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Mobile App Chapter 15: Problem 25
(a) Represent as an exponential Fourier transform the function
$$
f(x)=\left\\{\begin{array}{cl}
\sin x, & 0
Short Answer
Expert verified
Represent \( \text{sin }x \) using exponential form and calculate the Fourier transform. Integrate terms and adjust to get the final result.
Step by step solution
01
Convert \(\text{sin }x\) to Exponential Form
The first step is to represent \(\text{sin }x\) in complex exponential form:\[\text{sin } x = \frac{e^{ix} - e^{-ix}}{2i}\]
02
Write Down Exponential Fourier Transform
The exponential Fourier transform of \(f(x)\) is given by:\[F(k) = \int_{-\infty}^{\infty} f(x) e^{-ikx} dx\]Since \(f(x) = 0\) for \(x otin [0, \pi]\), rewrite it for the range where \(f(x) = \text{sin }x\):\[F(k) = \int_{0}^{\pi} \frac{e^{ix} - e^{-ix}}{2i} e^{-ikx} dx\]
03
Simplify the Integral
Combine the exponents and simplify the integral:\[F(k) = \frac{1}{2i} \int_{0}^{\pi} \(e^{i(1-k)x} - e^{i(-1-k)x}\)dx\]
04
Integrate Exponential Terms
Integrate each term separately:\[\int_{0}^{\pi} e^{i(1-k)x} dx = \left[\frac{e^{i(1-k)x}}{i(1-k)}\right]_0^\pi = \frac{e^{i(1-k)\pi} - 1}{i(1-k)}\]Similarly,\[\int_{0}^{\pi} e^{i(-1-k)x} dx = \left[\frac{e^{i(-1-k)x}}{i(-1-k)}\right]_0^\pi = \frac{e^{i(-1-k)\pi} - 1}{i(-1-k)}\]
05
Combine Results
Combine the results to get the final Fourier transform:\[F(k) = \frac{1}{2i}\left(\frac{e^{i(1-k)\pi} - 1}{i(1-k)} - \frac{e^{i(-1-k)\pi} - 1}{i(-1-k)}\right)\]
06
Represent Function in Terms of Integral (Part b)
Using the inverse Fourier transform, \(f(x)\) can be represented as:\[f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(k) e^{ikx} dk\]Substitute \(F(k)\) from the above steps:\[f(x) = \frac{1}{2i\pi} \int_{-\infty}^{\infty} \left[\frac{e^{i(1-k)\pi} - 1}{i(1-k)} - \frac{e^{i(-1-k)\pi} - 1}{i(-1-k)}\right] e^{ikx} dk\]Recognize symmetry and adjust bounds of integration to get the final form:\[f(x) = \frac{1}{\pi} \int_{0}^{\infty} \frac{\cos \alpha x + \cos \alpha(x-\pi)}{1 - \-\alpha^2} d \alpha\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Exponential Form
The complex exponential form is a convenient way to express trigonometric functions using exponential functions of imaginary arguments. Specifically, Euler's formula tells us that \(e^{ix} = \text{cos}(x) + i \text{sin}(x)\). To express \(\text{sin}(x)\) in terms of exponentials, we utilize this and write:
\(\text{sin } x = \frac{e^{ix} - e^{-ix}}{2i}\)
This transformation is useful because it allows for easier manipulation in integrals, especially when dealing with Fourier transforms. By representing \( \text{sin } x \) this way, we convert a trigonometric problem into one that can be tackled using the properties and theorems of exponential functions.
\(\text{sin } x = \frac{e^{ix} - e^{-ix}}{2i}\)
This transformation is useful because it allows for easier manipulation in integrals, especially when dealing with Fourier transforms. By representing \( \text{sin } x \) this way, we convert a trigonometric problem into one that can be tackled using the properties and theorems of exponential functions.
Fourier Integral
The Fourier integral is a tool used to transform a time-domain function into its frequency-domain representation. For a function \(f(x)\), the Fourier transform \(F(k)\) is given by the integral:
\[ F(k) = \int_{-\infty}^{\infty} f(x) e^{-ikx} dx \]
This integral essentially breaks down the function \(f(x)\) into its constituent frequencies. For the given function, which is \( \text{sin } x \) in the range \( 0 < x < \pi \), this becomes:
\[ F(k) = \int_{0}^{\pi} \frac{e^{ix} - e^{-ix}}{2i} e^{-ikx} dx \]
By expressing \( \text{sin } x \) in its complex exponential form, it turns the Fourier integral into a more manageable form where the exponents can be integrated separately.
\[ F(k) = \int_{-\infty}^{\infty} f(x) e^{-ikx} dx \]
This integral essentially breaks down the function \(f(x)\) into its constituent frequencies. For the given function, which is \( \text{sin } x \) in the range \( 0 < x < \pi \), this becomes:
\[ F(k) = \int_{0}^{\pi} \frac{e^{ix} - e^{-ix}}{2i} e^{-ikx} dx \]
By expressing \( \text{sin } x \) in its complex exponential form, it turns the Fourier integral into a more manageable form where the exponents can be integrated separately.
Inverse Fourier Transform
The inverse Fourier transform is used to reconstruct the original time-domain function from its frequency-domain representation. It is given by:
\[ f(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(k) e^{ikx} dk \]
For our problem, after calculating the Fourier transform \(F(k)\), we use this formula to return to the function \(f(x)\). Plugging in the result from our Fourier integral, we get:
\[ f(x) = \frac{1}{2i \pi} \int_{-\infty}^{\infty} \left[\frac{e^{i(1-k)\pi} - 1}{i(1-k)} - \frac{e^{i(-1-k)\pi} - 1}{i(-1-k)}\right] e^{ikx} dk \]
This step ensures that the function is expressed in terms of integral bounds, thus confirming the correct inverse Fourier transform and ensuring it matches our initial function.
\[ f(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(k) e^{ikx} dk \]
For our problem, after calculating the Fourier transform \(F(k)\), we use this formula to return to the function \(f(x)\). Plugging in the result from our Fourier integral, we get:
\[ f(x) = \frac{1}{2i \pi} \int_{-\infty}^{\infty} \left[\frac{e^{i(1-k)\pi} - 1}{i(1-k)} - \frac{e^{i(-1-k)\pi} - 1}{i(-1-k)}\right] e^{ikx} dk \]
This step ensures that the function is expressed in terms of integral bounds, thus confirming the correct inverse Fourier transform and ensuring it matches our initial function.
Sin Function Representation
The original function is a piecewise one, defined as \( \text{sin } x \) for \( 0 < x < \pi \). By representing this in the Fourier domain, we ultimately aim to express it as an integral:
\[ f(x) = \frac{1}{\pi} \int_{0}^{\infty} \frac{\cos \alpha x + \cos \alpha(x-\pi)}{1-\alpha^2} d \alpha \]
This form is significant because it demonstrates how complex Fourier transforms can simplify and represent trigonometric functions in an integrated form. Notice the use of cosine functions in the final integral. Cosine naturally arises due to the even symmetry properties, converting the expression back into a more intuitive trigonometric form. Such representations are invaluable for signal processing and understanding frequency components of a given time-domain function.
\[ f(x) = \frac{1}{\pi} \int_{0}^{\infty} \frac{\cos \alpha x + \cos \alpha(x-\pi)}{1-\alpha^2} d \alpha \]
This form is significant because it demonstrates how complex Fourier transforms can simplify and represent trigonometric functions in an integrated form. Notice the use of cosine functions in the final integral. Cosine naturally arises due to the even symmetry properties, converting the expression back into a more intuitive trigonometric form. Such representations are invaluable for signal processing and understanding frequency components of a given time-domain function.
