For a function f(t), the Laplace transformation is defined...

For a function $f(t)$, the Laplace transformation is defined as $F(s) = \mathcal{L}(f(t)) = \int_0^\infty f(t)e^{-st}dt$. Prove that for two functions $f$ and $g$ we have $\mathcal{L}(f*g) = \mathcal{L}(f)\mathcal{L}(g)$. where $f*g$ is the convolution of two function $f$ and $g$.

Transcript

0:00 Hi there.

00:01 So for this problem, we state improved the convolution theorem for laplace transform of two functions and using it.

00:08 We need to find inverse laplace transform of the following function.

00:12 And that function is that f of x is equal to x divided by x square plus four to the square.

00:27 So with this, we need to use the combo illusion theorem to find the laplace, the inverse laplace transform.

00:41 So we can write this equation as the following, as the peribur between x divided by x squared plus 4, and this times 1 over x squared plus 4.

00:55 And that's because we know the inverse laplace transform of each of one of this.

01:06 We know that inverse transform, we're going to call this capital f of x and this one capital g of x.

01:18 So this is going to be the product between these two functions that depends on x.

01:25 Now, we know that the inverse transform of f of x is f of d, where this is going to be, in this case, is going to be the cosine of theta.

01:48 Well, cosine of 2 times theta, because in here we will have a 4.

01:56 So that 4, we will include that the square root of that in here, and we know that the square root of 4 is 2.

02:05 Now, for the other term in here, you can search that for this one in here, well, we are going to call this g of d, and that is equal to the sign of 2 times the time t.

02:24 Now, with that set, therefore, we can, find that q of s is equal to the laplace transform of q of s.

02:46 Where this is, we're going to call this better q of x.

02:50 So we will have this q of x.

02:53 So the answer for that is the laplace transform of q of x.

02:58 So we will have the laplace transform of this is equal to the laplace transform of the product between f of s and g of x.

03:07 So to obtain this, we use the convolution of the functions f and g with respect to the parameter t.

03:18 So using the convolution theorem, we know that that is equal to the integral from zero to the parameter t.

03:29 And this, the first function, which is f, so we will put in here, sign of four times the time, and this minus b, and this cosine of cosine of b.

04:03 So this is the integral that we need to solve.

04:07 And in this case, what we are going to do is a trigonometry identity.

04:15 We know that the sign of a times the sign of b is equal to, well, sorry, in here we will have cosine of b.

04:30 So that product, the product that we have in here, as you can see, and we can write it as the sign of a plus b, plus this times the sign of a minus b...

Question

For a function $f(t)$, the Laplace transformation is defined as $F(s) = \mathcal{L}(f(t)) = \int_0^\infty f(t)e^{-st}dt$. Prove that for two functions $f$ and $g$ we have $\mathcal{L}(fg) = \mathcal{L}(f)\mathcal{L}(g)$. where $fg$ is the convolution of two function $f$ and $g$.

Question

For a function $f(t)$, the Laplace transformation is defined as $F(s) = \mathcal{L}(f(t)) = \int_0^\infty f(t)e^{-st}dt$. Prove that for two functions $f$ and $g$ we have $\mathcal{L}(f*g) = \mathcal{L}(f)\mathcal{L}(g)$. where $f*g$ is the convolution of two function $f$ and $g$.

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For a function $f(t)$, the Laplace transformation is defined as $F(s) = \mathcal{L}(f(t)) = \int_0^\infty f(t)e^{-st}dt$. Prove that for two functions $f$ and $g$ we have $\mathcal{L}(fg) = \mathcal{L}(f)\mathcal{L}(g)$. where $fg$ is the convolution of two function $f$ and $g$.