→ maxwellyann:上面PO文有誤，因為用word編輯的，問題如下↓ 02/15 13:11
→ maxwellyann:5*BER≦5*10^-5.左式的5是由那一串英文敘述出來的嗎? 02/15 13:12
推 wolf0000:發射端+四個repeater，總共會傳送五次故有五次BER 02/15 13:25
→ maxwellyann:我了解了，感恩你唷！ 給你一個推！ 02/15 13:31
煩請各位大大撥冗來幫我看一下這個題目吧！
題目如下：
A digital communication system consists of a transmission line with 4
regenerative
repeaters(excluding the last receiver). The communication environment and
design
of all receivers are identical. The channel has an ideal frequency response
over
320MHz≦f≦325MHz. The modulation scheme is OQPSK with coherent detection and
the channel noise is AWGN with No=10^-10W/Hz.
(b)If the bit-error-rate of the whole system≦5*10^-5 is required, the
minimum received
Eb/No at each receiver is .
(Note: It is required Eb/No=12.6dB for BFSK signal with coherent detection
and Pb=10^-5. )
解答為：
5*BER≦5*10^-5
BER≦10^-5
Eb/No=9.6dB
我的疑問在於，5*BER≦5*10^-5，畫底線的5是從哪裡來的？是從題目敘述中的
transmission line with 4 regenerative repeaters(excluding the last receiver)得
知的嗎？請各位大大幫我解答，感恩！
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