1. ECA1212 Introduction to Electrical & Electronics Engineering Chapter 8: Digital Electronics – Combinational Logic by Muhazam Mustapha, November 2011

2. Learning Outcome
By the end of this chapter, students are expected to understand
binary number concepts
Boolean algebra and minimization

3. Chapter Content Motivation Number Systems Negative Numbers
Boolean Algebra
Gates
Karnaugh Map

4. Motivations
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5. Definition & Nature
Field of electronics that represents signals by DISCRETE bands of analog levels, rather than by a continuous range
Work of digital electronics:
estimates analog values as numbers (binary)
math is more in boolean and integer mode
deals more with algorithms and timing
Opposite: Analog electronics
represents signals as continuous range
works on manipulating signal shape
math is real valued and may involve calculus
deals more with transfer functions and models
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6. Definition & Nature Digital amplifier Analog amplifier DSP Analog view
32 bits
32 bits
Analog view
Digital view F X X F Y Y
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7. Number Systems
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8. Binary System (Base-2)
Binary (short as ‘bin’) system is the positional number system that uses only 2 symbols (b): 0 and 1 (N = 2)
...b3b2b1b0 . b−1b−2b−3...
Number value = b3×23 + b2×22 + b1×21 + b0×20
+ b−1×2−1 + b−2×2−2 + b−3×2−
For binary system, bn, which is either 0 or 1, is also called BIT
The above formula is used to get the decimal (base-10) values from non-base-10 numbers
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9. Octal & Hexadecimal
Base-2 (binary) is too small which makes the number tends to be very long
Larger bases are needed so that the number is shorter and more handy
Octal (short as ‘oct’) numbers are base-8, using 8 symbols: 0, 1, 2, 3, 4, 5, 6, 7
Hexadecimal (short as ‘hex’) numbers are base-16, using 16 symbols (10 digits and 6 first alphabets either lower or upper case): 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F
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10. Oct & Hex to Dec Conversion
Similar conversion rule as bin to dec (short for decimal) applies
Octal:
...d3d2d1d0 . d−1d−2d−3...
Number value = d3×83 + d2×82 + d1×81 + d0×80
+ d−1×8−1 + d−2×8−2 + d−3×8−
Hexadecimal:
...d3d2d1d0 . d−1d−2d−3...
Number value = d3×163 + d2×162 + d1×161 + d0×160
+ d−1×16−1 + d−2×16−2 + d−3×16−
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11. Conversion from Dec (Integer)
If the dec number has fraction part, we have to split it into integer and fraction
To convert from integer dec, M, to any base do the following algorithm:
DIVIDE the integer M with the base (N)
Keep the REMAINDER of the division (in sequence) – these are the digits of the base-N number
Also keep the integer value of the division – this will be the new value of M
If M is more than 0, repeat step 1-4
The base-N number is the remainders kept in step 2 written from the last one to the first one (reversed)
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12. Conversion from Dec (Integer)
Example: Convert 62 dec to bin
Remainder 2 62 2 31 1 2 15 1
6210 = 2 7 1 2 3 1 2 1 1
Try yourself to convert 62 dec to oct and hex
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13. Conversion from Dec (Fraction)
To convert from dec fraction, m, to any base do the following algorithm:
MULTIPLY the fraction m with the base (N)
Keep and remove the INTEGER PART of the multiplication (in sequence) – these are the digits of the base-N number
Also keep the fraction value of the multiplication – this will be the new value of m
If m is more than 0, repeat step 1-4, or continue to step 5 until the required no. fraction digits are obtained
The base-N number is the integer parts kept in step 2 written from the first one to the last one (in order)
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14. Conversion from Dec (Fraction)
Example: Convert 0.62 dec to bin
Integer Part
= 2 × 0.62
= 2 × 0.24
= 2 × 0.48 =
= 2 × 0.96
= 2 × 0.92
= 2 × 0.84
Try yourself to convert 0.62 dec to oct and hex
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15. Bin, Oct & Hex Conversions
Groups mapping:
Bin
Hex
0000
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 8
1001 9
1010 A
1011 B
1100 C
1101 D
1110 E
1111 F
Bin
Oct
000
001 1
010 2
011 3
100 4
101 5
110 6
111 7
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16. Bin, Oct & Hex Conversions
Example: Convert bin to oct and hex
= 15138 1 5 1 3
= 34B16 3 4 B
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17. Bin, Oct & Hex Conversions
Example: Convert 526 oct to bin
5268
5268 =
Example: Convert 52C hex to bin
52C16
52C16 =
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18. Negative Numbers
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19. Representing Negative Number
For our course we will only consider integer negative numbers
Sign-magnitude representation:
the magnitude is represented like positive number
the sign will be kept as 1 bit flag
straight-forward but requires more circuit
Complement method:
less direct but results in simpler circuit
we will study complement methods for bin numbers – 1’s and 2’s complement
there are complement methods for dec numbers – 9’s and 10’s complement, but we won’t go into details
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20. Motivation
In digital processing we will be doing a lot of binary arithmetic
Most of the arithmetic will be additive (addition or subtraction)
because multiplicative arithmetic can always be converted to additive ones
The motivation to have complement method in binary numbers is to convert subtraction into addition
so that both operations (and all operations) can homogeneously handled as one type of operation
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21. 1’s and 2’s Complement
1’s and 2’s complements (comp) are negative number representation for binary numbers
The size of the number must be known
The negative number (1’s or 2’s comp) is then indicated by a 1 in the MSB (most significant bit)
1’s complement is constructed by:
toggling (change 1 to 0, or 0 to 1) each bit in the number
there will be 2 ways to represent 0 (zero):
– positive zero, or
– negative (1’s comp) zero
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22. 1’s and 2’s Complement 2’s complement is constructed by:
Constructing the 1’s complement, then
Add 1 to the result
Differences from 1’s comp:
there is only 1 way to represent zero:
zero is considered positive since the MSB is 0
the number with only MSB as 1 and the rest is zero ( ) is the largest negative number that has NO matching positive number (check for yourself why)
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23. 1’s and 2’s Complement 4 bits 1’s and 2’s comp: CO3 1’s comp Dec 0000
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 −7
1001 −6
1010 −5
1011 −4
1100 −3
1101 −2
1110 −1
1111 −0
2’s comp
Dec
0000
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 −8
1001 −7
1010 −6
1011 −5
1100 −4
1101 −3
1110 −2
1111 −1
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24. 1’s and 2’s Complement
Example: Compute 53 dec 1’s and 2’s comp for bin number of 8 bits long
53 dec = bin
−53 dec = − bin = ’s comp = ’s comp
Example: Compute the following 1’s and 2’s comp numbers’ sign-magnitude dec
’s comp = bin = -100 dec
’s comp = ’s comp = bin = -811 dec
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25. 2’s Comp Arithmetic We shall cover only 2’s complement arithmetic
Take place only when it involves SUBTRACTION or NEGATIVE numbers
Represent any negative number with the corresponding 2’s comp
Replace subtraction with addition of its negative
Subtraction of 2’s complement is the addition with the corresponding positive number
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26. 2’s Comp Arithmetic
Both addition and subtraction are performed as ADDITION
Even though the MSB can be considered as sign flag, but the addition is done disregarding that fact, i.e. MSB is involved in the addition
Ignore any carry beyond MSB:
OVERFLOW happens when addition of 2 positive numbers results in a negative number
UNDERFLOW happens when addition of 2 negative numbers results in a positive number
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27. 2’s Comp Arithmetic 2’s comp example:
−3−2 dec = −0011−0010 bin = bin = bin = −5 dec
5+5 dec = bin = bin = −6 dec ⇒ wrong answer due to overflow
overflow indication
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28. Boolean Algebra
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29. Definition
Boolean algebra has 2 similar but different definitions for mathematician and engineers
Mathematician: Boolean algebra is part of the so called abstract mathematics that involves structures and lattice
Wikipedia:
Engineers: Boolean algebra is calculus of truth values that is used as a means to facilitate the design of logic and digital systems
Wikipedia:
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30. Boolean Values
Boolean expressions represent the 2 levels of binary system in the digital systems
In real digital systems the levels can be ON/OFF states, LOW/HIGH voltage levels
In Boolean expressions the levels (values) are TRUE/FALSE or 1/0 1
HIGH
LOW
True
False ON
OFF
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31. Boolean Operation
Boolean algebra has 3 fundamental operations: AND, OR, NOT (also called inverter)
AND:
Operates on 2 operands
Gives true (1) if both operands are true
Symbol: ‘·’ (dot) - optional
Example:
Assume A = True, B = False Then A AND B = False Written as: A·B = AB = False = 0
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32. Boolean Operation OR: Example: Operates on 2 operands
Gives true (1) if either or both operands are true
Symbol: ‘+’
Example:
Assume A = True, B = False Then A OR B = True Written as: A+B = True = 1
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33. Boolean Operation NOT: Example: Operates on 1 operand
Gives 1 if the operand is 0, otherwise it gives 0
symbol: ‘‾’ (bar / overline)
Example:
Assume A = True Then NOT A = False Written as: A = False = 0
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34. Boolean Function & Expression
Boolean Expressions: Composed of any combination of Boolean variables and operations
Example:
Boolean Functions: Function with Boolean values and defined by Boolean valued variables (parameters)
Example: Function f of a, b and c
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35. Truth Table
Since Boolean functions are Boolean valued, and composed of Boolean valued parameters, all possible values of the function with respect to all possible combinations of the parameter can be listed in a table
Such a table is called Truth Table
Truth table is a way to define a Boolean function that is not available in other mathematical functions – truth table can be applied to Boolean functions only
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36. Truth Table Truth table of the 3 fundamental Boolean operations:
NOT OR
AND A 1 A B
A+B 1 A B AB 1
Truth table can also be built for any Boolean function
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37. Truth Table
Example of step by step construction of truth table of function: A B C AC F 1
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38. Boolean Expression from Truth Table
Since truth table is constructed from Boolean functions, it is also possible to construct Boolean function from a given truth table
By MINTERM
Sum of Product (SOP)
Create minterms by AND-ing all variables where the outputs (function values) are 1
The variables that have value of 0 will be inverted
The variables that have value of 1 will not be inverted
The minterms will then be combined by OR operation
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39. Boolean Expression from Truth Table
By MAXTERM
Product of Sum (POS)
Create maxterms by OR-ing all variables where the outputs (function values) are 0
The variables that have value of 1 will be inverted
The variables that have value of 0 will not be inverted
The maxterms will then be combined by AND operation
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40. Boolean Expression from Truth Table
The SOP and POS of minterms and maxterm, respectively, are also called the CANONICAL form of the boolean function
The form is the one with the smallest no. terms and all variables appear in each term
Canonical also means ‘standard’
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41. Boolean Expression from Truth Table
Minterm example:
Truth Table A B C F 1
Sum of Product (SOP)
short notation
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42. Boolean Expression from Truth Table
Maxterm example:
Truth Table A B C F 1
Product of Sum (POS)
short notation
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43. DeMorgan’s Theorem Used to split groups of large inversion CO3 A B AB1 A B
A+B AB 1
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44. Other Boolean Operations
NOR
OR followed by NOT
A NOR B = A+B
NAND
AND followed by NOT
A NAND B = AB
XOR
true if either one, but not both, is true
symbol: ‘⊕’
XNOR
true if both parameters are the same
XOR followed by NOT
symbol: ‘⊙’
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45. Other Boolean Operations
NAND A B AB 1
XOR A B 1
NOR A B
A+B 1 A B 1
XNOR
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46. Gates
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47. Boolean Operators as Gates
NOT A A
A XOR B
A AND B A A
A OR B B B B
A XNOR B A
A NAND B
A NOR B A A B B B
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48. Gate Tree For more general and complicated boolean functions CO3 B B+C
A(B+C) A
A(B+C)+AD A D AD
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49. Karnaugh Map
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50. Karnaugh Map
K-Map is a transformation of truth table into a table that arranges the terms in such a way that the adjacent terms have ONLY 1 VARIABLE CHANGE
The mapping from truth table to K-map is a one-to-one mapping, but the counting sequence in K-map is changed to maintain 1 variable change between the neighboring (adjacent) box
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51. Karnaugh Map 1 variable truth table 1 variable K-Map
Term No. 1 A A B
Term No. 1 2 3 A 1 B 1 1 2 1 3 1 A B C
Term No. 1 2 3 4 5 6 7
3 variable truth table
3 variable K-Map AB C 00 01 11 10 2 6 4 1 3 7 5 1
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52. Karnaugh Map AB 4 variable K-Map CD 5 variable K-Map A = 0 A = 1 BC BC00 01 11 10 00 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 01 11 10
5 variable K-Map
A = 0
A = 1 BC BC DE 00 01 11 10 DE 00 01 11 10 00 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 00 16 20 28 24 17 21 29 25 19 23 31 27 18 22 30 26 01 01 11 11 10 10
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53. Karnaugh Map Example 2: Put the following function into K-map
F(A, B, C, D) = Σm(2, 3, 6, 10, 11, 12) AB CD 00 01 11 10 00 1 01 11 10
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54. Implicants
Implicants are coverings of the K-map that maintain adjacency in ALL of its variables
1 minterm is an implicant by itself
2 adjacent minterms can be combined to form a bigger (2-minterm) implicant
This new implicant will lose the changing variable
2 adjacent implicants, with all its variables are also adjacent, can also be combined to form a yet bigger implicant
This bigger implicant will lose one more changing variable
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55. Implicants Example 3: F(A,B,C,D)=Σm(2,3,6,10,11,12,14,15) AB AB CD CD00 01 11 10 CD 00 01 11 10 1 1 00 00 01 01 11 11 10 10
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56. Implicants
An implicant that is contained entirely within another implicant can be eliminated
An implicant that cannot be grown anymore is called PRIME implicant
An essential prime implicant is the prime implicant that cover some area that is NOT covered by any other implicant
The simplified function is the one that contains the minimum no. implicants that cover all 1-s
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57. Final Simplification Example 3 (continued):
F(A,B,C,D)=Σm(2,3,6,10,11,12,14,15) AB CD 00 01 11 10 1 00 01 11 10
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