机械工程测试技术基础第三版课后题答案
H(?)?110?3j??1
A(?)?111?(??)2?1?(10?3?)
?(?)??arctan(??)??arctan(10?3?)
Uy?10A(1000)sin(1000t??(1000))?10?0.707sin(1000t?450)?7.07sin(1000t?450)
4.11 解：A(?)?11?(??)2
?(?)??arctan(??) 1
??10时，A(10)?1?(0.05?10)?0.816
?(10)??arctan(0.05?10)?26.56?
1
??100时，A(100)?1?(0.05?100)?0.408
?(100)??arctan(0.05?100)?78.69?
y(t)?0.5?0.816cos(10t?26.56?)?0.2?0.408cos(100t?45??78.69?)?0.408cos(10t?26.56?)?0.0816cos(100t?33.69?)
5.1 h(t)?? ?e??t;(t?0,??0)?0;(t?0) R??x(?)??h(t)?h(t??)dt??????0e??te??(t??)dt ??????????0ee?2?tdt?e2?
5.2 x(t)?A1sin(?1t???1?2)?A2sin(?2t???2?2)
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机械工程测试技术基础第三版课后题答案
由同频相关，不同频不相关得：
RA21A22x(?)?2cos?1??2cos?2?
5.3：由图可写出方波的基波为x1(t)?4?sin(?t??2)
Rxy(?)?2?cos(????2)
5.4: Sxy(f)?H(f)Sx(f)
H(f)?Sxy(f)/Sx(f)
Sxy(f)?F[Rxy(?)]
Sx(f)?F[Rx(?)]?F[Rxy(??T)]?F[Rxy(?)]ej?T H(f)?e?j?T
5.5:见图5-16
5.6:由自相关函数的性质可知:
?2x?Rx(0)?Acos0?A x?2rms?x?A
5.7:由对称性性质:
F{sinc2(t)}?1 f??2?f??2
??2(t)dt?df??
??sinc2????2
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