求助 大家帮忙解答一下这道题，谢谢！题目见图片 10
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我认为是y(t)=cos(3(t-1/2))
解：x(t)=exp(j2t) -> y(t)=exp(j3t), x(t)=exp(-j2t) -> y(t)=exp(-j3t);
=> x(t)=exp(j2(t-1/2)) -> y(t)=exp(j3(t-1/2)), x(t)=exp(-j2(t-1/2)) -> y(t)=exp(-j3(t-1/2))
x(t)=cos(2(t-1/2))=[exp(j2(t-1/2))+exp(-j2(t-1/2))]/2,
=> y(t)=[exp(j3(t-1/2))+exp(-j3(t-1/2))]/2=cos(3(t-1/2))
解：x(t)=exp(j2t) -> y(t)=exp(j3t), x(t)=exp(-j2t) -> y(t)=exp(-j3t);
=> x(t)=exp(j2(t-1/2)) -> y(t)=exp(j3(t-1/2)), x(t)=exp(-j2(t-1/2)) -> y(t)=exp(-j3(t-1/2))
x(t)=cos(2(t-1/2))=[exp(j2(t-1/2))+exp(-j2(t-1/2))]/2,
=> y(t)=[exp(j3(t-1/2))+exp(-j3(t-1/2))]/2=cos(3(t-1/2))
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