Math Problem Statement
Define the functions 𝑓 , 𝑔: R → R as 𝑓 (𝑥) := 1 if0 <𝑥 < 1, 0 else, and 𝑔(𝑥) := 𝑥 if 0 < 𝑥 ≤ 1, 2 −𝑥 if1<𝑥 <2, 0 else. Compute the Fourier transform of 𝑔.
Solution
To compute the Fourier transform of the function , we will proceed by analyzing as defined in different intervals. The function is defined as:
\begin{cases} x & \text{if } 0 < x \leq 1, \\ 2 - x & \text{if } 1 < x < 2, \\ 0 & \text{elsewhere}. \end{cases}$$ ### Fourier Transform Definition The Fourier transform of $$g(x)$$, denoted $$\hat{g}(k)$$, is given by: $$\hat{g}(k) = \int_{-\infty}^{\infty} g(x) e^{-2 \pi i k x} \, dx.$$ Since $$g(x)$$ is zero outside $$(0, 2)$$, we can restrict the limits of integration: $$\hat{g}(k) = \int_0^1 x e^{-2 \pi i k x} \, dx + \int_1^2 (2 - x) e^{-2 \pi i k x} \, dx.$$ ### Step 1: Compute the First Integral We compute: $$\int_0^1 x e^{-2 \pi i k x} \, dx.$$ Using integration by parts, let: - $$u = x \Rightarrow du = dx$$, - $$dv = e^{-2 \pi i k x} dx \Rightarrow v = \frac{e^{-2 \pi i k x}}{-2 \pi i k}$$. Then, $$\int_0^1 x e^{-2 \pi i k x} \, dx = \left. \frac{-x e^{-2 \pi i k x}}{2 \pi i k} \right|_0^1 + \int_0^1 \frac{e^{-2 \pi i k x}}{2 \pi i k} \, dx.$$ Evaluating at $$x = 1$$ and $$x = 0$$, we get: $$= \frac{-e^{-2 \pi i k}}{2 \pi i k} + \frac{1}{(2 \pi i k)^2} \left( e^{-2 \pi i k x} \right)_0^1.$$ Substitute and simplify this result. ### Step 2: Compute the Second Integral Now, we compute: $$\int_1^2 (2 - x) e^{-2 \pi i k x} \, dx.$$Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Fourier Transform
Piecewise Functions
Complex Exponentials
Integration by Parts
Formulas
Fourier Transform formula: \( \hat{g}(k) = \int_{-\infty}^{\infty} g(x) e^{-2 \pi i k x} \, dx \)
Integration by parts: \( \int u \, dv = uv - \int v \, du \)
Theorems
Fourier Transform theorem for piecewise functions
Suitable Grade Level
University Level (Calculus and Fourier Analysis)
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