A????1????1???????n????2?????4?2?????n?2
????2 ?12??400?2?2?400????4??0.14????1??800800????????2
?1.31
???2??????n?? ??????arctg 2???1???????n??400?2?0.14???800?? ??arctg 2?400?1???800?? ??10.6 当阻尼比改为??0.7时 A?????12??400?2?2?400????4??0.7????1??800800????????2?0.97
?400?2?0.7???800??? ??????arctg ??432?400?1???800?? 即阻尼比变化时，二阶振荡系统的输出副值变小，同时相位角也变化剧烈，相位
差变大。
2.12对一个可视为二阶系统的装置输入一单位阶跃函数后，测得其响应中产生了数值为1.5的第一个超调量峰值。同时测得其振荡周期为6.28s。设已知该装置的静态增益为3，试求该装置的传递函数和该装置在无阻尼固有频率处的频率响应。
解： 最大超调量
??????1?1??2
M?e?????1.5
6
即 ??1?2?0.13
????ln1.5???1 且 Td?2???6.28
d ∴ ??22?d?n1???6.28?1 ?1n?1??2?1.01
1??0.13?2?1 系统的传递函数 H?s??Y?s?kX?s??S22?S
?2?n??1n ?3S2
?1.01?2?2?0.13?S1.01?1该装置在无阻尼固有频率处的频率响应 由H?j???Y???X????K?2 ??j???2??j????n?????1n ?K???1?????2j??
??2n???n ∴ H?j?K3n???? ??1?????2??0.26j?????n?????2j???n ?d为有阻尼固有频率 M=0.5，?2?d?T?1 ???????? M?e?1??2?????1?2?0.215????lnM???17
? ?????2dn1 ，∴ ?n?d1??2?1.02
S=3
∴H?s???2nS2?2??2?S nS??n ?1.04S2?0.44?S?1.04?3
A??1n??34?2??6.98 （???n时代入得）
A????12?,??????90? ????n???arctg???2
y?t??6.98sin??1.02t????2?? 4.1解 ：?=2?m时，
单臂，U?Ry?4RU0 0
USg?R??y?4RU0
2?120?2?10?66
Uy?4?120*3?3?10?(V)双臂，URy??2RU0 0
USg?R??y?2RU0
U2?120?2?10?6?6y?2?120*3?6?10(V)
：?=2000?m时，
8
单臂，Uy??RU0 4R0Sg?R??4RU0
Uy?
2?120?2000?10?6Uy?*3?3?10?3(V) 4?120双臂，Uy??RU0 2R0Sg?R??2RU0
Uy?2?120?2000?10?6Uy?*3?6?10?3(V) 2?120双臂的灵敏度比单臂的提高一倍。
4.4解：Uy??RU0 R0Sg?R??RU0
Uy?Uy?Sg?(Acos10t?Bcos100t)?Esin10000t
?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)9
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)?1?s?1?1RCs?1?110?3s?1
H(?)?110?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?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) ?? Rx(?)??h(t)?h(t??)dt??????0e??te??(t??)dt
??????????0ee?2?tdt?e2?
10
??5.2 x(t)?A1sin(?1t??1?2)?A2sin(?2t??2?2)
由同频相关，不同频不相关得：
R?A212cos?A22x(?)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(?)]
Sj?Tx(f)?F[Rx(?)]?F[Rxy(??T)]?F[Rxy(?)]e 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
11