Chapter 4: Problem 3
Let
$$
f(t)=\left\\{\begin{array}{llr}
0 & \text { if } & -1
Short Answer
Expert verified
Fourier series: \( \frac{1}{4} - \frac{2}{\pi^2} \cos(\pi t) - \frac{1}{\pi} \sin(\pi t) - \frac{2}{9\pi^2} \cos(3\pi t) - \frac{1}{3\pi} \sin(3\pi t) \).
Step by step solution
01
Understand the Function
The function \( f(t) \) is defined piecewise. It is 0 for \(-1 < t \leq 0\) and \( t \) for \( 0 < t \leq 1 \). Since it's extended periodically, the period \( T \) is 2, from \(-1\) to 1.
02
Calculate Fourier Coefficients
The Fourier coefficients are calculated using:\[a_0 = \frac{1}{T} \int_{-1}^{1} f(t) \, dt,\]\[a_n = \frac{2}{T} \int_{-1}^{1} f(t) \cos\left(\frac{n \pi t}{L}\right) \, dt,\]\[b_n = \frac{2}{T} \int_{-1}^{1} f(t) \sin\left(\frac{n \pi t}{L}\right) \, dt.\]Here, \( L = \frac{T}{2} = 1 \). Calculate \( a_0 \), \( a_n \), and \( b_n \) for the relevant piece of the function.
03
Calculate a0
\[a_0 = \frac{1}{2} \left( \int_{-1}^{0} 0 \, dt + \int_{0}^{1} t \, dt \right)= \frac{1}{2} \left( 0 + \left[ \frac{t^2}{2} \right]_0^1 \right)= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.\]
04
Calculate an
\[a_n = \frac{2}{2} \int_{0}^{1} t \cos(n\pi t) \, dt.\]Integrate by parts: let \( u = t \) and \( dv = \cos(n\pi t) \, dt \). Thus, \( du = dt \) and \( v = \frac{1}{n\pi} \sin(n\pi t) \).\[a_n = \left.\frac{t \sin(n\pi t)}{n\pi}\right|_0^1 - \int_{0}^{1} \frac{\sin(n\pi t)}{n\pi} \, dt= \frac{1 \cdot 0 - 0}{n\pi} - \frac{1}{n\pi} \left[ -\frac{1}{n\pi} \cos(n\pi t)\right]_0^1\]\[= 0 - \frac{1}{n^2\pi^2}(\cos(n\pi) - 1)= -\frac{2}{n^2\pi^2} \quad \text{for odd } n= 0 \quad \text{for even } n.\]
05
Calculate bn
\[b_n = \frac{2}{2} \int_{0}^{1} t \sin(n\pi t) \, dt.\]Integrate by parts: let \( u = t \) and \( dv = \sin(n\pi t) \, dt \). Thus, \( du = dt \) and \( v = -\frac{1}{n\pi} \cos(n\pi t) \).\[b_n = \left.-\frac{t \cos(n\pi t)}{n\pi}\right|_0^1 + \int_{0}^{1} \frac{\cos(n\pi t)}{n\pi} \, dt\]\[ = \frac{-1 \cdot (-1) + 0}{n\pi} + \frac{1}{n\pi} \left[ \frac{1}{n\pi} \sin(n\pi t)\right]_0^1= \frac{-1}{n\pi}, \quad \text{for odd } n\= 0 \quad \text{for even } n.\]
06
Assemble the Fourier Series
Using the coefficients calculated:\[f(t) = \frac{1}{4} + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{n\pi t}{1}\right) + b_n \sin\left(\frac{n\pi t}{1}\right) \right)\]Substitute \(a_n\) and \(b_n\) into the formula to get:\[f(t) = \frac{1}{4} - \sum_{n\, \text{odd}} \frac{2}{n^2\pi^2} \cos(n\pi t) + \sum_{n\, \text{odd}} \left( \frac{-1}{n\pi} \right) \sin(n\pi t)\]
07
Write the Series Up to 3rd Harmonic
For the first three harmonics, use the odd values of \( n = 1, 3 \).\[f(t) \approx \frac{1}{4} - \frac{2}{1^2\pi^2} \cos(\pi t) - \frac{1}{1\pi} \sin(\pi t) - \frac{2}{3^2\pi^2} \cos(3\pi t) - \frac{1}{3\pi} \sin(3\pi t)\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Piecewise Function
A piecewise function is a function defined by multiple sub-functions, each applied to a particular part of the domain. For the function \( f(t) \) given in the exercise, it is defined as follows:
\[ \begin{align*} \text{If } -1 < t \leq 0, & \quad f(t) = 0, \ \text{If } 0 < t \leq 1, & \quad f(t) = t. \end{align*} \]
This means for different intervals of \( t \), the function behaves differently. These functions are very important in mathematics because they allow us to handle situations with abrupt changes or those that are non-continuous.
\[ \begin{align*} \text{If } -1 < t \leq 0, & \quad f(t) = 0, \ \text{If } 0 < t \leq 1, & \quad f(t) = t. \end{align*} \]
This means for different intervals of \( t \), the function behaves differently. These functions are very important in mathematics because they allow us to handle situations with abrupt changes or those that are non-continuous.
- The piecewise function provided is zero on one interval and linear on another.
- It is useful to visualize these functions using graphs to understand their behavior over different intervals.
Periodic Function
A periodic function is a function that repeats its values at regular intervals or periods. The function \( f(t) \) is extended periodically with a period \( T \) equal to 2. This means:
Relating it to real-world examples, think of periodic functions like the sound waves in music, which repeat consistently over time.
- Every value of \( f(t) \) repeats itself every 2 units along the \( t \)-axis.
- In terms of an equation, \( f(t + T) = f(t) \) for all \( t \).
Relating it to real-world examples, think of periodic functions like the sound waves in music, which repeat consistently over time.
Fourier Coefficients
Fourier coefficients are integral to determining how a periodic function can be expressed as a sum of sinusoids. In our exercise, the coefficients represent the weights of the sine and cosine functions in the Fourier series expansion. The coefficients \( a_0, a_n, \text{ and } b_n \) are calculated as:
\[ a_0 = \frac{1}{T} \int_{-1}^{1} f(t) \, dt, \]
\[ a_n = \frac{2}{T} \int_{-1}^{1} f(t) \cos\left(\frac{n \pi t}{L}\right) \, dt, \]
\[ b_n = \frac{2}{T} \int_{-1}^{1} f(t) \sin\left(\frac{n \pi t}{L}\right) \, dt. \]These coefficients are essential in constructing the Fourier series expansion which recreates the original function through infinite sine-cosine terms. Each coefficient gives the amplitude of its corresponding harmonic component.
\[ a_0 = \frac{1}{T} \int_{-1}^{1} f(t) \, dt, \]
\[ a_n = \frac{2}{T} \int_{-1}^{1} f(t) \cos\left(\frac{n \pi t}{L}\right) \, dt, \]
\[ b_n = \frac{2}{T} \int_{-1}^{1} f(t) \sin\left(\frac{n \pi t}{L}\right) \, dt. \]These coefficients are essential in constructing the Fourier series expansion which recreates the original function through infinite sine-cosine terms. Each coefficient gives the amplitude of its corresponding harmonic component.
- \( a_0 \) gives the average value of the function over one period.
- \( a_n \) and \( b_n \) measure the contribution of cosine and sine terms, respectively.
Integral Calculus
Integral calculus is a fundamental part of mathematics that deals with integrals, essentially capturing accumulation of quantities, such as areas under a curve. In this exercise, integral calculus is used to compute the Fourier coefficients.
This involves integrating the sub-function over specific intervals:
This involves integrating the sub-function over specific intervals:
- For \( a_0 \), the integral captures the average value of the function over its period.
- For \( a_n \) and \( b_n \), integrals calculate contributions of the cosine and sine terms, which represent harmonic frequencies.