Post on 05-Jan-2016
Unit 1Minimization Techniques and
Logic Gates
Introduction to Digital Systems• Analog devices and systems process time-varying
signals that can take on any value across a continuous range.
• Digital systems use digital circuits that process digital signals which can take on one of two values, we call:
0 and 1 (digits of the binary number system)
or LOW and HIGH or FALSE and TRUE
• Digital computers represent the most common digital systems.• Once-analog Systems that use digital systems today:
– Audio recording (CDs, DAT, mp3) – Phone system switching– Automobile engine control – Movie effects– Still and video cameras….
High
Low
Digital circuit
inputs outputs: :
Analog Signal
Digital Signal
3
Advantages of Digital Systems Over Analog Systems
1. Reproducibility of the results 2. Accuracy of results3. More reliable than analog systems due to better immunity to
noise.
4. Ease of design: No special math skills needed to visualize the behavior of small digital (logic) circuits.
5. Flexibility and functionality.
6. Programmability.
7. Speed: A digital logic element can produce an output in less than 10 nanoseconds (10-8 seconds).
8. Economy: Due to the integration of millions of digital logic elements on a single miniature chip forming low cost integrated circuit (ICs).
4
Boolean AlgebraBoolean Algebra
• Boolean Algebra named after George Boole who used it to study human logical reasoning – calculus of proposition.
• Elements : true or false ( 0, 1)
• Operations: a OR b; a AND b, NOT a
e.g. 0 OR 1 = 1 0 OR 0 = 0
1 AND 1 = 1 1 AND 0 = 0
NOT 0 = 1 NOT 1 = 0
What is an Algebra? (e.g. algebra of integers)set of elements (e.g. 0,1,2,..)set of operations (e.g. +, -, *,..)postulates/axioms (e.g. 0+x=x,..)
Boolean function
• Boolean function: Mapping from Boolean variables to a Boolean value.
• Boolean algebra: Deals with binary variables and logic operations operating on those variables.
BASIC IDENTITIES OF BOOLEAN ALGEBRA
• Postulate 1 (Definition): A Boolean algebra is a closed algebraic system containing a set K of two or more elements and the two operators · and + which refer to logical AND and logical OR
Basic Identities of Boolean Algebra(Existence of 1 and 0 element)
(1) x + 0 = x
(2) x · 0 = 0
(3) x + 1 = 1
(4) x · 1 = 1
Basic Identities of Boolean Algebra (Existence of complement)
(5) x + x = x
(6) x · x = x
(7) x + x’ = x
(8) x · x’ = 0
Basic Identities of Boolean Algebra (Commutativity):
(9) x + y = y + x
(10) xy = yx
Basic Identities of Boolean Algebra (Associativity):
(11) x + ( y + z ) = ( x + y ) + z
(12) x (yz) = (xy) z
Basic Identities of Boolean Algebra (Distributivity):
(13) x ( y + z ) = xy + xz
(14) x + yz = ( x + y )( x + z)
Basic Identities of Boolean Algebra (DeMorgan’s Theorem)
(15) ( x + y )’ = x’ y’
(16) ( xy )’ = x’ + y’
Basic Identities of Boolean Algebra (Involution)
(17) (x’)’ = x
Function Minimization using Boolean
Algebra
• Examples:
(a) a + ab = a(1+b)=a
(b) a(a + b) = a.a +ab=a+ab=a(1+b)=a.
(c) a + a'b = (a + a')(a + b)=1(a + b) =a+b
(d) a(a' + b) = a. a' +ab=0+ab=ab
Try
• F = abc + abc’ + a’c
The other type of questionShow that;
1- ab + ab' = a2- (a + b)(a + b') = a
1- ab + ab' = a(b+b') = a.1=a2- (a + b)(a + b') = a.a +a.b' +a.b+b.b'
= a + a.b' +a.b + 0 = a + a.(b' +b) + 0 = a + a.1 + 0 = a + a = a
More Examples• Show that;
(a) ab + ab'c = ab + ac(b) (a + b)(a + b' + c) = a + bc
(a) ab + ab'c = a(b + b'c) = a((b+b').(b+c))=a(b+c)=ab+ac
(b) (a + b)(a + b' + c) = (a.a + a.b' + a.c + ab +b.b' +bc)
= …
DeMorgan's Theorem
(a) (a + b)' = a'b'
(b) (ab)' = a' + b'
Generalized DeMorgan's Theorem
(a) (a + b + … z)' = a'b' … z'
(b) (a.b … z)' = a' + b' + … z‘
DeMorgan's Theorem
• F = ab + c’d’
• F’ = ??
• F = ab + c’d’ + b’d
• F’ = ??
DeMorgan's Theorem
Show that: (a + b.c)' = a'.b' + a'.c'
More DeMorgan's exampleShow that: (a(b + z(x + a')))' =a' + b' (z' + x')
(a(b + z(x + a')))' = a' + (b + z(x + a'))' = a' + b' (z(x + a'))' = a' + b' (z' + (x + a')') = a' + b' (z' + x'(a')') = a' + b' (z' + x'a)=a‘+b' z' + b'x'a=(a‘+ b'x'a) + b' z' =(a‘+ b'x‘)(a +a‘) + b' z' = a‘+ b'x‘+ b' z‘ = a' + b' (z' + x')
More Examples
(a(b + c) + a'b)'=b'(a' + c')
ab + a'c + bc = ab + a'c
(a + b)(a' + c)(b + c) = (a + b)(a' + c)
Simplification of Switching Functions
Karnaugh Maps (K-Map)
A K-Map is a graphical representation of a logic function’s truth table
Two-Variable K-Map
0 1
0
1
ab
Three-Variable K-Map
abc 00 01 11 10
0
1
0m
1m
2m
3m
6m
7m
4m
5m
Three-Variable K-Map
abc 00 01 11 10
0
1
Three-Variable K-Map
abc 00 01 11 10
0
1
Edges are adjacent
Four-variable K-Map
abcd 00 01 11 10
00
01
11
10
0m
1m
2m
3m
6m
7m
4m
5m
12m
13m
14m
15m
10m
11m
8m
9m
Four-variable K-Mapab
cd 00 01 11 10
00
01
11
10
Four-variable K-Mapab
cd 00 01 11 10
00
01
11
10
Edges are adjacent
Edg
es a
re a
djac
ent
Plotting Functions on the K-map
SOP Form
Canonical SOP Form
Three Variable Example
F ABC ABC ABC ABC
using shorthand notation
6 3 1 5F m m m m
, , 1,3,5,6F A B C m
Three-Variable K-Map Example
abc 00 01 11 10
0
1
, , 1,3,5,6F a b c m
Plot 1’s (minterms) of switching function
1 1 1
1
Three-Variable K-Map Example
abc 00 01 11 10
0
1
, ,F a b c ab bc
Plot 1’s (minterms) of switching function
1 1 1
1 abbc
Four-variable K-Map Example
abcd 00 01 11 10
00
01
11
10
, , , 0, 2,9,12,14F a b c d m
1
1
1
1
1
Karnaugh Maps (K-Map)
Simplification of Switching Functions
using K-MAPS
Terminology/Definition• Literal
– A variable or its complement
• Logically adjacent terms– Two minterms are logically adjacent if
they differ in only one variable position– Ex: abc abcand
m6 and m2 are logically adjacent
Note: abc abc a a bc bc Or, logically adjacent terms can be combined
Terminology/Definition
• Implicant– Product term that could be used to cover minterms of
a function• Prime Implicant
– An implicant that is not part of another implicant• Essential Prime Implicant
– An implicant that covers at least one minterm that is not contained in another prime implicant
• Cover– A minterm that has been used in at least one group
Guidelines for Simplifying Functions
• Each square on a K-map of n variables has n logically adjacent squares. (i.e. differing in exactly one variable)
• When combing squares, always group in powers of 2m , where m=0,1,2,….
• In general, grouping 2m variables eliminates m variables.
Guidelines for Simplifying Functions
• Group as many squares as possible. This eliminates the most variables.
• Make as few groups as possible. Each group represents a separate product term.
• You must cover each minterm at least once. However, it may be covered more than once.
K-map Simplification Procedure
• Plot the K-map• Circle all prime implicants on the K-
map• Identify and select all essential prime
implicants for the cover.• Select a minimum subset of the
remaining prime implicants to complete the cover.
• Read the K-map
Example
• Use a K-Map to simplify the following Boolean expression
, , 1, 2,3,5,6F a b c m
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 1: Plot the K-map
1 1 1
1
, , 1, 2,3,5,6F a b c m
1
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 2: Circle ALL Prime Implicants
1 1 1
1
, , 1, 2,3,5,6F a b c m
1
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 3: Identify Essential Prime Implicants
1 1 1
1
, , 1, 2,3,5,6F a b c m
1
EPI
EPI
PI
PI
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 4: Select minimum subset of remaining Prime Implicants to complete the cover.
1 1 1
1
, , 1, 2,3,5,6F a b c m
1
EPIPI
EPI
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 5: Read the map.
1 1 1
1
, , 1, 2,3,5,6F a b c m
1
bcab
bc
Solution
, ,F a b c ab bc bc ab b c
Example
• Use a K-Map to simplify the following Boolean expression
, , 2,3,6,7F a b c m
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 1: Plot the K-map
11
11
, , 2, 4,5,7F a b c m
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 2: Circle Prime Implicants
11
11
, , 2,3,6,7F a b c m
Wrong!!We reallyshould drawA circle aroundall four 1’s
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 3: Identify Essential Prime Implicants
EPIEPI
, , 2,3,6,7F a b c m
11
11Wrong!!We reallyshould drawA circle aroundall four 1’s
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 4: Select Remaining Prime Implicants to complete the cover.
EPIEPI
11
11
, , 2,3,6,7F a b c m
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 5: Read the map.
abab
11
11
, , 2,3,6,7F a b c m
Solution
, ,F a b c ab ab b
Since we can still simplify the functionthis means we did not use the largestpossible groupings.
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 2: Circle Prime Implicants
11
11
, , 2,3,6,7F a b c m
Right!
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 3: Identify Essential Prime Implicants
EPI
, , 2,3,6,7F a b c m
11
11
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 5: Read the map.
b
11
11
, , 2,3,6,7F a b c m
Solution
, ,F a b c b
Special Cases
Three-Variable K-Map Example
abc 00 01 11 10
0
1 1 1 1
1
, , 1F a b c
11
1
1
Three-Variable K-Map Example
abc 00 01 11 10
0
1
, , 0F a b c
Three-Variable K-Map Example
abc 00 01 11 10
0
1 1
, ,F a b c a b c
1
1
1
Four Variable Examples
Example
• Use a K-Map to simplify the following Boolean expression
, , , 0, 2,3,6,8,12,13,15F a b c d m
Four-variable K-Mapab
cd 00 01 11 10
00
01
11
10
, , , 0, 2,3,6,8,12,13,15F a b c d m
1
1
1
1
11
1
1
Four-variable K-Mapab
cd 00 01 11 10
00
01
11
10
0,2,3,6,8,12,13,15F m
1
1
1
1
11
1
1
Four-variable K-Mapab
cd 00 01 11 10
00
01
11
10
F abd abc acd abd acd
1
1
1
1
11
1
1
Example
• Use a K-Map to simplify the following Boolean expression
, , , 0, 2,6,8,12,13,15
3,9,10
F a b c d m
d
D=Don’t care (i.e. either 1 or 0)
Four-variable K-Mapab
cd 00 01 11 10
00
01
11
10
1
1
d
1
11
1
1
, , , 0, 2,6,8,12,13,15 3,4,9F a b c d m d
d
d
Four-variable K-Mapab
cd 00 01 11 10
00
01
11
10
1
1
d
1
11
1
1
F ac ad abd
d
d
Five Variable K-Maps
, , , ,F a b c d e
Five variable K-map
A=1
A=0
Use two four variable K-maps
Use Two Four-variable K-Maps
bcde 00 01 11 10
00
01
11
10
bcde 00 01 11 10
00
01
11
10
A=0 map A=1 map
Five variable example
, , , , 5,7,13,15,21,23,29,31F a b c d e m
Use Two Four-variable K-Maps
bcde 00 01 11 10
00
01
11
10
bcde 00 01 11 10
00
01
11
10
A=0 map A=1 map
, , , , 5,7,13,15,21,23,29,31F a b c d e m
1
1
1
1
1
1
1
1
Use Two Four-variable K-Maps
bcde 00 01 11 10
00
01
11
10
bcde 00 01 11 10
00
01
11
10
A=0 map A=1 map
1
1
1
1
1
1
1
1
1F a ce 2F a ce
Five variable example
1 2F F F a ce a ce ce
Plotting POS Functions
K-map Simplification Procedure
• Plot the K-map for the function F• Circle all prime implicants on the K-map• Identify and select all essential prime implicants
for the cover.• Select a minimum subset of the remaining
prime implicants to complete the cover. • Read the K-map• Use DeMorgan’s theorem to convert F to F in
POS form
Example
• Use a K-Map to simplify the following Boolean expression
, , 1, 2,3,5,6F a b c M
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 1: Plot the K-map of F
1 1 1
11
, , 1, 2,3,5,6F a b c M
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 2: Circle ALL Prime Implicants
1 1 1
11
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 3: Identify Essential Prime Implicants
1 1 1
11
EPI
EPI
PI
PI
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 4: Select minimum subset of remaining Prime Implicants to complete the cover.
1 1 1
11
EPIPI
EPI
Three-Variable K-Map Example
abc 00 01 11 10
0
1
Step 5: Read the map.
1 1 1
11
bcab
bc
Solution
F ab bc bc
F ab bc bc
a b b c b c
, , 1, 2,3,5,6F a b c M
SOP and POS Forms
• SOP Given a Table of Combinations
– What is the SOP form for the following 3 input / 1 output digital device?
S A B f
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 0
1 1 1 1
• Computing the SOP (2)
– This SOP has 4 minterms:• f = S'AB' + S'AB + SA'B + SAB
S A B f minterm name
0 1 0 1 m2
0 1 1 1 m3
1 0 1 1 m5
1 1 1 1 m7
• Canonical SOP
– Boolean functions can use shorthand notation when in SOP form:
• f = S'AB' + S'AB + SA'B + SAB
f(S,A,B) = (m2,m3,m5,m7)
or
f(S,A,B) = m(2,3,5,7)
• Canonical SOP Example
– f(x1,x2,x3) = m(1,4,5,6)
– f =
minterm x1 x2 x3 f
0 0 0 0 0
1 0 0 1 1
2 0 1 0 0
3 0 1 1 0
4 1 0 0 1
5 1 0 1 1
6 1 1 0 1
7 1 1 1 0
x1'x2'x3 + x1x2'x3' + x1x2'x3 + x1x2x3'
• Product of Sums Form
– An alternate canonical “two-level” format
• “Product of sums” POS
• Two levels– OR level followed by AND level
– Again, NOT doesn’t count as a level
• Not a common as SOP, but can be useful in some situations– Which ones?
• Computing the POS
– Identify rows with “0” on output (f = 0)
– Represent the input for each 0 row as a maxterm• A logical “sum” of the input bits which guarantees that term
will be “0” (sum of literals)
A B f
0 0 0
0 1 1
1 0 0
1 1 0
• Canonical POS Example
– f(x1,x2,x3) = (M0,M2,M3,M7) = M(0,2,3,7)
– f =
maxterm x1 x2 x3 f
0 0 0 0 0
1 0 0 1 1
2 0 1 0 0
3 0 1 1 0
4 1 0 0 1
5 1 0 1 1
6 1 1 0 1
7 1 1 1 0
(x1+x2+x3)(x1+x2'+x3)(x1+x2'+x3')(x1'+x2'+x3')
NAND/NOR Circuits
• Completeness of NAND
• Any Boolean function can be implemented using just NAND gates. Why?– Need AND, OR, and NOT– NOT: 1-input NAND (or 2-input NAND with inputs tied
together)– AND: NAND followed by NOT– OR: NAND preceded by NOTs
• Likewise for NOR
• Using NAND as Universal Logic
– NOT
– AND
– OR
• SOP Using NORs & POS Using NANDs
– NANDs are natural for SOP networks• You can extend this idea to multi-level circuits as long as the
levels alternate AND/OR/AND/OR ending with OR• You can implement an SOP circuit using only NOR gates
– All gates become NORs; just add an extra “inverter” following the final NOR
– NORs are natural for POS networks• You can extend this idea to multi-level circuits as long as the
levels alternate OR/AND/OR/AND ending with AND• You can implement a POS circuit using only NAND gates
– All gates become NANDs; just add an extra inverter following the final NAND
• SOP Using NAND Networks
– SOP can be implementedwith just NAND gates
• “pushing the bubbles”• Every gate just becomes
a NAND!
x 1 x 2
x 3 x 4 x 5
x 1 x 2
x 3 x 4 x 5
x 1 x 2
x 3 x 4 x 5
• 2x1 MUX Using NANDs
– Implement f = S'A + SB with NAND gates only
– This one is complicated by the inverter on S!
• POS Using NOR Networks
– POS can be implementedwith just NOR gates
• Every gate just becomesa NOR
x 1
x 2
x 3
x 4
x 5
x 1
x 2
x 3
x 4
x 5
x 1
x 2
x 3
x 4
x 5
• Schematics of DeMorgan’s Laws
(x ∙ y)' = x' + y'
(x + y)' = x' ∙ y'
• Universal Logic Families
– Any logic function can be designed using only:
• AND, OR, NOT• NAND• NOR
– These are called “universal logic families”
– Actual components are often designed using either NAND or NOR gates only
• NAND and NOR require fewer transistors to build• Just having a single gate design is simpler than having 3!
• AND/OR Networks NAND/NAND
– Convert multi-level AND/OR net NAND/NAND
• And Again … But Be Careful
conserve the polarity of the input/output signals
Logic GatesLogic Gates
AND FunctionOutput Y is TRUE if inputs A AND B are TRUE, else it is FALSE.
Logic Symbol
Text Description
Truth Table
Boolean Expression
ANDA
BY
INPUTS OUTPUT
A B Y 0 0 0 0 1 0 1 0 0 1 1 1
AND Gate Truth Table
Y = A x B = A • B = AB
AND Symbol
OR FunctionOutput Y is TRUE if input A OR B is TRUE, else it is FALSE.
Logic Symbol
Text Description
Truth Table
Boolean Expression Y = A + B
OR Symbol
A
BYOR
INPUTS OUTPUT
A B Y 0 0 0 0 1 1 1 0 1 1 1 1
OR Gate Truth Table
NOT Function (inverter)Output Y is TRUE if input A is FALSE, else it is FALSE. Y is the inverse of A.
Logic Symbol
Text Description
Truth Table
Boolean Expression
INPUT OUTPUT
A Y 0 1 1 0
NOT Gate Truth Table
A YNOT
NOT Bar
Y = AY = A’
Alternative Notation
Y = !A
NAND FunctionOutput Y is FALSE if inputs A AND B are TRUE, else it is TRUE.
Logic Symbol
Text Description
Truth Table
Boolean Expression
A
BYNAND
A bubble is an inverterThis is an AND Gate with an inverted output
Y = A x B = AB
INPUTS OUTPUT
A B Y 0 0 1 0 1 1 1 0 1 1 1 0
NAND Gate Truth Table
NOR FunctionOutput Y is FALSE if input A OR B is TRUE, else it is TRUE.
Logic Symbol
Text Description
Truth Table
Boolean Expression Y = A + B
A
BYNOR
A bubble is an inverter.This is an OR Gate with its output inverted.
INPUTS OUTPUT
A B Y 0 0 1 0 1 0 1 0 0 1 1 0
NOR Gate Truth Table
Circuit-to-Truth Table Example
OR
A
Y
NOT
ANDB
CAND
2# of Inputs = # of Combinations
2 3 = 8
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
A B C Y
Circuit-to-Truth Table Example
OR
A
Y
NOT
ANDB
CAND
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
A B C Y
0
0
0
0
10
0
0
Circuit-to-Truth Table Example
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
A B C Y
0
OR
A
Y
NOT
ANDB
CAND
0
0
1
0
11
1
1
Circuit-to-Truth Table Example
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
A B C Y
010
OR
A
Y
NOT
ANDB
CAND
0
1
0
0
10
0
0
Circuit-to-Truth Table Example
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
A B C Y
010
0
OR
A
Y
NOT
ANDB
CAND
0
1
1
0
11
1
1
Circuit-to-Truth Table Example
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
A B C Y
0101
0
OR
A
Y
NOT
ANDB
CAND
1
0
0
0
00
0
0
Circuit-to-Truth Table Example
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
A B C Y
01010
0
OR
A
Y
NOT
ANDB
CAND
1
0
1
0
00
0
0
Circuit-to-Truth Table Example
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
A B C Y
010100
0
OR
A
Y
NOT
ANDB
CAND
1
1
0
1
00
1
1
Circuit-to-Truth Table Example
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
A B C Y
0101001
0
OR
A
Y
NOT
ANDB
CAND
1
1
1
1
00
1
1
Circuit-to-Boolean Equation
OR
A
Y
NOT
ANDB
CAND
A B
A C
A= A B + A C
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
A B C Y
0
0
00
0
11
}
1
1
}
A - O - I Logic
OR
A
Y
NOT
ANDB
CAND
AND Gates
INVERTER Gates
OR GatesOther Logic Arrangements:
NAND - NAND LogicNOR - NOR Logic
NAND Gate – Special Application
INPUTS OUTPUT
A B Y0 0 10 1 11 0 11 1 0
A
BYNAND
TNANDS
S T
00
1
0 1
1 0
Equivalent To An Inverter Gate
NOR Gate - Special Application
S T
00
1
0 1
1 0
Equivalent To An Inverter Gate
TS NOR
A
BYNOR
INPUTS OUTPUT
A B Y0 0 10 1 01 0 01 1 0