Thi t k logic s ế ế ố
(VLSI design)
Bộ môn KT Xung, số, VXL
quangkien82@gmail.com
https://sites.google.com/site/bmvixuly/thiet-ke-
logic-so
•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.
2/11
M c đích, n i dungụ ộ
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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
-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(2) (1)|0 0 0 1 0 1 0 1
2s(2) 0 |0 0 1 0 1 0 1
-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
+2^4d 0 |1 0 0 1 0 q2 = 0
------------------------------
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
------------------------------
S(5) = (1)|0 0 1 1 1 q0 = 1
s = 2s(5) = 0 1 1 1 = 7
q = 0 1 0 0 1 = 9
d = 1 1 1 0 = 14
-d = 1 0 0 1 0
z = 1 0 0 0 0 1 0 1 = 133
q = 0 1 0 0 1 = 9
S = 0 1 1 1 = 7
3/11
quangkien82@gmail.com
Non-restoring 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
-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 restor
-2^4d 0 |1 0 0 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
-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
-2^4d 0 |1 0 0 1 0 q2 = 0
------------------------------
= u
= -d
------------------------------
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)
2*(u-d) + d = 2*u -d
4/11
quangkien82@gmail.com
Restoring division VS Non-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
-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 0 |1 0 0 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
-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
+2^4d 0 |1 0 0 1 0 q2 = 0
------------------------------
….
------------------------------
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|
------------------------------
s(1) (0)|1 1 0 1 0|0 1 0 1
2s(1) 0 |1 0 0 0 0|1 0 1
+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
-2^4d 1 |1 0 0 1 0 q3 = 1
------------------------------
s(3) 0)|1 0 1 1 1 0 1
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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