Simple Laplace transform question

[GSR]

People said "1/s" is an integrator. Why it is an integrator? I guess it is because "1/s" ==> [u(t): step function] in time domain, so at t = 2, f(t) = u(t)+u(t-1)+u(t-2) which is adding up all the previous values. Therefore "1/s" is an integrator

However, for the case, "1/(s+a)" ==> [exp(-at)]. In time domain should just a exp curve, no add up.

Which part of understanding is wrong??

[vugie]

Integration properties of 1/s come directly from defition of Laplace transform and mean that for any F(s), inverse transform of F(s)/s is equal to integral of f(t), where f(t) is inverse transform of F(s). It has nothing to inverse transform of 1/s itself.

If you want to apply this rule to 1/(s+a) you have to extract 1/s from it: 1/(s+a)=(1/s)*(s/(s+a)). Inverse transform of s/(s+a) is -a*exp(-at), integral of which is exp(-at).

Laplace transorm may help you to calculate integral of given function f(t). You simply has to calculate its transform L{f(t)}=F(s), divide it by s (multiply by 1/s) and make inverse transform of the result. It will be equal to integral of f(t).

Edited January 16, 2010 by vugie