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ID:8423268
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时间:2018-03-27
《测试技术_(第二版)课后习题答案》由会员上传分享,免费在线阅读,更多相关内容在学术论文-天天文库。
1、测试技术第一章 习 题(P29)解:(1)瞬变信号-指数衰减振荡信号,其频谱具有连续性和衰减性。(2)准周期信号,因为各简谐成分的频率比为无理数,其频谱仍具有离散性。(3)周期信号,因为各简谐成分的频率比为无理数,其频谱具有离散性、谐波性和收敛性。解:x(t)=sin2的有效值(均方根值):11解:周期三角波的时域数学描述如下:0T0/2-T0/21x(t)t......(1)傅里叶级数的三角函数展开: ,式中由于x(t)是偶函数,是奇函数,则也是奇函数,而奇函数在上下限对称区间上的积分
2、等于0。故0。因此,其三角函数展开式如下:11(n=1,3,5,…)其频谱如下图所示:0wA(w)w03w05w00ww03w05w0j(w)单边幅频谱单边相频谱(2)复指数展开式复指数与三角函数展开式之间的关系如下:C0=a0CN=(an-jbn)/2C-N=(an+jbn)/2ReCN=an/2ImCN=-bn/2故ReCN=an/2 ImCN=-bn/2 =0有11双边相频谱虚频谱实频谱0wReCnw03w05w0-w0-3w0-5w00wImCnw03w05w0-w0-3w0-5w0解:利用频移特性来
3、求,具体思路如下:11A/2A/2当f04、1==2.48×108mV/Pc解:=2s,T=150s,=2π/T11300-×100=200.35℃300+×100=399.65℃故温度变化范围在200.35~399.65℃.解:(1)则 ≤7.71×10-4S(2)j(w)=-arctgwt=-arctg()=-13.62°解:=0.04S,(1)当f=0.5Hz时,(2)当f=1Hz时,(3)当f=2Hz时,11解:=0.0025S则 w<131.5(弧度/s)或 f<w/2π=20.9Hz相位差:j(w)=-arctgwt=-arctg()=-15、8.20°解:fn=800Hz,=0.14,f=400 解:由得11第五章 习 题(P162)解:均不能提高灵敏度,因为半桥双臂灵敏度,与供桥电压成正比,与桥臂上应变片数无关。AcknowledgementsMydeepestgratitudegoesfirstandforemosttoProfessoraaa,mysupervisor,forherconstantencouragementandguidance.Shehaswalkedmethroughallthestagesofthewritingoft6、histhesis.Withoutherconsistentandilluminatinginstruction,thisthesiscouldnothavereacheditspresentform.Second,IwouldliketoexpressmyheartfeltgratitudetoProfessoraaa,wholedmeintotheworldoftranslation.IamalsogreatlyindebtedtotheprofessorsandteachersattheDepartme7、ntofEnglish:Professordddd,Professorssss,whohaveinstructedandhelpedmealotinthepasttwoyears.Lastmythankswouldgotomybelovedfamilyfortheirlovingconsiderationsandgreatconfidenceinmeallthroughtheseyears.Ialsoowemysinceregratitudetomyfriendsandmyfellowclassmateswh8、ogavemetheirhelpandtimeinlisteningtomeandhelpingmeworkoutmyproblemsduringthedifficultcourseofthethesis.MydeepestgratitudegoesfirstandforemosttoProfessoraaa,mysupervisor,forherconstantencouragementandgu
4、1==2.48×108mV/Pc解:=2s,T=150s,=2π/T11300-×100=200.35℃300+×100=399.65℃故温度变化范围在200.35~399.65℃.解:(1)则 ≤7.71×10-4S(2)j(w)=-arctgwt=-arctg()=-13.62°解:=0.04S,(1)当f=0.5Hz时,(2)当f=1Hz时,(3)当f=2Hz时,11解:=0.0025S则 w<131.5(弧度/s)或 f<w/2π=20.9Hz相位差:j(w)=-arctgwt=-arctg()=-1
5、8.20°解:fn=800Hz,=0.14,f=400 解:由得11第五章 习 题(P162)解:均不能提高灵敏度,因为半桥双臂灵敏度,与供桥电压成正比,与桥臂上应变片数无关。AcknowledgementsMydeepestgratitudegoesfirstandforemosttoProfessoraaa,mysupervisor,forherconstantencouragementandguidance.Shehaswalkedmethroughallthestagesofthewritingoft
6、histhesis.Withoutherconsistentandilluminatinginstruction,thisthesiscouldnothavereacheditspresentform.Second,IwouldliketoexpressmyheartfeltgratitudetoProfessoraaa,wholedmeintotheworldoftranslation.IamalsogreatlyindebtedtotheprofessorsandteachersattheDepartme
7、ntofEnglish:Professordddd,Professorssss,whohaveinstructedandhelpedmealotinthepasttwoyears.Lastmythankswouldgotomybelovedfamilyfortheirlovingconsiderationsandgreatconfidenceinmeallthroughtheseyears.Ialsoowemysinceregratitudetomyfriendsandmyfellowclassmateswh
8、ogavemetheirhelpandtimeinlisteningtomeandhelpingmeworkoutmyproblemsduringthedifficultcourseofthethesis.MydeepestgratitudegoesfirstandforemosttoProfessoraaa,mysupervisor,forherconstantencouragementandgu
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