高频电子第五版
31解:f
1MHz
2Δf0.7110699010310(kHz)f01106
Q1003
2Δf0.71010
取R10Ω则L
C
QR
0
1
10010
159(H)6
23.1410
1
159(pF)
02L(23.14106)2159106
11
32解:(1)当01或ω时,产生并联谐振。02
L1C1L2C2
(2)当01(3)当01
1
或ω02L1C1
1
或ω02L1C1
1
时,产生串联谐振。L2C2
1
时,产生并联谐振。L2C2
1L1
(Rjω0L)(R)R2jω0LR(12)R2L
jω0CCω0LCR33证明:Z112RRjω0LR2Rjω0L(12)jω0Cω0LC 34解:1由15C16052450C5352得C40pF
由12C16052100C5352得C-1pF不合理舍去
故采用后一个。
2L3
L C C’
11
180μH2
CC23.1453510324504010120
35解:Q
11
2126-12
0C0R23.141.510100105
L0
11
112μH226-120C023.141.51010010
Vom110-3
Iom0.2mA
R5
VLomVComQ0VSm212110-3212mV
11
36解:L2253μH
0C23.1410621001012
Q0
VC10100VS0.1
CCX11
2100pFCX200pF
CCX0L23.141062253106RX
0L
Q
0L
Q0
23.1410625310623.1410625310647.7Ω
2..1100
11
47.7j47.7j796Ω612
0CX23.14102001011
37解:L220.2μH
ω0C23.145106501012
ZXRXj
f05106100
Q0
2Δf0.71501033
2Δf10025.5510620ξQ06
f035103
22f0.7,则Q因2Δf0.700.5Q0,故R0.5R,所以应并上21kΩ电阻。
2πf0CωC
38证明:4πΔf0.7C0g
f02Δf0.7Q
CC0C1520202018.3pF39解:CCi2
C2C0C1202020
f0
1141.6MHz612
2πLC23.140.81018.310
L
100C12
0.8106
20.9kΩ
20201012
2
2
RPQ0
C2C0C1202020
RRiRPR1020.955.88kΩ0
C201R5.88103
QL28.266
ω0L23.1441.6100.8102Δf0.7
f041.61061.48MHzQL28.2
312解:1Zf10
2Zf103Zf1R
313解:1L1L2
C1C2
ρ1
01
103
159μH6
23.1410
11
159pF226601L123.141015910
ηR120M13.18μH6
0123.1410
2Zf1
01M2
R2
23.1410
6
3.1810620Ω20
2
L1159106
ZP25kΩ12
R1Rf1C12020159103Q1
01L1
R1Rf1
23.1410615910625
2020
42Δf0.75C2
f0f010620
22228.2kHz
Qρ1R1103
1
2
02
L2
1
23.149501032
15910
6
177pF
1
Z22R2jL021C
022
1
20j23.1410615910620j10061223.141017710Zf1
01M2
Z22
23.1410
6
3.181060.768j3.84Ω20j100
6
2
315解:R
L15910
20R1
312
RPC501015910
Rf10M0
f0106
Q221003
2f0.71410
316解:1Rf1
01M2
R2
10
7
106205
2
Rab23Q
R
R1
01L2
1
Rf110
7
10010640k520
2
01M01L
R1
1071062
5107100106200
5
2f0.711
221222210.013f0Q200 I111
317解:Q22.5
22I01.252f10103
11Q300103Qf
0
11R11.8
Q0C22.523.1430010320001012
2f10103
1Q300103Qf
0
QQ3022.57.5
1
2ω串联联谐L375μHLC2
318解:1
1L2125μω并联联谐L1L2C
II0
1
2
1
2
1
Q302
45解:当f1MHz时,
β0β0f1fTβ0β0f1fTβ0β0ffT
2
50501061250106
50
2
49
当f20MHz时,
2
当f50MHz时,
2
5020101250106
50
6
2
12.1
5050106250106
2
5
47解:gbe
gmCbe
IE1
0.754mS
26β0126501β0
500.75410337.7mSrbe
gm37.710324pF6
2πfT23.1425010
a1rbbgbe1700.7541031bωCberbb23.14107241012700.1yie
gbejCbeajb0.754103
a2b2
0.895j1.41mS
j23.141072410121j0.1
120.12
j23.14107310121j0.1yre0.0187j0.187mS2222
ab10.1
gmajb37.71031j0.1yfe37.327j3.733mSa2b2120.12
gjCbcajbjC1rgajb
yoegcejCbcrbbgmbcbcbbm
a2b2a2b2
1j0.1
j23.141073101217037.710320.049j0.68mS210.1
Av
48解:令A
vo
Av令Avo
mm
gjCbcajbbc
4Q2Δf4Q2Δf
2
0.7
f04
1
m
得2Δf0.7
1mf0
421Q
2
0.1
f04
11010121
1m2m
m
得2Δf0.1
2mf0
4101Q
故Kr0.1
2Δf0.1
2Δf0.7
2m
4101
1m421
49解:p1
gp
N235
0.25N1320
p2
N455
0.25N1320
11
37.2μS66
ω0Q0L2π10.710100410
116
200102860106228.5μS2244
2
ggpp12goep2gie37.2106
Avo
p1p2yfe
gΣ
0.250.2545103
12.36
228.510
2
ApoAvo12.32151.3
QL2Δf0.7
11
16.3666
gΣω0L228.5102π10.710410f010.71060.657MHZQL16.3
2
2
QL16.3
K111.43Q0100
fere54o88.5o
ξtantan2.95
222
gpp2gie37.21060.252200106
g3008.8μSL
p120.252
yfeyre
26662
gsgiegoeg1ξ12286010200103008.81012.95LS
451030.31103
1gp410解:
11
0.037mS66
Q0ω0L10023.1410.710410
12
p12goep2gie0.0370.10.320.0820.320.150.158mSR5
0.30.33824.2221.78
0.158
ggpAvo
p1p2yfe
g
22Δf0.7ω0Lgf023.1410.7106241060.158103454.4kHz3Avo4Avo421.784225025.3842Δf0.7452Δf0.7
14
212Δf0.721454.4197.65kHz1044.6kHz
1414
2Δf0.7
21
2Δf0.72Δf0.71044.6454.4590.2kHzAvo
Avo2Δf0.721.78454.4
9.472Δf0.71044.6
4Avo49.4748042.66Avo
4225025.38-8042.66216982.72Avo4Avo
411解:CCp12Coe5000.3218501.62pF
L
1
2πf02C
1
23.141.510
62
501.6210
12
22.5μH
Kr0.11.9不能满足
AvoS414解:
yfe2.50Cre
26.4236.42
7.74
2.50.3
417解:L1
11
118μH22312ω0C12π46510100010
118118118
L36L2L34L566013.5120
737373
C12C1Co100041004pF
13.5
C36C2pCi1000401004pF
74.5
ω0C12π465103100010126
g12go201049μS
Q0100
22
2
ω0C213.52π465103100010123
g36pgi49μS0.6210
Q010074.5
初、次级回路参数相等。若为临界耦合,即1,则
22
2
Avo
p1p2yfe
g
1
13.5
40103
74249106
ω0C122π46510310041012
QL606
g1249102Δf0.7
f04651032210.9kHZ
QL60
Kr0.13.16
2
420解:vn4kTRΔfn41.381023290100010712.65μV
2
in4kTGΔfn41.38102329010-310712.65nA
2222
421解:vnvn1vn2vn34kT1R1Δfn4kT2R2Δfn4kT3R3Δfn
4kT1R1T2R2T3R3Δfn4kTR1R2R3ΔfnT
T1R1T2R2T3R3
R1R2R3
2222
又inin1in2in34kT1G1Δfn4kT2G2Δfn4kT3G3Δfn
4kT1G1T2G2T3G3Δfn4kTG1G2G3ΔfnT
T1G1T2G2T3G3R1R2T3R2R3T1R3R1T2
G1G2G3R1R2R2R3R3R1
418证明:1Ib1yieVbe1yreVce11
Ic1yfeVbe1yoeVce12Ib2yieVbe2yreVce2Ic2yfeVbe2yoeVce2
yieVce1yreVcb2Vce1yreVcb2yieyreVce13
yfeVce1yoeVcb2Vce1yoeVcb2yfeyoeVce14
23得Ic2yfeVbe1yieyreyoeVce1yreVcb2
Vce1
Ic2yreVcb2yfeVbe1
yieyreyoe
5
Ic2yreVcb2yfeVbe1
yieyreyoe
5代入4Ic2yoeVcb2yfeyoe
Ic2
2yfeyfeyoeyieyoeyreyfeyoe
Vbe1Vcb26yieyreyfe2yoeyieyreyfe2yoe
yf
yfeyfeyoeyfeyieyreyfe2yoe
由1乘yfeyoe与4乘yre后相加得
Ib1yfeyoeIc2yreyieyfeyoeVbe1yreyoeVcb2由6代入消去Ic2得Ib1
2yieyieyreyieyfe2yieyoeyreyfeyreyoeyreVbe1Vcb2
yieyreyfe2yoeyieyreyfe2yoe
2
yieyoeyreyfeyoe
yoyre
yieyreyfe2yoe
2yieyieyreyieyfe2yieyoeyreyfe
yiyie
yieyreyfe2yoe
yr
yreyoeyreyyyrereoeyieyreyfe2yoeyfe
略
同理可证2
2
422解:vbn4kTrbfn41.38102327319702001030.2261012V2
2
ien2qIEfn21.610191032001030.641016A2
0
f
f
2
0.95101061500106
2
0.95
2
icn2qIC10fn21.6101910310.952001030.321017A2
423证明:fn
A2fdfA2f0
1
2
424解:Fn高3dB1.995倍
Fn混1FnFn高
ff0
12Qf0
Fn中6dB3.981倍
df
f0
2Q
Ti6011.207T290
F1Fn中11.20713.9811n混1.99510
Ap高KpcAp高Ap高0.2Ap高
20lg1.8882.76dB
2
s
Ap高1.888
425解:Fn
PsiPniPPVRsRRR11
nos2s1s
PsoPnoPniApApPoPsPoVs4RsRRR PsiPniPsIs24GsGGLrCL1
426解:Fn21
PsoPnoApPoIs4GsGGLrCLGs
427解:A为输入级,B为中间级,C为输出级。
APA6dB3.981倍FnFnA
ApB12dB15.849倍
F1FnB12141
nC1.72ApAApAApB3.9813.98115.849
428解:不能满足要求。设A前置放大器,B为输入级,C为下一级。
PsiPni105F1F11011.9951Fn4FnAnBnCFnAFnA8.1
PsoPno10ApAApAApB10100.1
58解:ikvkV0Vmcos0t
2
2
1212kV022V0Vmcos0tVmVmcos220t
22
当VmV0时,ikV022V0Vmcos0t,该非线性元件就能近似当成线性元件来处理,即当V0较大时,静态工作点选在抛物线上段接近线性部分,然后当Vm很小时,根据泰勒级数原则,可认为信号电压在特性的线性范围内变化,不会进入曲线弯曲部分,故可只取其级数的前两项得到近似线性特性。
512解:为了使iC中的二次谐波振幅达到最大值,C应为60o。
cos60o
VBZVBB1
Vm2
1
VBBVmVBZ
2
gVcost
513解:im
0
iIncosnt
n0
当cost0当cost0
11
I0gVcostdtgVmm
211I1gVmcos2tdtgVm
2
121gVm2
IngVmcostcosntdtn1
0
515解:iiD1iD2
n为偶数n为奇数
gVcostiD1m
0
2
4
当cost0当cost0
gVcostiD2m
0
当cost0当cost0
k1
1k1,2,3,igVmgVm0t2
k12k1
516解:当V01msintsin0t0时,i0;
当V01msintsin0t0时,igV01msintsin0t
517解:v0RLiD1iD2RLkv1v2kv1v24kRLv1v2
2
2
cos2kω0tsinΩtcos2kω0t
igV01msinΩt22m22
4k14k1k1k1k1,2,3,
2
518解:v0RLi2i3RLi4i1RLi2i4i1i3
RbRbRb
LLL
RLb0b1v1v2b2v1v2b3v1v2
2
2
3
3
b1v1v2b2v1v2b3v1v2b1v1v2b2v1v2b3v1v2
2
2
3
00
b1v1v2b2v1v2b3v1v2
3
8RLb2v1v2
1gm523解:
diC23
b12b2vBE3b3vBE4b4vBE
dvBE
diCdvBE
b12b2V0mcos0t3b3V02mcos20t4b4V03mcos30t
vBEv0
gmt
b12b2V0mcos0tgm12b2V0m3b4V03m
3
b3V02m1cos202b4V03mcos0tb4V03mcos0tcos30t2
1
gm1b2V0m1.5b4V03m2
qvBEdiCaISqkT
2gmvBEe
dvBEkTgcdigmtC
dvBE
omcos0taIq
SVomcos0tekT
kT
q
vBEv0
23
aISqq1q1q
Vomcos0t1Vomcos0tVomcos0tVomcos0tkT2kT6kTkT
qVomIqVIqVqVIScos0tISomcos20tSomcos30tSomcos40t
kT2kT6kTkTgm1
qVom3ISqVomIS
kT8kT
3
234
IqV3ISqVom1
gcgm1Som
22kT16kT
525解:ii1i3i2i4
22
3
a0a1v0vsa2v0vsa3v0vsa4v0vs
3
4
a0a1v0vsa2v0vsa3v0vsa4v0vs
3
4
a0a1v0vsa2v0vsa3v0vsa4v0vs
2
3
4
a0a1v0vsa2v0vsa3v0vsa4v0vs
2
3
4
3
8a2v0vs16a4v0vs16a4vs3v0
529解:gc0.5
IE26
2
gicgbe
sIE1r26bbTI0.5E0.55mS2602635
0.5IE0.50.5
9.6mS2626
gocgce4SApcmax
2
gc9.621047340dB4gicgoc40.550.004
2
2
2
QL2fi2465122830.1dBApcApcmax1A1104731pcmaxQQ2f10010000.7
IE260.5IE0.50.08
530解:gc0.51.54mS22626sIE
126rbbT
gicgbe
IE0.080.1mS2602630
gocgce10SApcmax
2gc1.542592.928dB4gicgoc40.10.01
2
2
gcGL1.540.1ApcgGg0.010.10.119623dBLicoc
532解:ii1i2i3i4
aa
a0a1v0vsa2v0vsa3v0vsa4v0vs
2
3
4
2
3
4
2
3
a0a1v0vsa2v0vsa3v0vsa4v0vs
00
2
3
44
a1v0vsa2v0vsa3v0vsa4v0vsa1v0vsa2v0vsa3v0vsa4v0vs
3
8a2v0vs16a4v0vs16a4vs3v0
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fnf0fi
时产生镜像干扰。由于不存在三次项,不会产生交调干扰;有二次项,可能产生互调干扰;若有强干扰信号,则能产生阻塞干扰。
535解:1.此现象属于组合频率干扰。这是由于混频器的输出电流中,除需要的中频电流外,
还存在一些谐波频率和组合频率,如果这些组合频率接近于中频放大的通带内,它就能和有用中频一道进入中频放大器,并被放大后加到检波器上,通过检波器的非线性效应,与中频差拍检波,产生音频,最终出现哨叫声。
2.因fi465kHz,p、q为本振和信号的谐波次数,不考虑大于3的情况。所以落于535
~1605kHz波段内的干扰在fS930kHz和fS1395kHz附近,1kHz的哨叫声在929kHz、931kHz、1394kHz、1396kHz时产生。
3.提高前端电路的选择性,合理选择中频等。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、q2,2f1f2277410353.618MHz,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、q3,3f1f2377410355.427MHz,会产生互调干扰;p、q大于3谐波较小,可以不考虑。
3f2f021S537解:fSf00.8MHz2fS3f02fS2f02
fSf00.4MHz
2f3f2S0
fS0.2MHzf00.6MHz
2
3fs2f030
fsf012MHz
2f3f30s0
fs2f030
fsf020MHz
2fsf030fS4MHzf016MHz
539解:若满足pf1qf2fs,则会产生互调干扰。已知f119.6MHz、
f219.2MHz、fsf0fi23320MHz,故没有互调信号输出。
64解:PVCCICO240.256W
C
P05
83.3%P6
22VcmVCC242
Rp57.6
2P02P025
Icm1
2P02P025
0.417AVcmVCC24Icm10.42
1.67Ic00.25
gcc
查表得c77o
2ηVCC20.712
66解:gcθc1.56查表得θc91o
Vcm10.8
P0Ik2R2214W
11
PCPP01P141.7Wη0
0.7c
Ic090
67解:icmax282mAo
α0900.319
Ic1mα190oicmax0.5282141mA
11
RpIc21m2000.14122W22P02
ηc74%
VCCIc0300.09 P0
22222VcmIkmR20LIkmR20LRCIkmR20LRIkmR
68证:P02
2L2RP2L220LRC
i2.2
69解:VcmVCCvcminVCCcmax2421.25V
gcr0.8
2
2
2
Ic0icmax070o2.20.2530.5566A
Icm1icmax170o2.20.4360.9592APVCCIc0240.556613.36W11
P0VcmIcm121.250.959210.19W
22
PCPP013.3610.193.17W
C
P010.1976.3%P13.36
2Vcm21.252
Rp22.16
2P0210.19
22VcmVCC242
610解:R1Rp144
2P02P022
XC1C1
R1144
14.4QL10
11
221pF6
2fXC123.14501014.4
R2
R22
QL11R1
200200
10011144
16.95
XL1
L1
XL116.95
0.054H6
2f23.145010
2
1RP增加一倍,放大器工作于过压状态,Vcm变化不大,P0Vcm611解:/2RP0.5P0;
22RP减小一半,放大器工作于欠压状态,Icm变化不大,P0IcmRP/22P0。
10144QLR1R2200
112.572QL1QLXL110011016.9511
C21239pF
2fXC223.14501062.57 XC2
612解:k
r
r1r
22
VCCVCEsatVcm120.5613解:RP662P02P021
2
111
57.4%
L1L211
11122
QQk100150.032Q1Q2M12
设QL10C1
则XC1
RP66
6.6QL10
11
241pF2fXC123.141086.6
RL
R
1QR
2L
L
XC2
1
50
P
110501
66
2
5.5
C2
11
290pF2fXC223.141085.5
QLRPRL106650
1112.522QL1QLXC2101105.5
XL1L1
XL112.519.9nH2f23.14108
C1和Ra将R1C1和R2C2串联电路改为R1614证:2C2并联电路,并设XC1
2
XC
R1212R1
R1XC1
2XC
222R2R2
R2XC2
R1QL
2R2
22XCXC2
2
R2XC2
R12
12XCXC1
2
R1XC1
22
XCXC11
,即R12匹配时R1R2R1R1222R2
22
R1XC11QLR2XC2
XC2
R2
1QRR
2L
2
1
1
XL1
2
R12R2
1XC22XCXC12XC2
22
R1XC1R2XC2
2
XCR12R1R2R1QL1R1R2R1QL1
2XC122222
R1XCR12XCXC21QL1QLXC21QL11
R2
1QX
LC2
C1和Rb将R1C1和R2L1串联电路改为R12L1并联电路,并设XC1
2
XC
R1212R1
R1XC1
2XL1
2R2R2
2
R2XL1
R1QL
2R2
2XL1XL1
2
R2XL1
R12
12XCXC1
2
R1XC1
22
XC1X
,即R1212R1匹配时R1R2R12L12R2
2
R1XC11QLR2XL1
XL1
R2
1QRR
2L
2
1
1
XC2
2R2R12
1XC12XLXL12XC1
22
R2XL1R1XC1
R2
1QXLL1
1天线断开,工作于过压状态,集电极直流电表读数减小,天线电流表读数为0;618解:
2
XCRRRQRQR2RQRQ212121L21L21L21L2
R1XC1XL11QL1QLQLXL11QL1QL
2天线接地,工作于欠压状态,集电极直流电表读数略增,天线电流表读数增加;3中介回路失谐,工作于欠压状态,集电极直流电表读数略增,天线电流表读数减小。1PAPPCPk10316W619解:
2k3c
PA6
85.7%
PPC103PPC103
70%P10
PA6
60%P10
620解:当kkc时,k1
1若k
625解:
r1
50%则rr1
r1r
1
1Q2
kc1Q0
1QQ09
3
3kcQ
r1
90%则r9r1故kr1r
627解:260o160o,P0减小,工作于欠压状态。
V
V
628解:Ri99RL
I3I
VVBB0.61.45
629解:VbmBZ6Vo
coscos70
VBVbmVBB61.454.55V
iCmaxIcm1P0
VBVBZ4.550.6
1.98A22
iCmax170o1.980.4360.86A
VcmVCCgcriCmax241.9822.02VIcm1Vcm0.8622.01
9.47W22QL10
PA1P19.478.52W0Q0100
630解:Vbm
VBZVBB0.61.5
6.14Vcoscos70o
VBVbmVBB6.141.54.64ViCmaxIcm1P0
VBVBZ4.640.6
2.02A22
iCmax170o2.020.4360.88A
VcmVCC0.92421.6VIcm1Vcm0.8821.6
9.5W22QL10
PA1P19.58.8W0Q0100
a电路可能振荡,属于电感反馈式振荡电路;75解:
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,属于电容反馈式振荡电路,f1f2f0f3;76解:
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:
11
100MHz712
2πLCCd23.1420510
C1L3
6
2gd11
RPQ30.06~0.08V
1fq726解:
6
2051012
10
7
5.27mS
11.5~1.5001MHz727解:
1.657101.65710
4.14MHzd0.4
S200
Cq21.110521.11050.105pF
d0.4d30.43
Lq43.543.514mH
S200d0.4
rd42500B425000.2521.2Ω
S200S200
C03.961023.9610219.8pF
d0.4
1.051.05Qq104d1040.416800
B0.251.6571061.657106
2d0.11mm
fq15106
2不能
3不能,普通三极管没有负阻特性。
728解:恒温槽、稳压电源、高稳定度克拉泼振荡电路、共集电极缓冲级等。
729解:并联cb型(皮尔斯)晶体振荡电路。
93解:iI1macosΩtcosω0t
Icosω0tI有效值
II
macosω0Ωtmacosω0Ωt22
2
2
2
III
mama22222
2
maI
22
1v2510.7cos2π5000t0.3cos2π10000tsin2π106t94解:
25sin2π106t8.75sin2π1005000sin2π9950003.75sin2π1010000sin2π990000
2包络2510.7cos2π5000t0.3cos2π10000t
VV02510.70.325
峰值调幅度mmax0.4
上
V0
25
V0Vmin252510.70.31
V025
121
1ma1时95解:Pω0ΩPω0ΩmaP0T10025W
44121
2ma0.3时Pω0ΩPω0ΩmaP0T0.321002.25W
44
3
96解:ib1vb3v不包含平方项,不能产生调幅作用。
1Pω0ΩPω0Ω1ma2P0T10.725000612.5W97解:
44
Pω0Ω2Pω0Ω1225W
谷值调幅度m下
2PP0av
η
P0T5000
10kWη0.5
2ma0.72
P0T15000122P0av12.45kW3P
ηη0.5
1ma1时98解:
121
maP0T1000250W44
P0P0TPω0ΩPω0Ω10002502501500WPω0ΩPω0Ω
2ma0.7时
121
maP0T0.721000122.5W44
P0P0TPω0ΩPω0Ω1000122.5122.51245W Pω0ΩPω0Ω
99解:ff0f1f2f3f45202001780800010005kHz
910解:i1b0b1vvΩb2vvΩb3vvΩ
2
3
i2b0b1vvΩb2vvΩb3vvΩ
2
3
3
v0i1i2RR2b1vΩ4b2vvΩ6b3v2vΩ2b3vΩ
3
2b1RVΩcosΩt3b3RV02VΩcosΩt1.5b3RVΩcosΩt
2b2RV0VΩcosω0Ωt2b2RV0VΩcosω0Ωt1.5b3RV02VΩcos2ω0Ωt1.5b3RV02VΩcos2ω0Ωt
30.5b3RVΩcos3Ωt
输出端的频率分量:、3、0、20
P010.125
912解:m121210.5P90T
121
P0P0m2P0T10.1250.42910.845kW
22
9131vAtvtVcost
vBtvt
2若D1D2开路,则vAtvBtvt
vAB0
3若D1D2短路,则vAtvtVcostvB0vABtVcost
918解:RR1
R2ri247001000
5101335R2ri247001000
3Rd33.14100
0.57
R5104700Kdcos0.87
VKdmaVim0.870.30.50.13
2V0.132
P6.33W
2R21335
22VimVimKd0.520.87P41.7W2RidR5104700
AP
P6.330.152P41.7
R
R
R1R22
R22ri223501000
5102350
R22ri20.55R1R25104700
1中间位置919解:
2最高端R
R
R1
R2ri247001000
510
R2ri20.26R1R25104700
R2的触点在中间位置会产生负峰切割失真,而在最高端不会。
920解:由RR1R25~10k
11
R1~R2
510
取R26kR11.5k
R2ri262
RR11.53k
R2ri262
R31
ma
R93
取ma0.3
-ma20.32
C0.0187FmaRmax0.3900023.143000Ce
1minri2
1
0.26F
23.143002000
取C1C20.01F
取Ce20F
3Rd33.14100
0.5R60001500Kdcos0.9
R9000Rid5k能满足要求
2Kd20.9
0C23.144651032001012
921解:GP5.84S
Q0100f0465QL23.5
2f0.720
Q0100622
GPp24goeGP5.840.31005.8410Q23.5p34L0.153
2gid
4700
ikmV1costcos1tV0cos0t
1
kmV1V0cos10tcos10t4
cos10tcos10t
11
当01时,vSkmRLV1V0costcostkmRLV1V0coscost
42
无失真,只影响输出幅度。
1
当01时,vSkmRLV1V0cos10tcost
2
有失真。
1
2v1mV1cos1t
21
ikmV1cos1tV0cos0t21
kmV1V0cos10tcos10t41
vSkmRLV1V0cos10t4
当01时,只产生相移;当01时,有失真。
1v1mV1costcos1t924解:
高频电子第五版
31解:f
1MHz
2Δf0.7110699010310(kHz)f01106
Q1003
2Δf0.71010
取R10Ω则L
C
QR
0
1
10010
159(H)6
23.1410
1
159(pF)
02L(23.14106)2159106
11
32解:(1)当01或ω时,产生并联谐振。02
L1C1L2C2
(2)当01(3)当01
1
或ω02L1C1
1
或ω02L1C1
1
时,产生串联谐振。L2C2
1
时,产生并联谐振。L2C2
1L1
(Rjω0L)(R)R2jω0LR(12)R2L
jω0CCω0LCR33证明:Z112RRjω0LR2Rjω0L(12)jω0Cω0LC 34解:1由15C16052450C5352得C40pF
由12C16052100C5352得C-1pF不合理舍去
故采用后一个。
2L3
L C C’
11
180μH2
CC23.1453510324504010120
35解:Q
11
2126-12
0C0R23.141.510100105
L0
11
112μH226-120C023.141.51010010
Vom110-3
Iom0.2mA
R5
VLomVComQ0VSm212110-3212mV
11
36解:L2253μH
0C23.1410621001012
Q0
VC10100VS0.1
CCX11
2100pFCX200pF
CCX0L23.141062253106RX
0L
Q
0L
Q0
23.1410625310623.1410625310647.7Ω
2..1100
11
47.7j47.7j796Ω612
0CX23.14102001011
37解:L220.2μH
ω0C23.145106501012
ZXRXj
f05106100
Q0
2Δf0.71501033
2Δf10025.5510620ξQ06
f035103
22f0.7,则Q因2Δf0.700.5Q0,故R0.5R,所以应并上21kΩ电阻。
2πf0CωC
38证明:4πΔf0.7C0g
f02Δf0.7Q
CC0C1520202018.3pF39解:CCi2
C2C0C1202020
f0
1141.6MHz612
2πLC23.140.81018.310
L
100C12
0.8106
20.9kΩ
20201012
2
2
RPQ0
C2C0C1202020
RRiRPR1020.955.88kΩ0
C201R5.88103
QL28.266
ω0L23.1441.6100.8102Δf0.7
f041.61061.48MHzQL28.2
312解:1Zf10
2Zf103Zf1R
313解:1L1L2
C1C2
ρ1
01
103
159μH6
23.1410
11
159pF226601L123.141015910
ηR120M13.18μH6
0123.1410
2Zf1
01M2
R2
23.1410
6
3.1810620Ω20
2
L1159106
ZP25kΩ12
R1Rf1C12020159103Q1
01L1
R1Rf1
23.1410615910625
2020
42Δf0.75C2
f0f010620
22228.2kHz
Qρ1R1103
1
2
02
L2
1
23.149501032
15910
6
177pF
1
Z22R2jL021C
022
1
20j23.1410615910620j10061223.141017710Zf1
01M2
Z22
23.1410
6
3.181060.768j3.84Ω20j100
6
2
315解:R
L15910
20R1
312
RPC501015910
Rf10M0
f0106
Q221003
2f0.71410
316解:1Rf1
01M2
R2
10
7
106205
2
Rab23Q
R
R1
01L2
1
Rf110
7
10010640k520
2
01M01L
R1
1071062
5107100106200
5
2f0.711
221222210.013f0Q200 I111
317解:Q22.5
22I01.252f10103
11Q300103Qf
0
11R11.8
Q0C22.523.1430010320001012
2f10103
1Q300103Qf
0
QQ3022.57.5
1
2ω串联联谐L375μHLC2
318解:1
1L2125μω并联联谐L1L2C
II0
1
2
1
2
1
Q302
45解:当f1MHz时,
β0β0f1fTβ0β0f1fTβ0β0ffT
2
50501061250106
50
2
49
当f20MHz时,
2
当f50MHz时,
2
5020101250106
50
6
2
12.1
5050106250106
2
5
47解:gbe
gmCbe
IE1
0.754mS
26β0126501β0
500.75410337.7mSrbe
gm37.710324pF6
2πfT23.1425010
a1rbbgbe1700.7541031bωCberbb23.14107241012700.1yie
gbejCbeajb0.754103
a2b2
0.895j1.41mS
j23.141072410121j0.1
120.12
j23.14107310121j0.1yre0.0187j0.187mS2222
ab10.1
gmajb37.71031j0.1yfe37.327j3.733mSa2b2120.12
gjCbcajbjC1rgajb
yoegcejCbcrbbgmbcbcbbm
a2b2a2b2
1j0.1
j23.141073101217037.710320.049j0.68mS210.1
Av
48解:令A
vo
Av令Avo
mm
gjCbcajbbc
4Q2Δf4Q2Δf
2
0.7
f04
1
m
得2Δf0.7
1mf0
421Q
2
0.1
f04
11010121
1m2m
m
得2Δf0.1
2mf0
4101Q
故Kr0.1
2Δf0.1
2Δf0.7
2m
4101
1m421
49解:p1
gp
N235
0.25N1320
p2
N455
0.25N1320
11
37.2μS66
ω0Q0L2π10.710100410
116
200102860106228.5μS2244
2
ggpp12goep2gie37.2106
Avo
p1p2yfe
gΣ
0.250.2545103
12.36
228.510
2
ApoAvo12.32151.3
QL2Δf0.7
11
16.3666
gΣω0L228.5102π10.710410f010.71060.657MHZQL16.3
2
2
QL16.3
K111.43Q0100
fere54o88.5o
ξtantan2.95
222
gpp2gie37.21060.252200106
g3008.8μSL
p120.252
yfeyre
26662
gsgiegoeg1ξ12286010200103008.81012.95LS
451030.31103
1gp410解:
11
0.037mS66
Q0ω0L10023.1410.710410
12
p12goep2gie0.0370.10.320.0820.320.150.158mSR5
0.30.33824.2221.78
0.158
ggpAvo
p1p2yfe
g
22Δf0.7ω0Lgf023.1410.7106241060.158103454.4kHz3Avo4Avo421.784225025.3842Δf0.7452Δf0.7
14
212Δf0.721454.4197.65kHz1044.6kHz
1414
2Δf0.7
21
2Δf0.72Δf0.71044.6454.4590.2kHzAvo
Avo2Δf0.721.78454.4
9.472Δf0.71044.6
4Avo49.4748042.66Avo
4225025.38-8042.66216982.72Avo4Avo
411解:CCp12Coe5000.3218501.62pF
L
1
2πf02C
1
23.141.510
62
501.6210
12
22.5μH
Kr0.11.9不能满足
AvoS414解:
yfe2.50Cre
26.4236.42
7.74
2.50.3
417解:L1
11
118μH22312ω0C12π46510100010
118118118
L36L2L34L566013.5120
737373
C12C1Co100041004pF
13.5
C36C2pCi1000401004pF
74.5
ω0C12π465103100010126
g12go201049μS
Q0100
22
2
ω0C213.52π465103100010123
g36pgi49μS0.6210
Q010074.5
初、次级回路参数相等。若为临界耦合,即1,则
22
2
Avo
p1p2yfe
g
1
13.5
40103
74249106
ω0C122π46510310041012
QL606
g1249102Δf0.7
f04651032210.9kHZ
QL60
Kr0.13.16
2
420解:vn4kTRΔfn41.381023290100010712.65μV
2
in4kTGΔfn41.38102329010-310712.65nA
2222
421解:vnvn1vn2vn34kT1R1Δfn4kT2R2Δfn4kT3R3Δfn
4kT1R1T2R2T3R3Δfn4kTR1R2R3ΔfnT
T1R1T2R2T3R3
R1R2R3
2222
又inin1in2in34kT1G1Δfn4kT2G2Δfn4kT3G3Δfn
4kT1G1T2G2T3G3Δfn4kTG1G2G3ΔfnT
T1G1T2G2T3G3R1R2T3R2R3T1R3R1T2
G1G2G3R1R2R2R3R3R1
418证明:1Ib1yieVbe1yreVce11
Ic1yfeVbe1yoeVce12Ib2yieVbe2yreVce2Ic2yfeVbe2yoeVce2
yieVce1yreVcb2Vce1yreVcb2yieyreVce13
yfeVce1yoeVcb2Vce1yoeVcb2yfeyoeVce14
23得Ic2yfeVbe1yieyreyoeVce1yreVcb2
Vce1
Ic2yreVcb2yfeVbe1
yieyreyoe
5
Ic2yreVcb2yfeVbe1
yieyreyoe
5代入4Ic2yoeVcb2yfeyoe
Ic2
2yfeyfeyoeyieyoeyreyfeyoe
Vbe1Vcb26yieyreyfe2yoeyieyreyfe2yoe
yf
yfeyfeyoeyfeyieyreyfe2yoe
由1乘yfeyoe与4乘yre后相加得
Ib1yfeyoeIc2yreyieyfeyoeVbe1yreyoeVcb2由6代入消去Ic2得Ib1
2yieyieyreyieyfe2yieyoeyreyfeyreyoeyreVbe1Vcb2
yieyreyfe2yoeyieyreyfe2yoe
2
yieyoeyreyfeyoe
yoyre
yieyreyfe2yoe
2yieyieyreyieyfe2yieyoeyreyfe
yiyie
yieyreyfe2yoe
yr
yreyoeyreyyyrereoeyieyreyfe2yoeyfe
略
同理可证2
2
422解:vbn4kTrbfn41.38102327319702001030.2261012V2
2
ien2qIEfn21.610191032001030.641016A2
0
f
f
2
0.95101061500106
2
0.95
2
icn2qIC10fn21.6101910310.952001030.321017A2
423证明:fn
A2fdfA2f0
1
2
424解:Fn高3dB1.995倍
Fn混1FnFn高
ff0
12Qf0
Fn中6dB3.981倍
df
f0
2Q
Ti6011.207T290
F1Fn中11.20713.9811n混1.99510
Ap高KpcAp高Ap高0.2Ap高
20lg1.8882.76dB
2
s
Ap高1.888
425解:Fn
PsiPniPPVRsRRR11
nos2s1s
PsoPnoPniApApPoPsPoVs4RsRRR PsiPniPsIs24GsGGLrCL1
426解:Fn21
PsoPnoApPoIs4GsGGLrCLGs
427解:A为输入级,B为中间级,C为输出级。
APA6dB3.981倍FnFnA
ApB12dB15.849倍
F1FnB12141
nC1.72ApAApAApB3.9813.98115.849
428解:不能满足要求。设A前置放大器,B为输入级,C为下一级。
PsiPni105F1F11011.9951Fn4FnAnBnCFnAFnA8.1
PsoPno10ApAApAApB10100.1
58解:ikvkV0Vmcos0t
2
2
1212kV022V0Vmcos0tVmVmcos220t
22
当VmV0时,ikV022V0Vmcos0t,该非线性元件就能近似当成线性元件来处理,即当V0较大时,静态工作点选在抛物线上段接近线性部分,然后当Vm很小时,根据泰勒级数原则,可认为信号电压在特性的线性范围内变化,不会进入曲线弯曲部分,故可只取其级数的前两项得到近似线性特性。
512解:为了使iC中的二次谐波振幅达到最大值,C应为60o。
cos60o
VBZVBB1
Vm2
1
VBBVmVBZ
2
gVcost
513解:im
0
iIncosnt
n0
当cost0当cost0
11
I0gVcostdtgVmm
211I1gVmcos2tdtgVm
2
121gVm2
IngVmcostcosntdtn1
0
515解:iiD1iD2
n为偶数n为奇数
gVcostiD1m
0
2
4
当cost0当cost0
gVcostiD2m
0
当cost0当cost0
k1
1k1,2,3,igVmgVm0t2
k12k1
516解:当V01msintsin0t0时,i0;
当V01msintsin0t0时,igV01msintsin0t
517解:v0RLiD1iD2RLkv1v2kv1v24kRLv1v2
2
2
cos2kω0tsinΩtcos2kω0t
igV01msinΩt22m22
4k14k1k1k1k1,2,3,
2
518解:v0RLi2i3RLi4i1RLi2i4i1i3
RbRbRb
LLL
RLb0b1v1v2b2v1v2b3v1v2
2
2
3
3
b1v1v2b2v1v2b3v1v2b1v1v2b2v1v2b3v1v2
2
2
3
00
b1v1v2b2v1v2b3v1v2
3
8RLb2v1v2
1gm523解:
diC23
b12b2vBE3b3vBE4b4vBE
dvBE
diCdvBE
b12b2V0mcos0t3b3V02mcos20t4b4V03mcos30t
vBEv0
gmt
b12b2V0mcos0tgm12b2V0m3b4V03m
3
b3V02m1cos202b4V03mcos0tb4V03mcos0tcos30t2
1
gm1b2V0m1.5b4V03m2
qvBEdiCaISqkT
2gmvBEe
dvBEkTgcdigmtC
dvBE
omcos0taIq
SVomcos0tekT
kT
q
vBEv0
23
aISqq1q1q
Vomcos0t1Vomcos0tVomcos0tVomcos0tkT2kT6kTkT
qVomIqVIqVqVIScos0tISomcos20tSomcos30tSomcos40t
kT2kT6kTkTgm1
qVom3ISqVomIS
kT8kT
3
234
IqV3ISqVom1
gcgm1Som
22kT16kT
525解:ii1i3i2i4
22
3
a0a1v0vsa2v0vsa3v0vsa4v0vs
3
4
a0a1v0vsa2v0vsa3v0vsa4v0vs
3
4
a0a1v0vsa2v0vsa3v0vsa4v0vs
2
3
4
a0a1v0vsa2v0vsa3v0vsa4v0vs
2
3
4
3
8a2v0vs16a4v0vs16a4vs3v0
529解:gc0.5
IE26
2
gicgbe
sIE1r26bbTI0.5E0.55mS2602635
0.5IE0.50.5
9.6mS2626
gocgce4SApcmax
2
gc9.621047340dB4gicgoc40.550.004
2
2
2
QL2fi2465122830.1dBApcApcmax1A1104731pcmaxQQ2f10010000.7
IE260.5IE0.50.08
530解:gc0.51.54mS22626sIE
126rbbT
gicgbe
IE0.080.1mS2602630
gocgce10SApcmax
2gc1.542592.928dB4gicgoc40.10.01
2
2
gcGL1.540.1ApcgGg0.010.10.119623dBLicoc
532解:ii1i2i3i4
aa
a0a1v0vsa2v0vsa3v0vsa4v0vs
2
3
4
2
3
4
2
3
a0a1v0vsa2v0vsa3v0vsa4v0vs
00
2
3
44
a1v0vsa2v0vsa3v0vsa4v0vsa1v0vsa2v0vsa3v0vsa4v0vs
3
8a2v0vs16a4v0vs16a4vs3v0
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fnf0fi
时产生镜像干扰。由于不存在三次项,不会产生交调干扰;有二次项,可能产生互调干扰;若有强干扰信号,则能产生阻塞干扰。
535解:1.此现象属于组合频率干扰。这是由于混频器的输出电流中,除需要的中频电流外,
还存在一些谐波频率和组合频率,如果这些组合频率接近于中频放大的通带内,它就能和有用中频一道进入中频放大器,并被放大后加到检波器上,通过检波器的非线性效应,与中频差拍检波,产生音频,最终出现哨叫声。
2.因fi465kHz,p、q为本振和信号的谐波次数,不考虑大于3的情况。所以落于535
~1605kHz波段内的干扰在fS930kHz和fS1395kHz附近,1kHz的哨叫声在929kHz、931kHz、1394kHz、1396kHz时产生。
3.提高前端电路的选择性,合理选择中频等。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、q2,2f1f2277410353.618MHz,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、q3,3f1f2377410355.427MHz,会产生互调干扰;p、q大于3谐波较小,可以不考虑。
3f2f021S537解:fSf00.8MHz2fS3f02fS2f02
fSf00.4MHz
2f3f2S0
fS0.2MHzf00.6MHz
2
3fs2f030
fsf012MHz
2f3f30s0
fs2f030
fsf020MHz
2fsf030fS4MHzf016MHz
539解:若满足pf1qf2fs,则会产生互调干扰。已知f119.6MHz、
f219.2MHz、fsf0fi23320MHz,故没有互调信号输出。
64解:PVCCICO240.256W
C
P05
83.3%P6
22VcmVCC242
Rp57.6
2P02P025
Icm1
2P02P025
0.417AVcmVCC24Icm10.42
1.67Ic00.25
gcc
查表得c77o
2ηVCC20.712
66解:gcθc1.56查表得θc91o
Vcm10.8
P0Ik2R2214W
11
PCPP01P141.7Wη0
0.7c
Ic090
67解:icmax282mAo
α0900.319
Ic1mα190oicmax0.5282141mA
11
RpIc21m2000.14122W22P02
ηc74%
VCCIc0300.09 P0
22222VcmIkmR20LIkmR20LRCIkmR20LRIkmR
68证:P02
2L2RP2L220LRC
i2.2
69解:VcmVCCvcminVCCcmax2421.25V
gcr0.8
2
2
2
Ic0icmax070o2.20.2530.5566A
Icm1icmax170o2.20.4360.9592APVCCIc0240.556613.36W11
P0VcmIcm121.250.959210.19W
22
PCPP013.3610.193.17W
C
P010.1976.3%P13.36
2Vcm21.252
Rp22.16
2P0210.19
22VcmVCC242
610解:R1Rp144
2P02P022
XC1C1
R1144
14.4QL10
11
221pF6
2fXC123.14501014.4
R2
R22
QL11R1
200200
10011144
16.95
XL1
L1
XL116.95
0.054H6
2f23.145010
2
1RP增加一倍,放大器工作于过压状态,Vcm变化不大,P0Vcm611解:/2RP0.5P0;
22RP减小一半,放大器工作于欠压状态,Icm变化不大,P0IcmRP/22P0。
10144QLR1R2200
112.572QL1QLXL110011016.9511
C21239pF
2fXC223.14501062.57 XC2
612解:k
r
r1r
22
VCCVCEsatVcm120.5613解:RP662P02P021
2
111
57.4%
L1L211
11122
QQk100150.032Q1Q2M12
设QL10C1
则XC1
RP66
6.6QL10
11
241pF2fXC123.141086.6
RL
R
1QR
2L
L
XC2
1
50
P
110501
66
2
5.5
C2
11
290pF2fXC223.141085.5
QLRPRL106650
1112.522QL1QLXC2101105.5
XL1L1
XL112.519.9nH2f23.14108
C1和Ra将R1C1和R2C2串联电路改为R1614证:2C2并联电路,并设XC1
2
XC
R1212R1
R1XC1
2XC
222R2R2
R2XC2
R1QL
2R2
22XCXC2
2
R2XC2
R12
12XCXC1
2
R1XC1
22
XCXC11
,即R12匹配时R1R2R1R1222R2
22
R1XC11QLR2XC2
XC2
R2
1QRR
2L
2
1
1
XL1
2
R12R2
1XC22XCXC12XC2
22
R1XC1R2XC2
2
XCR12R1R2R1QL1R1R2R1QL1
2XC122222
R1XCR12XCXC21QL1QLXC21QL11
R2
1QX
LC2
C1和Rb将R1C1和R2L1串联电路改为R12L1并联电路,并设XC1
2
XC
R1212R1
R1XC1
2XL1
2R2R2
2
R2XL1
R1QL
2R2
2XL1XL1
2
R2XL1
R12
12XCXC1
2
R1XC1
22
XC1X
,即R1212R1匹配时R1R2R12L12R2
2
R1XC11QLR2XL1
XL1
R2
1QRR
2L
2
1
1
XC2
2R2R12
1XC12XLXL12XC1
22
R2XL1R1XC1
R2
1QXLL1
1天线断开,工作于过压状态,集电极直流电表读数减小,天线电流表读数为0;618解:
2
XCRRRQRQR2RQRQ212121L21L21L21L2
R1XC1XL11QL1QLQLXL11QL1QL
2天线接地,工作于欠压状态,集电极直流电表读数略增,天线电流表读数增加;3中介回路失谐,工作于欠压状态,集电极直流电表读数略增,天线电流表读数减小。1PAPPCPk10316W619解:
2k3c
PA6
85.7%
PPC103PPC103
70%P10
PA6
60%P10
620解:当kkc时,k1
1若k
625解:
r1
50%则rr1
r1r
1
1Q2
kc1Q0
1QQ09
3
3kcQ
r1
90%则r9r1故kr1r
627解:260o160o,P0减小,工作于欠压状态。
V
V
628解:Ri99RL
I3I
VVBB0.61.45
629解:VbmBZ6Vo
coscos70
VBVbmVBB61.454.55V
iCmaxIcm1P0
VBVBZ4.550.6
1.98A22
iCmax170o1.980.4360.86A
VcmVCCgcriCmax241.9822.02VIcm1Vcm0.8622.01
9.47W22QL10
PA1P19.478.52W0Q0100
630解:Vbm
VBZVBB0.61.5
6.14Vcoscos70o
VBVbmVBB6.141.54.64ViCmaxIcm1P0
VBVBZ4.640.6
2.02A22
iCmax170o2.020.4360.88A
VcmVCC0.92421.6VIcm1Vcm0.8821.6
9.5W22QL10
PA1P19.58.8W0Q0100
a电路可能振荡,属于电感反馈式振荡电路;75解:
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,属于电容反馈式振荡电路,f1f2f0f3;76解:
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:
11
100MHz712
2πLCCd23.1420510
C1L3
6
2gd11
RPQ30.06~0.08V
1fq726解:
6
2051012
10
7
5.27mS
11.5~1.5001MHz727解:
1.657101.65710
4.14MHzd0.4
S200
Cq21.110521.11050.105pF
d0.4d30.43
Lq43.543.514mH
S200d0.4
rd42500B425000.2521.2Ω
S200S200
C03.961023.9610219.8pF
d0.4
1.051.05Qq104d1040.416800
B0.251.6571061.657106
2d0.11mm
fq15106
2不能
3不能,普通三极管没有负阻特性。
728解:恒温槽、稳压电源、高稳定度克拉泼振荡电路、共集电极缓冲级等。
729解:并联cb型(皮尔斯)晶体振荡电路。
93解:iI1macosΩtcosω0t
Icosω0tI有效值
II
macosω0Ωtmacosω0Ωt22
2
2
2
III
mama22222
2
maI
22
1v2510.7cos2π5000t0.3cos2π10000tsin2π106t94解:
25sin2π106t8.75sin2π1005000sin2π9950003.75sin2π1010000sin2π990000
2包络2510.7cos2π5000t0.3cos2π10000t
VV02510.70.325
峰值调幅度mmax0.4
上
V0
25
V0Vmin252510.70.31
V025
121
1ma1时95解:Pω0ΩPω0ΩmaP0T10025W
44121
2ma0.3时Pω0ΩPω0ΩmaP0T0.321002.25W
44
3
96解:ib1vb3v不包含平方项,不能产生调幅作用。
1Pω0ΩPω0Ω1ma2P0T10.725000612.5W97解:
44
Pω0Ω2Pω0Ω1225W
谷值调幅度m下
2PP0av
η
P0T5000
10kWη0.5
2ma0.72
P0T15000122P0av12.45kW3P
ηη0.5
1ma1时98解:
121
maP0T1000250W44
P0P0TPω0ΩPω0Ω10002502501500WPω0ΩPω0Ω
2ma0.7时
121
maP0T0.721000122.5W44
P0P0TPω0ΩPω0Ω1000122.5122.51245W Pω0ΩPω0Ω
99解:ff0f1f2f3f45202001780800010005kHz
910解:i1b0b1vvΩb2vvΩb3vvΩ
2
3
i2b0b1vvΩb2vvΩb3vvΩ
2
3
3
v0i1i2RR2b1vΩ4b2vvΩ6b3v2vΩ2b3vΩ
3
2b1RVΩcosΩt3b3RV02VΩcosΩt1.5b3RVΩcosΩt
2b2RV0VΩcosω0Ωt2b2RV0VΩcosω0Ωt1.5b3RV02VΩcos2ω0Ωt1.5b3RV02VΩcos2ω0Ωt
30.5b3RVΩcos3Ωt
输出端的频率分量:、3、0、20
P010.125
912解:m121210.5P90T
121
P0P0m2P0T10.1250.42910.845kW
22
9131vAtvtVcost
vBtvt
2若D1D2开路,则vAtvBtvt
vAB0
3若D1D2短路,则vAtvtVcostvB0vABtVcost
918解:RR1
R2ri247001000
5101335R2ri247001000
3Rd33.14100
0.57
R5104700Kdcos0.87
VKdmaVim0.870.30.50.13
2V0.132
P6.33W
2R21335
22VimVimKd0.520.87P41.7W2RidR5104700
AP
P6.330.152P41.7
R
R
R1R22
R22ri223501000
5102350
R22ri20.55R1R25104700
1中间位置919解:
2最高端R
R
R1
R2ri247001000
510
R2ri20.26R1R25104700
R2的触点在中间位置会产生负峰切割失真,而在最高端不会。
920解:由RR1R25~10k
11
R1~R2
510
取R26kR11.5k
R2ri262
RR11.53k
R2ri262
R31
ma
R93
取ma0.3
-ma20.32
C0.0187FmaRmax0.3900023.143000Ce
1minri2
1
0.26F
23.143002000
取C1C20.01F
取Ce20F
3Rd33.14100
0.5R60001500Kdcos0.9
R9000Rid5k能满足要求
2Kd20.9
0C23.144651032001012
921解:GP5.84S
Q0100f0465QL23.5
2f0.720
Q0100622
GPp24goeGP5.840.31005.8410Q23.5p34L0.153
2gid
4700
ikmV1costcos1tV0cos0t
1
kmV1V0cos10tcos10t4
cos10tcos10t
11
当01时,vSkmRLV1V0costcostkmRLV1V0coscost
42
无失真,只影响输出幅度。
1
当01时,vSkmRLV1V0cos10tcost
2
有失真。
1
2v1mV1cos1t
21
ikmV1cos1tV0cos0t21
kmV1V0cos10tcos10t41
vSkmRLV1V0cos10t4
当01时,只产生相移;当01时,有失真。
1v1mV1costcos1t924解: