−=
dB16
=
.0
158
=
H53.4
μ
10
12
−
1
×
2
7
)
56
=
67.66
×
=
.0
124
−=
dB13.18
1
26.0
×
10
×
67.66
2
⎞
⎟
⎠
⎛
⎜
⎝
第一章
1-1
=S
f
= f
0 ±
100
将
得
1-2
(1)
1
26.0(1
×
+
10
640
0 =f
×
2
)67.66
代入
kHz
f
0
Q
=
640
200
=
32
kHz
及
kHz
Q=20
BW
=
dB3
S
=
1
+
L
=
Q
0
=
1
C
ω
2
0
f
0
BW
1
=
dB3
2(
10
×
π
10
15.0
=
0
Q
⎛
⎜⎜
⎝
300
=
(2
f
0
f
−
f
0
时
=
kHz
f
0
BW
dB3
=
2
)
⎞
⎟⎟
⎠
1
+
10
3.0
=
33.33
(2)当
BW
dB3 =
Qe
=
回路谐振电导
G
1
Q
ρ
回路空载谐振电导
1
Q
ρ
G
=
0
e
并联电导
并联电阻
1-3
L
1
=
1
)
2(
2
f
π
1
C
1
1
Cf
(
)
π
2
2
1
)
f
π
C
2
2
3
3
2(
L
2
=
L
3
=
H06.2
μ
=
=
.
742
H
μ
2
=
H68.0
μ
=
ω
0
Q
C
e
=
2
π
×
10
×
10
12
−
56
×
7
33.33
=
55.10
×
10
−
5
(s)
e
=
ω
0
Q
C
0
0
=
2
π
×
10
×
10
12
−
56
×
7
67.66
=
27.5
×
10
−
5
(s)
GGG
−
=
e
0
55.10(
−
)27.5
R
=
1
G
=
28.5
=
−
5
10
K9.18
Ω
=
1
×
10
×
5
−
=
28.5
×
10
5
−
(s)
L2
C2
L3
C3
v1
v3
C1
L1
2
2’
1-4
x
x
f0
f
f0
f
(a)
f
0 =
1
π2
LC
(b)
f
0 =
1
π2
LC
(c)
f
1
=
2
π
f
2
=
1
LC
(
1
1
CL
1
2
π
+
L
2
)
(d)
f
1
=
f
2
=
π
2
2
1
LC
1
CC
1
2
C
C
+
1
2
2
π
L
1-5
由于回路为高 Q,所以回路谐振频率
=
≈
1
f
0
1
12
−
π LC
2
2
π
300
×
10
×
390
回路的损耗电阻
=
465
5.
kHz
−
6
×
10
Ω
)
6
−
×
×
0
r
×
=
=
r
=
+
=
L
Ω
1(
5.
10
390
465
3
10
100
2
×
π
RP
2
0Q
4.11
K114
ω
0
Q
回路的谐振阻抗
=
考虑信号源内阻及负载后回路的总谐振阻抗为
R
S
回路的有载 Q 值为
R
×
3
10
42
= Σ
Lf
πρ
2
0
5.
465
37
处的选择性为:
通频带
10=Δf
f
0
eQ
56.12
kHz
K42
Qe
kHz
BW
=Σ
37
dB3
Ω
=
=
R
R
R
=
=
=
=
||
P
||
L
在
1-6
回路特性阻抗
回路谐振阻抗
由
+
=
2
P
2
R
L
1
R
P
2
P
1
R
S
S
=
=
1
1
+
Q
e
⎛
⎜⎜
⎝
f
2
Δ
f
0
2
⎞
⎟⎟
⎠
1
1
+
⎛
⎜
⎝
37
×
2
20
5.465
⎞
⎟
⎠
=
.0
532
−→
dB47.5
7
1
×
100
k9.15
Ω
10
=
=
159
Ω
−
12
10
×
π
×
2
100
×
=
ρ
=
1
Cf
π
2
0
RP ρ
Q
159
=
=
可求得
2 =P
336.0
R
S
2
P
1
=
=
.0(
8.12
2
)8.0(
1
336
)
2
=
k20
Ω
=
k86.8
Ω
R
'
L
=
R
L
2
P
2
1
86.8
=
.0
0629
+
05.0
+
112.0
=
.0
226
ms
信号源内阻 SR 折合到回路两端为:
R
'
S
=
负载电阻 LR 折合到回路两端为:
回路总谐振阻抗 ΣR 为
1
1
'
'
R
R
S
L
即
k43.4
=Σ
Ω
回路有载 Q 值为
1
9.15
1
20
43.4
1
R
1
R
P
R
+
=
+
+
=
=
Σ
+
Qe
R
= Σ
ρ
回路的通频常
BW
dB3
=
f
0
eQ
3
10
×
159
10
10
×
8.27
=
=
8.27
6
=
.0
359
MHz
1-7
由于
BW
=
dB3
f
0
eQ
所以回路有载
Qe
=
f
0
BW
dB3
=
6
10
×
10
20
=
50
3
回路谐振时的总电导为
=
ω eLQ
回路的空载电导为
G
=
1
Σ
0
×
π
2
1
159
6
10
×
−
6
×
10
×
50
=
.020
ms
(即
R
=Σ
50
K
)Ω
=
G p ω
0
1
LQ
0
=
.010
ms
(即
RP
100=
K
)
信号源内阻折合到回路两端的电导值为
由于
G
'
S
,所以电容接入系数为:
GG
010=
= Σ
=⇒
GP
01.0
ms
' =
S
G
−
P
P
=
=
'
S
.
p
−
3
2
S
2
1.0
01.0
10
×
3
−
10
S
L
=
1
26
)
28.6(
×
10
×
159
×
10
−
6
=
159
PF
回路总电容
C
∵接入系数
P
=
=
=
G
G
1
2
ω
0
1
C
ω
1
C
ω
=C
2
1
1
−
CP =
C
1
,所以
1-8
+
因此回路的总电容为
+
C
C
C
C
=
2
2
'
=Σ
i
=
C
C
2
所示
C
=
2
=
1590
PF
C
P
159
9.0
=
176
PF
C
0
=
PF40
CC
'
1
C
C
1
⋅
+
+=
5
2
'
2
20
20
×
+
40
40
=
PF3.18
回路谐振频率
ω
0
=
1
ΣLC
=
1
×
8.0
×
10
−
6
3.18
×
10
12
−
=
26
×
10
7
rad/s
回路的空载谐振阻抗为
Q
ωρ
0
RP
=
=
0
0
电阻 0R 对回路的接入系数为
C
1
考虑了 iR 与 0R 后的谐振阻抗 ΣR 为
P
=
=
26
×
LQ
7
10
C
1
C
+
×
8.0
=
'2
1
R
Σ
=
1
R
P
+
1
R
i
+
P
R
2
0
=
1
9.20
+
1
10
+
6
−
×
100
=
k9.20
Ω
=
17.0
ms(5.9k
Ω
)
10
×
1
3
2
)
1(
3
5
回路有载品质因数为
=
=
Qe
R
Σ
ωρ
0
BW ω
eQ
=
dB
0
3
1
17.0
×
×
26
28
L
=
10
7
×
10
≈
28
−
3
回路通频常
=
.
930
7
×
10
rad/s
=
1.48MHz
1-9
设回路的空载 ∞=0Q ,设 P 为电容接入系数
率传输,
P
=
C
1
C
+
2
C
1
,由于有最大功
可得:
L
=
R
S
=→=
P
.0
333
R
P
L
2
BW
dB3
=→=
10
e
Q
f
0
Q
→
e
Qe
=
Σ
=
R
Σ
L
0ω
R
eQ
ω
0
CC
⋅
2
C
C
+
C
Σ
P
PF33
=
1
1
=
=
C
Σ
C
2
1 =C
5.4
16
2
π
×
1
ω
2
0
=
L
2
22
333.0
=
2(
×
π
=
PF66
R
=Σ
R
S
||
R
L
2P
=
k5.4
Ω
×
×
10
10
3
6
×
10
=
H48.4
μ
1
)
2
16
×
10
6
×
48.4
×
10
−
6
=
PF22
∴
∵
∵
1-10
由
C
=Σ
BW
dB3
=→=
Q
e
f
0
Q
e
R
X
i
f
0
BW
dB3
50
40
=
25
9
10
=
40
,
6
10
×
∵
e
=
→
=ΣCX
Q
,由 2R 与 2C 组成的并联支路 Q 大于 4 以上,则
25.1
=
C
Σ
2 >>Q ,
1
25.1
2
x
2
=
<
=
10
20
Ω
LX
⋅=
Qr
=×
当 Q=2 时要求与 r=10 串联的电抗值为
8.62
Ω
因此在匹配网络中采用电容 1C 的容抗与 0.1μH 的电抗部分抵消,见
图示。
=→Ω
PF2.37
0.1μH
=−
8.62
8.42
10Ω
20
C2
C1
C
X
X
=
=
−
=
x
C
1
L
由于
Q
=
50
X
C
2
→
X
C
2
=→Ω=
25
C
2
25
×
1
28.42
×
π
8
10
×
PF7.63
=
10
8
1
1
2
π
×
1-16
(a)
Vi = , I
4= ,
I i
V
Ri
=
V
则
VL
R
R
(b) 由图(b):
V
+
1
4= , I
I L = ,
V
I
V =
1
1
I i = ,
1
16
I = ,
2
2V
, 1I
,
2I
Vi
Z
=
=
=
C
L
2
i
RL
1
4
2V
=
则
I L = ,
22I
2V
VL = ,
R
R
16
1
=
i
L
V
I
4
V
4
=
I
R
L
2
Ri =
V
2
4I
1
V
24
I
1
RL =
I1
I1
V1
I2
V2
RL
V1
V2
I2
(b)
=
1
2
R
i
=
8
R
L
=
2
R
L
Z
C
1
Z
C
2
=
=
V
1
I
1
V
2
I
V
2
I
V
I
2
Ri
= ,
RL
= ,
V
I
2
R
R
L
i
=
Z
C
=
2=
R
L
Vi
R
R
i
L
Z
C
=
V
3=
9
1
V
I
=
I
VL = ,
I i = , V
3=
R
L
Ri
V
V
4
1
RL
2
RL
2
I L
3=
I
Ri
(c)
I
V
V
I
I
RL
I
(d)
V =
3
3V
1
I
I = ,
3
V
I
RL
I = , 2
I
1
Ri = ,
Ri ,
R
3
V
1
3I
1
9
Z
Z
C
=
=
=
C
1
1
2
Z
C
3
=
3
=
V
I
3
V
3
1
I
1
3
3
=
V = ,
1 V
2
V
3
1
I
1
V
1
I
=
=
R
1
2
L
=
1
3
R
L
Ri
V1
V2
V3
I3
I3
RL
I1
I1
I2
I2
(e)
(c)
(d)
(e)
则
第二章
2-1
10
-23
×
290
×
510
×
10
3
×
10
5
=
16.8
×
10
10
−
)V(
2
2
=nV
kTR4
1
=nI
2
4B
kT4
=
×
B
1.38
×=
1
R
1
250
510
×
250
510
+
kTRB
68.2
=
=
1 RR
||
2
2
=nV
4
3.14
×
10
21-
)A(
2
=
167
k7.
Ω
×
10
10
−
)V(
2
2-2
则
由于匹配,所以输出额定噪声功率
kTB=nP
2-4
先把 dB 数化为自然数
G=15dB=31.62;NF=2dB=2;
T
+=→
800
=
k
1
F
2
2
e
2
+=
1
T
e
T
0
800
290
=
.
763
放大器的等效噪声温度为
F
=
(
Te
1
−
T
)1
0
=
)12(
×−
290
=
290
k
系统的等效噪声温度为
T
e
=
T
e
1
+
系统的噪声系数为
FF
1
=
+
T
e
2
G
=
290
+
800
62.31
=
k3.315
1
F
−
2
G
+=
2
1
76.3
−
62.31
=
.2
087
2-5
噪声底数为
−=tF
−=
174
174
dBm/Hz
10
++
3
+
log
NF(dB)
10logB
+
10
121
dBm
−=
5
放大器的线性动态范围可以定义为它的 1dB 压缩点的输入功率与噪声底数之比,即
若
−=lDR
dB
SNR
(
则灵敏度
dBm
dB
111
F
=
t
)121(
P
in
SNR o
,)
(
20=min
101−=
10
min
min
+
=
−
)
o
,
,
−=lDR
放大器的线性动态范围也可以定义为它的 1dB 压缩点的输入功率与灵敏度之比,
则线性动态范围为
(
101
dB
10
91
=
−
)