(t−1）×（t+1)的傅立叶变换
解：X(ω)=∫_-∞^∞x(t)e^(-jωt)dt
X(ω)=(1/2π)∫_-∞^∞ (t-1)(t+1)e^(-jωt)dt
= (1/2π)[-(2/jω)+((2/jω^3)-(2/jω))]
= (1/(2π))[-(4/(jω^2))]
= -(2/(π ω^2))
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