篇一：高频电子线路课后习题答案(第五版)张肃文
高频电子第五版
3?1解：f
?1MHz
2Δf0.7?1?106?990?103?10(kHz)f01?106
Q???1003
2Δf0.710?10
取R?10Ω则L?C?
QR
?01
?
100?10
?159(?H)6
2?3.14?10
1
?159(pF)
?02L(2?3.14?106)2?159?10?6
11
3?2解：(1)当?01?或ω?时，产生并联谐振。02
L1C1L2C2
?
(2)当?01?(3)当?01?
1
或ω02?L1C1
1
或ω02?L1C1
1
时，产生串联谐振。L2C2
1
时，产生并联谐振。L2C2
1L1
)R2??jω0LR(1?2)R2?Ljω0CCω0LC?R3?3证明：Z???112RR?jω0L?R?2R?jω0L(1?2)jω0Cω0LC
3?4解：1?由?15?C??16052??450?C?5352得C?40pF
(R?jω0L)(R?
由?12?C??16052??100?C?5352得C?-1pF?不合理舍去?
故采用后一个。
2?L?3?
LCC’
11
??180?μH?
?02C?C?2?3.14?535?1032??450?40??10?12
3?5解：Q
?
11
??2126-12
?0C0R2?3.14?1.5?10?100?10?5
L0?
11
??112?μH?226-12?0C02?3.14?1.5?10100?10
Vom1?10-3
Iom???0.2?mA?
R5
VLom?VCom?Q0VSm?212?1?10-3?212?mV?
11
3?6解：L?2??253?μH?
?0C2?3.14?1062?100?10?12
Q0?
VC10??100VS0.1
C?CX11
?2??100?pF??CX?200pF
C?CX?0L2?3.14?1062?253?10?6
2?3.14?106?253?10?62?3.14?106?253?10?6
RX?????47.7?Ω?
QQ02.0.1100
?0L?0L
11
?47.7?j?47.7?j796?Ω?6?12
?0CX2?3.14?10?200?1011
3?7解：L?2??20.2?μH?
ω0C2?3.14?5?106?50?10?12
ZX?RX?j
f05?106100
Q0???
2Δf0.7150?1033
2Δf1002??5.5?5??10620ξ?Q0???6
f035?103
??2?2?f0.7，则Q??因2Δf0.7，所以应并上21kΩ电阻。 0?0.5Q0，故R?0.5R
2πf0CωC
3?8证明：4πΔf0.7C??0?g?
f02Δf0.7Q
?C?C0?C1?5??20?20?20?18.3?pF?3?9解：C?Ci?2
C2?C0?C120?20?20
f0?
11??41.6?MHz??6?12
2πLC2?3.140.8?10?18.3?10
L
?100?C12
0.8?10?6
?20.9?kΩ?
20?20?10?12
2
2
RP?Q0
?C2?C0?C1??20?20?20?
?R??RiRP?R?1020.9???5?5.88?kΩ???0
C20??1??R?5.88?103
QL???28.26?6
ω0L2?3.14?41.6?10?0.8?102Δf0.7
f041.6?106???1.48?MHz?QL28.2
3?12解：1?Zf1?0
2?Zf1?03?Zf1?R
3?13解：1?L1?L2?
C1?C2?
ρ1
103
??159?μH?6
2?3.14?10
?01
11
??159?pF?226?6?01L12?3.14?10?159?10
ηR1?20M?1??3.18?μH?6
?012?3.14?10
2?Zf1
??01M?2?
R2
?2?3.14?10?
6
?3.18?10?6??20?Ω?20
2
L1159?10?6
ZP???25?kΩ??12
R1?Rf1C120?20?159?103?Q1?
?01L1R1?Rf1
2?3.14?106?159?10?6??25
20?20
4?2Δf0.7??5?C2
f0f0106?20
?2?2?2??28.2?kHz?
Qρ1R1103
1
2
??02
L2
?
1
2?3.14?950?1032
?159?10
?6
?177?pF?
?1?
Z22?R2?j??L??021?C???
022??
1??
?20?j?2?3.14?106?159?10?6??20?j1006?12?2?3.14?10?177?10??Zf1
??01M?2
?
Z22
?2?3.14?10?
?6
?3.18?10?6??0.768?j3.84?Ω?20?j100
6
2
3?15解：?R?
L159?10
??20????R1
RPC50?103?159?10?12
?Rf1?0?M?0
f0106
Q?2?2?100
2?f0.714?103
3?16解：1?Rf1
??01M?2
?
R2
?10?
7
?10?6??20???5
2
Rab?2???3?Q?
R
R1
??01L?2
1
?Rf1?10?
7
?100?10?6??40?k??5?20
2
?01M?01LR1
107?10?6??2
5107?100?10?6??200
5
2?f0.711
?2?2??1??22?2?2?1??0.013f0Q200 I111
3?17解：????Q??22.5
22I01.25?2?f??10?103?
1??1???Q?300?103???Q?f??
??0??
11R???11.8
Q?0C22.5?2?3.14?300?103?2000?10?12
?2?f??10?103?
??1???Q300?103???Qf??
??0??
Q?Q??30?22.5?7.5
1?
2ω?串联联谐??L1?375μHL2C?
3?18解：???
1?L2?125μ?ω?并联联谐L1?L2C??
I?I0
1
2
?
1
2
?
1
?Q?302
4?5解：当f?1MHz时，??
β0?β0f1???f?Tβ0?β0f1???f?Tβ0?β0f???f?T
????????????
2
?
50?50?106????250?106????
50
2
?49
当f?20MHz时，??
2
?
当f?50MHz时，??
2
?
?50?20?101???250?106
?
50
6
????
2
?12.1
?50?50?106?1???250?106????
2
?5
4?7解：gb?e?
gm?Cb?e
IE1
??0.754?mS?
26β0?126?50?1β0
?50?0.754?10?3?37.7?mS?rb?e
gm37.7?10?3???24?pF?6
2πfT2?3.14?250?10
a?1?rbb?gb?e?1?70?0.754?10?3?1b?ωCb?erbb??2?3.14?107?24?10?12?70?0.1yie
?gb?e?j?Cb?e??a?jb??0.754?10?3???
a2?b2
?0.895?j1.41?mS?
j2?3.14?107?24?10?12???1?j0.1?
12?0.12
j2?3.14?107?3?10?12??1?j0.1?yre??0.0187?j0.187?mS?2222
a?b1?0.1
gm?a?jb?37.7?10?3??1?j0.1?yfe???37.327?j3.733?mS?a2?b212?0.12
?g?j?Cb?c??a?jb??j?C?1?rga?jb?
yoe?gce?j?Cb?c?rbb?gmb?c?b?c?bb?m
a2?b2a2?b2??
1?j0.1??
?j2?3.14?107?3?10?12?1?70?37.7?10?3?2?0.049?j0.68?mS?2?1?0.1??
?Av?
4?8解：令??A??
?vo?
?Av?令??A???vo?
mm
?g?j?Cb?c??a?jb?????b?c
?????
4?Q2Δf4?Q2Δf
2
0.7
f04
?
??1?2?
m
得2Δf0.7
1?m?f0
???4?2?1??Q
??
?
????
2
0.1
f04
?
??1?10?10?12?1
1m2m
m
得2Δf0.1
2?m?f0
???4?10?1??Q
??
故Kr0.1?
2Δf0.1
?2Δf0.7
2?m?
?4??1??
??1?m??4?2?1????
?
篇二：高频电子线路习题答案(第五版)张肃文
看不惯那些家伙。 高频电子第五版
3?1解：f
?1MHz
2Δf0.7?1?106?990?103?10(kHz)f01?106
Q???100
2Δf0.710?103
取R?10Ω则L?C?
QR
?01
?
100?10
?159(?H)
2?3.14?106
1
?159(pF)
?02L(2?3.14?106)2?159?10?6
11
3?2解：(1)当?01?或ω?时，产生并联谐振。02
L1C1L2C2
?
(2)当?01?(3)当?01?
1
或ω02?L1C1
1
或ω02?L1C1
1
时，产生串联谐振。L2C2
1
时，产生并联谐振。L2C2
1L1
)R2??jω0LR(1?2)R2?Ljω0CCω0LC?R3?3证明：Z???2RR?jω0L?R?2R?jω0L(1?2)jω0Cω0LC 22
3?4解：1?由?15?C??1605??450?C?535得C?40pF
(R?jω0L)(R?
由?12?C??16052??100?C?5352得C?-1pF?不合理舍去?
故采用后一个。
2?L?3?
LCC’
11
??180?μH?
?02C?C?2?3.14?535?1032??450?40??10?12
3?5解：Q
?
11
??2126-12
?0C0R2?3.14?1.5?10?100?10?5
L0?
11
??112?μH?226-12?0C02?3.14?1.5?10100?10
Vom1?10-3
Iom???0.2?mA?
R5
VLom?VCom?Q0VSm?212?1?10-3?212?mV?
11
3?6解：L?2??253?μH?
?0C2?3.14?1062?100?10?12
Q0?
VC10??100VS0.1
C?CX11
?2??100?pF??CX?200pF
C?CX?0L2?3.14?1062?253?10?6
2?3.14?106?253?10?62?3.14?106?253?10?6
RX?????47.7?Ω?
QQ02.0.1100
?0L?0L
11
?47.7?j?47.7?j796?Ω?6?12
?0CX2?3.14?10?200?1011
3?7解：L?2??20.2?μH?
ω0C2?3.14?5?106?50?10?12
ZX?RX?j
f05?106100
Q0???
2Δf0.7150?1033
2Δf1002??5.5?5??10620ξ?Q0???6
f035?103
??2?2?f0.7，则Q??因2Δf0.7，所以应并上21kΩ电阻。 0?0.5Q0，故R?0.5R
2πf0CωC
3?8证明：4πΔf0.7C??0?g?
f02Δf0.7Q
?C?C0?C1?5??20?20?20?18.3?pF?3?9解：C?Ci?2
C2?C0?C120?20?20
f0?
11??41.6?MHz??6?12
2πLC2?3.140.8?10?18.3?10
L
?100?C12
0.8?10?6
?20.9?kΩ?
20?20?10?12
2
2
RP?Q0
?C2?C0?C1??20?20?20?
?R??RiRP?R?1020.9???5?5.88?kΩ???0
C20??1??R?5.88?103
QL???28.26?6
ω0L2?3.14?41.6?10?0.8?102Δf0.7
f041.6?106???1.48?MHz?QL28.2
3?12解：1?Zf1?0
2?Zf1?03?Zf1?R
3?13解：1?L1?L2?
C1?C2?
ρ1
103
??159?μH?6
2?3.14?10
?01
11
??159?pF?226?6?01L12?3.14?10?159?10
ηR1?20M?1??3.18?μH?6
?012?3.14?10
2?Zf1
??01M?2?
R2
?2?3.14?10?
6
?3.18?10?6??20?Ω?20
2
L1159?10?6
ZP???25?kΩ??12
R1?Rf1C120?20?159?103?Q1?
?01L1R1?Rf1
2?3.14?106?159?10?6??25
20?20
4?2Δf0.7??5?C2
f0f0106?20
?2?2?2??28.2?kHz?
Qρ1R1103
1
2
??02
L2
?
1
2?3.14?950?1032
?159?10
?6
?177?pF?
?1?
Z22?R2?j??L??021?C???
022??
1??
?20?j?2?3.14?106?159?10?6??20?j1006?12?2?3.14?10?177?10??Zf1
??01M?2
?
Z22
?2?3.14?10?
?6
?3.18?10?6??0.768?j3.84?Ω?20?j100
6
2
3?15解：?R?
L159?10
??20????R1
RPC50?103?159?10?12
?Rf1?0?M?0
f0106
Q?2?2?100
2?f0.714?103
3?16解：1?Rf1
??01M?2
?
R2
?10?
7
?10?6??20???5
2
Rab?2???3?Q?
R
R1
??01L?2
1
?Rf1?10?
7
?100?10?6??40?k??5?20
2
?01M?01LR1
107?10?6??2
5107?100?10?6??200
5
2?f0.711
?2?2??1??22?2?2?1??0.013f0Q200 I111
3?17解：????Q??22.5
22I01.25?2?f??10?103?
1??1???Q?300?103???Q?f??
??0??
11R???11.8
Q?0C22.5?2?3.14?300?103?2000?10?12
?2?f??10?103?
??1???Q300?103???Qf??
??0??
Q?Q??30?22.5?7.5
1?
2ω?串联联谐??L1?375μHL2C?
3?18解：???
1?L2?125μ?ω?并联联谐L1?L2C??
I?I0
1
2
?
1
2
?
1
?Q?302
4?5解：当f?1MHz时，??
β0?β0f1???f?Tβ0?β0f1???f?Tβ0?β0f???f?T
????????????
2
?
50?50?106????250?106????
50
2
?49
当f?20MHz时，??
2
?
当f?50MHz时，??
2
?
?50?20?101???250?106
?
50
6
????
2
?12.1
?50?50?106?1???250?106????
2
?5
4?7解：gb?e?
gm?Cb?e
IE1
??0.754?mS?
26β0?126?50?1β0
?50?0.754?10?3?37.7?mS?rb?e
gm37.7?10?3???24?pF?6
2πfT2?3.14?250?10
a?1?rbb?gb?e?1?70?0.754?10?3?1b?ωCb?erbb??2?3.14?107?24?10?12?70?0.1yie
?gb?e?j?Cb?e??a?jb??0.754?10?3???
a2?b2
?0.895?j1.41?mS?
j2?3.14?107?24?10?12???1?j0.1?
12?0.12
j2?3.14?107?3?10?12??1?j0.1?yre??0.0187?j0.187?mS?2222
a?b1?0.1
gm?a?jb?37.7?10?3??1?j0.1?yfe???37.327?j3.733?mS?a2?b212?0.12
?g?j?Cb?c??a?jb??j?C?1?rga?jb?
yoe?gce?j?Cb?c?rbb?gmb?c?b?c?bb?m
a2?b2a2?b2??
1?j0.1??
?j2?3.14?107?3?10?12?1?70?37.7?10?3?2?0.049?j0.68?mS?2?1?0.1??
?Av?
4?8解：令??A??
?vo?
?Av?令??A???vo?
mm
?g?j?Cb?c??a?jb?????b?c
?????
4?Q2Δf4?Q2Δf
2
0.7
f04
?
??1?2?
m
得2Δf0.7
1?m?f0
???4?2?1??Q
??
?
????
2
0.1
f04
?
??1?10?10?12?1
1m2m
m
得2Δf0.1
2?m?f0
???4?10?1??Q
??
故Kr0.1?
2Δf0.1
?2Δf0.7
2?m?
?4??1??
??1?m??4?2?1????
?
篇三：高频电子线路习题答案(第五版)张肃文
高频电子第五版
3?1解：f
?1MHz
2Δf0.7?1?106?990?103?10(kHz)f01?106
Q???1003
2Δf0.710?10
取R?10Ω则L?C?
QR
?01
?
100?10
?159(?H)6
2?3.14?10
1
?159(pF)
?02L(2?3.14?106)2?159?10?6
11
3?2解：(1)当?01?或ω?时，产生并联谐振。02
L1C1L2C2
?
(2)当?01?(3)当?01?
1
或ω02?L1C1
1
或ω02?L1C1
1
时，产生串联谐振。L2C2
1
时，产生并联谐振。L2C2
1L1
)R2??jω0LR(1?2)R2?Ljω0CCω0LC?R3?3证明：Z???112RR?jω0L?R?2R?jω0L(1?2)jω0Cω0LC
3?4解：1?由?15?C??16052??450?C?5352得C?40pF
(R?jω0L)(R?
由?12?C??16052??100?C?5352得C?-1pF?不合理舍去?
故采用后一个。
2?L?3?
LCC’
11
??180?μH?
?02C?C?2?3.14?535?1032??450?40??10?12
3?5解：Q
?
11
??2126-12
?0C0R2?3.14?1.5?10?100?10?5
L0?
11
??112?μH?226-12?0C02?3.14?1.5?10100?10
Vom1?10-3
Iom???0.2?mA?
R5
VLom?VCom?Q0VSm?212?1?10-3?212?mV?
11
3?6解：L?2??253?μH?
?0C2?3.14?1062?100?10?12
Q0?
VC10??100VS0.1
C?CX11
?2??100?pF??CX?200pF
C?CX?0L2?3.14?1062?253?10?6
2?3.14?106?253?10?62?3.14?106?253?10?6
RX?????47.7?Ω?
QQ02.0.1100
?0L?0L
11
?47.7?j?47.7?j796?Ω?6?12
?0CX2?3.14?10?200?1011
3?7解：L?2??20.2?μH?
ω0C2?3.14?5?106?50?10?12
ZX?RX?j
f05?106100
Q0???
2Δf0.7150?1033
2Δf1002??5.5?5??10620ξ?Q0???6
f035?103
??2?2?f0.7，则Q??因2Δf0.7，所以应并上21kΩ电阻。 0?0.5Q0，故R?0.5R
2πf0CωC
3?8证明：4πΔf0.7C??0?g?
f02Δf0.7Q0
3?9解：C?Ci?
f0?
?C2?C0?C1
C2?C0?C1
?5?
?20?20?20
20?20?20
?18.3?pF?
11??41.6?MHz??6?12
2πLC2?3.140.8?10?18.3?10
L
?100?C12
0.8?10?6
?20.9?kΩ?
20?20?10?12
2
2
RP?Q0
?C2?C0?C1??20?20?20?
?R??RiRP?R?1020.9???5?5.88?kΩ???0
C20??1??R?5.88?103
QL???28.26?6
ω0L2?3.14?41.6?10?0.8?102Δf0.7
f041.6?106???1.48?MHz?QL28.2
3?12解：1?Zf1?0
2?Zf1?03?Zf1?R
3?13解：1?L1?L2?
C1?C2?
ρ1
103
??159?μH?6
2?3.14?10
?01
11
??159?pF?226?6?01L12?3.14?10?159?10
ηR1?20M?1??3.18?μH?6
?012?3.14?10
2?Zf1
??01M?2?
R2
?2?3.14?10?
6
?3.18?10?6??20?Ω?20
2
L1159?10?6
ZP???25?kΩ??12
R1?Rf1C120?20?159?103?Q1?
?01L1R1?Rf1
2?3.14?106?159?10?6??25
20?20
4?2Δf0.7??5?C2
f0f0106?20
?2?2?2??28.2?kHz?
Qρ1R1103
1
2
??02
L2
?
1
2?3.14?950?1032
?159?10
?6
?177?pF?
?1?
Z22?R2?j??L??021?C???
022??
1??
?20?j?2?3.14?106?159?10?6??20?j1006?12?2?3.14?10?177?10??Zf1
??01M?2
?
Z22
?2?3.14?10?
?6
?3.18?10?6??0.768?j3.84?Ω?20?j100
6
2
3?15解：?R?
L159?10
??20????R1
RPC50?103?159?10?12
?Rf1?0?M?0
f0106
Q?2?2?100
2?f0.714?103
3?16解：1?Rf1
??01M?2
?
R2
?10?
7
?10?6??20???5
2
Rab?2???3?Q?
R
R1
??01L?2
1
?Rf1?10?
7
?100?10?6??40?k??5?20
2
?01M?01LR1
107?10?6??2
5107?100?10?6??200
5
2?f0.711
?2?2??1??22?2?2?1??0.013f0Q200 I111
3?17解：????Q??22.5
22I01.25?2?f??10?103?
1??1???Q?300?103???Q?f??
??0??
11R???11.8
Q?0C22.5?2?3.14?300?103?2000?10?12
?2?f??10?103?
??1???Q300?103???Qf??
??0??
Q?Q??30?22.5?7.5
1?
2ω?串联联谐??L1?375μHL2C?
3?18解：???
1?L2?125μ?ω?并联联谐L1?L2C??
I?I0
1
2
?
1
2
?
1
?Q?302
4?5解：当f?1MHz时，??
β0?β0f1???f?Tβ0?β0f1???f?Tβ0?β0f???f?T
????????????
2
?
50?50?106????250?106????
50
2
?49
当f?20MHz时，??
2
?
当f?50MHz时，??
2
?
?50?20?101???250?106
?
50
6
????
2
?12.1
?50?50?106?1???250?106????
2
?5
4?7解：gb?e?
gm?Cb?e
IE1
??0.754?mS?
26β0?126?50?1β0
?50?0.754?10?3?37.7?mS?rb?e
gm37.7?10?3???24?pF?6
2πfT2?3.14?250?10
a?1?rbb?gb?e?1?70?0.754?10?3?1b?ωCb?erbb??2?3.14?107?24?10?12?70?0.1yie
?gb?e?j?Cb?e??a?jb??0.754?10?3???
a2?b2
?0.895?j1.41?mS?
j2?3.14?107?24?10?12???1?j0.1?
12?0.12
j2?3.14?107?3?10?12??1?j0.1?yre??0.0187?j0.187?mS?2222
a?b1?0.1
gm?a?jb?37.7?10?3??1?j0.1?yfe???37.327?j3.733?mS?a2?b212?0.12
?g?j?Cb?c??a?jb??j?C?1?rga?jb?
yoe?gce?j?Cb?c?rbb?gmb?c?b?c?bb?m
a2?b2a2?b2??
1?j0.1??
?j2?3.14?107?3?10?12?1?70?37.7?10?3?2?0.049?j0.68?mS?2?1?0.1??
?Av?
4?8解：令??A??
?vo?
?Av?令??A???vo?
mm
?g?j?Cb?c??a?jb?????b?c
?????
4?Q2Δf4?Q2Δf
2
0.7
f04
?
??1?2?
m
得2Δf0.7
1?m?f0
???4?2?1??Q
??
?
????
2
0.1
f04
?
??1?10?10?12?1
1m2m
m
得2Δf0.1
2?m?f0
???4?10?1??Q
??
故Kr0.1?
2Δf0.1
?2Δf0.7
2?m?
?4??1??
??1?m??4?2?1????
?
《高频电子线路张肃文第五版课后答案》出自：百味书屋
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