Laplace transform of derivative of convolution

In summary, the convolution h(t) of two functions f(t) and g(t) has the Laplace transform given by:$$s H(s) = G(s) f(0) + G(s)(s F(s) - f(0) ) = s F(s)G(s)$$
  • #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?
 
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  • #2
Gallo said:
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
Ray Vickson said:
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 which is
$$ s H(s) = G(s) f(0) + G(s)(s F(s) - f(0) ) = s F(s)G(s) $$

Thank you!
 

Related to Laplace transform of derivative of convolution

1. What is the Laplace transform of a derivative of a convolution?

The Laplace transform of a derivative of a convolution is equal to the product of the Laplace transform of the convolution and the Laplace transform of the derivative. This can be written as L{d(f*g)/dt} = L{f*g}*L{d/dt}.

2. How can the Laplace transform of a derivative of a convolution be computed?

The Laplace transform of a derivative of a convolution can be computed using the convolution theorem, which states that the Laplace transform of a convolution is equal to the product of the individual Laplace transforms. In this case, the convolution is between the functions f and g, so the Laplace transform can be written as L{d(f*g)/dt} = L{f}*L{g}.

3. What is the significance of the Laplace transform of a derivative of a convolution in signal processing?

The Laplace transform of a derivative of a convolution is a useful tool in signal processing as it allows for the analysis and manipulation of signals in the frequency domain. This can provide insights into the behavior and characteristics of signals and can be used to design filters and other signal processing techniques.

4. Can the Laplace transform of a derivative of a convolution be used to solve differential equations?

Yes, the Laplace transform of a derivative of a convolution can be used to solve linear differential equations with constant coefficients. By taking the Laplace transform of both sides of the equation, the differential equation can be transformed into an algebraic equation, which can then be solved for the unknown function.

5. Are there any limitations or restrictions when using the Laplace transform of a derivative of a convolution?

One limitation of using the Laplace transform of a derivative of a convolution is that it can only be used for linear differential equations with constant coefficients. Additionally, the functions f and g must be well-behaved and satisfy certain conditions for the Laplace transform to exist. In some cases, the inverse Laplace transform may also be difficult to compute, requiring the use of tables or numerical methods.

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