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Thiết kế logic số (VLSI design)
Bộ môn KT Xung, số, VXL
quangkien82@gmail.com
https://sites.google.com/site/bmvixuly/thiet-ke-
logic-so
Mục đích, nội dung
• Nội dung: Khối chia số nguyên có dấu và không dấu. Phương pháp tiết kiệm tài nguyên thiết kế bằng cấu trúc lặp cứng
• Thời lượng: 3 tiết bài giảng
• Yêu cầu: Sinh viên có sự chuẩn bị sơ bộ trước nội dụng bài học.
quangkien82@gmail.com 2/11
Restoring division
------------------------------
z 1 0 0 0 0 1 0 1 2^d 1 1 1 0 s(0) 0 |0 0 1 0 0 0 0 1 0 1 2s(0) 0 |0 1 0 0 0|0 1 0 1
------------------------------s(4) (0)|1 1 1 0 0 1 q1 = 0 2s(4) 1 |1 0 1 0 1 restore +2^4d 0 |1 0 0 1 0
-2^4d 1 |1 0 0 1 0| ------------------------------
------------------------------s(1) (0)|1 1 0 1 0|0 1 0 1 2s(1) 0 |1 0 0 0 0|1 0 1 restore
-2^4d 1 |1 0 0 1 0 q4 = 0
S(5) = (1)|0 0 1 s = 2s(5) = 0 1 1
d = 1 1= 1 0 = -d = 1 0 0 1 z = 1 0 0 0 0
1 1 q0 = 1 1 = 7
041 = 9 0
1 0 1 = 133
------------------------------s(2) (1)|0 0 0 1 0 1 0 1 2s(2) 0 |0 0 1 0 1 0 1
q = 0 1 0 0 1 = 9 S = 0 1 1 1 = 7
-2^4d 1 |1 0 0 1 0 q3 = 1
------------------------------s(3) (0)|1 0 1 1 1 0 1
2s(3) 0 |0 1 0 1 0 1 restore
quangkien82@gmail.com 0 q2 = 0 3/11
Nonrestoring division principle
------------------------------ ------------------------------z 1 0 0 0 0 1 0 1
2^d 1 1 1 0 s(0) 0 |0 0 1 0 0 0 0 1 0 1
2s(0) 0 |0 1 0 0 0|0 1 0 1 = u -2^4d 1 |1 0 0 1 0| = -d
------------------------------s(1) (0)|1 1 0 1 0|0 1 0 1 2s(1) 0 |1 0 0 0 0|1 0 1 restor -2^4d 0 |1 0 0 1 0 q4 = 0
------------------------------s(2) (1)|0 0 0 1 0 1 0 1
------------------------------u –d
= 2*(u-d) (u-d >0) | 2u (u-d <0) = -d |
----------------------------
2*(u-d)–d (u-d >0) | 2u–d(u-d <0)
2s(2) 0 |0 0 1 0 1 0 1 2*(u-d) + d = 2*u -d -2^4d 1 |1 0 0 1 0 q3 = 1
------------------------------s(3) (0)|1 0 1 1 1 0 1
2s(3) 0 |0 1 0 1 0 1 restore
quangkien82@gmail.com 0 q2 = 0 4/11
Restoring division VS NonRestoring division
------------------------------ ------------------------------
z 1 0 0 0 0 1 0 1 2^d 1 1 1 0 s(0) 0 |0 0 1 0 0 0 0 1 0 1
z 1 0 0 0 0 1 0 1 2^d 1 1 1 0 s(0) 0 |0 0 1 0 0 0 0 1 0 1
2s(0) 0 |0 1 0 0 0|0 1 0 1 -2^4d 1 |1 0 0 1 0|
2s(0) 0 |0 1 0 0 0|0 1 0 1 -2^4d 1 |1 0 0 1 0|
------------------------------ ------------------------------s(1) (0)|1 1 0 1 0|0 1 0 1 s(1) (0)|1 1 0 1 0|0 1 0 1 2s(1) 0 |1 0 0 0 0|1 0 1 restore 2s(1) 0 |1 0 0 0 0|1 0 1
-2^4d 0 |1 0 0 1 0 q4 = 0 +2^4d 0 |0 1 1 1 0 q4 = 0 ------------------------------
s(2) (1)|0 0 0 1 0 1 0 1 ------------------------------2s(2) 0 |0 0 1 0 1 0 1 s(2) (1)|0 0 0 1 0 1 0 1
-2^4d 1 |1 0 0 1 0 q3 = 1 2s(2) -2^4d
0 |0 0 1 0 1 0 1
1 |1 0 0 1 0 q3 = 1
------------------------------
s(3) (0)|1 0 1 1 1 0 1 ------------------------------2s(3) 0 |0 1 0 1 0 1 restore s(3) 0)|1 0 1 1 1 0 1
+2^4d 0 |1 0 0 1 0 q2 = 0 2s(3) 1 |0 1 1 1 0 1
------------------------------….
+2^4d 0 |0 1 1 1 0 q2 = 0 5/11
------------------------------
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