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电子电路基础第六章习题答案

第六章习题6. 1求习题图6・1所示的电路的传递函数H(a)) = V o/V t•习题图6. 1•• RH ——解：（%-匕）-MC 八V jwLjcoL - 3、RLCR + jcoL-cerRLCjcoCR一ja）CR — e f CL6・2对于习题图6. 2所示的电路，求传递函数二丿6. 3串联RLC网络有R二5G, L二10川H, C二1 “F,求该电路的谐振角频率.特征阻抗和品质因数。

当外加电压有效值为24V时，求谐振电流、电感和电容上的电压值。

解：电路的谐振角频率卧忌一皿特征阻抗0二姑= 100。

品质因数0二二20谐振电流人吕十A电感和电容上的电压值U L = U c = U a Q = 480V 6・4设计一个串联RLC电路，使其谐振频率q 二50m〃/“品质因数为80,且谐振时的阻抗为10Q,并求英带宽。

解:B 二色=0. 625rad / 56. 5对于习题图6. 5所示的电路，求和，（/）为同相时的频率解••阶S叫+盏血5将厶二1乩厶二\H、C =一"+ y(w_ ♦W 1 + M rIQ F卜I w谐振时虚部为零，沙——+——二0W 1 +讥厂得出，W二0・7861 6. 6并联RLC网络有R二50G, L二4〃M・C二160 “八求并联电路谐振频率和品质因数。

若外接电流源有效值为2A,求谐振时电阻、电感及电容上的电流值。

解：电路的谐振角频率％二二I・25xl0rad/s4LC品质因数Q二毬CR ==10谐振时电阻、电感及电容上的电流值h二2AJ L二Ic二I K・Q = 2OA6. 7并联谐振电路,其品质因数为120,谐振频率是6x10%/ 〃/ “计算其带宽。

6・8计算习题图6. 8所示的电路的谐振角频率叫，品质因数Q和带宽Bo3好20m H 冒2jfcQ i6“戸〒习题图6・81 1 1 c C 1解：y 二 >・(5//C2)+一+— = 一+7 (=_)谐振时Y的虚部为0沙• g-一丄=0 C] + C、wL得出w =Q = qRC = CD O R(C( //CJ 二20B二八二2^0rad/.y 6. 9习题图6. 9所示的电路，已知电容值C为固圧，欲使电路在© 时发生并联谐振，而在©时发生串联谐振，求厶、乙的值。

清华大学电子电路与系统基础(2)第一讲

• 鼓励参加
– 一些思路的探讨，对理论课的加强
李国林电子电路与系统基础清华大学电子工程系 2016年秋季学期 3
运算放大器大纲
• 运算放大器及其外端口特性
• 理想运算放大器特性
• 负反馈线性应用
– 四种负反馈放大器 – 其他…
• 非线性应用
– 开环 – 闭环
• 负反馈 • 正反馈
李国林电子电路与系统基础
二阶模型的解
Rout R2 Rin

R2 R1 1
一阶模型的解
Av 0
Rin , Rout 0
Av 0 1
理想模型的解 R2 Rin , Rout 0 R1 Av 0 仅和外接电阻有关，和运放无关 20 这是我们希望的：随便换个运放，得到相同的结果
vid
O
vip
二端口网络描述方程
vid
vout Vsat
vid vin
负饱和区等效电路
李国林电子电路与系统基础
Vsat
iid 0 vout Vsat
vid
Vsat
10
二端口网络描述方程
清华大学电子工程系 2016年秋季学期
线性区外端口特性等效电路
vn1
1 vout Av 0
vout
舍入误差 vn1 0.0001 vin v out 9.9994
李国林电子电路与系统基础
R2 R1
R 1 vin 2 vin 10vin R1 1 R2 1 1 公式变得极度简单 Av 0 R 1
– 助教联系方式和作业批改班级分配情况见网络学堂
李国林电子电路与系统基础清华大学电子工程系 2016年秋季学期 2

清华大学电子电路第五讲

电子电路与系统基础
理论课第五讲单端口线性网络的等效电路和线性受控源
李国林闻和清华大学电子工程系
等效电路补充说明本节考察单端口线性网络
等效电路
端口特性等效电路范畴
两者具有完全一致的端口描述方程（端口特性）如纯电阻网络、线性电源等
功能特性等效器件范畴
两者端口特性不完全相同，但近似，具有一致的功能特性如理想电源、各种形式电阻、二极管等效电路模型等
itest
vtest
李国林电子电路与系统基础
P
R1 R3 R R R3 R4 2 1
v s
17
解的解析（1）
R1
vs
P
R3
vtest
itest
R1 || R2 R3 || R4
R2
R4
P
R1 R3 R R R3 R4 2 1
i
v v1 v 2
i
v1 v R1 R R1 R2 i0 O
iR iR1 iR2
v i0 R1 i0 R2 i0 R
v2
R2
串联同一电流下电压相加
单端口纯电阻网络（不含独立源）等效电路？ …… i
i
i1 v G1
i2 G2 G G1 G 2
i i1 i2 v vG vG1 vG 2
i1 i3
1 R2 vs itest R1 R2 R1 R2 1 R4 vs itest R3 R4 R3 R4
i3 R3 i3 itest R4 vs
i1 R1 i3 R3 vtest 0
李国林电子电路与系统基础
请同学自行练习用叠加定理计算获得最终结果

电子电路基础习题册参考答案-第一章

电子电路基础习题册参考答案（第三版）全国中等职业技术第一章常用半导体器件§1-1 晶体二极管一、填空题1、物质按导电能力的强弱可分为导体、绝缘体和半导体三大类，最常用的半导体材料是硅和锗。

2、根据在纯净的半导体中掺入的杂质元素不同，可形成N 型半导体和P 型半导体。

3、纯净半导体又称本征半导体，其内部空穴和自由电子数相等。

N型半导体又称电子型半导体，其内部少数载流子是空穴；P型半导体又称空穴型半导体，其内部少数载流子是电子。

4、晶体二极管具有单向导电性，即加正向电压时，二极管导通，加反向电压时，二极管截止。

一般硅二极管的开启电压约为0.5 V，锗二极管的开启电压约为0.1 V；二极管导通后，一般硅二极管的正向压降约为0.7 V,锗二极管的正向压降约为0.3 V。

5.锗二极管开启电压小，通常用于检波电路，硅二极管反向电流小，在整流电路及电工设备中常使用硅二极管。

6.稳压二极管工作于反向击穿区，稳压二极管的动态电阻越小，其稳压性能好。

7在稳压电路中，必须串接限流电阻，防止反向击穿电流超过极限值而发生热击穿损坏稳压管。

8二极管按制造工艺不同，分为点接触型、面接触型和平面型。

9、二极管按用途不同可分为普通二极管、整流二极管、稳压二极管、开关、热敏、发光和光电二极管等二极管。

10、二极管的主要参数有最大整流电流、最高反向工作电压、反向饱和电流和最高工作频率。

11、稳压二极管的主要参数有稳定电压、稳定电流和动态电阻。

12、图1-1-1所示电路中，二极管V1、V2均为硅管，当开关S与M 相接时，A点的电位为无法确定V,当开关S与N相接时，A点的电位为0 V.13图1-1-2所示电路中，二极管均为理想二极管，当开关S打开时，A点的电位为10V 、流过电阻的电流是4mA ；当开关S闭合时，A点的电位为0 V，流过电阻的电流为2mA 。

14、图1-1-3所示电路中，二极管是理想器件，则流过二极管V1的电流为0.25mA ，流过V2的电流为0.25mA ,输出电压U0为+5V。

清华电子系电子电路(2)第一讲习题课电路抽象

I out I in Vin
B
Vout
ib
E
NPN BJT
D
G
这里，我们以 BJT 的发射极，以 MOSFET 的源极作为公共端，构成二端口网络
• 一句话：电路设计中，没有掌握极致化原则的，会淹没在一大堆混乱的公式或混乱的思维中，没有头绪，不知如何进行电路设计 • 而掌握了极致化原则的，电路设计将变得十分的简单
– 因为我们抓住了主要矛盾
电路基础
清华大学电子工程系李国林
20
限定性原则
• 不管是离散化原则，还是极致化原则，抽象最根本的思想就是抓住主要矛盾，舍弃次要矛盾
5
考评
• 期末考试 • 期中考试 • 作业 45分 40分 15分
– 不要抄袭，理解之后自己做
• 关于作业
– 本周布置的作业，下周上课时提交 – 本周提交的作业，下周上课时取走 – 助教批改，随时汇报作业问题，习题课教师汇总讲解
李国林电子电路与系统基础清华大学电子工程系 2014年秋季学期 6
课程内容规划
一阶RC、RL电路；滤波、延时、移相、积分、微分电路、张弛振荡、…
二阶RLC谐振电路；二阶滤波、延时；匹配，阻抗变换；正弦波振荡器、… 触发器，…
谐振振荡第二学期下半学期
第二学期动态特性上半学期 (6)电容、电感基本分析方法 (5)电路抽象
分压器、分流器、衰减器、电桥、理想放大器、理想变压器、理想回旋器、…
教材
• 讲义
– 2012版
• 根据2011年小班试讲课件与录像资料整理而成
– 2013版
• 根据2011年小班试讲、大班推广实践整理而成
– 上册430页，每周30页内容 – 下册490页，每周30页内容

电子行业-电子电路基础第六章习题答案精品

第六章习题6.1 求习题图6.1所示的电路的传递函数()/o i H V V ω=。

习题图6.1解：1//()i o oR V V jwCjwLV -=22()oi V j L RLCH R j L RLCV ωωωωω-==+- 6.2 对于习题图6.2所示的电路，求传递函数()o iI H I ω=。

习题图6.2解：2()11o iI R j CRH j CR CL I jwL R jwCωωωω===-+++ 6.3 串联RLC 网络有R=5Ω，L=10mH ,C=1F μ,求该电路的谐振角频率、特征阻抗和品质因数。

当外加电压有效值为24V 时，求谐振电流、电感和电容上的电压值。

解：电路的谐振角频率40110/rad s LCω== 特征阻抗100LCρ==Ω 品质因数020LQ Rω==谐振电流0 4.8mU I A R== 电感和电容上的电压值L 480V C m U U U Q ===6.4 设计一个串联RLC 电路，使其谐振频率050/rad s ω=，品质因数为80，且谐振时的阻抗为10Ω，并求其带宽。

解：00.625rad /B s Qω==6.5 对于习题图6.5所示的电路，求()v t 和()i t 为同相时的频率ω。

习题图6.5解：12()1Z (//)()v t jwL R L i t jwC==++ 121,1,1,1L H L H C F R ====Ω将代入2221Z ()11w w j w w w w-=+-+++谐振时虚部为零，2101w w w w -+=+ 0.7861w =得出,6.6 并联RLC 网络有R=50Ω，L 4mH =,C=160F μ,求并联电路谐振频率和品质因数。

若外接电流源有效值为2A ，求谐振时电阻、电感及电容上的电流值。

解：电路的谐振角频率3011.2510rad /s LCω==⨯ 品质因数010LQ CR RCω=== 谐振时电阻、电感及电容上的电流值2A,20A R L C R I I I I Q ====6.7 并联谐振电路，其品质因数为120，谐振频率是6610/rad s ⨯，计算其带宽。

数字电路与系统设计课后习题答案

解：略
4.8在双轨输入信号下，用最少或非门设计题4.7的组合电路。
解：将表达式化简为最简或与式：
(1)F=(A+C)(A+B+C)= A+C+A+B+C
(2)F=(C+D)(B+D)(A+B+C)= C+D+B+D+A+B+C
(3)F=(A+C)(A+B+D)(A+B+D)=A+C+A+B+D+A+B+D
解：F=AB+BC+AC
或：F=AB+AC+BC
(4) F=ACD+BC+BD+AB+AC+BC
解：F=AD+C+B
(5) F=AC+BC+B(AC+AC)
解：F=AC+BC
2.12用卡诺图把下列函数化简为最简与或式
（1）F(A,B,C)=m(0,1,2,4,5,7)
解：F=B+AC+AC
图略
（2）F(A,B,C,D)=m(0,2,5,6,7,9,10,14,15)
(3)F（ABC）=∏M(1,3,4,5,7)
2.10试写出下列各函数表达式F的F和F的最小项表达式。
（1）F=ABCD+ACD+BCD
（2）F=AB+AB+BC
解：（1）F=∑m(0,1,2,3,5,6,7,8,9,10,13,14)
F'=∑m(1,2,5,6,7,8,9,10,12,13,14,15)

清华电路基础答案第16章

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators Chapter 16, Problem 1.Determine i (t ) in the circuit of Fig. 16.35 by means of the Laplace transform.Figure 16.35For Prob. 16.1.Chapter 16, Solution 1.Consider the s-domain form of the circuit which is shown below.222)23()21s (11s s 1s 1s 1s 1)s (I ++=++=++=⎟⎟⎠⎞⎜⎜⎝⎛=t 23sin e 32)t (i 2t -=)t (i A )t 866.0(sin e 155.1-0.5t1/s1sPROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsFind v x in the circuit shown in Fig. 16.36 given v s .= 4u (t )V.Figure 16.36For Prob. 16.2.PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsV )t (u )e 2e 24(v 38j 34s 125.038j 34s 125.0s 25.016)8s 8s 3(s 2s 16V s 32s 16)8s 8s 3(V 0V s V )s 4s 2(s)32s 16()8s 4(V 0s840V 20V s s 4V t )9428.0j 3333.1(t )9428.0j 3333.1(x 2x 2x x 2x 2x x x x −−+−++−=⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎝⎛−+−+++−+−=+++−=+=++=++++−+=+−+−+−v x = V t 322sin e 26t 322cos e )t (u 43/t 43/t 4⎟⎟⎠⎞⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛−−−4s 8/s s 4PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsFind i (t ) for t > 0 for the circuit in Fig. 16.37. Assume i s = 4u (t ) + 2δ(t )mA. (Hint: Canwe use superposition to help solve this problem?)Figure 16.37For Prob. 16.3.PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsIn the s-domain, the circuit becomes that shown below.42s+We transform the current source to a voltage source and obtain the circuit shown below.284s +84204030.2(15)15s A B s I s s s s s ++===++++40815204052,153153x A B −+====− 8/352/315I s s =++ 15852()()33t i t e u t −⎡⎤=+⎢⎥⎣⎦PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsThe capacitor in the circuit of Fig. 16.38 is initially uncharged. Find v 0(t ) for t > 0.Figure 16.38For Prob. 16.4.Chapter 16, Solution 4.The circuit in the s-domain is shown below.5451/o o V I I I sV s+=⎯⎯→= But 52o V I −=512.5525/2o o o V sV V s −⎛⎞=⎯⎯→=⎜⎟+⎝⎠2.5()12.5 V t o v t e −=PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators If i s (t ) = e t 2−u (t ) A in the circuit shown in Fig. 16.39, find the value of i 0(t).Figure 16.39For Prob. 16.5.Chapter 16, Solution 5.()()A )t (u t 3229.1sin 7559.0e orA)t (u e e e 3779.0e e e 3779.0e )t (i 3229.1j 5.0s )646.2j )(3229.1j 5.1()3229.1j 5.0(3229.1j 5.0s )646.2j )(3229.1j 5.1()3229.1j 5.0(2s 1)3229.1j 5.0s )(3229.1j 5.0s )(2s (s 2Vs I )3229.1j 5.0s )(3229.1j 5.0s )(2s (s 22s s s 22s 12s 21s 112s 1V t 2t 3229.1j 2/t 90t 3229.1j 2/t 90t 2o 222o 2−=++=−++++−+++−−−−++=−++++==−++++=⎟⎟⎠⎞⎜⎜⎝⎛+++=⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎝⎛+++=−−°−−°−−or i o (t) = A )t (u t 27sin 72e t 2⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛−−s 2 2s 1+ I oPROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsFind v (t ), t > 0 in the circuit of Fig. 16.40. Let v s =20 V.Figure 16.40For Prob. 16.6.Chapter 16, Solution 6.For t<0, v(0) = v s = 20 VFor t>0, the circuit in the s-domain is as shown below.1101000.1mF F sC s=⎯⎯→= 20210110s I s s==++ 20101V I s ==+ ()20()t v t e u t −=10 Ω 20sPROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsFind v 0(t ), for all t > 0, in the circuit of Fig. 16.41.Figure 16.41For Prob. 16.7.PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators The circuit in the s-domain is shown below. Please note, i L (0) = 0 and v o (0) = o becauseboth sources were equal to zero for all t<0.1 111112/2(21/)11o o V V s V V V s V s s−−=+⎯⎯→=+− (1) At node O,111(1/2)1/12/2o o o o V V V s V V s V s s s −+==⎯⎯→=+− (2) Substituting (2) into (1) gives112/(21/)(1/2)(2)o o s s s V V s s=++−+− 22(41)( 1.51) 1.51o s A Bs C V s s s s s s ++==+++++2241( 1.51)s A s s Bs Cs +=++++ We equate coefficients.s 2 :0 = A+ B or B = - A s: 4=1.5A + Cconstant: 1 = A, B=-1, C = 4-1.5A = 2.52222241 2.513/441.51(3/4)(3/4)44o x s s V s s s s s s −++=+=−+++⎛⎞⎛⎞++++⎜⎟⎜⎟⎝⎠⎝⎠3/43/4()()cos 4.9135sin 44t t v t u t e t e t −−=−+PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsIf v 0(0) = -1V,obtain v 0(t ) in the circuit of Fig. 16.42.Figure 16.42For Prob. 16.8.PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators 1122F sC s⎯⎯→= We analyze the circuit in the s-domain as shown below. We apply nodal analysis.0314140211(4)o o o V V V s s s s V s s s−−−−−++=⎯⎯→=+4o A B V s s =++14187/2,9/244A B ====−− 7/29/24o V s s =−+ 479()()22t o v t e u t −⎛⎞=−⎜⎟⎝⎠4s 3sPROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators Find the input impedance Z in (s) of each of the circuits in Fig. 16.43.Figure 16.43For Prob. 16.9.Chapter 16, Solution 9.(a) The s-domain form of the circuit is shown in Fig. (a). =+++=+=s 1s 2)s 1s (2)1s (||2Z in 1s 2s )1s (222+++(b) The s-domain equivalent circuit is shown in Fig. (b).2s 3)2s (2s 23)s 21(2)s 21(||2++=++=+ 2s 36s 5)s 21(||21++=++ =⎟⎠⎞⎜⎝⎛+++⎟⎠⎞⎜⎝⎛++⋅=⎟⎠⎞⎜⎝⎛++=2s 36s 5s 2s 36s 5s 2s 36s 5||s Z in 6s 7s 3)6s 5(s 2+++ 2(a) (b)2/s 1PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsUse Thevenin’s theorem to determine v 0(t ), t > 0 in the circuit of Fig. 16.44.Figure 16.44For Prob. 16.10.Chapter 16, Solution 10.11Hs ⎯⎯→ and i L (0) = 0 (the sources is zero for all t<0). 1144F sC s ⎯⎯→= and v C (0) = 0 (again, there are no source contributions for all t<0).To find Z Th , consider the circuit below.21//(2)3Th s Z s s +=+=+PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators To find V Th , consider the circuit below.V244104044236123o Th Th s s V V s s s s Z s s s ====+++++++ ()o v t =Solve for the mesh currents in the circuit of Fig. 16.45. You may leave your results in the s-domain.Figure 16.45For Prob. 16.11.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsPROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators In the s-domain, the circuit is as shown below.12101(1)44s I sI s=+− (1) 1215(4)044sI I s −++= (2) In matrix form,12110144150444s s I s I s s ⎡⎤+−⎡⎤⎢⎥⎡⎤⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦⎢⎥−+⎢⎥⎣⎦⎢⎥⎣⎦ 4s 49s 412++=∆ 11014050454044s s s s −∆==++ 2101541204s s s +∆==− )16s 9s (s 160s 504s 25.2s 25.0225s 40I 2211+++=+++=∆∆=16s 9s 104s 25.2s 25.05.2I 2222++=++=∆∆=1 s 10sPROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsFind v o (t ) in the circuit of Fig. 16.46.Figure 16.46For Prob. 16.12.PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators We apply nodal analysis to the s-domain form of the circuit below.o o o sV 24V s 3s V 1s 10+=+−+ 1s 15s 1510151s 10V )s s 25.01(o 2+++=++=++1s 25.0s C Bs 1s A )1s 25.0s )(1s (25s 15V 22o +++++=++++=740V )1s (A 1-s o =+==)1s (C )s s (B )1s 25.0s (A 25s 1522++++++=+Equating coefficients :2s : -AB B A 0=⎯→⎯+= 1s :C -0.75A C B A 25.015+=++= 0s : C A 25+=740A =, 740-B =, 7135C =4321s 233271554321s 21s 7401s 17404321s 7135s 740-1s 740V 222o +⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛⋅++⎟⎠⎞⎜⎝⎛++−+=+⎟⎠⎞⎜⎝⎛++++=⎟⎟⎠⎞⎜⎜⎝⎛+⎟⎟⎠⎞⎜⎜⎝⎛−=t 23sin e )3)(7()2)(155(t 23cos e 740e 740)t (v 2t -2t -t -o =)t (v o V )t 866.0sin(e 57.25)t 866.0cos(e 714.5e 714.52-t 2-t -t +−s 3/sPROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators Determine i 0(t) in the circuit of Fig. 16.47.Figure 16.47For Prob. 16.13.Chapter 16, Solution 13.Consider the following circuit.Applying KCL at node o,oo o V 1s 21s 12V 1s 2V 2s 1++=+++=+ )2s )(1s (1s 2V o +++=2s B 1s A )2s )(1s (11s 2V I o o +++=++=+=1A =, -1B =2s 11s 1I o +−+==)t (i o ()A )t (u e e -2t -t −2s11/s I oPROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators Chapter 16, Problem 14.* Determine i 0(t) in the network shown in Fig. 16.48.Figure 16.48 For Prob. 16.14.* An asterisk indicates a challenging problem.Chapter 16, Solution 14.We first find the initial conditions from the circuit in Fig. (a).A 5)0(i o =−, V 0)0(v c =−We now incorporate these conditions in the s-domain circuit as shown in Fig.(b).4 Ω1 Ω(a)4/s41(b)PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators At node o,0s440V s 5s 2V 1s 15V o o o =+−+++− o V )1s (4s s 211s 5s 15⎟⎠⎞⎜⎝⎛+++=− o 2o 22V )1s (s 42s 6s 5V )1s (s 4s 2s 2s 4s 4s 10+++=+++++= 2s 6s 5)1s (40V 2o +++=s 5)4.0s 2.1s (s )1s (4s 5s 2V I 2o o ++++=+=4.0s 2.1s C Bs s A s 5I 2o +++++=s C s B )4.0s 2.1s (A )1s (4s 2++++=+Equating coefficients :0s : 10A A 4.04=⎯→⎯=1s : -84-1.2A C C A 2.14=+=⎯→⎯+=2s : -10-A B B A 0==⎯→⎯+=4.0s 2.1s 8s 10s 10s 5I 2o +++−+=2222o 2.0)6.0s ()2.0(102.0)6.0s ()6.0s (10s 15I ++−+++−==)t (i o ()[]A )t (u )t 2.0sin()t 2.0cos(e 10150.6t -−−PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators Find V x (s) in the circuit shown in Fig. 16.49.Figure 16.49 For Prob. 16.15.Chapter 16, Solution 15.First we need to transform the circuit into the s-domain.2s 5V V 2s 5V V ,But 2s s5V 120V )40s s 2(02s s 5sV V s 2V 120V 400102s 5V s /50V 4/s V 3V x o o x x o 2o o 2x o o o x o ++=→+−=+−−++==+−++−=+−+−+−We can now solve for V x .)40s 5.0s )(2s ()20s (5V 2s )20s (10V )40s 5.0s (202s s 5V 1202s 5V )40s s 2(22x 2x 2x x 2−+++−=++−=−+=+−−⎟⎠⎞⎜⎝⎛++++10 3V xs/42s 5+PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators* Find i 0(t ) for t > 0 in the circuit of Fig. 16.50.Figure 16.50 For Prob. 16.16.* An asterisk indicates a challenging problem.PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsWe first need to find the initial conditions. For 0t <, the circuit is shown in Fig. (a).To dc, the capacitor acts like an open circuit and the inductor acts like a short circuit. Hence,A 1-33-i )0(i o L ===, V 1-v o =V 5.221-)1-)(2(-)0(v c =⎟⎠⎞⎜⎝⎛−=We now incorporate the initial conditions for 0t > as shown in Fig. (b).For mesh 1,02V s 5.2I s 1I s 122s 5-o21=++−⎟⎠⎞⎜⎝⎛+++(a)3 V2 ΩV o-1 V2V oPROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators But,2o o I I V ==s5.22s 5I s 121I s 1221−+=⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛+(1)For mesh 2,0s5.22V 1I s 1I s 1s 1o 12=−−+−⎟⎠⎞⎜⎝⎛++ 1s5.2I s 1s 21I s 1-21−=⎟⎠⎞⎜⎝⎛+++(2)Put (1) and (2) in matrix form.⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−+=⎥⎥⎦⎤⎢⎢⎣⎡⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡++−+1s 5.2s 5.22s 5I I s 1s 21s 1-s 121s 1221s 32s 2++=∆, )2s (s 5s 42-2+++=∆3s 2s 2CBs 2s A )3s 2s 2)(2s (132s -I I 22222o +++++=++++=∆∆==)2s (C )s 2s (B )3s 2s 2(A 132s -222++++++=+Equating coefficients :2s :B A 22-+= 1s :C B 2A 20++= 0s : C 2A 313+=Solving these equations leads to7143.0A =, -3.429B =, 429.5C =5.1s s 714.2s 7145.12s 7143.03s 2s 2429.5s 429.32s 7143.0I 22o ++−−+=++−−+=25.1)5.0s ()25.1)(194.3(25.1)5.0s ()5.0s (7145.12s 7143.0I 22o ++++++−+==)t (i o []A )t (u )t 25.1sin(e 194.3)t 25.1cos(e 7145.1e 7143.0-0.5t -0.5t -2t +−PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators Chapter 16, Problem 17.Calculate i 0(t ) for t > 0 in the network of Fig. 16.51.Figure 16.51 For Prob. 16.17.Chapter 16, Solution 17.We apply mesh analysis to the s-domain form of the circuit as shown below.For mesh 3,0I s I s1I s 1s 1s 2213=−−⎟⎠⎞⎜⎝⎛+++ (1)For the supermesh,0I s s 1I )s 1(I s 11321=⎟⎠⎞⎜⎝⎛+−++⎟⎠⎞⎜⎝⎛+ (2)1PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators Adding (1) and (2) we get, I 1 + I 2 = –2/(s+1) (3)But –I 1 + I 2 = 4/s(4)Adding (3) and (4) we get, I 2= (2/s) – 1/(s+1) (5)Substituting (5) into (4) yields, I 1 = –(2/s) – (1/(s+1))(6)Substituting (5) and (6) into (1) we get,1s 2I s 1s 1s s 2)1s (s 1s 2322+−=⎟⎟⎠⎞⎜⎜⎝⎛++++−++js j5.05.1j s j 5.05.1s 2I 3−+++−+−=Substituting (3) into (1) and (2) leads to)1s (s )2s 2s (2I s 1s I s 1s -2232+++−=⎟⎠⎞⎜⎝⎛++⎟⎠⎞⎜⎝⎛+(4)232s)1s (4I s 1s I s 1s 2+−=⎟⎠⎞⎜⎝⎛+−⎟⎠⎞⎜⎝⎛++ (5) We can now solve for I o . I o = I 2 – I 3 = (4/s) – (1/(s+1)) + ((–1.5+0.5j)/(s+j)) + ((–1.5–0.5)/(s–j)) or i o (t) = [4 – e –t + 1.5811e –jt+161.57˚ + 1.5811e jt–161.57˚]u(t)AThis is a challenging problem. I did check it with using a Thevenin equivalent circuit and got the same exact answer.PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators (a) Find the Laplace transform of the voltage shown in Fig. 16.52(a). (b) Using that value of v s (t ) in the circuit shown in Fig. 16.52(b), find the value of v 0(t).Figure 16.52 For Prob. 16.18.Chapter 16, Solution 18.v s (t) = 3u(t) – 3u(t–1) or V s = )e 1(s3s e s 3s s −−−=−V)]1t (u )e 22()t (u )e 22[()t (v )e 1(5.1s 2s 2)e 1()5.1s (s 3V V V )5.1s (02VsV 1V V )1t (5.1t 5.1o ss o s o o o s o −−−−=−⎟⎠⎞⎜⎝⎛+−=−+==+→=++−−−−−−V s 2 ΩPROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsIn the circuit of Fig. 16.53, let i (0) = 1 A, v 0(0) and v s = 4e t 2−u (t ) V. Find v 0(t ) for t > 0.Figure 16.53 For Prob. 16.19.PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsAt the supernode,o 11sV s 1s V 22V ))2s (4(++=+−+o 1V s s1V s 12122s 2++⎟⎠⎞⎜⎝⎛+=++ (1)But I 2V V 1o += and s1V I 1+= 2s 2V s s )2s (s 2V V s )1V (2V V o o 111o +−=+−=⎯→⎯++= (2)Substituting (2) into (1)o o V s 2s 2V 2s s s 22s s 122s 2+⎥⎦⎤⎢⎣⎡+−⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛+=−++ o V s 21s 1s 122s 2⎥⎦⎤⎢⎣⎡+⎟⎠⎞⎜⎝⎛=+−++ o V )2/1s (2s 6s 2)2s (24s 2+=++=+++2s B2/1s A )2/1s )(2s (6s 2V o +++=+++=333.3)25.0/()61(A =+−+−=, 3333.1)2/12/()64(B −=+−+−=2s 3333.12/1s 333.3V o +−+= Therefore,=)t (v o (3.333e -t/2 – 1.3333e -2t )u(t) V2PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsFind v 0(t ) in the circuit of Fig. 16.54 if v x (0) = 2 V and i (0) = 1A.Figure 16.54 For Prob. 16.20.PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsWe incorporate the initial conditions and transform the current source to a voltage source as shown.At the main non-reference node, KCL givess 1s V 1V s 11V s 2)1s (1o o o ++=+−−+s 1s V )s 11)(1s (V s 21s s o o ++++=−−+ o V )s 12s 2(2s1s 1s s ++=−+−+ )1s 2s 2)(1s (1s 4s 2-V 22o +++−−=5.0s s CBs 1s A )5.0s s )(1s (5.0s 2s -V 22o +++++=+++−−=1V )1s (A 1-s o =+==)1s (C )s s (B )5.0s s (A 5.0s 2s -222++++++=−−Equating coefficients :2s : -2B B A 1-=⎯→⎯+=1s : -1C C B A 2-=⎯→⎯++=0s :-0.515.0C A 5.00.5-=−=+=222o )5.0()5.0s ()5.0s (21s 15.0s s 1s 21s 1V +++−+=+++−+==)t (v o []V )t (u )2t cos(e 2e 2-t -t −1/sPROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsFind the voltage v 0(t ) in the circuit of Fig. 16.55 by means of the Laplace transform.Figure 16.55 For Prob. 16.21.PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsThe s-domain version of the circuit is shown below. 1 sAt node 1, o o o V s V s V s s V V V s )12()1(1021102111−++=⎯→⎯+−=− (1)At node 2,)12(2211++=⎯→⎯+=−s sV V sV V sV V o o o o(2)Substituting (2) into (1) giveso o o V s s s V s V s s s )5.12()12()12/)(1(10222++=−++++=5.12)5.12(1022++++=++=s s C Bs s A s s s V oCs Bs s s A ++++=22)5.12(10B A s +=0:2C A s +=20:-40/3C -20/3,B ,3/205.110:constant ===⎯→⎯=A A ⎥⎦⎤⎢⎣⎡++−+++−=⎥⎦⎤⎢⎣⎡+++−=222227071.0)1(7071.0414.17071.0)1(113205.1221320s s s s s s s s V o Taking the inverse Laplace tranform finally yields[]V )t (u t 7071.0sin e 414.1t 7071.0cos e 1320)t (v t t o −−−−=PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No partof this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators Find the node voltages v 1 and v 2 in the circuit of Fig. 16.56 using the Laplace transform technique. Assume that i s = 12e t −u (t )A and that all initial conditions are zero.For Prob. 16.22.Chapter 16, Solution 22.The s-domain version of the circuit is shown below. 4sAt node 1,s4V s 411V 1s 12s 4V V 1V 1s 1221211−⎟⎠⎞⎜⎝⎛+=+⎯→⎯−+=+ (1)At node 2,⎟⎠⎞⎜⎝⎛++=⎯→⎯+=−1s 2s 34V V V 3s 2V s 4V V 2212221 (2)Substituting (2) into (1),2222V 23s 37s 34s 41s 4111s 2s 34V 1s 12⎟⎠⎞⎜⎝⎛++=⎥⎦⎤⎢⎣⎡−⎟⎠⎞⎜⎝⎛+⎟⎠⎞⎜⎝⎛++=+)8s 4s (CBs )1s (A )8s 4s )(1s (9V 222+++++=+++=)1s (C )s s (B )89s 47s (A 922++++++=PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators Equating coefficients:B A 0:s 2+= A 43C CA 43C B A 470:s −=⎯→⎯+=++=-18C -24,B ,24A A 83C A 899:constant ===⎯→⎯=+=6423)87s (36423)87s ()8/7s (24)1s (2489s 47s (18s 24)1s (24V 2222++++++−+=+++−+=Taking the inverse of this produces:[])t (u )t 5995.0sin(e 004.5)t 5995.0cos(e 24e 24)t (v t 875.0t 875.0t 2−−−+−=Similarly,)89s 47s (F Es )1s (D )89s 47s )(1s (1s 2s 349V 2221+++++=+++⎟⎠⎞⎜⎝⎛++=)1s (F )s s (E )89s 47s (D 1s 2s 349222++++++=⎟⎠⎞⎜⎝⎛++Equating coefficients:E D 12:s 2+= D 436F F D 436or F E D 4718:s −=⎯→⎯+=++=0F 4,E ,8D D 833or F D 899:constant ===⎯→⎯=+=6423)87s (2/76423)87s ()8/7s (4)1s (8)89s 47s (s 4)1s (8V 2221++−+++++=++++=Thus,[])t (u )t 5995.0sin(e 838.5)t 5995.0cos(e 4e 8)t (v t 875.0t 875.0t 1−−−−+=Consider the parallel RLC circuit of Fig. 16.57. Find v(t) and i(t) given that v(0) = 5 and i(0) = -2 A.Figure 16.57For Prob. 16.23.PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsPROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorsThe s-domain form of the circuit with the initial conditions is shown below.At the non-reference node,sCV sL V R V C 5s2s 4++=++⎟⎠⎞⎜⎝⎛++=+LC 1RC s s s CV s sC 562LC1RC s s C6s 5V 2+++=But 880101RC 1==, 208041LC 1==222222)4s ()2)(230(2)4s ()4s (520s 8s 480s 5V ++++++=+++==)t (v V )t (u ))t 2sin(e 230)t 2cos(e 5(-4t -4t + )20s 8s (s 4480s 5sL V I 2+++==20s 8s C Bs s A )20s 8s (s 120s 25.1I 22++++=+++=6A =, -6B =, -46.75C =222222)4s ()2)(375.11(2)4s ()4s (6s 620s 8s 75.46s 6s 6I ++−+++−=+++−==)t (i 0t ),t (u ))t 2sin(e 375.11)t 2cos(e 66(-4t -4t >−−V 5C。

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vs t
R1 26
R3 26
R2 35
RL 50
L 20 log 2vL ？？？
vS
i0 0.0100
i1
0.0100
i2
0.0068
v
S
i3 0.0032
i4 0.0032
vL i4RL 0.1577vS
李国林电子电路与系统基础
清华大学电子工程系 2013年春季
10
0.025 0 0.1577
vL v 3 0.1577vS
李国林电子电路与系统基础
清华大学电子工程系 2013年春季
清华大学电子工程系 2013年春季
2
作业1：T型电阻衰减网络
RS 50
R1 26
R3 26
vs t
R2
RL
35
50
• 用支路电压电流法、支路电流法、回路电流法、结点电压法列写上述电路的电路方程
– 选取其中的一种方法，矩阵求逆求解，获得负载电压与源电压之间的比值关系，说明衰减系数为多大
李国林电子电路与系统基础
vs t
R2
RL
35
50
结点电压法
GS G1
G1
0
G1 G1 G2 G3
G3
0 G3
v v
1 2
GS vS
0
G3
GL v
3
0
58.46 38.46 0
38.46 105.49 38.46
0 38.46
v v
1 2
20vS
0
58.46
v
3
6
Matlab辅助计算
v0 0.4998
v1
0.2601
v2
v3
0.2397
0.0820
v4 i0
0.1577
0.0100vs
i1
0.0100
i2
i3
0.0068
0.0032
i4 0.0032
李国林电子电路与系统基础
清华大学电子工程系 2013年春季
7
结果分析
R3 26
R2 35
RL 50
v1 i1R1 0 v2 i2R2 0 v3 i3R3 0 v4 i4R4 0
n-1=3个KCL方程
i0 i1 0
b-n+1=2个KVL方程
i1 i2 i3 0 v0 v1 v2 0
i3 i4 0
v2 v3 v4 0
李国林电子电路与系统基础
RS 50
vs t
R1 26
R3 26
R2 35
RL 50
n-1=3个KCL方程
i0 i1 0
i1 i2 i3 0
-1
-1
0
0
1
-1
0
0
0
- 50
26
35
0
0
-35
b-n+1=2个KVL方程
i0Rs i1R1 i2R2 vs
i3 i4 0
i2R2 i3R3 i4R4 0
电子电路与系统基础
习题课第六讲
1、第四周作业讲解 2、戴维南定理简化方程列写
3、简单电路的参量描述 4、二端口线性网络噪声分析
李国林清华大学电子工程系
习题课第六讲
• 第四周作业讲解
大纲
• 戴维南定理简化方程列写例
• 简单二端口网络的网络参量简单列写方法
• 二端口网络噪声分析
李国林电子电路与系统基础
V2 S ,rms
L PS,max 4RS
PL
V2 L,rms
RL 4RS
VS ,rms VL,rms
2
1 1 2
4 0.1577
10.05 10dB
RL
10dB的衰减器：负载电阻只获得了信源额定功率的10%
李国林电子电路与系统基础
清华大学电子工程系 2013年春季
8
支路电流法
李国林电子电路与系统基础
清华大学电子工程系 2013年春季
0
0 i0 0
-1
0
i1
0
1
-1
i2
0
0
0 i3 vs
26
50i4 0
i0 0.0100
i1
0.0100
i2
0.0068
vS
i3 0.0032
i4 0.0032
9
Matlab辅助计算
RS 50
RS 50
R1 26
R3 26
vs t
il1
R2
RL
35
il 2
50
回路电流法
RS R1 R2
R2
R2
R2 R3
R4
il1
il
2
vs
050 26 35353535 26
il1 50 il 2
vs
0
111 35
35il1
111
il
2
清华大学电子工程系 2013年春季
3
电路结构分析
RS 50
R1 26
R3 26
vs t
R2
RL
35
50
b=5
5条支路
n=4 4个结点
李国林电子电路与系统基础
清华大学电子工程系 2013年春季
4
支路电压电流法
b=5个支路元件约束方程
v0 i0 Rs vs
RS 50
vs t
R1 26
v0 0.4998
v1
0.2601
v2
v3
0.2397
0.0820
v4 i0
0.1577
0.0100vs
i1
0.0100
i2
i3
0.0068
0.0032
i4 0.0032
RS 50
vs t
R1 26
R3 26
R2 35
RL 50
vL v4 0.1577vs
0
v v
1 2
58.46 38.46
v
3
0
38.46 105.49 38.46
0 1 20vS 0.0250
38.46
0
0.0120
58.46 0 0.0079
0.0120 0.0182 0.0120
0.007920vS 0.4998
0.0120
0
0.2397vS
vs
0
il1 il 2
111 35
351
111
vs
0
0.0100 0.0032
0.0032vs 0.0100 0
0.0100 0.0032vs
vL il2RL 0.1577vS
李国林电子电路与系统基础
清华大学电子工程系 2013年春季
11
RS 50
R1 26
R3 26
0 0
0
0
0
0
1
0
0
0
0
-50
v4
0
0
0
0
0
0
-1
-1
0
0
0 i0 0
0
0
0
0
0
0
1
-1
-1
0
i1
0
0
0
-1
1
0 1
0 0
0 0
0 0
0 0
0 0
1 0
-1 0
i2 i3
0
0
0
0
-1
1
1
0
0
0
0
0 i4 0
李国林电子电路与系统基础
清华大学电子工程系 2013年春季
清华大学电子工程系 2013年春季
5
支路电压电流法矩阵方程
RS 50
vs t
R1 26
R3 26
R2 35
RL 50
1 0
0 1
0 0
0 0
0 0
50 0
0
0
-26 0
0 0
0 v0
0
v1
vs
0
0 0
0 0
1 0
0 1
0 0
0 0
0 0
-35 0
0 -26
0 0
v2 v3