高频电子线路习题答案(第四版)
3-1解:f
=1MHz
0.7
2ΔfQ=
=1⨯10f0
=
0.7
6
-990⨯10
63
3
=10(kHz)
1⨯10
2Δf10⨯10
=100
取R=10Ω则L=C=
QR
ω01
2
=
100⨯102⨯3.14⨯10
6
=159(μH)
ω0L
=
1
(2⨯3.14⨯10
6
)⨯159⨯10
2-6
=159(pF)
3-2解:(
1)当ω01=
1L1C11L1C11L1C1
或ω02=
1L2C21L2C21L2C2
时,产生并联谐振。
(2)当ω01=或ω02=或ω02=
时,产生串联谐振。
(3)当ω01=
时,产生并联谐振。
1ω0LC1
)
2
(R+jω0L)(R+
3-3证明:Z
=
R+jω0L+R+
1jω0C1jω0C
)=
R
2
+
LC
+jω0LR(1-
)=
R
2
+
L=R
2R+jω0L(1-
2R
ω0LC
2
)
=180(μH
3-4解:1)由(15+C)⨯1605
2
=(450+C)535得C=40pF
22
由(12+C)⨯1605故采用后一个
2)L=3)
1
2
2
=(100+C)535得C=-1pF(不合理舍去
。
=
1
ω0(C+C')
(2⨯3.14
⨯535⨯10
32
)
⨯(450+40)⨯10
-12
)
3-5解:Q
=
1
ω0C0R1
2
=
1
2⨯3.14⨯1.5⨯10
1
26
⨯100⨯10
-12
⨯5
=212
L0=
ω0C0VomR
=
=
(2⨯3.14
-3
⨯1.5⨯10
6
)100
⨯10
-12
=112(μH)
Iom=
1⨯10
5
=0.2(mA)
-3
VLom=VCom=Q0VSm=212⨯1⨯10
=212(mV)
)
3-6解:L=
1
ω0CVCVS
=
2
=10
1
(2⨯3.14
=100
⨯10
6
)
2
⨯100⨯10
-12
=253(μH
Q0=
0.11
2
C⋅CXC+CXRX=Z
=
ω0L-
=
1
(2⨯3.14
=
⨯10
6
)
2
⨯253⨯10⨯253⨯10
-6-6
=100(pF)→
6
CX=200pF⨯253⨯10
-6
ω0LQ
ω0LQ0
1
2⨯3.14⨯10
6
2.50.1
=47.7-j
1
2⨯3.14⨯10
6
-
2⨯3.14⨯10
100
-12
=47.7(Ω)
X
=RX-j
ω0CX
⨯200⨯10
=47.7-j796(Ω)
3-7解:L=
1ω0Cf02Δf
2
=
1
2⨯3.14
=
5⨯10
⨯5⨯10
63
6
⨯50⨯10
-12
=20.2(μH)
Q0=
0.7
150⨯101003
⨯
=
1003
6
ξ=Q0
2Δff0
=
2⨯(5.5-5)⨯10
5⨯10
6
=
203
21kΩ电阻。
'=2⨯2∆f0.7,则Q'因2Δf0.7=0.5Q0,故R'=0.5R,所以应并上0
2πf0Cf02Δf
0.7
3-8证明:4πΔf
C=0.7
=
ω0CQ
=g∑
3-9解:C
=Ci+
12π
(C2
=
+C0)C1
C2+C0+C1
=5+
(20
1
+20)20
20+20+20
=18.3(pF)
)
f0=
LCLC12
2⨯3.14=100⨯
0.8⨯10
-6
⨯18.3⨯10
-6
-12
-12
=41.6(MHz
RP=Q0
0.8⨯10
(20+20)⨯10
2
=20.9(kΩ)
2
⎛C2+C0+C1⎫⎛20+20+20⎫
⎪R0=1020.9 R∑=RiRP ⎪⨯5=5.88(kΩ) ⎪C20⎝⎭1⎝⎭QL=
R∑ω0L=
=f0QL
5.88⨯10
2⨯3.14⨯41.6⨯10=
41.6⨯1028.2
2
6
63
⨯0.8⨯10
-6
=28.2
2Δf
0.7
=1.48(MHz
2
)
2
3.12解:1)Z 2) Z 3) Z
f1
=
(ωM)Z22
=
(ωM)jωL2
=-j
ωM
L2
f1
=0=(ωM)1QG2
2
2
f1
=(ωM)QG2=
22
ωMC2
G2
422
(Q=
ωC2
G2
)
3-13解:1)L1=L2=
C1=C2=
ηR1
ρ1
ω01
1
2
=
10
3
6
2⨯3.14⨯10=
=159(μH1
)
=159(pF
ω01L1=
(2⨯3.14
6
⨯10
6
)
2
⨯159⨯10
-6
)
M=
1⨯202⨯3.14⨯10
=
ω01
=3.18(μH⨯10
6
)
-6
2)Z
f1
=
(ω01M)2
R2L1
1
(2⨯3.14
=
1
⨯3.18⨯10
)
2
20159⨯10
-6
-12
=20(Ω)
Z
P
=
R
+R
f1
C
=
(20+20)⨯159⨯10
6
=25(kΩ)
3)Q1=4)2Δf
ω01L1R1+R=1')(ω02
2f1
2⨯3.14⨯10⨯159⨯10
-6
20+20
2
f0ρ1R1
=
2⨯110
6
=25
0.7
2
f0Q
=
⨯20
3
10
=28.2(kHz)
)
'=5)C2
L2
=
(2⨯3.14
⨯950⨯10⎫
⎪⎪⎭
3
)
2
⨯159⨯10
-6
=177(pF
⎛1
Z22=R2+j ωL- 021ωC'
022⎝⎛
=20+j 2⨯3.14⨯10
⎝Z
f1
6
⨯159⨯10⨯10
6
-6
-
1
2⨯3.14⨯10
-6
6
⨯177⨯10
-12
⎫
⎪=20+j100⎭
=
(ω01M)2
Z22
=
(2⨯3.14
159⨯10
3
⨯3.18⨯10
)
2
20+j100
-6
-12
=0.768-j3.84(Ω)
3-15解: R=
LRPC
=
50⨯10⨯159⨯10
=20(Ω)=R1
∴Rf1=0→M=0Q=
2
f02∆f0.7
=
2
10
6
3
14⨯10
=100
3-16解:1)Rf1=
(ω01M)2
R2
=
(10(10
7
⨯105
-6
)
2
=20(Ω)
-6
Rab=2)η=3)Q=
(ω01L)2
7
R
R1
1
+Rf1=10
7
=
⨯100⨯105+20
)
2
=40(kΩ)
ω01Mω01L
R1f0
=
⨯105
-6
=2
-6
=
10
7
⨯100⨯10
5
=200
2
2∆f0.7
1
2
+2η-1⋅
1Q
=2+2⨯2-1⨯
1200=
1
=0.013
3-17解:
II0
=
⎛2∆f
'1+ Q f0⎝1Qω0C
=1⎛2∆f1+ Qf
0⎝
⎫⎪⎪⎭
2
⎫⎪⎪⎭
2
=
1
⎛10⨯10
'1+ Q
300⨯10⎝
1
3
33
⎫
⎪⎪⎭
2
1.25
→Q'=22.5
R=II0
22.5⨯2⨯3.14⨯300⨯10
=
1
⨯2000⨯10
=
12
-12
=11.8
=
⎛10⨯101+ Q
⎝300⨯10
33
⎫⎪⎪⎭
2
→Q=30
Q-Q'=30-22.5=7.5
⎧L1=375μH→⎨
⎩L2=125μ
1⎧
2ω=串联联谐⎪
L2C⎪
3-18解:⎨
1
⎪ω=并联联谐
(L1+L2)C⎪⎩
4-5解:当f=1MHz时,β=
β0
⎛β0f
1+ f
⎝T
当f=20MHz时,β=
β0
⎛β0f1+ f
⎝T
当f=50MHz时,β=
β0
⎛β0f1+ f
⎝T
⎫⎪⎪⎭
2
⎫⎪⎪⎭
2
=
50
⎛50⨯101+
⎝250⨯10
66
⎫⎪⎪⎭
2
=49
⎫⎪⎪⎭
2
=
50
⎛50⨯20⨯101+ 6
⎝250⨯10
6
⎫⎪⎪⎭
2
=12.1
=
50
⎛50⨯50⨯10+ 6
⎝250⨯10
6
⎫⎪⎪⎭
2
=5
4-7解:gb'e=
gm=
IE
26(β0
+1)
=
126⨯(50
-3
+1)
=0.754(mS)
β0rb'e
=50⨯0.754⨯10
=37.7(mS)
)
Cb'e=
gm2πfT
=
37.7⨯10
-3
6
2⨯3.14⨯250⨯10
=24(pF
-3
a=1+rbb'gb'e=1+70⨯0.754⨯10b=ωCyie=
b'ebb'
≈1
-12
r=2⨯3.14⨯10
7
⨯24⨯10=
⨯70≈0.1
-3
(gb'e+jωCb'e)(a-jb)a
2
(0.754
⨯10+j2⨯3.14⨯10
2
7
⨯24⨯10
2
-12
)⨯(1-
j0.1)
+b
2
1+0.1
7
-12
=0.895+j1.41(mSyre=-y
=
)
≈-
j2⨯3.14⨯10
⨯3⨯10
2
2
(gb'c+jωCb'c)(a-jb)a
2
⨯(1-j0.1)
+b=
2
-3
1+0.1
2
=-0.0187-j0.187(mS)
gm(a-jb)a
2
37.7⨯10
2
⨯(1-j0.1)
fe
+b
2
1+0.1
=37.327-j3.733(mS)
yoe=gce+jωCb'c+rbb'gm
≈j2⨯3.14⨯10
7
(gb'c
-12
+jωCb'c)(a-jb)a
2
+b
2
a-jb⎫⎛
≈jωCb'c 1+rbb'gm2
2⎪
a+b⎝⎭
-3
⨯3⨯10
⎛
1+70⨯37.7⨯10⎝
⎫⎪⎪⎭
m
⨯
1-j0.1⎫
=0.049+j0.68(mS22⎪
1+0.1⎭
⎛⎫f4 2m-1⎪⋅0 ⎪Q⎝⎭
2m
1
)
⎛Av⎫
⎪4-8解:令 A⎪⎝vo⎭
⎛Av⎫
⎪令 A⎪⎝vo⎭
m
m
⎛= ⎝
2
4+(Q2Δf
2
0.7
f0)
4
=
12
得2Δf
0.7
=
4
⎛= ⎝
4+(Q2Δf
0.10.7
0.1
f0)
4
⎫⎪⎪⎭
m
=
110
得2Δf
0.1
=
⎛4 10 ⎝⎫f-1⎪⋅0
⎪Q⎭
故Kr0.1=
2Δf2Δf
=
⎛4 10 ⎝
2m
⎫-1⎪⎪⎭
2
=
10
m1
-1
⎛⎫4 2m-1⎪ ⎪⎝⎭
1
2m-1
4-9解:p1=
gp=
N23N131
=
520=
2
=0.25
p2=1
N45N13
=
520
=0.25=37.2(μS
-6
ω0Q0L
2π⨯10.7⨯10
2
6
⨯100⨯4⨯10
-6
-6
)
+14
2
g∑=gp+p1goe+p2gie=37.2⨯10
p1p2ygΣ
2
fe
+
-3
14
2
⨯200⨯10⨯2860⨯10
-6
=228.5(μS)
Avo=
=
2
0.25⨯0.25⨯45⨯10
228.5⨯10
-6
=12.3
Apo=Avo=12.3=151.3QL=
1gΣω0L=
f0QL
=
1
228.5⨯1010.7⨯1016.3
-2
6
-6
6
-6
⨯2π⨯10.7⨯10
Z
⨯4⨯10
=16.3
2Δf
0.7
=
=0.657(MH
-2
)
⎛QL⎫
⎪K= 1- Q0⎪⎝⎭ξ=tan
16.3⎫⎛
= 1-⎪
100⎭⎝=tan
-54
o
=1.43-88.52
-6
o
ϕ
fe
+ϕre2
22
=-2.95
22
g'=L
gp+p2gie
p1
=
37.2⨯10+0.250.25
⨯200⨯10
-6
3008.8(μS)
-6
S=
(gs
+gie
)(goe
y
fe
)(1+ξ+g'L
yre
2
)
=
2⨯2860⨯10
-6
(200
⨯10
-6
+3008.8⨯100.31⨯10
-3
)(1+2.95)
2
45⨯10
-3
>>1
(1)gp=4-10解:
1Q0ω0L
1
=
1
100⨯2⨯3.14⨯10.7⨯10
2
2
6
⨯4⨯10
-6
=0.037(mS)
2
g∑=gp+
R5
fe
+p1goe+p2gie=(0.037+0.1+0.3⨯0.082+0.3⨯0.15)=0.158(mS
2
)
Avo=
p1p2yg∑=ω0Lg=(Avo
=2Δf
1
=
0.3⨯0.3⨯38
2
+4.2
2
0.158
6
=21.78
(2)2Δf0.7(3)(Avo)4
∑
f0=2⨯3.14⨯(10.7⨯10=21.78
4
)
2
⨯4⨯10
-6
⨯0.158⨯10
-3
=454.4(kHz)
)4
1
=225025.38
1
0.7
(4)(2Δf0.7)4(5)2Δf0'.7
=
24-1⋅2Δf
0.7
=
24-1⨯454.4=197.65(kHz)
=1044.6(kHz)
)
24-1
2Δf0'.7-2Δf'=Avo')4(Avo
0.7
=1044.6-454.4=590.2(kHz=
21.78⨯454.4
1044.6
4
Avo2Δf'=(Avo'-(Avo
2
0.7
2Δf0'.7
=9.47
)4)4
=9.47=8042.66
(Avo)4
1
=225025.38-8042.66=216982.72
2
=22.5(μH
4-11解:C∑=C+p1Coe=500+0.3⨯18=501.62(pF
L=
)
-12
(2πf0)
2
C∑
=
1
(2⨯3.14⨯1.5⨯10)
62
2
)
⨯501.62⨯10
Kr0.1
=
26.4+36.42.5⨯0.3
2
(Avo4-14解:)S
=
y
fe
2.5ω0Cre
=7.74
4 - 17解: L 1 =
= = 118( μH ) 2 2 - ω0 C 1 (2π ⨯ 465 ⨯ 10 3 ) ⨯ 1000 ⨯ 10 12
+ ⨯ 60 + ⨯ 13. 5 = 120 73 73 73
C 12 = C 1 + C o = 1000 + 4 = 1004( pF ) L 36 = L 2 + L 34 + L 56 =
2 2 i
⎫ 2 ⎛C 36 = C 2 + pC= 1000 + pF⎪ ⨯ 40 = 1004( )
⎭ ⎝ 74 .5
3 ⨯ ⨯ ⨯ ⨯ -12
- 6 = + = ⨯ + = 49 g12 g o 20 10 (μS ) Q 0 100
∴ 初、次级回路参数相等 。若为临界耦合,即η = 1,则 Avo =
p p y g
= 1 ⨯
⨯ 40 ⨯ 10 - 3 = 74
- 6 2 ⨯ 49 ⨯ 10
2 ⨯ ⨯ 3 ⨯ ⨯ -12⎛⎫ - 2 3 = = 49 g 36 = p 2 g i + 62 ⨯ 10 + μS⎪ ⨯ 0.( )
Q 0 74 .5 100 ⎝ ⎭
⨯ ⨯ 3 ⨯ ⨯ -12
= = 60 Q L =
6 g 49 ⨯ 10 -
12
= 2 2 Δf 0 .7 = 3.K r 0 .1 16
⨯ 3 = 2 ⨯= 10. 9 (kH Z ) Q L 60
4-20解:
2
vn=in=
2
2
4kTRΔfn=4kTGΔfn=
2
2
2
4⨯1.38⨯104⨯1.38⨯10
-23
-23
⨯290⨯1000⨯10
-3
7
=12.65(μV)
⨯290⨯10⨯10
7
=12.65(nA)
4-21解: vn=vn1+vn2+vn3=4kT1R1Δf
=4k(T1R1+T2R2+T3R3)Δf=4kT(R1+R2+R3)Δf∴T=
2
n
n
n
+4kT2R2Δf
n
+4kT3R3Δf
n
T1R1+T2R2+T3R3
R1+R2+R3
2
2
2
n
又 in=in1+in2+in3=4kT1G1Δf
=4k(T1G1+T2G2+T3G3)Δf=4kT(G1+G2+G3)Δf∴T=
T1G1+T2G2+T3G3
G1+G2+G3
n
n
+4kT2G2Δf
n
+4kT3G3Δf
n
=
R1R2T3+R2R3T1+R3R1T2
R1R2+R2R3+R3R1
(4 - 18证明: 1) I b1V + y re V ce11 = y )ie be1
(2I c 1 = y fe V be1V 1 + y ) oe ce
∙ ∙
∙
∙
∙ ∙ ∙
∙ ∙ ∙ ∙ ∙ ⎛ ∙ - ∙ ⎫ =
= + = - + - + (I b 2 y ie V be 2 y re V ce 2 y ie V ce y re )V V cb 2 V ce1 ⎪ y re V cb 2 (y 1 y ce1re ie ⎝ ⎭ ∙ ∙ ∙ ∙ ∙ ∙ ⎛ ∙ - ∙ ⎫ =
= + = - + - + I c 2 y fe V be 2 y oe V ce 2 y fe V ce1y V cb 2 (y V cb 2 V ce fe y )V ce1 y 1 ⎪ oe oe ⎝ ⎭ oe
+ y + y )V 得 - y V cb 2 + (y(2 ) - (3 ) I c 2 = y fe V be1 ie re oe ce1re I + y V - y V = (5V ce1 )y + y + y
∙
ie
re
oe
∙
∙
∙
∙ ∙ ∙ ∙
2 ∙ ∙ - + + V cb2 + (6V be1 )
y ie + y re + y fe + 2 y oe y ie + y re + y fe + 2 y oe
+ ∴ y =≈ y
f fe y ie + y re + y fe + 2 y oe
I + y V - y V ∙ ∙ 代入 fe + y (5 ) (4 ) I c 2 = y oe V cb 2 - (yoe y ie + y re + y oe I c 2 =
∙
∙ ∙ ∙
2 - + ≈ - y
re y ie + y re + y fe + 2 y oe
乘由( 1与 4乘 re 后相加得 (y fe + y oe ) ( ) ) y
y o =
∙
I b1y + y oe )VV fe + y )+ I c 2 y re = y (y + y re y be1oe ie ( fe oe cb2 由( 6 I c 2 得 )代入消去
+ + + - ∙ ∙ 2+ VV + I b1 = cb2 be1
y ie + y re + y fe + 2 y oe y ie + y re + y fe + 2 y oe
∙
2 + + + - ≈ y y i = ie y ie + y re + y fe + 2 y oe
+ + ≈ - y r =
y ie + y re + y fe + 2 y oe y fe
∙
∙ ∙ ∙
同理可证2 )
略
4-22解:vbn=4kTrb∆fn=4⨯1.38⨯10
ien=2qIE∆fn=2⨯1.6⨯10
2
-19
2-23
⨯(273+19)⨯70⨯200⨯10
-3
3
=0.226⨯10
-12
(V)
2
⨯10⨯200⨯10=0.95
3
=0.64⨯10
-16
(A)
2
=
α0⎛f⎫
⎪1+ f⎪⎝α⎭
2
=
0.95⎛10⨯101+
⎝500⨯10=2⨯1.6⨯10
66
⎫⎪⎪⎭
2
icn=2qI
2
C
(1-α0)∆fn
∞
-19
⨯10
-3
⨯(1-0.95)⨯200⨯10
3
=0.32⨯10
-17
(A)
2
4-23证明:∆fn=
⎰
AA
2
(f)df(f0)
=
∞
2
⎰
1
⎛f-f0⎫
⎪1+ 2Q ⎪f0⎝⎭
2
df=
πf0
2Q
4-24解:Fn高=3dB(1.995倍)
Fn混=1+
TiT=1+
60290+
Fn中=6dB(3.981倍)=1.207Fn中-1
=1.995+
1.207-1Ap高
+
3.981-10.2⨯Ap高
=10
Fn=Fn高+Ap高=1.888
Fn混-1Ap高
K
pc
Ap高
20lg1.888=2.76(dBPnoPniAp1Ap
1Ap
1
)
=PsPo
2
=
VsVs
2
2
4-25解:Fn=
PsiPniPsoPnoPsiPniPsoPno
===
4Rs
PoPs
4(Rs+R)
=
Rs+RR
=1+
RsR
4-26解:Fn=
==
PsPo
=
Is4Gs
Is4(Gs+G+GL+rCL)
2
=1+
G+GL+rCL
Gs
4-27解:A为输入级,B为中间级,C为输出级。
APA=6dB(3.981倍)Fn=FnA+
FnB-1ApA
+
ApB=12dB(15.849倍)FnC-1ApA⋅ApB
=1.7+
2-13.981
+
4-13.981⨯15.849
=2
→FnA≤8.1
4-28解:不能满足要求。设
Fn=
PsiPniPsoPno
=1010
54
A前置放大器,≥FnA+
FnB-1ApA
+
B为输入级,FnC-1
=FnA+
C为下一级。10-110
+
1.995-110⨯0.1
ApA⋅ApB
5-8解:i=kv
2
=k(V0+Vmcosω0t)
2
1212⎛2
=k V0+2V0Vmcosω0t+Vm+Vmcos
22⎝
2
2
⎫
2ω0t⎪
⎭
似当成线性元件来处理
Vm很小时,根据泰线弯曲部分,故
,
当Vm
5-12解:为了使
cos60
o
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。
最大值,12
围内变化,不会进入曲
iC中的二次谐波振幅达到=
VBZ+VBB
Vm
=12
θC应为60。
o
∴VBB=Vm-VBZ
⎧gVmcosωt
5-13解:i=⎨
⎩0
∞
当cosωt>0当cosωt
i=
∑
n=0
Incosnωt1
I0=I1=
2π1
⎰π
-+π-
+π
gVmcosωtd(ωt)=
2
1
π
12
gVmgVm
n为偶数n为奇数
π
1
⎰π
-
gVmcos
ωtd(ωt)=
In=
π
⎰π
+π
gVm
1⎧2
⎪gVm2
cosωtcosnωtd(ωt)=⎨πn-1
⎪⎩0
5-15解:i=iD1+iD2
iD1
⎧gVmcosωt
=⎨⎩02
4
∞
当cosωt>0当cosωt
iD2
⎧gVmcosωt=⎨⎩0
当cosωt0
(-1)k-1
i=gVm+gVm∑ω0t2
ππk=1(2k)-1
5-16解:当V0(1+msinΩt)sinω0t
2
(k=1,2,3, )
当V0(1+msinΩt)sinω0t>0时,i=gV0(1+msinΩt)sinω0t
∞∞
cos2kω0tsinΩtcos2kω0t⎫⎛
i=gV0 1+msinΩt-2∑-2m⎪∑22
π4k-14k-1k=1k=1⎝⎭
(k=1,2,3, )
L
)=
R
L
5-17解:v0=R
(iD1
-iD2
[k(v
1
+v2)-k(v1-v2)
22
]=4kR
L
v1v2
5-18解:v0=RL(i2-i3)+RL(i4-i1)=RL(i2+i4-i1-i3)
=RLb0+b1(v1-v2)+b2(v1-v2)+b3(v1-v2)+RL-RL-RL
2
[[b[b[b
23
]
3
+b1(-v1+v2)+b2(-v1+v2)+b3(-v1+v2)+b1(v1+v2)+b2(v1+v2)+b3(v1+v2)
2
2
3
]]
]
3
+b1(-v1-v2)+b2(-v1-v2)+b3(-v1-v2)
=-8RLb2v1v2
2
3
(1)gm=5-23解:
diCdv
BE
=b1+2b2vBE+3b3vBE+4b4vBE
2
2
3
3
gm(t)=
diCdv
BE
vBE=v0
=b1+2b2V0mcosω0t+3b3V0mcos
32
ω0t+4b4V0mcosω0t
3
=b1+2b2V0mcosω0t+gm1=2b2V0m+3b4V0mgc=
12
3
b3V0m(1+cos2ω0)+2b4V0mcosω0t+b4V0m(cosω0t+cos3ω0t)
2
3
gm1=b2V0m+1.5b4V0m
=diCdvVom
BE
vBE=v0
3
(2)gm
=
diCdv
BE
aISqkT
q
vBE⋅ekT
aISqkT
vBE
q
gm(t)=aISqkT
=
Vomcosω0t⋅ekT
omcosω0t
=
⎡⎤q1⎛q1⎛q⎫⎫
cosω0t⎢1+Vomcosω0t+ Vomcosω0t⎪+ Vomcosω0t⎪+ ⎥
kT2⎝kT6⎝kT⎭⎭⎢⎥⎣⎦⎛qVom⎫
cosω0t+αIS ⎪cos
kT⎝⎭
3αIS⎛qVom⎫
+ ⎪
8⎝kT⎭
32
2
23
=αIS
qVomkT
ω0t+
αIS⎛qVom⎫
2⎝kT
3
3
⎪cos⎭
ω0t+
αIS⎛qVom⎫
6⎝kT
4
4
⎪cos⎭
ω0t
gm1≈αIS
12
qVomkT
gc=gm1=
αISqVom
2kT
3αIS⎛qVom⎫+ ⎪
16⎝kT⎭
3
5-25解:i∑=i1-i3+i2-i4
=a0+a1(v0+vs)+a2(v0+vs)+a3(v0+vs)+a4(v0+vs)+ +a0+a1(v0-vs)+a2(v0-vs)+a3(v0-vs)+a4(v0-vs)+ -a0-a1(-v0+vs)-a2(-v0+vs)-a3(-v0+vs)-a4(-v0+vs)- -a0-a1(-v0-vs)-a2(-v0-vs)-a3(-v0-vs)-a4(-v0-vs)- =8a2v0vs+16a4v0vs+16a4vsv0+
3
3
2
3
4
2
3
4
2
3
4
2
3
4
5-29解:gc=0.5
IE26⎛ωsIE⎫
⎪+ ⋅rbb'⎪ ω
⎝T26⎭
2
≈
0.5IE26
=
0.5⨯0.526
=9.6(mS)
gic≈gb'e=
IE26β0
2
=
0.526⨯35
=0.55(mS)
goc≈gce=4(μS)Apcmax=
gc
=
9.6
2
4gicgoc
4⨯0.55⨯0.004
2
=10473
(≈
⎫
⎪⎪⎭
40dB
2
)
⎫⎪⎪⎭
2
Apc=Apcmax
⎛Q⎫ 1-L⎪ Q0⎪⎝⎭
IE26
=Apcmax
⎛2fi
1- Q02∆f0.7⎝0.5IE26
=
⎛2⨯465
=10473⨯ 1-
100⨯10⎝=1.54(mS
=1228(≈30.1dB)
5-30解:gc=0.5
⎛ωsIE⎫
1+ ⋅rbb'⎪⎪⎝ωT26⎭
gic≈gb'e=
IE26β0
2
2
≈
0.5⨯0.08
26
)
=
0.0826⨯30
=0.1(mS)
goc≈gce=10(μS)Apcmax=
gc
=
2
1.54
2
4gicgoc
4⨯0.1⨯0.01
=592.9(≈28dB
2
)
)
Apc
⎛gc
= g+G
L⎝oc⎫GL1.540.1⎛⎫⎪⋅=⨯=196(≈23dB ⎪⎪gic0.1⎝0.01+0.1⎭⎭
4
5-32解:i∑=i1-i2-i3+i4
=a0+a1(v0+vs)+a2(v0+vs)+a3(v0+vs)+a4(v0+vs)+
2
3
[
-[a+[a
-a0+a1(v0-vs)+a2(v0-vs)+a3(v0-vs)+a4(v0-vs)+
2
3
234
]
44
+a1(-v0+vs)+a2(-v0+vs)+a3(-v0+vs)+a4(-v0+vs)+ +a1(-v0-vs)+a2(-v0-vs)+a3(-v0-vs)+a4(-v0-vs)
3
3
2
3
]+ ]
=8a2v0vs+16a4v0vs+16a4vsv0+
5-34解:因存在二次项,能
进行混频。只要满足
fn=fi就会产生中频干扰;当
生交调干扰;有二次项扰。
fn=f0+fi
,可能产生
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
5-35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fi=465kHz,p、q为本振和信号的谐波次
~1605kHz波段内的干扰在
3的情况。所以落于535
fS=930kHz和fS=1395kHz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。
5-36解:若满足
±pf1±qf
2
=fs,则会产生互调干扰:
p=1、q=1,f1+f2=774+1035=1.809p=1、q=2,f1+2f
2
(MHz),不会产生互调干扰;
(MHz),会产生互调干扰;
(MHz),会产生互调干扰;(MHz),会产生互调干扰;
=774+2⨯1035=2.844
p=2、q=1,2f1+f2=2⨯774+1035=2.583(MHz),会产生互调干扰;p=2、q=2,2(f1+f2)=2⨯(774+1035p=2、q=3,2f1+3fp=3、q=2,3f1+2f
22
)=
3.618
=2⨯774+3⨯1035=4.653(MHz),会产生互调干扰;=3⨯774+2⨯1035=4.392
p=3、q=3,3(f1+f2)=3⨯(774+1035p、q大于3谐波较小,可以不考虑
。
)=
5.427
(MHz),会产生互调干扰;
))
⎧3f+2f0
(1)⎨S5-37解:⇒fS+f0
2f+3f
⎧-fS+2f0
⇒-fS+f0
⎩-2fS+3f0
f0
⎧-3fs+2f0>30(2)⎨⇒-fs+f0>12(MHz
-2f+3f>30s0⎩
⎧fs+2f0>30
⇒fs+f0>20(MHz⎨
2f+f>300⎩s
∴fS=4MHz
5-39解:若满足
±pf1±qf
2
)
)
f0>16MHz
=fs,则会产生互调干扰。
已知f1=19.6MHz、
。
f2=19.2MHz、fs=f0-fi=23-3=20(MHz),故没有互调信号输出
6-4解:P==VCCICO=24⨯0.25=6(W
)
ηC=
Rp=Icm1=
P0P=Vcm
2
=
56=
=83.3%VCC2P02P0VCC=
o2
2P02P0Vcm
=
24
2
2⨯52⨯524
=57.6(Ω)=0.417(A)
==
gc(θc)=
Icm1Ic0
0.420.25
=1.67
查表得θc=77
=
2⨯0.7⨯12
10.8
=1.56查表得θc=91
o
6-6解:gc(θc)=
2
2ηVCCVcm
2
P0=IkR=2⨯1=4(W
)
)
⎛1⎫⎛1⎫ PC=P=-P0= -1⎪P=-1 ⎪⨯4=1.7(W⎪0η0.7⎝⎭⎝c⎭
6-7解:icmax=
α090
o
Ic0
o
=
900.319
=282(mA)
))
Ic1m=α1(90P0=ηc=
12
2
)i
cmax
=0.5⨯282=141(mA⨯200⨯0.1412
=74%
2
RpIc1m=P0
=
12
=2(W
VCCIc0Vcm2RP
2
30⨯0.09
2
IkmR
2
6-8证:P0=
=
IkmR
[
2
+(ω0L)2LRC
2
]=
[
2
+(ω0L)RC2L
2
]
=
IkmR
2
[
2
+(ω0L)R
2
2
]
2(ω0L)
≈
IkmR2
2
6-9解:Vcm=VCC-vcmin=VCC-
Ic0=icmaxα0(70Icm1=icmaxα1
o
icmaxgcr
=24-
2.20.8
=21.25(V)
)=2.2⨯0.253=0.5566(A)(70)=2.2⨯0.436=0.9592(A)
o
P==VCCIc0=24⨯0.5566=13.36(WP0=
12
VcmIcm1=
12
)
)
⨯21.25⨯0.9592=10.19(W
PC=P=-P0=13.36-10.19=3.17(W
)
ηC=
Rp=
P0P=Vcm
2
=
10.1913.36=
=76.3%
2
21.25
2P0
2⨯10.19
=22.16(Ω)
6-10解:R1=Rp=
X
=R1QL12πfX=
R2R1
L1=
X
L1
C1
Vcm2P014410=
2
≈
VCC2P0
2
=
24
2
2⨯2
=144(Ω)
C1
=
=14.4(Ω)
1
=221(pF
C1=X
2⨯3.14⨯50⨯10
=
+1)-116.95
200144
6
⨯14.4200
)
R2
L1
=16.95(Ω)
(Q
2L
⨯(100+1)-1=0.054(μH
2πf=
=
2⨯3.14⨯50⨯10
6
)
X
C2
QLR1⎛R2
2
QL+1 ⎝QLX1
=
C2
L1
⎫10⨯144⎛200⎫
-1⎪=-1 ⎪=2.57(Ω)⎪100+110⨯16.95⎝⎭⎭
1
6
C2=
2πfX2⨯3.14⨯50⨯10⨯2.57
=1239
(pF)
2
P0'=Vcm/2RP0'=IcmR
2
P
P
(1)RP增加一倍,放大器工作6-11解:
(2)RP减小一半,放大器工作
6-12解:ηk=
r'r1+r'
=1+
1
于过压状态,于欠压状态,
11+
1Q1Q2k
2
Vcm变化不大,Icm变化不大,=
2
=0.5P0;
/2=2P0。
ωL1ωL2Q1Q2(ωM
=
1
1+
1
100⨯15⨯0.03
2
=57.4%
)
2
6-13解:RP=
Vcm2P0
2
=
(V
CC
-VCE(sat))2P0
2
=6610
(12-0.5)2⨯1
=66(Ω)
设QL=10C1=X
12πfX=
C1
则X
C1
=
RPQL1
8
=
=6.6(Ω)=241(pF50
=
2⨯3.14⨯10
=
RLRP
-1
1
2⨯3.14⨯10
⨯6.6
)
=5.5(Ω)
RL
C2
(1+Q)
2L
(1+10)⨯
2
8
5066
-1
C2=
12πfX=X
C2
=
⨯5.5
=290(pF)
X
L1
QLRP⎛RL
1+2
QL+1 QLXC2
⎝
L1
⎫10⨯66⎛50⎫⎪=1+ ⎪=12.5(Ω)2⎪10+110⨯5.5⎝⎭⎭
=19.9(nH
L1=
2πf
=
12.52⨯3.14⨯10
8
)
(a)将R1C1和R2C2串联电路改为6-14证:
R1'=
X
2
2C1
2C1
'C1'和R'R1C'并联电路,并设22X
2C2
2C2
XC1=
R1QL'2=XC
R2
2
2
2C2
R1+X
R1'=R2
R2+XX
2C1
2C1
2
R2
11+QL
2
'1=XC
R1
2
2C2
2
2C1
R1+XX
2
2
X
C1
R2+X
X
C2
',即R1'=匹配时R1'=R2∴X
=
R2
R1+X
2
R1=R1=
R2+X
R2
C2
C2
(1+Q)R
2L
2
R1
2
-1R1
2
2C12C1
2C1
X
L1
'1+XC'2==XC
R1
2
2
2C1
R1+X+
X
2
X
C1
+
R2
2
2
2C2
R2+X=
X
C2
=
R1+X
X
C1
R1+X
⋅
R1R2X
C2
R1QL1+QL
2
+
11+QL
2
⋅
R1R2X
C2
=
R1QL⎛R2
1+2
1+QL QLXC2
⎝
R1QL
'=XL1
R2
2
2
⎫
⎪⎪⎭
(b)将R1C1和R2L1串联电路改为
R1'=
X
2
2C1
2C1
'C1'和R''并联电路,并设R1L1
2X
2
2L1
2L1
XC1=
R1+X
R1'=R2
R2+XX
2C1
2C1
R2
11+QL
2
'1=XC
R1
2
2L1
2
2C1
R1+XX
2
2
X
C1
R2+X
2L1
X
L1
',即R1'=匹配时R1'=R2∴X
=
R2
R1+X
2
R1=R1=
R2+X
R2
L1
L1
(1+Q)R
2L
2
R1
-1R2
2
2
2L1
X
C2
'1-XC'1==XL
XC1R1+X
2
2C12
R2+X-
X
L1
-
R1
2
2
2C1
R1+X
2
X
C1
=⋅
R1R2X
L1
R1QL1+QL
2
=
R1QL1+QL
⋅
R2QLX
L1
-
R1QL1+QL
2
=
R1QL⎛R2
2
1+QL⎝QLX
L1
⎫-1⎪⎪⎭
读数为0;读数增加;流表读数减小。
(1)天线断开,工作于过压6-18解:
(2)天线接地,工作于欠压(3)中介回路失谐,工作于
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
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
(1)PA=P=-PC-Pk=10-3-1=6(W6-19解:
(2)ηk(3)ηc
=
PAP=-PCP=-PC
P=
=
610-310-310
=85.7%
)
===70%η=
PAP=
=
610
=60%
6-20解:当k=kc时,ηk=1-
'=1-若ηk
r1r1+r''
r1r1+r'
=50%则r'=r1
1Q1Q2
=
kc=
1Q0
1Q
=Q09
3Q=3k
c
=90%则r''=9r1故k=
6-25解:
6-27解: α2(60
o
)
o
1
。
V
6-28解:Ri=
V
=93=9RL
I3I
o
6-29解:Vbm=
VBZ+VBB
cosθ
=
0.6+1.45cos70
=6(V)
VB=Vbm-VBB=6-1.45=4.55(ViCmax=
VB-VBZ
2
=
o
)
4.55-0.6
2
=1.98(A)=0.86(A)
Icm1=iCmaxα1(70
)=1.98⨯0.436
Vcm=VCC-gcriCmax=24-1.98=22.02(VP0=
Icm1Vcm
2
=
0.86⨯22.01
2
=9.47(W
)
)
)
⎛QL⎫10⎫⎛
⎪P0= 1-PA= 1-⎪⨯9.47=8.52(W ⎪Q100⎝⎭0⎭⎝
6-30解:Vbm=
VBZ+VBB
cosθ
=
0.6+1.5cos70
o
=6.14(V)
VB=Vbm-VBB=6.14-1.5=4.64(ViCmax=
VB-VBZ
2
=
o
)
4.64-0.6
2
=2.02(A)
=0.88(A)
Icm1=iCmaxα1(70
)=2.02⨯0.436
)
Vcm=ξVCC=0.9⨯24=21.6(VP0=
Icm1Vcm
2
=
0.88⨯21.6
2
=9.5(W)
)
⎛QL⎫10⎫⎛
⎪P0= 1-PA= 1-⎪⨯9.5=8.8(W ⎪Q0⎭100⎭⎝⎝
(a)电路可能振荡,属于电7-5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L2C2
Cbe可能振荡,属于电容反
容反馈式振荡电路
馈式振荡电路。
(1)有可能振荡,属于电容7-6解:
(2)有可能振荡,属于电感(4)有可能振荡;属于电容(3)(5)(6)不可能。
7-7解:
反馈式振荡电路,反馈式振荡电路,反馈式振荡电路,
f1f2>f0>f3;f1=f2
1
1
2⨯3.14⨯13⨯
-7
2π1RP
L(C+Cd=1Q
CL
(1)f0=7-21解:
(2)gd
=
)
=
⨯(20+5)⨯10=5.27(mS
-12
=100(MHz)
=
(20+5)⨯1010
-7
-12
)
(3)0.06~0.08V
(1)fq=7-26解:
1.657⨯10
d
6
=Sd
1.657⨯10
0.4=21.1⨯10
0.4
3
6
=4.14(MHz
-5
)
)
Cq=21.1⨯10Lq=43.5
dS
3
-5
⨯
2000.4
=0.105(pF
=43.5⨯dS
-2
200
=14(mH)
=21.2(Ω)=19.8(pF
rd=42500B
=42500⨯0.25⨯Sd
=3.96⨯101.050.25
⨯10
64-2
0.4200⨯2000.4
C0=3.96⨯10Qq=
1.05Bfq
)
⨯10d=
6
4
⨯0.4=16800
(2)d
=
1.657⨯10
=
1.657⨯1015⨯10
6
=0.11(mm)
(1)1.5~1.5001(MHz7-27解:
(2)不能
)
负阻特性。
(3)不能,普通三极管没有
7-28解:恒温槽、稳压电源7-29解:并联
电路、共集电极缓冲级
等。
、高稳定度克拉泼振荡
电路。
c-b型(皮尔斯)晶体振荡
9-3解:i=I(1+macosΩt)cosω0t
=Icosω0t+
I2
2
macos(ω0+Ω)t+
2
I2
macos(ω0-Ω)t
2
I有效
值
=
⎛I⎫⎛I⎫⎛I⎫
ma⎪+ ma⎪ ⎪+
⎝2⎭⎝22⎭⎝22⎭I2
1+
ma2
2
=
(1)v=25(1+0.7cos2π5000t-0.3cos2π10000t)sin2π10t9-4解:
6
=25sin2π10t+8.75(sin2π1005000
6
+sin2π995000
)
-3.75(sin2π1010000
+sin2π990000
)
(2)包络25(1+0.7cos2π5000t-0.3cos2π10000t)
m上=m下=
Vmax-V0
V0V0-Vmin
V0
)
峰值调幅度=
25⨯(1+0.7-0.3)-25
25
25-25(1-0.7-0.3)
25
=0.4
谷值调幅度==1
)
)
(1)ma=1时9-5解:
(2)ma
=0.3时
3
P(ω
+Ω
=P(ω
-Ω
)
=
14
)
maP0T=14
2
2
14
⨯100=25(W14
2
P(ω
0+Ω
)
=P(ω
0-Ω
=
maP0T=
⨯0.3⨯100=2.25(W
9-6解:i=b1v+b3v不包含平方项,不能产
生调幅作用。
2
)
(1)P(ω9-7解:
P(ω
0+Ω
))
=P(ω
0+Ω
)
=
)
14
maP0T=
2
14
⨯0.7⨯5000=612.5(W
±Ω
=2P(ωP0avη
=
+Ω
=1225(W=50000.5
2
)
)
2
(2)P=
=
P0TηP0T
=10(kW
(3)P=
=
P0avη
=
⎛m⎫
1+a⎪ 2⎪⎝⎭η
=
⎛0.7
5000⨯ 1+
2⎝
0.5
⎫
⎪⎪⎭
=12.45(kW)
(1)ma=1时9-8解:
P(ω
+Ω
)
=P(ω
-Ω
)
=
)
14
maP0T=
2
14
⨯1000=250(W)
)
P0=P0T+P(ω
+Ω
+P(ω
-Ω
)
=1000+250+250=1500(W
(2)ma
P(ω
=0.7时
)
0+Ω
=P(ω
0-Ω
)
=
)
14
maP0T=
2
14
⨯0.7⨯1000=122.5(W
2)
)
P0=P0T+P(ω
+Ω
+P(ω
-Ω
)
=1000+122.5+122.5=1245(W
(kHz)
9-9解:f=f0+f1+f2+f3+f4=5+20+200+1780+8000=10005
9-10解:i1=b0+b1(v+vΩ)+b2(v+vΩ
i2=b0+b1(v-vΩ)+b2(v-vΩ
)2)2
+b3(v+vΩ+b3(v-vΩ
2
)3)3
3
v0=(i1-i2)R=R2b1vΩ+4b2vvΩ+6b3vvΩ+2b3vΩ
=2b1RV
Ω
[]
cosΩt+3b3RV0VΩcosΩt+1.5b3RV
2
2
23Ω
cosΩt
+2b2RV0VΩcos(ω0+Ω)t+2b2RV0VΩcos(ω0-Ω)t+1.5b3RV0VΩcos(2ω0+Ω)t+1.5b3RV0VΩcos(2ω0-Ω)t+0.5b3RV
3Ω
cos3Ωt
Ω、3Ω、ω0±Ω、2ω0±Ω
输出端的频率分量:
9-12解:m1=
⎛P0⎫
⎪=2 -1 P⎪
⎝0T⎭12
2
⎛10.125⎫
2⨯ -1⎪=0.5
9⎝⎭
12
⨯0.4⨯9=10.845(kW
2
P0'=P0+m2P0T=10.125+
)
9-13(1)vA(t)=vΩ(t)=VΩcosΩt
vB(t)=v(t)
(2)若D1D2开路,则
vAB=0
vA(t)=vB(t)=vΩ(t)vA(t)=vΩ(t)=VΩcosΩt
(3)若D1D2短路,则
vB=0
vAB(t)=VΩcosΩt
9-18解:RΩ=R1+
3πRdR
R2ri2R2+ri2=
=510+
4700⨯10004700+1000=0.57
=1335(Ω)
θ=
3⨯3.14⨯100510+4700
Kd=cosθ=0.87
VΩ=KdmaVim=0.87⨯0.3⨯0.5=0.13PΩ=
VΩ
2
2
2RΩVim2RidPΩPω
=
2
=
0.13
2⨯1335VimKd
R
2
=6.33(μW0.5⨯0.87510+4700
2
)
=41.7(μW
Pω=AP=
==
)
6.3341.7
=0.152
R22⋅ri2R22+ri2
=
2350⨯1000
=0.55
(1)中间位置9-19解:
RΩR
R1+R22+=
510+2350+
R1+R2R2ri2R2+ri2
=510+
2350+1000
510+4700
(2)最高端
RΩR
R1+=
4700⨯1000
=0.26最高端不会。
R1+R2
510+4700
∴R2的触点在中间位置会产
生负峰切割失真,而在
9-20解:由R=R1+R2=(5~10)kΩ
RΩ=R1+
R2ri2R2+ri2
2
⎛11R1= ~
⎝510=3(kΩ)
2
⎫⎪R2⎭
RΩR=
取R2=6kΩ39=13
R1=1.5kΩ
=1.5+
6⨯26+2
ma
C
-ma
1
maRΩmax
Ωminri23πRdR
=
=
1-0.3
0.3⨯9000⨯2⨯3.14⨯3000
1
2⨯3.14⨯300⨯20003⨯3.14⨯1006000+1500
=0.5
=0.0187(μF)
取C1=C2=0.01μF
Ce>>
=
=0.26(μF)
取Ce=20μF
θ=
Kd=cosθ=0.9Rid=
R2K
d
=
90002⨯0.9
=5(kΩ)能满足要求
3
-12
=5.84(μS)
9-21解:GP=
QL=
ω0CQ0f0
=
=
2⨯3.14⨯465⨯10⨯200⨯10
100
=23.5
46520
2∆f0.7
Q0
p34=
QL
GP-p24goe-GP
gid
=
2
⎛100⎫2-6
⨯5.84-0.3⨯100-5.84⨯10 ⎪⎝23.5⎭
=0.153
2
4700
(1)v1=mV1cosΩtcosω1t9-24解:
i=kmV1cosΩtcosω1tV0cos(ω0t+ϕ=14
)
kmV1V0{cos[(ω1+ω0+Ω)t+ϕ]+cos[(ω1-ω0+Ω)t-ϕ]
+cos[(ω1+ω0-Ω)t+ϕ]+cos[(ω1-ω0-Ω)t-ϕ]}当ω0=ω1时,vS=
1412kmR
L
V1V0[cos(Ωt-ϕ)+cos(Ωt+ϕ
)]=
12
kmR
L
V1V0cosϕcosΩt
无失真,ϕ只影响输出幅度。当ω0≠ω1时,vS=有失真。
kmR
L
V1V0cos[(ω1-ω0)t-ϕ]cosΩt
(2)v1
=
1212
mV1cos(ω1+Ω)t
i==14
kmV1cos(ω1+Ω)tV0cos(ω0t+ϕ
)
kmV1V0{cos[(ω1+ω0+Ω)t+ϕ]+cos[(ω1-ω0+Ω)t-ϕ]}14kmR
V1V0cos[(ω1-ω0+Ω)t-ϕ]
vS=
L
当ω0=ω1时,ϕ只产生相移;当ω0≠ω1时,有失真。