CHAPTER 1
DIGITAL SYSTEMS AND BINARY
NUMBERS
1
LECTURE 1
OUTLINE OF CHAPTER 1
1.1
 1.2
 1.3
 1.4
 1.5
 1.6
 1.7
 1.8
 1.9

Digital Systems
Binary Numbers
Number-base Conversions
Octal and Hexadecimal Numbers
Complements
Signed Binary Numbers
Binary Codes
Binary Storage and Registers
Binary Logic
2
2
DIGITAL SYSTEMS AND BINARY
NUMBERS

Digital computers
General purposes
 Many scientific, industrial and commercial
applications


Digital systems
Digital Telephone
 Digital camera
 Electronic calculators
 Digital TV

3
3
ANALOG AND DIGITAL SIGNAL

Analog system


The physical quantities or signals may vary continuously over
a specified range.
Digital system

The physical quantities or signals can assume only discrete
values.
X(t)
X(t)
t
Analog signal
t
Digital signal
4
4
BINARY DIGITAL SIGNAL

For digital systems, the variable takes on
discrete values.


Two level, or binary values.
Binary values are represented abstractly by:
Digits 0 and 1
 Words (symbols) False (F) and True (T)
 Words (symbols) Low (L) and High (H)
 And words Off and On

5
DECIMAL NUMBER SYSTEM

Base (also called radix) = 10


Digit Position


Weight = (Base)
1
0
5 1 2
-1
-2
7 4
Position
Magnitude


2
Integer & fraction
Digit Weight


10 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
100
10
1
0.1 0.01
10
2
0.7 0.04
Sum of “Digit x Weight”
Formal Notation
500
d2*B2+d1*B1+d0*B0+d-1*B-1+d-2*B-2
(512.74)10
6
6
OCTAL NUMBER SYSTEM

Base = 8


Weights


Weight = (Base)
Position
Magnitude


8 digits { 0, 1, 2, 3, 4, 5, 6, 7 }
Sum of “Digit x Weight”
Formal Notation
64
8
1
1/8 1/64
5 1 2
7 4
2
-1
1
0
-2
5 *82+1 *81+2 *80+7 *8-1+4 *8-2
=(330.9375)10
(512.74)8
7
7
BINARY NUMBER SYSTEM

Base = 2


Weights


2 digits { 0, 1 }, called binary digits or “bits”
Weight = (Base)
Position
Magnitude

Sum of “Bit x Weight”
Formal Notation
 Groups of bits
8 bits = Byte

11000101
4
2
1
1/2 1/4
1 0 1
0 1
2
-1
1
0
-2
1 *22+0 *21+1 *20+0 *2-1+1 *2-2
=(5.25)10
(101.01)2
8
8
HEXADECIMAL NUMBER SYSTEM

Base = 16


Weights


Weight = (Base)
Position
Magnitude


16 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F }
Sum of “Digit x Weight”
Formal Notation
256
16
1
1/16 1/256
1 E 5
7 A
2
-1
1
0
-2
1 *162+14 *161+5 *160+7 *16-1+10 *16-2
=(485.4765625)10
(1E5.7A)16
9
9
THE POWER OF 2
n
2n
n
2n
0
20=1
8
28=256
1
21=2
9
29=512
2
22=4
10
210=1024
3
23=8
11
211=2048
4
24=16
12
212=4096
5
25=32
20
220=1M
Mega
6
26=64
30
230=1G
Giga
7
27=128
40
240=1T
Tera
Kilo
10
10
ADDITION

Decimal Addition
1
+
1
1
Carry
5
5
5
5
1
0
= Ten ≥ Base
➔ Subtract a
Base
11
11
BINARY ADDITION

Column Addition
1 1 1 1 1 1
1 1 1 1 0 1
= 61
1 0 1 1 1
= 23
1 0 1 0 1 0 0
= 84
+
≥ (2)10
12
12
BINARY SUBTRACTION

Borrow a “Base” when needed
1
2
0 2 2 0 0 2
−
= (10)2
1 0 0 1 1 0 1
= 77
1 0 1 1 1
= 23
0 1 1 0 1 1 0
= 54
13
13
BINARY MULTIPLICATION

Bit by bit
1 0 1 1 1
x
1 0 1 0
0 0 0 0 0
1 0 1 1 1
0 0 0 0 0
1 0 1 1 1
1 1 1 0 0 1 1 0
14
14
DECIMAL (INTEGER) TO BINARY CONVERSION
Divide the number by the ‘Base’ (=2)
 Take the remainder (either 0 or 1) as a coefficient
 Take the quotient and repeat the division

Example: (13)10
Quotient Remainder
13 / 2 =
6/2=
3/2=
1/2=
Answer:
6
3
1
0
1
0
1
1
Coefficient
a0 = 1
a1 = 0
a2 = 1
a3 = 1
(13)10 = (a3 a2 a1 a0)2 = (1101)2
15
MSB
LSB
15
DECIMAL (FRACTION) TO BINARY CONVERSION
Multiply the number by the ‘Base’ (=2)
 Take the integer (either 0 or 1) as a coefficient
 Take the resultant fraction and repeat the multiplication

Example: (0.625)10
Integer Fraction Coefficient
0.625 * 2 =
0.25 * 2 =
0.5
*2=
Answer:
1
0
1
.
.
.
25
5
0
a-1 = 1
a-2 = 0
a-3 = 1
(0.625)10 = (0.a-1 a-2 a-3)2 = (0.101)2
MSB
LSB
16
16
DECIMAL TO OCTAL CONVERSION
Example: (175)10
Quotient Remainder
175 / 8 =
21 / 8 =
2 /8=
21
2
0
Answer:
Coefficient
a0 = 7
a1 = 5
a2 = 2
7
5
2
(175)10 = (a2 a1 a0)8 = (257)8
Example: (0.3125)10
Integer Fraction Coefficient
0.3125 * 8 =
0.5
*8=
Answer:
2
4
.
.
5
0
a-1 = 2
a-2 = 4
(0.3125)10 = (0.a-1 a-2 a-3)8 = (0.24)8
17
17
BINARY − OCTAL CONVERSION
8 = 23
 Each group of 3 bits represents an
octal digit

Assume Zeros
Example:
( 1 0 1 1 0 . 0 1 )2
( 2
6
. 2 )8
Octal
Binary
0
000
1
001
2
010
3
011
4
100
5
101
6
110
7
111
Works both ways (Binary to Octal & Octal to
Binary)
18
18
BINARY − HEXADECIMAL CONVERSION
16 = 24
 Each group of 4 bits represents a
hexadecimal digit

Assume Zeros
Example:
( 1 0 1 1 0 . 0 1 )2
(1
6
. 4 )16
Hex
Binary
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Works both ways (Binary to Hex & Hex to
Binary)
19
19
OCTAL − HEXADECIMAL CONVERSION

Convert to Binary as an intermediate step
Example:
( 2
6
.
2 )8
Assume Zeros
Assume Zeros
( 0 1 0 1 1 0 . 0 1 0 )2
(1
6
.
4 )16
Works both ways (Octal to Hex & Hex to
Octal)
20
20
DECIMAL, BINARY, OCTAL AND HEXADECIMAL
Decimal
Binary
Octal
Hex
00
01
02
03
04
05
06
07
08
09
10
11
12
13
14
15
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
00
01
02
03
04
05
06
07
10
11
12
13
14
15
16
17
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
21
21