Exam Solutions for ECE 2030 C - Spring 2001 - Computer Engineering, Exams of Computer Science

The solutions to exam one for the computer engineering course ece 2030 c in spring 2001. The exam covers topics such as boolean algebra, incomplete circuits, and karnaugh maps. Students are required to complete truth tables, transform boolean expressions, and implement circuits using various logic gates.

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2012/2013

Uploaded on 04/08/2013

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ECE 2030 C Computer Engineering Spring 2001
4 problems, 5 pages Exam One Solutions 8 February 2001
1
Problem 1 (2 parts, 25 points) Boolean Algebra and Standard Forms
Part A (15 points) For each expression, complete the truth table that represents its behavior.
OUTX =
CBACBACBACBA โ‹…โ‹…+โ‹…โ‹…+โ‹…โ‹…+โ‹…โ‹…
OUTY = )()( CBACBA ++โ‹…++ OUTZ = CBA
โŠ•
โŠ•
ABCOUTXABCOUTYABCOUTZ
0 0 0 0 0 0 0 0 0 0 0 0
0 0 1 1 0 0 1 1 0 0 1 1
0 1 0 0 0 1 0 1 0 1 0 1
0 1 1 0 0 1 1 1 0 1 1 0
1 0 0 1 1 0 0 1 1 0 0 1
1 0 1 1 1 0 1 1 1 0 1 0
1 1 0 1 1 1 0 1 1 1 0 0
1 1 1 0 1 1 1 0 1 1 1 1
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.
))(( EDCBAOutX++โ‹…โ‹…= = ))()(())()(())(( EDCBAEDCBAEDCBA +โ‹…++=+โ‹…โ‹…+=++โ‹…+
Note that this is not the same as ))(( EDCBA +โ‹…++ which is incorrect.
FEDCBAOutYโ‹…โ‹…+++= ))()(( = FEDCBAFEDCBA โ‹…++โ‹…+=โ‹…++++ ))()(())()((
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4 problems, 5 pages Exam One Solutions 8 February 2001

Problem 1 (2 parts, 25 points) Boolean Algebra and Standard Forms Part A (15 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 ) OUTZ =^ A^ โŠ• B โŠ• C

A B C OUTX A B C OUTY A B C OUTZ

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 โ‹… C +( D + E )) = A +( B โ‹… C +( D + E ))= A +(( B โ‹… C )โ‹…( D + E ))= A +(( B + C )โ‹…( D + E )) Note that this is not the same as A + ( B + C โ‹…( D + E )) which is incorrect.

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

4 problems, 5 pages Exam One Solutions 8 February 2001

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, 5 pages Exam One Solutions 8 February 2001

Problem 4 (2 parts, 30 points) Karnaugh Maps Part A (15 points) Given the following Karnaugh Map, circle and list all the prime implicants, indicating which are essential and write the simplified SOP expression.

A B C

A

A

B B

C

C

C

D

D D

prime implicants

essential? yes no A D

A C D

B C D

A B D

simplified SOP expression (^) A โ‹… D + A โ‹… C โ‹… D + A โ‹… B โ‹… C Part B (15 points) Simplify the following POS expression using a Karnaugh Map. Circle and list all the prime implicants, indicating which are essential and write the simplified POS expression. Out =( A + C + D )( B + C + D )( A + B + C + D )( A + B + C + D )

A

A

B B

C

C

C

D

D D

prime implicants

essential? yes no A + C + D A + C + D B + C + D A + B + D

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