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第14章 线性动态电路的复频域分析
一、选择题
图所示电路的谐振角频率为( )。[北京交通大学2009研]
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image697.jpg?sign=1739249729-FGvs13mGlTRBEJlJIMvUJalSbKJ3izir-0-ac1737cafa73fd4aa66da2bba922cf7f)
图14-1
A.
B.
C.
D.
【答案】B
【解析】设受控源两端电压为u,由,所以受控源等效电感为
,电感与受控源并联电感为
,所以谐振角频率为
二、填空题
1.图14-2所示电路在单位阶跃电流(mA)激励下的零状态响应
=_____。[华南理工大学2012研]
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image708.jpg?sign=1739249729-DFfLVOWfKeBJPk5cagjD4MkaIAUnjfvk-0-98597f3398655473091b3c79e8bfb8c6)
图14-2
【答案】
【解析】在复频域分析,运算电路如图14-3所示。
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image710.jpg?sign=1739249729-fp7qjjcNX5Ree0zUhleDfpy8GV45lDCW-0-2fd44d63886fc88291218b1d373a9c06)
图14-3
RC并联部分复阻抗:
电压:
零状态响应:
2.图14-4所示电路的输入阻抗=_____。[华南理工大学2012研]
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image715.jpg?sign=1739249729-9K1927wuuJg1GblOhnXWvukhCP95Ot3X-0-a0be62b2079a1ca1a541324287f15ecc)
图14-4
【答案】
【解析】
二、计算题
1.图14-5(a)所示电路中的电压的波形如图14-5(b)所示,试求电流
。[华南理工大学2012研]
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image719.jpg?sign=1739249729-BPvjUaFOhyLCMl19m1emW1qyit4KAFTE-0-1a67c7fa95b27b9480fbe99fa43ee104)
(a) (b)
图14-5
解:在复频域分析电路。
求电路复频域下的等效阻抗
电流
用拉式反变换后得:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image722.png?sign=1739249729-zUb3W0W0Ckf0ugfXblPxYadNxS0htK6z-0-190027bdc76c7789ae822883a1d4e22c)
2.图14-6所示为RLC串联电路,其中方块1、2、3表示各元件。已知t=0时电路的总储能为25(J);当时,
,
,试求R、L和C的值。[华南理工大学2012研]
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image726.jpg?sign=1739249729-LR8GQqL9dTaXm9TYq0MZaFqvCJ4N0QkR-0-651744aeced9b9be85467c5785b6ee23)
图14-6
解:t=0时刻,电路电流,则电感储能
在复频域下进行分析,有:;
元件2、3串联阻抗,则有:
。
t=0时刻电容上电压
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image733.png?sign=1739249729-tVLsfISmx9BLfeS8fq2zWqHntsPVVyu3-0-0ae7cf2d0d5b105b196cc6da1f8e2c61)
则仅有电感上储能,所以,得
。
3.已知图14-7所示电路中,,
,
。试用拉氏变换法求RLC并联电路的响应
。[华南理工大学2012研]
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image740.jpg?sign=1739249729-dvrbgsVQwIoo8ntQSjD6zPmLR9GNFyGb-0-58d3448a226c1716a9c86796dbefd302)
图14-7
解:用拉式变换法,运算电路如图14-8所示。
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image741.png?sign=1739249729-Nd3LFlDIIGoNPMznbrmmln3yPA1UgyT8-0-ccd2671176c88c8dbdf2232ec5d425e0)
图14-8
左端电流源单独作用时,电容端电压为:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image742.png?sign=1739249729-1OZAQ7riXMbH6BZDTn9S4oaSSlki9waE-0-41b61e92f742a55879fe560b36ebe1c9)
与电容串联的电压源单独作用时:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image743.png?sign=1739249729-XsNAFXiCDEREDpvKLyksxHNXckWSq0DD-0-4a3fbe649fb899c4f0721ffc03a7a122)
右端电流源单独作用时:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image744.png?sign=1739249729-tRpz52hUVrZLKBWJmYzHtxZmElYSxheC-0-ebd12d400ea56751032d6cac4e93af79)
叠加有:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image745.png?sign=1739249729-DxXX4BwR03gakjKKHdSxnD6tUlEeMkWL-0-81f99250769f60ea262222d9dfcd4a4e)
拉式反变换后有:
。
4.如图14-9所示,电路在t<0时开关S闭合且处于稳态,当t=0时开关S打开,画出其运算电路图并用运算法求u2(t)。[同济大学2007研]
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image747.jpg?sign=1739249729-f6yiO18hilw6G69veV8KC8tS2sBfu51d-0-031aebc5b09f3eb3de787fbadf0ce38e)
图14-9
解:当开关S闭合时,解耦后电路如图14-10所示,在直流电源作用下,电感可视为短路,则图中各电流为:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image748.png?sign=1739249729-nK3unhwY8WAfmHUtUIHIA4bZIOgwQzjx-0-23cd8f6500f2e1f914fdbe23b119faba)
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image749.png?sign=1739249729-YkEIyxNN1hMrMGQ71NhweYVisjFApTDY-0-40956820eb5876e52a83b937c44b2bd1)
所以
i1(0-)=2A,i2(0-)=i3(0-)=1A
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image750.jpg?sign=1739249729-zmYFD3d2Y3ars1pIykXd8hKZFIux2gjT-0-ffd9a9687bc5b8a4461971ae66c7e77b)
图14-10
当开关S打开后,运算电路如图14-11所示。
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image751.jpg?sign=1739249729-s16Bem3151BnQCAPJTo6w6TTaieGM1CD-0-6a81a561f9ac1143fbfe415aa1f4601d)
图14-11
用节点电压法可得:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image752.png?sign=1739249729-6OVVH6IQUMEKVdqIXUydurmb4QevswbI-0-90b541171ee66b95ff0950816c775ee5)
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image753.png?sign=1739249729-Hnk9V9mgryVn7VCpp8Mr43lz4noy3K6H-0-1f9a0b2816723c839d3cc7ee35a1ea38)
V
5.如图14-12所示网络电容两端原无电压,当uS为2e-t(V)时,电容两端电压uC为5e-2t,当uS为0.6e-2t时,求uC。[北京交通大学2004研]
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image755.jpg?sign=1739249729-m7Tjlb0GDu80QyASPblUmehWpXGFuHip-0-e2035bef14090b3dd52e76391437d794)
图14-12
解:令,因为线性无源电阻网络传递函数是不变的,所以
。
将0.6e-2t进行拉普拉斯变换得:
将2e-t进行拉普拉斯变换得:
将5e-2t进行拉普拉斯变换得:
所以
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image761.jpg?sign=1739249729-qWx46kbjjMlCPTVX3qZMYHrKpfF6d99E-0-e75dca88824505a07b541afe1eb1285e)
将UC2(s)进行拉普拉斯反变换得:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image762.jpg?sign=1739249729-ojGEZUawuyFIURccgMM1b5HhdCxj0PZm-0-316e19a9085cb0a0119e52e49224eb08)