In trying to derive a control law for a linear system so as to make it track a reference signal, I came up with the following and ended up with a few question. Let the following be the dynamic equation of a linear system, where $x$ is the output and $u$ the input:
$$x^{(n)}+a_{n-1}x^{(n-1)}+\dots+a_0x=u$$
We wish to determine some control law $u$ so that $x$ will track some reference signal $r$. Define the error $e=x-r$. Now we have $x=e+r$. The dynamics of the new system whose output is $e$ are given by:
$$e^{(n)}+a_{n-1}e^{(n-1)}+\dots+a_0e=u-r^{(n)}-a_{n-1}r^{(n-1)}\dots-a_0r$$
Let $u$ be
$$u=-K_{n-1}e^{(n-1)}-K_{n-2}e^{(n-2)}-\dots-K_{0}e+r^{(n)}+r^{(n-1)}+\dots+r$$
Making the substitution, we get
$$e^{(n)}+(a_{n-1}+K_{n-1})e^{(n-1)}+\dots+(a_0+K_0)e^{(n)}=0$$
As can be seen, the dynamics of the error don't depend on the particular reference signal, only on the initial conditions. Taking the Laplace Transform and assuming zero initial conditions:
$$E(s^n+(a_{n-1}+K_{n-1})s^{n-1}+\dots+a_0+K_0)=0$$
So that $E=0$, i.e., if the initial conditions are of zero error, the error will remain zero.
Substituting these results back into the original equation, we get
$$x^{(n)}+(a_{n-1}+K_{n-1})x^{(n-1)}+\dots+(a_0+K_0)x=r^{(n)}+(a_{n-1}+K_{n-1})r^{(n-1)}+\dots+(a_0+K_0)r$$
Taking Laplace Transforms and assuming zero initial conditions, we obtain the transfer function of the system:
$$\frac{X}{R}=\frac{s^n+(a_{n-1}+K_{n-1})s^{n-1}+\dots+a_o+K_0}{s^n+(a_{n-1}+K_{n-1})s^{n-1}+\dots+a_o+K_0}$$
As expected, the numerator and denominator cancel entirely, giving us $\frac{X}{R}=1$, which, once again, indicates that if the system has zero initial error, it will always have zero error, as $x$ will track $r$ perfectly.
My question is how this relates to the controllability of the system for the input $r$. From what I've learned, if there is any pole-zero cancellation in the transfer function of a system, the system isn't controllable along the cancelled mode. Since everything cancelled out in this case, I'd assume that would mean nothing could be controlled. However, conceptually, it seems, by definition of the results, as if we can make the system reach any state we'd like with a suitable input $r$: given any initial condition $x_0$, let $r$ be an adequately differentiable signal whose value and whose derivates up to order $n-1$ are the same as those of the output at $t_0$, and that eventually reaches $r(t_1)=0$, $r\dot(t_1)=0$, $\dots$, $r^{(n-1)}(t_1)=0$ at some time $t_1$. Being able to find a suitable $r$, the system would be controllable. Why does the contradiction arise?
If there are no zero-pole cancellations in the transfer function of a single-input single-output system, then the system is both controllable and observable. If a zero-pole cancellation occurs, then the system is either uncontrollable or unobservable or both uncontrollable and unobservable.
There are two possible state-space representations for the transfer function you are considering. The first one is the observable canonical form given by
$$\dot{x}_o(t)=A_ox_o(t),\quad y(t)=e_{n}^Tx_o(t)+r(t)$$
whereas the second one is the controllable canonical form
$$\dot{x}_c(t)=A_cx_c(t)+e_{n}r(t),\quad y(t)=r(t).$$
The first realization is observable but completely uncontrollable whereas the second one is controllable but completely unobservable.
In both cases, under zero initial conditions, we get that the transfer is simply $y(t)=r(t)$.