[理工] [通訊系統] 98成大通訊 Bit error rate小問題

看板Grad-ProbAsk作者 (不彥其煩)時間13年前 (2011/02/15 13:09), 編輯推噓1(103)
留言4則, 2人參與, 最新討論串1/1
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! -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.248.14.224

02/15 13:11, , 1F
上面PO文有誤,因為用word編輯的,問題如下↓
02/15 13:11, 1F

02/15 13:12, , 2F
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, 2F

02/15 13:25, , 3F
發射端+四個repeater,總共會傳送五次故有五次BER
02/15 13:25, 3F

02/15 13:31, , 4F
我了解了,感恩你唷! 給你一個推!
02/15 13:31, 4F
文章代碼(AID): #1DMWgE33 (Grad-ProbAsk)