高频电子线路张肃文pdf版

由 (12 + C )×16052 = (100 + C)5352得C = -1pF (不合理舍去 )
故采用后一个。
( ) 2)L
=
ω02
1
(C +
C′)
=
1
2 × 3.14 × 535 ×103 2 × (450 + 40)×10−12
= 180(μH )
3)
L C C’
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=
1 2 × 3.14 ×106 2 × 253 ×10−6
= 100(pF ) → CX
= 200 pF
RX
= ω0L Q
− ω0L Q0
=
2 × 3.14 ×106 × 253 ×10−6 2.5 0.1
− 2 × 3.14 ×106 × 253 ×10−6 100
= 47.7(Ω)
ZX
= RX
(R +
= R+
jω0L)(R +
1 jω0C
)
jω0L + R +
1 jω0C
=
R2
+
L C
+
jω0LR(1 −
1 ω02 LC
2R +
jω0L(1 −
1
ω2 0
LC
)
)
=
R2 + L C
2R
=R
3 − 4解：1)由(15 + C )×16052 = (450 + C )5352得C = 40 pF
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页码，4/5
3 − 13解：1)L1
= L2
ρ =1
ω01
=
103 2 × 3.14 ×106
= 159(μH )
( ) C1
=
C2
=
f0
=
2π
1 LC
=
2 × 3.14
1 0.8 ×10−6 ×18.3 ×10−12
= 41.6(MHz)
RP = Q0
L = 100 × C12
0.8 × 10−6
(20 + 20)×10−12
= 20.9(kΩ)
R∑
=
Ri
RP
C2
+ C0 C1
+ C1 2 R0
= 10
20.9
2
20
+
20 20
= 24(pF )
a = 1 + rbb′gb′e = 1 + 70 × 0.754 × 10−3 ≈ 1
b = ωCb′erbb′ = 2 × 3.14 ×107 × 24 × 10−12 × 70 ≈ 0.1
( ) ( yie = gb′e +
jωCb′e )(a −
a2 + b2
jb) =
0.754 ×10−3 +
页码，1/5
3 − 1解：f 0 = 1MHz
2Δf0.7 = 1 ×106 − 990 ×103 = 10(kHz)
Q
=
f0 2 Δf0.7
=
1 ×106 10 ×103
= 100
取R = 10Ω
则L
=
QR ω0
=
100 ×10 2 × 3.14 ×106
= 159(µH)
C
=
1 ω02 L
=
1 ( 2 × 3.14 ×106 )2
R1
5
2∆f0.7 = η 2 + 2η − 1 ⋅ 1 = 22 + 2 × 2 − 1 × 1 = 0.013
f0
Q
200
3 − 17解： I = I0
1
=
1 + Q′
2∆f f0
2
1
= 1 → Q′ = 22.5
1
+
Q
′
10 ×103 300 ×103
2
1.25
R=
1 Qω0C
=
1 22.5 × 2 × 3.14 × 300 ×103 × 2000 ×10−12
= 100 3
ξ
=
Q0
2 Δf f0
=
100 3
×
2
×
(5.5 − 5) ×106
5 ×106
= 20 3
因2Δf 0′.7
=
2
×
2∆f
，
0.7
则Q′0
=
0.5Q
，故
0
R′ =
0.5R，
所以应并上21kΩ电阻。
3 − 8证明：4πΔf0.7C =
2πf0C f 0 2 Δf0.7
= ω0C Q
= g∑
= 11.8
I=
1
=
1
= 1 → Q = 30
I0
1+
Q
2∆f f0
2
1
+
Q
10 ×103 300 ×103
2
2
Q − Q′ = 30 − 22.5 = 7.5
3 − 18解：
ω
2ω =
= (L1
1 串联联谐
L2C
1
并联联谐
+ L2 )C
→
LL1 2==317255μμH
页码，5/5
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页码，2/5
3 − 5解：Q
0
=
1 ω0C0 R
=
1 2 × 3.14 ×1.5 ×106 ×100 ×10-12
×5
= 212
( ) L0
=
1 ω02C0
=
1 2 × 3.14 ×1.5 ×106 2100 ×10-12
= 112(μH )
I om
= Vom R
=
1 × 10-3 5
=
0.2(mA)
−
1 j
ω0C X
= 47.7 −
1 j 2 × 3.14 ×106 × 200 ×10−12
= 47.7 −
j 796(Ω )
( ) 3 − 7解：L
=
1 ω02C
=
1 2 × 3.14 × 5 ×106
× 50 × 10−12
= 20.2(μH )
Q0
=
f0 2 Δf 0.7
=
5 ×106 150 ×103
=
−0.0187 −
j0.187(mS )
y fe
=
gm (a − jb)
a2 + b2
=
37.7 ×10−3 × (1 −
12 + 0.12
j0.1)
=
37.327 −
j3.733(mS )
( )( ) yoe = gce + jωCb′c + rbb′gm
gb′c + jωCb′c a − jb a2 + b2
页码，1/6
4 − 5解：当f = 1MHz时，β = 当f = 20MHz时，β = 当f = 50MHz时，β =
β0
=
1 +
β0 f fT
2
β0
=
1+
β0 f fT
2
β0
=
1+
β0 f fT
2
50
= 49
1
+
50 ×106 250 ×106
2
50
= 12.1
1
+
50 × 20 250 ×
× 10 6 106
=
28.2(kHz )
( ) 5)C2′
=
1
(ω0′2 )2 L2
=
1 2 × 3.14 × 950 ×103 2 ×159 ×10−6
= 177(pF )
Z 22
=
R2
+
jω02 L1
−
1 ω02C2′
=
20 +
j 2 × 3.14 ×106
× 159 ×10−6
−
1 2 × 3.14 ×106 ×177 ×10−12
× 159 ×10−6
= 159(pF)
3 − 2解：( 1)当ω01 =
1 L1C1
或ω 02
=
1 时，产生并联谐振。 L2C2
( 2 )当ω01 =
1 L1C1
或ω 02
=
1 时，产生串联谐振。 L2C2
( 3 )当ω01 =
1 L1C1
或ω02
=
1 时，产生并联谐振。 L2C2
3 − 3证明：Z
∴Rf1 = 0→ M = 0
Q=
2 f0 = 2∆f 0.7
2 106 14 ×103
= 100
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( ) 3 − 16解：1)R f 1
=
(ω01M )2
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( ) 4 − 8解：令
Av Avo
m
=
2
4 + (Q2Δf0.7
)f0 4
m
=
1 2
( ) 令
Av Avo
m
=
2
4 + (Q2Δf0.1
)f0 4
m
=
1 10
故K r0.1
=
2 Δf 0.1 2 Δf 0.7
=
4
410
2 m
− 1
4
4
2
1 m
− 1
=4
2
10m − 1
1
R2
=
107 × 10−6 5
2
= 20(Ω)
( ) ( ) Rab =
(ω01 L )2
R1 + R f 1
= 107 ×100 ×10−6 2 = 40(kΩ)
5 + 20
2)η = ω01M = 107 × 10−6 = 2
R1
5
3)Q = ω01L = 107 ×100 ×10−6 = 200
2m −1
得2Δf 0.7
=
4
4
2
1 m
− 1
⋅
f0 Q得2Δf 0.1=4410
2 m
− 1
⋅
f0 Q
4 − 9解：p1
=
N 23 N 13
=
5 20
= 0.25
p2
=
N 45 N13
=
5 20
= 0.25
gp
=
1 ω0Q0 L
=
1 2π ×10.7 ×106 ×100 × 4 ×10−6
=
37.2(μS )
=
20 +
j100
( ) ( ) Z f 1 =
ω01 M Z 22
2
=
2 × 3.14 ×106 × 3.18 ×10−6 20 + j100
2
= 0.768 − j3.84(Ω)
3 − 15解：Θ
R=
L RPC
=
159 ×10−6 50 ×103 ×159 ×10−12
= 20(Ω) = R1
j2 × 3.14 ×107 × 24 ×10−12 12 + 0.12
× (1 −
j0.1)
= 0.895 + j1.41(mS )
y re
=
(− gb′c
+
jωCb′c )(a −
a2 + b2
jb)
≈
−
j2 × 3.14 ×107 × 3 ×10−12 12 + 0.12
× (1 −
j0.1)
1 ω021L1
=
1 2 × 3.14 ×106 2 ×159 ×10−6
= 159(pF )
M
=
ηR1 ω01
=
1 × 20 2 × 3.14 ×106
= 3.18(μH )
( ) 2)Z
f1
=
(ω01 M
R2
)2
=
2 × 3.14 ×106 × 3.18 ×10−6 2 20
= 20(Ω)
( ) Z = P
R1
L1 + Rf1
C1
=
159 × 10−6
(20 + 20) ×159 ×10−12
= 25(kΩ)
3)Q1
=
ω01 L1 R1 + R f 1
=
2 × 3.14 ×106 ×159 ×10−6 20 + 20
=
25
4)2Δf0.7 =
2 f0 = Q
2
f0 ρ1 R1
=
2
×
106 × 20 103
VLom = VCom = Q0VSm = 212 ×1 × 10-3 = 212(mV )
( ) 3 − 6解：L
=
1 ω02C
=
1 2 × 3.14 × 106 2 × 100 × 10−12
= 253(μH )
Q0
= VC VS
=
10 = 100 0.1
( ) C ⋅ CX
C + CX
=
1 ω02 L
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3 − 9解：C
=
Ci
+
(C2 + C0 )C1
C2 + C0 + C1
=5+
(20 + 20)20
20 + 20 + 20
= 18.3( pF )
2
50
=5
1
+
50 × 50 ×106 250 ×106
2
4
−
7解：g b′e
=
IE
26(β0
+ 1)
=
1
26 × (50
+ 1)
=
0.754(mS )
gm
=
β 0
rb′e
= 50 × 0.754 ×10−3
= 37.7(mS )
Cb′e
=
gm 2πfT
=
2
×
37.7 × 10−3 3.14 × 250 ×106
≈
jωCb′c
1
+
rbb′
gm
a − jb a2 + b2
≈
j2 × 3.14 ×107
× 3 ×10−12 1 + 70 × 37.7 ×10−3
×
1− 12 +