Single Part - Computer Engineering - Solved Exam, Exams of Computer Science

Main points of this past exam are: Single Part, Incomplete Circuits, Switching Network, Boolean Expression, Mixed Logic Design, Transistors, Design Technique, Switch-Ready Expressions, Boolean Expressions, Karnaugh Map

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2012/2013

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ECE 2030 B 1:00pm Computer Engineering Fall 2004
4 problems, 4 pages Exam One Solutions 15 September 2004
1
Problem 1 (3 parts, 30 points) Incomplete Circuits
For each partial switch circuit below, complete the complementary switching network so the
circuit contains no floats or short. Also write the Boolean expression computed by the completed
circuit. Assume the inputs and their complements are available.
B
DC
Out
x
A
Out
y
A
B E
C
D
F
A
DB
Out
z
E
C
no dot
E
A
D
B
E
C
A
B
C
D
E
F
A B
CD
OUTx = CEBDA ++โ‹…+ )()(
OUTy =
FEDCBA โ‹…โ‹…โ‹…โ‹…โ‹…
OUTz =
DCBA โ‹…+โ‹…
pf3
pf4

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4 problems, 4 pages Exam One Solutions 15 September 2004

Problem 1 (3 parts, 30 points) Incomplete Circuits

For each partial switch circuit below, complete the complementary switching network so the circuit contains no floats or short. Also write the Boolean expression computed by the completed circuit. Assume the inputs and their complements are available.

B

C D

Out x

A

Outy A

B E

C

D

F

A

B D

Outz

E

C

no dot (^) E

A

D

B

E

C

A B C D E F

A B

C D

OUTx = ( A + D )โ‹…( B + E )+ C

OUTy = A โ‹… B โ‹… C โ‹… D โ‹… E โ‹… F

OUTz = A โ‹… B + C โ‹… D

4 problems, 4 pages Exam One Solutions 15 September 2004

Problem 2 (3 parts, 26 points) Mixed Logic Design

Implement the following expressions to minimize total transistors (switches) required. Where two expressions are specified for a single part, your implementation should provide both outputs. Use proper mixed logic design technique. Any combination of Two, three, and four input AND, OR, NAND, NOR, and NOT gates may be used. Do not simplify the expression. Part A (6 points) OUT (^) X = A โ‹… B + C โ‹… D + E โ‹… F

B C

E F

D OUTx

A

switches = 3 x 4 + 1 x 6 = 18T

Part B (8 points) OUTY = A โ‹… B + C โ‹… D

A B

C

OUTy

D

switches = 3 x 4 + 1 x 2 = 14T

Part C (12 points) OUTZ 1 (^) = A โ‹… B + C , OUTZ (^) 2 = B + C + D + E

A

B C

E

OUTZ

D OUTZ

switches = 2 x 4 + 1 x 6 + 3 x 2 = 20T

4 problems, 4 pages Exam One Solutions 15 September 2004

Problem 4 (2 parts, 24 points) Karnaugh Maps

Part A (12 points) For the follow expression, derive a simplified sum of products expression using a Karnaugh Map. Circle and list the prime implicants, indicating which are essential.

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

A

A

B B

C

C

C

D

D D

prime implicants

essential? yes no

A B

C D

A C

A D

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

Part B (12 points) For the follow expression, derive a simplified product of sums expression using a Karnaugh Map. Circle and list the prime implicants, indicating which are essential.

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

A

A

B B

C

C

C

D

D D

prime implicants

essential? yes no

B + D

B + C

A + B

A + C + D

A + B + C

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