Question

Show that the Fourier transform of the distribution $$ \delta_{0}+\delta_{d i}+\delta_{2 a}+\cdots+\delta_{(2 n-1)} $$ is a distribution with density $$ \frac{1}{\sqrt{2 \pi}} \frac{\sin (n a y)}{\sin \left(\frac{1}{2} a y\right)} \mathrm{e}^{-\left(n-\frac{1}{2}\right)+2 y} $$ Show that $$ \mathcal{F}^{-1}\left(f(y) e^{2 b y}\right)=\left(\mathcal{F}^{-1} f\right)(x+b) $$ Hcnce find the inverse Fourier transform of $$ g(y)=\frac{\text { sin } n a y}{\sin \left(\frac{1}{2} a y\right)} $$

Short answer

The Fourier transform of the given distribution is \( \frac{1}{\sqrt{2 \pi}} \frac{\sin (n a y)}{\sin \left(\frac{1}{2} a y\right)} e^{-\left(n-\frac{1}{2}\right)+2 y} \). Moreover, the Fourier transform of \( f(y)e^{2by} \) is \( \mathcal{F}^{-1}(f)(x+b) \). Thus, the inverse Fourier transform of \( g(y) = \frac{\sin(nay)}{\sin(ay/2)} \) is \( \mathcal{F}^{-1}(f)(x+b) \).

Step-by-step solution

Fourier Transform of Distributions

To find the Fourier transform of the given distribution, observe that the Fourier transform of \( \delta (x-a) \) is \( e^{iay} / \sqrt{2 \pi} \). Then, the Fourier transform of the given distribution is the sum for k = 0 to n, of \( e^{-ia(2k-1)y} / \sqrt{2 \pi} \). Now, this sum can be rewritten as \( \frac{1}{\sqrt{2 \pi}} \frac{\sin (n a y)}{\sin \left(\frac{1}{2} a y\right)} e^{-\left(n-\frac{1}{2}\right)+2 y} \) by using the formula for sum of geometric series and trigonometric identities.

Fourier Transform of Function times Exponential

This requires proving that the Fourier transform of \( f(y)e^{2by} \) is \( \mathcal{F}^{-1}(f)(x+b) \). Given that the Fourier transform of \( f(y) \) is \( \mathcal{F}(f)(x) \), we can apply the shift theorem to \( f(y)e^{2by} \). The shift theorem states that the Fourier transform of \( e^{bx}f(x) \) is \( \mathcal{F}(f)(x-b) \). Applying this result with a shift -b in frequency domain, we get \( \mathcal{F}^{-1}(f)(x+b) \) as required.

Inverse Fourier Transform of a Function

We have to find the inverse Fourier transform of \( g(y) = \frac{\sin(n a y)}{\sin\left(a y /2 \right)} \). Instead of taking the inverse Fourier transform directly, we find it easier to write g(y) as \( g(y) = f(y)e^{2by} \), with \( f(y) = \frac{1}{\sqrt{2 \pi}} \frac{\sin (n a y)}{\sin \left(\frac{1}{2} a y\right)} \) and \( b = -\left(n-\frac{1}{2}\right)+2 \). Now, we can use the result in Step 2 to find that \( \mathcal{F}^{-1}(g)(x) = \mathcal{F}^{-1}(f)(x+b) \). This yields the inverse Fourier transform of g(y).

Key concepts

Distributions

The concept of distributions in mathematics, particularly in the context of Fourier transforms, refers to generalized functions. These functions can include entities like Dirac delta functions, which are not traditional functions but still convey information about point values. In our exercise, distributions like \( \delta_{0}, \delta_{d i}, \delta_{2 a}, \cdots, \delta_{(2 n-1)} \) stand for sequences of impulse points along an axis.
The Fourier transform applied to these distributions allows us to analyze signals that are constructed from impulses. Each spike, represented by a delta function \( \delta(x-a) \), can be transformed using the Fourier transform. This gives a function in frequency space proportional to \( \frac{e^{iay}}{\sqrt{2\pi}} \).
These transformations help us shift from a temporal or spatial domain into the frequency domain, where these concepts can be manipulated algebraically — an essential process in signal processing and physics.
Understanding distributions and their transformations is fundamental, particularly when combined with the sum of geometric terms and trigonometric identities for real-world applications.

Inverse Fourier Transform

The inverse Fourier transform is the process of converting data from the frequency domain back to the original domain, be it time or space.
For our problem, we examine the formula \( \mathcal{F}^{-1}(g)(x) = \left( \mathcal{F}^{-1}(f) \right)(x+b) \). When functions like \( g(y) = \frac{\sin(n a y)}{\sin(a y /2)} \) are considered, the inverse transforms rely on using predefined functions and adding exponential components as shifts.
One of the challenges is simplification. We denote \( g(y) \) with the elaborated expression of \( f(y)e^{2by} \). By understanding the relationship between functions and their exponential shifts, we convert the problem into manageable parts. This reduces the complexity of going from the frequency domain to the time domain.

• Identify the base function \( f(y) \)
• Use transformation theorems strategically
• Apply shifts to analyze results appropriately

Understanding this inverse relation is pivotal in reconstructing signals from frequency analyses.

Trigonometric Identities

Trigonometric identities play an essential role in simplifying expressions in mathematics. When dealing with Fourier transforms, they become indispensable.
In the specific exercise, you use trigonometric identities to express and simplify complex sums. The expression \( \sin(n a y) / \sin(a y / 2) \) appears complicated at first, but these identities provide a mathematical backbone that makes simplification possible.
Some common identities include:

• Sum-to-product identities, such as \( \sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \)
• Product-to-sum identities for converting products into sums

These identities allow the transformation of sums into simpler forms, often necessary for smooth applications in signal processing, oscillations, and waves.
Employing these identities correctly allows mathematicians and engineers to decompose and reconstruct expressions effectively.

Shift Theorem

The Shift Theorem is a fundamental property in the study of Fourier transforms. It states that shifting a function in the time domain corresponds to a multiplication in the frequency domain, and vice versa.
For our context, consider \( f(y)e^{2by} \), a shifted function. The theorem guides us in understanding how this affects the outcomes:\[ \mathcal{F}(f)(x-b) \]. This leads directly to \( \mathcal{F}^{-1}(f)(x+b) \), as proven in the exercise.
Understanding the Shift Theorem helps in managing how signals behave when altered in the frequency domain or in time, which is critical for applications like signal encryption or data compression.

• Predict and control signal transformations effectively
• Explore phase shifts and their implications
• Facilitate the design of electronic filters

By utilizing the shift theorem, one can manage signal behavior with a profound leverage in applied science and engineering procedures.