高频电子线路习题答案(第五版)张肃文[1]
高频电子第五版
31解:f
1MHz
6
3
2Δf0.711099010Q
f02Δf0.7
QR
110
63
10(kHz)
1010
100
取R10Ω则LC
32解:(
01
2
1001023.1410
6
159(H)
159(pF)
1
(23.1410)15910
1L1C11L1C11L1C1
6
2
6
0L
1)当01或ω02
1L2C21L2C21L2C2
时,产生并联谐振。时,产生串联谐振。
(2)当01(3)当01
或ω02或ω02
1jω0C1jω0C
时,产生并联谐振。
2
(Rjω0L)(R
33证明:Z
Rjω0LR
34解:1由15C1605
2
)
R
LC
jω0LR(1
1ωLC1
)
20
)
R2R
2
LCR
2Rjω0L(1
2
ωLC
20
450C535得C40pF
2
由12C1605故采用后一个2L3
1
20
2
100C535得C-1pF不合理舍去
180μH
。
1
CC
23.1453510450
32
4010
12
L C C’
35解:Q
1
0C0R
1
20
1
23.141.51010010
1
62
-12
6
-12
5
212
L0
C0
VomR1
23.141.510100
-3
112μH
10
Iom
1105
0.2mA
-3
VLomVComQ0VSm212110
36解:L
212mV
12
0C
VCVS
2
1
23.1410
100
62
253μH
10010
Q0
100.11
2
CCXCCXRX
1
0L
23.1410
626
25310
6
100pFCX200pF23.141025310
100
12
6
6
0L
Q
0L
Q0
1
23.141025310
2.50.1
47.7j
6
47.7Ω
ZXRXj
1
23.141020010
12
0CX
6
47.7j796Ω
37解:L
1ωCf0
20
1
23.145105010
6
20.2μH
Q0
2Δf0.72Δff0
510
63
150101003
1003
6
6
ξQ0
25.5510
510
203
21kΩ电阻。
22f0.7,则Q因2Δf0.70.5Q0,故R0.5R,所以应并上0
38证明:4πΔf
C0.7
2πf0Cf02Δf0.7
ω0CQ
g
39解:C
Ci
12π
C2
C0C1
C2C0C1
23.14100
5
202020
2020201
6
18.3pF
f0
LCLC12
0.81018.310
6
12
12
41.6MHz
RPQ0
0.810
202010
2
20.9kΩ
2
C2C0C1202020
RRiRPR1020.955.88kΩ0
C201QL
Rω0L
f0QL
5.8810
6
6
3
6
23.1441.6100.810
41.61028.2
1.48MHz
28.2
2Δf0.7
312解:1Zf10
2Zf103Zf1R
313解:1L1L2
C1C2M
ηR1
ρ1
10
3
6
01
1
2
01
23.1410
159μH1
159pF
L1
23.1410
6
62
15910
6
12023.1410
01
3.18μH
6
62
2Z
f1
01M2
R2L1
23.1410
3.1810
6
2015910
20Ω
ZP3Q1
R
1
Rf1C1
202015910
6
12
25kΩ
01L1
R1Rf1
2102
2
23.141015910
2020
2
f0ρ1R1
6
6
2528.2kHz
42Δf0.75C2
f0Q
21
102010
32
3
L2
23.14950
10
15910
6
177pF
1
Z22R2jL021C
022
166
20j23.14101591020j100612
23.141017710Z
f1
01M2
Z22LRPC
23.1410
15910
3
6
6
3.1810
62
20j100
12
0.768j3.84Ω
315解:R
501015910
20R1
Rf10M0Q
2
f02f0.7
2
10
6
3
1410
100
316解:1Rf1
01M2
R2
10
7
105
62
20
62
Rab23Q
R
R1
01L2
1
Rf17
10
7
10010520
40k
01M01LR1f0
1010
5
7
6
2
6
1010010
5
200
2221
1
2
2f0.7
II0
21
1
2
1Q
1200
1
0.013
Q22.5
317解:
2f1Qf01Q0C
12f1Qf
0
1L2C1L2C
2
3
10101Q300103
2
1.25
RII0
1
22.523.1430010200010
2
3
12
11.8
1
1010Q300103
3
2
12
Q30
QQ3022.57.5
2ω
318解:
ω
L1
L1375μH
L2125μ
串联联谐并联联谐
β0
β0ff
T
2
45解:当f1MHz时,
50
501061250106
2
49
当f20MHz时,
β0
β0ff
T
2
50
5020101250106
6
2
12.1
当f50MHz时,
β0
β0f1f
T
2
50
50501061250106
2
5
47解:gbe
gmCbe
IE
26β01
126501
3
0.754mS
β0rbe
500.75410
37.7mS
gm2πfT
37.710
3
6
23.1425010
24pF
3
a1rbbgbe1700.75410
7
1
12
bωCberbb23.14102410yie
700.1
3
gbejCbeajbab
2
2
0.754
10j23.14102410
10.1
2
2
712
1
j0.1
0.895j1.41mSyreyfe
j23.1410310
10.1
2
2
2
7
12
gbcjCbcajbab
2
2
3
1j0.1
0.0187j0.187mS
gmajbab
2
2
37.710
2
1j0.1
10.1
37.327j3.733mS
yoegcejCbcrbbgm
7
gbc
12
jCbcajbab
2
2
ajb
jCbc1rbbgm2
2
ab
j23.1410310
Av
48解:令Avo
Av
令Avo
mm
1j0.13
0.049j0.68mS17037.71022
10.1
m
2
4Q2Δf
2
0.7
f0
4
12
得2Δf0.7
1f0
m421Q
2m
4Q2Δf
0.1
f0
4
m
110
得2Δf0.1
4
410f10
Q
故Kr0.1
2Δf0.12Δf0.7
410
2m
1
2
10
m1
1
1m421
2m1
49解:p1
gp
N23N131
520
2
0.25
p21
N45N13
520
0.2537.2μS
6
ω0Q0L
2π10.710100410
2
6
66
14
2
ggpp1goep2gie37.210Avo
p1p2yfe
gΣ
2
3
14
2
20010286010
6
228.5μS
2
0.250.254510
228.510
6
12.3
ApoAvo12.3151.3QL
1gΣω0L
f0QL
1
228.51010.71016.3
2
6
6
2π10.710410
Z
66
16.3
2Δf0.7
0.657MH
2
QL
K1Q0ξtangLS
16.3
1
100
o
1.43
o
fe
re2
2
tan
5488.5
2
6
2.95
2
6
gpp2gie
p
2
1
37.2100.25200100.25
2
6
3008.8μS
6
gs
1ξgiegoegL
yfeyre
2
2286010
200
10
6
3008.810
3
4510
3
0.3110
12.951
2
1gp410解:
1Q0ω0L
1
1
10023.1410.710410
2
2
6
6
0.037mS
2
ggp
R5
p1goep2gie0.0370.10.30.0820.30.150.158mS
2
Avo
p1p2yfe
g
0.30.3384.2
22
0.158
21.78
62
22Δf0.73Avo4
ω0Lgf023.1410.710Avo21.78
41
4
410
6
0.15810
3
454.4kHz
225025.38
1
42Δf0.7452Δf0.7
2
4
12Δf0.71044.6kHz
241454.4197.65kHz
2Δf0.7
1
241
2Δf0.72Δf0.71044.6454.4590.2kHzAvo4Avo
Avo4
Avo2Δf0.72Δf0.7
4
21.78454.41044.6
4
9.47
9.47Avo
2
8042.66
4225025.38-8042.66216982.72Avo
2
411解:CCp1Coe5000.318501.62pF
L
1
12
2πf02C
1
23.141.510
2
62
22.5μH
501.6210
Kr0.11.9不能满足
AvoS414解:
yfe2.50Cre
26.436.42.50.3
2
7.74
417解:L1
1ωC1
20
1
2π46510
11873
73
32
10001060
11873
12
118μH
L36L2L34L56
118
13.5120
C12C1Co100041004pF
2
12
C36
13.5
C2pCi1000401004pF
74.5
22
g12go
22
ω0C1Q0
2010
6
2
2π46510100010
100
3
49μS
3
12
g36pgi
ω0C2Q02π4651010001013.53
0.6210
10074.5
。若为临界耦合,即4010
3
49μS
初、次级回路参数相等
p1p2yfe
gω0C12g12
2f0QL
1
13.5
1,则
AvoQL
74.5
6
24910
3
74
12
2π46510100410
4910
2
46510
60
63
60
2Δf0.7
10.9kH
Z
Kr0.13.16
420解:
2
vnin
2
2
4kTRΔfn4kTGΔfn
2
2
2
41.381041.3810
23
23
290100010
-3
7
12.65μV
2901010
7
12.65nA
421解:vnvn1vn2vn34kT1R1Δfn4kT2R2Δfn4kT3R3Δfn
4kT1R1T2R2T3R3Δfn4kTR1R2R3Δfn
T
2
T1R1T2R2T3R3
R1R2R3
2
2
2
又inin1in2in34kT1G1Δfn4kT2G2Δfn4kT3G3Δfn
4kT1G1T2G2T3G3Δfn4kTG1G2G3Δfn
T
T1G1T2G2T3G3
G1G2G3
R1R2T3R2R3T1R3R1T2
R1R2R2R3R3R1
418证明:1Ib1yieV
be1
yreV
ce1
12
yieVce1yreV
cb2Vce1yreV
cb2Vce1yoeV
ce1
cb2
be1
Ic1yfeV
yoeV
ce1
Ib2yieV
be2
yreV
ce2
yieyreV
ce1
34
Ic2yfeV
be2
yoeV
ce2
yfeVce1yoeV
yieyreyoeV
be1
cb2
yfeyoeV
ce1
23得
be1
Ic2yfeV
cb2
yreV
cb2
V
ce1
Ic2yreVyfeV
yieyreyoe
5
cb2
5代入4Ic2
yoeV
cb2yfeyoe
Ic2yreVyfeV
2
be1
yieyreyoe
Ic2yfyo
yfeyfeyoe
2
yieyreyfe2yoe
yfeyfeyoe
V
be1
yieyoeyreyfeyoeyieyreyfe2yoe
V
cb2
6
yieyreyfe2yoe
yfe
yieyoeyreyfeyoeyieyreyfe2yoe
yre
由1乘yfeyoe与4乘yre后相加得Ib1yfeyoeIc2yreyieyfeyoeV由6代入消去
2
be1
yreyoeV
cb2
Ic2得
Ib1
yieyieyreyieyfe2yieyoeyreyfe
yieyreyfe2yoe
yieyieyreyieyfe2yieyoeyreyfe
yieyreyfe2yoe
yreyoeyre
略
23
V
be1
yreyoeyre
yieyreyfe2yoe
V
cb2
2
yiyr
yie
yieyreyfe2yoe
yreyoeyre
yfe
同理可证2
2
422解:vbn4kTrbfn41.3810
ien2qIEfn21.610
2
19
273197020010
3
3
0.22610
12
V
2
10200100.95
3
0.6410
16
A
2
0
f
1f
2
0.95101061500106
2
icn2qIC10fn21.610
2
19
10
3
10.9520010
3
0.3210
17
A
2
423证明:fn
AA
2
fdff0
2
1
ff012Qf0
2
df
f0
2Q
424解:Fn高3dB1.995倍
Fn混1
TiT1
60290
Fn中6dB3.981倍1.207Fn中1
1.995
1.2071Ap高
FnFn高Ap高1.888
Fn混1Ap高
3.98110.2Ap高
KpcAp高
10
20lg1.8882.76dB
2
425解:Fn426解:Fn
PsiPniPsoPnoPsiPniPsoPno
PnoPniAp1Ap
PsPo
1Ap
1PoPs
PsPo
VsVs
2
2
4Rs
4RsR
RsRR
1
RsR
Is4Gs
Is4GsGGLrCL
2
1
GGLrCL
Gs
427解:A为输入级,B为中间级,C为输出级。
APA6dB3.981倍FnFnA
FnB1ApA
ApB12dB15.849倍FnC1ApAApB
1.7
213.981
413.98115.849
2
1.9951100.1
428解:不能满足要求。设
Fn
PsiPniPsoPno
2
A前置放大器,FnA
FnB1ApA
2
B为输入级,FnC1
FnA
C为下一级。10110
FnA8.1
1010
54
ApAApB
58解:ikv
kV0Vmcos0t
121222
kV02V0Vmcos0tVmVmcos20t
22
当VmV0时,ikV02V0Vmcos0t,该非线性元件就能近
2
似当成线性元件来处理
Vm很小时,根据泰线弯曲部分,故
,
即当V0较大时,静态工作点选勒级数原则,可认为信可只取其级数的前两项
在抛物线上段接近线性部分,然后当
号电压在特性的线性范得到近似线性特性。
围内变化,不会进入曲
o
512解:为了使
cos60
o
iC中的二次谐波振幅达到
VBZVBB
Vm
12
最大值,C应为60。12
VmVBZ
VBB
gVmcost
513解:i
0
当cost0当cost0
i
I
n0
n
cosnt
I0I1
121
gVmcostdt
2
1
12
gVmgVm
n为偶数n为奇数
当cost0当cost0
1
gVmcostdt
In
gVm
12
gVm2
costcosntdtn1
0
当cost0当cost0
515解:iiD1iD2
gVmcost
iD1
0
2
4
gVmcostiD2
0
1k1
k1,2,3,igVmgVm0t2
k12k1
516解:当V01msintsin0t0时,i0;
当V01msintsin0t0时,igV01msintsin0t
cos2kω0tsinΩtcos2kω0t
igV01msinΩt22m22
4k14k1k1k1
2
k1,2,3,
517解:v0RLiD1iD2RLkv1v2kv1v2
2
2
4kR
3
L
518解:v0RLi2i3RLi4i1RLi2i4i1i3
RLb0b1v1v2b2v1v2b3v1v2
2
v1v2
RLRLRL
bbb
3
000
b1v1v2b2v1v2b3v1v2
2
b1v1v2b2v1v2b3v1v2
2
2
3
3
b1v1v2b2v1v2b3v1v2
8RLb2v1v2
1gm523解:
diCdvBE
b12b2vBE3b3vBE4b4vBE
b12b2V0mcos0t3b3V0mcos0t4b4V0mcos0t
vBEv0
2
2
3
3
23
gmt
diCdvBE
b12b2V0mcos0tgm12b2V0m3b4V0mgc
12
3
32
b3V0m1cos202b4V0mcos0tb4V0mcos0tcos30t
2
3
3
gm1b2V0m1.5b4V0m
aISqkT
q
3
2gm
diCdvBE
vBEekTaISqkT
vBE
gmtaISqkT
diCdvBEVom
vBEv0
q
Vomcos0tekT
omcos0t
23
q1q1q
cos0t1Vomcos0tVomcos0tVomcos0t
kT2kT6kT
IS
qVomkT
ISqVomISqVomqVom234
cos0tIScos0tcos0tcos0t
2kT6kTkTkT
3ISqVom
8kT
3
234
gm1IS
12
qVom
gcgm1
ISqVom
2kT
3ISqVom
16kT
2
3
525解:ii1i3i2i4
a0a1v0vsa2v0vsa3v0vsa4v0vs
3
4
a0a1v0vsa2v0vsa3v0vsa4v0vs
2
3
4
a0a1v0vsa2v0vsa3v0vsa4v0vs
2
3
4
a0a1v0vsa2v0vsa3v0vsa4v0vs
2
3
4
8a2v0vs16a4v0vs16a4vsv0
33
529解:gc0.5
IE26sIE
1rbb
T26
2
0.5IE26
0.50.526
9.6mS
gicgbe
IE260
2
0.52635
0.55mS
gocgce4SApcmax
gc4gicgoc
9.6
2
2
40.550.004
1047340dB
2
122830.1dB
2
Apc
QL
ApcmaxApcmax1Q0
IE26sIE
1rbb26T
2
530解:gc0.5
2fi2465
1104731
Q02f0.710010
0.5IE0.50.081.54mS2626
gicgbe
IE260
2
0.082630
0.1mS
gocgce10SApcmax
gc4gicgoc
2
1.54
2
40.10.01
592.928dB
2
4
Apc
gc
gG
LocGL0.11.5419623dBg0.10.010.1ic
2
3
532解:ii1i2i3i4
a0a1v0vsa2v0vsa3v0vsa4v0vs
aa
a0a1v0vsa2v0vsa3v0vsa4v0vs
2
3
4
2
3
00
44
a1v0vsa2v0vsa3v0vsa4v0vsa1v0vsa2v0vsa3v0vsa4v0vs
2
3
3
3
fnf0fi
,可能产生
8a2v0vs16a4v0vs16a4vsv0
534解:因存在二次项,能进行混频。只要满足
fnfi就会产生中频干扰;当
生交调干扰;有二次项扰。
时产生镜像干扰。由于互调干扰;若有强干扰
不存在三次项,不会产信号,则能产生阻塞干
535解:1.此现象属于组合频率干
还存在一些谐波频率和它就能和有用中频一道的非线性效应,与中频
扰。这是由于混频器的组合频率,如果这些组进入中频放大器,并被差拍检波,产生音频,
输出电流中,除需要的合频率接近于中频放大放大后加到检波器上,最终出现哨叫声。数,不考虑大于
中频电流外,的通带内,通过检波器
2.因fi465kHz,p、q为本振和信号的谐波次
~1605kHz波段内的干扰在
931kHz、1394kHz、1396kHz时产生。3.提高前端电路的选择性
,合理选择中频等。
3的情况。所以落于535
fS930kHz和fS1395kHz附近,1kHz的哨叫声在
929kHz、
536解:若满足
pf1qf2fs,则会产生互调干扰:
p1、q1,f1f277410351.809MHz,不会产生互调干扰;p1、q2,f12f2774210352.844MHz,会产生互调干扰;p2、q1,2f1f2277410352.583MHz,会产生互调干扰;p2、q2,2f1f227741035
3.618MHz,会产生互调干扰;
p2、q3,2f13f22774310354.653MHz,会产生互调干扰;p3、q2,3f12f23774210354.392MHz,会产生互调干扰;p3、q3,3f1f237741035p、q大于3谐波较小,可以不考虑
。
5.427MHz,会产生互调干扰;
3f2f02
1S537解:fSf00.8MHz
2fS3f02
fS2f02
fSf00.4MHz
2f3f2S0
fS0.2MHz
f00.6MHz
3fs2f0302fs3f030
2fsf012MHz
fs2f030
fsf020MHz
2fsf030fS4MHz
539解:若满足
已知f119.6MHz、
。
f016MHz
pf1qf2fs,则会产生互调干扰。
f219.2MHz、fsf0fi23320MHz,故没有互调信号输出
64解:PVCCICO240.256W
CRpIcm1
P0PVcm
2
56
83.3%VCC2P02P0VCC
o2
2P02P0Vcm
24
2
252524
57.60.417A
gcc
Icm1Ic0
0.420.25
1.67
查表得c77
20.712
10.8
1.56查表得θc91
o
66解:gcθc
2
2ηVCCVcm
2
P0IkR214W
11
P0PCPP01141.7Wη
0.7c
Ic090
67解:icmax282mAo
α0900.319
Ic1mα190P0ηc
68证:P0
o
i
cmax
0.5282141mA2000.1412
74%
2
12
RpIc1mP0
2
12
2
2W
VCCIc0Vcm2RP
2
300.09
2
IkmR0L
2
2LRC
IkmR0LRC
2
2
2
2L2.20.8
IkmR0LR
2
2
2
20L
2
IkmR2
2
icmaxgcr
24
21.25V
69解:VcmVCCvcminVCC
Ic0icmax070Icm1icmax1
12
o
2.20.2530.5566A702.20.4360.9592A
o
PVCCIc0240.556613.36WP0
VcmIcm1
12
21.250.959210.19W
PCPP013.3610.193.17W
CRp
P0PVcm
2
10.1913.36
76.3%
2
21.25
2P0
210.19
22.16
610解:R1Rp
XC1C1X
R1QL12fXC1
R2R1
L1
X
L1
Vcm2P014410
2
VCC2P0
2
24
2
22
144
14.4
1
221pF
23.145010
1116.95
6
14.4200
R2
L1
Q
2L
200144
6
16.95
10011
2f
23.145010
0.054H
XC2C2
QLR1R2
2
QL1QLX1
L1
10144200
112.5710011016.95
1
6
2fXC2
23.1450102.57
1239
pF
2
cm
1RP增加一倍,放大器工作611解:
2RP减小一半,放大器工作
612解:k
rr1rVcm2P0
2
于过压状态,于欠压状态,
11
1Q1Q2k
2
Vcm变化不大,P0V
2
/2R
P
0.5P0;
Icm变化不大,P0IcmRP/22P0。
2
1
11
1
1
100150.03
2
L1L2Q1Q2MVCEsat2P0
RPQL1
82
57.4%
2
613解:RP
V
CC
120.521
66
设QL10C1
12fXC1
则XC1
6610
6.6241pF50
23.14106.6
L
5.5
XC2
RL
1QR
2L
RP
1
1
11050
2
8
66
1
C2X
12fXC2
23.14105.5
290pF
L1
X
QLRPRL
12
QL1QLXC2
L1
1066
1021
19.9nH
50112.5105.5
L1
2f
12.523.1410
8
a将R1C1和R2C2串联电路改为614证:
R1
XC1R1XC1
2
22
2
C1和RR1C并联电路,并设22
R2
11QL
2
XC1XC1
R1QL2XC
R2
2
2
2
R1R2
XC2R2XC2XC1
2
22
2
1XCR1
R1
2
2
2
R1XC1XC2
22
2
R2XC2
XC2
,即R1匹配时R1R2XC2
R2
R1XC1
2
R1
R2XC2
R2
1QR
2
L
2
R1
21
1R1
2
2C12
X
L1
1XC2XC
R1
2
2
2
RX
XC1
R2
2
2
2
2C2
RX
XC2
11QL
2
R1XC1
XC1
XC1R1XC1
2
2
R1R2XC2
R1QL1QL
2
R1R2XC2
R1QLR2
12
1QLQLXC2
R1QL
XL1
R2
2
2
b将R1C1和R2L1串联电路改为
R1
XC1R1XC1
2
22
C1和R并联电路,并设R1L1
2XL1
2
XC1XC1
R1R2
R2XL1XC1
2
2
C1
22
R2
11Q
2L
1XCR1
R1
2
2
2
R1XC1
XL1RX
2
2
2L12
R2XL1
2
XL1
,即R1匹配时R1R2XL1
R2
RX
2
1
R1R2
1QR
2L
2
R1
1R2
22
2
2L1
1XC1XC2XL
XC1R1XC1
2
22
RX
XL1
R1
2
1
2
2C1
RX
XC1
R1QL1QL
2
R1R2XL1
R1QL1QL
2
R1QL1QL
2
R2QLXL1
R1QLR2
121QLQLXL1
1天线断开,工作于过压618解:
2天线接地,工作于欠压3中介回路失谐,工作于1PAPPCPk10619解:
2k3c
PAPPCPPC
P
610310310
状态,集电极直流电表状态,集电极直流电表欠压状态,集电极直流
316W
读数减小,天线电流表读数略增,天线电流表电表读数略增,天线电
读数为0;读数增加;流表读数减小。
85.7%70%
PAP
610
60%
620解:当kkc时,k1
1若k
r1r1r
r1r1r
50%则rr1
1Q1Q2
kc
1Q0
1Q
3Q3kc
90%则r9r1故k
Q09
625解:
627解:o
260
60o
1
,P0
减小,工作于欠压状态
V
628解:RV
i
I933I9RL
629解:Vbm
VBZVBB0.6cos1.45cos70
o
6V
VBVbmVBB61.454.55V
iBVBZ
4.550.6
Cmax
V2
2
1.98AIo
cm1iCmax170
1.980.436
0.86A
VcmVCCgcriCmax241.9822.02V
PIcm1Vcm
.8622.01
0
2
02
9.47W
PQLA1P10019.478Q.52W0100 630解:VVBZVBB0.61.5
bmcoscos70
o
6.14VVBVbmVBB6.141.54.64V
iVBVBZ
4.640.6
Cmax
2
2
2.02AIo
cm1iCmax170
2.020.436
0.88A
VcmVCC0.92421.6V
PIcm1Vcm
.8821.6
0
2
02
9.5W
PQLA1P10019.58.Q8W0100
。
a电路可能振荡,属于电75解:
感反馈式振荡电路;
e电路可能振荡,属于电容反馈式振荡电路;h电路可能振荡,属于电容反馈式振荡电路;b、c、d电路不可能振荡;
f电路在L2C2L3C3时有可能振荡,属于电容反馈式振荡电路g电路计及Cbe可能振荡,属于电容反馈式振荡电路。1有可能振荡,属于电容76解:反馈式振荡电路,f1f2f0f3;2有可能振荡,属于电感反馈式振荡电路,f1f2f0f3;4有可能振荡;属于电容反馈式振荡电路,f1f2f0f3;356不可能。
77解:
1f0721解:
2gd
1
2π1RP
LCCd1Q
6
23.1413
1
7
205105.27mS
12
100MHz
CL
2051010
7
12
30.06~0.08V
1fq726解:
1.65710
d
5
Sd
1.65710
0.421.1100.4
3
6
4.14MHz
5
Cq21.110Lq43.5
d
3
2000.4
0.105pF
S
43.5dS
2
200
14mH
21.2Ω19.8pF
rd42500B
425000.25Sd
3.96101.050.25
42
0.42002000.4
C03.9610Qq
1.05Bfq
10d
6
4
100.416800
6
2d
1.65710
1.657101510
6
0.11mm
11.5~1.5001MHz727解:
负阻特性。
、高稳定度克拉泼振荡
电路。
2不能
3不能,普通三极管没有
728解:恒温槽、稳压电源
电路、共集电极缓冲级
等。
729解:并联cb型(皮尔斯)晶体振荡
93解:iI1macosΩtcosω0t
Icosω0t
I2
2
macosω0Ωt
2
I2
macosω0Ωt
2
I有效值
III
mama
22222I2
1
ma2
2
1v2510.7cos2π5000t0.3cos2π10000tsin2π106t94解:
25sin2π10t8.75sin2π1005000sin2π995000
6
3.75sin2π1010000sin2π990000
2包络2510.7cos2π5000t0.3cos2π10000t
m上m下
VmaxV0
V0V0Vmin
V0
2510.70.325
25
252510.70.3
25
0.4
峰值调幅度
谷值调幅度1
14
10025W1
2
1ma1时95解:
2ma
0.3时
3
Pω0ΩPω0Ω
14
maP0T14
2
2
Pω0ΩPω0Ω
maP0T
96解:ib1vb3v不包含平方项,不能产
4
生调幅作用。
0.31002.25W
1Pω0ΩPω0Ω97解:
P0avη
P0Tη
14
maP0T
2
14
0.75000612.5W
2
Pω0Ω2Pω0Ω1225W
2P
50000.5
10kW
3P
P0avη
1ma1时98解:
22
ma0.7P0T50001212
12.45kW
η0.5
Pω0ΩPω0Ω
14
maP0T
2
14
1000250W
P0P0TPω0ΩPω0Ω10002502501500W
2ma
0.7时
14
maP0T
2
Pω0ΩPω0Ω
14
0.71000122.5W
2
P0P0TPω0ΩPω0Ω1000122.5122.51245W
99解:ff0f1f2f3f45202001780800010005kHz
910解:i1b0b1vvΩb2vvΩ
i2b0b1vvΩb2vvΩ
22
b3vvΩ
3
3
3
b3vvΩ
2
v0i1i2RR2b1vΩ4b2vvΩ6b3vvΩ2b3vΩ
2
3
2b1RVΩcosΩt3b3RV0VΩcosΩt1.5b3RVΩcosΩt2b2RV0VΩcosω0Ωt2b2RV0VΩcosω0Ωt1.5b3RV0VΩcos2ω0Ωt1.5b3RV0VΩcos2ω0Ωt
2
2
3
0.5b3RVΩcos3Ωt输出端的频率分量:
、3、0、20
912解:m1
P0P0
P0
21P
0T12
2
10.125
210.5
9
12
0.4910.845kW
2
m2P0T10.125
9131vAtvtVcost vBtvt 2若D1D2开路,则vAtvBtvt
vAB0
3若D1D2短路,则vAtvtVcost
vB0
vABtVcost
918解:RR1
R2ri2R2ri2
510
47001000470010000.57
1335
3RdR
33.141005104700
Kdcos0.87
VKdmaVim0.870.30.50.13PPAP
V
2
2
2RVim2RidPP
2
0.13
2
21335VimKd
R
6.33W0.50.875104700
2
41.7W
6.3341.7RR
0.152R1R22
R22ri2R22ri2
23501000
5102350
1中间位置919解:
R1R2R2ri2R2ri2
510
0.55
5104700
2最高端
RR
R1
47001000
R1R2
0.265104700
最高端不会。
R2的触点在中间位置会产
生负峰切割失真,而在
920解:由
RR1R25~10k
R2ri2R2ri2
2
11R1~R2
5103k
2
取R26kRR3913
R11.5k
RR1
1.5
6262
ma取ma0.3
C
1-ma
1
maRmax
minri23RdRR2Kd
10.31
0.3900023.14300023.14300200033.1410060001500
0.5
0.0187F
取C1C20.01F
Ce
0.26F
取Ce20F
Kdcos0.9Rid
900020.9
5k能满足要求
921解:GP
QL
0CQ0f0
23.144651020010
100
2023.5
312
5.84S
465
2f0.7Q0QL
GPp24goeGP
gid
2
p34
10026
5.840.31005.841023.50.153
2
4700
1v1mV1costcos1t924解:
ikmV1costcos1tV0cos0t14
kmV1V0cos10tcos10t
cos10tcos10t当01时,vS
1412kmR
VVcostcostL10
12
kmRLV1V0coscost
无失真,只影响输出幅度。当01时,vS有失真。
kmRLV1V0cos10tcost
2v1
1212
mV1cos1t
i
kmV1cos1tV0cos0t
14
kmV1V0cos10tcos10t14
kmRLV1V0cos10t
vS
当01时,只产生相移;当
01时,有失真。