where we have $$\frac{L}{C}= R = 1$$

and develops the systems equations choosing $$x_1 = \phi_L :\ \text{magnetic flux in the inductor}$$ and $$x_2 = Q_c:\ \text{capacitor eletric charge}$$ \ After some simple circuit manipulations we get the system's state equations as $$\dot x_1 = -\frac{1}{L}x_1+u_1$$ $$\dot x_2 = -\frac{1}{C}x_2+u_1$$ and output equation as $$y_1 = \frac{1}{L}x_1+\frac{1}{C}x_2+u_1$$ After that, he changes the state and output equations through an equivalence transformation $$\overline{x}_1 = \frac{x_1+x_2}{2}$$ $$\overline{x}_2 = \frac{x_1-x_2}{2}$$ and so we get another set of state equations $$\dot{\overline{x}}_1 = -\frac{1}{L}\overline{x}_1+u_1$$ $$\dot{\overline{x}}_2 = -\frac{1}{L}\overline{x}_2$$ and another output equation $$y_1 = \frac{2}{L}\overline{x}_2+u_1$$

Question: Why the system is uncontrollable and unobservable if in the first set of equations the input u is affecting both state variables and they're also present in the output equation? I see that when Kalman changes the system state and output equations, it becomes clear that $$x_2\ \text{is uncontrollable}$$ $$x_1\ \text{is unobservable}$$

but I coundn't see that looking at the first set of equations.

Controllability is not just being able to affect the state, it is actually to be able to guide the state from any initial point $$\x_0\$$ back to the origin $$\ x(t_1) = 0\$$ in a finite amount of time. And observability is, more or less, being able to deduce the state from previous measurements of the output.
For your system, $$\dot x_1 = -\frac{1}{L}x_1+u_1 \\ \dot x_2 = -\frac{1}{C}x_2+u_1$$ with $$\u = 0\$$, any state would go back to the origin in an infinite amount of time (it would exponentially decay and go to zero but never reach it in a finite amount of time). If you think of the initial point $$\x_1 = 1, x_2 = -1\$$, you could definitely use the control $$\u\$$ to drive one of them to zero in a finite time (say $$\x_1\$$), but then, to also get the other one to zero you would have to use some nonzero $$\u\$$, which would get $$\x_1\$$ to stop being $$\0\$$, and waiting for $$\x_2\$$ to decay to zero would take infinite time. That would mean that the system is not controllable.
Now, regarding observability, for $$y_1 = \frac{1}{L}x_1+\frac{1}{C}x_2+u_1$$ we can ignore $$\u_1\$$ (because we will be the ones determining it, so we can just subtract it from $$\y_1\$$) and look at
$$y_1 = \frac{1}{L}x_1+\frac{1}{C}x_2,$$ and its derivatives $$\dot{y_1} = \frac{1}{L}\dot{x_1}+\frac{1}{C}\dot{x_2} = -\frac{1}{L^2}x_1-\frac{1}{C^2}x_2, \\ \ddot{y_1} = -\frac{1}{L^3}x_1-\frac{1}{C^3}x_2, \\ \vdots$$
since $$\C=L\$$, we have $$y_1 = \frac{1}{L}x_1+\frac{1}{L}x_2 = \frac{1}{L}(x_1+x_2), \\ \dot{y_1} = -\frac{1}{L^2}x_1-\frac{1}{L^2}x_2 = \frac{1}{L^2}(x_1+x_2), \\ \ddot{y_1} = -\frac{1}{L^3}x_1-\frac{1}{L^3}x_2 = \frac{1}{L^3}(x_1+x_2), \\ \vdots$$
meaning that the derivatives of the output will add no information to what we know about the state and from some $$\y(t) = c/L\$$ we can never know if $$\x_1=0,x_2=c\$$, or $$\x_1=c,x_2=0\$$, or $$\x_1=-c,x_2=2c\$$. Hence we are not able to deduce the state $$\x\$$ from the output $$\y\$$, and the system is not observable.