1. XMLDTD Transparency No. 1 Java Programming assignment 1 Cheng-Chia Chen

2. XML DTD Transparency No. 2 ToDo works 1. Implement FA1 ( 1-bit full adder) 2. Implement FA8 ( 8-bit full adder) 3. Implement FS1 (1-bit full subtractor ) 4. Implement FS8 (b-bit full subtractor ) 5. Implement Multiplier8 ( 8-bit x 8-bit multiplication) 6. Implement Divider8 (8-bit by 8-bit division + remainder)

3. XML DTD Transparency No. 3 Full-Adder

4. XML DTD Transparency No. 4 The Binary Adder

5. XML DTD Transparency No. 5 The n-bit Ripple-Carry Adder : FAn

6. XML DTD Transparency No. 6 The Binary Multiplication

7. XML DTD Transparency No. 7 The Array Multiplier

8. XML DTD Transparency No. 8 Multiplier4 FA4 DeMUX4 0000 A 0..A 3 B0B0 DeMUX4 0000 A 0..A 3 B1B1 DeMUX4 0000 A 0..A 3 B2B2 DeMUX4 0000 A 0..A 3 B3B3 FA4 0 0 0 0 m7m7 m5m5 m6m6 m0m0 m4m4 m3m3 m2m2 m1m1

9. XML DTD Transparency No. 9 Subtracter Subtracter circuits take two binary numbers as input and subtract one binary number input with other binary number input. Similar to to adders it gives out two output, difference and borrow (Carry in the case of Adder). There are two types of subtracters. Half Subtracter. Full Subtracter.

10. XML DTD Transparency No. 10 Half Substractor IO relations: D = X (~Y) \/ (~X) Y Bout = (~X) Y

11. XML DTD Transparency No. 11 Full Subtractor Symbol: D = (X xor Y) xor Bin Bout = X'.Y + X'.Bin + Y.Bin

12. XML DTD Transparency No. 12 Full substrator : implementations

13. XML DTD Transparency No. 13 Parallel binary n-bit substrator

14. XML DTD Transparency No. 14 Divider How division is computed ? Ex: 101101 / 000110 = ? | 000111 |-------------- 000110|zzzzz101101 -000110-- 0001010- -000110 0001001 -000110 000011 remainder quotient

15. XML DTD Transparency No. 15 The rules for divider4 A 0..A 3 / B 1..B 3 = Q 0..Q 3 -- R 0.. R 3 1. T 0 = zzzA 0 2. if (T 0 Q 0 = 0; T 1 = T 0 A 1 ; else => Q 0 = 1; T 1 = (T 0 -B) A 1 ; 3. if (T 1 Q 1 = 0; T 2 = T 1 A 2 ; else => Q 1 = 1; T 2 = (T 1 -B) A 2 ; … if T k Q k = 0; T k+1 = T k A k+1 ; else Q k = 1; T k+1 = (T k -B) A k+1 ; if T 3 Q 3 = 0; R 4 = T 3 ; else Q 3 = 1; R 4 = (T 3 -B);

16. XML DTD Transparency No. 16 The building block M4 for divider4 T k – BDeMUX T k - S < 0 0000 B S TkTk B TkTk D0D0 QkQk T k+1 TkTk B A k+1

17. XML DTD Transparency No. 17 Divider4 M4 B[0:3] 000A 0 A1A1 M4 B[0:3] M4 B[0:3] A2A2 M4 B[0:3] A3A3 Q0Q0 Q1Q1 Q2Q2 Q3Q3 X R 0..R 3

18. XML DTD Transparency No. 18 References: http://www.asic-world.com/digital/arithmetic.html http://bwrc.eecs.berkeley.edu/IcBook/slides.htm chapter 11.