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Solve the given problems.A pulsating electric current \(i\) (in mA) with a period of 2 s can be described by \(i=e^{-t}\) for \(-1 \leq t<1\) s for one period (only positive values of \(t\) have physical significance). Find the Fourier seriec that represents this current.

Short Answer

Expert verified
The Fourier series for the current is obtained by calculating \( a_0 \), \( a_n \), and \( b_n \) using integrals over one period, then combining them into a series form.

Step by step solution

01

Define the Periodic Function

Given the function is \( i(t) = e^{-t} \) for \( -1 \leq t < 1 \) with a period \( T = 2 \) seconds. To expand this into a Fourier series, recognize that this function can be described over one complete cycle, repeating every 2 seconds.
02

Calculate the Fourier Coefficients

For a function defined on \([-1, 1]\), the Fourier coefficients are given by:\[ a_0 = \frac{1}{T} \int_{-1}^{1} e^{-t} \, dt \]\[ a_n = \frac{2}{T} \int_{-1}^{1} e^{-t} \cos\left(\frac{n\pi t}{L}\right) \, dt \]\[ b_n = \frac{2}{T} \int_{-1}^{1} e^{-t} \sin\left(\frac{n\pi t}{L}\right) \, dt \]where \( L = 1 \), since half the period is 1.
03

Calculate the DC Component, \( a_0 \)

Calculate the integral for \( a_0 \):\[ a_0 = \frac{1}{2} \int_{-1}^{1} e^{-t} \, dt = \frac{1}{2} \left[ -e^{-t} \right]_{-1}^{1} = \frac{1}{2} (1 - e^{-2}) \]
04

Calculate the Coefficient \( a_n \)

Calculate the coefficients \( a_n \):\[ a_n = \int_{-1}^{1} e^{-t} \cos\left(n\pi t\right) \, dt \]Insert the limits and compute this integral for each \( n \). This typically requires integration by parts or direct computation using known integral tables.
05

Calculate the Coefficient \( b_n \)

Calculate the coefficients \( b_n \):\[ b_n = \int_{-1}^{1} e^{-t} \sin\left(n\pi t\right) \, dt \]Similar to \( a_n \), evaluate these integrals for each \( n \) using appropriate methods such as integration by parts or integral tables.
06

Write the Full Fourier Series

Combine the results from \( a_0 \), \( a_n \), and \( b_n \) to write the full Fourier series representation of the function. The Fourier series is:\[ i(t) = a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos\left(\frac{n\pi t}{L}\right) + b_n \sin\left(\frac{n\pi t}{L}\right) \right] \]

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Key Concepts

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

Fourier coefficients
In the quest to express periodic functions, like an electric current, as a sum of sines and cosines, Fourier coefficients are crucial. They help in converting a complicated periodic function into an infinite series that can be easily analyzed and understood. There are three types of coefficients:
  • DC Component \( a_0 \): This is the constant term in the Fourier series and represents the average value of the function over one period. In our example, it is calculated as \( a_0 = \frac{1}{2} \int_{-1}^{1} e^{-t} \, dt = \frac{1}{2} (1 - e^{-2}) \).

  • Cosine Coefficients \( a_n \): These coefficients are related to the even symmetry of the function and are calculated using the integral: \( a_n = \frac{2}{T} \int_{-1}^{1} e^{-t} \cos\left(\frac{n\pi t}{L}\right) \, dt \).

  • Sine Coefficients \( b_n \): Corresponding to the odd symmetry, these coefficients capture the variation in terms of sine functions: \( b_n = \frac{2}{T} \int_{-1}^{1} e^{-t} \sin\left(\frac{n\pi t}{L}\right) \, dt \).
Together, these Fourier coefficients enable us to express any periodic current as a series of cosine and sine functions.
Periodic function
A periodic function is one that repeats its values in regular intervals or periods. In other words, for a function \( f(t) \), it satisfies: \( f(t + T) = f(t) \) for all \( t \). The smallest positive value of \( T \) for which this holds true is called the period.

In electrical engineering, periodic functions describe various phenomena, such as alternating currents. Our given function \( i(t) = e^{-t} \), defined on the interval \( -1 \leq t < 1 \), repeats every 2 seconds. This makes it ideal for expressing through a Fourier series as it cycles regularly and does not drift over time. Recognizing and analyzing these repeating patterns allows us to simplify complex signals into predictable, oscillating parts.
Integration by parts
Integration by parts is a technique used to integrate the product of two functions. It's particularly useful when dealing with integrals that involve a product of an algebraic function and an exponential function, like in our Fourier series expansion.

The formula for integration by parts is: \(\int u \, dv = uv - \int v \, du\).Choosing the appropriate \( u \) and \( dv \) is crucial:
  • Select \( u \) to be a function that simplifies when you differentiate it.
  • Choose \( dv \) such that it can be easily integrated to find \( v \).


In our problem, to find Fourier coefficients \( a_n \) and \( b_n \), you might need to apply this technique to handle the cumbersome integrals involving exponential and trigonometric components concisely, effectively breaking down over-complicated expressions into simpler, solvable units.
Electric current function
Electric current is the flow of electric charge, and in this context, it is modeled using a mathematical function that varies over time. Specifically, the current we're working with is represented as \( i(t) = e^{-t} \). It is a decaying exponential function within a specified interval, repeating with a cycle of 2 seconds.

This current function abides by the principles of periodicity, allowing it to be expressed using Fourier series. Doing so breaks down the complex waveforms of the current into simpler sinusoidal components. This separation helps in understanding how current behaves, assisting engineers in analyzing circuits and predicting their responses to periodic inputs. By reconstructing the function using its Fourier series, insights into resonance, harmonics, and system stability are revealed, providing a more in-depth understanding of the electric current flow.

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Most popular questions from this chapter

Calculate the value of each of the given functions. Use the expansion for \(\sqrt{1+x},\) and in Exercises 23 and \(24,\) use the expansion for \(\sqrt[3]{1+x} .\) Use three terms. $$\sqrt[3]{0.9628}$$

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Solve the given problems as indicated. If term \(a_{1}\) is given along with a rule to find term \(a_{n+1}\) from term \(a_{1},\) the sequence is said to be defined recursively. If \(a_{1}=2\) and \(a_{n+1}=(n+1) a_{n},\) find the first five terms of the sequence.

Find the indicated series by the given operation. Find the first four terms of the Maclaurin expansion of the function \(f(x)=\frac{2}{1-x^{2}}\) by adding the terms of the series for the functions \(\frac{1}{1-x}\) and \(\frac{1}{1+x}\)

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