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Let $$ f(x)= \begin{cases}A \sin \omega_{0} t, & 0

Short Answer

Expert verified
Compute coefficients \( C_n \) using integrals; sum them for the series.

Step by step solution

01

Understand the Problem Statement

We need to find the complex Fourier series of the piecewise function \( f(t) \). The function \( f(t) \) is defined as \( A \sin \omega_0 t \) for \( 0 < t < \frac{T}{2} \) and as 0 for \( \frac{T}{2} \leq t < T \). The parameter \( \omega_0 = \frac{2\pi}{T} \) is the fundamental angular frequency.
02

Determine the Formula for the Complex Fourier Series

The complex Fourier series for a function on the interval \([0, T]\) is defined as: \[ f(t) = \sum_{n=-\infty}^{\infty} C_n e^{i n \omega t}, \] where \( C_n \) are the Fourier coefficients given by \[ C_n = \frac{1}{T} \int_{0}^{T} f(t) e^{-i n \omega t} \, dt. \] Here, \( \omega = \frac{2\pi}{T} \).
03

Calculate Fourier Coefficients \( C_n \)

We calculate \( C_n \) by breaking the integral into two parts due to the piecewise definition of \( f(t) \): \[ C_n = \frac{1}{T} \left( \int_{0}^{\frac{T}{2}} A \sin \left( \frac{2\pi}{T}t \right) e^{-i n \omega t} \right. dt + \left. \int_{\frac{T}{2}}^{T} 0 \, dt \right). \] The second integral is zero due to the zero function between \( \frac{T}{2} \) and \( T \). Thus, \[ C_n = \frac{A}{T} \int_{0}^{\frac{T}{2}} \sin \left( \frac{2\pi}{T}t \right) e^{-i n \omega t} \, dt. \]
04

Simplify the Integral

The integral can be solved by using the identity for sine in terms of exponentials: \[ \sin(\omega_0 t) = \frac{e^{i \omega_0 t} - e^{-i \omega_0 t}}{2i}. \]This leads to: \[ C_n = \frac{A}{2iT} \int_{0}^{\frac{T}{2}} \left( e^{i \omega_0 t} - e^{-i \omega_0 t} \right) e^{-i n \omega t} \, dt. \] Breaking this into two integrals yields:\[ C_n = \frac{A}{2iT} \left( \int_{0}^{\frac{T}{2}} e^{i(\omega_0 - n\omega)t} \, dt - \int_{0}^{\frac{T}{2}} e^{-i(\omega_0 + n\omega)t} \, dt \right). \]
05

Evaluate the Integrals

Each integral is of the form \( \int e^{i \alpha t} \, dt \), which results in:\[ \int e^{i \alpha t} \, dt = \frac{e^{i \alpha t}}{i \alpha}. \] Applying this, we compute:\[ \int_{0}^{\frac{T}{2}} e^{i(\omega_0 - n\omega)t} \, dt = \frac{e^{i(\omega_0 - n\omega)\frac{T}{2}} - 1}{i(\omega_0 - n\omega)}, \] \[ \int_{0}^{\frac{T}{2}} e^{-i(\omega_0 + n\omega)t} \, dt = \frac{1 - e^{-i(\omega_0 + n\omega)\frac{T}{2}}}{i(\omega_0 + n\omega)}. \]
06

Simplify the Coefficient Expression

Substitute the results back into the expression for \( C_n \):\[ C_n = \frac{A}{2iT} \left( \frac{e^{i(\omega_0 - n\omega)\frac{T}{2}} - 1}{i(\omega_0 - n\omega)} - \frac{1 - e^{-i(\omega_0 + n\omega)\frac{T}{2}}}{i(\omega_0 + n\omega)} \right). \]Simplify to obtain the final expression for \( C_n \).
07

Write the Final Fourier Series Expression

Combine all the coefficients to write the final form of the complex Fourier series:\[ f(t) = \sum_{n=-\infty}^{\infty} C_n e^{i n \omega t}, \] using the expressions derived for \( C_n \). Simplify if possible, using trigonometric identities or symmetries to express the series in the simplest form.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Fourier Series
The complex Fourier series is a powerful tool in mathematical analysis used to express periodic functions as an infinite sum of complex exponentials. It holds significance because it provides a comprehensive way to represent functions in the frequency domain.
The basic form of the complex Fourier series on an interval \([0, T]\) is:
  • \( f(t) = \sum_{n=-\infty}^{\infty} C_n e^{i n \omega t} \)
Here, \( C_n \) are complex Fourier coefficients, and \( \omega = \frac{2\pi}{T} \) is the fundamental frequency. Each term \( e^{i n \omega t} \) in the series represents a sinusoidal function from the Euler's formula \( e^{ix} = \cos(x) + i \sin(x) \).
By using complex exponentials, a broader range of applications is facilitated, particularly in engineering and physics, where understanding phase shifts and oscillations is crucial.
Fourier Coefficients
Fourier coefficients \( C_n \) are the building blocks of Fourier series, serving to adjust the amplitude and phase of each term to match the function being represented. They are calculated using the integral formula:
  • \( C_n = \frac{1}{T} \int_{0}^{T} f(t) e^{-i n \omega t} \, dt \)
This formula integrates the product of the function \( f(t) \) and a complex exponential over one period.
The presence of the exponential \( e^{-i n \omega t} \) acts like a filter, allowing only the specific frequency component corresponding to \( n \) to be analyzed and extracted. In essence, it captures how much of this sinusoidal component is present within the function. The concept is about finding a match in terms of magnitude and phase between the function and pure sine waves.
Piecewise Function
A piecewise function is a function defined by multiple sub-functions, each applying to a certain interval of the domain. In the context of Fourier series, piecewise functions often introduce discontinuities or changes in behavior over the domain of interest.
The exercise involves a piecewise function given by:
  • \( f(x)= \begin{cases}A \sin \omega_{0} t, & 0
This particular function reinforces the importance of understanding how to handle different segments separately while integrating or constructing series expressions. Handling these multiple cases often leads to more complex Fourier coefficients and series, necessitating additional calculations specific to each segment.
Sine Function Representation
Representing a function using only sine (or cosine) functions is a common application of Fourier series, particularly for periodic signals. Sine functions have properties that make them ideal for capturing oscillatory behaviors, which is fundamental in analyzing sound and waveforms.
In scenarios where the function is piecewise and partly a sine function, as in:
  • \( A \sin \omega_0 t \) for \( 0 < t < \frac{T}{2} \)
The complex Fourier series leverages sine components by expressing them with exponential functions. Thus, the sine function is represented using Euler's identity, \( \sin(x) = \frac{e^{ix} - e^{-ix}}{2i} \).
This conversion is crucial in simplifying the integration process when deriving Fourier coefficients and creating more uniform formulas that are applicable in digital signal processing and electrical engineering.

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Most popular questions from this chapter

Let \(f(x)=x^{2}\) and $$ f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n} \cos n x+b_{n} \sin n x\right] $$ denote the Fourier series of \(f\) on \([\pi, 3 \pi]\). (a) Calculate the \(a_{n}\) and \(b_{n}\). (b) Set \(g(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos \frac{n x}{2},-\pi \leq x \leq \pi\). Determine \(g\) and sketch the graph of \(g\) on \([-\pi, \pi]\). (c) Set \(h(x)=\sum_{n=1}^{\infty} b_{n} \sin \frac{n x}{2},-\pi \leq x \leq \pi\). Determine \(h\) and sketch the graph of \(h\) on \([-\pi, \pi]\).

Set \(f(x)=1-x^{2}\) in the interval \([-\pi, \pi]\) and let $$ f(x) \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n} \cos n x+b_{n} \sin n x\right] $$ be the Fourier series of \(f\). (a) Calculate the \(a_{n}\) and \(b_{n}\). (b) To what values does the Fourier series of \(f\) converge at the points \(x=5 \pi\) and \(x=6 \pi\) ? Explain.

Let \(f \in E\) and assume $$ \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left[a_{n} \cos n x+b_{n} \sin n x\right] $$ is the Fourier series of \(f\). Show that if there exist constants \(c\) and \(d\) such that $$ \left|a_{n}\right| \leq \frac{c}{n^{2}}, \quad\left|b_{n}\right| \leq \frac{d}{n^{2}} $$ for all \(n\), then \(f\) may be considered to be continuous on \([-\pi, \pi]\), satisfying \(f(-\pi)=f(\pi)\), and the Fourier series of \(f\) converges uniformly to \(f\) on \([-\pi, \pi]\).

We define the function $$ f(x)= \begin{cases}\sin 2 x, & -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \\ 0, & \text { otherwise }\end{cases} $$ on the interval \([-\pi, \pi]\). (a) Determine the Fourier series of \(f\) on \([-\pi, \pi]\). (b) Determine the Fourier series of \(f^{\prime}\) on \([-\pi, \pi]\). (c) To what values does the Fourier series of \(f^{\prime}\) converge at the points \(x=\pm \frac{\pi}{2} ?\) (d) Calculate the sums $$ \sum_{k=1}^{\infty} \frac{1}{(2 k-3)^{2}(2 k+1)^{2}}, \quad \sum_{k=1}^{\infty} \frac{(2 k-1)(-1)^{k}}{(2 k-3)(2 k+1)} $$

Let \(f\) be a \(2 \pi\)-periodic piecewise continuous function satisfying $$ \int_{-\pi}^{\pi} f(x) d x=0 . $$ \(\operatorname{Set} g(x)=\int_{0}^{x} f(t) d t\). (a) Prove that \(g\) is \(2 \pi\)-periodic. (b) Let \(\sum_{n=-\infty}^{\infty} c_{n} e^{i n x}\) be the complex Fourier series of the function \(f\) and \(\sum_{n=-\infty}^{\infty} d_{n} e^{i n x}\) the complex Fourier series of the function \(g\). Prove that for all real \(x\) we have the equality $$ g(x)=\sum_{n=-\infty}^{\infty} d_{n} e^{i n x} $$ where \(d_{n}=\frac{c_{n}}{i n}\) for every integer \(n \neq 0\).

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