Question

Let \(f(t)=e^{i \omega t}\) on \((-\pi, \pi) .\) Expand \(f(t)\) in a complex exponential Fourier series of period 2 \(\pi . \text { (Assume } \omega \neq \text { integer. })\)

Short answer

The Fourier series expansion of \(f(t)\) is \[ f(t) = \sum_{k=-\infty}^{\infty} \frac{\sin(\pi \, (\omega - k))}{\pi \, (\omega - k)} e^{i k t}\]

Step-by-step solution

Understand the problem

The task is to expand the function \(f(t) = e^{i \, \omega t}\) into a complex exponential Fourier series with a period of \(2 \pi\). We assume \(\omega\) is not an integer.

Determine the Fourier coefficients

To expand \(f(t)\) in a Fourier series, we need to find the Fourier coefficients \(c_k\) which are given by the formula: \[ c_k = \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{i \, \omega t} e^{-i k t} \, dt\] where \(k\) is an integer.

Compute the integral for the coefficients

Compute the integral: \[ c_k = \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{i \, (\omega - k) t} \, dt\] Consider the case when \(\omega = k\), the integral simplifies to: \[ c_k = \frac{1}{2\pi} \int_{-\pi}^{\pi} 1 \, dt = 1 \, (if \, \omega = k)\] If \(\omega \eq k\): \[ c_k = \frac{1}{2\pi} \frac{e^{i \, (\omega - k)t}}{i \, (\omega - k)} \bigg|_{-\pi}^{\pi} \right = \frac{e^{i \pi \, (\omega - k)} - e^{-i \pi \, (\omega - k)}}{2\pi i \,(\omega - k)} = \frac{2i \, \sin(\pi \, (\omega - k))}{2 \, \pi i \,(\omega - k)} = \frac{\sin(\pi \, (\omega - k))}{\pi \, (\omega - k)}\]

Summarize the coefficients

Therefore, the Fourier coefficients \(c_k\) are: \[ c_k = \begin{cases} 1 & \text{if } k = \omega \ \frac{\sin(\pi \, (\omega - k))}{\pi \, (\omega - k)} & \text{if } k \eq \omega \end{cases} \]

Write the Fourier series expansion

With the coefficients, the function \(f(t)\) can be written as: \[ f(t) = \sum_{k=-\infty}^{\infty} c_k e^{i k t} \] Substituting \(c_k\): \[ f(t) = e^{i \omega t} = \sum_{k=-\infty}^{\infty} \frac{\sin(\pi \, (\omega - k))}{\pi \, (\omega - k)} e^{i k t}\]

Key concepts

Fourier series

A Fourier series is a way to represent a function as a sum of periodic components. In other words, you can think of it like breaking down a complex waveform into simpler sine and cosine waves. Fourier series are essential in signal processing, physics, and many other areas of science and engineering.
For a function defined on a specific interval, a Fourier series allows for the decomposition into trigonometric functions with different frequencies. The key idea is to find the coefficients that, when multiplied with these basis functions, provide an accurate representation of the original function.

Complex exponentials

Complex exponentials are expressions of the form \(e^{i \, \theta}\), where \(i\) is the imaginary unit and \(\theta\) is a real number. This form is incredibly powerful in signal processing and other fields because it simplifies the manipulation of sinusoidal functions.
Euler's formula \(e^{i \, \theta} = \cos(\theta) + i \sin(\theta)\) links complex exponentials with trigonometric functions. When dealing with Fourier series, complex exponentials provide a cleaner, more algebraic approach compared to trigonometric functions alone.

Fourier coefficients

Fourier coefficients are the weights given to each term in the Fourier series. They tell us how much of each sine and cosine wave is present in the original function. In formulas, these weights are often represented as \(a_n\), \(b_n\) for real-valued series, or \(c_k\) in the complex form.
The key part of solving Fourier series problems is determining these coefficients, which typically involves integral calculus. For a complex exponential Fourier series, the coefficients \(c_k\) are given by:

• \[ c_k = \frac{1}{2\pi} \int_{-\text{\pi}}^{\text{\pi}} f(t) \, e^{-i \ k t} dt \]

Integral calculus

Integral calculus provides the tools to calculate areas under curves. In the context of Fourier series, it is used to compute the Fourier coefficients. Essentially, you integrate the product of the given function and a complex exponential over a defined interval.
To determine the Fourier coefficients \(c_k\) for a function \(f(t)\), you use the integral:

• \[ c_k = \frac{1}{2 \ \pi} \int_{-\text{\pi}}^{\text{\pi}} f(t) \, e^{-i \ k t} dt \]

This process involves evaluating integrals, which might require techniques like substitution or integration by parts.

Trigonometric functions

Trigonometric functions, namely sines and cosines, form the basis of Fourier series in their real form. Recognizing them in the complex exponential format can make solving Fourier series problems easier.
Recall that using Euler's formula, a complex exponential can be decomposed into sine and cosine terms:

• \[ e^{i \, \theta} = \cos(\theta) + i \sin(\theta) \]

This linkage allows you to handle periodic functions more conveniently, whether you are working directly with sine and cosine terms or translating between them and complex exponentials. Understanding this relationship is critical for solving problems involving Fourier series.