# Laplace transform of derivative of convolution

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1. Apr 22, 2017

### Gallo

Prelude
Consider the convolution h(t) of two function f(t) and g(t):
$$h(t) = f(t) \ast g(t)=\int_0^t f(t-\tau) g(\tau) d \tau$$
then we know that by the properties of convolution
$$\frac{d h(t)}{d t} = \frac{d f(t)}{d t} \ast g(t) = f(t) \ast \frac{d g(t)}{d t}$$

Intermezzo
We also know that the Laplace Transform of the convolution is given :
$$H(s) = F(s) G(s)$$
and that of the derivative of for example f(t) is
$$s F(s) - f(0)$$

Finale
Therefore I should have for the Laplace transform of the derivative of the convolution:
$$s H(s) = s F(s) G(s) - f(0) G(s) = F(s) s G(s) - F(s) g(0)$$

But in general we don't have :
$$f(0) G(s) = F(s) g(0)$$

Are this formulas correct? What is going on? Are they equivalent?

2. Apr 22, 2017

### Ray Vickson

Your formula $d h(t)/dt = (f'*g)(t)$ is incorrect. Instead of trying to (mis-)apply canned formulas, just start with the definition $h(t) = (f*g)(t) = \int_0^t f(t-\tau) g(\tau) \, d\tau$ and differentiate it using standard results in calculus.

3. Apr 22, 2017

### Gallo

Gotcha!

$$\frac{d h(t)}{d t} =\frac{d}{ d t}\int_0^t f(t-\tau) g(\tau) d \tau$$
and using Leibniz integral rule
$$\frac{d h(t)}{d t} = f(0)g(t) + \int_0^t \frac{d f(t-\tau)}{ d t} g(\tau) d \tau$$

The laplace transform of wich is
$$s H(s) = G(s) f(0) + G(s)(s F(s) - f(0) ) = s F(s)G(s)$$

Thank you!