Exam Solution for ECE 2030 Computer Engineering Fall 2006 - Problem 1 to 4, Exams of Computer Science

The solutions to exam one of the ece 2030 computer engineering course offered in fall 2006. The solutions include completing incomplete circuits, implementing expressions using gates, transforming boolean expressions for switch level implementation, and deriving simplified expressions using karnaugh maps.

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ECE 2030 B 1:00pm Computer Engineering Fall 2006
4 problems, 4 pages Exam One Solution 20 September 2006
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.
Outx
A
CB
DE
A B D
CE
Outy
A
B
E
C
D
C
A
B
D
E
Outz
AC
BD
A
C
B
D
OUTx = ECDBA โ‹…+โ‹…โ‹…
OUTy = CEDBA โ‹…+โ‹…+ )(
OUTz = )()( DBCA +โ‹…+
pf3
pf4

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4 problems, 4 pages Exam One Solution 20 September 2006

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.

Out x

A

B C

D E

A B D

C E

Out y A

B

E

C

D

C

A

B

D

E

Outz

A C

B D

A

C

B

D

OUTx = A โ‹… B โ‹… D + C โ‹… E

OUTy = ( A + B โ‹… D + E )โ‹… C

OUTz = ( A + C )โ‹…( B + D )

4 problems, 4 pages Exam One Solution 20 September 2006

Problem 2 (2 parts, 20 points) Mixed Logic Design

Implement the following expressions using gates 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 multi-input input AND, OR, NAND, NOR, and NOT gates may be used. Do not simplify the expression.

Part A (8 points) OUTX = A โ‹… B + C โ‹… D + E โ‹… F

A B C D E F

OUTX

switches = 5 x 4 + 2 x 2 = 24T

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

A B C D E

OUTY

OUTZ

switches = 4 x 4 + 2 x 2 = 20T

4 problems, 4 pages Exam One Solution 20 September 2006

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

A

A

B B

C

C

C

D

D D

prime implicants

essential? yes no

B C

C D

A C

A B D

A B C

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

A

A

B B

C

C

C

D

D D

prime implicants

essential? yes no

A + D

C + D

C + D

A + C

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