PPT

Report
EELE 367 – Logic Design
Module 1 – Classic Digital Design
•
Agenda
1. Number Systems
2. Combinational Logic
3. Sequential Logic
Number Systems
• Base Notation We’ll Use
- We will use the same notation as the HC12 Assembler.
Decimal:
Binary:
Octal:
Hexadecimal:
ASCII:
nothing
%
@
$
‘
ex)
ex)
ex)
ex)
ex)
11
%1011
@13
$BB
‘a
Base 10 = 10 Symbols
Base 2 = 2 Symbols
Base 4 = 4 Symbols
Base 16 = 16 Symbols
Code for Text
Module 1: Classic Digital Design
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Number Systems
• Base Conversion – Binary to Decimal
- each digit has a “weight” of 2n that depends on the position of the digit
- multiply each digit by its “weight”
- sum the resultant products
ex) Convert %1011 to decimal
23
%1
22
0
21
1
20
1
(weight)
= 1•(23) + 0• (22) + 1• (21) + 1• (20)
= 1•(8) + 0• (4) + 1• (2) + 1• (1)
= 8
+ 0
+ 2
+ 1
= 11 Decimal
Module 1: Classic Digital Design
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Number Systems
• Base Conversion – Binary to Decimal with Fractions
- the weight of the binary digits have negative positions
ex) Convert %1011.101 to decimal
23
%1
22
0
21
1
20
1
= 1•(23) + 0• (22) + 1• (21) + 1• (20)
= 1•(8) + 0• (4) + 1• (2) + 1• (1)
= 8
+ 0
+ 2
+ 1
= 11.625 Decimal
2-1
. 1
2-2
0
2-3
1
+ 1• (2-1) + 0• (2-2) + 1• (2-3)
+ 1• (0.5) + 0• (0.25) + 1• (0.125)
+ 0.5
+ 0
+ 0.125
Module 1: Classic Digital Design
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Number Systems
• Base Conversion – Decimal to Binary
- the decimal number is divided by 2, the remainder is recorded
- the quotient is then divided by 2, the remainder is recorded
- the process is repeated until the quotient is zero
ex) Convert 11 decimal to binary
Quotient
2
2
2
2
11
5
2
1
5
2
1
0
Remainder
1
1
0
1
LSB
MSB
= %1011
Module 1: Classic Digital Design
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Number Systems
• Base Conversion – Decimal to Binary with Fractions
- the fraction is converted to binary separately
- the fraction is multiplied by 2, the 0th position digit is recorded
- the remaining fraction is multiplied by 2, the 0th digit is recorded
- the process is repeated until the fractional part is zero
ex) Convert 0.375 decimal to binary
Product
0.375•2
0.75•2
0.5•2
0.75
1.50
1.00
0th Digit
0
1
1
MSB
LSB
0.375 decimal = % .011
finished
Module 1: Classic Digital Design
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Number Systems
• Base Conversion – Hex to Decimal
- the same process as “binary to decimal” except the weights are now BASE 16
- NOTE ($A=10, $B=11, $C=12, $D=13, $E=14, $F=15)
ex) Convert $2BC to decimal
162 161
$2
B
160
C
(weight)
= 2• (162) + B• (161) + C• (160)
= 2•(256) + 11• (16) + 12• (1)
= 512
+ 176
+ 12
= 700 Decimal
Module 1: Classic Digital Design
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Number Systems
• Base Conversion – Hex to Decimal with Fractions
- the fractional digits have negative weights (BASE 16)
- NOTE ($A=10, $B=11, $C=12, $D=13, $E=14, $F=15)
ex) Convert $2BC.F to decimal
162 161
$2
B
160
16-1
C . F
(weight)
= 2• (162) + B• (161) + C• (160) + F• (16-1)
= 2•(256) + 11• (16) + 12• (1) + 15• (0.0625)
= 512
+ 176
+ 12
+ 0.938
= 700.938 Decimal
Module 1: Classic Digital Design
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Number Systems
• Base Conversion – Decimal to Hex
- the same procedure is used as before but with BASE 16 as the divisor/multiplier
ex) Convert 420.625 decimal to hex
1st, convert the integer part…
Quotient
16 420
16 26
16 1
Remainder
26
1
0
4
10
1
LSB
MSB
= $1A4
2nd, convert the fractional part…
Product
0.625•16
0th Digit
10.00
10
MSB
= $ .A
420.625 decimal = $1A4.A
Module 1: Classic Digital Design
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Number Systems
• Base Conversion – Octal to Decimal / Decimal to Octal
- the same procedure is used as before but with BASE 8 as the divisor/multiplier
Module 1: Classic Digital Design
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Number Systems
• Base Conversion – Hex to Binary
- each HEX digit is made up of four binary bits
ex) Convert $ABC to binary
$A
= % 1010
B
1011
C
1100
= %1010 1011 1100
Module 1: Classic Digital Design
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Number Systems
• Base Conversion – Binary to Hex
- every 4 binary bits for one HEX digit
- begin the groups of four at the LSB
- if necessary, fill the leading bits with 0’s
ex) Convert $ABC to binary
= % 11 0010 1111
$3
2
F
Module 1: Classic Digital Design
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Number Systems
• Binary Addition
- same as BASE 10 addition
- need to keep track of the carry bit for a given system size (n), i.e., 4-bit, 8-bit, 16-bit,…
ex) Add %1011 and $1001
1
1
% 1011
% 1001
+________
1 0100
Carry Bit
Module 1: Classic Digital Design
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Number Systems
• Two’s Complement
- we need a way to represent negative numbers in a computer
- this way we can use “adding” circuitry to perform subtraction
- since the number of bits we have is fixed (i.e., 8), we use the MSB as a sign bit
Positive #’s :
Negative #’s :
MSB = 0
MSB = 1
(ex: %0000 1111, positive number)
(ex: %1000 1111, negative number)
- the range of #’s that a two’s compliment code can represent is:
(-2n-1) < N < (2n-1 – 1) :
n = number of bits
ex) What is the range of #’s that an 8-bit two’s compliment code can represent?
(-28-1) < N < (28-1 – 1)
(-128) < N < (+127)
Module 1: Classic Digital Design
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Number Systems
• Two’s Complement Negation
- to take the two’s compliment of a positive number (i.e., find its negative equivalent)
Nc = 2n – N
Nc = Two’s Compliment
N = Original Positive Number
ex) Find the 8-bit two’s compliment representation of –52dec
N = +52dec
Nc = 28 – 52 = 256 – 52 = 204 = %1100 1100
Note the Sign Bit
Module 1: Classic Digital Design
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Number Systems
• Two’s Complement Negation (a second method)
- to take the two’s compliment of a positive number (i.e., find its negative equivalent)
1) Invert all the bits of the original positive number (binary)
2) Add 1 to the result
ex) Find the 8-bit two’s compliment representation of –52dec
N = +52dec
= %0011 0100
Invert:
Add 1
= %1100 1011
= %1100 1011
+
1
-------------------%1100 1100
(-52dec)
Module 1: Classic Digital Design
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Number Systems
• Two’s Complement Addition
- Addition of two’s compliment numbers is performed just like
standard binary addition
- However, the carry bit is ignored
Module 1: Classic Digital Design
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Number Systems
• Two’s Compliment Subtraction
- now we have a tool to do subtraction using the addition algorithm
ex) Subtract 8dec from 15dec
15dec = %0000 1111
8dec = %0000 1000 -> two’s compliment -> invert %1111 0111
add1 %1111 0111
+
1
----------------%1111 1000 = -8dec
Now Add: 15 + (-8) =
%0000 1111
%1111 1000
+_________
% 1 0000 0111 = 7dec
Disregard Carry
Module 1: Classic Digital Design
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Number Systems
• Two’s Compliment Overflow
- If a two’s compliment subtraction results in a number that is outside
the range of representation (i.e., -128 < N < +127), an “overflow”
has occurred.
ex) -100dec – 100dec- = -200dec (can’t represent)
- There are three cases when overflow occurs
1) Sum of like signs results in answer with opposite sign
2) Negative – Positive = Positive
3) Positive – Negative = Negative
- Boolean logic can be used to detect these situations.
Module 1: Classic Digital Design
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Number Systems
• Binary Coded Decimal
- Sometimes we wish to represent an individual decimal digit as
a binary representation (i.e., 7-segment display to eliminate a decoder)
- We do this by using 4 binary digits.
Decimal
0
1
2
3
4
5
6
7
8
9
BCD
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
ex) Represent 17dec
Binary = %10111
BCD = %0001 0111
Module 1: Classic Digital Design
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Number Systems
• ASCII
- American Standard Code for Information Interchange
- English Alphanumeric characters are represented with a 7-bit code
ex) ‘A’ = $40
‘a’ = $61
Module 1: Classic Digital Design
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Number Systems
• Important Things to Remember
Know the Code :
A computer just knows 1’s and 0’s. It is up to you to keep trace of whether
the bits represent unsigned numbers, two’s complement, ASCII, etc…
Two’s Complement Size :
2’s Complement always need a “size” associated with it. We always say
“n-bit, two’s complement”
Module 1: Classic Digital Design
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Basic Gates
Combinational Logic
Combinational Logic Gates :
- Output depends on the logic value of the inputs
- no storage
Module 1: Classic Digital Design
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Basic Gates
NOT
out = in’ = in
f(in) = in’ = in
OR
out = a+b
f(a,b) = a+b
AND
out = a·b
f(a,b) = a·b
Module 1: Classic Digital Design
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Basic Gates
XOR
out = ab
f(a,b) = ab
NOR
out = a+b
f(a,b) = a+b
NAND
out = a·b
f(a,b) = a·b
Module 1: Classic Digital Design
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Basic Gates
XNOR
out = ab
f(a,b) = ab
Also remember about XOR Gates:
f(a,b) = ab = (a’b + b’a)
Also remember the priority of logic operations
NOT, AND, OR
Module 1: Classic Digital Design
(without parenthesis) is:
26
Boolean Algebra
• Boolean Algebra
- formulated by mathematician George Boole in 1854
- basic relationships & manipulations for a two-value system
• Switching Algebra
- adaptation of Boolean Logic to analyzer and describe behavior of relays
- Claude Shannon of Bell Labs in 1938
- this works for all switches (mechanical or electrical)
- we generally use the terms "Boolean Algebra" & "Switching Algebra" interchangeably
Module 1: Classic Digital Design
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Boolean Algebra
• What is Algebra
- the basic set of rules that the elements and operators in a system follow
- the ability to represent unknowns using variables
- the set of theorems available to manipulate expressions
• Boolean
- we limit our number set to two values (0, 1)
- we limit our operators to AND, OR, INV
Module 1: Classic Digital Design
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Boolean Algebra
• Axioms
- also called "Postulates"
- minimal set of basic definitions that we assume to be true
- all other derivations are based on these truths
- since we only have two values in our system, we typically define an axiom and then its
complement (A1 & A1')
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Boolean Algebra
• Axiom #1 "Identity"
- a variable X can only take on 1 or 2 values (0 or 1)
- if it isn't a 0, it must be a 1
- if it isn't a 1, it must be a 0
(A1) X = 0, if X ≠ 1
(A1') X = 1, if X ≠ 0
• Axiom #2 "Complement"
- a prime following a variable denotes an inversion function
(A2) if X = 0, then X' = 1
(A2') if X = 1, then X' = 0
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Boolean Algebra
• Axiom #3 "AND"
- also called "Logical Multiplication"
- a dot (·) is used to represent an AND function
• Axiom #4 "OR"
- also called "Logical Addition"
- a plus (+) is used to represent an OR function
• Axiom #5 "Precedence"
- multiplication precedes addition
(A3) 0·0 = 0
(A4) 1·1 = 1
(A5) 0·1 = 1·0 = 0
(A3') 1+1 = 1
(A4') 0+0 = 0
(A5') 0+1 = 1+0 = 1
Module 1: Classic Digital Design
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Boolean Algebra
• Theorems
- Theorems use our Axioms to formulate more meaningful relationships & manipulations
- a theorem is a statement of TRUTH
- these theorems can be proved using our Axioms
- we can prove most theorems using "Perfect Induction"
- this is the process of plugging in every possible input combination and observing the
output
Module 1: Classic Digital Design
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Boolean Algebra
•
Theorem #1 "Identity"
(T1) X+0 = X
•
Theorem #2 "Null Element"
(T2) X+1 = 1
•
(T2') X·0 = 0
Theorem #3 "Idempotency"
(T3) X+X = X
•
(T1') X·1 = X
(T3') X·X = X
Theorem #4 "Involution"
(T4) (X')' = X
•
Theorem #5 "Complements"
(T5) X+X' = 1
(T5') X·X' = 0
Module 1: Classic Digital Design
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Boolean Algebra
• Theorem #6 "Commutative"
(T6) X+Y = Y+X
(T6') X·Y = Y·X
• Theorem #7 "Associative"
(T7) (X+Y)+Z= X+(Y+Z)
(T7') (X · Y) · Z= X · (Y · Z)
• Theorem #8 "Distributive"
(T8) X·(Y+Z) = X·Y + X·Z
(T8') (X+Y)·(X+Z) = X + Y·Z
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Boolean Algebra
• Theorem #9 "Covering"
(T9) X + X·Y = X
(T9') X·(X+Y) = X
• Theorem #10 "Combining"
(T10) X·Y + X·Y' = X
(T10') (X+Y)·(X+Y') = X
• Theorem #11 "Consensus"
(T11) X·Y + X'·Z + Y·Z= X·Y + X'·Z
(T11') (X+Y)·(X'+Z)·(Y+Z) = (X+Y) ·(X'+Z)
Module 1: Classic Digital Design
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Boolean Algebra
• Notes on the Theorems
- T9/T9' and T10/T10' are used heavily in logic minimization
- these theorems can be useful for making routing more reasonable
- these theorems can reduce the number of gates in a circuit
- they can also change the types of gates that are used
Module 1: Classic Digital Design
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Boolean Algebra
• More Theorem's
- there are more generalized theorems available for large number of variables
T13, T14, T15
- one of the most useful is called "DeMorgan's Theorem"
• DeMorgan's Theorem
- this theorem states a method to convert between AND and OR gates using inversions on the
input / output
Module 1: Classic Digital Design
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Boolean Algebra
• DeMorgan's Theorem
Part 1: an AND gate whose output is complemented is equivalent to
an OR gate whose inputs are complemented
=
Part 2: an OR gate whose output is complemented is equivalent to
an AND gate whose inputs are complemented
=
Module 1: Classic Digital Design
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Boolean Algebra
• Complement
- complementing a logic function will give outputs that are inverted versions of the original function
ex)
AB
0 0
0 1
1 0
1 1
F
0
0
0
1
F'
1
1
1
0
- DeMorgan's Theorem also gives us a generic formula to complement any expression:
- for a Logic function F, we can get F' by :
1) Swapping all + and ·
2) Complementing all Variables
- KEEP THE PARENTHESIS ORDER OF THE ORGINAL FUNCTION !!!
Module 1: Classic Digital Design
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Boolean Algebra
• Complement
- Example: Complement the Function F
ex)
AB
0 0
0 1
1 0
1 1
F
0
0
0
1
F'
1
1
1
0
We know:
F=A·B
We swap + and · first, then we complement all variables
F' = A' + B'
- This is the same as putting an inversion
bubble on the output of the logic diagram
Module 1: Classic Digital Design
→
40
Boolean Algebra
• Duality
- An Algorithm to switch between Positive Logic and Negative Logic
- Duality means that the logic expression is exactly the same even though the circuitry has been
altered to produce Complementary Logic
- The steps are:
- for a Logic function F, we can get FD by :
1) Swapping all + and ·
2) Swapping all 0's and 1's
- Ex)
F=A·B
(Positive Logic)
We swap + and · first, then swap any 0's and 1's
FD = A + B
(Negative Logic or "Dual")
Module 1: Classic Digital Design
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Boolean Algebra
• Complement vs. Duality, What is the difference?
Module 1: Classic Digital Design
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Minterms
• Truth Tables
Row
0
1
2
3
4
5
6
7
ABC
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
F
1
0
0
1
1
0
1
1
Row
We assign a "Row Number" for each entry starting at 0
Variables We enter all input combinations in ascending order.
We use straight binary with the MSB on the left
Function
We say the output is a function of the input variables
F(A,B,C)
n = the number of input variables
2n = the number of input combinations
Module 1: Classic Digital Design
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Minterms
• Let's also define the following terms
Literal
= a variable or the complement of a variable
ex) A, B, C, A', B', C'
Product Term
= a single literal or Logical Product of two or more literals
ex) A
A·B
B'·C
Sum or Products = (SOP), the Logical Sum of Product Terms
ex) A + B
A·B + B'·C
Module 1: Classic Digital Design
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Minterms
• Minterm
- a normal product term w/ n-literals
- a Minterm is written for each ROW in the truth table
- there are 2n Minterms for a given truth table
- we write the literals as follows:
- if the input variable is a 0 in the ROW, we complement the Minterm literal
- if the input variable is a 1 in the ROW, we do not complement the Minterm literal
- for each ROW, we use a Logical Product on all of the literals to create the Minterm
Module 1: Classic Digital Design
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Minterms
• Minterm
Row
0
1
2
3
4
5
6
7
ABC
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Minterm
A'·B'·C'
A'·B'·C
A'·B·C'
A'·B·C
A·B'·C'
A·B'·C
A·B·C'
A·B·C
F
F(0,0,0)
F(0,0,1)
F(0,1,0)
F(0,1,1)
F(1,0,0)
F(1,0,1)
F(1,1,0)
F(1,1,1)
• Canonical Sum
- we Logically Sum all Minterms that correspond to a Logic 1 on the output
- the Canonical Sum represents the entire Logic Expression when the Output is TRUE
- this is called the "Sum of Products" or SOP
Module 1: Classic Digital Design
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Minterms
• Minterm List
- we can also describe the full logic expression using a list of Minterms corresponding to a Logic 1
- we use the Σ symbol to indicate we are writing a Minterm list
- we list the Row numbers corresponding to a Logic 1
Row
0
1
2
3
4
5
6
7
ABC
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Minterm
A'·B'·C'
A'·B'·C
A'·B·C'
A'·B·C
A·B'·C'
A·B'·C
A·B·C'
A·B·C
F = ΣA,B,C (1,2,6)
F
0
1
1
0
0
0
1
0
= (A'·B'·C) + (A'·B·C') + (A·B·C')
- this is also called the "ON-set"
- this list is very verbose and NOT minimized using our Axioms and Theorems (more on this later…)
Module 1: Classic Digital Design
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Maxterms
• Let's define the following terms
Sum Term
= a single literal or a Logical Sum of two or more literals
ex) A
A + B'
Product of Sums = (POS), the Logical Product of Sum Terms
ex) (A+B)·(B'+C)
Normal Term
= a term in which no variable appears more than once
ex) "Normal
A·B
A + B'
ex) "Non-Normal"
A·B·B'
A + A'
Module 1: Classic Digital Design
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Maxterms
• Maxterm
- a Normal Sum Term w/ n-literals
- a Maxterm is written for each ROW in the truth table
- there are 2n Maxterms for a given truth table
- we write the literals as follows:
- if the input variable is a 0 in the ROW, we do not complement the Maxterm literal
- if the input variable is a 1 in the ROW, we complement the Maxterm literal
- for each ROW, we use a Logical Sum on all of the literals to create the Maxterm
Module 1: Classic Digital Design
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Maxterms
• Maxterm
Row
0
1
2
3
4
5
6
7
ABC
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Minterm
A'·B'·C'
A'·B'·C
A'·B·C'
A'·B·C
A·B'·C'
A·B'·C
A·B·C'
A·B·C
Maxterm
A+B+C
A+B+C'
A+B'+C
A+B'+C'
A'+B+C
A'+B+C'
A'+B'+C
A'+B'+C'
F
F(0,0,0)
F(0,0,1)
F(0,1,0)
F(0,1,1)
F(1,0,0)
F(1,0,1)
F(1,1,0)
F(1,1,1)
• Canonical Product
- we Logically Multiply all Maxterms that correspond to a Logic 0 on the output
- the Canonical Product represents the entire Logic Expression when the Output is TRUE
- this is called the "Product of Sums" or POS
Module 1: Classic Digital Design
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Maxterms
• Maxterm List
- we can also describe the full logic expression using a list of Maxterms corresponding to a Logic 0
- we use the π symbol to indicate we are writing a Maxterm list
- we list the Row numbers corresponding to a Logic 0
Row
0
1
2
3
4
5
6
7
ABC
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Minterm
A'·B'·C'
A'·B'·C
A'·B·C'
A'·B·C
A·B'·C'
A·B'·C
A·B·C'
A·B·C
Maxterm
A+B+C
A+B+C'
A+B'+C
A+B'+C'
A'+B+C
A'+B+C'
A'+B'+C
A'+B'+C'
F
0
1
1
0
0
0
1
0
F = πA,B,C (0,3,4,5,7) = (A+B+C) · (A+B'+C') · (A'+B+C) · (A'+B+C') · (A'+B'+C')
- this is also called the "OFF-set”, this list is very verbose and NOT minimized
Module 1: Classic Digital Design
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Maxterms
• Maxterm vs. Minterm
- a Maxterm is the Dual of a Minterm
- this implies an inversion
- however, by writing a POS for when the Maxterm is a Logic 0, we perform another inversion
- these two inversions yield the original logic expression for when the function is 1
- SOP = POS
Module 1: Classic Digital Design
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Maxterms
• Minterms & Maxterm
- we now have 5 ways to describe a Logic Expression
1) Truth Table
2) Minterm List
3) Canonical Sum
4) Maxterm List
5) Canonical Product
- these all give the same information
• Converting Between Minterms and Maxterms
- Minterms and Maxterms are Duals
- this means we can convert between then easily using DeMorgan's duality theorem
- converting a SOP to its dual gives Negative Logic
- converting a POS to its dual gives Negative Logic
Module 1: Classic Digital Design
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Circuit Synthesis
• Circuit Synthesis
- there are 5 ways to describe a Logic Expression
1) Truth Table
2) Minterm List
3) Canonical Sum
4) Maxterm List
5) Canonical Product
- we can directly synthesis circuits from SOP and POS expressions
SOP = AND-OR structure
POS = OR-AND structure
Module 1: Classic Digital Design
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Circuit Synthesis
• Circuit Synthesis
- For the given Truth Table, synthesize the SOP and POS Logic Diagrams
Row
0
1
2
3
Minterm List & SOP
F = ΣA,B (0,1) = A'·B' + A'·B
AB
0 0
0 1
1 0
1 1
Minterm
A'·B'
A'·B
A·B'
A·B
Maxterm
A+B
A+B'
A'+B
A'+B'
F
1
1
0
0
Maxterm List & POS
F = πA,B (2,3) = (A'+B) · (A'+B')
Module 1: Classic Digital Design
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Circuit Synthesis
• Circuit Manipulation
- we can manipulate our Logic Diagrams to give the same logic expression but use
different logic gates
- this can be important when using technologies that:
- only have certain gates (i.e., INV, NAND, NOR)
- have certain gates that are faster than others (i.e., NAND-NAND, NOR-NOR)
- we can convert a SOP/POS logic diagram into a NAND-NAND or NOR-NOR structure
Module 1: Classic Digital Design
56
Circuit Synthesis
• Common Circuit Manipulation
"DeMorgan's"
=
=
Module 1: Classic Digital Design
57
Circuit Synthesis
• Common Circuit Manipulation
"Moving Inversion Bubbles"
Module 1: Classic Digital Design
58
Circuit Synthesis
• Common Circuit Manipulation
"Inserting Double Inversion Bubbles"
Module 1: Classic Digital Design
59
Circuit Synthesis
• Common Circuit Manipulation
"Bubbles can be moved to either side of an Inverter"
Module 1: Classic Digital Design
60
Logic Minimization
• Logic Minimization
- We've seen that we can directly translate a Truth Table into a SOP/POS and in turn a Logic Diagram
- However, this type of expression is NOT minimized
ex)
Row
0
1
2
3
Minterm List & SOP
F = ΣA,B (0,1) = A'·B' + A'·B
AB
0 0
0 1
1 0
1 1
Minterm
A'·B'
A'·B
A·B'
A·B
Maxterm
A+B
A+B'
A'+B
A'+B'
F
1
1
0
0
Maxterm List & POS
F = πA,B (2,3) = (A'+B) · (A'+B')
Module 1: Classic Digital Design
61
Logic Minimization
• Logic Minimization
- using our Axioms and Theorems, we can manually minimize the expressions…
Minterm List & SOP
Maxterm List & POS
F = A'·B' + A'·B
F = (A'+B) · (A'+B')
F = A'·(B'+B) = A'
F = A' + (B'·B) = A'
- doing this by hand can be difficult and requires that we recognize patterns associated with
our 5 Axioms and our 15+ Theorems
• Karnaugh Maps
- a graphical technique to minimize a logic expression
Module 1: Classic Digital Design
62
Logic Minimization
• Karnaugh Map Creation
- we create an array with 2n cells
- each cell contain the value of F at a particular input combination
- we label each Row and Column so that we can easily determine the input combinations
- each cell only differs from its adjacent neighbors by 1-variable
List Variables Top-to-Bottom
List Input Combinations for the
Variables for each Row/Column
A
B
0
1
0
1
Module 1: Classic Digital Design
63
Logic Minimization
• Karnaugh Map Creation
A
A
B
0
1
0
2
1
3
0
B
1
We can put redundant labeling for when a
variable is TRUE. This helps when creating
larger K-maps
We can also put the Truth Table Row
number so that copying the Truth Table
values into the K-map is
straight-forward
Module 1: Classic Digital Design
64
Logic Minimization
• Karnaugh Map Creation
A
A
- we can now copy in the Function values
B
0
0
Row
0
1
2
3
AB
0 0
0 1
1 0
1 1
Minterm
A'·B'
A'·B
A·B'
A·B
Maxterm
A+B
A+B'
A'+B
A'+B'
F
1
1
0
0
2
1
0
1
B
1
1
0
3
1
0
- at this point, the K-map is simply the same
information as in the Truth Table
- we could write a SOP or POS directly from the K-map if we wanted to
Module 1: Classic Digital Design
65
Logic Minimization
• Karnaugh Map Creation
- we can create 3 variable K-Maps
- notice the input combination numbering in order to achieve no more than 1-variable
difference between cells
AB
C
A
00
01
11
10
0
2
6
4
1
3
7
5
0
C
1
B
Module 1: Classic Digital Design
66
Logic Minimization
• Karnaugh Map Creation
- we can create 4 variable K-Maps
AB
CD
A
00
01
11
10
0
4
12
8
1
5
13
9
3
7
15
11
2
6
14
10
00
01
D
11
C
10
B
Module 1: Classic Digital Design
67
Logic Minimization
• Karnaugh Map Minimization
- we can create a minimized SOP logic expression by performing the following:
1) Circle adjacent 1's in groups of power-of-2
- powers of 2 means 1,2,4,8,16,….
- adjacent means neighbors above, below, right, left (not diagonal)
- we can wrap around the ends to form group
- this is called "Combining Cells"
2) We then write a Product Term for each circle following:
- if the circle covers areas where the variable is 0,
we enter a complemented literal in the product term
- if the circle covers areas where the variable is 1,
we enter an non-complemented literal in the product term
- if the circle covers areas where the variable is both a 0 and 1,
we can exclude the variable from the product term
3) We then Sum the Product Terms
Module 1: Classic Digital Design
68
Logic Minimization
• Karnaugh Map Minimization
- let's write a minimized SOP for the following K-map
A
A
B
0
0
2
1
0
1
B
1
1
0
3
1
0
- this circle covers 2 cells
- the circle covers:
- where A=0, so the literal is A'
- where B=0 and 1, so the literal is excluded
- our final SOP expression is
F = A'
Module 1: Classic Digital Design
69
Logic Minimization
• Karnaugh Map Minimization
AB
C
A
00
0
1
1
11
2
0
0
C
01
6
1
4
0
3
1
10
7
0
0
- this circle covers 1 cell
- the circle covers:
- where A=0, so the literal is A'
- where B=1, so the literal is B
- where C=0, so the literal is C'
- The product term for this circle is : A'·B·C'
5
1
1
- this circle covers 2 cells
- the circle covers:
B
- where A=1, so the literal is A
- where B=0 and 1, exclude the literal
- where C=1, so the literal is C
- this circle covers 2 cells
- the circle covers:
- where A=0 and 1, exclude the literal where B=0, so the literal is B'
- where C=1, so the literal is C
- The product term for this circle is : A·C
- our final minimized SOP expression is
- The product term for this circle is : B'·C
Module 1: Classic Digital Design
F = A'·B·C' + A·C + B'·C
70
Logic Minimization
• Karnaugh Map Minimization
- the original Canonical SOP for this Map would have been
F = A'·B'·C + A'·B·C' + A·B'·C + A·B·C
- our minimized SOP expression is now:
F = A'·B·C' + A·C + B'·C
- this minimization resulted in:
- one less input on the OR gate
- one less AND gate
- two AND gates having 2 inputs instead of 3
AB
C
A
00
0
1
1
11
2
0
0
C
01
6
1
4
0
3
1
10
7
0
0
5
1
1
B
Module 1: Classic Digital Design
71
Logic Minimization
• 2-Variable K-Map Example
A
A
- write a minimal SOP for the following truth table
B
0
0
Row
0
1
2
3
AB
0 0
0 1
1 0
1 1
Minterm
A'·B'
A'·B
A·B'
A·B
Maxterm
A+B
A+B'
A'+B
A'+B'
F
0
1
1
1
2
0
0
1
B
1
1
1
3
1
1
- we first copy in the output values into the K-map
Module 1: Classic Digital Design
72
Logic Minimization
• 2-Variable K-Map Example
- we then circle groups of 1's in order to find the Product Term for each circle
A
A
B
0
0
2
0
0
1
B
1
1
- this circle covers 2 cells
- the circle covers:
1
- where A=1, so the literal is A
- where B=0 and 1, exclude the literal
3
1
1
- The product term for this circle is : A
- this circle covers 2 cells
- the circle covers:
- where A=0 and 1, exclude the literal
- where B=1, so the literal is B
- The product term for this circle is : B
Module 1: Classic Digital Design
73
Logic Minimization
• 2-Variable K-Map Example
- we then write a SOP expression for each circles' Product Term
A
A
B
0
0
2
0
0
1
B
1
1
1
3
1
1
The minimized SOP is : A + B
Module 1: Classic Digital Design
74
Logic Minimization
• 3-Variable K-Map Example
- write a minimal SOP expression for the following truth table
Row
0
1
2
3
4
5
6
7
ABC
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Minterm
A'·B'·C'
A'·B'·C
A'·B·C'
A'·B·C
A·B'·C'
A·B'·C
A·B·C'
A·B·C
Maxterm
A+B+C
A+B+C'
A+B'+C
A+B'+C'
A'+B+C
A'+B+C'
A'+B'+C
A'+B'+C'
F
0
1
1
0
1
1
1
1
AB
C
A
00
0
1
1
11
2
0
0
C
01
6
1
4
1
3
1
10
7
0
1
5
1
1
B
Module 1: Classic Digital Design
75
Logic Minimization
• 3-Variable K-Map Example
AB
C
A
00
0
1
1
11
2
0
0
C
01
6
1
4
1
3
1
10
7
0
1
- this circle covers 2 cells
- the circle covers:
- where A=0/1, so exclude literal
- where B=1, so the literal is B
- where C=0, so the literal is C'
- The product term for this circle is : B·C'
5
1
1
- this circle covers 4 cells
- the circle covers:
B
- where A=1, so the literal is A
- where B=0/1, so exclude the literal
- where C=0/1, so exclude the literal
- The product term for this circle is : A
- this circle covers 2 cells
- the circle covers:
- where A=0 and 1, exclude the literal where B=0, so the literal is B'
- where C=1, so the literal is C
- our final minimized SOP expression is
- The product term for this circle is : B'·C
Module 1: Classic Digital Design
F = A + B·C' + B'·C
76
Logic Minimization
• 4-Variable K-Map Example
- write a minimal SOP expression for the following truth table
AB
Row
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
ABCD
00 0 0
00 0 1
00 1 0
00 1 1
01 0 0
01 0 1
01 1 0
01 1 1
10 0 0
10 0 1
10 1 0
10 1 1
11 0 0
11 0 1
11 1 0
11 1 1
F
0
1
0
1
0
0
0
0
1
1
1
1
0
0
0
0
CD
A
00
0
4
0
00
1
11
1
9
0
15
0
6
0
0
0
1
10
8
13
7
2
10
12
5
3
11
0
1
01
C
01
1
0
14
0
D
11
1
10
0
1
B
Module 1: Classic Digital Design
77
Logic Minimization
• 4-Variable K-Map Example
AB
CD
A
00
0
01
4
0
00
11
12
0
10
8
0
- this circle covers 4 cells
- the circle covers:
- where A=1, so the literal is A
- where B=0, so the literal is B'
- where C=0/1, so exclude the literal
- where D=0/1, so exclude the literal
1
- The product term for this circle is : A·B'
1
5
1
01
3
11
C
0
7
1
2
10
13
0
15
0
6
0
9
1
11
0
14
0
1
10
0
1
- this circle covers 4 cells
D - the circle covers:
- where A=0/1, so exclude the literal
- where B=0, so the literal is B'
- where C=0/1, so exclude the literal
- where D=1, so the literal is D
- The product term for this circle is : B'·D
B
- our final minimized SOP expression is
F = A·B' + B'·D
Module 1: Classic Digital Design
78
Logic Minimization
• K-Map Logic Minimization
- we can use K-maps to write a minimal SOP (and POS)
- however, we've seen that there is a potential for redundant Product Terms
AB
C
A
00
0
2
0
0
1
C
1
01
6
0
3
0
11
4
1
7
1
10
1
5
1
0
Is this circle necessary?
- we need to define what it is to be "Minimized"
Module 1: Classic Digital Design
79
Logic Minimization
• K-Map Logic Minimization
Minimal Sum
- No other expression exists that has
- fewer product terms
- fewer literals
Imply
- a logic function P "implies" a function F if
- every input combination that causes P=1
- also causes F=1
- may also cause more 1's
- we say that:
- "F includes P"
- "F covers P"
- "F => P"
Module 1: Classic Digital Design
80
Logic Minimization
• K-Map Logic Minimization
Prime Implicant
- a Normal Product Term of F (i.e., a P that implies F) where if any
variable is removed from P, the resulting product does NOT imply F
- K-maps: a circled set of 1's that cannot be larger without circling
1 or more 0's
Prime Implicant Theorem
- a Minimal Sum is a sum of Prime Implicants
BUT
Does not need to include ALL prime Implicants
Module 1: Classic Digital Design
81
Logic Minimization
• K-Map Logic Minimization
Complete Sum
- the product of all Prime Implicants (not minimized)
Distinguished 1-Cell
- an "input combination" that is covered by only ONE Prime Implicant
Essential Prime Implicant
- a Prime Implicant that covers one or more "Distinguished 1-Cells"
NOTE:
- the sum of Essential Prime Implicants is the Minimal Sum
- this means we're done minimizing
Module 1: Classic Digital Design
82
Logic Minimization
• K-Map Logic Minimization
Steps for Minimization
1) Identify All Prime Implicants
2) Identify the Distinguished 1-Cells
3) Identify the Essential Prime Implicants
4) Create the SOP using Essential Prime Implicants
Module 1: Classic Digital Design
83
Logic Minimization
• Don't Cares
- sometimes it doesn't matter whether the output is a 1 or 0 for a certain input combination
- we can take advantage of this during minimization by treating the don't care as a 1 or 0
depending on whether it makes finding Prime Implicants easier
- We denote X = Don't Care
Row
0
1
2
3
4
5
6
7
ABC
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
Minterm
A'·B'·C'
A'·B'·C
A'·B·C'
A'·B·C
A·B'·C'
A·B'·C
A·B·C'
A·B·C
Maxterm
A+B+C
A+B+C'
A+B'+C
A+B'+C'
A'+B+C
A'+B+C'
A'+B'+C
A'+B'+C'
F
0
0
0
X
1
1
1
X
AB
C
A
00
0
1
1
11
2
0
0
C
01
6
0
4
1
3
0
10
7
X
1
5
X
1
B
Module 1: Classic Digital Design
84
Timing Hazards
• Hazards
- we've only considered the Static (or steady state) values of combination logic
- in reality, there is delay present in the gates
- this can cause different paths through the circuit which arrive at different times at the
input to a gate
- this delay can cause an unwanted transition or "glitch" on the output of the circuit
- this behavior is known as a "Timing Hazard"
- a Hazard is the possibility of an input combination causing a glitch
Module 1: Classic Digital Design
85
Timing Hazards
• Static-1
- when we expect the output to produce a steady 1, but a 0-glitch occurs
- this occurs in SOP (AND-OR) structures
Definition
A pair of input combinations that
(a) differ in only one input variable
(b) both input combinations produce a 1
There is a possibility that a change between these input combinations will cause a 0
Module 1: Classic Digital Design
86
Timing Hazards
• Static-0
- when we expect the output to produce a steady 0, but a 1-glitch occurs
- this occurs in POS (OR-AND) structures
Definition
A pair of input combinations that
(a) differ in only one input variable
(b) both input combinations produce a 0
There is a possibility that a change between these input combinations will cause a 1
Module 1: Classic Digital Design
87
Timing Hazards
• Hazards and K-maps
- K-maps graphically show input combinations that vary by only one variable
- it is easy to see when adjacent cells have 1's and have a potential Timing Hazard
AB
C
A
00
0
1
1
11
2
0
0
C
01
6
0
4
1
3
0
10
7
1
1
5
1
0
B
- this is a Minimal Sum, BUT, what about the transition from A·B·C to A·B·C'?
- there is a Timing Hazard present!!!
Module 1: Classic Digital Design
88
Timing Hazards
• Hazards and K-maps
- the solution is to add an additional product term (Prime Implicant) to cover the transition
- this ensures that the output is valid while transitioning between any input combination
AB
C
A
00
0
1
1
11
2
0
0
C
01
6
0
4
1
3
0
10
7
1
1
5
1
0
B
- this is NOT a Minimal Sum, but it is Hazard Free
Module 1: Classic Digital Design
89
Timing Hazards
• Dynamic Hazards
- when we undergo a transition on the output but multiple transitions occur
- this is again due to multiple paths w/ different delays from input to output
- typically is larger leveled logic
- Solution: If the circuit is Static Hazard Free, then it is Dynamic Hazard Free
Module 1: Classic Digital Design
90
Timing Hazards
• Hazard Prevention
- adding redundant Prime Implicants will prevent Hazards but can sometime add too much logic
- we can also perform delay matching through the circuit by inserting buffers so that the delay
is the same at each level of logic
Module 1: Classic Digital Design
91
Design Flow
• Combinational Logic Design Flow
- We now have all the pieces for a complete design process
1) Design Specifications
: description of what we want to do
2) Truth Table
: listing the logical operation of the system
3) Describe using
SOP/POS/Minterm/Maxterm
: creating the logic expression
4) Logic Minimization
: K-maps
5) Logic Manipulation
: Convert to desired technology (NAND/NAND, …)
6) Hazard Prevention
Module 1: Classic Digital Design
92
Sequential Logic
Sequential Logic
- Concept of “Storage Element”
- With Storage, logic functions can depend on current & past values of inputs
- Sequential State Machines can be created
D-Flip-Flop
- on timing event (i.e., edge of clock input), D input goes to Q output
CLK
D
Q
D
Q
Q
Q
tc2q
Module 1: Classic Digital Design
93
Finite State Machines
• Synchronous
- we now have a way to store information on an edge (i.e., a flip-flop)
- we can use these storage elements to build "Synchronous Circuitry"
- Synchronous means that events occur on the edge of a clock
• State Machine
- a generic name given to sequential circuit design
- it is sequential because the outputs depend on :
1) the current inputs
2) past inputs
- state machines are the basis for all modern digital electronics
Module 1: Classic Digital Design
94
Finite State Machines
• State Memory
- a set of flip-flops that store the Current State of the machine
- the Current State is due to all of the input conditions that have existed since the machine started
- if there are "n" flip-flops, there can be 2n states
- a state contains everything we need to know about all past events
- we define two unique states in a machine
1) Current State
2) Next State
• Current State
- the state that the machine is currently in
- this is found on the outputs of the flip-flops (i.e., Q)
Module 1: Classic Digital Design
95
Finite State Machines
• Next State
- the state that the machine "will go to" upon a clock edge
- this is found on the inputs of the flip-flops (i.e., D)
- the Next State depends on
- the Current State
- any Inputs
- we call the Next State Logic "F"
Module 1: Classic Digital Design
96
Finite State Machines
• State Transition
- upon a clock edge, the machine changes from the "Current State" to the "Next State"
- After the clock edge, we reassign back the names (i.e., Q=Current State, D= Next State)
• State Table
- a table where we list which state the machine will transition to on a clock edge
- this table depends on the "Current State" and the "Inputs"
- we can use the following notation when describing Current and Next States
Current
S
Scur
SC
Acur
QC
Next
S*
Snxt
SN
Bnxt
QN
Module 1: Classic Digital Design
97
Finite State Machines
• State Table cont…
- we typically give the states of our machine descriptive names (i.e., Reset, Start, Stop, S0)
ex) State Table for a 4-state counter, no inputs
Current State
Next State
S0
S1
S1
S2
S2
S3
S3
S0
Module 1: Classic Digital Design
98
Finite State Machines
• State Table cont…
- when there are inputs, we list them in the table
ex) State Table for a 4-state counter with directional input
Current State
Dir
Next State
S0
Up
Down
Up
Down
Up
Down
Up
Down
S1
S3
S2
S0
S3
S1
S0
S2
S1
S2
S3
- we don't need to exhaustively write all of the Current States for each Input combination (i.e., Direction)
which makes the table more readable
Module 1: Classic Digital Design
99
Finite State Machines
• State Variables
- remember that the State Memory is just a set of flip-flops
- also remember that the state is just the current/next binary number on the in/out of the flip-flops
- we use the term State Variable to describe the input/output for a particular flip flop
- let's call our State Variables Q1, Q0, ….
Current State
Q1 Q0
Dir
Next State
Q1* Q0*
0
0
0
1
1
0
1
1
Up
Down
Up
Down
Up
Down
Up
Down
0
1
1
0
1
0
0
1
1
1
0
0
1
1
0
0
Module 1: Classic Digital Design
100
Finite State Machines
• State Variables
- we call the assignment of a State name to a binary number "State Encoding"
ex)
S0 = 00
S1 = 01
S2 = 10
S3 = 11
- this is arbitrary and up to use as designers
- we can choose the encoding scheme based on our application and optimize for
1) Speed
2) Power
3) Area
Module 1: Classic Digital Design
101
Finite State Machines
• Next State Logic "F"
- for each "Next State" variable, we need to create logic circuitry
- this logic circuitry is combinational and has inputs:
1) Current State
2) any Inputs
ex)
Q1* = F(Current State, Inputs) = F(Q1, Q0, Dir)
Q0* = F(Current State, Inputs) = F(Q1, Q0, Dir)
- the logic expression for the Next State Variable is called the "Characteristic Equation"
- this is a generic term that describes the logic for all flip-flops
- when we write the logic for a specific flip-flop, we call this the "Excitation Equation"
- this is a specific term that describes logic for your flip-flop (i.e., DFF)
- there will be an Excitation Equation for each Next State Variable (i.e., each Flip-Flop)
Module 1: Classic Digital Design
102
Finite State Machines
• State Variables and Next State Logic "F"
- we already know how to describe combinational logic (i.e., K-maps, SOP, SOP)
- we simply apply this to our State Table
ex) State Table for a 4-state counter, no inputs
F inputs
Current State
Q1 Q0
Next State
Q1* Q0*
0
0
0
1
0
1
1
0
1
0
1
1
1
1
0
0
Next
State
Variable
F outputs
- we put these inputs and outputs in a K-map and come up with Q1* = Q1  Q0
Module 1: Classic Digital Design
103
Finite State Machines
• State Variables and Next State Logic "F"
- we continue writing Excitation Equations for each Next State Variable in our Machine
- let's now write an Excitation Equation for Q0*
Next
State
Variable
ex) State Table for a 4-state counter, no inputs
F inputs
Current State
Q1 Q0
Next State
Q1* Q0*
0
0
0
1
0
1
1
0
1
0
1
1
1
1
0
0
F outputs
- we simply put these values in a K-map and come up with Q0* = Q0'
Module 1: Classic Digital Design
104
Finite State Machines
• Logic Diagram
- the state machine just described would be implemented like this:
Module 1: Classic Digital Design
105
Finite State Machines
• Excitation Equations
- we designed this State Machine using D-Flip-Flops
- this is the most common type of flip-flop and widely used in modern digital systems
- we can also use other flip-flops
- the difference is how we turn a "Characteristic Equation" into an "Excitation Equation"
- we will look at State Memory using other types of Flip-Flops later
Module 1: Classic Digital Design
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Finite State Machines (Mealy vs. Moore)
• Output Logic "G"
- last time we learned about State Memory
- we also learned about Next State Logic "F"
- the last part of our State Machine is how we create the output circuitry
- the outputs are determined by Combinational Logic that we call "G"
- there are two basic structures of output types
1) Mealy
2) Moore
= the outputs depend on the Current State and the Inputs
= the outputs depend on the Current State only
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Finite State Machines (Mealy vs. Moore)
• State Machines
“Mealy Outputs” – outputs depend on the Current State and the Inputs
- G(Current State, Inputs)
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Finite State Machines (Mealy vs. Moore)
• State Machines
“Moore Outputs” – outputs depend on the Current State only
- G(Current State)
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Finite State Machines (Mealy vs. Moore)
• Output Logic "G"
- we can create an "Output Table" for our desired results
- most often, we include the outputs in our State Table and form a hybrid "State/Output Table"
ex)
Current State
Q1 Q0
In
0 0
0
1
0
1
0
1
0
1
0 1
1 0
1 1
Next State
Q1* Q0*
0
0
1
0
0
1
0
0
0
1
0
0
0
1
0
0
Out
0
0
0
0
0
0
0
1
Module 1: Classic Digital Design
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Finite State Machines (Mealy vs. Moore)
• State Machine Tables
- officially, we use the following terms:
State Table
- list of the descriptive state names and how they transition
Transition Table
- using the explicitly state encoded variables and how they transition
Output Table
- listing of the outputs for all possible combinations of Current States and Inputs
State/Output Table - combined table listing Current/Next states and corresponding outputs
Module 1: Classic Digital Design
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Finite State Machines (Mealy vs. Moore)
• Output Logic "G"
- we simply use the Current State and Inputs (if desired) as the inputs to G
and form the logic expression (K-maps, SOP, POS) reflecting the output variable
- this is the same process as creating the Excitation Equations for the Next State Variables
ex)
Current State
Q1 Q0
In
0 0
0
1
0
1
0
1
0
1
0 1
1 0
G inputs
1 1
Next State
Q1* Q0*
0
0
1
0
0
1
0
0
0
1
0
0
0
1
0
0
Out
0
0
0
0
0
0
0
1
Output
Variable
G outputs
- plugging these inputs/outputs into a K-map, we get G=Q1·Q0·In
Module 1: Classic Digital Design
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Finite State Machines
• State Diagrams
- a graphical way to describe how a state machine transitions states depending on :
- Current State
- Inputs
- we use a "bubble" to represent a state, in which we write its descriptive name or code
- we use a "directed arc" to represent a transition
- we can write the Inputs next to a directed arc that causes a state transition
- we can write the Output either near the directed arc or in the state bubble (typically in parenthesis)
Module 1: Classic Digital Design
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Finite State Machines
• State Diagrams
Ex)
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Finite State Machines
• Timing Diagrams
- notice we don't show any information about the clock in the State Diagram
- the clock is "implied" (i.e., we know transitions occur on the clock edge)
- note that the outputs can change asynchronously in a Mealy machine
Clock
In
Found
STATE
S0
S1
S2
t
Module 1: Classic Digital Design
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Finite State Machines
• State Machines
- there is a basic structure for a Clocked, Synchronous State Machine
1) State Memory
2) Next State Logic “F”
3) Output Logic “G”
(i.e., flip-flops)
(combinational logic)
(combinational logic)
Module 1: Classic Digital Design
116
Finite State Machines
• State Machines
- the steps in a state machine design are:
1) Word Description of the Problem
2) State Diagram
3) State/Output Table
4) State Variable Assignment
5) Choose Flip-Flop type
6) Construct F
7) Construct G
8) Logic Diagram
Module 1: Classic Digital Design
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Finite State Machines
• State Machine Example “Simple Gray Code Counter”
1) Design a 2-bit Gray Code Counter. There are no inputs (other than the clock).
Use Binary for the State Variable Encoding
2) State Diagram
Module 1: Classic Digital Design
118
Finite State Machines
• State Machine Example “Simple Gray Code Counter”
3) State/Output Table
Current State
Next State
Out
CNT0
CNT1
00
CNT1
CNT2
01
CNT2
CNT3
11
CNT3
CNT0
10
Module 1: Classic Digital Design
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Finite State Machines
• State Machine Example “Simple Gray Code Counter”
4) State Variable Assignment – binary
Current State
Q1 Q0
Next State
Q1* Q0*
Out
0
0
0
1
00
0
1
1
0
01
1
0
1
1
11
1
1
0
0
10
5) Choose Flip-Flops
- let's use DFF's
Module 1: Classic Digital Design
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Finite State Machines
• State Machine Example “Simple Gray Code Counter”
Q1
6) Construct Next State Logic “F”
Q0
0
0
2
0
0
1
Q1* = Q1  Q0
1
1
1
3
1
0
0
1
Q1
Q0
0
1
0
Q0* = Q0’
2
1
1
1
3
0
Module 1: Classic Digital Design
0
121
Finite State Machines
• State Machine Example “Simple Gray Code Counter”
Q1
7) Construct Output Logic “G”
Q0
0
0
2
0
0
1
Out(1) = Q1
1
1
1
3
0
1
0
1
Q1
Q0
0
0
0
Out(0) = Q1  Q0
2
1
1
1
3
1
Module 1: Classic Digital Design
0
122
Finite State Machines
• State Machine Example “Simple Gray Code Counter”
8) Logic Diagram
Module 1: Classic Digital Design
123
Finite State Machines
• State Machines
- there is a basic structure for a Clocked, Synchronous State Machine
1) State Memory
2) Next State Logic “F
3) Output Logic “G”
(i.e., flip-flops)
(combinational logic)
(combinational logic)
Module 1: Classic Digital Design
124
Finite State Machines
• State Machines
- the steps in a state machine design are:
1) Word Description of the Problem
2) State Diagram
3) State/Output Table
4) State Variable Assignment
5) Choose Flip-Flop type
6) Construct F
7) Construct G
8) Logic Diagram
Module 1: Classic Digital Design
125
Finite State Machines
• State Machine Example “Sequence Detector”
1) Design a machine by hand that takes in a serial bit stream and looks for the pattern “1011”.
When the pattern is found, a signal called “Found” is asserted
2) State Diagram
Module 1: Classic Digital Design
126
Finite State Machines
• State Machine Example “Sequence Detector”
3) State/Output Table
Current State
In
Next State
Out
(Found)
S0
0
1
0
1
0
1
0
1
S0
S1
S2
S0
S0
S3
S0
S0
0
0
0
0
0
0
0
1
S1
S2
S3
Module 1: Classic Digital Design
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Finite State Machines
• State Machine Example “Sequence Detector”
4) State Variable Assignment – let’s use binary
Current State
Q1 Q0
In
0 0
0
1
0
1
0
1
0
1
0 1
1 0
1 1
Next State
Q1* Q0*
0
0
1
0
0
1
0
0
0
1
0
0
0
1
0
0
Out
Found
0
0
0
0
0
0
0
1
5) Choose Flip-Flop Type
- 99% of the time we use D-Flip-Flops
Module 1: Classic Digital Design
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Finite State Machines
• State Machine Example “Sequence Detector”
Q1
Q1 Q0
6) Construct Next State Logic “F”
In
00
0
2
0
0
1
Q1* = Q1’∙Q0∙In’ + Q1∙Q0’∙In
In
01
6
1
3
0
1
11
10
4
0
7
0
0
5
0
1
Q0
Q1
Q1 Q0
In
00
0
2
0
0
Q0* = Q0’∙In
1
In
1
01
11
6
0
3
1
10
4
0
7
0
0
5
0
1
Q0
Module 1: Classic Digital Design
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Finite State Machines
• State Machine Example “Sequence Detector”
7) Construct Output Logic “G”
Q1
Q1 Q0
In
00
0
Found = Q1∙Q0∙In
2
0
0
1
In
1
01
11
6
0
3
0
10
4
0
7
0
0
5
1
0
Q0
8) Logic Diagram
Module 1: Classic Digital Design
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State Variable Encoding
• State Variable Encoding
- we design State Machines using descriptive state names
- when we’re ready to synthesize the circuit, we need to assign values to the states
- we can arbitrarily assign codes to the states in order to optimize for our application
Module 1: Classic Digital Design
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State Variable Encoding
• Binary
- this is the simplest and most straight forward method
- we simply assign binary counts to the states
S0 = 00
S1 = 01
S2 = 10
S3 = 11
- for N states, there will be log(N)/log(2) flip flops required to encode in binary
- the advantages to Binary State Encoding are:
1) Efficient use of area (i.e., least amount of Flip-flops)
- the disadvantages are:
1) Multiple bits switch at one time leading to increased Noise and Power
2) The next state logic can be multi-level which decreases speed
Module 1: Classic Digital Design
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State Variable Encoding
• Gray Code
- to avoid multiple bits switching, we can use a Gray Code
S0 = 00
S1 = 01
S2 = 11
S3 = 10
- for N states, there will be log(N)/log(2) flip flops required
- the advantages of Gray Code Encoding are:
1) One-bit switching at a time which reduces Noise/Power
- the disadvantages are:
1) Unless doing a counter, it is hard to guarantee the order of state execution
2) The next state logic can again be multi-level which decreases speed
Module 1: Classic Digital Design
133
State Variable Encoding
• One-Hot
- this encoding technique uses one flip-flop for each state.
- each flip-flop asserts when in its assigned state
S0 = 0001
S1 = 0010
S2 = 0100
S3 = 1000
- for N states, there will be N flip flops required
- the advantages of One-Hot Encoding are:
1) Speed, the next state logic is one level (i.e., a Decoder structure)
2) Suited well for FPGA’s which have LUT’s & FF’s in each Cell
- the disadvantages are:
1) Takes more area
Module 1: Classic Digital Design
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Sequential Logic
State Machine Example: Design a 2-bit Gray Code Counter
with “State Encoded Outputs”
00
1) Number of States?
2) Number of bits to encode states?
3) Moore or Mealy?
:4
: 2n=4, n=2
: Moore
01
For this counter, we can make the outputs be the state codes
11
10
Module 1: Classic Digital Design
135
Sequential Logic
State Machine Example: Design a 2-bit Gray Code Counter
00
01
11
STATE
Current Next
Acur Bcur Anxt Bnxt
0
0
1
1
0
1
1
0
0
1
1
0
Anxt Logic
Bnxt Logic
Bcur
0 1
Acur 0
1
1
1
0
0
0
1
0
1
Bcur
0 1
Acur 0
1
1
1
0
0
Bnxt = Acur’
Anxt = Bcur
10
A
B
D
Q
A
D
counter
output
Q
B
Q
Q
CLK
Module 1: Classic Digital Design
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