1. Decimal Binary Octal Hex
Number systems, Binary Logic, Gates
Number systems:
Roman vs Arabic
The importance of "zero“:
significance of position of digits in a number
Decimal Binary Octal Hex A B C D E F

2. Number systems Equivalence of numbers in different systems
Convert 258 to Decimal
Convert 2510 to Octal

3. Coding: Binary representation of non-numbers
ASCII
Octal characters
|000 NUL|001 SOH|002 STX|003 ETX|004 EOT|005 ENQ|006 ACK|007 BEL|
|010 BS |011 HT |012 NL |013 VT |014 NP |015 CR |016 SO |017 SI |
|020 DLE|021 DC1|022 DC2|023 DC3|024 DC4|025 NAK|026 SYN|027 ETB|
|030 CAN|031 EM |032 SUB|033 ESC|034 FS |035 GS |036 RS |037 US |
|040 SP |041 ! |042 " |043 # |044 $ |045 % |046 & |047 ' |
|050 ( |051 ) |052 * | |054 , | | |057 / |
| | | | | | | | |
| | |072 : |073 ; |074 < |075 = |076 > |077 ? |
|100 @ |101 A |102 B |103 C |104 D |105 E |106 F |107 G |
|110 H |111 I |112 J |113 K |114 L |115 M |116 N |117 O |
|120 P |121 Q |122 R |123 S |124 T |125 U |126 V |127 W |
|130 X |131 Y |132 Z |133 [ |134 \ |135 ] |136 ^ |137 _ |
|140 ` |141 a |142 b |143 c |144 d |145 e |146 f |147 g |
|150 h |151 i |152 j |153 k |154 l |155 m |156 n |157 o |
|160 p |161 q |162 r |163 s |164 t |165 u |166 v |167 w |
|170 x |171 y |172 z |173 { |174 | |175 } |176 ~ |177 DEL|
Total symbols: 128
Total octal digits required ?
Total binary digits required ?

4. Basis: Boolean Algebra
Binary Logic and Gates
- We will use electronic circuits to control “logic”
- We can construct circuits that work based on presence/absence of voltage
- Logic circuits (complex) are unreliable if multiple voltage levels have
different interpretations
 We use BINARY LOGIC CIRCUITS
Basis: Boolean Algebra

5. Basic operators of Boolean Algebra
AND OR
NOT
Relation of Boolean Algebra and Discrete Logic:

6. Common Logic Gates

7. Truth Tables: relation of inputs, output of an operator

8. Gates basics Functioning of an NPN transistor 5V A is ON 
LED
A is ON 
Small current at transistor base 
Collector-Emitter conducts freely 
Current through LED 
LED is ON
NPN transistor R1 5V R2 R3
NOR( A, B) A B
A Resistor-Transistor Logic (RTL) NOR gate
…you will learn these in detail in ELEC xxx

9. Some simple binary number operations
ADDITION:
2’s COMPLEMENT:
- start from the least significant digit
- copy bit-by-bit until (including) the first ‘1’
- inverted the remaining digits ( ‘0’  ‘1’, and ‘1’  ‘0’)
Example:
Binary number, b =
2's complement of b =

10. Binary number operations: motivation
Computers have ‘limited’ representations [fixed length numbers]
Consider the following 4-bit additions:
borrow
overflow
- how to represent negative numbers?
- what does the ‘borrow’ mean?

11. Binary number operations: motivation..
Conventions:
- Use of one extra bit, to identify overflow in addition
- Reserve Most-Significant-Bit (MSB) to identify negative number
- Store –ve numbers as 2’s complement of its absolute value
Consequence:
- MSB is 1  -ve number
- replace addition by addition of 2’s complement!!

12. Binary number operations: motivation...
number (nd)
binary (nb)
-2's complement ( - n ) 3
0 : 0011
1 : 1101 5
0 : 0101
1 : 1011 7
0 : 0111
1 : 1001 8
0 : 1000
1 : 1000
CALCULATE:
5 - 7
-3 - 3
8 - 3

13. Boolean Algebra - Developed by George Boole
- Wanted to develop a language-based logic (“and” “or” “not”)
- Boolean Algebra was used by Claude Shannon for design of logic circuits
- What is algebra ?
- Boolean Algebra:
two constants: 0, 1
variable can take value of any constant
rules to construct ‘expressions’
rules to construct statements ( expr1 = expr2)
logic for testing truth of a statement

14. Boolean Algebra constants: 1, 0 symbols: (, ), , , =, ¬
variables: A, B, …
Expressions:
a series of symbols
rules:
constant
variable
constant OPb constant
OPu constant
( expression )
( expression ) OPb ( expression )
OPu ( expression )

15. Boolean Algebra Evaluating an expression:
- substitute the value of each variable
- use truth table of operators to evaluate
if X = 1, value of expression X  1 =
Statement:
expression = expression
X  (X  Y) = 1
Truth of a statement:
if the equality is true for all possible instantiations of variables
X  (X  Y) = 1

16. Boolean Algebra Theorems: Statements that are always true.
Some theorems of one variable

17. Boolean Algebra
Some theorems of two variables
de Morgan’s theorems

18. Boolean Algebra Use of theorems: ‘simplification’ of expressions
How many gates would we need to implement the original expression?
How many are needed to implement the equivalent expression?

19. Karnaugh Maps (K-Maps)
Graphical method to simplify expressions
K-Map:
matrix of 2n elements
each element corresponds to one distinct instantiation of the variables

20. K-Maps, example T = X  (X  Y)
- Write T as a disjunction of eight conjunction terms (OR of eight AND-terms)
T = X  (X  Y) = X (X  Y) = X X  XY
- T = 1 if any AND-term = 1 [WHY ?]
- Put a ‘1’ in each cell corresponding to each AND-term X Y 1 XY
X X  XY = X  XY X Y 1 XY
- Loop all 1’s into rectangular BLOCKS
Loop = X = T

21. K-Maps, example
- T is a disjunction of eight conjunction terms (OR of eight AND-terms)
- T = 1 if any AND-term = 1 [WHY ?]
- Put a ‘1’ in each cell corresponding to each AND-term A B C D 1
NOTE: Which cell corresponds to BCD ?

22. K-Maps, example How to make ‘loops’: - Must be rectangular or square
- Must contain 2n cells (1, 2, 4, 8, …)
- Must contain no ‘0’
- Can join leftmost column with rightmost
- Can join top row with bottom
- All 1’s must belong to at least one loop
- Try to maximize the size of loops

23. K-Maps construction: f cells
Example: Elevator control
Elevator is going up: Switch A is ON
Elevator is going down: Switch B is ON
denotes that the elevator is going up
denotes that the elevator is neither going up, nor down
What about the state AB ?
State that is physically impossible is filled with f in K-map
f in K-map: may be used as 1 or 0

24. K-Maps construction
prime implicant: a block which is not entirely included in a larger block
essential prime implicant: a prime implicant with at least one cell that
cannot be included in any other prime implicant
1. Identify input, output variables; formulate the logic expression
2. Re-write the expression: disjunction of purely conjunctive terms
3. Make the K-map
4. Fill a 1 in the cell(s) corresponding to each term, and F where possible
5. Identify and "block" each essential prime implicant.
6. If any un-blocked "1" cells remain, create corresponding prime implicant
7. Write output = disjunction of conjunctive terms (1 term for each implicant)