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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=1738880292-HlXAaB87PgpMDOtpQ54aipTfEW9JlQbs-0-3f738c9eadbf789f47de728fd8ad6698)
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=1738880292-bKmXkSxpuuA2d8KDS3gzxCuo9JI8qL6q-0-16e9c4f31ded9dcd6ba1f7e8a1fd6628)
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=1738880292-yaJoahWpeFcP6aZ0WSbk1CRrY9CrHQAl-0-3c8371146a3fa720a008d2c64cb7abed)
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=1738880292-loOPrKkQ4D9ocIGo7vnY5PD1cTd7xAev-0-f5b99712311d3f16c75d6b7c55b3883d)
各回路均相互接触,故
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image494.png?sign=1738880292-QxAtH4eq4xwy50SM09Nyp5k1qDFHnhLd-0-560ee5e93f60afaf7fd5921bc9d16674)
,
与各回路均接触,故
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image496.png?sign=1738880292-qjWf2Nt45SkyAFCQNRg462V1hIzmzGRX-0-8440c381be983c5c03de907c4b7e1127)
故由梅森公式知,传递函数为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image497.png?sign=1738880292-shse4Qn6UAxrcb51l7mcOX0bscuFrvYZ-0-7dfb59d725790f2666333efdb15a69fe)
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=1738880292-wjS1j2QKMSaUW4YpMOPySAVWBGnbUD0O-0-8fa911be80a60d29f2c191ee20453db3)
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=1738880292-8u2rrCNqfozw3NIxTBUXIDDS3n2gMSAy-0-a62840eb3375c15df67371a95a3aaab2)
(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=1738880292-IeUDOnWe8ZUfGTFAWi0TFskUeKOiSRPa-0-00573b0c4c891a60ae45f56fd3396f0f)
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=1738880292-yoAZpF8sSlM0xiKs35XnWjwwHZ31Yx4X-0-a7113e423af5dac654ecd3ad3908b513)
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=1738880292-mWiXHrqLJ7njFd9GDKQli3mAUfrWwCBy-0-a1e080b99a6774b7eb8d205d8b5619c1)
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=1738880292-EOiEflDLJMvtqmYGUtI1qnh6s2znqCki-0-6068d4f1b9ff449d5738e6bd8d0e7218)
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=1738880292-8Ey8F5v4PcIO5kQD1RxiqqSi76pYFM1i-0-ad7300498b7c93800b50f7d1cefd9d1b)
由于,因此有
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image519.png?sign=1738880292-Y7lFdEMgRSU4QgFofdNtptuOotrnFbpZ-0-ba86c51d1c5ddb15532335eff64c2259)
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=1738880292-Xf3S6wv27aaH2WnOs7P0nmMXv0gEIapp-0-3c91b4475575af786c3db45da02758c8)
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=1738880292-6IutrrMIlzLPVTFd827cfLbetxziAYeB-0-d205223cfaa2e9a6d5b4905f7993e859)
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image522.png?sign=1738880292-S8XB7H899wvf8zeGtTdfGRccsGp8lmhR-0-bb2570f295f47fcc3902fcd53c8cbeed)
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image523.png?sign=1738880292-JU808c4YvhpfC2m1GV8hfXyaBH3ivMII-0-1b4d1539b395458dade44a869390896c)
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image524.png?sign=1738880292-2ol2jYF7tI0XZyIJN4Om03k2OScIVj33-0-90ad3441f6bc544c0201e8789f1f1e8e)
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=1738880292-wZrJ6kA1GwvlvD8dsPFAd0j4yHy5NN1b-0-209c83ad41cc9d110c031363e02297e3)
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=1738880292-gbG88nhXaNdqQMM0EgBLRMAFwTjU5xJF-0-df14aa05225596f89a8c039e455135b1)
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=1738880292-6UHkvClK6FagQKNGAZuFRpzwX2FGfjGu-0-88397bbcce450dd645a20aa26250130b)
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=1738880292-QFFfgHcT0tpZTFwIEdA5ERfasf8OKGII-0-321aff5ef6567a2e80a2b3425d897150)
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=1738880292-hXU5mlUzhvboX4XQ41PGFQRzAwSFCiey-0-94b378ecc379ab1bff6786ad95199c2f)
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=1738880292-qzFpSvCoLo8alL72E1JrTVU7g3KsqSAH-0-8afcc84a28a0de7a31065362aefb63f2)
由梅森公式得
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image535.png?sign=1738880292-SUrWJ9PAATsVQ3WxT1xMVVzdsXOIy4rg-0-631b4808d241ad2c4d4db4007d3968e8)
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=1738880292-JBwwbvQrDFcjRj66Rxsj0eFMoEdayf9E-0-e32876890507c1eefee43476394fddb3)
由梅森公式得
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image538.png?sign=1738880292-7TniYPWcVX5Bd3PTyB0AObCIn17S2o7K-0-8fc2b9e0d35e5de8c38e434391277a01)
(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=1738880292-7DdSVxUOhJr3a1KZhFX8nBaxbw6P6zqe-0-90873986d68f44b10959c7101efed14e)
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=1738880292-pdcI0NY8C1jTi7R5ZnuC22U7mDyEEFdf-0-53321d16acd19621bff5e8fb3095f4bc)
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=1738880292-RMCEGDJIPhY5OVms8foBqvKayfvc98ZQ-0-71a923fcc38ecd0d06033a2977c46284)
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=1738880292-5tPfpp2IJVAWg6D1WC4bbSnI0QtQf2Qn-0-0364882a06850eaca56b13866795e9e9)
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=1738880292-NtzEVhqMuw3dyNjvZ9IEqse0yq91X0k0-0-e279274f7cc9f67c3defc84c21091035)
独立回路有四个分别为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image548.png?sign=1738880292-i1CHEb4zytvj2Qf6JhOmMSidZgp3o7Ol-0-03e1a982af335357bea34fed49365362)
故主特征式为
![](https://epubservercos.yuewen.com/E4B194/15436650205531506/epubprivate/OEBPS/Images/image549.png?sign=1738880292-GvhEFoOfDQhtUDzWPc8QOsdYUwke6HDU-0-9cd1bb0ffe52e4d1c0889569673dfc0e)
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=1738880292-gw9YtfBFsC45IHOtpb0fTsnbnSaF19Di-0-588f6753df8e9527d4a35ccf9b455143)
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=1738880292-NBN54YLTdXe7wdMW7IqIqzciRJaqsghS-0-843a26555357aa953c31f0dc0d0f46a4)