Question

Prove the convolution theorem: $$\int_{-\infty}^{\infty} f(x) g(y-x) d x=\int_{-\infty}^{\infty} \tilde{f}(k) \tilde{g}(k) e^{i k y} d k$$ What will this give when \(y=0\) ?

Short answer

The convolution theorem is proved. When \(y=0\), the result is \(\int_{-\infty}^{\infty} \tilde{f}(k) \tilde{g}(k) dk\), the Fourier Transform of the product of the Fourier Transforms \(f(x)\) and \(g(x)\).

Step-by-step solution

- Fourier Transformation

The first step is to convert the convolution integral \(\int_{-\infty}^{\infty} f(x) g(y-x) d x\) into their Fourier transformations: \(\int_{-\infty}^{\infty} \tilde{f}(k) \tilde{g}(k) e^{i k y} d k\). Let's define: \(f(x) = \int_{-\infty}^{\infty} \tilde{f}(k) e^{ikx} dk \) and \(g(x) = \int_{-\infty}^{\infty} \tilde{g}(k) e^{ikx} dk \). Where \(\tilde{f}(k)\) and \(\tilde{g}(k)\) are the Fourier transforms of \(f(x)\) and \(g(x)\) respectively.

- Convolution Operation

Substitute these in and then use the convolution operation: \(\int_{-\infty}^{\infty} f(x) g(y-x) dx = \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dk \int_{-\infty}^{\infty} dl \tilde{f}(k) \tilde{g}(l) e^{ikx} e^{-il(y-x)}. Changing the order of integrals and then using delta function property, we arrive at:\(\int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dk \tilde{f}(k) \tilde{g}(k) e^{ikx} e^{-ik(y-x)} = \int_{-\infty}^{\infty} \tilde{f}(k) \tilde{g}(k) e^{iky} dk.\)

- Calculating for \(y = 0\)

Setting \(y = 0\), this results in an integral:\(\int_{-\infty}^{\infty} \tilde{f}(k) \tilde{g}(k) dk\). Here, there's essentially no exponential term \(e^{iky}\) to consider. This is the Fourier Transform of the multiplied transformations.

Key concepts

Fourier Transformation

The Fourier Transformation is a mathematical technique that transforms a function of time into a function of frequency. This is incredibly powerful for analyzing the frequencies present in a signal. When you apply the Fourier Transform to a time-domain function, you essentially break it down into its constituent sine and cosine waves. This can provide insights that are not easily discernible in the time domain alone. In the context of our exercise, the functions \(f(x)\) and \(g(x)\) are transformed into \(\tilde{f}(k)\) and \(\tilde{g}(k)\) respectively. Here:

• \(\tilde{f}(k) = \int_{-\infty}^{\infty} f(x) e^{-ikx} dx\) represents the Fourier transform of \(f(x)\).
• \(\tilde{g}(k) = \int_{-\infty}^{\infty} g(x) e^{-ikx} dx\) represents the Fourier transform of \(g(x)\).

For these transformations, the original functions are reconstructed using an inverse Fourier Transform. This is particularly useful when working with signals in different domains, such as converting between time and frequency domains.

Convolution Operation

Convolution is a key operation in signal processing, particularly for functions that overlap in time or space. This operation involves sliding one function over another and integrating their product. In our original problem, the convolution integral \(\int_{-\infty}^{\infty} f(x) g(y-x) dx\) is transformed using the Fourier Transform into \(\int_{-\infty}^{\infty} \tilde{f}(k) \tilde{g}(k) e^{iky} dk\). The convolution operation allows us to simplify complex expressions significantly, especially when dealing with Fourier-transformed functions. By working in the Fourier domain, multiplication of transforms replaces the convolution of functions in the time domain:

• The Fourier Transform of a convolution is the product of their individual transforms.
• This property simplifies analysis and calculations for many practical problems.

A key benefit of convolution is its ability to filter signals, allowing certain frequencies to pass while blocking others.

Delta Function Property

The delta function, often denoted as \(\delta(x)\), is a fascinating mathematical concept that simplifies many types of problems, especially in integrals involving Fourier Transforms. This function is defined to be zero everywhere except at \(x = 0\), where it is infinitely high with a total integral of one. In our exercise, when computing the convolution using Fourier Transforms, you change the order of integration. The delta function simplifies the integral. Specifically, due to its sifting property:

• It selects only the values where the arguments of the delta equal zero.
• This significantly reduces complexity, making computations involved in transformations more straightforward.

When resolving the convolution theorem, this property ensures that \(\int_{-\infty}^{\infty} dk \tilde{f}(k) \tilde{g}(k) e^{iky}\), for \(y = 0\), boils down to a straightforward multiplication of transformed variables without an exponential term.