- #1

Gallo

- 9

- 1

**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?