?Sg?AEcos10tsin10000t?Sg?BEcos100tsin10000t11SgAE(sin10010t?sin9990t)?SgBE(sin10100t?sin9900t)221100101001099909990Uy(f)?jSgAE[?(f?)??(f?)??(f?)??(f?)]42?2?2?2?1101001010099009900?jSgBE[?(f?)??(f?)??(f?)??(f?)]42?2?2?2??
4.5解：xa?(100?30cos?t?20cos3?t)(cos?ct)
?100cos2000?t?30cos1000?tcos2000?t?20cos3000?tcos2000?t
?100cos2000?t?15(cos3000?t?cos1000?t)?10(cos5000?t?cos1000?t)Xa(f)?50[?(f?10000)??(f?10000)]?7.5[?(f?10500)??(f?10500)]?7.5[?(f?9500)??(f?9500)]?5[?(f?11500)??(f?11500)]?5[?(f?8500)??(f?8500)]4.10 解：H(s)?
H(?)?111 ???3?s?1RCs?110s?11
10?3j??1
A(?)?11?(??)2?11?(10?)?3
?(?)??arctan(??)??arctan(10?3?)
Uy?10A(1000)sin(1000t??(1000))?10?0.707sin(1000t?450)?7.07sin(1000t?45)0
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?
??100时，A(100)?11?(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
??
?e??t;(t?0,??0)h(t)???0;(t?0)Rx(?)??h(t)?h(t??)dt??e??te??(t??)dt??0??????e0?????e?2?tedt?2????5.2 x(t)?A1sin(?1t??1??2)?A2sin(?2t??2??2)
由同频相关，不同频不相关得：
2A2Rx(?)?cos?1??cos?2?
22A125.3：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:
2?x?Rx(0)?Acos0?A
2xrms??x?A
5.7:由对称性性质:
F{sinc2(t)}?1 f???2?2?f??2
???sinc2(t)dt??2??df??
sk0机械工程测试技术基础课后题
