Confguration - Computer Engineering - Exam, Exams of Computer Science

Main points of this exam paper are: Confguration, Transistors, Standard, Truth Table, K-Map, Boolean Expression, Switch Level Circuit, Bottom Half, Boolean Algebra Manipulations, Truth Table Representation

Typology: Exams

2012/2013

Uploaded on 04/08/2013

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GEORGIA INSTITUTE OF TECHNOLOGY
School of Electrical and Computer Engineering
EE 2030
QUIZ #1
Thursday, September 16, 1999
Name:
Last, First
Closed book, closed notes.
None of the problems require involved calculations. Reconsider your approach before
doing something tedious.
Clearly identify each answer.
Part pts Score
1 16
2 15
3 15
4 14
5 15
6 10
7 15
Total 100
Some useful Boolean identities.
1. X+ 0 = X2. X·1 = X
3. X+ 1 = 1 4. X·0 = 0
5. X+X=X6. X·X=X
7. X+X= 1 8. X·X= 0
9. X=X
10. X+Y=Y+X11. XY =Y X Commutative
12. X+ (Y+Z) = (X+Y) + Z13. X(Y Z) = (X Y )ZAssociative
14. X(Y+Z) = (XY +X Z) 15. X+Y Z = (X+Y)(X+Z) Distributive
16. X+Y=X·Y17. X·Y=X+YDeMorgan’s
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GEORGIA INSTITUTE OF TECHNOLOGY

School of Electrical and Computer Engineering

EE 2030

QUIZ

Thursday, September 16, 1999

Name:

Last, First

• Closed book, closed notes.

• None of the problems require involved calculations. Reconsider your approach before

doing something tedious.

• Clearly identify each answer.

Part pts Score

Total 100

Some useful Boolean identities.

  1. X + 0 = X 2. X · 1 = X
  2. X + 1 = 1 4. X · 0 = 0
  3. X + X = X 6. X · X = X
  4. X + X = 1 8. X · X = 0
  5. X = X
  6. X + Y = Y + X 11. XY = Y X Commutative
  7. X + (Y + Z) = (X + Y ) + Z 13. X(Y Z) = (XY )Z Associative
  8. X(Y + Z) = (XY + XZ) 15. X + Y Z = (X + Y )(X + Z) Distributive
  9. X + Y = X · Y 17. X · Y = X + Y DeMorgan’s

Implement the following expressions using n-type and p-type transistors in the standard CMOS configuration that we used in class.

  1. F = a b + c
  2. F = x( y + z)

Complete the following switch level circuit by designing the bottom half (the pull-down net- work). Provide the expressions for F and G = F.

F =

F =

AA

AA

BB

AA

DD

FF

+V

By using Boolean algebra manipulations, express the following function as a sum of products (SOP) and as a product of sums (POS). You are not required to simplify but you may if you want.

F = (A + BC ) D + A D (C + B )

F = (SOP)

F = (POS)

Draw a gate implementation of the following function that exhibits a minimum amount of propagation delay. F = D(A( B C + B) + C A )

For the following functions, derive a simplified sum of products (SOP) expression using a Kar- naugh map. Circle the prime implicants used in the simplified expression.

F :

AB

@CD

@ @@

00

A

A

  

    

 

B

B

B

}

  

  }

C C

︷ ︸︸ ︷︷ ︸︸ ︷

D D D

︸ ︷︷ ︸︸ ︷︷ ︸︸ ︷︷ ︸

F =

G :

AB

@CD

@ @@

00

A

A

  

    

 

B

B

B

}

  

  }

C C

︷ ︸︸ ︷︷ ︸︸ ︷

D D D

︸ ︷︷ ︸︸ ︷︷ ︸︸ ︷︷ ︸

G =