2. Chapter 1 2
Overview
Digital Systems and Computer Systems
Information Representation
Number Systems [binary, octal and hexadecimal]
Arithmetic Operations
Base Conversion
Decimal Codes [BCD (binary coded decimal),
parity]
Gray Codes
Alphanumeric Codes
3. Chapter 1 3
Digital System
Takes a set of discrete information inputs and discrete
internal information (system state) and generates a set
of discrete information outputs.
System State
Discrete
Information
Processing
System
Discrete
Inputs Discrete
Outputs
4. Chapter 1 4
Types of Digital Systems
No state present
• Combinational Logic System
• Output = Function(Input)
State present
• State updated at discrete times
=> Synchronous Sequential System
• State updated at any time
=>Asynchronous Sequential System
• State = Function (State, Input)
• Output = Function (State)
or Function (State, Input)
5. Chapter 1 5
Digital System Example:
A Digital Counter (e. g., odometer):
1 3
0 0 5 6 4
Count Up
Reset
Inputs: Count Up, Reset
Outputs: Visual Display
State: "Value" of stored digits
Synchronous or Asynchronous?
6. Chapter 1 6
A Digital Computer Example
Synchronous or
Asynchronous?
Inputs:
Keyboard,
mouse, modem,
microphone
Outputs: CRT,
LCD, modem,
speakers
Memory
Control
unit Datapath
Input/Output
CPU
7. Chapter 1 7
Signal
An information variable represented by physical
quantity.
For digital systems, the variable takes on discrete
values.
Two level, or binary values are the most prevalent
values in digital systems.
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 On and Off.
Binary values are represented by values or ranges of
values of physical quantities
8. Chapter 1 8
Signal Examples Over Time
Analog
Asynchronous
Synchronous
Time
Continuous
in value &
time
Discrete in
value &
continuous
in time
Discrete in
value & time
Digital
10. Chapter 1 10
What are other physical quantities
represent 0 and 1?
•CPU Voltage
•Disk
•CD
•Dynamic RAM
Binary Values: Other Physical Quantities
Magnetic Field Direction
Surface Pits/Light
Electrical Charge
11. Chapter 1 11
Number Systems – Representation
Positive radix, positional number systems
A number with radix r is represented by a
string of digits:
An - 1An - 2 … A1A0 . A- 1 A- 2 … A- m 1 A- m
in which 0 Ai < r and . is the radix point.
The string of digits represents the power series:
(Number)r=
j = - m
j
j
i
i = 0
i r
A
r
A
(Integer Portion) + (Fraction Portion)
i = n - 1 j = - 1
12. Chapter 1 12
Number Systems – Examples
General Decimal Binary
Radix (Base) r 10 2
Digits 0 => r - 1 0 => 9 0 => 1
0
1
2
3
Powers of 4
Radix 5
-1
-2
-3
-4
-5
r0
r1
r2
r3
r4
r5
r -1
r -2
r -3
r -4
r -5
1
10
100
1000
10,000
100,000
0.1
0.01
0.001
0.0001
0.00001
1
2
4
8
16
32
0.5
0.25
0.125
0.0625
0.03125
13. Chapter 1 13
Special Powers of 2
210
(1024) is Kilo, denoted "K"
220
(1,048,576) is Mega, denoted "M"
230
(1,073, 741,824)is Giga, denoted "G"
15. Chapter 1 15
To convert to decimal, use decimal arithmetic
to form (digit × respective power of 2).
Example:Convert 110102 to N10:
Converting Binary to Decimal
16. Chapter 1 16
Method 1
• Subtract the largest power of 2 (see slide 14) that gives a
positive remainder and record the power.
• Repeat, subtracting from the prior remainder and recording
the power, until the remainder is zero.
• Place 1’s in the positions in the binary result corresponding
to the powers recorded; in all other positions place 0’s.
Example: Convert 62510 to N2
Converting Decimal to Binary
17. Chapter 1 17
Commonly Occurring Bases
Name Radix Digits
Binary 2 0,1
Octal 8 0,1,2,3,4,5,6,7
Decimal 10 0,1,2,3,4,5,6,7,8,9
Hexadecimal 16 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
The six letters (in addition to the 10
integers) in hexadecimal represent:
19. Chapter 1 19
Conversion Between Bases
Method 2
To convert from one base to another:
1) Convert the Integer Part
2) Convert the Fraction Part
3) Join the two results with a radix point
20. Chapter 1 20
Conversion Details
To Convert the Integral Part:
Repeatedly divide the number by the new radix and
save the remainders. The digits for the new radix are
the remainders in reverse order of their computation.
If the new radix is > 10, then convert all remainders >
10 to digits A, B, …
To Convert the Fractional Part:
Repeatedly multiply the fraction by the new radix and
save the integer digits that result. The digits for the
new radix are the integer digits in order of their
computation. If the new radix is > 10, then convert all
integers > 10 to digits A, B, …
21. Chapter 1 21
Example: Convert 46.687510 To Base 2
Convert 46 to Base 2
Convert 0.6875 to Base 2:
Join the results together with the
radix point:
22. Chapter 1 22
Additional Issue - Fractional Part
Note that in this conversion, the fractional part
became 0 as a result of the repeated
multiplications.
In general, it may take many bits to get this to
happen or it may never happen.
Example: Convert 0.6510 to N2
• 0.65 = 0.1010011001001 …
• The fractional part begins repeating every 4 steps
yielding repeating 1001 forever!
Solution: Specify number of bits to right of
radix point and round or truncate to this
number.
23. Chapter 1 23
Checking the Conversion
To convert back, sum the digits times their
respective powers of r.
From the prior conversion of 46.687510
1011102 = 1·32 + 0·16 +1·8 +1·4 + 1·2 +0·1
= 32 + 8 + 4 + 2
= 46
0.10112 = 1/2 + 1/8 + 1/16
= 0.5000 + 0.1250 + 0.0625
= 0.6875
24. Chapter 1 24
Why Do Repeated Division and
Multiplication Work?
Divide the integer portion of the power series on
slide 11 by radix r. The remainder of this division
is A0, represented by the term A0/r.
Discard the remainder and repeat, obtaining
remainders A1, …
Multiply the fractional portion of the power series
on slide 11 by radix r. The integer part of the
product is A-1.
Discard the integer part and repeat, obtaining
integer parts A-2, …
This demonstrates the algorithm for any radix r
>1.
25. Chapter 1 25
Octal (Hexadecimal) to Binary and
Back
Octal (Hexadecimal) to Binary:
• Restate the octal (hexadecimal) as three
(four) binary digits starting at the radix
point and going both ways.
Binary to Octal (Hexadecimal):
• Group the binary digits into three (four) bit
groups starting at the radix point and going
both ways, padding with zeros as needed in
the fractional part.
• Convert each group of three bits to an octal
(hexadecimal) digit.
26. Chapter 1 26
Octal to Hexadecimal via Binary
Convert octal to binary.
Use groups of four bits and convert as above to
hexadecimal digits.
Example: Octal to Binary to Hexadecimal
6 3 5 . 1 7 7 8
Why do these conversions work?
27. Chapter 1 27
A Final Conversion Note
You can use arithmetic in other bases if
you are careful:
Example: Convert 1011102 to Base 10
using binary arithmetic:
Step 1 101110 / 1010 = 100 r 0110
Step 2 100 / 1010 = 0 r 0100
Converted Digits are 01002 | 01102
or 4 6 10
28. Chapter 1 28
Binary Numbers and Binary Coding
Flexibility of representation
• Within constraints below, can assign any binary
combination (called a code word) to any data as long
as data is uniquely encoded.
Information Types
• Numeric
Must represent range of data needed
Very desirable to represent data such that simple,
straightforward computation for common arithmetic
operations permitted
Tight relation to binary numbers
• Non-numeric
Greater flexibility since arithmetic operations not applied.
Not tied to binary numbers
29. Chapter 1 29
Given n binary digits (called bits), a binary code
is a mapping from a set of represented elements
to a subset of the 2n
binary numbers.
Example: A
binary code
for the seven
colors of the
rainbow
Code 100 is
not used
Non-numeric Binary Codes
Binary Number
000
001
010
011
101
110
111
Color
Red
Orange
Yellow
Green
Blue
Indigo
Violet
30. Chapter 1 30
Given M elements to be represented by a
binary code, the minimum number of
bits, n, needed, satisfies the following
relationships:
2n
> M > 2(n – 1)
n = log2 M where x , called the ceiling
function, is the integer greater than or
equal to x.
Example: How many bits are required to
represent decimal digits with a binary
code?
Number of Bits Required
31. Chapter 1 31
Number of Elements Represented
Given n digits in radix r, there are rn
distinct elements that can be represented.
But, you can represent m elements, m < rn
Examples:
• You can represent 4 elements in radix r = 2
with n = 2 digits: (00, 01, 10, 11).
• You can represent 4 elements in radix r = 2
with n = 4 digits: (0001, 0010, 0100, 1000).
• This second code is called a "one hot" code.
32. Chapter 1 32
Binary Codes for Decimal Digits
Decimal 8,4,2,1 Excess3 8,4,-2,-1 Gray
0 0000 0011 0000 0000
1 0001 0100 0111 0100
2 0010 0101 0110 0101
3 0011 0110 0101 0111
4 0100 0111 0100 0110
5 0101 1000 1011 0010
6 0110 1001 1010 0011
7 0111 1010 1001 0001
8 1000 1011 1000 1001
9 1001 1100 1111 1000
There are over 8,000 ways that you can chose 10 elements
from the 16 binary numbers of 4 bits. A few are useful:
33. Chapter 1 33
Binary Coded Decimal (BCD)
The BCD code is the 8,4,2,1 code.
This code is the simplest, most intuitive binary
code for decimal digits and uses the same
powers of 2 as a binary number, but only
encodes the first ten values from 0 to 9.
Example: 1001 (9) = 1000 (8) + 0001 (1)
How many “invalid” code words are there?
What are the “invalid” code words?
34. Chapter 1 34
What interesting property is common
to these two codes?
Excess 3 Code and 8, 4, –2, –1 Code
Decimal Excess 3 8, 4, –2, –1
0 0011 0000
1 0100 0111
2 0101 0110
3 0110 0101
4 0111 0100
5 1000 1011
6 1001 1010
7 1010 1001
8 1011 1000
9 1100 1111
35. Chapter 1 35
What special property does the Gray code have
in relation to adjacent decimal digits?
Gray Code
Decimal 8,4,2,1 Gray
0 0000 0000
1 0001 0100
2 0010 0101
3 0011 0111
4 0100 0110
5 0101 0010
6 0110 0011
7 0111 0001
8 1000 1001
9 1001 1000
36. Chapter 1 36
B0
111
110
000
001
010
011
100
101
B1
B2
(a) Binary Code for Positions 0 through 7
G0
G1
G2
111
101
100 000
001
011
010
110
(b) Gray Code for Positions 0 through 7
Gray Code (Continued)
Does this special Gray code property
have any value?
An Example: Optical Shaft Encoder
37. Chapter 1 37
Gray Code (Continued)
How does the shaft encoder work?
For the binary code, what codes may be
produced if the shaft position lies
between codes for 3 and 4 (011 and 100)?
Is this a problem?
38. Chapter 1 38
Gray Code (Continued)
For the Gray code, what codes may be
produced if the shaft position lies
between codes for 3 and 4 (010 and 110)?
Is this a problem?
Does the Gray code function correctly for
these borderline shaft positions for all
cases encountered in octal counting?
39. Chapter 1 39
Warning: Conversion or Coding?
Do NOT mix up conversion of a decimal
number to a binary number with coding
a decimal number with a BINARY
CODE.
1310 = 11012 (This is conversion)
13 0001|0011 (This is coding)
40. Chapter 1 40
Binary Arithmetic
Single Bit Addition with Carry
Multiple Bit Addition
Single Bit Subtraction with Borrow
Multiple Bit Subtraction
Multiplication
BCD Addition
41. Chapter 1 41
Single Bit Binary Addition with Carry
Given two binary digits (X,Y), a carry in (Z) we get the
following sum (S) and carry (C):
Carry in (Z) of 0:
Carry in (Z) of 1: Z 1 1 1 1
X 0 0 1 1
+ Y + 0 + 1 + 0 + 1
C S 0 1 1 0 1 0 1 1
Z 0 0 0 0
X 0 0 1 1
+ Y + 0 + 1 + 0 + 1
C S 0 0 0 1 0 1 1 0
42. Chapter 1 42
Extending this to two multiple bit
examples:
Carries 0 0
Augend 01100 10110
Addend +10001 +10111
Sum
Note: The 0 is the default Carry-In to the
least significant bit.
Multiple Bit Binary Addition
43. Chapter 1 43
Given two binary digits (X,Y), a borrow in (Z) we
get the following difference (S) and borrow (B):
Borrow in (Z) of 0:
Borrow in (Z) of 1:
Single Bit Binary Subtraction with Borrow
Z 1 1 1 1
X 0 0 1 1
- Y -0 -1 -0 -1
BS 11 1 0 0 0 1 1
Z 0 0 0 0
X 0 0 1 1
- Y -0 -1 -0 -1
BS 0 0 1 1 0 1 0 0
44. Chapter 1 44
Extending this to two multiple bit examples:
Borrows 0 0
Minuend 10110 10110
Subtrahend - 10010 - 10011
Difference
Notes: The 0 is a Borrow-In to the least significant
bit. If the Subtrahend > the Minuend, interchange
and append a – to the result.
Multiple Bit Binary Subtraction
46. Chapter 1 46
BCD Arithmetic
Given a BCD code, we use binary arithmetic to add the digits:
8 1000 Eight
+5 +0101 Plus 5
13 1101 is 13 (> 9)
Note that the result is MORE THAN 9, so must be
represented by two digits!
To correct the digit, subtract 10 by adding 6 modulo 16.
8 1000 Eight
+5 +0101 Plus 5
13 1101 is 13 (> 9)
+0110 so add 6
carry = 1 0011 leaving 3 + cy
0001 | 0011 Final answer (two digits)
If the digit sum is > 9, add one to the next significant digit
47. Chapter 1 47
BCD Addition Example
Add 2905BCD to 1897BCD showing
carries and digit corrections.
0001 1000 1001 0111
+ 0010 1001 0000 0101
0
48. Chapter 1 48
Error-Detection Codes
Redundancy (e.g. extra information), in the
form of extra bits, can be incorporated into
binary code words to detect and correct errors.
A simple form of redundancy is parity, an extra
bit appended onto the code word to make the
number of 1’s odd or even. Parity can detect all
single-bit errors and some multiple-bit errors.
A code word has even parity if the number of
1’s in the code word is even.
A code word has odd parity if the number of 1’s
in the code word is odd.
49. Chapter 1 49
4-Bit Parity Code Example
Fill in the even and odd parity bits:
The codeword "1111" has even parity and the
codeword "1110" has odd parity. Both can be
used to represent 3-bit data.
Even Parity Odd Parity
Message- Parity Message - Parity
000 - 000 -
001 - 001 -
010 - 010 -
011 - 011 -
100 - 100 -
101 - 101 -
110 - 110 -
111 - 111 -
50. Chapter 1 50
ASCII Character Codes
American Standard Code for Information
Interchange
This code is a popular code used to represent
information sent as character-based data. It uses
7-bits to represent:
• 94 Graphic printing characters.
• 34 Non-printing characters
Some non-printing characters are used for text
format (e.g. BS = Backspace, CR = carriage return)
Other non-printing characters are used for record
marking and flow control (e.g. STX and ETX start
and end text areas).
(Refer to Table 1-4 in the text)
51. Chapter 1 51
ASCII Properties
ASCII has some interesting properties:
Digits 0 to 9 span Hexadecimal values 3016 to 3916 .
Upper case A-Z span 4116 to 5A16 .
Lower case a-z span 6116 to 7A16 .
• Lower to upper case translation (and vice versa)
occurs by flipping bit 6.
Delete (DEL) is all bits set, a carryover from when
punched paper tape was used to store messages.
Punching all holes in a row erased a mistake!
52. Chapter 1 52
UNICODE
UNICODE extends ASCII to 65,536
universal characters codes
• For encoding characters in world languages
• Available in many modern applications
• 2 byte (16-bit) code words
• See Reading Supplement – Unicode on the
Companion Website
http://www.prenhall.com/mano
#6:Answer: Part of specification for a PC is in MHz. What does that imply? A clock which defines the discrete times for update of state for a synchronous system. Not all of the computer may be synchronous, however.
#9:The HIGH range typically corresponds to binary 1 and LOW range to binary 0. The threshold region is a range of voltages
for which the input voltage value cannot be interpreted reliably as either a 0 or a 1.
#15:Powers of 2: 43210
110102 =>
1 X 24 = 16
+ 1 X 23 = 8
+ 0 X 22 = 0
+ 1 X 21 = 2
+ 0 X 20 = 0
2610
#16:625 – 512 = 113 => 9
113 – 64 = 49 => 6
49 – 32 = 17 => 5
17 – 16 = 1 => 4
1 – 1 = 0 => 0
Placing 1’s in the result for the positions recorded and 0’s elsewhere,
9 8 7 6 5 4 3 2 1 0
1 0 0 1 1 1 0 0 0 1
#17:Answer: The six letters A, B, C, D, E, and F represent the digits for values
10, 11, 12, 13, 14, 15 (given in decimal), respectively, in hexadecimal.
Alternatively, a, b, c, d, e, f are used.
#21:Answer 1: Converting 46 as integral part: Answer 2: Converting 0.6875 as fractional part:
46/2 = 23 rem = 0 0.6875 * 2 = 1.3750 int = 1
23/2 = 11 rem = 1 0.3750 * 2 = 0.7500 int = 0
11/2 = 5 remainder = 1 0.7500 * 2 = 1.5000 int = 1
5/2 = 2 remainder = 1 0.5000 * 2 = 1.0000 int = 1
2/2 = 1 remainder = 0 0.0000
1/2 = 0 remainder = 1 Reading off in the forward direction: 0.10112
Reading off in the reverse direction: 1011102
Answer 3: Combining Integral and Fractional Parts:
101110. 10112
#24:Divide integral part equation on slide 21 by r. Note that the powers of r decline by 1 in the quotient and that a fraction a0/r results, giving a0 as the remainder.
The next time, fraction a1/r results giving a1 as the remainder.
In general, the remainder will be ai for the ith step.
At the nth step, an/r will result with the quotient = 0.
Similarly, for multiply, a–1 will emerge as the integral part first since r x r –1 = 1.
#26:Answer 1:
6 3 5 . 1 7 7 8
110|011|101 . 001|111|111 2
Regroup:
1|1001|1101 . 0011|1111|1(000)2
Convert:
1 9 D . 3 F 816
Answer 2: Marking off in groups of three (four) bits corresponds to dividing or multiplying by 23 = 8 (24 = 16) in the binary system.
#30:M = 10
Therefore n = 4 since:
24 =16 is 10 > 23 = 8
and the ceiling function for log2 10 is 4.
#34:Answer: Both of these codes have the property that the codes for 0 and 9, 1 and 8, etc. can be obtained from each other by replacing the 0’s with the 1’s and vice-versa in the code words. Such a code is sometimes called a complement code.
#35:Answer: As we “counts” up or down in decimal, the code word for the Gray code changes in only one bit position as we go from decimal digit to digit including from 9 to 0.
#36:Yes, as illustrated by the example that followed.
#37:Answer 1: The encoder disk contains opaque and clear areas.
Opaque represents a 0 and clear a 1. Light shining through each ring of the encoder corresponding to a bit of the code strikes a sensor to produce a 0 or a 1.
Answer 2: In addition to the correct code, either 011 or 100, the codes 000, 010, 001, 110, 101, or 111 can be produced.
Answer 3: Yes, the shaft position can be completely UNKNOWN!
#38:Answer 1: Only the correct codes, either 010 or 110
Answer 2: No, the shaft position is known to be either 3 or 4 which is OK since it is halfway in between.
Answer 3: Yes, since an erroneous code cannot arise. This includes between 0 and 7 (000 and 100).
![Chapter 1 2
Overview
Digital Systems and Computer Systems
Information Representation
Number Systems [binary, octal and hexadecimal]
Arithmetic Operations
Base Conversion
Decimal Codes [BCD (binary coded decimal),
parity]
Gray Codes
Alphanumeric Codes](https://image.slidesharecdn.com/lcdf3chap01-250509164223-f59b80bd/85/LCDF3_Chap_01-computer-engineering-01-ppt-2-320.jpg)
