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Chapter 76: Problem 1

The voltage from a square wave generator is of the form: $$ v(t)= \begin{cases}0, & -4

Short Answer

Expert verified
The Fourier series is \(v(t) = 5 + \frac{20}{\pi}\left[\sin\left(\frac{\pi t}{4}\right) + \frac{1}{3}\sin\left(\frac{3\pi t}{4}\right) + \cdots\right]\).

Step by step solution

01

Understanding the Problem

The problem involves finding the Fourier series representation for a given square wave function that operates over a period of 8 ms. The function is defined as 0 for the interval -4 to 0 ms and 10 for the interval 0 to 4 ms.
02

Basics of Fourier Series

A Fourier series represents periodic functions as a sum of sines and cosines. It comprises a constant term \(a_0\), and infinite sums involving sine and cosine terms with coefficients \(a_n\) and \(b_n\). This series attempts to 'fit' periodic functions by using trigonometric functions with varying frequencies.
03

Finding \(a_0\)

To find \(a_0\), compute the average value of the function over its period. The formula is \(a_0 = \frac{1}{8} \int_{-4}^{4} v(t) \, dt\). Given the function definition: the integral is 0 over the interval -4 to 0 ms and 10 from 0 to 4 ms. This evaluation yields \(a_0 = 5\).
04

Calculating \(a_n\) Coefficients

The \(a_n\) coefficients represent the cosine terms and can be found using: \(a_n = \frac{2}{8} \int_{-4}^{4} v(t) \cos(\frac{\pi n t}{4}) \, dt\). Evaluating this integral results in \(a_n = 0\) for all \(n\), as the sine terms \(\sin(\pi n)\) cancel each other.
05

Calculating \(b_n\) Coefficients

The coefficients \(b_n\) represent the sine components and are computed as \(b_n = \frac{2}{8} \int_{-4}^{4} v(t) \sin(\frac{\pi n t}{4}) \, dt\). Calculation shows that \(b_n = 0\) for even \(n\), while for odd \(n\), the coefficients take the form: \(b_1 = \frac{20}{\pi}, b_3 = \frac{20}{3\pi}, b_5 = \frac{20}{5\pi}\), and so forth.
06

Constructing the Fourier Series

Summarizing the series, the Fourier series is expressed as: \[ v(t) = 5 + \frac{20}{\pi} \left[ \sin(\frac{\pi t}{4}) + \frac{1}{3} \sin(\frac{3\pi t}{4}) + \frac{1}{5} \sin(\frac{5\pi t}{4}) + \cdots \right] \]. This equation represents the square wave function as an infinite series of sine functions.

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Key Concepts

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

Square Wave
A square wave is a specific type of periodic waveform which alternates between two levels. In the given problem, these levels are 0 and 10 volts. Unlike a sine wave, a square wave changes from maximum to minimum value instantaneously, forming a sequence of rectangles along the time axis.
Commonly, square waves are used in electronics and signal processing. One characteristic of this wave type is its harmonic-rich content, producing a distinct sound or signal.
  • A square wave has straight-edged transitions.
  • These waves repeat after a fixed interval, known as the period.
  • Given the rapid transition between states, they can represent binary data or clock pulses in digital circuits.
Periodic Function
A periodic function is a function that repeats its values at regular intervals or periods. In our square wave example, the period is 8 ms, meaning every 8 milliseconds, the waveform pattern repeats.
Periodic functions can be characterized by their amplitude, frequency, and period. These functions are central in modeling oscillatory and wave-like behavior in physical systems.
  • The function is defined as 'periodic' if there exists some positive constant \(T\) such that \(f(t + T) = f(t)\) for all values of \(t\).
  • Periodicity is essential for Fourier analysis, allowing complex waveforms to be broken down into simpler sine and cosine components.
  • Periodic phenomena appear in various domains, including sound waves, light waves, and mechanical vibrations.
Trigonometric Series
A trigonometric series is a series of sine and cosine functions that can represent more complex periodic functions.
The Fourier series is a specific example of a trigonometric series that allows the decomposition of any periodic waveform into a sum of sines and cosines.
  • In the Fourier series, coefficients are used to weigh the contributions of each sine and cosine term to the overall function.
  • Trigonometric series leverage the orthogonality properties of sine and cosine functions.
  • This method is powerful in signal processing, enabling the transformation of time-domain signals into frequency-domain representations.
Harmonic Analysis
Harmonic analysis is the study of representing functions or signals as overlapping waves of different frequencies.
In particular, Fourier series is a tool within harmonic analysis that decomposes periodic functions into sums of sine and cosine terms. This is essential in understanding the frequency components present in a waveform.
  • Each term in a Fourier series corresponds to a harmonic, defined as an integral multiple of the fundamental frequency.
  • Harmonic analysis aids in identifying dominant frequencies within signals that may be used to reconstruct the original waveform accurately.
  • This analysis is crucial for music applications, where understanding harmony and tone involves analyzing sound wave harmonics.

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Most popular questions from this chapter

Obtain the Fourier series for the function defined by: $$ f(x)= \begin{cases}0, & \text { when } & -2

Find the half-range Fourier sine series for the function \(f(x)=x\) in the range \(0 \leq x \leq 2\). Sketch the function within and outside of the given range. A half-range Fourier sine series indicates an odd function. Thus the graph of \(f(x)=x\) in the range 0 to 2 is shown in Fig. \(76.5\) and is extended outside of this range so as to be symmetrical about the origin, as shown by the broken lines. Figure \(76.5\) From para. (c), for a half-range sine series: $$ \begin{aligned} f(x) &=\sum_{n=1}^{\infty} b_{n} \sin \left(\frac{n \pi x}{L}\right) \\ b_{n} &=\frac{2}{L} \int_{0}^{L} f(x) \sin \left(\frac{n \pi x}{L}\right) \mathrm{d} x \\ &=\frac{2}{2} \int_{0}^{2} x \sin \left(\frac{n \pi x}{L}\right) \mathrm{d} x \\\ &=\left[\frac{-x \cos \left(\frac{n \pi x}{2}\right)}{\left(\frac{n \pi}{2}\right)}+\frac{\sin \left(\frac{n \pi x}{2}\right)}{\left(\frac{n \pi}{2}\right)^{2}}\right]_{0}^{2} \\ &\left.=\left[\frac{-2 \cos n \pi}{\left(\frac{n \pi}{2}\right)}+\frac{\sin n \pi}{\left(\frac{n \pi}{2}\right)^{2}}\right)-\left(0+\frac{\sin 0}{\left(\frac{n \pi}{2}\right)^{2}}\right)\right] \\ &=\frac{-2 \cos n \pi}{\frac{n \pi}{2}}=\frac{-4}{n \pi} \cos n \pi \end{aligned} $$ Hence \(b_{1}=\frac{-4}{\pi}(-1)=\frac{4}{\pi}\) $$ b_{2}=\frac{-4}{2 \pi}(1)=\frac{-4}{2 \pi} $$ $$ b_{3}=\frac{-4}{3 \pi}(-1)=\frac{4}{3 \pi} \text { and so on. } $$ Thus the half-range Fourier sine series in the range 0 to 2 is given by: $$ \begin{array}{r} f(x)=\frac{4}{\pi}\left[\sin \left(\frac{\pi x}{2}\right)-\frac{1}{2} \sin \left(\frac{2 \pi x}{2}\right)\right. \\ \left.\quad+\frac{1}{3} \sin \left(\frac{3 \pi x}{2}\right)-\frac{1}{4} \sin \left(\frac{4 \pi x}{2}\right)+\cdots\right] \end{array} $$
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