1. Basic Digital Logic Systems
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Basic Digital Logic Systems
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2. Outline Number Systems / Binary Encoding Boolean Algebra
Combinational Circuits
Sequential Circuits
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3. 1. Number Systems
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4. Positional Number System
For base-10 numbers:
289 = 2x x x100
Works for other bases
Base 6: (we indicate base explicitly) = 2x62 + 3x61 + 5x60
Fractional Numbers:
= 7x x x x x10-3
52.18 = 5x81 + 2x80 + 1x8-1 = /8 =
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5. Binary Numbers Digital Systems
Everything must be represented internally using {0,1}
Base 2 or “binary” = 1x23 + 1x21 + 1x20 = = 1110
Binary digits = “bits”
k-bit unsigned number represents values 0 to 2k-1
Convention:
Most significant bit (MSB)
Least significant bit (LSB)
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6. Conversion Between Bases
Like long division: Example: convert to base 6
112
(3x62) 4
(0x61) => 3046
(4x60)
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7. Conversion Between Bases
12.625
(1x23)
4.625
(1x22)
0.625
(0x21)
(0x20)
(1x2-1)
0.125
(0x2-2)
(1x2-3) =>
Other bases Example: convert to binary
Should be able to do this to understand meaning
In practice: use a calculator (except simple conversions)
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8. Hexadecimal
Base 16 is Hexadecimal (“Hex” for short)
Run out of numbers for digits, so use letters
For example 1A2F16 = 1x x x x160 =
Why are hex numbers useful? Shorthand for binary.
Going back and forth equally easy
Worthwhile to memorize hex digits and binary equivalents
Hex 1 2 3 4 5 6 7 8 9 A B C D E F
Dec Equiv. 10 11 12 13 14 15
0000|1010|0110| binary
0 A hex
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9. Other Codes Binary Coded Decimal (BCD) Gray Codes ASCII
Single decimal digit maps to 4 bits
Simplifies display / conversion of decimal numbers
Example: = BCD
More bits needed, but it is still often useful
Gray Codes
Adjacent numbers differ only by one bit
ASCII
“American Standard Code for Information Interchange”
7-bit code for alphabet and some control characters ‘A’ = 4116 or 6510, ‘.’ = 2E
Decimal
Gray
000 1
001 2
011 3
010 4
110 5
111 6
101 7
100
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10. Unsigned Arithmetic
Addition of unsigned (non-negative) of any base can be done with “elementary school” method
Add 2 binary numbers, just need simple rules
0+0+0=00, 0+0+1=01, 0+1+1=10, 1+1+1= C = “carry” bit
Addition of two k-bit numbers Can lead to k+1 bit number
Carry in the left-most position means “overflow” has occurred 1
1 1 1
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11. Signed Magnitude How do we represent negative numbers?
How about usual thing with written numbers?
Use one bit to mean + (0) or – (1)
Digital systems: called the “sign” bit
Example:
+310 =+ (0112) = 0011SM
-310 =- (0112) = 1011SM
Advantages: Simple. Sign and number separate
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12. Signed Magnitude Disadvantages Arithmetic with signed magnitude
Two representations of 0
Arithmetic more complicated
Arithmetic with signed magnitude
5 + (-2) = 5 – 2 = 3
5 – (-2) = = 7
-5 – 1 = - (5 + 1) = -6
Must consider signs and do add/sub accordingly
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13. 2s Complement
Use the usual positional notation, but we invert sign of leftmost weight
Note: MSB can still be considered a “sign” bit indicating if number is + or –
For + number, remaining bits can be interpreted as unsigned number
11112C = -1x23 + 1x22 + 1x21 + 1x20
= = -110
10112C = -1x23 + 0x22 + 1x21 + 1x20
= = -510
01112C = -0x23 + 1x22 + 1x21 + 1x20
= = +710
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14. 2s Comp: Sign Extension
How do we make 2’s complement number longer? (maybe need more bits)
Answer: duplicate the sign bit.
00112C = C = C = 310
10112C = C = C = -510
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15. 2s Comp: Negating How do we flip the sign of a 2s complement number?
With signed magnitude: just flip bit.
For 2s complement, invert all bits and add 1 (use modulo unsigned arithmetic)
Why does this last conversion fail?
+510 = 01012C =>(inv)=> = 10112C = -510
-510 = 10112C =>(inv)=> = 01012C = -510
+0 = 00002C =>(inv)=> = 00002C = 0
-810 = 10002C =>(inv)=> = 10002C = !?
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16. 2s Comp: Addition This is the main advantage of 2s complement
Just do normal arithmetic (regardless of sign bit)
We can also have overflow when we add 2s complement numbers.
How do we detect it?
Only possible for same sign. Check sign bit, if it changes => overflow.
How do we avoid it?
Sign extend, then add
1 1 1
1 1 1
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17. 2. Boolean Algebra
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18. Boolean Algebra Named after George Boole in 19th century
Map: TRUE  1, FALSE  0
Operators: AND a·b or just ab OR a+b NOT a’ or a
DeMorgan’s (DM) Rules
(AB)’ = A’ + B’
(A+B)’ = A’B’
Easily proven using a truth table ...
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19. Truth Tables Simple 2-variable truth tables
Can obviously extend to more than 2 input variables (output variables) A B F 1 A B F 1 A B F 1 A B F 1
A=B
F=1 OR
F=AB’
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20. Truth Tables  Boolean Alg.
Sum of Products (SOP) form
OR the true rows in the truth table
F = A’B + A B’ A B F 1
A’B’
A’B
AB’ AB
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21. Truth Tables  Boolean Alg.
Product of Sums (POS) form
AND the false rows in the truth table
F’ = A’B’ + AB
F’ = [(A’B’)’ (AB)’]’ (DM)
F = (A+B) (A’ + B’) (DM) A B F 1
A’B’
A’B
AB’ AB
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22. Boolean Eq.  Truth Table
Repeatedly evaluate expression for all combinations.
Can be more efficient to fill just TRUE terms F = A’B + A B
F = A + A’B
A’B AB A B F 1
A’B A A B F 1
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23. Boolean Identities Single Variable
Two Variable useful for reducing # of terms on complex exp. Simplification Theorems Consensus Theorem difficult to find opportunities to use cons. thm.
A · 0 = 0
A · 1 = A
A + 0 = A
A + 1 = 1
A · A’ = 0
A + A’ = 1
A · A = A
A + A = A
A + A’B = A + B
A(A’ + B) = AB
AB’ + AB = A
(A+B’)(A+B) = A
AC + A’B + BC = AC + A’B
Redundant
(A+C)(A’+B)(B+C) = (A+C)(A’+B)
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24. Boolean Identities (cont’d)
Communitive, Associative, Distributive Thms
AB = BA A+B=B+A Communitive
(AB)C = A(BC) (A+B)+C = A+(B+C) Associative
A(B+C)=AB + AC Distributive
A+BC = (A+B)(A+C) Distributive
Duality: For a valid equality, its “dual” is also true Dual: Replace AND ↔ OR ↔ Example: A(B+C) = AB + AC know this is valid, then A + BC = (A+B)(A+C) Note: An expression is not equivalent to its dual! Be careful of precedence.
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25. Canonical Forms Purpose SOP form “minterm” expansion
Means: “Write a an expression in most standard way.”
Trivial to tell if two expressions are identical
SOP form “minterm” expansion
“minterm” = product term having all variables A B F 1 m0 m1 m2 m3
F = A’B’ + A’B + AB = m0 + m1 + m3 = ∑m(0,1,3)
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26. Canonical Forms POS form “maxterm” expansion A B F 1 M0 M1 M2 M31 M0 M1 M2 M3
F = (A’ + B)(A’ + B’) = M2 · M3 = π M(2,3)
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27. Proving Boolean Expressions
Methods
Truth table
Minterm/maxterm expansion (same terms?)
Manipulate one side to = other (identities)
Manipulate both sides with identical operations
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28. Boolean Minimization
Make expression as “simple” as possible (low number of terms)
AB’ = AB = A(B’ +B) = A· 1 = A
Less complicated expression  Fewer gates, less area, runs faster
Attempt to find minimal cost solution
What is minimal representation? Depends on technology and contraints.
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29. Boolean Minimization Example technologies MSI - 7400
Min. chip count One type of gate preferable
NMOS
NOR gates
Literal count
(count # of terms)
*CMOS
NAND gates
FPGA
Minimum number of
Lookup Tables (LUTs)
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30. Incompletely Specified FuncsA B C F X Y Z W A B C F 1 X
Suppose we know N1 will never generate 101 or 111
Can pick don’t cares to help minimize logic.
“Don’t cares”
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31. K-Maps Karnaugh Map (K-Map for short)
Graphical way of obtaining minimal solutions
Take advantage of logical adjacency to reduce exp.
Consider: F = A’B’ + A’B = A’ From simplification theorem AB’ + AB = A
Terms A’B’ and A’B are adjacent in gray code (differ by 1 bit)
See adj. terms  Immediately use simp. thm.
F = ABCD + AB’CD = ACD A B F 1
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32. K-Maps (cont’d) 2 Variable Rearrange truth table (gray code) B A
K-map more compact (can see adjacent bits) A B F 1 A B F 1
F = B
F = B
B A 1
F = A’B + AB = B
Can write F=B from K-map without using simplification theorem
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33. K-Maps (cont’d)
Minterm (SOP) Method - Draw rectangles (ovals) to group as many 1s as possible. - Repeat until all 1s circled - Now OR the resulting terms (1 per oval) A B F 1
B A 1
F = B A B F 1
B A 1
F = A + B
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34. K-Maps: Don’t cares Can put X’s (don’t cares) in K-map
What do we do with them?
Doesn’t matter! We can circle them or not, in order to have as few terms (biggest ovals) as possible.
B A 1 X
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35. K-Maps: 3 Variable BCA BCA BCA 1 0 0 m0 m4 0 1 m1 m5 1 1 m3 m7 1 0 m21
0 0 m0 m4
0 1 m1 m5
1 1 m3 m7
1 0 m2 m6
BCA 1
0 0 X
0 1
1 1
1 0
X = 0  F = A’C + BC A B C F X 1
BCA 1
0 0 X
0 1
1 1
1 0
Exploit don’t cares  F = C
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