How to use Laplace with this example?

Question

i am studying the book "Linear Systems" by Kailath and on page 32 I do not unsterstand how this author uses laplace to get from

$$ \left[ \begin{matrix} \dot x_1 \\ \dot x_2 \end{matrix} \right] =\left[ \begin{matrix} -1 & 0 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x_1 \\ x_2 \end{matrix} \right] +\left[ \begin{matrix} -2 \\ 1 \end{matrix} \right]v, ~~~ \left[ \begin{matrix} x_1(0) \\ x_2(0) \end{matrix} \right] =\left[ \begin{matrix} x_{10} \\ x_{20} \end{matrix} \right] $$

$\dot x_1 = - x_1 - 2v, ~~~ x_1(0)=x_{10}$ (I do understand this)

to

$ x_1(t)=e^{-t}x_{10}-2e^{-t} * v $ (* denotes convolution)

I cannot find a laplace theorem like http://www.rapidtables.com/math/calculus/laplace_transform.htm that would explain this.

Answer 1

Since convolution in the time domain is just multiplication in the Laplace domain, it is easier to solve this using the Laplace domain.

In general a inhomogeneous linear time invariant differential equation can be written as,

$$ \dot{x}(t) = A\,x(t) + B\,u(t), \tag{1} $$

with $x\in\mathbb{R}^{n\times 1}$, $A\in\mathbb{R}^{n\times n}$, $\dot{x}$ the time derivative of $x$, $B\in\mathbb{R}^{n\times m}$ and $u\in\mathbb{R}^{m\times 1}$. Transforming equation $(1)$ into the Laplace domain yields,

$$ s\,X(s) - x(0) = A\,X(s) + B\,U(s), \tag{2} $$

with $X(s)=\mathcal{L}\{x(t)\}$ and $U(s)=\mathcal{L}\{u(t)\}$. This can be rewritten to,

$$ X(s) = (s\,I - A)^{-1} x(0) + (s\,I - A)^{-1} B\,U(s). \tag{3} $$

Similar to $\mathcal{L}^{-1}\! \left\{\frac{1}{s-a}\right\}=e^{a\,t}$ the inverse Laplace transform of $(s\,I - A)^{-1}$ yields a matrix exponential $e^{A\,t}$. This mean that the inverse Laplace transform of equation $(3)$ can be written as,

$$ x(t) = e^{A\,t} x(0) + \mathcal{L}^{-1}\!\left\{\mathcal{L}\{e^{A\,t}\}\,\mathcal{L}\{B\,u(t)\}\right\}. \tag{4} $$

But as stated in the beginning of my answer, multiplication in the Laplace domain is the same as convolution in the time domain. In your case you have $A=-1$ and $B=-2$. In general when dealing with matrices the order of multiplication matters, since they do not commute, but in this case we are dealing with scalars, which do commute, so solving for $x_1(t)$ using equation $(4)$ would yield,

$$ x_1(t) = e^{-t} x_{10} - 2\, [e^{-t}\,*\,u(t)](t), $$

with $[f\,*\,g](t)$ meaning the convolution between $f$ and $g$.

Answer 2

This is actually pretty easy to prove if you work only in Laplace domain. $$\dot{x}_1(t)=-x_1(t)-2v(t)$$ Taking Laplace transform on both sides we get: $$sX_1(s)-x_{10}=-X_1(s)-2\mathcal{L}(v)$$ $$\Rightarrow X_1(s)=\frac{1}{s+1} x_{10} - \frac{2}{s+1}\mathcal{L}(v)$$ $$\Rightarrow X_1(s) = \mathcal{L}(e^{-t}) - \mathcal{L}(2e^{-1})\mathcal{L}(v)$$ Notice that the last term is a product of two functions in $s-$domain, and so taking inverse Laplace would replace the product with a convolution as follows: $$x_1(t)=e^{-t}-2e^{-t}*v(t)$$