Supose I have a function f(t) which corresponding Laplace Transform is F(s)
Now I want to calculate the Laplace transform of $f(a (t-b))$ using Laplace Transform properties.
If $L(g(t))= G(s)$
Property A: Time shift $L( g(t-a)) = e^{-a s} G(s)$
Property B. Time scaling $L(g(at) = \frac{1}{a} G(\frac{s}{a}) )$
Then I have a question about the order in which I can use them.
For example if I go A then B:
$L(g(a(t-b))) = e^{-b s} L(g(at))$
$e^{-b s} L(g(at)) = e^{-b s} L( g(t-a)) = e^{-b s} \frac{1}{a} G(\frac{s}{a}) $
If I go B then A:
$L(g(a(t-b))) = \frac{1}{a} L( g(t-b) ) $ the transform must be evaluated with $\frac{s}{a}$
$\frac{1}{a} L( g(t-b) ) = \frac{1}{a} e^{-bs} L( g(t) ) = \frac{1}{a} e^{-b \frac{s}{a}} G(\frac{s}{a}) $
Which one is correct?
Please correct me if anything is wrong, even language :D (not native here)
For the second one $$L(g(a(t-b))) =L(g(at-ab))) = \frac{1}{a} L( g(t-ab) )$$ the transform must be evaluated with $s/a$ $$L(g(a(t-b))) = \frac{1}{a} e^{-abs/a}G(\frac s a)$$ $$L(g(a(t-b))) = \frac{1}{a} e^{-bs}G(\frac s a)$$ It gives the same answer in both cases.