(爱班)《电网络分析》习题参考答案

习题答案

50 3

2-1 Z =

40 340

3 50 3

(30 j40) j40

2-2 Z = j10 j40

R(rm1+R) 2-3 Z=

(rm+R)R 2

2 1+j

2-4 Y=

1

1 2 1 j

1 j(ωC ) ωL

2-5 Y=

1 j ωL 1.5 0.5

2-6 H=

4 0.5 1

R+jωC1

2-7 Y=

n R 5 4

2-8 T=

3 40

45

6 5 4

1

ωL

11 jRωL j

2

n

+jωC2 R

nR

1.53 j0.4990.843 j0.167

2-9 T=

1.589 j1.4971.530 j0.499

2-10 T=

2-11 =R2.4Β ;Pmax=1.35W;P=18.9W US

1

(1) 2

ωC1L

2-12 T=

1

j

ωL

2-13

1 ω2L(C1+C2)

j

ω3LC1C2

1

(1 2)

ωC2L

Z1

2-14 Y=

GZ1 Z1

2-15 H=

0

2-16

0

1 Z2

3Z1 2Z3 Z2

2 2Z3 Z2

Z2(Z1+Z3)

+Z5+Z6

Z1+Z2+Z3 Z1Z2

+Z6

Z1+Z2+Z3

Z1(Z2+Z3)

Z+Z+Z+Z4+Z6

123

2-17 Z=

Z1Z2

Z+Z+Z+Z6

123 1 Y+ a11Z

1

2-18 Y=

1

Y a21Z 1

n

2-19 T=

n Za

11 Ya22++

Z1Z2 Ya12

Zbn

1Z1

Zb 1

(1+)nZa

r2

2-20 Rab=

R 1 2-21 T= 0

jωC

g2

1

U1 9 0.25 U2

2-22 = 4 0

I1 I2

0 100111 01 110 10

3-1 Aa=

100 100 1 1010100

0 100111

0111010Aa= 100 100 1

1 0

3-2 Bf=

0 1

011111011000010000100 0 0 1 10 00 00

0 1

010 0 1 1

3-3 Bm=

101

100

000 11001

1001001 0100 1 1 1

=3-4 Qf

001 1 10 1 0 11100 [B 1]

3-6 Bf= 1 10010=t

11 1001 10001 1

[1 Q]

=Qf = 010111t

001 101

100 1 10

010111=Qf 3-7

001 10 1

111

++

R1R3R4 1

3-8

R4

1

R1

1

R4

111++R2R4R5

1R2

US3US1 +

U1 R3R1

1

IS2 U2 =

R2 U U

3 S1 I

111 S2

R++ 1

R1R2R6

1

R1

R1+R4+R5+R6 R5 R63-9 R4 R6

11 j

ωL1 R1

3-10 Y =0

gm

R5 R6R2+R5+R6

R5

R4 R6 I1

I =R5

2

R3+R4+R5 I3

0 0 1 R3

US1

I×R

2 S2

U3

01R2 j

1

ωC2

0 j1ωL2

111 1 j j+ RωL1ωL2R j 12

ωC2

Yn=

11

gm+j

ωL2R j 2

ωC2 11

+jωC 1 R+jωLjωL211

3-11

1

jωL2

jωL1 jωM 03-13 Z= 0 0 0

jωM

jωL20000

0001

0jωC

R4000

00

111

j+ R3ωL2R j

2

ωC2 j

11

ωL2R j2

ωC2

0 U 1 =

11 U2 US +jωC2

R2 R2jωL2

0000

R50

0 0 0 0 0 R6

1jωL2

1

+ ++R42ωωωjLjLjM12 jωC 1

=Zn jωL2 jωM+

jωC

1

+ωωjLjM1

jωC

11

+jωCjωL2jωMjωL2+1

+R5

jωC

1

jωM+

jωC

jωL1 jωM+

1 jωC

1

jωM+

jωC

1

+R6 jωL1+jωC

jωL1

jωM

12 0

3-14 Z= 0

0 0

jωM12

jωL2 jωM23

000

0 jωM23jωL3000

000R400

0000R50

0 0 0 0 0 1 jωC6

R1+R2+R4

R43-15 R2 1

jωC

1

0 0

3-18 K= 0

0 0 0 1 0 0

F= 0

0 0 0

R4

R3+R4+R5

R50 jωL2jωM000000100000001000000010000000100

R2 Il1 US1+R2×IS2

I =

R5U23lS R2+R5+R6 Il3 R2×IS2

00

0 R4000

0000000

000001jωC20

0 0 0 0 0 0 1

jωM

jωL3

0000

0100000

0

0 0 0

0

0 0 0 V=

0 US5 0 0

0 IS7