Question

求助！！！解答下这道题，谢谢！

Answer 1

我认为是y(t)=cos(3(t-1/2))

解：慎敬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/誉启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))]/宽虚慎2,
=> y(t)=[exp(j3(t-1/2))+exp(-j3(t-1/2))]/2=cos(3(t-1/2))

Answer 2

望采纳，谢谢，

Answer 3

题目在哪里？

Answer 4

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