eem16 lecture 7 ucla

8/14/2019 EEM16 Lecture 7 UCLA

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EEM16/CSM51A:Logic Design of Digital Systems

Lecture #7Arithmetic Circuits

Prof. Danijela Cabric

Fall 2013


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2(c) 2005 - 2012 W. J. Dally

Numbers

Digital logic works with binary numbers

Each bit has a value based on its position


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3(c) 2005 - 2012 W. J. Dally

From Decimal To Binary

Example: 17 10 Convert number to binary base:

17 2 = 8 remainder 1

8 2 = 4 remainder 0

4 2 = 2 remainder 0

2 2 = 1 remainder 0

1 2 = 0 remainder 1

17 10 = 10001 2

MSB

LSB


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From Decimal To Binary To Hexadecimal

Example: 1963 10

1963 10 = 0111 1010 1011 2

7 A B hex


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Binary addition

1-bit 0+0 = 0, 1+0 = 1, 1+1 = 10

1

6 110+ 3 011

5 101+7 111

0 _ 10 _

01001

110 _

10019 0

1 _

00

11 _

100

111 _

110012

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6(c) 2005 - 2012 W. J. Dally

Half Adder

HA

a

b

c

s

b

ac

s


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One bit of an adder

Counts the number of 1 bits on its input Outputs the result in binary For a half-adder, 2 inputs, output can be 0, 1, or 2 For a full-adder, 3 inputs, output can be 0, 1, 2, or 3

c[i] b[i] a[i] count c[i+1] s[i]0 0 0 0 0 00 0 1 1 0 10 1 0 1 0 10 1 1 2 1 01 0 0 1 0 11 0 1 2 1 01 1 0 2 1 01 1 1 3 1 1


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Full adder from a truth table

cin

Majority

a

b

s

cout

c[i] b[i] a[i] count c[i+1] s[i]0 0 0 0 0 00 0 1 1 0 1

0 1 0 1 0 10 1 1 2 1 01 0 0 1 0 11 0 1 2 1 01 1 0 2 1 01 1 1 3 1 1


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Majority circuit

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Multi-bit Adder

FA

a[0]

b[0]

a

bcin

cout

s s[0]

cin

FA

a[1]

b[1]

a

bcin

cout

s s[1]

c [ 1 ]

FA

a[n-1]

b[n-1]

a

bcin

cout

s s[n-1]

cout


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Negative Integers

Thus far we have only addressed positive integers. What aboutnegative numbers?

3 ways to represent negative integers in binary: Sign Magnitude Ones complement Twos complement

Example: Consider +23 23

Sign Magnitude 0 10111 1 10111 Ones complement 0 10111 1 01000 Twos complement 0 10111 1 01001


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Twos Complement System


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Twos complement for n=4

0000

0111

0011

1011

1111

1110

1101

1100

1010

1001

1000

0110

0101

0100

0010

0001

+0

+1

+2

+3

+4

+5

+6

+7-8

-7

-6

-5

-4

-3

-2

-1

0 10 0 = + 4

1 10 0 = - 4

+

-


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Mapping in Twos complement system


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Converse Mapping


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Example of Converse Mapping


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Change of Sign in 2s Complement System


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Simpler way


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Change of Sign Procedure


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2s complement makes subtraction easy Represent negative number, x as 2 n x

All arithmetic is done modulo 2 n so no adjustments are necessaryx + ( y) = x + (2 n y) (mod 2 n)consider 4-bit numbers

4 3 = 4 + (16 3) (mod 16)

= 4 + (15 3 + 1)= 0100 + (1111 0011) + 0001= 0100 + 1100 + 0001= 0001

Why do we all use 2s complement?


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Overflow Detection in Addition


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Adder

a

nn

b n

sub

a

bcin

cout

s out

n

ovf

Adder a[n-2:0]

n-1b[n-2:0]n-1

sub

a

bcin

cout

s out[n-2:0]

Adder a

bcin

cout

s

n-1n-1

out[n-1]a[n-1]b[n-1]

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Compare two numbers with subtraction

diff = a b

if diff is negative (sign bit = 1), a a a=b

Compare

a

b

lt

eqn

n


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Multiplication

Shifting left multiplies by 2e.g., 5 = 101, 10 = 1010, 20 = 10100

To multiply by 3, compute 3x = x + 2xExample, 7 x 5

1 1 1

x 1 0 11 1 10 0 0

1 1 1

1 0 0 0 1 1

Multiplication


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Multipliers

25


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Multiplication Bit Matrix

26


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a 3 a 2 a 1

b0

b1

b2

b3

a 0

First generate partial products: p ij = a i b j has weight i+j

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a 3 a 2 a 1

b0

b1

b2

b3

a 0

FA 0

p 2

p 1

p 0

FA

FA

FA

FA

FA

p 3

FA

FA

p 4

FAFA

FA

p 5

p 6

FA

0

0

0

p 7

Then sum partial products to get sum

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Summary

Binary number representation Add numbers a bit at a time 2s complement x = (2 n x) = (2 n 1) x + 1 = neg(x) + 1 Subtract by 2s complement and add Multiply form partial products p ij and sum

eem16 lecture 7 ucla

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