|
发表于 2006-3-25 22:57:00
|
显示全部楼层
设发射功率为<FONT face="宋体, MS Song">PT</FONT>,发射天线增益为<FONT face="宋体, MS Song">GT</FONT>,工作频率为<FONT face="宋体, MS Song">f . </FONT>接收功率为<FONT face="宋体, MS Song">PR</FONT>,接收天线增益为<FONT face="宋体, MS Song">GR</FONT>,收、发天线间距离为<FONT face="宋体, MS Song">R</FONT>,那么电波在无环境干扰时,传播途中的电波损耗<FONT face="宋体, MS Song"> L0 </FONT>有以下表达式:<FONT face="宋体, MS Song">
L0 (dB) = 10 Lg</FONT>(<FONT face="宋体, MS Song"> PT / PR </FONT>)<FONT face="宋体, MS Song">
= 32.45 + 20 Lg f ( MHz ) + 20 Lg R ( km ) - GT (dB) - GR (dB)
[ </FONT>举例<FONT face="宋体, MS Song">] </FONT>设:<FONT face="宋体, MS Song">PT = 10 W = 40dBmw </FONT>;<FONT face="宋体, MS Song">GR = GT = 7 (dBi) </FONT>;<FONT face="宋体, MS Song"> f = 1910MHz
</FONT>问:<FONT face="宋体, MS Song">R = <st1:chmetcnv UnitName="m" SourceValue="500" HasSpace="True" Negative="False" NumberType="1" TCSC="0">500 m</st1:chmetcnv> </FONT>时,<FONT face="宋体, MS Song"> PR = </FONT>?<FONT face="宋体, MS Song">
</FONT>解答:<FONT face="宋体, MS Song"> (1) L0 (dB) </FONT>的计算<FONT face="宋体, MS Song">
L0 (dB) = 32.45 + 20 Lg 1910( MHz ) + 20 Lg 0.5 ( km ) - GR (dB) - GT (dB)
= 32.45 + 65.<st1:chsdate Year="1962" Month="6" Day="7" IsLunarDate="False" IsROCDate="False">62 - 6 - 7</st1:chsdate></FONT><FONT face="宋体, MS Song"> - 7 = 78.07 (dB)
</FONT>(<FONT face="宋体, MS Song">2</FONT>)<FONT face="宋体, MS Song">PR </FONT>的计算<FONT face="宋体, MS Song">
PR = PT / ( 10 7.807 ) = 10 ( W ) / ( 10 7.807 ) = 1 ( μW ) / ( 10 0.807 )
= 1 ( μW ) / 6.412 = 0.156 ( μW ) = 156 ( mμW )
</FONT>顺便指出,<FONT face="宋体, MS Song">1.9GHz</FONT>电波在穿透一层砖墙时,大约损失<FONT face="宋体, MS Song"> (10~15) dB </FONT><BR line-break"><BR line-break">[br]<p align=right><font color=red>+3 RD币</font></p> |
|