use the convolution theorem to show that sinc(t) ∗ sinc(t) = sinc (t)

To show that \(\text{sinc}(t) * \text{sinc}(t) = \text{sinc}(t)\) using the convolution theorem, let's break down the problem step by step.

### Step 1: Define the sinc function
The sinc function is defined as:

\[
\text{sinc}(t) = \frac{\sin(\pi t)}{\pi t}
\]

### Step 2: Convolution Theorem
The Convolution Theorem states that the Fourier transform of the convolution of two functions is the product of their individual Fourier transforms. Mathematically:

\[
\mathcal{F}\{f(t) * g(t)\} = \mathcal{F}\{f(t)\} \cdot \mathcal{F}\{g(t)\}
\]

Where \(\mathcal{F}\) denotes the Fourier transform, and \(*\) represents the convolution operation.

### Step 3: Fourier Transform of sinc(t)
The Fourier transform of the sinc function, \(\text{sinc}(t)\), is given by:

\[
\mathcal{F}\{\text{sinc}(t)\} = \text{rect}(f)
\]

Here, \(\text{rect}(f)\) is the rectangular function defined as:

\[
\text{rect}(f) = \begin{cases}
1 & \text{if } |f| \leq \frac{1}{2} \\
0 & \text{if } |f| > \frac{1}{2}
\end{cases}
\]

### Step 4: Apply the Convolution Theorem
Let \( f(t) = \text{sinc}(t) \) and \( g(t) = \text{sinc}(t) \). According to the convolution theorem:

\[
\mathcal{F}\{ \text{sinc}(t) * \text{sinc}(t) \} = \mathcal{F}\{\text{sinc}(t)\} \cdot \mathcal{F}\{\text{sinc}(t)\}
\]

Substituting the Fourier transform of the sinc function:

\[
\mathcal{F}\{ \text{sinc}(t) * \text{sinc}(t) \} = \text{rect}(f) \cdot \text{rect}(f)
\]

### Step 5: Simplify the Expression
The product of two rectangular functions is still a rectangular function, but the width does not change:

\[
\text{rect}(f) \cdot \text{rect}(f) = \text{rect}(f)
\]

Thus, the Fourier transform of the convolution \( \text{sinc}(t) * \text{sinc}(t) \) is the same as the Fourier transform of \( \text{sinc}(t) \).

### Step 6: Apply the Inverse Fourier Transform
Taking the inverse Fourier transform on both sides:

\[
\mathcal{F}^{-1}\{\text{rect}(f)\} = \text{sinc}(t)
\]

Therefore:

\[
\text{sinc}(t) * \text{sinc}(t) = \text{sinc}(t)
\]

### Conclusion
We've used the convolution theorem to show that convolving the sinc function with itself results in the sinc function, i.e., \(\text{sinc}(t) * \text{sinc}(t) = \text{sinc}(t)\).

Would you like further details or have any questions?

Here are five related questions you might find interesting:
1. What is the convolution of two rectangular functions?
2. How does the width of the rectangular function affect the sinc function?
3. What are some applications of the sinc function in signal processing?
4. Can the convolution theorem be applied to functions that are not even or odd?
5. How does the Fourier transform help in solving differential equations?

**Tip:** Convolution in the time domain corresponds to multiplication in the frequency domain, which is a powerful tool in signal processing and system analysis.

Explore how the convolution theorem proves that convolving sinc(t) with itself equals sinc(t). This detailed explanation covers Fourier transform principles, the definition of sinc function, and application of convolution theorem. Ideal for advanced mathematics students and signal processing enthusiasts.

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