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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=1738927003-zcGJkcrfluH2yMuayGAic3T3wnJJfgnA-0-d4171346a3b4d688d6d500a2d8930641)
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=1738927003-5KzvEf0Dnrh45Q5GPRnfxTn6LEv4M7IC-0-b05622e2434fa20557ae5c33423596e1)
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=1738927003-4iiOrtBTUJR5OAv5Syr1KyRCV87YKBeg-0-bac1785d1851533210a438ca5ef1dd98)
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=1738927003-FQ0tMXVQ2KX0EHyjXEQ14BNkbBl42uLL-0-ead5acfd384db254becb7e1792fa72ee)
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=1738927003-OBBIHtv5gnuGpNs1vAaFY3bpPp53sxrF-0-97958f770fbff5be4ae062bd28b29e90)
(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=1738927003-Rjg1atTuZtUgtlKP6y61QhrgPLf88rAR-0-76e9a127af477506046be2bdc914965c)
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=1738927003-qEHs2R1qssmnxuQ9lDmxt78ngjX0amEn-0-df5dd3223c40cb47473d3d5f2e988464)
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=1738927003-fXwnVIa8ZoYWJJiq7iueun08bwQ9GqYQ-0-2dee3ff89bd026bd86f4be0ca9fad80d)
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=1738927003-NSyd4NC78TiUozGsPsuSbMEehvU9O4DY-0-1ad57bbeb83a8b2d190ff33d36d2bad3)
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=1738927003-dlABPRB9i6dfCx0ueGcrEi0Ivc9RWvsF-0-19bb0ae6889819a5b0cd3d0bd232d16d)
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=1738927003-ARvPieJPqDvA5uqvr4PQAZILQxA4gMqk-0-a0c44cde35ba71018a281070e8e9ed28)
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=1738927003-IJiQphrxoGjiM1UEATEnbeXV4rWR4sJt-0-6d0ce12a5a60e09a5bf0e5e6c7d968ce)
右端电流源单独作用时:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image744.png?sign=1738927003-6jRYbeeVhcKmWLoTW1n6F4U1EmOvZc9z-0-929c6104f38df77e614ab2ecbc27ca8b)
叠加有:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image745.png?sign=1738927003-IwTF4fDFmDVFHKdp9J2JBxpPITHsPhgp-0-232193c2f8f90be1b1083c8c1c74ce3c)
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=1738927003-zclZjl15mRAPQxTw6I7FGvLgZxitglgo-0-a78d9261a43fb9c24a027ecfdc875c31)
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=1738927003-4fuxPjeJw2DOZY0H0xEZiuAY444wHWoN-0-7201ab96f8f735db3a2a660079a6a3f7)
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image749.png?sign=1738927003-xroiUEepcGzmBV4iwTocImPTw6tsdYLZ-0-4f0ff4ffb3af2a13f35568673e5586d4)
所以
i1(0-)=2A,i2(0-)=i3(0-)=1A
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image750.jpg?sign=1738927003-KHtB7ZOIaPnjb4xoWIKbtMfaB0x683Tp-0-2472444b646e790af16331e0f598f33a)
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=1738927003-mcGJ9S4deTvJrGhhCxDyMyl2rnKfJNQG-0-b1d0fdfc5366a0360e3e4a378c4b72f2)
图14-11
用节点电压法可得:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image752.png?sign=1738927003-T6jiw3FmPFnMGlYw2UTHxncBmlsfjnv4-0-21eff15c5bc4bfc33e0a2d4175cbea3c)
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image753.png?sign=1738927003-TmuTt6rTdGKz0C80wGlvrstuvWc2YlvT-0-f3d403d816c20bab5edfb697a4a60c24)
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=1738927003-nfVpoAuzjCjZmN2fJpUa8HPJkPLHlqkZ-0-5a1868b404c91f994fb54b7f101c5a81)
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=1738927003-QNio3TQAHNuRiZh6P41LmDFQQ0MJswpw-0-b863cdfa17bea7a2f02fb245ed7e4806)
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=1738927003-yz5oZNmOOwdKkh9ETN5h0oOLNa9SIzit-0-e7e05450ec562f6f92f0d1798107269b)