What is Convolution? State and prove Time convolution theorem

Convolution is a mathematical operation that combines two functions to produce a third function. It is commonly used in signal processing and image processing to analyze and manipulate signals.

The Time Convolution Theorem states that the convolution of two functions in the time domain is equivalent to the multiplication of their Fourier transforms in the frequency domain.

To prove the Time Convolution Theorem, we start by considering two functions, f(t) and g(t), and their respective Fourier transforms F(ω) and G(ω).

The convolution of f(t) and g(t) is defined as:

(f * g)(t) = ∫[f(τ) * g(t-τ)] dτ

where * denotes the convolution operation.

Now, let's take the Fourier transform of both sides of the convolution equation:

F(ω) = ∫[f(t) * e^(-jωt)] dt G(ω) = ∫[g(t) * e^(-jωt)] dt

Using the convolution theorem, we know that the Fourier transform of the convolution of two functions is equal to the product of their individual Fourier transforms:

F(ω) * G(ω) = ∫[f(t) * e^(-jωt)] dt * ∫[g(t) * e^(-jωt)] dt

Expanding the right side of the equation:

F(ω) * G(ω) = ∫∫[f(τ) * g(t-τ) * e^(-jωt)] dτ dt

Now, let's change the order of integration by substituting t-τ = τ':

F(ω) * G(ω) = ∫∫[f(τ) * g(τ') * e^(-jω(t-τ))] dτ dτ'

Simplifying the equation:

F(ω) * G(ω) = ∫∫[f(τ) * g(τ') * e^(-jωt) * e^(jωτ)] dτ dτ'

Since e^(-jωt) * e^(jωτ) = e^(-jω(t-τ)) = e^(-jωτ') = e^(-jωτ') * e^(jωτ') = 1, the equation becomes:

F(ω) * G(ω) = ∫[f(τ) * g(τ')] dτ

which is the Fourier transform of the convolution of f(t) and g(t).

Therefore, we have proven that the convolution of two functions in the time domain is equivalent to the multiplication of their Fourier transforms in the frequency domain, which is known as the Time Convolution Theorem.