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2.3 名校考研真题详解
1.某系统结构如图2-3-1所示,为输入,
为扰动,
为输出。试求:
(1)画出系统的信号流图;
(2)求传递函数及
;
(3)说明在什么条件下,输出不受扰动
的影响。[华南理工大学2012研]
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image488.jpg?sign=1738949341-A7uCrri7WnmU47PvHmmPe03RS7pEJCsh-0-ef74dead4476885993dffeadcdeea1d3)
图2-3-1
答:(1)信号流图如图2-3-2所示。
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image489.jpg?sign=1738949341-UsNI6nLLPZhK5haR0CfpkC5qMsP62w4D-0-c0590375c2581d00176aa4acc42a7004)
图2-3-2
(2)①对于
由信号流图知有两条前向通道,增益分别为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image491.png?sign=1738949341-u6hNBLEniaYvNGivUGRZHkgHs1HkHnsZ-0-beb40f5606783f8af5794e9d3281e019)
由信号流图知有四条回路,增益分别为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image493.png?sign=1738949341-1YR8j0FqC3AweuP8ahAp5mohyddfJRK7-0-3329963464d2b1d594d842e72a72186b)
各回路均相互接触,故
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image494.png?sign=1738949341-9PtAJ1LKZsXTk7ugin9g0VAad3f5g9tm-0-3f8916fd5e12ff6c9b54f89e92968617)
,
与各回路均接触,故
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image496.png?sign=1738949341-06U7JHH8TevY5ZzbCXEDS30VHOeQukcj-0-529d447676497aa2b4098dc013eda776)
故由梅森公式知,传递函数为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image497.png?sign=1738949341-l2p9fGMj5hUCLuYrLujI5rH2O6KLm1zN-0-29366954cec323a3e5c1f229c77d3494)
②对于
由信号流图知有两条前向通道,增益分别为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image499.png?sign=1738949341-4YNYOvhXZuyX6CB3WQ8DckfyeRCFRXR1-0-b3d6d34f0436f7a8673a4d0af85ea862)
回路与上相同,与
不接触,故
;
与各回路均接触,故
;
故由梅森公式知,传递函数为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image505.png?sign=1738949341-fDSb0xEekjvekzyHnfAmXjF7lKetyJRi-0-6fc57cbffd5af3cc8dd8212b02d37d64)
(3)由可知,当
时,即
时,
,输出
不受扰动
的影响。
2.已知某系统的结构图如图2-3-3所示,求传递函数和
。[北京航空航天大学2012研]
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image513.jpg?sign=1738949341-b1xOw8zLUJoBFe4bCNL6rBt97kphT0kz-0-5dc6bf564fc428f6a08d41daee5580ef)
图2-3-3
答:由系统的结构图可知:
单独回路有2个,即
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image514.png?sign=1738949341-U6RHQcnl1BfuIZpY9ZH1vNDda7imNxUF-0-132c17f1251adee29f2656408dabf70b)
没有互不接触的回路,于是信号流图特征式为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image515.png?sign=1738949341-qDzNQPF3MVv6s1SyrHiKRLcPddFMI5RR-0-cc216f9e20f6b1c2e8d7642ce7e3053a)
从源节点R到阱节点C的前向通路共有2条,其前向通路总增益以及余子式分别为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image516.png?sign=1738949341-tE5aRBfQtu35tYtokNtq63ZlV2qszqa3-0-4398ab175a3c083424fd6257cf5235b5)
因此由梅森公式求得系统的传递函数为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image517.png?sign=1738949341-yNWWpNGRrtT5XplnhkqUkgU2kCTtPYFw-0-1a8cc17fe84a9caa1dd0c4f721f004ee)
由于,因此有
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image519.png?sign=1738949341-4a2qfkXNGlbJk95zWjiE8nNbobcMUVav-0-aad1efabe5a194bcb0f3728091d8b4a2)
3.已知某系统的结构图如图2-3-4所示,求传递函数。[北京航空航天大学2010]
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image520.jpg?sign=1738949341-6HoOuMcXgwchPJ3q7It0pQxAVgqQ4vZP-0-a6b0f35cfeabc4ed6342c33e04d53c2b)
图2-3-4
解:由系统结构图可知,本系统有三条前向通道,五个单独回路,没有互不接触回路,因此有
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image521.png?sign=1738949341-Uw2Duu8ZsGcYPb0vVv4NZetwFvlUZxZW-0-418ca12f2187eee31ecfbc66157640ff)
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image522.png?sign=1738949341-26aHoi0NENMgdnTFGwmwCCQTilEyL3k1-0-25c5903a60f5691e389e25b4aa470d37)
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image523.png?sign=1738949341-0Qahej5GYW64ARMFxkf4wGCPjki1VU5D-0-81aa83a1dfa352c295c3be2145d10aa0)
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image524.png?sign=1738949341-ghjm0VBZj2xfWT1xo8b2bdknalEMCYQx-0-13e28bd99713c72c290e0118659d18e2)
由梅森增益公式可得,传递函数为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image525.png?sign=1738949341-zlYoOIAdVearA2dAqfEFzzebA2uwC7D7-0-66b928f64731e41076a2238b560607df)
4.已知系统结构图如图2-3-5所示,图中为输入信号,
为干扰信号,
为输出信号。
(1)试求传递函数和
;
(2)若此时系统特征方程为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image528.png?sign=1738949341-hWm36oTN7EDGBLVu9bllyvDiU1m28UR0-0-e736193b425b5935eb6185c84c14c5c5)
试用劳斯判据判断系统的稳定性,系统是否有在s右半平面和虚轴上的特征根?若有,请求出这些根。[南京航空航天大学2008研]
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image529.jpg?sign=1738949341-5qLN6Zqc8nKLdRRbSmEqwTisNiwA4vTR-0-2f678c7c69e75b96fd47867ca557ba22)
图2-3-5
解:(1)由图2-3-5可知,共有3个单独回路分别为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image530.png?sign=1738949341-PIGqco4EETVJ0fKx6OBmfiNwi3O2LBXe-0-096381ad3724ec7c0dc3f4c053895605)
其中,互不接触的回路为。
则系统主特征式为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image532.png?sign=1738949341-2va6voJD6g3ciFpq2LDr1txutxzXISLL-0-fc0450f03dc1d50576801d71f73390f8)
①求
从输入到输出共有2条前向通道分别为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image534.png?sign=1738949341-NRtQVj0CgyDxsdSA4tbMlnsqF32qRG6M-0-6a4a07887d9b6b04067bc546f28c3d87)
由梅森公式得
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image535.png?sign=1738949341-V811ZFADjVexlUkFCun3wPCegQvXgYOP-0-4290b5b0d86a5f1d9ad440ea30e8b339)
②求
从干扰到输出共有3条前向通道分别为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image537.png?sign=1738949341-X96Zv0JdzgPZaIf0KSFZninl59LHTfxh-0-a297fa00a4c6c0f909feaf44a22a0027)
由梅森公式得
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image538.png?sign=1738949341-dveu8uKCsAzkGwTN7ubjEA2ThyF1HoVL-0-44c78803cb3ed360eac53f4f28739d0f)
(2)由于会出现全零行,取行系数做辅助方程
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image540.png?sign=1738949341-lUNjqDdRmzAZXLvgWz9koAJ5UfcAuYeU-0-22b241da1679a2faaa31e072e7288a84)
取辅助方程对变量s的导数,得导数方程
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image541.png?sign=1738949341-zNTa6orgQkdNaL2JM4kqoEFwUhfsMvOV-0-4a503840fb32ac18b15617243bd1a640)
用导数方程的系数取代全零行相应的元得劳斯表如下
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image542.png?sign=1738949341-t2yoQAhrf6GsXpJ6QR8MuKgOKIRadEsK-0-797d50b15b452b388c57d1aa3d8c7097)
由劳斯稳定判据可知该系统不稳定,且劳斯表中第一列符号改变两次,即有2个根在s右半平面,又劳斯表中出现全零行即有虚轴上的根。
解辅助方程,则虚轴上的根为
。
5.某负反馈控制系统框图如图2-3-6所示,试求系统的传递函数。[电子科技大学2010研]
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image546.jpg?sign=1738949341-VVVZn0GXoyckpXXz0gRejJfVgJqW5BAS-0-ba39758c8f5814be79b9c81322361c08)
图2-3-6
解:由图可知,系统包含九条前向通道分别为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image547.png?sign=1738949341-VoTd0WIjeRuSD1LESvqi2relm6COA5iS-0-ee18dc75ac94c9bd3bfda581f86425a2)
独立回路有四个分别为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image548.png?sign=1738949341-I2TczCahZosh5NLQqL63T0YDnM42ycoq-0-e0032870e37721e70f2f642b090f9ce5)
故主特征式为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image549.png?sign=1738949341-ND9P42hI1uYVfTuIBLKu9e00QgU4c0P1-0-24496488ec721cfcedb0f2361bb7bac0)
因为前向通道与各单独回路均接触,所以各余子式为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image550.png?sign=1738949341-IvsyPllJK3ZNcReSqs0Mx5tigkY2SwlG-0-e9176daa083a4e75d68cdc9c1625c386)
由梅森公式可知,系统传递函数为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image551.png?sign=1738949341-UAhCTZGK5t4SlqjuwJ1lPfjh4bbJuuAq-0-2262a5fba42ee2117c4c757c0419b5a7)