Problem 10
Question
Find the Fourier transforms of the functions $$ f(x)= \begin{cases}1 & \text { if }-a \leq x \leq a \\ 0 & \text { otherwise }\end{cases} $$ and $$ g(x)= \begin{cases}1-\frac{\mid x}{2} & \text { if }-a \leq x \leq a \\ 0 & \text { otherwisc }\end{cases} $$
Step-by-Step Solution
Verified Answer
The Fourier transforms of \(f(x)\) and \(g(x)\) will be complex exponential functions of \(k\), obtained by computing their respective integrals on the interval \(-a \leq x \leq a\). These integrals will need to be solved using standard rules and methods of integration.
1Step 1: Compute the Fourier Transform of Function f(x)
The Fourier transform of a function \(f(x)\) is given by the formula: \[ F(k) = \int^{+\infty}_{-\infty} f(x) \; e^{-2\pi ikx} \; dx \] But our function is a piecewise function that is only non-zero in the interval \(-a \leq x \leq a\). So, the computation is: \\[ F_f(k) = \int^{+a}_{-a} e^{-2\pi ikx} \; dx\]Which can be solved using standard methods of integration.
2Step 2: Compute the Fourier Transform of Function g(x)
Next, compute the Fourier transform of function \(g(x)\). Again, this function is a piecewise function that is only non-zero in the interval \(-a \leq x \leq a\). So, the computation becomes:\[ F_g(k) = \int^{+a}_{-a} (1-\frac{|x|}{2}) e^{-2\pi ikx} \; dx\]This integral is more complex due to the \(|x|\) part of the function. It would need to be split into two parts and solved accordingly, one for the range \(-a \leq x \leq 0\) and the other for the range \(0 \leq x \leq a\).
3Step 3: Evaluate the Results
Finally, evaluate the integrals to obtain the Fourier transforms of the functions f(x) and g(x). Solve the integrals using standard integration rules and methods, which should result in complex exponential functions of \(k\), which denote the frequency domain representation of our original functions.
Key Concepts
Piecewise FunctionsIntegration MethodsFrequency Domain
Piecewise Functions
A piecewise function is one that has different expressions for different intervals of the input. These functions are commonly used in mathematics to represent phenomena with distinct behaviors over specified regions. This can include functions defined by segments, like in our exercise where:
When dealing with piecewise functions in a Fourier transform context, focus is given to determining how each individual piece contributes to the overall transformation.
- For function \(f(x)\), the value is 1 for \(-a \leq x \leq a\) and 0 otherwise.
- For function \(g(x)\), the value gradually changes according to \(1 - \frac{|x|}{2}\) within the same interval, but is 0 outside \([-a, a]\).
When dealing with piecewise functions in a Fourier transform context, focus is given to determining how each individual piece contributes to the overall transformation.
Integration Methods
Integration is the process of finding the integral of a function, and it's an essential part of computing the Fourier transform. For piecewise functions, integration can sometimes be straightforward or slightly complex, depending on the mathematical functions involved.
In our exercise, we can see:
In some cases, numerical methods might also come into play especially when manual integration proves to be computationally heavy or exhaustive.
In our exercise, we can see:
- For \(f(x)\), the integration is less complex within the bounds \(-a \leq x \leq a\), because it's a constant piecewise function.
- For \(g(x)\), we encounter the absolute value \(|x|\), necessitating dividing the integral into two segments: one from \(-a\) to 0, and another from 0 to \(a\).
In some cases, numerical methods might also come into play especially when manual integration proves to be computationally heavy or exhaustive.
Frequency Domain
The frequency domain represents a crucial aspect of Fourier transforms. It's a way to view and understand signals in terms of their frequency components rather than time or space.
When you apply the Fourier transform to a function, like \(f(x)\) or \(g(x)\), you essentially break down a signal into sinusoidal (wave-like) components, each with a different frequency and amplitude.
The result of the transform — represented usually by \(F(k)\) — depicts the signal's behavior across the frequency spectrum:
In essence, moving from the time domain to the frequency domain can simplify the analysis and manipulation of complex functions or signals.
When you apply the Fourier transform to a function, like \(f(x)\) or \(g(x)\), you essentially break down a signal into sinusoidal (wave-like) components, each with a different frequency and amplitude.
The result of the transform — represented usually by \(F(k)\) — depicts the signal's behavior across the frequency spectrum:
- \(F(k)\) provides insight into all the frequencies present in the original function.
- By analyzing the magnitude and phase of these frequencies, one can comprehend how drastically each frequency impacts the overall waveform.
In essence, moving from the time domain to the frequency domain can simplify the analysis and manipulation of complex functions or signals.
Other exercises in this chapter
Problem 8
Show that for a monotone function \(f(x)\) such that \(f(\pm \infty)=\pm \infty\) with \(f(a)=0\) $$ \int_{-\infty}^{\infty} \varphi(x) \delta^{\prime}(f(x)) \m
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Show the identities $$ \frac{d}{d x}(\delta(f(x)))=f^{\prime}(x) \delta^{\prime}(f(x)) $$ and $$ \delta(f(x))+f(x) \delta^{\prime}(f(x))=0 $$ Hence show that \(
View solution Problem 12
Evaluate Fourier transforms of the following distributional functions: (a) \(\delta(x-a)\). (b) \(\delta^{\prime}(x-a)\). (c) \(\delta^{(n)}(x-a)\). (d) \(\delt
View solution Problem 13
Prove that $$ x^{m} \delta^{(n)}(x)=(-1)^{m} \frac{n !}{(n-m) !} \delta^{(n-m)}(x) \quad \text { for } n \geq m $$ Hence show that the Fourier trarsform of the
View solution