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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?

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Beware: the Laplace transform of the derivative is not always $s$ times the Laplace transform of the function. –  Did Nov 12 '11 at 20:19
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4 Answers

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 $f$..

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One way to remember it is that if you think of the integral in the inverse formula as a sort of sum (completely non-rigorous here):

$$ f(t) = c_1 e^{s_1 t} + c_2 e^{s_2 t} + c_3 e^{s_3 t} + \cdots $$

Here the coefficient $c_1$ is proportional to the Laplace transform at $s_1$, and so on. Then since the derivative of $e^{st}$ with respect to $t$ is $se^{st}$, you get that the coefficients in the sum are multiplied by $s$ when you take the derivative:

$$ f'(t) = c_1 s_1 e^{s_1 t} + c_2 s_2 e^{s_2 t} + c_3 s_3 e^{s_3 t} + \cdots $$

Conversely, integrating $f$ multiplies the coefficients by $1/s$. Again, this is not in any way a proof, but hopefully it gives some intuition.

Another way to think about it is that the smoothness of a function is related to its decay in the frequency domain as $s \to \infty$. Differentiating makes a function less smooth, (a twice differentiable function becomes onces differentiable, etc.) while integrating makes a function more smooth (a bounded discontinuous function becomes continuous, etc.). So it makes sense that differentiation would make a function decay slower in the frequency domain, in the sense that $1/s$ decays slower than $1/s^2$.

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I had a similar question here and came to the following conclusion:

The Laplace transform involves the exponential function: e^n*x.

Differentiating this means n*e^n*x - which is simply multiplication by n.

Integrating this means 1/n*e^n*x - which is simply division by n.

This holds true for power series which are the discrete form of integral transforms (sort of) and originally stems from the power rule (for differentiation) - although you would have a pesky division by the base using "ordinary" power terms. This is prevented by using the exponential function.

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Do you mean?

$$ \mathcal{L} \{ t^n f(t) \} = (-1)^n \frac{d^n}{ds^n} \mathcal{L}\{f(t)\}$$

and

$$ \mathcal{L}^{-1} \left\{ \frac{F(s)}{s^n} \right\} = \int_0^t {\underbrace \cdots _n} \int_0^t {f\left( t \right)dt^n} $$

i.e, multplying by $t$ means a derivation in terms of $s$ and dividing by $s$ means an integration in terms of $t$.

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