5.2: The Laplace Transform
The Laplace Transform is typically credited with taking dynamical problems into static problems. Recall that the Laplace Transform of the function \(h\) is
\[\mathscr{L} (h(s)) \equiv \int_{0}^{\infty} e^{-(st)} h(t) dt \nonumber\]
MATLAB is very adept at such things. For example:
The Laplace Transform in MATLAB
>> syms t >> laplace(exp(t)) ans = 1/(s-1) >> laplace(t*(exp(-t)) ans = 1/(s+1)^2
The Laplace Transform of a matrix of functions is simply the matrix of Laplace transforms of the individual elements.
\[\mathscr{L} (\begin{pmatrix} {e^{t}}\\ {te^{-t}} \end{pmatrix}) = \begin{pmatrix} {\frac{1}{s-1}}\\ {\frac{1}{(s+1)^2}} \end{pmatrix} \nonumber\]
Now, in preparing to apply the Laplace transform to our equation from the dynamic strang quartet module:
\[\textbf{x}' = B \textbf{x}+\textbf{g}\]
we write it as
\[\mathscr{L} (\frac{dx}{dt}) = \mathscr{L}(B \textbf{x}+\textbf{g})\]
and so must determine how \(\mathscr{L}\) acts on derivatives and sums. With respect to the latter it follows directly from the definition that
\[ \begin{align*} \mathscr{L}(B \textbf{x}+\textbf{g}) &= \mathscr{L}(B \textbf{x})+\mathscr{L}(\textbf{g}) \\[4pt] &= B \mathscr{L}(\textbf{x})+\mathscr{L}(\textbf{g}) \end{align*}\]
Regarding its effect on the derivative we find, on integrating by parts, that
\[\begin{align} \mathscr{L} \left(\frac{d \textbf{x}}{dt}\right) &= \int_{0}^{\infty} e^{-(st)} \frac{d \textbf{x}(t)}{dt} dt \\[4pt] &= \textbf{x}(t) \left. e^{-(st)} \right|_{0}^{\infty}+s\int_{0}^{\infty} e^{-(st)} \textbf{x}(t) dt \end{align}\]
Supposing that \(x\) and \(s\) are such that \(x(t) e^{-(st)} \rightarrow 0\) as \(t \rightarrow \infty\) we arrive at
\[\mathscr{L} (\frac{d \textbf{x}}{dt}) = s\mathscr{L} (\textbf{x})-x(0) \nonumber\]
Now, upon substituting Equation 2 and Equation 3 into Equation 1 we find
\[s \mathscr{L} (\textbf{x})- \textbf{x}(0) = B \mathscr{L}(\textbf{x})+\mathscr{L}(\textbf{g}) \nonumber\]
which is easily recognized to be a linear system for \(\mathscr{L}(\textbf{x})\)
\[(\textbf{s}I-B) \mathscr{L}(\textbf{x}) = \mathscr{L}(\textbf{g})+x(0) \nonumber\]
The only thing that distinguishes this system from those encountered since our first brush with these systems is the presence of the complex variable \(s\). This complicates the mechanical steps of Gaussian Elimination or the Gauss-Jordan Method but the methods indeed apply without change. Taking up the latter method, we write
\[\mathscr{L}(\textbf{x}) = (sI-B)^{-1} (\mathscr{L}(\textbf{g})+x(0)) \nonumber\]
The matrix \((sI-B)^{-1}\) is typically called the transfer function or resolvent , associated with \(B\), at \(s\). We turn to MATLAB for its symbolic calculation. (for more information, see the tutorial on MATLAB's symbolic toolbox). For example,
>> B = [2 -1; -1 2] >> R = inv(s*eye(2)-B) R = [ (s-2)/(s*s-4*s+3), -1/(s*s-4*s+3)] [ -1/(s*s-4*s+3), (s-2)/(s*s-4*s+3)]
We note that \((sI-B)^{-1}\) well defined except at the roots of the quadratic, \(s^{2}-4s+3\) determinant of \((sI-B)\) and is often referred to as the characteristic polynomial of \(B\). Its roots are called the eigenvalues of \(B\).
Let us take the \(B\) matrix of the dynamic Strang quartet module with the parameter choices specified in fib3.m, namely
\[B = \begin{pmatrix} {-0.135}&{0.125}&{0}\\ {0.5}&{-1.01}&{0.5}\\ {0}&{0.5}&{-0.51} \end{pmatrix} \nonumber\]
The associated \((sI-B)^{-1}\) is a bit bulky (please run fig3.m) so we display here only the denominator of each term, i.e.,
\[s^3+1.655s^2+0.4078s+0.0039 \nonumber\]
Assuming a current stimulus of the form \(i_{0}(t) = \frac{t^{3}e^{-\frac{t}{6}}}{10000}\) and \(E_{m} = 0\) brings
\[\mathscr{L}(\textbf{g})(s) = \begin{pmatrix} {\frac{0.191}{(s+\frac{1}{6})^{4}}}\\ {0}\\ {0}\\ {0} \end{pmatrix} \nonumber\]
and so Equation persists in
\[\begin{align*} \mathscr{L}(\textbf{x}) &= (sI-B)^{-1} \mathscr{L}(\textbf{g}) \\[4pt] &= \frac{0.191}{(s+\frac{1}{6})^{4}(s^3+1.655s^2+0.4078s+0.0039)} \begin{pmatrix} {s^2+1.5s+0.27}\\ {0.5s+0.26}\\ {0.2497} \end{pmatrix} \end{align*}\]
Now comes the rub. A simple linear solve (or inversion) has left us with the Laplace transform of \(\textbf{x}\). The accursed No Free Lunch Theorem
We shall have to do some work in order to recover \(\textbf{x}\) from \(\mathscr{L}(\textbf{x})\) confronts us. We shall face it down in the Inverse Laplace module.