作者BF3716 (JS)
看板Grad-ProbAsk
標題Re: [理工] 計算機組織與系統
時間Fri Apr 17 16:50:19 2009
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
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1F：推 sak1346:原PO計組神 快拜 04/17 16:52
2F：推 newfantasy:推神手~幫我看一下18068^^ 04/17 17:08
3F：推 beautyanzi:step234 : max(3*40ns,220ns) = 220ns ，有寫錯嗎? 04/17 21:24
4F：→ beautyanzi:3*40ns=? 04/17 21:25
5F：→ Westind:感謝 又懂一題了 04/17 22:03
6F：→ BF3716:Step2+Step3+Step4 跟記憶體存取會並行，要取比較久的時間 04/17 22:44
7F：→ BF3716:可以參考交握協定的時序圖 04/17 22:45