i*****r
发帖数: 318 | 1 10年没碰通信原理了，都忘光了。
帮忙看一下这道题目怎么解，很简单。多谢。
A receiver requires a minimum S/N ratio of 14 dB. The average noise power at
the receiver is 1 x 10-5 （10的5次方）Watts. The transmit power at the
transmitter is 1.5 Watts. Between the transmitter and the receiver there is
a 13 dB gain amplifier. The transmitter and receiver are connected by a
cable. This cable has a cable loss of 0.7 dB/km. Determine the maximum
possible cable distance between the transmitter and receiver. |
w*******e
发帖数: 16 | 2 noise power at receiver = 1e-5 watts = 10log10(1e-5) = -50dB
minimum signal power required at receiver = -50+14=-36dB
transmit power = 1.5 watts = 10log10(1.5) = 1.76dB
signal attenuation by cable is 0.7L dB
therefore, 1.76 + 13 - 0.7L = -36, L = 72.5(km) |
z*****n
发帖数: 7639 | 3 10log(S/N) = 14dB --->
N = 10e-5W = -50dBW---> S@receiver = -36dBW
Ptx = 1.5W = 1.76dBW with 13dB gain so
D*0.7dB/km = Ptx + 13 - S
D = (1.76 + 13 + 36)/0.7 = 71.76km
at
is
【在 i*****r 的大作中提到】
: 10年没碰通信原理了，都忘光了。
: 帮忙看一下这道题目怎么解，很简单。多谢。
: A receiver requires a minimum S/N ratio of 14 dB. The average noise power at
: the receiver is 1 x 10-5 （10的5次方）Watts. The transmit power at the
: transmitter is 1.5 Watts. Between the transmitter and the receiver there is
: a 13 dB gain amplifier. The transmitter and receiver are connected by a
: cable. This cable has a cable loss of 0.7 dB/km. Determine the maximum
: possible cable distance between the transmitter and receiver.
|
