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Homework Help: Laplace transform of derivative of convolution

  1. Apr 22, 2017 #1
    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
    $$\frac{d h(t)}{d t} = \frac{d f(t)}{d t} \ast g(t) = f(t) \ast \frac{d g(t)}{d t}$$

    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
    $$s F(s) - f(0)$$

    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. jcsd
  3. Apr 22, 2017 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    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.
  4. Apr 22, 2017 #3

    $$\frac{d h(t)}{d t} =\frac{d}{ d t}\int_0^t f(t-\tau) g(\tau) d \tau $$
    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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