1、8 第二章作业 2.1-2 1 = 2 28 4 2 2 = 2 32432 4 0.443 辐射近区界限条件为 0.443 2 将 = 10, = 7.6带入得 50.4 1316 2.1-4 题号 () () () () 1.6() () 最小距离 （ 1） 1 5 300 47.7 480 6.6103 480m （ 2） 0.8 4 16 0.027 0.2667 7.68 7.68m （ 3） 13 65 0.05 0.008 0.08 6760 6.76km 2.1-8 = 302 + (2 2)+(2 2) 1) = 2 4 = 2 = 302(2)+Si2Si(2) 查表得 =
2、 1.852 (2) = 1.371 (2) = +ln(2)(2) 其中 C = 0.57721 (2) = 0.472 6.7076 2) = 2 = 2 = 3022 +(22 4) = 3042 4 9 2 = +2 2 其中 2 = 0.0227 4 = +4 4 其 中 4 = 0.006 199.1 2.2-1 = 60 = 60 10 20037590 103 9.2/ 采用无方向性天线 = 0 = 200 10 = 2000 2.2-4 1) = 4 2(,)020= 4 sin2 sin4 00= 4 sin3 sin4 00= 44338= 8 2) = 2 ， 令 (2
3、,) = sin(2)sin2 =22 ， 解得 = 2(90) = 65.54 = 2， 令 (,2) = sin()sin2 (2) = 22 ， 解得 = 2(90) = 90 3) 41253= 4125365.5490 = 6.99 2.2-7 = = 30 = 103 = 0.03 1) 由题 2.18 算得 = 6.7 = = 4 107 1075.7107 = 0.83 103 = 8 = 0.0331 = + 99.51% 10 由图 2.24 对称振子方向系数曲线得 1.5 () = 10lg() = 1.74 2) Rr = 200 = = 4 107 1075.7107
4、 = 0.83 103 = 8 = 0.0331 = + 99.98% 由图 2.24 对称振子方向系数曲线得 2.5 () = 10lg() = 3.98 2.3-1 1) = = sin2 = 73.13sin2(2 4)= 73.13 = Zin += 73.13 5073.13 +50 = 0.188 = 1+1 = 1.463 = 12 = 96.5% 2) Rr由图 2.19 对称振子辐射电阻查得为 100 = = sin2 = 100sin2(2 34 )= 100 = Zin += 100 50100 +50 = 13 = 1+1 = 2 = 12 = 88.9% 2.3-5
5、= = 3108200 106 = 1.5 = 0.750.015 = 25 由图 2.37 估计 = 25 时半波振子波长缩短系数 1 1.06 20 = 21= 21.06 0.7075 11 = sin2 = 73.13sin2(1.06 2 )= 73.1 2.4-2 1) = 1(1 +22 )2+12 cos2 当 = 0 时 ， 取得最大值 = 1(1 22 )2当 = 2时，取得最小值 = 1(1 +22 )22) 1 = 1 2 = 10 320 = (2 +1)22(22 +1) = 97.16% 2.5-3 1) = = 5000 = 105000 = 37 2) 2 =
6、 602 = 12 2120 =12601202 = 0.293/2 0 = = 0.586/2 2.5-4 = 24 = 100 ( 31089.375 109)24 = 8.151032 = 10.580% 8.15 103 = 3.26 2.5-5 = 22(4)34 = 2106 (103.37)2 0.112 3(4)310134 = 211751 当天线增益增大两倍，距离增大 2倍 ， 即 299462m 2.5-6 = 22(4)34 =8000(102.95)2 ( 31089.3109)25(4)3 (37000)4 = 8.91011 () = 10lg(8.9108) = 70.5
