Midterm 1 Revision

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Midterm 1 Revision
Lecture 8

• Prof. Sin-Min Lee
• Department of Computer Science

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Truth table to K-Map
B A 0 1
0 1 1
1 1
A B P
0 0 1
0 1 1
1 0 0
1 1 1
minterms are represented by a 1 in the
corresponding location in the K map.
The expression is A.B A.B A.B
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K-Maps

• Adjacent 1s can be paired off
• Any variable which is both a 1 and a zero in this
  pairing can be eliminated
• Pairs may be adjacent horizontally or vertically

B A 0 1
0 1 1
1 1
a pair
B is eliminated, leaving A as the term
another pair
A is eliminated, leaving B as the term
The expression becomes A B
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• Two Variable K-Map

A B C P
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 0
A.B.C A.B.C A.B.C
BC A 00 01 11 10
0 1
1 1 1
One square filled in for each minterm.
Notice the code sequence 00 01 11 10 a Gray
code.
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Grouping the Pairs
BC A 00 01 11 10
0 1
1 1 1
equates to B.C as A is eliminated.
Here, we can wrap around and this pair equates
to A.C as B is eliminated.
Our truth table simplifies to A.C B.C as
before.
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Groups of 4
Groups of 4 in a block can be used to eliminate
two variables
BC A 00 01 11 10
0 1 1
1 1 1
The solution is B because it is a 1 over the
whole block (vertical pairs) BC BC B(C C)
B.
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Karnaugh Maps

• Three Variable K-Map
• Extreme ends of same row considered adjacent

A BC
00
01
11
10
0
1
10
00
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Karnaugh Maps

• Three Variable K-Map example

A BC
00
01
11
10
0
1
X
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The Block of 4, again
A BC
00
01
11
10
0
1
1
1
1
1
X C
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Returning to our car example, once more

• Two Variable K-Map

A B C P
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 0
A.B.C A.B.C A.B.C
AB C 00 01 11 10
0 1 1 1
1
There is more than one way to label the axes of
the K-Map, some views lead to groupings which are
easier to see.
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Karnaugh Maps

• Four Variable K-Map
• Four corners adjacent

AB CD
00
01
11
10
00
01
11
10
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Karnaugh Maps

• Four Variable K-Map example

AB CD
00
01
11
10
00
01
11
10
F
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Product-of-Sums
We have populated the maps with 1s using
sum-of-products extracted from the truth
table. We can equally well work with the 0s
A B C P
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 0
AB C 00 01 11 10
0 0
1 0 0 0
AB C 00 01 11 10
0 1 1 1
1 1
P (A B).(A C) P A.B A.C
equivalent
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Inverted K Maps

• In some cases a better simplification can be
  obtained if the inverse of the output is
  considered
• i.e. group the zeros instead of the ones
• particularly when the number and patterns of
  zeros is simpler than the ones

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Karnaugh Maps

• Example Z5 of the Seven Segment Display

X1 X2 X3 X4 Z5
0 1 2 3 4 5 6 7 8 9
0 0 0 0 1
0 0 0 1 0
X1X2 X3 X4
00
01
11
10
0 0 1 0 1
0 0 1 1 0
00
0 1 0 0 0
01
0 1 0 1 0
11
0 1 1 0 1
0 1 1 1 0
10
1 0 0 0 1
1 0 0 1 0
Z5
1 0 1 0 X
1 0 1 1 X

• Better to group 1s or 0s?

1 1 0 0 X 1 1 0 1 X 1 1 1 0 X 1 1 1 1 X
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Inverter

• Input- output behavior given by table called
  truth table
• Symbols used in circuits to denote an inverter is
  given
• Bubble at the end is called the inversion bubble
• Forms a basic function in Boolean algebra and
  circuits

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NAND GATES

• Symbol above is for NAND
• A, B input, X output
• Notice the inversion bubble just before the input
• Output is 0 iff both inputs are 1
• Output is 1 if at least one input 0
• Will see NAND is NotAND

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AND Gates

• Has corresponding electronic circuit
• A,B inputs, X output
• Output 1 iff both inputs 1
• Output 0 iff either input 1
• Notice diagram is NAND without the inversion
  bubble
• NAND NOTAND

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OR Gates

• Two inputs A,B,
• one output X
• X is 1 iff either of A, B is 1
• X is 0 if both A, B are 0
• Symbol on top
• Truth Table on bottom
• Note No inversion bubble

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Nor Gates

• Symbol on top
• Truth table on bottom
• Notice inversion bubble in front of OR symbol
• Get output value by switching the value coming
  out of OR gates

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Boolean Algebra

• Boolean Algebra A function has only two outcomes
  (1true), (o-false)
• N variable boolean function can be described has
  only 2N possible outputs. They can be written
  as a table
• If agree on order of listing argument
  combinations such as base 2 , can write output as
  2n bit binary number

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Example Majority Function

• Three inputs A, B, C
• One output M
• Output takes truth value of majority inputs. I.e.
• M is 1 iff two of A,B,C is 1
• M is 0 iff two of A, B, C is 0
• Notice writing large truth tables is cumbersome

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Alternative Representation

• Collect the combinations of variable that give 1
  for output.
• Write the function as a SUM of these terms
• In terms, write variable name for value 1, and a
  bar over the name for 0.
• EG M ABCABCABCABC

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Rationale for New Notation

• Consider ABC The product is for AND
• Consider ABCABC The sum is for OR
• So we are writing the function as a sum of
  products
• I.e. AND-ing OR-terms Called conjunctive normal
  form.
• Consider ABC This is 1 iff A0, B1 and C1
• A function of N variables can be given as sum of
  2N n-variable products

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Creating Circuits for Boolean Functions

• MABCABCABCABC
• 1,2,3 are NOT gates feeding lines A,B,C
• 4,5,6,7 are AND gates corresponding to the four
  product terms
• 8 is an OR term corresponding to the sum
• A,B,C have been inserted to avoid clutter they
  could be connected directly out of NOT gate

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Implementing Boolean Functions

• Write the truth table
• Provide inverters for complementing inputs
• Draw an AND gate for each term with 1in output
  column
• Wire the AND gates to appropriate inputs
• Feed the outputs of all AND gates into an OR gate

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Using A Single Gate Type

• It is desirable to use only one type of gate
  generate the whole circuit.
• Can use NAND or NOR gate.
• In order to do so, enough to show that
• NOT, AND, OR NAND can be generated by NOR gates
• NOT, AND, OR, NOR ca be generated by NAND gates.
• We say that NAND, NOR are complete for Boolean
  circuits

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Completeness of NAND
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Completeness of NOR
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Circuit Equivalence

• Sometimes need to minimize number of elements on
  a board
• get minimum number of gates
• Two input gates instead of four input gates
• Need to find an equivalent circuit for the given
  circuit
• Equivalent having same input output behavior
  computing same Boolean function
• Use Boolean Algebra

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Example Using ABAC A(BC)
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Some Laws of Boolean Algebra
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Consequences of De Morgans Law
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Using De Morgans Laws to covert sum of products
to NAND
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De Morgan again

• A NAND gate
• Y A.B A B
• is the same as an OR gate with two NOT gates
• Similarly a NOR gate is the same as an AND gate
  with two inverters
• Y A B A.B
• not the individual terms
• change the sign
• not the lot

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Dual gates
not the individual inputs change the gate not the
output
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Truth Tables and Boolean Notation

• NAND Gate Representation
• It is possible to implement any boolean
  expression using only NAND gates

NOT
X
X
AND
A.B
A
A.B
B
OR
A
AB
B
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Truth Tables and Boolean Notation

• NAND Gate representation
• Implement the following circuit using only NAND
  gates

x2
x4
x3
De Morgan can also be represented visually
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Exercise

• Implement NOT, AND and OR using NOR gates
• Example AND gate dual
  circuit

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Solution

• Similar pattern to using NAND gates (not
  surprising)
• NOT
• AND
• OR

X
X
A.B
A
A
A.B
B
A.B
B
A
AB
A.B
A
AB
B
B
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Truth Tables and Boolean Notation

• NOR Gate representation
• It is also possible to implement any boolean
  expression using only NOR gates
• Implement the following circuit using only NOR
  gates

X4
X3
X
2
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Solution

• Two NOR gates in sequence acting as NOTs can be
  eliminated

X4
X3
X
2
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Examples

• The half adder
• The half adder is a circuit for adding two single
  bit numbers
• Develop a truth table and Boolean expressions for
  the half adder

A B S C
0 0
0 1
1 0
1 1
S and C are the Sum and Carry
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Half adder

• The sum is XOR operation and the carry an AND

C
A
A B S C
0 0 0 0
0 1 1 0
1 0 1 0
1 1 0 1
B
S
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Examples

• The full adder
• Develop a truth table and Boolean expressions for
  the full adder, this circuit also includes a
  carry in.

Sum
A
full adder
B
Cout
Cin
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Truth table for full adder
Exercise Complete the Karnaugh maps for the Sum
and the Carry out columns
Cin A B S Cout
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 0 1
1 0 0 1 0
1 0 1 0 1
1 1 0 0 1
1 1 1 1 1
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Full adder circuit
A
B
Sum
Cout
Cin
Sum Cin xor A xor B
Cout A.B A.Cin B.Cin
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Karnaugh Map Method of Multiplexer Implementation
Consider the function
A is taken to be the data variable and B,C to be
the select variables.
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Example of MUX combo circuit

• F(X,Y,Z) Sm(1,2,6,7)

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