F1 =
Hi FFU,
Let's take a look at these one by one.
syms t real
f1 = 2^(-t);
I've never seen a Laplace transform table that includes an explicit entry for this form of a function, but I'd like to. Maybe that's why laplace fails?
F1 = laplace(f1)
But we can do
F1 = laplace(rewrite(f1,'exp'))
f2 = exp(-log(2)*t);
f2(t) is just a form for exp(-a*t), which has transform F2(s) = 1/(s + a), which laplace handles easily.
laplace(f2)
Note that log(2) is being approximated in floating point and then expressed as a rational number on conversion to sym. It might be more clear to say
f2 = exp(-log(sym(2))*t)
then
F2 = laplace(f2)
Turning to f3
f3 = exp(1)^(-t)
Again, exp(1) is being evaluated in floating point with numeric exp, so not exactly equal to e, and then converted to a sym rational expression. So now we are back to the same situation as for f1, i.e., a function of the form a^(-t) (with a ~= e)
laplace(f3)
laplace(rewrite(f3,'exp'))
Putting sym on the inside uses the symbolic function exp (which, amazingly, doesn't have its own doc page I don't believe)
f3 = exp(sym(1))^(-t)
which works fine
laplace(f3)
and is, of course, the same as f4
f4 = exp(-t);
laplace(f4)
