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Exercise 3 - Solutions

This document contains solutions to problems from a digital logic course. Problem I involves simplifying logic expressions using Boolean algebra. Problem II involves designing a combinational circuit to detect when two or three of four switches are activated. The circuit is specified using truth tables and logic gates. Problem III discusses advantages and disadvantages of the Quine-McCluskey algorithm compared to K-maps for deriving minimum sum-of-products expressions. Problem IV involves deriving a logic expression from a K-map.

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Wirnkar Basil Nsanyuy
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79 views5 pages

Exercise 3 - Solutions

This document contains solutions to problems from a digital logic course. Problem I involves simplifying logic expressions using Boolean algebra. Problem II involves designing a combinational circuit to detect when two or three of four switches are activated. The circuit is specified using truth tables and logic gates. Problem III discusses advantages and disadvantages of the Quine-McCluskey algorithm compared to K-maps for deriving minimum sum-of-products expressions. Problem IV involves deriving a logic expression from a K-map.

Uploaded by

Wirnkar Basil Nsanyuy
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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EEL 3701C - Digit Logic & Computer Systems

Exercise III
Solutions
Problem I (Logic Simplifications)

Multiply and simplify each of the following expression to obtain a minimal sum of products

1. ( A+ B ) ( C+ B ) ( D ' + B ) ( AC D ' + E )= ( AC + B ) ( D ' +B )( AC D ' + E )


¿ ( AC D + B ) ( AC D + E )
' '

¿ ACD ' + BE
2. ( A '+ B+C ' )( A ' +C ' + D)(B ' + D ')=(A '+C '+ BD )(B ' + D' )
' ' ' '
¿ A B + B C + B ' BD+ A ' D '+C ' D' +BDD '
¿ A ' B ' + A ' D' +C ' B' +C ' D '
Problem II (Design of Combinational Circuit)

Design a circuit with four switches and one output, detecting if two or three switches have been
activated on any of its inputs.

1. Define the inputs and outputs

Solution: four inputs x 1 , x 2 , x 3 , x 4 for the switches and one output Y,

2. Provide the Truth table and the logical function

Solution: Truth table


X1 X2 X3 X4 Y
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 1 0 0 0
0 1 0 1 1
0 1 1 0 1
0 1 1 1 1

1 0 0 0 0
1 0 0 1 1
1 0 1 0 1
1 0 1 1 1
1 1 0 0 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 0
Y =x 1 x3 x 4 + x 1 x 2 x 4 + x2 x3 x 4 + x 1 x 2 x 4+ x1 x 2 x 3 + x1 x 2 x 3

3. Implement the circuit with gates (and, or, etc.)

Problem III: (Quine-McCluskey Algorithm)


1. What are the advantages to the Quine-McCluskey Algorithm? What are the disadvantages?
The Quine-McCluskey algorithm can easily scale to many more variables and can often find a
better minimal form than a k-map. The disadvantages are that it has many more steps and is
more complex.
2. Why do computers use the Quine-McCluskey Algorithm instead of k-maps?
Computers are bad at k-maps because it requires spatial reasoning, intuition, and lots of
possibilities. Computers are good at QM because it is heuristic based, and computers are also
good at handling large amounts of data and combinations.
3. What are the limitations of k-maps?
K-maps are only effective for up to 4 variables. 5-variable k-maps are doable but tedious.
4. Find the minimum sum-of-products expression of F using the Quine-McCluskey method.
Underline the essential prime-implicants in this expression
F ( A , B , C , D , E)=∑ m ( 0 , 2 , 6 ,7 ,8 , 10 , 11, 12 ,13 , 14 , 16 , 18 ,19 , 29 , 30 ) + ∑ d ( 4 , 9 , 21)
: Essential
: Row or column to be removed

F = A’BC’ + B’C’E’ + A’B’CD + AB’C’D+BCDE’

0 2 12 13 29
13, 29 BCD’E X X
Row dominance: BCD’E
21, 29 ACD’E X
8, 9, 12, 13 A’BD’ X X
0, 2, 4, 6, 8, 10, 12, 14 A’E’ X X X

0 2 12 13 29
13, 29 BCD’E X X Column dominance: 13
8, 9, 12, 13 A’BD’ X X
0, 2, 4, 6, 8, 10, 12, 14 A’E’ X X X

0 2 12 29
13, 29 BCD’E X Row dominance: A’E’
8, 9, 12, 13 A’BD’ X
0, 2, 4, 6, 8, 10, 12, 14 A’E’ X X X

0 2 12 29
13, 29 BCD’E X
0, 2, 4, 6, 8, 10, 12, 14 A’E’ X X X

F = A’BC’ + B’C’E’ + A’B’CD + AB’C’D+BCDE’ + BCD’E + A’E’

its state.
Problem IV (K-map):

1. (no solution provided, this should be easy)

2.

AB\CD 00 01 11 10
00 1
01 1
11 1 1 1 1
10 1

F= AB + CD

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