Re: [理工] 計算機組織與系統

看板Grad-ProbAsk作者 (JS)時間16年前 (2009/04/17 16:50), 編輯推噓3(304)
留言7則, 5人參與, 最新討論串2/2 (看更多)
We want to compare the maximum bandwidth for a synchronous and asynchronous bus. The synchronous bus has a clock cycle time of 50ns,and each bus transmission takes 1 clock.The asynchronous bus requires 40 ns per handshake.The data portion of both buses is 32 bits wide.Find the bandwidth for each bus when performing one-word reads from a 220-ns memory. [同步] 1. 傳送位址給記憶體: 50ns 2. 記憶體存取: 220ns 3. 傳送資料: 50ns 傳送one-word 合計需要320ns one-word 4bytes 匯流排頻寬: ---------- = --------- = 12.5 MB/Sec 320ns 320ns [非同步] 交握協定具有7個step 1. step1 : 40ns 2. step234 : max(3*40ns,220ns) = 220ns 3. step5 : 40ns 4. step6 : 40ns 5. step7 : 40ns 傳送one-word 合計需要380ns one-word 4bytes 匯流排頻寬: ---------- = --------- = 10.52 MB/Sec 380ns 380ns -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 211.74.186.222

04/17 16:52, , 1F
原PO計組神 快拜
04/17 16:52, 1F

04/17 17:08, , 2F
推神手~幫我看一下18068^^
04/17 17:08, 2F

04/17 21:24, , 3F
step234 : max(3*40ns,220ns) = 220ns ,有寫錯嗎?
04/17 21:24, 3F

04/17 21:25, , 4F
3*40ns=?
04/17 21:25, 4F

04/17 22:03, , 5F
感謝 又懂一題了
04/17 22:03, 5F

04/17 22:44, , 6F
Step2+Step3+Step4 跟記憶體存取會並行,要取比較久的時間
04/17 22:44, 6F

04/17 22:45, , 7F
可以參考交握協定的時序圖
04/17 22:45, 7F
文章代碼(AID): #19w4BCwk (Grad-ProbAsk)
文章代碼(AID): #19w4BCwk (Grad-ProbAsk)