哈工大测试技术大作业(锯齿波)

HarbinInstituteofTechnology课程大作业说明书课程名称：机械工程测试技术基础设计题目：信号的分析与系统特性院系：班级：设计者：学号：指导教师：设计时间：2013/07/05哈尔滨工业大学目录1题目：...............................................................................................................12幅频谱和相频谱...............................................................................................13频率成分分布...................................................................................................33.1H(s)伯德图...............................................................................................43.1.1一阶系统伯德图.........................................................................43.1.2二阶系统伯德图.........................................................................44讨论减小失真的措施.......................................................................................54.1一阶系统对特定频率影响....................................................................54.1.1一阶系统Simulink仿真..............................................................64.2二阶系统输出响应分析........................................................................94.2.1二阶阶系统Simulink仿真..........................................................94.2.2二阶系统响应输出...................................................................10参考文献..............................................................................................................121/141题目：写出下列信号中的一种信号的数学表达通式，求取其信号的幅频谱图（单边谱和双边谱）和相频谱图，若将此信号输入给特性为传递函数为)(sH的系统，试讨论信号参数的取值，使得输出信号的失真小。（选其中一个信号）1-1信号参数名称)(sH、n、波形图锯齿波11)(ssH0.005，0.010.015，0.02等22240)(nnnsssH教师指定0002=tan,=45,=1w2KTsT假设锯齿波的斜取周期，则圆周率，A=12幅频谱和相频谱00()(+nT)(tT)wtwtKtt0将其分解为三角函数表示形式的傅里叶级数，0200-002111=(t)==2TTTawdttdtTT02000-00222()cos()cos()0TTTnawtnwtdttnwtdtTT02000-00222()sin()sin()1=(123)TTTnbwtnwtdttnwtdtTTnn、、……式中002==2wT。A0T0t)(tx2/14所以0001111(t)=(sin(wt)+sin(2wt)+sin(3wt)+223w…)转换为复指数展傅里叶级数：0000000-2021-0--1000-022220001=(t)e=e11=ee|11=e(2)TjnwtTnjnwtjnwtjnwtjnwtcwdtTtdttjnwjnwjnwnwnww其中当n=0时，01==22Ac，0=0；=1,2,3,n当…时，22111222nnnncabAn，1,2,32=1,2,32nnn等等用Matlab做出其双边频谱-20-15-10-50510152000.020.040.060.080.10.120.140.160.180.2A=1T0=1图1锯齿波双边幅频谱3/14-20-15-10-50510152000.050.10.150.2A=1;T0=1-20-15-10-505101520-2-1012图2锯齿波双边相频谱单边频谱：0246810121416182000.10.20.30.4单边幅频谱02468101214161820-1-0.500.51单边相频谱图3锯齿波单边频谱3频率成分分布由信号的傅里叶级数形式及可以看出，锯齿波是由一系列正弦波叠加而成，4/14正弦波的频率由0w到20w，30w……，其幅值由A到2A，3A，……依次减小，各频率成分的相位都为0。3.1H(s)伯德图3.1.1一阶系统1()1Hss伯德图`3.1.2二阶系统2240()2nnnHsss-40-30-20-100Magnitude(dB)10-1100101102103104-90-450Phase(deg)BodeDiagramFrequency(rad/s)=0.005-80-60-40-200Magnitude(dB)10-1100101102103104-90-450Phase(deg)BodeDiagramFrequency(rad/s)=0.1-80-60-40-200Magnitude(dB)10-1100101102103104-90-450Phase(deg)BodeDiagramFrequency(rad/s)=0.5-80-60-40-200Magnitude(dB)10-1100101102103104-90-450Phase(deg)BodeDiagramFrequency(rad/s)=0.7-150-100-50050Magnitude(dB)10-1100101102103104-180-135-90-450Phase(deg)BodeDiagramFrequency(rad/s)=0.7,_n=10-100-80-60-40-20020Magnitude(dB)10-1100101102103104-180-135-90-450Phase(deg)BodeDiagramFrequency(rad/s)=0.7,_n=385/144讨论减小失真的措施4.1一阶系统对特定频率影响频率成分由0001111(t)=(sin(wt)+sin(2wt)+sin(3wt)+223w…)构成，对于每一个频率成分，一阶系统的响应为：-/(t)=A'[sin(wt+)-esin]ty式中21'=1+()A，=-arctan()，2sin=-1+()由于T0=1s，所以0=2w。对于=0.005,0.01,0.015,0.02，w=0w，20w，30w…，A=A，2A，3A…的频率成分，可以得到其相应的响应表1幅值变化AW0.0050.10.50.70w0.31820.26950.09650.070620w0.15880.09910.02500.01830w0.10560.04970.01120.008wAT-100-80-60-40-20020Magnitude(dB)10-1100101102103104-180-135-90-450Phase(deg)BodeDiagramFrequency(rad/s)=0.7,_n=40-80-70-60-50-40-30-20Magnitude(dB)10-1100101102103104-180-135-90-450Phase(deg)BodeDiagramFrequency(rad/s)=0.7,_n=6006/14表2相角变化φW0.0050.10.50.70w-1.7994-32.1419-72.3423-77.190820w-3.5953-51.4881-85.9569-83.514330w-5.3841-62.0533-83.9434-85.66594.1.1一阶系统Simulink仿真图4一阶系统simulink方框图4.1.2一阶系统响应输出wAT7/1400.51图1 =0.00500.51图1=0.100.51图1=0.500.511.522.533.544.5500.511.5图1=0.7图5一阶系统输出对于一阶系统，为了实现近似不失真，要求1w，由上面的响应输出图像也可以看出这一结果。下图绘制出了在不同的时间常数下一阶系统对于不同的w下幅值和相位被放大和滞后的变化趋势。一阶系统的幅频和相频：1()1+()()arctan()A2Matlab程序：%%求一阶系统的幅频谱t1=[0.0050.10.50.7];forn=1:4w=0:0.01:200;A=1./sqrt(1+(t1(n)*w).^2);plot(w,A)holdonend%%求一阶系统的相频谱8/14forn=1:4w=0:0.01:200;P=-atan(t1(n)*w)/pi*180;plot(w,P)holdonend02040608010012014016018020000.10.20.30.40.50.60.70.80.91A()=0.005=0.1=0.5=0.7图6一阶系统不同常数下幅值变化020406080100120140160180200-90-80-70-60-50-40-30-20-100w(w)=0.005=0.1=0.5=0.7图7一阶系统不同常数下相角变化9/144.2二阶系统输出响应分析-3(t)=Asin(wt+)-sin(+)nwtddyet2221=[1-]+4nnA，22()=-arctan()1-()nn2=1-dw是系统在阻尼比为时（1）做有阻尼振荡时的圆频率232221-=-arctan()1-()-2nww4.2.1二阶阶系统Simulink仿真图84.2.1二阶阶系统Simulink仿真10/144.2.2二阶系统响应输出-0.100.1图1n=600,=0.7024图2n=10,=0.700.511.5图3n=38,=0.700.511.522.533.544.5500.51图4n=40,=0.7图9二阶系统在不同参数下响应对于二阶系统，为了实现近似不失真，阻尼比=(0.65~0.7)，此二阶系统取，因为此时不产生谐振A(w)曲线无峰值，输入信号中不可忽视的最高频率应小于0.6~0.8nw（），以使A(w)=1尽量接近，(w)尽量与w成线性关系。从以上图可以看出当=0.74038n，和时,二阶系统可以很好的检测锯齿波，=0.710600n，和时，锯齿波幅值和相位都有失真现象。二阶系统的幅频