欢迎访问ic37.com |
会员登录 免费注册
发布采购

PM73121-RI 参数 Datasheet PDF下载

PM73121-RI图片预览
型号: PM73121-RI
PDF下载: 下载PDF文件 查看货源
内容描述: AAL1分段重组处理器 [AAL1 Segmentation And Reassembly Processor]
分类和应用: ATM集成电路SONET集成电路SDH集成电路电信集成电路电信电路异步传输模式
文件页数/大小: 223 页 / 2148 K
品牌: PMC [ PMC-SIERRA, INC ]
 浏览型号PM73121-RI的Datasheet PDF文件第48页浏览型号PM73121-RI的Datasheet PDF文件第49页浏览型号PM73121-RI的Datasheet PDF文件第50页浏览型号PM73121-RI的Datasheet PDF文件第51页浏览型号PM73121-RI的Datasheet PDF文件第53页浏览型号PM73121-RI的Datasheet PDF文件第54页浏览型号PM73121-RI的Datasheet PDF文件第55页浏览型号PM73121-RI的Datasheet PDF文件第56页  
PM73121AAL1gator II  
Data Sheet  
PMC-Sierra, Inc.  
PMC-980620  
,VVXHꢀꢁ  
AAL1 SAR Processor  
Figure 23 shows how the CSD assigns credits to determine in which frames cells should be sent.  
Frame Boundaries  
TFTC  
CSD  
RL_SER(0)  
FR_ADVANCE_FIFO  
The CSD reads  
Set  
frame advances  
and determines  
cells to be sent  
RL_FSYNC(0)  
The TFTC sees  
frame advance and  
records this in the  
FR_ADVANCE_FIFO  
NEXT_  
SERV  
RL_SER(1)  
RL_FSYNC(1)  
Figure 23. Frame Advance FIFO Operation  
The following is an example of the calculations the CSD circuit performs. This example assumes  
a structured line with four channels allocated to one queue.  
1. The TFTC writes Line 3 and Frame 4 to the FR_ADVANCE_FIFO.  
2. The CSD circuit determines the queue for which a cell is ready by finding a set bit in the  
Transmit Calendar. In this example, it is queue number 100.  
3. The CSD circuit reads the number of credits for queue number 100. The number of credits is  
always greater than 47 because it is ready for service. In this example, QUE_CREDITS =  
59.375.  
4. The CSD circuit subtracts AVG_SUB_VALU, the average number of credits spent per cell.  
(Remember: For structured lines, the average number of credits per cell is 46-7/8. For unstruc-  
tured lines, the average number of credits per cell is 47.)  
Credits = 59.375 - 46.875  
Credits = 12.5  
5. The frame differential for the next service is computed from the number of credits needed to  
exceed 47 and NUM_CHAN, the number of channels allocated per frame.  
47 - 12.5 = 34.5  
34.5 ÷ 4 = 8.625  
Round 8.625 up, so the frame differential is 9.  
6. Therefore, the next cell will be sent nine frames ahead of the current cell.  
Next frame = present frame number + 9  
ꢂꢄ  
 
 复制成功!