《射频通信电路》陈邦媛著课后答案详细版
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' GS = G Σ − G p = 0.01 ms
' = P 2 G ，所以电容接入系数为： 由于 G S S
回路总电容
' GS 0.01 × 10 −3 = = 0.01 ⇒ P = 0.1 GS 10 −3 1 1 C= 2 = = 159PF 6 2 ω 0 L (6.28 × 10 ) × 159 × 10 −6
在 ∆f = 10kHz 处的选择性为： 1 S= 2∆f 1+ Qe f 0 1-6 回路特性阻抗 回路谐振阻抗 由 P22 RL = 1 + RP RS P12
2
=
1 20 1 + 37 × 465.5
2
= 0.532 → −πf 2 ) 2 C 2
L3 =
1-4
1 (2πf 3 ) 2 C 3
= 0.68µH
x
x
f0
f
f0
f
(a) f 0 = x
1 2π LC
(b) f 0 = x
1 2π LC
f1
f2
f
f1
f2
f
(c)
f1 =
1 2π C ( L1 + L2 ) 1 2π CL1
1 1 = = 159 Ω 7 2 πf 0 C 2π × 10 × 100 × 10 −12
可求得 P2 = 0.336
R P = ρQ = 159 × 100 = 15.9kΩ
' 信号源内阻 R S 折合到回路两端为： R S = ' 负载电阻 R L 折合到回路两端为： R L =
RS P12 =
第一章 1-1 S= 1 0.6 × 2 × 66.67) 2 1+ ( 10 = −16dB = 0.158
将 f = f 0 ± 100kHz 及 f 0 = 640kHz 代入 得 Q=20
BW3dB = f 0 640 = = 32kHz Q 200
1-2 （1） 1 1 = = 4.53µH 2 7 2 ω 0 C (2π × 10 ) × 56 × 10 −12 f0 10 = = 66.67 Q0 = BW3dB 0.15 1 1 S= = = 0.124 = −18.13dB 2 2 × 0 . 6 2 2( f − f 0 ) × 66.67 1+ 1+ Q0 10 f0 L=
P2 =
1 ωC 2 C C ∵接入系数 P = 所示 C 2 = = 1590PF = 1 C2 P ωC 159 C ，所以 C1 = 1− P = = 176PF C1 0.9 1-8
C ' 2 = C 2 + C 0 = 40PF
因此回路的总电容为
CΣ = Ci +
回路谐振频率
C1 ⋅ C ' 2 20 × 40 =5+ = 18.3PF 20 + 40 C1 + C ' 2
ω0 =
回路的空载谐振阻抗为
1 LC Σ
=
1 0.8 × 10
−6
× 18.3 × 10
−12
= 26 × 10 7 rad/s
R P = ρQ0 = ω 0 LQ0 = 26 × 10 7 × 0.8 × 10 −6 × 100 = 20.9kΩ
电阻 R0 对回路的接入系数为 P =
C1 1 = C1 + C ' 2 3
(d)
f1 =
f2 =
1 2π LC 2
1 2π L C1C 2 C1 + C 2
f2 =
1-5 由于回路为高 Q，所以回路谐振频率 1 1 f0 ≈ = = 465.5kHz − 12 2π LC 2π 300 × 10 × 390 × 10 −6 回路的损耗电阻
r=
回路的谐振阻抗
ω0L
Q0
=
2π × 465.5 × 10 3 × 390 × 10 −6 = 11.4Ω 100
2 R P = r (1 + Q0 ) = 114KΩ
考虑信号源内阻及负载后回路的总谐振阻抗为 2
RΣ = R S || R P || R L = 42KΩ
回路的有载 Q 值为
Qe =
通频带
RΣ
ρ
=
42 × 10 3 = 37 2πf 0 L
BW3dB =
f 0 465.5 = = 12.56kHz Qe 37
=
12.8 (0.8) 2 1
= 20kΩ = 8.86kΩ
RL P22
(0.336) 2
回路总谐振阻抗 RΣ 为
1 1 1 1 1 1 1 = + ' + ' = + + = 0.0629 + 0.05 + 0.112 = 0.226ms RΣ R P R S R L 15.9 20 8.86
即
RΣ = 4.43kΩ
RΣ
回路有载 Q 值为 Qe = 回路的通频常 1-7 由于 BW3dB =
ρ
=
4.43 × 10 3 = 27.8 159
BW3dB =
f 0 10 × 10 6 = = 0.359MHz Qe 27.8 f0 10 6 = = 50 BW3dB 20 × 10 3
f0 所以回路有载 Qe
Qe =
（2）当 BW3dB = 300kHz 时
Qe =
回路谐振电导
f0 10 = = 33.33 BW3dB 0.3
Ge =
回路空载谐振电导
ω C 2π × 10 7 × 56 × 10 −12 1 = 0 = = 10.55 × 10 −5 （s ） 33.33 Qe ρQe ω C 2π × 10 7 × 56 × 10 −12 1 = 0 = = 5.27 × 10 −5 （s ） 66.67 Q0 ρQ0
回路谐振时的总电导为
GΣ =
1 1 = = 0.02 ms （即 R Σ = 50 KΩ) 6 ω 0 LQ e 2π × 10 × 159 × 10 −6 × 50
3
回路的空载电导为
Gp =
1 = 0.01 ms （即 R P = 100 K ） ω 0 LQ0
信号源内阻折合到回路两端的电导值为
考虑了 Ri 与 R0 后的谐振阻抗 RΣ 为
1 ( )2 1 1 1 P2 1 1 = + + = + + 3 = 0.17 ms(5.9kΩ) 5 RΣ R P Ri R0 20.9 10
G0 =
并联电导
并联电阻 1-3
L1 = 1
2
G = G e − G 0 = (10.55 − 5.27) × 10 −5 = 5.28 × 10 −5 （s ） 1 1 R= = = 18.9KΩ G 5.28 × 10 −5 L2 L3
(2πf 1 ) C1 = 2.06µH
2 v1 v3 1 C2 C3 C1 L1 2’
' = P 2 G ，所以电容接入系数为： 由于 G S S
回路总电容
' GS 0.01 × 10 −3 = = 0.01 ⇒ P = 0.1 GS 10 −3 1 1 C= 2 = = 159PF 6 2 ω 0 L (6.28 × 10 ) × 159 × 10 −6
在 ∆f = 10kHz 处的选择性为： 1 S= 2∆f 1+ Qe f 0 1-6 回路特性阻抗 回路谐振阻抗 由 P22 RL = 1 + RP RS P12
2
=
1 20 1 + 37 × 465.5
2
= 0.532 → −πf 2 ) 2 C 2
L3 =
1-4
1 (2πf 3 ) 2 C 3
= 0.68µH
x
x
f0
f
f0
f
(a) f 0 = x
1 2π LC
(b) f 0 = x
1 2π LC
f1
f2
f
f1
f2
f
(c)
f1 =
1 2π C ( L1 + L2 ) 1 2π CL1
1 1 = = 159 Ω 7 2 πf 0 C 2π × 10 × 100 × 10 −12
可求得 P2 = 0.336
R P = ρQ = 159 × 100 = 15.9kΩ
' 信号源内阻 R S 折合到回路两端为： R S = ' 负载电阻 R L 折合到回路两端为： R L =
RS P12 =
第一章 1-1 S= 1 0.6 × 2 × 66.67) 2 1+ ( 10 = −16dB = 0.158
将 f = f 0 ± 100kHz 及 f 0 = 640kHz 代入 得 Q=20
BW3dB = f 0 640 = = 32kHz Q 200
1-2 （1） 1 1 = = 4.53µH 2 7 2 ω 0 C (2π × 10 ) × 56 × 10 −12 f0 10 = = 66.67 Q0 = BW3dB 0.15 1 1 S= = = 0.124 = −18.13dB 2 2 × 0 . 6 2 2( f − f 0 ) × 66.67 1+ 1+ Q0 10 f0 L=
P2 =
1 ωC 2 C C ∵接入系数 P = 所示 C 2 = = 1590PF = 1 C2 P ωC 159 C ，所以 C1 = 1− P = = 176PF C1 0.9 1-8
C ' 2 = C 2 + C 0 = 40PF
因此回路的总电容为
CΣ = Ci +
回路谐振频率
C1 ⋅ C ' 2 20 × 40 =5+ = 18.3PF 20 + 40 C1 + C ' 2
ω0 =
回路的空载谐振阻抗为
1 LC Σ
=
1 0.8 × 10
−6
× 18.3 × 10
−12
= 26 × 10 7 rad/s
R P = ρQ0 = ω 0 LQ0 = 26 × 10 7 × 0.8 × 10 −6 × 100 = 20.9kΩ
电阻 R0 对回路的接入系数为 P =
C1 1 = C1 + C ' 2 3
(d)
f1 =
f2 =
1 2π LC 2
1 2π L C1C 2 C1 + C 2
f2 =
1-5 由于回路为高 Q，所以回路谐振频率 1 1 f0 ≈ = = 465.5kHz − 12 2π LC 2π 300 × 10 × 390 × 10 −6 回路的损耗电阻
r=
回路的谐振阻抗
ω0L
Q0
=
2π × 465.5 × 10 3 × 390 × 10 −6 = 11.4Ω 100
2 R P = r (1 + Q0 ) = 114KΩ
考虑信号源内阻及负载后回路的总谐振阻抗为 2
RΣ = R S || R P || R L = 42KΩ
回路的有载 Q 值为
Qe =
通频带
RΣ
ρ
=
42 × 10 3 = 37 2πf 0 L
BW3dB =
f 0 465.5 = = 12.56kHz Qe 37
=
12.8 (0.8) 2 1
= 20kΩ = 8.86kΩ
RL P22
(0.336) 2
回路总谐振阻抗 RΣ 为
1 1 1 1 1 1 1 = + ' + ' = + + = 0.0629 + 0.05 + 0.112 = 0.226ms RΣ R P R S R L 15.9 20 8.86
即
RΣ = 4.43kΩ
RΣ
回路有载 Q 值为 Qe = 回路的通频常 1-7 由于 BW3dB =
ρ
=
4.43 × 10 3 = 27.8 159
BW3dB =
f 0 10 × 10 6 = = 0.359MHz Qe 27.8 f0 10 6 = = 50 BW3dB 20 × 10 3
f0 所以回路有载 Qe
Qe =
（2）当 BW3dB = 300kHz 时
Qe =
回路谐振电导
f0 10 = = 33.33 BW3dB 0.3
Ge =
回路空载谐振电导
ω C 2π × 10 7 × 56 × 10 −12 1 = 0 = = 10.55 × 10 −5 （s ） 33.33 Qe ρQe ω C 2π × 10 7 × 56 × 10 −12 1 = 0 = = 5.27 × 10 −5 （s ） 66.67 Q0 ρQ0
回路谐振时的总电导为
GΣ =
1 1 = = 0.02 ms （即 R Σ = 50 KΩ) 6 ω 0 LQ e 2π × 10 × 159 × 10 −6 × 50
3
回路的空载电导为
Gp =
1 = 0.01 ms （即 R P = 100 K ） ω 0 LQ0
信号源内阻折合到回路两端的电导值为
考虑了 Ri 与 R0 后的谐振阻抗 RΣ 为
1 ( )2 1 1 1 P2 1 1 = + + = + + 3 = 0.17 ms(5.9kΩ) 5 RΣ R P Ri R0 20.9 10
G0 =
并联电导
并联电阻 1-3
L1 = 1
2
G = G e − G 0 = (10.55 − 5.27) × 10 −5 = 5.28 × 10 −5 （s ） 1 1 R= = = 18.9KΩ G 5.28 × 10 −5 L2 L3
(2πf 1 ) C1 = 2.06µH
2 v1 v3 1 C2 C3 C1 L1 2’
