《射频通信电路》习题答案全
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R P = ρQ = 159 × 100 = 15.9kΩ
可求得 P2 = 0.336
= RL P22 RS P12 = = 12.8 (0.8) 2 1 (0.336) 2 = 20kΩ = 8.86kΩ
' 信号源内阻 RS 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
回路的空载谐振阻抗为
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
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
1-8
C ' 2 = C 2 + C 0 = 40PF
因此回路的总电容为
CΣ = Ci + C1 ⋅ C ' 2 20 × 40 =5+ = 18.3PF C1 + C ' 2 20 + 40
回路谐振频率
ω0 =
1 LC Σ = 1 0.8 × 10
−6
× 18.3 × 10
−12
= 26 × 10 7 rad/s
回路有载品质因数为
Qe = RΣ
ρ
=
1
ω 0 L × 0.17 × 10 −3
≈ 28
回路通频常 BW3 dB = 1-9
ω0
Qe
=
26 × 10 7 = 0.93 × 10 7 rad/s = 1.48MHz 28
C1 C1 + C 2
设回路的空载 Q0 = ∞ ,设 P 为电容接入系数 P = 率传输,
Ri =
则
V1 V 3V , RL = 3 = 1 I3 I1 3I 1 V 1 Ri 1 = , Z C1 = Z C 2 = 1 = RL R2 9 I1 3
Z C3 = V3 3V1 = = RL I3 I1
I1 V1 Ri V2 I2
(e)
I1 I2
I3 V3 I3 RL
第二章 2-1
则
Vn2 = 4kTR 1 B = 4 × 1.38 × 10 -23 × 290 × 510 × 10 3 × 10 5 = 8.16 × 10 −10 ( V 2 ) 1 2 = 4 kT In B = 3.14 × 10 -21 (A 2 ) R1 510 × 250 R1 || R2 = = 167.7 kΩ 510 + 250 Vn2 = 4kTRB = 2.68 × 10 −10 (V 2 )
(2)当 BW3dB = 300kHz 时
Qe = f0 10 = = 33.33 BW3dB 0.3
回路谐振电导
Ge =
ω C 2π × 10 7 × 56 × 10 −12 1 = 0 = = 10.55 × 10 −5 (s) ρQe Qe 33.33 ω C 2π × 10 7 × 56 × 10 −12 1 = 0 = = 5.27 × 10 −5 (s) ρQ0 Q0 66.67
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
Qe = RΣ
回路有载 Q 值为 回路的通频常 1-7 由于 BW3dB =
回路空载谐振电导
G0 =
并联电导
G = Ge − G0 = (10.55 − 5.27) × 10 −5 = 5.28 × 10 −5 (s)
并联电阻 1-3
L1 = 1
R=
1 1 = = 18.9KΩ G 5.28 × 10 −5
L2
L3 2
(2πf 1 ) C1
2
= 2.06μH
v1 v3
L2 =
C1 = 33PF
1-10
f0 f0 10 9 → Qe = = = 40 , Qe BW3dB 25 × 10 6 R ∵ Qe = i → X CΣ = 50 = 1.25 40 X CΣ BW3dB =
则必有 X C
2
< 1.25 ,由 R2 与 C 2 组成的并联支路 C1 C1 + C 2 R2 P
Te = Te1 +
系统的噪声系数为
Te2 G
= 290 +
800 = 315.3k 31.62
F = F1 +
2-5 噪声底数为
F2 − 1 3.76 − 1 =2+ = 2.087 G 31.62
Ft = −174dBm/Hz + NF(dB) + 10logB
= −174 + 3 + 10 log 10 5 = −121dBm
,由于有最大功
∴ ∵ ∵ 可得:
RS =
RL → P = 0.333 P2 f BW3dB = 0 → Qe = 10 Qe
Qe =
R RΣ → RΣ = R S || L = 4 .5 k Ω P2 ω0L RΣ 4.5 × 10 3 = = 4.48μH L= ω 0 Qe 2π × 16 × 10 6 × 10 C ⋅C 1 1 = 22PF CΣ = 1 2 = 2 = C1 + C 2 ω 0 L (2π × 16 × 10 6 ) 2 × 4.48 × 10 − 6 C 22 C2 = Σ = = 66PF P 0.333
回路的谐振阻抗
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
RΣ = R S || R P || R L = 42KΩ
回路的有载 Q 值为
QБайду номын сангаас = RΣ
ρ
=
42 × 10 3 = 37 2 πf 0 L
通频带 在 Δf
BW3dB =
f 0 465.5 = = 12.56kHz 37 Qe
(c)
(c)
2V , RL = V I 2I V Z C = = 2RL I Ri =
,
Ri 4 = RL 1
V
I V Ri V I
(d)
(d)
Vi = 3V I i = I Ri 9 = RL 1
, V L = V , I L = 3I
I I
RL
V Z C = = 3RL I
(e)
I 1 = I 3 , I 3 = I 2 , V1 = V2 , V3 = 3V1
f2 =
1 2π LC 2
1 C1C 2 C1 + C 2
f2 =
2π L
1-5 由于回路为高 Q,所以回路谐振频率
f0 ≈ 1 2π LC = 1 2π 300 × 10
−12
× 390 × 10 −6
= 465.5kHz
回路的损耗电阻
r=
ω0L
Q0
=
2π × 465.5 × 10 3 × 390 × 10 −6 = 11.4Ω 100
2
Q 大于 4 以上,则 Q 2 >> 1 ,
1 10 = 0.316
此题可用高 Q 计算。 接入系数 P = ,由题意有
= 50 ,∵ R2 = 5 ,所以 P =
' R2 R / P2 50 = 2 →L= = 0.199nH 2π × 10 9 × 40 ω0L ω0L 1 1 = 127 PF CΣ = 2 = 9 2 ω 0 L (2π × 10 ) × 0.199 × 10 −9
2-2 由于匹配,所以输出额定噪声功率 Pn = kTB
2-4 先把 dB 数化为自然数 G=15dB=31.62;NF=2dB=2;
Te2 = 800 k → F2 = 1 +
Te 2 800 =1 + = 3.76 T0 290
放大器的等效噪声温度为 Te1 = ( F − 1)T0 = (2 − 1) × 290 = 290k 系统的等效噪声温度为
= 10kHz 处的选择性为: 1 1 = S= = 0.532 → −5.47dB 2 2 20 ⎛ ⎞ ⎛ ⎞ 2Δf 1 + ⎜ 37 × ⎟ ⎟ 1+ ⎜ ⎜ Qe f ⎟ 465.5 ⎠ ⎝ 0 ⎠ ⎝
1-6 回路特性阻抗 回路谐振阻抗 由
P22 RL = 1 + RP RS P12
ρ=
1 1 = = 159 Ω 7 2πf 0 C 2 π × 10 × 100 × 10 −12
Qe =
由 CΣ =
C1 ⋅ C 2 C1 + C 2
及P=
C1 C1 + C 2
求得: C 2 = 401.9PF , C1 = 185.7PF
Qe =
R1 || R2 30.8 × 10 3 = = 38 ' ω 0 L1 811
1-12 0.1μH 在 100MHz 时的阻抗为
X L = ω 0 L = 2π × 10 8 × 0.1 × 10 −6 = 62.8Ω
因此在匹配网络中采用电容 C1 的容抗与 0.1μH 的电抗部分抵消,见 图示。
C1 C2 0.1μH
10Ω
X C1 = X L − x = 62.8 − 20 = 42.8Ω → C1 =
1 = 37.2PF 42.8 × 2π × 10 8
由于
Q=
50 1 → X C2 = 25Ω → C 2 = = 63.7 PF X C2 25 × 2π × 10 8
1-16 (a)
V 4I 4V V L = 4V , I L = I , R L = I V 1 Ri 1 = , Z C = = RL R L 16 I 4
Vi = V
, Ii
= 4 I , Ri =
则
(b) 由图(b):
I 2 = 2 I 1 , V1 = 2V2
I1 V1 V1 I1 I2 V2 RL I2
第一章 1-1
S= 1 0.6 × 2 1+ ( × 66.67) 2 10 = −16dB = 0.158
将f 得
= f 0 ± 100kHz 及 f 0 = 640kHz 代入
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 Q0 = = = 66.67 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=
f0 Qe
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
所以回路有载
Qe =
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
由于输入阻抗 Ri = 50Ω ,大于放大器串联输入电阻 10Ω,所以采用的 匹配网络应是将串联的 10Ω化为并联的 50Ω。 匹配网络的 Q 值为
Q= 50 −1 = 2 10
当 Q=2 时要求与 r=10 串联的电抗值为
x = r ⋅ Q = 10 × 2 = 20Ω < X L = 62.8Ω
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
' GS = G Σ − G p = 0.01 ms
由于 G S'
= P 2 G S ,所以电容接入系数为:
P2 =
' GS 0.01 × 10 −3 = = 0.01 ⇒ P = 0.1 GS 10 −3 1 = = 159PF 回路总电容 C = 1 2 6 2 ω 0 L (6.28 × 10 ) × 159 × 10 −6 1 ωC ∵接入系数 P = 2 = C 所示 C 2 = C = 1590PF 1 P C2 ωC C 1− P = ,所以 C1 = 159 = 176PF C1 0 .9
L3 =
1 = 2.74 μH (2πf 2 ) 2 C 2
1 (2πf 3 ) 2 C 3 = 0.68μH
C2
C3 C1
L1 2’
1-4
x
x
f0
f
f0
f
(a) f 0 =
1 2π LC
(b) f 0 =
1 2π LC
(c)
f1 =
1 2π C ( L1 + L2 )
1 2π CL1
(d)
f1 =
(b)
则
4V Vi = V1 + 2V2 , I i = I 1 , Ri = 2 I1 V V L = V2 , I L = 2 I 2 , R L = 2 4I1 Ri 16 = RL 1
V2
Z C1 =
ZC2
V1 1 = Ri = 8RL I1 2 V = 2 = 2 RL I2
V Ri
RL 2 RL 2