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

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Presentation transcript:

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

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)

XML DTD Transparency No. 3 Full-Adder

XML DTD Transparency No. 4 The Binary Adder

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

XML DTD Transparency No. 6 The Binary Multiplication

XML DTD Transparency No. 7 The Array Multiplier

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

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.

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

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

XML DTD Transparency No. 12 Full substrator : implementations

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

XML DTD Transparency No. 14 Divider How division is computed ? Ex: / = ? | | |zzzzz remainder quotient

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);

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

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

XML DTD Transparency No. 18 References: chapter 11.