Why is the time-domain derivative equivalent to multiplication by frequency ($s$) in the Laplace transform?
Why is the time-domain integral equivalent to division by frequency ($\frac{1}{s}$) in the Laplace transform?
Intuitively, I thought the reason was that the frequency was a sort of "rate of change", so it was somehow equivalent to $\frac{df}{dt}$. The Laplace transform turns rate of change into a variable ($s$), and holds that rate of change constant throughout the problem, which is why algebraic manipulations in the frequency domain are possible.
Am I on the right track?
Best Answer
Dimension analysis!
If $t$ is a unit of time, $ts$ must be dimensionless for $\mathrm e^{-ts}$ to exist hence $s$ is the inverse of a time. Since the derivative of $f$ is a limit of ratios of values of $f$ divided by times, $f'$ has the dimension of $f$ times $s$ and its integral should have the dimension of the similar integral of $f$ times $s$. That is, the Laplace transform of $f'$ should have the dimension of $s$ times the Laplace transform of $f$. Likewise for the integral of $f$, which has the dimension of $t$ times the dimension of $f$, hence the Laplace transform of the integral of $f$ should have the dimension of $s^{-1}$ times the Laplace transform of $f$..