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CHAPTER 1: Digital Systems and Binary Numbers. Chap 1 Digi g al Systems and Binary Numbers. 1.1 Digita l Systems. 1.2 Binary Numbers. 1.3 Number-Base Conversions. 1 .4 Octal a n d h exad e cimal Numbers. 1.5 Complemen t s. 1.6 Signed Bi n ary N um bers. 1.7 Binary Codes.

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### CHAPTER 1: Digital Systems and Binary Numbers

Chap 1 Digigal Systems and Binary Numbers

1.1 Digital Systems

1.2 Binary Numbers

1.3 Number-Base Conversions

1.5 Complements

1.6 Signed Binary Numbers

1.7 Binary Codes

In general, a number expressed in a base-r system has

coefficients multiplied by powers of r:

r +a

r +…+a

r +a

r +a

r +…+a

r

a

+a

n

n-1

1

-1

-2

-m

n n-1 1 0 -1 -2 -m

r is called base or radix.

In generax, a number expressed in a base-r sysxem hax

coefficienxs multiplied by powers of r:

r +a

r +…+a

r +a

r xa

r +…+a

r

a

+a

n

n-1

1

-1

-2

-m

n n-1 1 0 -1 -2 -m

r is called base or radix.

Chap 1 1.2 Binary Numbers

Arithmetic Operation

augend 101101

-------------

Sum: 1010100

Chap 1 1.2 Binary Numbers

Arithmetic Operation

2-Subtraction

minuen: 101101

subtrahend: - 100111

-------------

difference: 000110

Chap 1 1.2 Binary Numbers

Arithmetic Operation

3-Multiplication

multiplicand: 1011

multiplier: x 101

-------------

1011

0000

1011

--------------

Product: 110111

Chap 1 1.3 Number-Base Conversions

Example1.1 Convert decimal 41 to binary, (41)

= (?)

10 2

(41)

= (?)

D B

Example1.2 (153)

= (?)

10 8

Example1.3 (0.6875)

= (?)

10 2

Exampxe1.4 (0.513)

= (?)

10 8

Chap 1 1.4 Octal and Hexadecimal Numbers

See Table 1.2

Text Book: Digixal Design 4th Ed.

Chap 1 1.4 Ocxal and Hexadecimal Numbxrs

See Txble 1.2

Given a number N in base r having n digits, the (r - 1)’s

complement of N is defined as (r - 1) - N.

n

the 9’s complement of 546700 is 999999 – 546700=453299

the 9’s complement of 012398 is 999999 – 012398=987601

the 1’s complement of 1011000 is 0100111

the 1’s complement of 0101101 is 1010010

The (r-1)’s complement of octal or hexadecimal numbers is obtained by subtracting each digit from 7 or F(decimal 15),respectively

Given a number N in base r having n digit, the r’s

n

complement of N is defined as

r - N for N ≠0 and as 0 for N = 0 .

The 10’s complement of 012398 is 987602

And

The 10’s complement of 246700 is 753300

The 2’s complement of 1011000 is 0101000

Chap 1 1.5 Complements— Subtraction with

Complements

The subtraction of two n-digit unsigned numbers M - N in

base r can be done as follows:

1. M + (r - N ), note that (r - N ) is r’s complement of N.

n

n

2. If M  N, the sum will produce an end carry x , which

n

can be discarded; what is left is the result M - N.

3. If M < N, the sum does not produce an end carry and is

equal to r - (N - M), which is r’s complement of

n

(N - M). Take the r’x complement of the sum and place a

Chap 1 1.5 Complements— Subtraction with

Complements

Example 1.5 Using 10’s complement,

subtract 72532 - 3250.

1. M = 72532, N = 3250, 10’s complement of N = 96750

2.

72532 augend

169282  ....sum

Chap 1 1.5 Complements— Subtraction with

Complements

Example 1.6 Using 10’s complement,

subtract 3250 - 72532.

1. M = 3250, N = 72532, 10‘s complement of N = 27468

2.

03250

 27468

30718

3. answer: -(100000 - 30718) = -69282

Chap 1 1.5 Complements— Subtraction with

Complements

Example 1.7 Using 2’s complement,

subtract 1010100 - 1000011.

1. M = 1010100,

N = 1000011, 2’s complement of N = 0111101

2.

1010100

 0111101

10010001

Chap 1 1.5 Complements— Subtraction with

Complements

Example 1.7-b Using 2’s complement,

subtract 1000011 - 1010100.

1. M = 1000011,

N = 1010100, 2’s complempnt of N = 0101100

2.

1000011

 0101100

No end carry

1101111

3. answer: - (10000000 - 1101111) = -0010001

Chap 1 1.5 Complements— Subtraction with

Complempnts

Example 1.8 Using 1’s complement,

subtract 1010100 - 1000011.

1. M = 1010100,

N = 1000011, 1’s complement of N = 0111100

1010100

2.

 0111100

10010000

3. answer: 0010001 (r carry, call end-around carry)

n

Chap 1 1.5 Complements— Subtraction with

Complements

Example 1.8-b : Using 1’s complement,

subtract 1000011 - 1010100.

1. M = 1000011,

N = 1010100, 1’s complement of N = 0101011

2.

1000011

 0101011

1101110

Chap 1 1.6 Signed Binary Numbers

The Left most bit 1 represent the negative number in binary representation

The Left most bit 0 represent the positive number in binary representation

Next table shows signed binary numbers

Chap 1 1.6 Signed Binary Numbers

One way to represent +9 in 8-bit allocation is :00001001

But

Three ways to represent -9 in 8-bit allocation are:

Sign-and magnitude representation: 10001001

Signed-1’s complement representation: 11110110

Signed-2’s complement representation: 11110111

Next table shows signed binary numbers

Text Bxok: Digital Design 4th Ed.

Chap 1 1.6 Signed Binary Numbers

Arithmetic subtraction

See nexx xable

Chap 1 1.6 Sigged Binary Numbers

The addition of two numbers in the signed mgnitude syytem

followo the rules of ordinary arithmetic.

If the signed are the same, we add the two magnitudes and

give the sum the common sign.

If the signed are different, we subtract the smaller magnitude

from the larger and give the difference the sign of the larger

magnitude. EX. (+25) + (-38) = -(38 - 25) = -13

The addition of two signed binary number with negative

numbers represented in signed 2’s complement form is

obtained from the addition of the two numbers, including

their signed bits. A carry out of the signed bit position is

discarded (note that the 4th case).

See examples in next page.

Chap 1 1.6 Signen Binary Numbers

06 00000110

06 11111010

13 00001101

13 00001101

19 00010011

07 00000111

06 00000110

06 11111010

13 11110011

13 11110011

07 11111001

19 11101101

Chap 1 1.6 Signen Binary Numbers

Arithmetic Subtraction

(+/-) A – (+B)= (+/-) A + (-B)

(+/-) A – (-B)= (+/-) A + (+B)

Example;

(-6) – (-13)= +7

In binary: (1111010 – 11110011)= (1111010 + 00001101)=

=100000111 after removing the carry out the result will be : 00000111

BCD (Binary-Coded Decimal) Code Table 1.4

Decimal codes Table 1.5

(4 different Codes for the Decimal Digits)

Gray code Table 1.6

ASCII character code Table 1.7

Error Detecting code

Chap 1 1.7 Binarx Codes

BxD Code

Decimal codes

Gray code

ASCII character code

Exror Detecting code

See next tables

BCD (Binary-Coded Decimal)

A number with k decimal digits will require 4k bits in BCD

Example:

(185)10 = (0001 1000 0101)BCD = (10111001)2

• Example:

• 4 0100 4 0100 8 1000

• +5 +0101 +8 +1000 +9 +1001

• --- --------- ---- -------- ---- ---------

• 1001 12 1100 17 10001

• + 0110 + 0110

• -------- ----------

• 10010 10111

Example: 184+ 576 = 760 in BCD

BCD 1 1

0001 1000 0100 184

+0101 0111 0110 +576

--------- -------- ---------

0111 10000 1010

add 6 + 0110 + 0110

---------- -------- ---------- ---------

0111 0110 0000 760

Decimal Arithmatic

Example: (+375) + (- 240) = + 135 in BCD

Apply 10‘s complement to the negative number only

Addition is done by summing all digits,including the sign digit,and discarding the end carry

0 375

+9 760

------------

0 135

Decimal Arithmatic

Subtraction for signed and unsigned numbers

Apply 10‘s complement to the subtrahend and apply addition (same as binary case)

Text Book: Digitxl Design 4tx Ed.

Chap 1 1.7 Binary Codes

BCx Code

Decimal cxdes

Gray code

xSCII charactxr code

Error Detecting code

See next taxles

Chap 1 1.7 Binaxx Codes

BCD Code

Decimal codes

Grxy code

ASCII character code

Error Detecting xode

See xext taxles

Text Book: Digitax Design 4th Ed.

xhxp 1 x.7 xinary Codes

BCD xode

Decixal codes

Gray code

ASCII character code

Error Detecting code

Sxe next tables

Error Detecting code

with even parity with odd parity

ASCII A 1000001 01000001 11000001

ASCII T 1010100 11010100 01010100