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ELCT201: DIGITAL LOGIC DESIGN Prof. Dr. Eng. Tallal El-Shabrawy, [email protected]
Dr. Eng. Wassim Alexan, [email protected]
Lecture 2
هــ 1441محرم
Spring 2020
Following the slides of Dr. Ahmed H. Madian
COURSE OUTLINE
1. Introduction
2. Gate-Level Minimization
3. Combinational Logic
4. Synchronous Sequential Logic
5. Registers and Counters
6. Memories and Programmable Logic
BASIC LOGIC GATES
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• We have defined three basic logic gates and operators
• We could build any digital circuit from those basic logic gates
• In digital logic, we are not using normal mathematics, we are using Boolean algebra
Z Y X
1
0
0
1
0
1
0
1
0
0
1
1
Z Y X
0
1
1
0
0
1
0
1
0
0
1
1
Z Y X
1
1
1
0
0
1
0
1
0
0
1
1
Z Y X
1
0
0
0
0
1
0
1
0
0
1
1
DERIVED GATES
NAND
AND-Invert
NOR
OR-Invert
XOR
Odd
XNOR
Even
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PRACTICAL EXAMPLES OF AND & OR GATES
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Floyd 11th edition
A seat belt alarm system An intrusion detection system
PRACTICAL ICS FOR LOGIC GATES
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74 Series Logic Gate Functions
PRACTICAL ICS FOR LOGIC GATES
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74 Series Logic Gate Functions
MEASUREMENT DEVICES
How to practically monitor the output of a gate?
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Oscilloscope
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Logic State Analyzer Floyd 11th edition
Probe
MEASUREMENT DEVICES
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A hand-held Logic State
Analyzer
MEASUREMENT DEVICES
• Any Boolean expression can be converted into a circuit by combining basic gates in a relatively straightforward way
• The diagram below shows the inputs and outputs of each gate
• The precedencies are explicit in a circuit. Clearly, we have to make sure that the hardware does the operations in the right order!
EXPRESSIONS AND LOGIC CIRCUITS
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SIMPLIFICATION OF THE LOGIC FUNCTION
F(A,B)=A’B’ + A’B + AB’
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A
B
A’
B’ F
SIMPLIFICATION OF THE LOGIC FUNCTION
F(A,B)=A’B’ + A’B + AB’
= A’ * (B’ + B) + A * B’ (Distributivity)
= A’ * (B + B’) + A * B’ (Commutativity)
= A’ * 1 + A * B’ (x + x’ = 1)
= (A’ + B’) (De Morgan’s)
= (A B)’ 1 GATE (NAND) ONLY
= A’ + (A * B’) (x +x’y)=(x+x’)(x+y)(Distributivity)
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By using simplification rules, we can optimize the design, so that it is implemented with a single gate, instead of 7 (Two Basic gates)
A
B F
A
B F
EXPRESSIONS AND LOGIC CIRCUITS
• Now that we are familiar with Boolean algebra and logic gates, how would that help us build logic circuits?
• If you want to build a logic circuit, you must have a Boolean expression to represent it with logic gates!
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ALGEBRAIC FORMS OF REPRESENTING BOOLEAN FUNCTIONS
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• Sum of Products (SOP)
• Product of Sums (POS)
SUM OF PRODUCTS (SOP)
Switching functions formed by:
SUMMING (ORing) PRODUCT (ANDed) terms.
Example:
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, , ,F A B C D ABC B D AC D
literals
(product terms)
sum terms
SUM OF PRODUCTS (SOP)
Product terms are known as minterms
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A B C F 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1
F =
F = 001 011 101 110 111
+ A'BC + AB'C + ABC' + ABC A'B'C
The apostrophe ’ here means invert
SUM OF PRODUCTS (SOP)
Product term (or minterm) ANDed product – input combination for which output is true
Each variable appears exactly once, in true or inverted form (but not both)
18 Short-hand notation for minterms of 3 variables
A B C F minterms 0 0 0 0 A'B'C‘ m0 0 0 1 1 A'B'C m1 0 1 0 0 A'BC' m2 0 1 1 1 A'BC m3 1 0 0 0 AB'C' m4 1 0 1 1 AB'C m5 1 1 0 1 ABC' m6 1 1 1 1 ABC m7
F(A, B, C) = m(1,3,5,6,7) = m1 + m3 + m5 + m6 + m7 = A'B'C + A'BC + AB'C + ABC' + ABC
This form is called the canonical form
F(A, B, C) = A'B'C + A'BC + AB'C + ABC + ABC' = (A'B' + A'B + AB' + AB)C + ABC' = ((A' + A)(B' + B))C + ABC' = C + ABC' = ABC' + C
= AB + C This form is called the minimal form
From the pervious example:
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Answer:
SOP AND/OR
Two-level Implementation
F = A'B'C + A'BC + AB'C + ABC' + ABC
HOW TO BUILD A CIRCUIT FROM THE SOP FUNCTION?
PRODUCT OF SUMS (POS)
Switching functions formed by taking the:
PRODUCT (ANDing) of SUM (ORed) terms.
Example:
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, , ,F A B C D A B C B D A C D
Products
Literals
(Sum terms)
PRODUCT OF SUMS (POS)
Sum terms are known as Maxterms
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A B C F 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1
F = 000 010 100
F = (A + B + C) (A + B' + C) (A' + B + C)
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PRODUCT-OF-SUMS (CONT’D)
Sum term (or maxterm)
ORed sum of literals – input combination for which output is false
each variable appears exactly once in a sum term, in true or inverted form (but not both)
A B C maxterms 0 0 0 A+B+C M0 0 0 1 A+B+C' M1 0 1 0 A+B'+C M2 0 1 1 A+B'+C' M3 1 0 0 A'+B+C M4 1 0 1 A'+B+C' M5 1 1 0 A'+B'+C M6 1 1 1 A'+B'+C' M7
short-hand notation for Maxterms of 3 variables
F(A, B, C) = M(0,2,4) = M0 • M2 • M4 = (A + B + C) (A + B' + C) (A' + B + C)
This form is called the canonical form
F(A, B, C) = (A + B + C) (A + B' + C) (A' + B + C)
= (A + C) (B + C) (remember: F=AB + C)
This form is called the minimal form
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Answer:
POS OR/AND
Two-level Implementation
F(A, B, C) = (A + B + C) (A + B' + C) (A' + B + C)
HOW TO BUILD A CIRCUIT FROM THE POS FUNCTION?
F=m(1,3,5,6,7) F=M(0,2,4)
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A B C F 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1
SOP AND POS REPRESENT THE SAME FUNCTION
POS VERSUS SOP
• Any expression can be written either way
• We can convert from one form to the other using theorems
• Sometimes SOP looks simpler
AB + CD = ( A + C )( B + C )( A + D )( B + D )
• Other times POS looks simpler
(A + B)(C + D) = BD + AD + BC + AC
• However, SOP is most commonly used
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MINIMIZATION OF LOGIC FUNCTIONS
We have chips with millions of gates
Why care about minimizing a function?
What do a few gates matter?
Basic logic functions are replicated thousands of times
Saving one gate for a memory cell pays off
What is the criterion for minimization?
Should we minimize the…
Number of product terms?
Number of logic operations?
Number of variables (literals)?
Number of wires?
…?
For implementation: minimize the number of gates!
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HOW TO MINIMIZE THE GATE COUNT?
Example: F=A’BC’+AB’C’+AB’C+ABC’= Σm(2,4,5,6)
How many gates do we need for implementation?
If AND gates have 3 inputs and OR gates have 4 inputs?
If all gates are binary (2 inputs)?
Are there any tricks we can use?
Combine minterms:
A’BC’+ABC’=BC’
AB’C’+AB’C=AB’
F = BC’+AB’
How many gates does F need now?
This mainly depends on your experience but we need a systematic
approach to minimize Boolean expressions
Answer: Karnaugh maps (K-maps) 27
KARNAUGH MAPS
• Karnaugh maps (K-maps) are graphical representations of Boolean functions
• One map cell corresponds to a row in the truth table
• Also, one map cell corresponds to a minterm or a maxterm in the Boolean expression
• Multiple-cell areas of the map correspond to standard terms
x y minterm
0 0 m0
0 1 m1
1 0 m2
1 1 m3
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2-VARIABLE K-MAP
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m3
m2
1
m1
m0
0
1 0
x
y
0 1
2 3
• The ordering of variables is important for 𝑓(𝑥, 𝑦), 𝑥 is the
row, 𝑦 is the column
• For the K map on the left, cell 0 represents 𝑥′𝑦′; cell 1
represents 𝑥′𝑦, etc…
• If a minterm is present in the function, then a 1 is placed
in the corresponding cell
m3
m1
1
m2
m0
0
1
0
y
x
0 2
1 3
OR
BOOLEAN FUNCTIONS IN A K-MAP
The1s and 0s represent a function in a K-map
A 1 represents the On-set (F=1), while a 0 represents the Off-set (F=0)
Similar to the truth table
0s are typically not shown
x y f
0 0 0
0 1 0
1 0 0
1 1 1
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x y f
0 0 0
0 1 1
1 0 1
1 1 1
2-VARIABLE K-MAP
Any two adjacent cells in the map differ by ONLY one variable, which appears complemented in one cell and uncomplemented in the other
Example
𝑚0(= 𝑥′ 𝑦′) is adjacent to 𝑚1(= 𝑥′ 𝑦),
this means that
𝑥′𝑦′ + 𝑥′𝑦 = 𝑥′ 𝑦′ + 𝑦 = 𝑥′
Also 𝑚0(= 𝑥′ 𝑦′) is adjacent to 𝑚2(= 𝑥𝑦′)
𝑥′𝑦′ + 𝑥𝑦′ = 𝑦′ 𝑥′ + 𝑥 = 𝑦′
but 𝑚0(= 𝑥′ 𝑦′) is NOT adjacent 𝑚3(= 𝑥𝑦)! 31
2-VARIABLE K-MAP: AN EXAMPLE
The 1s are placed in the K-map for specified minterms: 𝑚0, 𝑚1 and 𝑚2
Grouping (ORing) of 1s allows for simplification
What (simpler) function is represented by each dashed rectangle?
Note that m0 is covered twice! 32
𝐹 𝑥1, 𝑥2 = 𝑥1′𝑥2′+ 𝑥1′ 𝑥2+ 𝑥1 𝑥2′
= 𝑚0 + 𝑚1+ 𝑚2 = 𝑥1
′ + 𝑥2′
𝑥1′ = 𝑚0 + 𝑚1
𝑥2′ = 𝑚0 + 𝑚2
3-VARIABLE MAP
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• Note the order of the
minterms
• Gray code is used, so that
the difference between any
adjacent cells is still ONLY
one literal
3-VARIABLE MAP: EXAMPLE I
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Simplify the Boolean expression: F 𝑥, 𝑦, 𝑧 = Σ(2,3,4,5)
F 𝑥, 𝑦, 𝑧 = 𝑥𝑦′ + 𝑥′𝑦
3-VARIABLE MAP: EXAMPLE II
35
Simplify the Boolean expression: F 𝑥, 𝑦, 𝑧 = Σ(3,4,6,7)
F 𝑥, 𝑦, 𝑧 = 𝑦𝑧 + 𝑥𝑧′
3-VARIABLE MAP: EXAMPLE III
36
Simplify the Boolean expression: F 𝑥, 𝑦, 𝑧 = Σ(0,2,4,5,6)
F 𝑥, 𝑦, 𝑧 = 𝑧′ + 𝑥𝑦′
3-VARIABLE MAP: EXAMPLE IV
37
Let the Boolean function 𝐹(𝐴, 𝐵, 𝐶) = 𝐴′𝐶 + 𝐴′𝐵 + 𝐴𝐵′𝐶 + 𝐵𝐶
(a) Express this function as a sum of minterms
(b) Find the minimal SOP expression
𝐹 𝐴, 𝐵, 𝐶 = 𝛴(1,2,3,5,7) 𝐹 𝐴, 𝐵, 𝐶 = 𝐶 + 𝐴′𝐵
NOTES ON A 3-VARIABLE MAP
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• The number of adjacent cells that may be combined must always represent a number that is a power of two, such as 1, 2, 4 and 8
• As more adjacent cells are combined, we obtain a product term with fewer literals
• One cell represents one minterm, giving a term with 3 literals
• Two adjacent cell represent a term with 2 literals
• Four adjacent cells represent a term with 1 literal
• Eight adjacent cells encompass the entire map and produce a function that is always equal to logic 1
