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.

  • 1
    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 at 17:10
  • 1
    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 at 17:16
up vote 1 down vote accepted

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 $n$.

  • This really saves me a huge amount of time. Thanks (once again) Kwin. – user463102 Aug 10 at 13:02


$$ \mathcal{L}\left(e^{2 t} (t \sin (t)+5 \cos (t))\right) = \frac{2 (s-2)}{\left(s^2-4 s+5\right)^2}+\frac{5 (s-2)}{(s-2)^2+1} = \frac{2 (s-2)}{\left(s^2-4 s+5\right)^2}+\frac{5 (s-2)}{s^2-4 s+5} $$

  • 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 at 15:55
  • 1
    @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 at 16:10

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