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Binary Addition Binary Multiplication. Section 4.5 and 4.7 . Topics. Calculations Examples Signed Binary Number Unsigned Binary Number Hardware Implementation Overflow Condition Multiplication. Unsigned Number. (2-bit example). Unsigned Addition. 1+2=. Unsigned Addition. 1+3=.

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Binary addition binary multiplication

Binary AdditionBinary Multiplication

Section 4.5 and 4.7


Topics
Topics

  • Calculations Examples

    • Signed Binary Number

    • Unsigned Binary Number

  • Hardware Implementation

  • Overflow Condition

  • Multiplication


Unsigned number
Unsigned Number

(2-bit example)



Unsigned addition1
Unsigned Addition

  • 1+3=

(Indicates Overflow)

(Carry Out)


Unsigned subtraction 1
Unsigned Subtraction (1)

  • 1-2=

(1’s complement)

(2’s complement)



Summary for unsigned addition subtraction
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
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
Signed Addition (70+80)

(Indicates a negative number)

70=21+22+26=2+4+64

80=24+26=16+64

10010110→01101001 →01101010

21+23+25+26=2+8+32+64=106

10010110↔-106

010010110

010010110↔ 21+22+24+27=2+4+16+128=150

Conclusion: There is a problem of overflow

Fix: Use the end carry as the sign bit, and let b7 be

the extra bit.


Signed subtraction 70 80
Signed Subtraction (70-80)

(Indicates a negative number)

70=21+22+26=2+4+64

80=24+26=16+64

11110110→00001001 →00001010

21+23=10

11110110↔-10

(No Problem)


Signed subtraction 70 801
Signed Subtraction (-70-80)

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

70=21+22+26=2+4+64

80=24+26=16+64

101101010 →010010101 → 010010110

010010110 ↔21+22+24+27=2+4+16+128=150

101101010 ↔-150

Conclusion: There is a problem of overflow

Fix: Use the end carry as the sign bit, and let b7 be

the extra bit.


Observations
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
Four-Bit Adder-Subtractor


The mode input 1
The Mode Input (1)

If M=0, =

If M=1, =

B0


The mode input 2
The Mode Input (2)

If M=0,

If M=1,


M=0

B3

B2

B0

B1

0


M=1

1

2’s complement is generated of B is generated!


Unsigned addition2
Unsigned Addition

When two unsigned numbers are added,

an overflow is detected from the end carry.


Detect overflow in signed addition
Detect Overflow in Signed Addition

Observe

The cary into the sign bit

The carry out of the sign bit

If they are not equal,

they indicate an overflow.


Two bit binary multiplier
Two-Bit Binary Multiplier

(multiplicand)

(multiplier)





Four bit by three bit binary multiplier
Four-bit by three-bit Binary Multiplier

S10=A0B1+A1B0

S11=A0B2+A1B1+C1

S12=A0B3+A1B2+C2

S13=0+A1B3+C3

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



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