Question

求信号的傅里叶变换

Answer 1

∫(1,0)(1-t^2)e^(-jwt)dt=∫(1,0)e^(-jwt)dt -∫(1,0)[-t^2e^(-jwt)]dt
=(-1/jw)e^(-jwt)|(1,0) - (1/jw)∫(1,0)(t^2 de^(-jwt)
=(-1/jw)[e^(-jw)-1] -(1/jw)[t^2 e^(-jwt) - ∫(1,0) 2te^(-jwt)dt]
=[1-e^(-jw)]/(jw) -(1/jw)t^2e^(-jwt)|(1,0) + (2/jw)∫(1,0) tde^(-jwt)
=[1-e^(-jw)]/(jw) -(1/jw)[e^(-jw)] + (2/jw)[te^(-jwt)|(1,0)-∫尺做(1,0)e^(-jwt)dt]
=[1-2e^(-jw)]/(jw) + (2/jw)[e^(-jw)] +1/敬辩(jw)e^(-jwt)|(1,0)
=1/(jw) + 1/(jw)[e^(-jw) - 1]
=(1/jw)e^(-jw)
最后得到：x(t)的傅立叶变换等于 ＝∫(1,0)x(t)e^(-jwt)dt = (1/jw) e^(-jw)
请验证一下吧。亮困缺