Laplace transform of derivative of convolution

  • #1
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?
 

Answers and Replies

  • #2
Ray Vickson
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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?

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
9
1
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.
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!
 

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