Knowee
Questions
Features
Study Tools

find the fourier series for the piecewise n(t) = 2t+1 for 0 <= t <1, 3e^(-3(t-1)) 1 <= t <2

Question

Find the Fourier series for the piecewise

n(t) = 2t + 1 for 0 <= t < 1,

3e^{-3(t-1)} for 1 <= t < 2.

🧐 Not the exact question you are looking for?Go ask a question

Solution

1. Break Down the Problem

To find the Fourier series of the piecewise function n(t) n(t) , we will:

  1. Define the function n(t) n(t) explicitly over the intervals.
  2. Determine the period of the function.
  3. Calculate the Fourier coefficients.

2. Relevant Concepts

The function is defined as: n(t)={2t+1for 0t<13e3(t1)for 1t<2 n(t) = \begin{cases} 2t + 1 & \text{for } 0 \leq t < 1 \\ 3e^{-3(t-1)} & \text{for } 1 \leq t < 2 \end{cases}

The period of the function seems to be T=2 T = 2 .

The Fourier coefficients a0,an, a_0, a_n, and bn b_n are calculated using the formulas: a0=1T0Tn(t)dt a_0 = \frac{1}{T} \int_0^T n(t) \, dt an=1T0Tn(t)cos(2πntT)dt a_n = \frac{1}{T} \int_0^T n(t) \cos\left(\frac{2 \pi n t}{T}\right) \, dt bn=1T0Tn(t)sin(2πntT)dt b_n = \frac{1}{T} \int_0^T n(t) \sin\left(\frac{2 \pi n t}{T}\right) \, dt

3. Analysis and Detail

  1. Calculate a0 a_0 : a0=12(01(2t+1)dt+123e3(t1)dt) a_0 = \frac{1}{2} \left( \int_0^1 (2t + 1) \, dt + \int_1^2 3e^{-3(t-1)} \, dt \right)

    • First integral: 01(2t+1)dt=[t2+t]01=1+1=2 \int_0^1 (2t + 1) \, dt = \left[t^2 + t\right]_0^1 = 1 + 1 = 2
    • Second integral (using substitution u=t1 u = t - 1 ): 123e3(t1)dt=013e3udu=[e3u]01=(e31)=1e3 \int_1^2 3e^{-3(t-1)} \, dt = \int_0^1 3e^{-3u} \, du = \left[-e^{-3u}\right]_0^1 = -(e^{-3} - 1) = 1 - e^{-3}
    • Now combine: a0=12(2+1e3)=3e32 a_0 = \frac{1}{2} (2 + 1 - e^{-3}) = \frac{3 - e^{-3}}{2}
  2. Calculate an a_n for n=1,2, n = 1, 2, \ldots : an=12(01(2t+1)cos(πnt)dt+123e3(t1)cos(πnt)dt) a_n = \frac{1}{2} \left( \int_0^1 (2t + 1) \cos(\pi n t) \, dt + \int_1^2 3e^{-3(t-1)} \cos(\pi n t) \, dt \right)

    • First integral (by parts or direct): 01(2t+1)cos(πnt)dt \int_0^1 (2t + 1) \cos(\pi n t) \, dt
    • Second integral: 123e3(t1)cos(πnt)dt \int_1^2 3e^{-3(t-1)} \cos(\pi n t) \, dt
  3. Calculate bn b_n for n=1,2, n = 1, 2, \ldots : bn=12(01(2t+1)sin(πnt)dt+123e3(t1)sin(πnt)dt) b_n = \frac{1}{2} \left( \int_0^1 (2t + 1) \sin(\pi n t) \, dt + \int_1^2 3e^{-3(t-1)} \sin(\pi n t) \, dt \right)

    • First integral: 01(2t+1)sin(πnt)dt \int_0^1 (2t + 1) \sin(\pi n t) \, dt
    • Second integral: 123e3(t1)sin(πnt)dt \int_1^2 3e^{-3(t-1)} \sin(\pi n t) \, dt

4. Verify and Summarize

The calculations for an a_n and bn b_n are more complex and may involve integration by parts or numerical methods for specific n n .

Final Answer

The Fourier series representation of n(t) n(t) can be given as: n(t)=a0+n=1(ancos(πnt)+bnsin(πnt)) n(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(\pi n t) + b_n \sin(\pi n t) \right) Where:

  • a0=3e32 a_0 = \frac{3 - e^{-3}}{2}
  • an a_n and bn b_n require further computation based on the integrals detailed above.

This problem has been solved

Similar Questions

Find the value of the real number 𝑎0 of the Fourier series if f(x) = 𝑥2 for 0 ≤ x ≤ 2𝜋.Question 1Select one:-8𝜋23-4𝜋234𝜋238𝜋23

The goal of this exercise is to determine the Fourier transform of the function𝑓(𝑥)={𝑥if −1<𝑥<10otherwisef(x)={ x0​ if −1<x<1otherwise​

Find the Fourier series of the odd periodic expansion of f(x) = x^2 ; 0 ≤ x ≤ π. Using the expansion, prove that 1 − 1/3^3 + 1/5^3 − 1/7^3 + · · · = π^3/32

Obtain the Fourier series of the function f x given as follows:𝑓(𝑥) = { 𝑥, 0 ≤ 𝑥 ≤ 𝜋0, 𝜋 ≤ 𝑥 ≤ 2𝜋 }

In a half range sine series f(x)=e^x,0<x<1, the Fourier coefficient bn is given by

1/1

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.