# Impulse response to Jordan form.

Which matrix $A$ in real Jordan from is such that, for suitable choices of the matrices $B$ and $C$, continuous-time state-space model $(A,B,C)$ of the form

$\frac{d}{dt}x(t)=Ax(t)+Bu(t), \quad y(t)=Cx(t), \quad t \in R^+$

is minimal and has its impulse response given by $h(t)=e^{2t}(t \sin(t)+5\cos(t))$?

$A) \quad A=\begin{bmatrix} 2&1\\-1&2 \end{bmatrix}$

$B) \quad A=\begin{bmatrix} 1&2&0&0\\-2&1&0&0\\0&0&1&2\\0&0&-2&1\end{bmatrix}$

$C) \quad A=\begin{bmatrix} 1&2&1&0\\-2&1&0&1\\0&0&1&2\\0&0&-2&1\end{bmatrix}$

$D) \quad A=\begin{bmatrix} 2&1&1&0\\-1&2&0&1\\0&0&2&1\\0&0&-1&2\end{bmatrix}$

$E) \quad A=\begin{bmatrix} 2+i&1&0&0\\0&2+i&0&0\\0&0&2-i&1\\0&0&0&2-i \end{bmatrix}$

I think i'm making a mistake with the laplace transformation. First I expand the brackets to get:

$e^{2t}t\sin(t)+5e^{2t}\cos(t)$ taking the laplace transform we get $\frac{1}{s-2} \cdot \frac{1}{s^2} \cdot \frac{1}{s^2+1}+5 \cdot \frac{1}{s-2} \cdot \frac{s}{s^2+1}$

Which is the same as: $\frac{1+5s^3}{s^2(s-2)(s^2+1)}$. This is a fifth order system and not among the possible answers. So something has to cancel out. But I can't see what or how.

• Look at this table to see how you can find the Laplace transform of sinusoidal functions times an exponential. – Kwin van der Veen Aug 8 '18 at 17:10
• Number 30 of that table also shows how to find the Laplace transform when it is multiplied by $t^n$. – Kwin van der Veen Aug 8 '18 at 17:16

You do not need to find the Laplace transform to find the answer to the question. Namely since you only need to find the real Jordan form of the $A$ matrix and thus only need to find the poles/eigenvalues of the system. The eigenvalues can be identified by listing all the $e^{\rho\,t}\sin(\omega\,t+\phi)\,t^n$ terms. Namely the poles corresponding to it are $\rho\pm j\,\omega$ with an algebraic multiplicity of $n+1$ but a geometric multiplicity of one. The geometric multiplicity of one means that the Jordan blocks associated with it will be of size $n+1$. It can be noted that those eigenvalues can also generate additional terms with lower powers in time than $n$.
$$\mathcal{L}\left(e^{2 t} (t \sin (t)+5 \cos (t))\right) = \frac{2 (s-2)}{\left(s^2-4 s+5\right)^2}+\frac{5 (s-2)}{(s-2)^2+1} = \frac{2 (s-2)}{\left(s^2-4 s+5\right)^2}+\frac{5 (s-2)}{s^2-4 s+5}$$
• Thanks Cesareo, I can see that $\mathcal{L}(5e^{2t}\cos(t))=\frac{5(s-2)}{(s-2)^2+1}$ but how did you transform $e^{2t}t\sin(t)$? do you take $\mathcal{L}(e^{2t}) \cdot \mathcal{L}(t) \cdot \mathcal{L}(\sin(t))$ or $\mathcal{L}(e^{2t}) \cdot \mathcal{L}(t \sin(t)))$? – user463102 Aug 8 '18 at 15:55
• @user463102 Think that $\int_0^{\infty}e^{-s t}t f(t) dt = -\frac{d}{ds}\int_0^{\infty}e^{-s t}f(t) dt$ – Cesareo Aug 8 '18 at 16:10