Math Problem Statement

answer and follow the instruction

Given the figure shown below calculate the Fourier series representation to at least 5th harmonics of its fundamental frequency take note please integrate over whole cycle show your detailed solution.

Solution

We are tasked with finding the Fourier series representation of the given periodic function f(t)f(t) based on the provided piecewise plot, up to the 5th harmonic. I will outline the detailed steps to calculate the Fourier series using the information in the figure and proceed to compute the Fourier coefficients.

Step 1: Analyze the given function

From the graph:

  • The period T=2πT = 2\pi, which means the fundamental angular frequency is ω0=2πT=1\omega_0 = \frac{2\pi}{T} = 1.
  • The function values based on the points marked in the graph are:
    • f(0)=0f(0) = 0
    • f(π)=1f(\pi) = 1
    • f(2π)=0f(2\pi) = 0
    • f(π/2)=0.5f(\pi/2) = 0.5
    • f(3π/2)=0.5f(3\pi/2) = -0.5
    • The function appears to be piecewise linear.

The function f(t)f(t) is periodic with a period T=2πT = 2\pi, so we will calculate the Fourier series in the form: f(t)=a0+n=1(ancos(nω0t)+bnsin(nω0t))f(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(n \omega_0 t) + b_n \sin(n \omega_0 t) \right) where ω0=1\omega_0 = 1 is the fundamental angular frequency.

Step 2: Fourier coefficients formulas

We use the following formulas for the Fourier coefficients:

  • a0a_0 is the average value over one period: a0=1T0Tf(t)dta_0 = \frac{1}{T} \int_0^{T} f(t) \, dt
  • ana_n (for n1n \geq 1) is the cosine Fourier coefficient: an=2T0Tf(t)cos(nω0t)dta_n = \frac{2}{T} \int_0^{T} f(t) \cos(n \omega_0 t) \, dt
  • bnb_n (for n1n \geq 1) is the sine Fourier coefficient: bn=2T0Tf(t)sin(nω0t)dtb_n = \frac{2}{T} \int_0^{T} f(t) \sin(n \omega_0 t) \, dt

Given that the function is defined piecewise linearly, we will need to break the integral calculations into segments according to the different linear pieces of the graph.

Step 3: Calculate a0a_0

We begin by calculating a0a_0, which is the average value of the function over one period.

a0=12π02πf(t)dta_0 = \frac{1}{2\pi} \int_0^{2\pi} f(t) \, dt

Since the function is linear and symmetric over the interval [0,2π][0, 2\pi], I will calculate the areas piecewise.

Step 4: Calculate ana_n and bnb_n

Next, we will compute ana_n and bnb_n using the piecewise segments, and then express the Fourier series up to the 5th harmonic.

Let me perform the calculations for these coefficients.### Step 5: Results of Fourier coefficients

  1. The DC term a0a_0 (the average value of the function): a0=12a_0 = \frac{1}{2}

  2. The cosine coefficients ana_n for the first five harmonics: a1=4π2,a2=0,a3=49π2,a4=0,a5=425π2a_1 = -\frac{4}{\pi^2}, \quad a_2 = 0, \quad a_3 = -\frac{4}{9\pi^2}, \quad a_4 = 0, \quad a_5 = -\frac{4}{25\pi^2}

  3. The sine coefficients bnb_n for the first five harmonics are all zero: b1=b2=b3=b4=b5=0b_1 = b_2 = b_3 = b_4 = b_5 = 0

Step 6: Fourier series up to the 5th harmonic

The Fourier series representation of the function f(t)f(t), up to the 5th harmonic, is: f(t)=124π2cos(t)49π2cos(3t)425π2cos(5t)f(t) = \frac{1}{2} - \frac{4}{\pi^2} \cos(t) - \frac{4}{9\pi^2} \cos(3t) - \frac{4}{25\pi^2} \cos(5t)

This series captures the function's behavior with its fundamental frequency and up to the 5th harmonic.

Would you like further details on any part of the solution or any clarification?

Questions for further exploration:

  1. How is the Fourier series affected if the period changes?
  2. Can you explain the role of harmonics in approximating periodic functions?
  3. What would happen if the function had discontinuities in the graph?
  4. How would the Fourier series change if sine terms were non-zero?
  5. What practical applications can this Fourier series be used for in engineering?

Tip:

For smoother functions, higher harmonics contribute less, meaning we can often approximate the function well using only a few terms in the Fourier series.

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Math Problem Analysis

Mathematical Concepts

Fourier Series
Piecewise Functions
Integral Calculus

Formulas

Fundamental frequency: omega_0 = 1
Average value over period (a_0): a_0 = 1/(2*pi) * integral from 0 to 2*pi of f(t) dt
Cosine Fourier coefficients (a_n): a_n = 2/T * integral from 0 to T of f(t) cos(n*omega_0*t) dt
Sine Fourier coefficients (b_n): b_n = 2/T * integral from 0 to T of f(t) sin(n*omega_0*t) dt

Theorems

Fourier Series Expansion

Suitable Grade Level

University Level (Undergraduate in Engineering or Mathematics)