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Determine the Transistor - Computer Engineering - Solved Exam, Exams of Computer Science

Main points of this past exam are: Determine the Transistor, Incomplete Circuits, Computer Engineering, Switch Level, Mixed Logic, Millennium Time, Capsule That Contains, Tech Engineer, Georgia Tech, Logic Notation

Typology: Exams

2012/2013

Uploaded on 04/08/2013

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ECE 2030 2:00pm Computer Engineering Spring 2001
4 problems, 5 pages Exam One Solution 9 February 2001
1
Problem 1 (3 parts, 30 points) Incomplete Circuits
For each expression below, create a switch level implementation using N and P type switches.
Assume both inputs and their complements are available. Your design should contain no shorts
or floats. Use as few transistors as possible, but do not simplify the expression.
A
BB
A
A
BA
B
Outx
D
C
Outy
B
A
A
BC
D
D
C
Outz
BA
A
B
C
E
E
D
OUTx = BABA + OUTy = DCBA ++ )( OUTz = )()( EDCBA +++
pf3
pf4

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4 problems, 5 pages Exam One Solution 9 February 2001

Problem 1 (3 parts, 30 points) Incomplete Circuits

For each expression below, create a switch level implementation using N and P type switches. Assume both inputs and their complements are available. Your design should contain no shorts or floats. Use as few transistors as possible, but do not simplify the expression.

A

B B

A

A

A B

B

Outx

D C Outy

B A

A

B C

D

D

C

Outz

A B

A

B

C

E

E

D

OUTx = AB + AB OUTy = A + ( BC )+ D OUTz = ( A + B + C )⋅( D + E )

4 problems, 5 pages Exam One Solution 9 February 2001

Problem 2 (2 parts, 25 points) Mixed Logic

In the year 3000, a Georgia Tech engineer opens a millennium time capsule that contains a gate design using mixed logic notation. “Great use of a common sub-expression!” she exclaims, “But you should have used NAND gates.” Extract the two output expressions from the circuit below and then reimplement using NAND gates. Determine the transistor savings of the new design. A B C D

E

OUTx

OUTy F

OUTx = (^) A + BCD

OUTy = (^) BCDEF

NAND gate implementation (use correct mixed logic notation)

A B C D

E

OUTx

OUTy F

Transistors in original implementation 2 x 6 + 1 x 4 + 6 x 2 = 28T

Transistors in NAND implementation 2 x 6 + 1 x 4 + 2 x 2 = 20T

4 problems, 5 pages Exam One Solution 9 February 2001

Part B (10 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.

A

A

B B

C

C

C

D

D D

prime implicants

essential? yes no

C

B D

B D

simplified SOP expression (^) C + BD + BD

Part C (10 points) For the following behavior (in map format), derive a simplified product of sums expression using a Karnaugh Map. Circle and list the prime implicants, indicating which are essential. Then write the simplified POS expression.

A

A

B B

C

C

C

D

D D

prime implicants

essential? yes no

A + D

A + C

C + D

A + C

C + D

A + D

simplified POS expression (^) ( C + D )⋅( A + C )⋅( A + D )or ( A + D )⋅( C + D )⋅( A + C )