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Output is Pulled - Computer Engineering - Solved Exam, Exams of Computer Science

Main points of this past exam are: OutputPulled, Standard Forms, Boolean Algebra, Truth Table, Boolean Expressions, Implemented Using Switches, Incomplete Circuit, Mixed Logic, Expression Using, Karnaugh Maps

Typology: Exams

2012/2013

Uploaded on 04/08/2013

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ECE 2030 Computer Engineering Fall 2000
4 problems, 4 pages Exam One Solutions 13 September 2000
1
Problem 1 (2 parts, 31 points) Boolean Algebra and Standard Forms
Part A (21 points) For each expression, complete the truth table that represents its behavior.
OUTX =
CBACBACBACBA โ‹…โ‹…+โ‹…โ‹…+โ‹…โ‹…+โ‹…โ‹…
OUTY =
)()()( CBACBACBA ++โ‹…++โ‹…++
OUTZ =
CBA โŠ•โŠ•
ABCOUT
XABCOUT
YABCOUT
Z
0001 0000 0000
1001 1000 1001
0100 0101 0101
1100 1101 1100
0011 0011 0011
1011 1010 1010
0110 0111 0110
1110 1111 1111
Part B (10 Points) Transform each of the following Boolean expressions to a form where they
can be implemented using switches (i.e., there should be no bars in the expression except for
complements of the inputs A, B, C, etc.). The behavior of the expression should remain
unchanged.
))(( DCBAOutXโ‹…+โ‹…= )( DCBA +โ‹…+
FEDCBAOutYโ‹…++โ‹…+= )()( FEDCBA โ‹…โ‹…+++
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4 problems, 4 pages Exam One Solutions 13 September 2000

Problem 1 (2 parts, 31 points) Boolean Algebra and Standard Forms

Part A (21 points) For each expression, complete the truth table that represents its behavior.

OUTX = A โ‹… B โ‹… C + A โ‹… B โ‹… C + A โ‹… B โ‹… C + A โ‹… B โ‹… C

OUTY =

( A + B + C )โ‹…( A + B + C )โ‹…( A + B + C )

OUTZ =

A โŠ• B โŠ• C

A B C OUT X A B C OUT Y A B C OUT Z

Part B (10 Points) Transform each of the following Boolean expressions to a form where they can be implemented using switches (i.e., there should be no bars in the expression except for complements of the inputs A, B, C, etc.). The behavior of the expression should remain unchanged.

Out (^) X = A โ‹…( B +( C โ‹… D )) A + B โ‹…( C + D )

Out (^) Y = ( A + B )โ‹… C +( D + E )โ‹… F A^ + B + C + D โ‹… E โ‹… F

4 problems, 4 pages Exam One Solutions 13 September 2000

Problem 2 (1 part, 15 points) Incomplete Circuit

For the incomplete circuit are shown below. Complete each circuit by adding the needed switching network so the output is pulled high or low for all combinations of inputs (i.e., no floats or shorts). Then write the expression. Assume both the inputs and their compliments are available.

OUTx = ( A โ‹… B + C )( E + F )+ D โ‹… G

4 problems, 4 pages Exam One Solutions 13 September 2000

Problem 4 (2 parts, 25 points) Karnaugh Maps

Part A (10 points) Describe the behavior of the following expression by completing the entries in the Karnaugh Map. You only need to put a 1 and 0 in each box. Do not simplify.

Out = A โ‹… C โ‹… B + B โ‹… D + C โ‹… D + A โ‹… B โ‹… D + A โ‹… B โ‹… C โ‹… D

Part B (15 points) For the following behavior (in map format), derive a simplified sum of products expression using a Karnaugh Map. Circle and list the prime implicants, indicating which are essential. Then write the simplified SOP expression.

simplified SOP expression (^) A โ‹… D + B โ‹… C + A โ‹… B โ‹… C + A โ‹… C โ‹… D or

A โ‹… D + B โ‹… C + B โ‹… C โ‹… D + A โ‹… C โ‹… D or A โ‹… D + B โ‹… C + B โ‹… C โ‹… D + A โ‹… B โ‹… D