Math Problem Statement

[ \mathcal{F}[f(t)] = \int_{0}^{\infty} e^{-2t} \sin t \cdot e^{-i \omega t} , dt = \int_{0}^{\infty} e^{-(2 + i \omega)t} \sin t , dt ]写成公式

Solution

当然,可以将傅里叶变换的步骤写成公式的形式:

F[f(t)]=0e2tsinteiωtdt=0e(2+iω)tsintdt\mathcal{F}[f(t)] = \int_{0}^{\infty} e^{-2t} \sin t \cdot e^{-i \omega t} \, dt = \int_{0}^{\infty} e^{-(2 + i \omega)t} \sin t \, dt

其中,f(t)=e2tu(t)sintf(t) = e^{-2t} u(t) \sin t,且 u(t)u(t) 是单位阶跃函数,所以积分从 00\infty 进行。

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Math Problem Analysis

Mathematical Concepts

Fourier Transform
Exponential Decay
Unit Step Function
Sine Function

Formulas

Fourier Transform formula

Theorems

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Suitable Grade Level

Undergraduate