yuh-jzer joungdigital systems1 number systems decimal number : 7397=7×10 3 +3×10 2 +9×10 1 +7×10...

Yuh-Jzer Joung Digital Systems 1

Number SystemsNumber Systems

decimal number :7397=7×103+3×102+9×101+7×100

a4a

3a

2a

1a

0 . a

-1a

-2

= a4×104 +a

3×103+a

2×102+a

1×101+a

0×100+a

-1x10-1+a

-2×10-2

ai : coefficient

10 : base or radix

a4 : most significant bit ( msb )

a-2

: least significant bit ( lsb )


In general, number N in a base-r system is

represented as N = (ana

n-1......a

1a

0 .a

-1a

-2......a

-m)

which has the value

N = (an.rn + a

n-1.rn-1 +......+a

1.r +a

0

+a-1.r-1 + a

-2 r-2 +......+a

-m.r-m)

In binary system﹐r = 2﹐ai {0, 1}.

In octal system﹐r = 8﹐ai {0,1,…,7}.

In decimal system﹐r = 10﹐ai {0,1,…,9}.

In hexadecimal system﹐r = 16﹐ai {0,1,…9,A,B,C,D,E,F}.


Conversion between different Bases• Base-r to base-10 conversion﹕

Straightforward﹗

• Base-r to base-s conversion﹕ By repeated division for integers and repeated multiplication for fractions.

Ex1. Convert (153.513)10 to octal number.

Integer﹕153 (153)10 = (231)8

19 1

2 3

0 2


fractional:

0.513×8 = 4.104

0.104×8 = 0.832

0.832×8 = 6.656 (0.513)10 = (0.4065…)8

0.656×8 = 5.248

(153.513)10 = (231.4065…)8

Ex2. Convert (12A)16 to octal.

1. Convert (12A)16 to (298)10

2. Convert (298)16 to octal as in Ex1.


Conversions between binary octal and hexadecimal can be do﹐ ﹐ne in a simpler way.

Ex1. Convert (10111010011)2 to octal and hexadecimal.

10 111 010 011 (2723)8

2 7 2 3

101 1101 0011 (5D3)16

5 D 3

Ex2. Convert (12A7F)16 to binary and octal.

1 2 A 7 F

0001 0010 1010 0111 1111 binary 00 010 010 101 001 111 111

0 2 2 5 1 7 7 octal


Complement for simplifying subtraction﹕ .

Two types of complements for each base-r system﹕• (r-1)’s complement

• r’s complement

The (r-1)’s complement of an n-digital number N﹕(rn 1) NEx. In decimal system the 9’s complement of 546700 is ﹐ (106 1) 546700 = 999999 546700 = 453299

The 9’s complement of 012398 is

999999 012398 = 987601

The 1’s complement of 01011000 is

(28 1) 01011000

= 11111111 01011000

= 10100111

The 1’s complement of 0101101 is 1010010.


The r’s complement of an n-digital number N﹕ rnN N ≠0

0 N = 0

Ex. The 10’s complement of 012398 is 987602. The 10’s complement of 2467000 is 7533000.



Comments﹕1.To compute the complement of a number having radix point, fi

rst remove the radix point compute the complement of the ﹐ ﹐new number and restore the radix point.﹐

Ex. The 1’s complement of 01101.101 is 10010.010

The 2’s complement of 01101.101 is 10010.011


2.The complement of the complement of N is N.

(r1)’s complement of (2n 1) N = (2n 1) ((2n 1) N)

= N

Signed Binary NumbersAddition﹕

9 1001

+ 13 + 1101

22 10110

Subtraction﹕ 9 1001

13 1101

4 ?


How do we represent signed numbers in digital systems?representation of +9 with 8 bits: 00001001

representation of 9:

signed magnitude: 10001001

signed -1’s-complement: 11110110

signed -2’s-complement: 11110111

Numbers that can be represented in n bits:

signed magnitude: (2 n1+1) ~ (2 n 1 1)

signed-1’s-complement: ( 2 n 1 +1) ~ (2 n 1 1)

signed-2’s-complement: ( 2 n1) ~ (2 n1 1)


Addition and Subtraction on signed-magnitude numbers + 9 00001001 + 13 + 00001101 + 22 00010110 9 10001001 00001101 + 13 + 00001101 10001001 + 4 00000100

+ 9 00001001 10001101 13 10001101 00001001

4 10000100 9 10001001

13 10001101

22 10010110

Summary: to perform arithmetic operations on signed magnitude numbers, we need to compare the signs and the magnitudes of the two numbers, and then perform either addition or subtraction.


Addition and subtraction on signed-2’s-complement numbers:Addition: add the corresponding digits ( including the sign digits) and

ignore any carry out.

+ 9 00001001 9 11110111

+ 13 + 00001101 + 13 + 00001101

+ 22 00010110 + 4 100000100

discarded carryout


+ 9 00001001

13 + 11110011 4 11111100 the 2’s complement of 00000100 (4)

9 11110111 13 11110011 22 111101010 the 2’s complement of 00010110 (22)

Discussion: ( N) + (M) (2n N) + M

= 2n (N M)Case N > M: there will be no carry out, and the result is the 2’s

complement of (N M) N M: = 2n + (MN)

there will be a carry out, and the result after discarding the carry out is (M N)


overflow: + 64 01000000 64 11000000 + 96 + 01100000 96 +10100000+ 160 10100000 160 101100000 negative! positive!

using 8 bits, we can represent only -128 ~ +127!

When add two n-bits (including the sign bit) numbers N and M, overflow occurs when:

1. N, M 0, N+M 2n1 12. N, M < 0, N+M 2n1

Note that overflow cannot occur if one of N and M is positive and the other is negative.


Subtraction:N M = N+ ( M) = N + (2’s complement of M)

9 00001001 00001001 13 00001101 + 11110011 4 11111100 2’s complement of 00000100


Addition and Subtraction on Signed-1’s-complement numbers: Assume that M, N 0, and they have n bits (including sign)Case M + N : normal binary addition.

+ 9 00001001 + 13 + 00001101 + 22 00010110

Case N + M : (1’s complement of N) + M= [(2n1) N] + M= (2n 1) (N M)

Sub-case: N M 0: there will be no carry. Sub-case: N M 0:

(2n 1) (N M) = 2n + (M N) 1

carry bit 0 So there will be a carry. To obtain (MN), we discard the carry and add 1 to the result.


Ex. 9 11110110 + 13 + 00001101 end-around carry + 4 100000011

+ 1 00000100

Case ( N) + ( M) = (N+M) :(1’s complement of N) + (1’s complement of M)

= [(2n1) N] + [(2n 1) M] = 2n 1 + (2n 1) (N+M)

carry 1’s complement of (N+M)

We obtain (2n 1) (N M) by discarding the carry and adding 1 to the result. 9 11110110 13 + 11110010 22 111101000

+ 1 11101001 1’s complement of 00010110 (22)


Overflow:

Similar to the case of 2’s complement, an overflow occurs whenN+M 2n1 1 or (N) + ( M) 2n1 +1

An overflow can be detected by examining the most significant bit of the result.

+ 64 01000000 (positive) + 96 + 01100000 (positive) 160 10100000 (negative)

64 10111111 (negative) 96 10011111 (negative) 160 101011110 (positive)

carry


Comparison of the three signed numbers systems :• signed-magnitude:

useful in ordinary arithmetic, awkward in computer arithmetic• signed-1’s-complement:

used in old computers, but is now seldom used.• signed-2’s-complement:

used in most computers.


Binary CodesBinary codes can be established for any set of discrete elements.

Using n bits, we can represent at most 2n distinct elements.

So, to represent m distinct objects, we need at least log2m bits.

For example, we need log210 = 4 bits to represent { 0,1,…9}.

MAN(1)

WOMAN (0)

WHITE(00)

RED(01) BLUE(10)

BLACK(11)


Decimal BCD Excess-3 84-2-1 2421 Bi-quinary digits (8421) (5043210)

0 0000 0011 0000 0000 0100001 1 0001 0100 0111 0001 0100010 2 0010 0101 0110 0010 0100100 3 0011 0110 0101 0011 0101000 4 0100 0111 0100 0100 0110000 5 0101 1000 1011 1011 1000001 6 0110 1001 1010 1100 1000010 7 0111 1010 1001 1101 1000100 8 1000 1011 1000 1110 1001000 9 1001 1100 1111 1111 1010000

Some possible binary codes are illustrated below.


Alphanumeric Codes-- ASCII (American Standard Code for Information Interchange)

• originally use 7 bits to code 128 characters• since most digital systems handle 8-bit (byte) more efficiently, an 8 bit version ASCII has also been developed.

-- EBDIC (Extended Binary-Coded Decimal Interchange Code)• used in most IBM systems• use 8 bits to code the same 128 character symbols as ASCII

Gray Code each binary code differs from either its successor or predecessor by only one bit.


Decimal Gray Code 0 0000 1 0001 2 0011 3 0010 4 0110 5 0111 6 0101 7 0100 8 1100 9 1101 10 1111 11 1110 12 1010 13 1011 14 1001 15 1000

4-bit Gray Code


Gray Code is often used in situations where other binary code might produce erroneous or ambiguous results during transitions from one code to another.

If we use BCD code to transmit 0111 (7) and then 1000 (8), then, an erroneous intermediate code 1001 may be generated if the rightmost bit takes more time to change.The Gray code eliminates this problem since only one bit changes between two consecutive numbers.

A/DConverter

Transmitter Receiver


Generation of Gray Codes:an n-bits code is generated by reflecting the (n-1)-bit code.

0 0 00 000 0000 1 1 01 001 0001

1 11 011 0011 0 10 010 0010 10 110 0110 11 111 0111

01 101 010100 100 0100

100 1100101 1101111 1111110 1110010 1010011 1011001 1001000 1000


Error-Detecting CodeBinary code that can detect errors during data transmission.The most common ways to achieve error-detection is by means of a parity bit.A parity bit is an extra bit included in a binary code to make the total number of 1’s transmitted either odd (odd parity) or (even parity). odd parity even parity message parity bit message parity bit 0010 0 0010 1 0110 1 0110 0 1110 0 1110 1 1010 1 1010 0

Transmitter

source

Paritygenerator ERROR

destination

Paritydetector

Receiver


4-bits Adder

FAFAFA FA

x3 x1 y1 x0 y0x2 y2y3

S3 S0S1S2

c0c4

c2 c1c3

Full Adder

x y

c z

S

Full Adder

Digital CircuitsDigital Circuits


X

Y F

Name Truth TableAlgebraic Function

AND F = xy

XY

FOR F = x+y

X FInverter F = x’

X FBuffer F = x

x y F

0 0 00 1 01 0 01 1 1

x y F

0 0 00 1 11 0 11 1 1

X

Y F

NAND F = (xy)’

x y F

0 0 10 1 11 0 11 1 0

x F

0 11 0

x F

0 01 1

Gate


XY

FNOR F = (x+y)’x y F

0 0 10 1 01 0 01 1 0

X

Y F

Exclusive-OR (XOR)

F = xy’+x’y = xy

x y F

0 0 00 1 11 0 11 1 0

X

Y FExclusive-NOR

(equivalence)F = xy+x’y’ = xy

x y F

0 0 10 1 01 0 01 1 1


Extension to multiple inputs

xyz

xyz

xyzxyz

(xyz)'

x y z

x+y+z (x+y+z)' x y z

xyz

x y z

x y z


In real implementation, the three-input Exclusive-OR is usually implemented by 2-input Exclusive-OR:

xyz

X

Y

Z

Even a two input Exclusive-OR is usually constructed with other types of gates.

X

Y

X'

Y

X

Y'

X'Y

XY'

x’y+y’x

xy z

xy z

xy = x’y+xy’


This is how a gate adds a binary 1 and 0 in the “ones” place. If you feed a 1 and a 0 to the gate, it puts out a 1, which is the correct result of adding a binary1 and 0.

1

1

0

0

Place 1 1 +0 1


0 1's place

1

1

12's place

Place Place

2 1 1 1+ 1 1 0

Addition with a carryover is a little more difficult, for instance, adding 1 plus 1. If you feed the gate a 1 and 1, it will put a 0 in the “ones” place and put a carryover of 1 in the “twos” place. This produces the correct result for adding 1 and 1 in binary.


x y z c s

0 0 0 0 00 0 1 0 10 1 0 0 10 1 1 1 01 0 0 0 11 0 1 1 01 1 0 1 01 1 1 1 1

c = f(x,y,z) = (3,5,6,7)s = (1,2,4,7)

c: s: y y z

x

c = xy+yz+xz s = x’y’z+x’yz’+xy’z’+xyz

0 0 0 1 1 1 1 0

0 11 1 1 1

y zx

0 0 0 1 1 1 1 0

0 1 11 1 1

y


S

C

X

Y'

X'

Y

X

Y

X

Y

X'

Y'

X

YC

S

(a) S = xy’+x’y C = xy

(b) S = (x+y)(x’+y’) C = xy

Various Implementations of a half-adder


S

C

X'

Y'

X

Y

S

C

X

Y

X'

Y'

X

YS

C

(c) S = (C + x’y’)’ C = xy

(d) S = (x+y)(x’+y’) C = xy

(e) S = xy C = xy


Binary Adder and Subtractor

XY

Z

S

C

= x y z

= xy + yz + xz

FA FA FA FA

B4 A 4 B3 A3 B2 A2 B1 A1

C1 C2 C3 C4

C5 S4 S3 S2 S1

4-bit parallel adder


FA FAFA FA

C5 S4 S3 S2 S1

C1

M

C2 C3 C4

A 1B1 B2 A2 B3 A 3 B4 A 4

4-bit adder-subtractor4-bit adder-subtractor

yuh-jzer joungdigital systems1 number systems decimal number : 7397=7×10 3 +3×10 2 +9×10 1 +7×10...

Documents

complement of n

ndigital number n rnn

n bits

value n

ndigital number nrn

bases conversion

octal number

baser systemr