微波技術(shù)與天線4smith

1、Smith Chart 史密斯圓圖0)(ZzZin歸一化輸入阻抗歸一化輸入阻抗: 0)()(ZzZzZininjvu)z(jXR)z(Zin)(1z)(1z)(1)(1zz)(1)(1zzjXR222)R11(v)R1Ru(222)X1()X1v() 1u(jvujvu11R=1圓心坐標(biāo)?半徑?R=2圓心坐標(biāo)?半徑?222)R11(v)R1Ru(222)X1()X1v() 1u(Resistance circles Reactance circles zj2Le)z(Smith Chart Applications: 1.Input impedance(Zin) determination u

2、sing a known load(ZL) 已知負(fù)載已知負(fù)載(ZL) 確定輸入阻抗確定輸入阻抗 (Zin) Example: ZL=50+j50W, Z0=50W, l=l/8 1) 歸一化. ZL=1+j2)畫等VSWR圓 zj2Le)z()(1)(1zzVSWR1+1jzj2Le)z(l/8-?度jjzj2822lZin=(2-1j)*50 )tan()tan()(000zjZZzjZZZzZLLin2,Input impedance(Zin) determination using the input reflection coefficient( in) 由反射系數(shù)確定輸入阻抗例: |

3、in|=1/sqrt(5) =0.4472 , q =Arg(in)=0.3524pi Zin=?)(1)(1)(0zzZzZinininZin=1+1j 3.Determination of admittance from impedance value 已知阻抗確定導(dǎo)納例: ZL=50+j50W, Z0=50W, find YL? YL = (0.5-j0.5)* Y0 = (0.5-j0.5)/50 S方法1:直接計(jì)算YL = 1/ ZL =方法2:在Smith圓圖上找中心對稱點(diǎn)1/( 50+j50)=(50-j50)/5000=0.01-j0.01S4. Determination o

4、f value and location of Zmax and Zmin Zmax 和和 Zmin的值和位置確定的值和位置確定)z(1)z(1)z(Zin| )(|1| )(|1maxzzZ|) z(|(1|) z(|(1Zmin1Z*Zmaxmin例: ZL=50+j50W, Z0=50W find Zmax and Zmin? Zmax=2.62, Zmin=0.382 | )(|z| )(|1| )(|1maxzzZ|)(|(1|)(|(1minzzZ618. 255550.3819665555 5010050jj0ZZL0ZZL551250025005. Determination

5、of VSWR using a known load(ZL) 已知負(fù)載確定已知負(fù)載確定VSWR VSWRminZ習(xí)題1.6maxZ| )(|1| )(|1zz| )(|1| )(|1zz| )(|1| )(|1zzVSWR16.Input impedance(Zin) determination using single stubs 單支節(jié)輸入阻抗的確定 Parallel stub Series stub 并聯(lián)支節(jié) 串聯(lián)支節(jié)例：50W 傳輸線，負(fù)載為250W，串聯(lián)短路支節(jié)實(shí)現(xiàn)匹配，求：負(fù)載到支節(jié)的長度d1，支節(jié)的長度d2。d1d2250W1)Series stubs串聯(lián)支節(jié)Single Stu

6、b Matching 單支節(jié)匹配0.20.51.02.05.0+0.2-0.2+0.5-0.5+1.0-1.0+2.0-2.0+5.0-5.00.0 a) 找到阻抗為5+j0的點(diǎn)Ab) 畫 VSWR 圓c) 沿VSWR圓順時(shí)針旋轉(zhuǎn),與電阻為1的電阻圓相交于Bd) 變量z從A到B的差值就為 d1 (0.3168-0.25=0.0668l)e) B 點(diǎn)歸一化阻抗為 1-j1.8, 所以支節(jié)的電抗應(yīng)為j1.8（點(diǎn)C）。 d2為從短路點(diǎn)到點(diǎn)C的長度。(0. 1691l)BACd1d2250W例：50W 傳輸線，負(fù)載為250W，并聯(lián)短路支節(jié)實(shí)現(xiàn)匹配，求：負(fù)載到支節(jié)的長度d1，支節(jié)的長度d2。2）Para

7、llel (or shunt) stubs: 并聯(lián)支節(jié)并聯(lián)支節(jié) f) C at 1+j1.76, the susceptance of the stub is j1.76,find the distance d2 from G= to B(susceptance)=-1.76,(0.82l)a) Find point 5+j0 at A Ab) Draw a VSWR circlec) Find admittances at Bd) Move toward generator (clockwise) from B until reach point C at R circle(G=1)e) t

8、he distance d1 from B to C is (0.182l)250W3) l l /4 transformer matching l l /4 變換線匹配變換線匹配Example:. 一段50W 的傳輸線，負(fù)載阻抗 ZL=100-j50W，用特性阻抗為Z0，長度為l/4的另一段傳輸線進(jìn)行阻抗變換而實(shí)現(xiàn)匹配。求 (1) d1, (2) Z0, (3) VSWR on d1 line and (4) VSWR on l/4 line 1 j25050j100ZLZ0a) Find point A :2-j1,then get VSWR on d1 b) Draw a VSWR circle from A to B(電抗為0 ), 38. 0ZBZB=19, Z0 =sqrt(19*50)=30.8 c) 62. 08 .3019BZd) Find point 0.62