?0 3?22??2?t
3?9?26?0 3?2?t
9?23?0 3?26?10??t
第三章 测试系统的基本特性
3.2 已知某测试系统静态灵敏度为4V/kg。如果输入范围为1kg到10kg，确定输出的范围。
解：
已知S＝4V/kg，x?1~10kg所以输出y?Sx?4~40(V)
3.6求周期信号x(t)?0.5cos10t?0.2cos(100t?45?)通过传递函数为H(s)?的装置后所得到的稳态响应。√ 解：
（1）由H(s)?
（2）令x1(t)?0.5cos10t，则?1?10(rad/s) 那么：A(?1)?10.005s?11，可知??0.005(s)
0.005s?111?(??1)2?11?(0.005?10)2?0.9988
?(?1)??arctan(??1)??arctan(0.005?10)??2.8624o
y1(t)?0.5A(?1)cos(10t??(?1))?0.4994cos(10t?2.8624?)
（3）令x2(t)?0.2cos(100t?45o)，则?2?100(rad/s) 那么：A(?2)?11?(??2)2?11?(0.005?100)2?0.8944
?(?2)??arctan(??2)??arctan(0.005?100)??26.5650o
y2(t)?0.2A(?2)cos(100t-45???(?2))?0.1789cos(10t?71.5650?)
（4）所以稳态响应为：
y(t)?y1(t)?y2(t)?0.4994cos(10t?2.8624?)?0.1789cos(10t?71.5650?)
3.10频率函数为
3155072的系统对正弦输入2(1?0.01j?)(1577536?176j???)x(t)?10sin(62.8t)的稳态响应的均值显示。√
解：
?H(j?)?3155072(1?0.01j?)(1577536?176j???2)112562?2??1?0.01j?12562?2?1256?0.07j??(j?)21?H1(j?)?,1?0.01j?12562H2(j?)?12562?2?1256?0.07j??(j?)2?系统为一阶系统和二阶系统串联，灵敏度S?2由x(t)?10sin(62.8t),可知x0?10,?=62.8(rad/s)1，有??0.01(s)1?0.01j?11?A1(?)???0.8469221?(??)1?(0.01?62.8)对于H1(j?)?12562对于H2(j?)?12562?2?1256?0.07j??(j?)2有??0.07,?A2(?)??n?125612????2?????1??????2???n????n?????1222?＝0.99862.82??62.8??1?()?2?0.07????1256?1256?????稳态输出的幅值为：y0?x0SA1(?)A2(?)?10?2?0.8496?0.998?16.9580?稳态输出的均值为：yrms?2/2?y0?11.9893
21.541?n3.11试求传递函数分别为和2的两环节串联后组成的系统的总灵23.5s?0.5s?1.4?ns??n敏度。√ 解： 由H1(s)?1.53?，可知K1?3
3.5s?0.57s?1241?n由H2(s)?2，可知K2?41 2s?1.4?ns??n所以，两环节串联后组成的系统的总灵敏度S?K1?K2?3?41?123