Find output given state space representation

Question

This is usually easy but there are things I haven't encountered in this one. This is the problem. $$\dot x=\begin{bmatrix}0&1\\-2&-3\end{bmatrix}x+\begin{bmatrix}0\\2\end{bmatrix}u=Ax+Bu\\y=\begin {bmatrix}0&1\end{bmatrix}x=Cx\\x(0)=\begin{bmatrix}1\\0\end{bmatrix}$$

First I'm asked to calculate $$x(t)=e^{At}x(0)$$

An exponential matrix in a control theory course seems a bit too much but I solved that. $$x(t)=\begin{bmatrix}1&1\\-1&-2\end{bmatrix}\begin{bmatrix}e^{-1t}&0\\0&e^{-2t}\end{bmatrix}\begin{bmatrix}2&1\\-1&-1\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}$$ The second and final question asks me to find the expression of the output y(t) for any input u(t).

Looking at the solution manual this is all that's written : $$y(t)=\begin{bmatrix}0&1\end{bmatrix}\int_0^t\begin{bmatrix}-1&-1\\-1&-2\end{bmatrix}\begin{bmatrix}e^{-1(t-τ)}&0\\0&e^{-2(t-τ)}\end{bmatrix}\begin{bmatrix}2&1\\-1&-1\end{bmatrix}\begin{bmatrix}0\\2\end{bmatrix}u(τ)dτ$$

That reminds me of convolution but the integral limits are different. Plus the result from the first question is used. Why would we use the exponential? I can't make the connection and understand the second part. Any ideas?

Answer 1

In Laplace space:

$$ sX(s)-x(0)=AX(s)+BU(s) $$

so:

$$ X(s)=(sI-A)^{-1}BU(s)+(sI-A)^{-1}x(0) $$

the second term, due to the initial condition, produces:

$$ x_0(t)=e^{At}~x(0) $$

while the fist term, due to the incoming signal, produces:

$$ x_u(t)=\int_0^te^{A\tau}~B~u(t-\tau)d\tau=\int_0^te^{A(t-\tau)}~~B~u(\tau)~d\tau $$

(where calculation of \$ e^{At} \$ can be seen here )

Finally:

$$ y(t)=Cx(t)=C(x_u(t)+x_0(t)) $$