Laplace transform simplification

Question

If $w$ is input signal and $t$ time. How come the following holds? $$ L(tw) = -L'(w)$$

I dont understand this at all. Could this be due to some kind of distribution quirk of the two-sided Laplacetransform that I am not aware of?

Answer 1

Pending your response on the strange notation, I will assume you mean the property $t * f(t) \to - \frac{d F(s)}{ds} $, where $F(s)$ denotes the Laplace transform of $f(t)$.

Recall the definition of the Laplace transform, $$ F(s) \equiv \mathcal{L}(f(t)) = \int_0^{\infty} f(t) e^{-st} dt $$ Supplying $f(t) = t f(t)$ into the definition, we have $$ \int_0^{\infty} tf(t) e^{-st} dt. $$ Let us use the definition of $\mathcal{L}(f(t))$ and differentiate by $s$ to see that $$ \frac{d}{ds} F(s) = \frac{d}{ds} \int_0^{\infty} f(t) e^{-st} dt = \int_0^{\infty} \frac{\partial}{\partial s} f(t) e^{-st} dt = \int_0^{\infty} - tf(t) e^{-st} dt = - \mathcal{L}(tf(t)) $$ Collecting left and right hand sides and rearranging we see $\mathcal{L}(tf(t)) = -\frac{d}{ds} \mathcal{L}(f(t))$, which I believe is the same as your notation of $L(tw) = -L'(w)$.