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Magazine Chapter 4: Problem 3
Suppose \(f(t)\) is defined on \([-\pi, \pi]\) as \(\sin (5 t)+\cos (3 t) .\) Extend periodically and compute the Fourier series of \(f(t)\).
Short Answer
Expert verified
The Fourier series of \(f(t)\) is \(\cos(3t) + \sin(5t)\).
Step by step solution
01
Identify the function and its properties
The function given is \(f(t) = \sin(5t) + \cos(3t)\), which is defined on the interval \([-\pi, \pi]\). We need to periodic extend it, which means \(f(t + 2\pi) = f(t)\) for all \(t\). Next, compute its Fourier series.
02
Determine the Fourier series formula
A periodic function \(f(t)\) with period \(2\pi\) can be expressed as a Fourier series: \(f(t) = a_0 / 2 + \sum_{n=1}^{\infty} \left(a_n \cos(nt) + b_n \sin(nt)\right)\). Compute \(a_n\) and \(b_n\) using the formulas \(a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(t) \cos(nt) \, dt\) and \(b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(t) \sin(nt) \, dt\).
03
Compute the Fourier coefficients - \(a_0\)
Since the function involves only sine and cosine terms, \(a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} f(t) \, dt = 0\). This is because integrating one complete period of sine and cosine over their respective intervals results in zero.
04
Compute the cosine coefficients \(a_n\)
Calculate \(a_n\) for \(n > 0\):\[\begin{align*}a_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} (\sin(5t) + \cos(3t)) \cos(nt) \, dt\&= \frac{1}{\pi} \left( \int_{-\pi}^{\pi} \sin(5t) \cos(nt) \, dt + \int_{-\pi}^{\pi} \cos(3t) \cos(nt) \, dt \right)\end{align*}\]Using orthogonality properties, only the term where \(n = 3\) contributes (as \(\cos^2(3t)\) integrates to \(\pi\) while others result in zero), so \(a_3 = 1\).
05
Compute the sine coefficients \(b_n\)
Calculate \(b_n\) for \(n > 0\):\[\begin{align*}b_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} (\sin(5t) + \cos(3t)) \sin(nt) \, dt\&= \frac{1}{\pi} \left( \int_{-\pi}^{\pi} \sin(5t) \sin(nt) \, dt + \int_{-\pi}^{\pi} \cos(3t) \sin(nt) \, dt \right)\end{align*}\]Similarly, only the term where \(n = 5\) contributes (as \(\sin^2(5t)\) integrates to \(\pi\) while others result in zero), so \(b_5 = 1\).
06
Final Step: Construct the Fourier series
With \(a_3 = 1\) and \(b_5 = 1\), the Fourier series becomes:\[ f(t) = \cos(3t) + \sin(5t) \]Thus, the function is directly expressed as its Fourier series with only the contributing terms.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Periodic Extension
When dealing with functions like \( f(t) = \sin(5t) + \cos(3t) \), initially defined on a specific interval such as \([-\pi, \pi]\), the concept of periodic extension plays a crucial role. To form a periodic function, we take this original definition and extend it infinitely in both directions. This means generating identical copies of \( f(t) \) every \( 2\pi \) interval.
- In essence, periodically extending a function means that the value of the function repeats itself at regular intervals. For our function \( f(t + 2\pi) = f(t) \).
- This seamless repetition ensures the function remains defined over every real number \( t \), not just the initial interval.
- The extended version retains the original function's periodicity, which is key for further analysis, particularly when calculating the Fourier series.
Fourier Coefficients
Fourier coefficients are essential components in building a Fourier series. In mathematics, they help express any periodic function as a sum of sine and cosine terms. To find these coefficients for \( f(t) = \sin(5t) + \cos(3t) \), we use specific integral formulas.
The key coefficients are \( a_n \) and \( b_n \):
The key coefficients are \( a_n \) and \( b_n \):
- \( a_0 \) is the average value. For pure sine and cosine functions like ours, \( a_0 = 0 \).
- Cosine coefficients \( a_n \) are calculated as \( \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \cos(nt) \ dt \).
- Sine coefficients \( b_n \) are calculated as \( \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin(nt) \ dt \).
- \( a_3 = 1 \): Contribution from \( \cos(3t) \)
- \( b_5 = 1 \): Contribution from \( \sin(5t) \)
Orthogonality Properties
Orthogonality is a fundamental property of trigonometric functions that simplifies the computation of Fourier coefficients. When integrated over \([-\pi, \pi]\), products of different sine or cosine functions yield zero, unless they're identical.
Using these properties:
Using these properties:
- \( \int_{-\pi}^{\pi} \sin(mt) \sin(nt) \ dt = 0 \) for \( m eq n \)
- \( \int_{-\pi}^{\pi} \cos(mt) \cos(nt) \ dt = 0 \) for \( m eq n \)
- Only \( \int_{-\pi}^{\pi} \cos^2(nt) \ dt = \pi \) and \( \int_{-\pi}^{\pi} \sin^2(nt) \ dt = \pi \) contribute for terms where \( m = n \).
- \( a_3 \) arises only from \( \cos(3t) \)
- \( b_5 \) arises only from \( \sin(5t) \)
Trigonometric Functions
Trigonometric functions, namely sine and cosine, are foundational in expressing periodic behavior in functions. For the function \( f(t) = \sin(5t) + \cos(3t) \), understanding their properties is vital.
These functions are naturally periodic:
These functions are naturally periodic:
- \( \sin(n t) \) and \( \cos(n t) \) both have period \( \frac{2\pi}{n} \).
- They are commonly used in Fourier series due to their repeating nature.
- Their integration within a full period often results in valuable orthogonality properties, beneficial in coefficient calculation.
