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Smithchart匹配习题

Smithchart匹配习题
Smithchart匹配习题

[作业]设负载阻抗为Z l =100+j50Ω接入特性阻抗为Z 0=50Ω的传输线上。f =1000MHz

1)用单支节调配法实现负载与传输线匹配,通过Smith 圆图求解。 2)用LC 匹配网络实现负载与传输线匹配,通过Smith 圆图求解。 解: 1)

解一:(开路支节)

负载阻抗归一化0

21L Z Z Z j ==+Ω ,在阻抗圆图上找到负载对应的A 点,在驻波比圆上转动直到与圆1+jb 相交,得到B 点(绿色线路径1)和B ’点(红色线路径1),可读出负载的归一化导纳分别为11L

Y j S =+ ,11L

Y j S =- ,由径向线在相位标尺上的角度可知,A 点到B 点旋转的角度为θ=144o ,A 点到B ’点旋转的角度为θ1=270o ,因此支节到负载的电长度分别为θ/2=72o 和θ1/2=135o ,即支节的接入位置分别为d=0.2λ ,d 1=0.375λ.

为了抵消B 点和B ’点中的电纳,截短线的归一化电纳为1j S 。由开路点D 点,沿ρ=∞圆顺时针转到1j S 的E 点(绿色线路径2)和E ’(红色线路径2)点。由相位标尺上的角度可知,D 点到E 点旋转的角度为θ=270o , D 点到E ’点旋转的角度为θ1=90o ,所以支节的长度为分别为θ/2=135o 和θ1/2=45o ,其对应的电长度分别为0.375λ、0.125λ,

因此支节线长度: E 点接入l =0.375λ。 E ’点接入l 1=0.125λ。

解二:(短路支节)

负载阻抗归一化0

21L Z Z Z j ==+Ω ,在阻抗圆图上找到负载对应的A 点,在驻波比圆上转动直到与圆1+jb 相交,得到B 点(绿色线路径1)和B ’点(红色线路径1),可读出负载的归一化导纳分别为11L

Y j S =+ ,11L

Y j S =- ,由径向线在相位标尺上的角度可知,A 点到B 点旋转的角度为θ=144o ,A 点到B ’

点旋转的角度为θ1=270o ,因此支节到负载的

电长度分别为θ/2=72o和θ1/2=135o,即支节的接入位置分别为d=0.2λ,d1=0.375λ.

为了抵消B点和B’点中的电纳,截短线的归一化电纳为1j S

。由短路点F点,沿ρ=∞圆顺时针转到1j S

的E点(绿色线路径2)和E’(红色线路径2)点。由相位标尺上的角度可知,F点到E点旋转的角度为θ=90o, F点到E’点旋转的角度为θ1=270o,所以支节的长度为分别为θ/2=45o和θ1/2=135o,其对应的电长度分别为0.125λ、0.375λ,因此支节线长度:

E点接入l=0.125λ。

E’点接入l1=0.375λ。

2)

解一:

负载导纳归一化00.40.2L

Y Z Z j S ==- ,在导纳圆图上找到负载对应的A 点,从负载朝源端看,并联一个电容,A 点从导纳圆上沿路径1旋转直到与1+jx 的阻抗圆相交,得到B 点,可读出B 点负载的归一化导纳为0.4+j0.49S ,,则ΔB=j0.69S ,由 0B cZ ω?=,得C=2.2pF ;然后串联一个电感,改变电抗,B 点在阻抗圆上沿路径2旋转直到1+j0Ω,B 点的归一化阻抗为1-j1.224Ω,所以电抗的该变量为ΔX=-j1.224Ω,由0

L X Z ω?=,得L=9.74nH

B

A 0.4-j 0.2S

0.4+j 0.49S/

1-j 1.224Ω

1

2

解二:

解:负载导纳归一化00.40.2L Y Z Z j S ==- ,在导纳圆图上找到负载对应的A 点,从负载朝源端看,并联一个电感,A 点从导纳圆上沿路径1旋转直到与1+jx 的阻抗圆相交,得到B 点,可读出B 点负载的归一化导纳为0.4-j0.49S ,,则ΔB=-j0.29S ,由 0Z B L

ω?=,得L=27.5nH ;

然后串联一个电容,改变电抗,B 点在阻抗圆上沿路径2旋转直到1+j0Ω,B 点的归一化阻抗为1+j1.224,所以电抗的该变量为ΔX=j1.224S ,由0

1X CZ ω?=

,得C=2.6pF

0.4-j 0.2S

0.4-j 0.49S/1+j1.224Ω

A B

1

2

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