If I have a linear state system:
$$\frac{dq(t)}{dt} = M \cdot q(t) + k \cdot i(t)$$
that lets me choose the inputs and I want an end state that depends only on some eigenvectors (e.g. $e1, e2, e3$) of the matrix $M$, I can just pick $q(0)$ and $i(t)$ to be some linear combination of $e1, e2$, and $e3$, assuming $M$ is invertible.
But what if I have no control over the input $i(t)$? I have two cases I want to figure out:
$i(t)$ is constant and is equal to some random constant $K$ at $i(0)$ and $0$ elsewhere
$i(t)$ is constant and equal to some random constant $K$ everywhere
How do I change the state $q$ so that I can guarantee an end state that is a linear combination of some eigenvectors?
Thanks.
The general response can be found with
$$ q(t) = e^{M\,t}\,q(0) + \int_0^t e^{M(t-\tau)}\,k\,i(\tau)\,d\tau $$
In your first case the integral part will be zero if $K$ is finite. However if would let $i(t)=K\,\delta(t)$, with $\delta(t)$ the Dirac delta function, then the integral equals a linear combination of the impulse responses, namely $e^{M\,t}\,k\,K$, so
$$ q(t) = e^{M\,t}\,(q(0) + k\,K) \tag{1} $$
In your second part you get the integral equals a linear combination of the step responses, namely $(e^{M\,t} - I)\,M^{-1}\,k\,K$, so
$$ q(t) = e^{M\,t}\,q(0) + (e^{M\,t} - I)\,M^{-1}\,k\,K \tag{2} $$
Since both equations are linear in both $q(0)$ and $K$, then it is straight forward to find $q(0)$ given a $K$ and $q(t_f)$.