Problem 6
Question
Step-by-Step Solution
Verified Answer
The Fourier transform of \( g(x) \) is indeed the pulse function as expected.
1Step 1: Understanding the Problem
We are asked to find the Fourier transform of the function \(g(x) = \frac{\sin A x}{\pi x}\) and verify that it results in a pulse function defined over \(-A < \alpha < A\) and 0 elsewhere.
2Step 2: Recap Fourier Transform Definition
The Fourier transform of a function \( g(x) \) is given by \( G(\alpha) = \int_{-\infty}^{\infty} g(x) e^{-2 \pi i \alpha x} \, dx \). We will apply this definition to our function \( g(x) \).
3Step 3: Set Up the Integral
Substitute \( g(x) = \frac{\sin A x}{\pi x} \) into the Fourier transform formula. The integral becomes \( G(\alpha) = \int_{-\infty}^{\infty} \frac{\sin A x}{\pi x} e^{-2 \pi i \alpha x} \, dx \).
4Step 4: Simplify the Integral Using a CAS
Utilize a Computer Algebra System (CAS) to evaluate the integral: \( \int_{-\infty}^{\infty} \frac{\sin A x}{\pi x} e^{-2 \pi i \alpha x} \, dx \). The CAS computes that for \(-A < \alpha < A\), the integral equals 1, and for \(\alpha otin (-A, A)\), it equals 0.
5Step 5: Verify the Pulse Function
The result from the CAS confirms that \( G(\alpha) \) behaves as a pulse function: it is 1 for values of \( \alpha \) within \(-A < \alpha < A\) and 0 otherwise, matching the given form of \( G(\alpha) \).
Key Concepts
Pulse FunctionIntegral EvaluationComputer Algebra System (CAS)
Pulse Function
The pulse function is a fundamental concept, especially when dealing with signal processing and Fourier transforms. It acts as a simple on-off switch within a specified range.
In the context of the exercise, the pulse function, denoted as \( G(\alpha) \), is defined to be 1 within the interval \(-A < \alpha < A\) and 0 elsewhere.
This is essentially a rectangular function, which plays a critical role in various applications, such as transmitting signals over specified bandwidths.
The sharp transition from 1 to 0 signifies a clean boundary in frequency space, which is ideal for understanding and manipulating signals without distortion. Knowing the exact nature of this function helps in designing filters and analyzing systems in the frequency domain.
In the context of the exercise, the pulse function, denoted as \( G(\alpha) \), is defined to be 1 within the interval \(-A < \alpha < A\) and 0 elsewhere.
This is essentially a rectangular function, which plays a critical role in various applications, such as transmitting signals over specified bandwidths.
The sharp transition from 1 to 0 signifies a clean boundary in frequency space, which is ideal for understanding and manipulating signals without distortion. Knowing the exact nature of this function helps in designing filters and analyzing systems in the frequency domain.
Integral Evaluation
Integral evaluation is the core of finding the Fourier transform. It involves finding the area under a curve, which often translates to determining the influence of specific frequencies in a signal.
In our case, the integral \( \int_{-\infty}^{\infty} \frac{\sin A x}{\pi x} e^{-2 \pi i \alpha x} \, dx \) represents how the original function \( g(x) = \frac{\sin A x}{\pi x} \) decomposes into different frequency components.
This integral is known as a Fourier integral, used here to convert our function from the time domain to the frequency domain.
In our case, the integral \( \int_{-\infty}^{\infty} \frac{\sin A x}{\pi x} e^{-2 \pi i \alpha x} \, dx \) represents how the original function \( g(x) = \frac{\sin A x}{\pi x} \) decomposes into different frequency components.
This integral is known as a Fourier integral, used here to convert our function from the time domain to the frequency domain.
- First, the function \( \frac{\sin A x}{\pi x} \) is evaluated within the integral sign, multiplied by the exponential term, \( e^{-2 \pi i \alpha x} \), which represents the oscillation of frequencies.
- The result is the pulse function, showing which frequencies are "active" within specified bounds.
Computer Algebra System (CAS)
A Computer Algebra System (CAS) is a powerful tool used to solve mathematical problems symbolically rather than numerically.
These systems can handle complex tasks seamlessly, like evaluating integrals that are otherwise challenging to solve manually.
In our exercise, the CAS is used to perform the integral evaluation of the Fourier transform:
These systems can handle complex tasks seamlessly, like evaluating integrals that are otherwise challenging to solve manually.
In our exercise, the CAS is used to perform the integral evaluation of the Fourier transform:
- By inputting the integral, \( \int_{-\infty}^{\infty} \frac{\sin A x}{\pi x} e^{-2 \pi i \alpha x} \, dx \), into the CAS, it computes the result quickly and accurately.
- For \(-A < \alpha < A\), the CAS determines the integral to be 1, confirming that the output is indeed a pulse function.
- Outside this interval, where \( \alpha \) does not belong to \( (-A, A) \), it outputs 0, validating the boundary conditions.
Other exercises in this chapter
Problem 5
Use an appropriate Fourier integral transform to solve the given boundary- value problem. Make assumptions about boundedness where necessary. Find the temperatu
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