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2.1.4.ak circuitsimplificationbooleanalgebra 2 (1)

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2.1.4 Circuit Simplification: Boolean Algebra
Procedure
Using the theorems and laws of Boolean algebra, simplify the following logic expressions.
Note the Boolean theorem/law used at each simplification step. Be sure to put your answer in
Sum-Of-Products (SOP) form.
1.
F 1  A( A  AB)
F 1  A( A  B )
F 1  AA  AB
F 1  0  AB
F 1  AB
2.
F 2  X Y Z  X Y Z  X YZ
F 2  Y Z ( X  X )  X YZ
F 2  Y Z (1)  X YZ
F 2  Y Z  X YZ
F2

Z (Y  Y X )
F2

Z (Y  X )
F 2  YZ  X Z
3.
F3  JK  J K
F3  J ( K  K )
F3  J (1)
F3  J
4.
F4  ( B  B)( A B  A BC )
F4  (1)( A B  A BC )
F4  A B (1  C )
F4  A B (1)
F4  A B
5.
F5  ( X  Y )( X  Y )
F5  XX  XY  X Y  YY
F5  X  XY  X Y  0
F5  X  X (Y  Y )
F5  X  X (1)
F5  X  X
F5  X
6.
F6  JK  ( J  K ) L  JK
F6  JK  J L  K L  JK
F6  JK  J L  K L
F6  JK  J L ( K  K )  K L
F6  JK  J KL  J K L  K L
F6  JK  J KL  K L ( J  1)
F6  JK  J KL  K L (1)
F6  JK  J KL  K L
F6  K ( J  J L )  K L
F6  K ( J  L )  K L
F6  JK  KL  K L
F6  JK  L ( K  K )
F6  JK  L (1)
F6  JK  L
An alternative solution for #6 utilizing DeMorgan’s theorem is shown below. Your teacher
may wish to spiral back to this exercise.
F 6  JK  ( J  K ) L  JK
F 6  JK  JK  ( JK ) L
F 6  JK  ( JK ) L
F 6  JK  L
7.
F 7  RS  R ( S  T )  S ( S  U )
F 7  RS  RS  RT  S S  SU
F 7  RS  RT  0  SU
F 7  RS  RT  SU
8.
F 8  ( N  N M )( N  NM )( N  M )
F 8  ( N  M )( N  M )( N  M )
F 8  ( N N  NM  N M  MM )( N  M )
F 8  (0  MN  M N  M )( N  M )
F 8  ( M  MN  M N )( N  M )
F 8  ( M  M N )( N  M )
F 8  ( M )( N  M )
F 8  MN  MM
F 8  MN  M
F8  M
Almost as important as being able to use the laws of Boolean algebra (i.e., associative,
commutative, or distributive) to simplify logic expressions, it is also critical that you are able to
identify them. Identify the law of Boolean algebra upon which the following equalities are
based.
9.
A B A C BC  A C  BC  A B
Commutative
10.
D E F G   D E FG
Associative
11.
R  S T  U  R  S  T  U
Associative
12.
J  K L  M   J L  J M  K L  K M
Distributive
13.
R ST SV  R ST  R SV


Distributive
Now that you’ve practiced simplifying logic expressions, apply your knowledge to simplifying
an actual circuit.
14. Shown below is a VERY poorly designed AOI circuit that is part of a coffee vending
machine. Write the UN-SIMPLIFIED logic expression for the output Brew.
15. Using the theorems and laws of Boolean algebra, simplify the logic expression
Brew Cut Off. Be sure to put your answer in Sum-Of-Products (SOP) form.
T
P
W
F
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
0
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
1
BrewCutOff ( P( TP  W )  T )  TP
BrewCutOff  TPP  PW  T  TP
BrewCutOff  0  PW  T
BrewCutOff  PW  T
16. In the space provided, draw an AOI circuit that implements the simplified logic
expression Brew Cut Off. For your implement, assume that only 2-input AND gates
(74LS08), 2-input OR gates (74LS32), and inverters (74LS04) are available. Draw this
circuit in the space provided.
If the temperature is too high (above 205 degrees F) or the pressure is not below the
safe value (15 Bar) with enough water present (2 cups), the brew sensor cuts off the
brew process.
U1
NC7S04_2V
U3A
U5A
4011BT_10V
4001BD_5V
U2
NC7S04_2V
U4A
4011BT_10V
Conclusion
1. Describe the process that you would use to simplify a logic expression using Boolean
algebra.
Use Boolean algebra to organize/group similar terms.
Then use Boolean algebra to simplify.
2. How do you know when you are finished simplifying and have arrived at the simplest
equation?
There are no Boolean theorems left to make the expression more condensed.
3. Other than using Boolean algebra, how could you prove that two circuits are
equivalent?
Compare the truth tables.
4. If you worked for a company that manufactured the coffee vending machine that used
the poorly designed circuit, how much money would your new design save the
company annually if each GATE cost 15¢ and the company made 500,000 vending
machines per year?
You would save the company $300,000 annually.
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