If you measure a flux density of 1.591×10^-5 W/m^2, and you are measuring that from 10,000 m from the transmitter, how much power (watts) is the transmitter outputting?
(assuming no gain)
如果在距离辐射信号源10000米的地方测量到的功率通量密度是
1.591 ×10^-5 W/m^2，假设辐射信号源没有增益，那辐射信号源的输出功率是多大呢？
The power flux density can be calculated from the following formula:
功率通量密度 的计算公式如下：
Where S = flux power density in W/m^2; Pt = power transmitted in W; R = distance to transmitter in m.
- 功率通量密度S 的单位是：W/m^2
- 辐射信号源的输出功率Pt的单位是：W
- 测量点与辐射信号源之间距离的单位是：m
rearranging terms:
公式可变换为：
plugging in our values:
带入相关数字：
The result is 20,000 Watts.
结果为 20,000Watts.
for a quick tool that can do this as well as compensate for gain and calculate electric and magnetic field strength see:
罗德施瓦茨公司提供了一个方便计算功率通量密度以及电磁场强度的软件，链接如下:
该软件的使用说明和相关计算的理论可参考以下文档：
以上计算结果使用该软件计算示例如下：
