约翰克劳斯《天线》一书的一道习题

在第二章的习题里面，有道关于火星地球间无线通信的题，不会做，我觉得题目比较有意思，很想弄清楚到底怎么分析，我认为题目好像条件不够啊，连天线是哪种类型都没说清楚，不过感觉可能指的是螺旋天线，因为后面有类似题目中就提到了。 
题目我贴上来了，麻烦大家帮我看看吧。 
 

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本书课后完整版答案： 
本部分内容设定了隐藏，需要回复后才能看到 
 
2-11-4. Mars and Jupiter links. 
(a) Design a two-way radio link to operate over earth-Mars distances for data and picture transmission with a Mars probe at 2.5 GHz with a 5 MHz bandwidth. A power of 10-19 
6 
W Hz-1 is to be delivered to the earth receiver and 10-17 W Hz-1 to the Mars receiver. The Mars antenna must be no larger than 3 m in diameter. Specify effective aperture of Mars and earth antennas and transmitter power (total over entire bandwidth) at each end. Take earth-Mars distance as 6 light-minutes. (b) Repeat (a) for an earth-Jupiter link. Take the earth-Jupiter distance as 40 light-minutes. 
2-11-4. continued 
Solution: 
(a) 
1961317611(earth)10510510 W(Mars)10510510 WrrPP−−−−=××=×=××=× 
Take )5.0( m 5.31.5(1/2)Mars)(ap22===επeA 
Take kW 1Mars)(=tP 
Take )5.0( m 35051(1/2)earth)(ap22===επeA 
MW 9.63505.312.0)103360(105)earth(Mars)(earth)(Mars)()earth(2281122=××××==−tetetrtPAArPPλ 
To reduce the required earth station power, take the earth station antenna 
22m 392750 )2/1(==πeA (ans.) 
so 
62(earth)6.910(15/50)620 kW ()tPans.=×= 
W10812.0)103360(39305.310earth)(Mars)(Mars)()earth(14228322−×=×××==λrAAPPerettr 
which is about 16% of the required 5 x 10−13 W. The required 5 x 10−13 W could be obtained by increasing the Mars transmitter power by a factor of 6.3. Other alternatives would be (1) to reduce the bandwidth (and data rate) reducing the required value of Pr or (2) to employ a more sensitive receiver. 
7 
As discussed in Sec. 12-1, the noise power of a receiving system is a function of its system temperature T and bandwidth B as given by P = kTB, where k = Boltzmann’s constant = 1.38 x 10−23 JK−1. 
For B = 5 x 106 Hz (as given in this problem) and T = 50 K (an attainable value), 
2-11-4. continued 
The received power (8 x 10−14 W) is about 20 times this noise power, which is probably sufficient for satisfactory communication. Accordingly, with a 50 K receiving system temperature at the earth station, a Mars transmitter power of 1 kW is adequate. 
(b) The given Jupiter distance is 40/6 = 6.7 times that to Mars, which makes the required transmitter powers 6.72 = 45 times as much or the required receiver powers 1/45 as much. 
Neither appears feasible. But a practical solution would be to reduce the bandwidth for the Jupiter link by a factor of about 50, making B = (5/50) x 106 = 100 kHz.

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想知道高手的回答                   

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有的 话就跟大家分享一下,何必绕这么大的湾子呢

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谢谢了zhl_gs1980，在这里又碰到你了，原来网上可以找到英文原版答案的。 

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好东西，瞧瞧看

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最近正在看着本书，非常感谢楼主提供

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希望有用！

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