Transcript Circuits II ENGG1015 Tutorial

ENGG 1203 Tutorial
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Combinational Logic (I)
1 Feb
Learning Objectives
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Recall Boolean Algebra (SoP/PoS, DeMorgan's Theorem,
grouping, redundant)
Simplify logic expressions
News
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Lab, TA office hour
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Tutorial: http://www.eee.hku.hk/~culei/ENGG1203.html
Ack.: HKU ELEC1008, Wikimedia Commons
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Quick quiz
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What is the only set of input conditions
that will produce a LOW output for any
OR gate?
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Any one of the input is LOW
Any one of the input is HIGH
All inputs are LOW
All inputs are HIGH
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Quick quiz
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What logic level should be applied to the
second input of a two-input AND gate if the
logic signal at the first input is to be inhibited
(prevented) from reaching the output?
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A LOW input will keep the output LOW
A LOW input will keep the output HIGH
A HIGH input will keep the output LOW
A HIGH input will keep the output HIGH
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Quick quiz
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What is the only input combination that
will produce a HIGH at the output of a
five-input AND gate?
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Any one of the input is LOW
Any one of the input is HIGH
All inputs are LOW
All inputs are HIGH
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Quick quiz
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What is the output expression of the following logiccircuit diagram?
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x  D   A  BC  E
x   D  A  BC   E
x   D  A  B  C   E
x  D   A  BC   E


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Boolean Algebra
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Boolean Algebra
A+B=B+A
A + (B + C) = (A + B) + C
A + BC = (A + B) (A + C)
A + AB = A
NOT (NOT (A)) = A
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AB = BA
A (BC) = (AB) C
A (B+C) = AB + AC
A (A + B) = A
More questions in Appendix
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De Morgan’s Theorem
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De Morgan's theorem
AB  A  B A  B  A  B
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Bubble pushing via De Morgan's theorem
 AND NOT  NOT OR
A B  A  B
 NOT OR  AND NOT
A  B  A B
 OR NOT  NOT AND
A  B  A B
 NOT AND  OR NOT
A B  A  B
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Solution 2
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Use DeMorgan's Theorem for simplification
 a  d   b  c    a  d   b  c   a  d  b  c
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 a  b  c    c  d    a  b  c . c  d   a  b  c  d
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a  d b  c c  d  a  d b c c  d  a b c  d
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Boolean Algebra Simplification
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Sum of Products
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Find out the “1”s
𝑌 = 𝐴∙𝐵 + 𝐴∙𝐵
Better if less “1”
Products of Sum
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Find out the “0”s
𝑌 = 𝐴+𝐵 ∙ 𝐴+𝐵
Better if less “0”
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Boolean Algebra Using SOP and POS
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Find an expression for F and F
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Solution
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Sum of Products for F
F  ABC  ABC  ABC  ABC  ABC
Product of Sums for F
FF
 ABC  ABC  ABC
 ABC  ABC  ABC


  A B  C A B  C  A B  C

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Solution
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Sum of Products for F
F  ABC  ABC  ABC
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Product of Sums for F



F  A B C  A B C 
 A  B  C A  B  C 
 A  B  C
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Representing logic operations
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Each function can be represented
equivalently in 3 ways:
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Truth table – Try every combinations of every
input variables
Boolean logic expression – SOP/POS + Simplify
the expression
Schematics – Construct from Boolean
expressions
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From logic equations
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Boolean expressions
 Truth table and logic circuit (AND/OR/NOT)
0
1
0
1
1
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From logic equations
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Boolean expressions
 Truth table and logic circuit
(AND/OR/NOT)
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From truth tables
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Derive the Boolean expression
of the output x in terms of the
input
Construct the logic circuit using
AND gates, OR gates, and
INVERTERs.
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Solution
x  ABC  ABC  ABC  ABC
Extra redundant
Terms
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 
 
 AC  B  B   AB  C  C   BC  A  A 
 ABC  ABC  ABC  ABC  ABC  ABC
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 AC  AB  BC
1. Construct A/B/C
2. Construct not A/B/C
3. Construct AND gates
4. Construct OR gate
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2
3
4
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From truth tables
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Truth table  Boolean expressions and logic
circuit
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From schematics
• Truth table first?
• SOP/POS first?
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From schematics
XOR
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Circuit representation of logic equations
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Show how x  ABC can be implemented with one
two-input NOR and one two-input NAND gate.
(How to convert ABC   AB   C ?)
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We need to apply De Morgan’s Theorem
x  ABC   AB  C   AB   C
A A
AB  A  B

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Circuit representation of digital logic
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a) Simplify the circuit shown in the figure using
Boolean algebra.
b) Change each NAND gate in the circuit of the figure
to a NOR gate,
MNQ
and simplify the
circuit using
M NQ
Boolean algebra.
M NQ
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Solution (a)

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
x  MNQ M NQ M NQ
 MNQ  M NQ  M NQ
 MNQ  M NQ  M NQ
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Procedure:
1) Obtain the Boolean expression from
the circuit
2) Check if we need NAND/NOR gate
3) Simplify the expression by Boolean
algebra
 M N  N Q  M NQ
 M 1 Q  M NQ
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 M MN Q
 M  N Q
• Less gate (power and resource)
• Shorter “longest path”
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Solution (b)
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First, we convert the circuit
M
N
Q
A
M  N Q
M  N Q
B
C
x
M  N Q
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Solution (b)
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Then, we simplify the Boolean expression
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
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 
X  M  N Q  M  N Q  M  N Q
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
 M  N  Q M  N  Q M  N  Q
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(DeMorgan's Theorem)
(Expand)
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
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  M  M N  MQ  NM  0  NQ  QM  QN  Q  M  N  Q  (Group, Group)
 MM  M N  MQ  NM  N N  NQ  QM  QN  QQ M  N  Q (Simplify)
A  B  A B
 A  B   A  B   AA  AB  BA  BB
AA  A
AA  0
A A 1
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Solution (b)
...


 M  M N  MQ  NM  0  NQ  QM  Q N  Q M  N  Q

 


(Group, Group)
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  M 1  N  Q  N  Q  Q N  N  1  M  N  Q (Simplify)




  M  Q  M  N  Q (Expand)
 M M  MN  MQ  QM  QN  QQ (Simplify)


 0  MN  Q M  M  Q  N  1 (Simplify)
1 A  1 0  A  A
1 A  A 0  A  0
 MN  Q  Q  MN  Q
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Conversion of three representations
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Describe the function using Boolean expressions
Draw the truth table and describe the function using
sum of product
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Solution

x  AB  B  CC

 AB  B  C
Approach 1: Boolean simplification  Find TT
Approach 2: Construct TT  Find POS
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(De Morgan)
 AB  B  C
(XOR expansion)
 ABBC  BC
(De Morgan)
 ABBC BC
(De Morgan)
(expansion)
  
(grouping,
 AB  BB  BC  C B  CC  expansion)
 AB B  C B  C
 ABBC  ABC B
 ABC
(cancellation)
POS: x  ABC
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Karnaugh map
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Draw the table  Fill in 0s
and 1s  Grouping
Group one/two/four/eight/
sixteen ‘1’(s) only
Use the least number of
groups to group all
numbers
To group as many numbers
as possible in every group
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Karnaugh map
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Yellow: Redundant
𝐴𝐵𝐶 𝐷 + 𝐴𝐵𝐶 𝐷 = 𝐵𝐶 𝐷
𝐴𝐵 𝐶𝐷 + 𝐶𝐷 + 𝐶 𝐷 + 𝐶 𝐷 = 𝐴𝐵
𝐴𝐵 𝐶 𝐷 + 𝐴𝐵 𝐶𝐷 + 𝐴𝐵𝐶 𝐷 + 𝐴𝐵 𝐶𝐷 = A𝐶
F(x) = 𝐵𝐶𝐷 + A𝐶 + 𝐴𝐵
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Examples of Karnaugh maps
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(Appendix) Questions for Boolean algebra
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Solutions
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(Appendix) Determining output level from
a diagram
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(Appendix) From logic equations
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Draw the circuit diagram to implement the expression
X  (A  B)(B  C)
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Draw the circuit diagram that implements the expression
x  ABC( A  D) using gates having no more than three
inputs.
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(Appendix) Circuit representation of
digital logic
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Construct the given circuit using NAND gates only
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Top down approach: ?
Bottom up approach: ?
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Solution (a)
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Top down: Expanding the Boolean expression
By DeMorgan’s Theorems, A  A  A  AA
X  ABC  ABC  ABD
 ABC  ABC  ABD
   
  AABBCC  ABBCC  AABBD 
 ABC ABC ABD
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Solution (b)
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Bottom-up: Construct NOT gate, AND gate and OR
gate from NAND gate
i) 1
0
1
0
iii)
ii)
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Solution (b)
X  ABC  ABC  ABD
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A A
(cancelled)
Top-down and Bottom-up: Same number of gate,
same configuration, different approach
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