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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?

  • 1
    Beware: the Laplace transform of the derivative is not always $s$ times the Laplace transform of the function.2011-11-12

3 Answers 3