**Convolution Theorem:**

The convolution theorem of Laplace transform states that, let f_{1} (t) and f_{2} (t) are the Laplace transformable functions and F_{1} (s), F_{2} (s) are the Laplace transforms of f_{1} (t) and f_{2} (t) respectively. Then the product of F_{1} (s) and F_{2} (s) is the Laplace transform of f(t) which is obtained from the convolution of f_{1} (t) and f_{2} (t). The convolution of f_{1} (t) and f_{2} (t) is denoted as f_{1} (t) * f_{2} (t) and is obtained by the equation,

where Ï„ is the dummy variable.

where f_{1} (t) * f_{2} (t) indicates convolution of f_{1} (t) and f_{2} (t).

**Proof:**

Let

Where

x and y are dummy variables.

As x and y are independent variables we can write,

Consider the new variables t and Ï„ such that,

Let us consider limits of integration interms of t and Ï„

The smallest value of y is zero hence t â‰¥ Ï„.

The equation t = Ï„ is straight line in t – Ï„ plane. So to integrate area between the line t = Ï„ and Ï„ = 0, we get the limits for t as 0 â†’âˆž while limits of Ï„ as 0 to t.

Now

From the definition of Laplace transform the equation (2) represents Laplace transform of integral

which is called convolution integral. So in the equation (1),

Thus explained.