通信原理概论课后习题答案 仅含马殿光布置的书后习题教材是数字与模拟通信系

本站小编 半岛在线注册/2019-03-28

1.1Use 1-6
d1 = 2h1 d 2 = 2h2
h1 is the height of transmitter antenna measured in feet
h2 is the height of receiver antenna measured in feet
d1 and d 2 are measured in miles
d = d1+ d 2
Since h1 =1200 ft d = 60miles then h2 = 60.6 ft
1.2 According to Figure1-3 2 2 2
d + re = (re + h)
Where re is the radius of the Earth, re = 5280miles
2 2 2 2 d + re = re + 2reh + h
Sinceh ı re , then 2 2 2 d + re = re + 2reh => 2 d = 2reh => d = 2reh
Since 1mile=5280feet, h is measured in feet, then
1
2 5280
5280
mile
d hfeet mile
feet
= ´ ´ ´ => d = 2hmile
1.3Use 1-6
d1 = 2h1 d 2 = 2h2
d = d1+ d 2 d1 = d 2
Sinced = 25miles , then 1 2 h = h = 78.125 ft

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