高频电子线路习题答案(第五版)张肃文

高频电子第五版

(pF)).(L C H)(.QR

则L Ω取R Δf f Q (kHz)Δf MHz

解：f

..159101*********

1

15910

143210

1001010010

101012101099010121136

26206

03

6

70036700

=⨯⨯⨯⨯=

=

=⨯⨯⨯=

===⨯⨯===⨯-⨯==--ωμω

时，产生并联谐振。C L 或ωC L )当(时，产生串联谐振。C L 或ωC L )当(时，产生并联谐振。C L 或ωC L )当解：(2

2021

1012

2021

1012

202

11011

1

31

1

21

1123==

====-ωωω

R R C L R )LC ωL(j ωR )LC ωLR(j ωC L R C j ωR L j ωR )C j ωL)(R j ω(R 证明：Z =+=-+-++=++++

+=-21121

11133220020020000

)()()()()())()()()())

318010

404501053514321

121535100160512405354501605151431223202222μH .C C L 故采用后一个不合理舍去pF -得C C C 由pF

得C C C 由解：=⨯+⨯⨯⨯⨯='+==+=⨯+=+=⨯+--ω。

L C C ’

()()()

()

mV V Q V V mA .R V I μH ..C L ..R C 解：Q

-Sm Com Lom -om om --212101212205

10111210

100105114321

12125

10100105114321

153303

1226020012

6000

=⨯⨯====⨯===⨯⨯⨯⨯==

=⨯⨯⨯⨯⨯⨯==

-ωω

()()()()()

()Ωj ..j .C j

R Z Ω.....Q L

Q L R pF C pF .L C C C C .V V Q μH .C 解：L X X X X X X X S C 796747102001014321

7471747100

1025310143210521025310143220010010

2531014321

11001

010

25310

1001014321

16312

606

666000626200122620-=⨯⨯⨯⨯-=-==⨯⨯⨯⨯-⨯⨯⨯⨯=-==→=⨯⨯⨯⨯==+⋅===

=⨯⨯⨯⨯==------ωωωωω()()

()。21k Ω0.5R ，R ，故0.5Q Q ，则f 22f 因2Δ3

20105105552310023

100

101501052220105010514321

173000.70.76

60036700012

620电阻所以应并上

='='∆⨯='=⨯⨯-⨯⨯===

⨯⨯===⨯⨯⨯⨯⨯==--.f Δf Q ξΔf f Q μH ..C ω解：L . ∑

===-g Q C

ωΔf f C πf C πΔf 证明：..070007022483

()()()()()()()()MHz ...Q f Δf .....L ωR Q k Ω..R C C C C R R R k Ω..C L

Q R MHz ....LC πf pF .C C C C C C C 解：C L .L P i P i 4812

281064122281080106411432108858855202020209201092010

20201080100641103181080143212131820

2020202020593607

06

63

02

02

110212

6

12

12

60102102

=⨯===⨯⨯⨯⨯⨯⨯===⨯⎪⎭⎫

⎝⎛++=⎪⎪⎭⎫ ⎝⎛++==⨯+⨯⨯===⨯⨯⨯⨯==

=++++=++++=--∑∑----

)))R Z 30Z 20

Z 1123f1f1f1===-解：

)()()()()

)()()()()()()))()

)()()

()

()()()Ω.j .j ..Z M Z j ..j C L j R Z pF .L C kHz .R ρf Q f Δf .R R L Q k ΩC R R L Z Ω..R M Z μH ..ηR M pF .L C C μH .ρL L 解：f .f f f P 8437680100

2010183101432100201017710143211015910143220117710

1591095014321

1

52281020

102222425

20

2010159101432325101592020101592020

10183101432218310143220115910

1591014321

11591014321011332

66

22

2

011

126662*********

2

32202

2

3

6110070661

11

01112

6

11112

66

2

2011

6

0116261201216

3

01

1

21-=+⨯⨯⨯⨯=

=

+=⎪⎭⎫

⎝⎛⨯⨯⨯⨯-⨯⨯⨯⨯+=⎪

⎪⎭⎫

⎝

⎛'-+==⨯⨯⨯⨯⨯=

'='=⨯⨯====+⨯⨯⨯⨯=+=

=⨯⨯+⨯=+==⨯⨯⨯⨯=

=

=⨯⨯⨯===⨯⨯⨯⨯==

==⨯⨯==

=----------ωωωωωωωωω

()100

1014102220

02010159105010

1591533

6

7.0011

12

36

=⨯=∆==→=∴=Ω=⨯⨯⨯⨯==

---f f Q M R R C R L R f P 解：