Binary AdditionBinary Multiplication

Section 4.5 and 4.7

Topics

• Calculations Examples–Signed Binary Number–Unsigned Binary Number

• Hardware Implementation• Overflow Condition• Multiplication

Unsigned Number

Decimal b1 b0

0 0 0

1 0 1

2 1 0

3 1 1

(2-bit example)

Unsigned Addition

• 1+2=

Decimal b1 b0

0 0 0

1 0 1

2 1 0

3 1 1

Decimal b1

b0

1 0 1

+ 2 1 0

3 1 1

Unsigned Addition

• 1+3=

Decimal b1 b0

0 0 0

1 0 1

2 1 0

3 1 1

Decimal b1

b0

1 1

1 0 1

+ 3 1 1

4 1 0 0

(Carry Out)

(Indicates Overflow)

Unsigned Subtraction (1)

• 1-2=

Decimal b1 b0

0 0 0

1 0 1

2 1 0

3 1 1

Decimal b1

b0

1 0 1

+ -2 1 0

1 1

0 0

-1 0 1

(1’s complement)

(2’s complement)

Unsigned Subtraction (2)

• 2-1=

Decimal b1 b0

0 0 0

1 0 1

2 1 0

3 1 1

Decimal b1 b0

1

2 1 0

+ -1 1 1

3 1 0 1

Summary for Unsigned Addition/Subtraction

• Overflow can be an issue in unsigned addition

• Unsigned Subtraction (M-N)– If M≥N, and end carry will be

produced. The end carry is discarded.– If M<N, • Take the 2’s complement of the sum• Place a negative sign in front

Signed Binary Numbers

• 4-bit binary number– 1 bit is used as a signed bit – -8 to +7– 2’s complement

Signed Addition (70+80)b7 b6 b5 b4 b3 b2 b1 b0

0 1

70 0 1 0 0 0 1 1 0

80 0 1 0 1 0 0 0 0

1 0 0 1 0 1 1 0

70=21+22+26=2+4+6480=24+26=16+64

10010110→01101001 →0110101021+23+25+26=2+8+32+64=106

10010110↔-106

(Indicates a negative number)

010010110

010010110↔ 21+22+24+27=2+4+16+128=150

Conclusion: There is a problem of overflowFix: Use the end carry as the sign bit, and let b7 bethe extra bit.

Signed Subtraction (70-80)b7 b6 b5 b4 b3 b2 b1 b0

70 0 1 0 0 0 1 1 0

-80 1 0 1 1 0 0 0 0

1 1 1 1 0 1 1 0

70=21+22+26=2+4+6480=24+26=16+64

11110110→00001001 →0000101021+23=10

11110110↔-10

(Indicates a negative number)

(No Problem)

Signed Subtraction (-70-80)b7 b6 b5 b4 b3 b2 b1 b0

1 0 1 1

-70 1 0 1 1 1 0 1 0

-80 1 0 1 1 0 0 0 0

0 1 1 0 1 0 1 0

70=21+22+26=2+4+6480=24+26=16+64

(Indicates a positive number! A negative number expected.)

101101010 →010010101 → 010010110

010010110 ↔21+22+24+27=2+4+16+128=150

101101010 ↔-150

Conclusion: There is a problem of overflowFix: Use the end carry as the sign bit, and let b7 bethe extra bit.

Observations• Given the similarity between addition

and subtraction, same hardware can be used.

• Overflow is an issue that needs to be addressed in the hardware implementation

• A signed number is not processed any different from an unsigned number. The programmer must interpret the results of addition and subtraction appropriately.

Four-Bit Adder-Subtractor

The Mode Input (1)

B0

If M=0, = If M=1, =

The Mode Input (2)

If M=0, If M=1,

M=0

0

B3 B2 B1 B0

M=1

1

2’s complement is generated of B is generated!

Unsigned Addition

When two unsigned numbers are added, an overflow is detected from the end carry.

Detect Overflow in Signed Addition

Observe1. The cary into the sign bit2. The carry out of the sign bit

If they are not equal, they indicate an overflow.

Two-Bit Binary Multiplier

(multiplicand)

(multiplier)

𝐴 0𝐵00 0 0

0 1 0

1 0 0

1 1 1

Use an AND gate to multiply A0 and B0

Hardware Correlation

A Four-Bit Adder

Four-bit by three-bit Binary Multiplier

B3 B2 B1 B0

A2 A1 A0

0 A0B3 A0B2 A0B1 A0B0

A1B3 A1B2 A1B1 A1B0

C14 S13 S12 S11 S10

A2B3 A2B2 A2B1 A2B0

C6 C5 C4 C3 C2 C1 C0

S10=A0B1+A1B0S11=A0B2+A1B1+C1S12=A0B3+A1B2+C2S13=0+A1B3+C3

(S1X, where 1 is the first 4-bit adder)

Four-bit by three-bit Binary Multiplier