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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=1738881853-DMWODmFY5c88ELnCuuGS9ZpthDkjuhbY-0-b1c5bd4c95a04b46f1181cf3fc05f8a0)
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=1738881853-J9AUY7sqSz7pg5qZ4N3B44UyoShwgpTx-0-ac43b8e9a2f78ef76905983931db5bcb)
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=1738881853-ZMQvQGEahRL0ETUMHuZoX7RIVqeBBjmu-0-61d49118331e8bd709476e9aae1dfe53)
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=1738881853-MabATQBmcTW3TFezVu5MzEwNAgP5Wm1G-0-0aa4206c74f6a566cc092eb82b588ba9)
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=1738881853-CozFfKyMAW39GdQt6lqKeLF1nsVY5Zou-0-5e46b75ba30e4e72d997dd65c6f4bf73)
(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=1738881853-hCnVbIos3bMRuz2TrBrbbrvC8iGkfqfn-0-c12284edef79ebab6e721713b3ef55e9)
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=1738881853-c9xiF0Vl4cPMM5wb6SCiIq2DOaML1CVx-0-00fb2c5ea121f981e5df1a5b11adc166)
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=1738881853-m5ZHCjlik13EtpCd55y4YSmpD3FPOaDa-0-cc2decd1804c4bb053f55c66121fd85c)
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=1738881853-OUhV4Qtsi64168oh9kyaIQkcgMmrVBH5-0-d9464d1b286e384d404f84bc29518579)
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=1738881853-66uC041XgzcodxpARKqVT10CaWl7QwSl-0-2deda345cadd5425226fa845d4bc38f6)
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=1738881853-daSXS636jqq27B2QtAWAy7bmSxem65ny-0-eb7c07543e97220c1b6515f2a463801f)
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=1738881853-N0cvFPBvSiltsd3z4EYfzQ3AtHNOH76o-0-b498f6b7d3190fee96a3497441bebbbb)
右端电流源单独作用时:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image744.png?sign=1738881853-D7s5krMTGkH85DKx66A6FipqENw2BIeK-0-b3ec05551518edd8b8b673b4cb6abf0e)
叠加有:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image745.png?sign=1738881853-5hkI07QDaMH8KDrbhnwAzqJikikuoISN-0-b82794583a088879d162bc4ef0b44308)
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=1738881853-adaVUk6XgOazwBxZj2eWxC5pkVvo9Bqe-0-ea0454c8bae1b90d88bf3fdf20a210e2)
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=1738881853-y9q7UCZcmN24RruYeSSgSB9KpbpwtUU1-0-985b7290ed5641ae0d10b549e35364b8)
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image749.png?sign=1738881853-2CP3eRD3yzDdMt3ZNcyHKXksAE0KPlZq-0-37ede637ca2fc40085ffb7e0ea04808c)
所以
i1(0-)=2A,i2(0-)=i3(0-)=1A
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image750.jpg?sign=1738881853-mzNoAAkK8YQDyijpiF3JNFHulNYC5QI5-0-3989025ca6ec4a08007f766a2203ca16)
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=1738881853-PYWrhUAq9GGc6A1pPsMSUkNrScCxR4cR-0-57bedb91b30a200c9d3c1a55d66b1bdb)
图14-11
用节点电压法可得:
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image752.png?sign=1738881853-xyfeAV5DvAfSdC4ktjq8Q03BKILmomPe-0-0f99d48fdcac78c570d87157d4a55f0f)
![](https://epubservercos.yuewen.com/B7597F/15436365804435306/epubprivate/OEBPS/Images/image753.png?sign=1738881853-IiKYUvXOumuA0WoLZuD4u1iJUplh9XH1-0-99abc2374cb67580182c47ac63c53475)
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=1738881853-VGSmHO18eCeR9KunL87yCYTuNgCANkXM-0-0a13619f559ad81e189ee8701a396f47)
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=1738881853-vlq0zuF7u4WdUhQxGDTZBnNo5Kl3okIy-0-4b92d5172db10e9aa894eeafcc8a91ea)
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=1738881853-xU1xoeVCDjj5FAC73goyq0weKkDNB1Ba-0-947d072b2e32fde6b39fd6f0a6007998)