+
=
从图中可以得到：
Y(jw)=g2(w+5)+g2(w+5)=2
=sa(t)gcos5t
p
解法2：
Y(jw)=g2(w+5)+g2(w+5)1
Sa(t)«g2(w)p
21
pg2(w)*p(d(w+5)+d(w-5))p2p
根据傅里叶变换的频移特性：ejw0tf(t)«F(j(w-w0))1
\e-j5tSa(t)«g2(w+5)
p1
ej5tSa(t)«g2(w-5)
p
1-j5t1j5t-1
\ y(t)=F[Y(jw)]eSa(t)+eSa(t)
pp1
=Sa(t)[e-j5t+ej5t]
p2
=Sa(t)sin(5t)
p