13.1.4. Proof of Convolution Theorem (Sketch)

Let ℎ(𝑡) = (𝑓∗𝑔)(𝑡) = ∫ 𝑓(𝜏)𝑔(𝑡−𝜏) 𝑑𝜏 0

In this step, we begin by defining the convolution of two functions, f and g, at time t, which is represented as h(t). The convolution is expressed as an integral from 0 to t, where we multiply the function f(τ) by the function g (reversed and shifted) at the point (t - τ) and integrate this product over the range from 0 to t.

Think of convolution like mixing two ingredients in a recipe, where one ingredient is gradually added while the other is adjusted based on the quantity already mixed. Just as combining these ingredients creates a new flavor, the convolution of two functions produces a new function representing their interaction over time.

Taking Laplace Transform on both sides: ℒ{ℎ(𝑡)} = ℒ{∫ 𝑓(𝜏)𝑔(𝑡−𝜏) 𝑑𝜏} 0

In this step, we take the Laplace transform of both sides of the equation defined earlier. The left side, ℒ{h(t)}, represents the Laplace transform of the convolution operation. The right side indicates that we are now finding the Laplace transform of the integral we defined earlier with f and g. This process is essential for switching from the time domain to the s-domain, which simplifies further manipulation.

Consider taking a photograph of a moving object. The photograph captures the state of the object at one moment in time. Similarly, the Laplace transform captures the behavior of a function at different frequencies, providing a snapshot of its transformation in the s-domain.

Using the property of Laplace Transform: ℒ{(𝑓∗𝑔)(𝑡)} = 𝐹(𝑠)⋅𝐺(𝑠)

Here, we apply a key property of Laplace transforms, which states that the Laplace transform of a convolution of two functions is equal to the product of their individual Laplace transforms. This means that (f ∗ g)(t) in the time domain translates into F(s)·G(s) in the s-domain. This property is crucial because it allows us to work with simpler algebraic expressions rather than convoluted integrals when dealing with Laplace transforms.

Think of it like multiplying the ingredients to make two different dishes. If you know the recipe for each dish (the transforms), you can quickly find out how to create a new dish when you combine them (the convolution in the s-domain). This shortcut makes cooking (doing math) faster and easier.

Hence, proved.

In this final statement, we conclude the proof of the Convolution Theorem. By showing that the Laplace transform of the convolution is indeed the product of the individual transforms, we confirm the theorem's validity. This closing statement emphasizes that we have logically and mathematically demonstrated the relationship between convolution in the time domain and multiplication in the s-domain.

Concluding a proof is akin to wrapping up a formal presentation. After arguing your points and presenting evidence (the steps of the proof), you finally summarize everything, confirming that the case has been made and the conclusion logically follows from the arguments presented.

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