Two separate experiments are performed for t ≥ 0 on the single-input single-output (SISO) linear system $\dot{x} = Ax + Bu, y = Cx + Du$, $x(0) = \begin{bmatrix}1\\2\\-1\end{bmatrix}$ (the initial condition is the same in each experiment)
In the first experiment, $u(t) = e^{-t}$ and the resulting output is y(t) = $e^{-3t} + e^{-2t}$
In the second, $u(t) = e^{-3t}$ and the resulting output is $y(t) = 3e^{-3t} − e^{-2t}$
(a) Can you determine the transfer function $C(sI−A)^{-1}$B+D from this information? If it is possible, do so. If not, find two linear systems consistent with all the data given which have different transfer functions.
(b) Can you determine A, B, C, or D?
i tried taking the laplace transform of each inputs and outputs :
$Y_1(s) = \frac{1}{s+3} + \frac{1}{s+2} , U_1(s) = \frac{1}{s+1}$
$Y_2(s) = \frac{3}{s+3} - \frac{1}{s+2} , U_2(s) = \frac{1}{s+3}$
from what i've known output is summation of the initial condition term and input term, so if the initial condition is non-zero.Is the transfer function of the system still $ \frac{Y(s)}{U{s}}$
I will post a beginning of a solution to help OP, and I will complete it bit by bit.
So, we can write the following system of equations
$$ \begin{bmatrix} Y_1(s)\\ Y_2(s) \end{bmatrix}=\begin{bmatrix} U_1(s) & 1\\ U_2(s) & 1 \end{bmatrix}\begin{bmatrix} H(s)\\ G(s) \end{bmatrix} $$ where $H(s)=C(sI-A)^{-1}B+D$ and $G(s)=C(sI-A)^{-1}x_0$. So, we get that $$ \begin{bmatrix} H(s)\\ G(s) \end{bmatrix}=\begin{bmatrix} U_1(s) & 1\\ U_2(s) & 1 \end{bmatrix}^{-1}\begin{bmatrix} Y_1(s)\\ Y_2(s) \end{bmatrix}=\dfrac{1}{U_1(s)-U_2(s)}\begin{bmatrix} 1 & -1\\ -U_2(s) & U_1(s) \end{bmatrix}\begin{bmatrix} Y_1(s)\\ Y_2(s) \end{bmatrix}. $$ Then, the idea would be to see whether the transfer function is realizable, which means that $H(s)$ is rational and proper (i.e. the degree of the numerator is at most that of the denominator).
Edit. Construction of a possible realization for the $H(s)$ of order 3.
So, after some tedious algebra we get (as @nap also correctly obtained)
$$H(s)=\dfrac{s+1}{s+2}=1-\dfrac{1}{s+2}$$ and $$G(s)=\dfrac{1}{s+3}.$$
Since, there is constant term in $H(s)$, we can put this in the direct feedthrough $D=1$, and then we get that $C(sI-A)^{-1}B=-1/(s+2)$.
The dimension of the system is 3 (from the initial condition), so the model will not be minimal and a mode will be uncontrollable and/or observable.
One way to build the system is to fix an eigenvalue of $A$ to be $-3$ and to build $A$ such that $x_0$ is an eigenvector. Similarly, we can set an eigenvalue to be -2 and to set $B$ to be an eigenvector (i.e. this eigenvalues is chosen this way because we need to have $s+2$ at the denominator in one way or another). So, we pick for instance,
$$B=\begin{bmatrix}1\\0\\0\end{bmatrix},$$
and note that this choice is completely arbitrary. We just need to make sure that $B$ is linearly independent of $x_0$. And we can just a last eigenvalue at random, say -1, and pick the eigenvector to be $(0,1,0)$ which is linearly independent of the others.
We can now build the matrix of eigenvectors
$$V=\begin{bmatrix}1 & 1 & 0\\0 & 2 & 1\\0 & -1 & 0\end{bmatrix}$$ which is associated with the diagonal matrix $$D=\begin{bmatrix}-2 & 0 & 0\\0 & -3 & 0\\0 & 0 & -1\end{bmatrix}$$ which is the diagonal form of $A$ in the basis $V$. We can now find $A$ using the expression $A=VDV^{-1}$ and we get
$$A=\begin{bmatrix} -2 & 0 & 1\\ 0 & -1 & 4\\ 0& 0 & -3\end{bmatrix}$$
Now, we need to select $C$. To do so, we evaluate the following expressions
$$(sI-A)^{-1}x_0=\dfrac{1}{s+3}x_0$$ as expected and $$(sI-A)^{-1}B=\dfrac{1}{s+2}B$$.
So, if we pick $C=\begin{bmatrix}-1 & 1 & 0\end{bmatrix}$, we get that
$$C(sI-A)^{-1}B+D=1-\dfrac{1}{s+2}$$
and
$$C(sI-A)^{-1}x_0=\dfrac{1}{s+3}$$
as desired.
Update. What if the order of the system is 2. In this case, $H(s)$ and $G(s)$ remain the same but one can define now $$D=\begin{bmatrix}-2 & 0\\0 & -3\end{bmatrix}$$ and $$V=\begin{bmatrix}b & x_0\end{bmatrix}$$
where $b$ must be such that $V$ is invertible; i.e. $b$ and $x_0$ are linearly independent. Then, the rest follows from the same lines as in the previous case.