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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
bonus 4
Total 100 104
Some useful Boolean identities.
- X + 0 = X 2. X · 1 = X
- X + 1 = 1 4. X · 0 = 0
- X + X = X 6. X · X = X
- X + X = 1 8. X · X = 0
- X = X
- X + Y = Y + X 11. XY = Y X Commutative
- X + (Y + Z) = (X + Y ) + Z 13. X(Y Z) = (XY )Z Associative
- X(Y + Z) = (XY + XZ) 15. X + Y Z = (X + Y )(X + Z) Distributive
- 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.
- F = a b + c
F = ( a + b) c
+V
aa
bb
cc
aa bb
cc
FF
- F = x( y + z)
F = x + y z
+V
zz
yy
FF
xx
yy zz
xx
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 = (A + A )D + A B = D + A B
F = (A + B ) D
AA
AA
BB
AA
DD
FF
+V
DD
AA BB
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 = A BCD + A C D + A B D = A BC + A B D (SOP)
F = A ( B + C)(B + D ) (POS)
There are several ways to convert an expression to a product of sums form. The direct way is to make a truth table and then extract the maxterms. Another way is to simplify the SOP expression and then use DeMorgan’s theorems as shown below. A method that we have not covered in class, but that was in your reading, is to use K-maps directly—working with the 0’s. Note, a POS expression is of this form: (A + B)(D + C ), not this: (AB + C)(D + E ), nor this: (A + B)(C + D).
F = A BC + A B D
= A ( ( B + C ) + (B + D) )
= A ( ( B + C )(B + D) )
= A ( B D + B C + C D)
= A ( B D + B C )
= A ( B D) (B C )
= A (B + D )( B + C)
Draw a gate implementation of the following function that exhibits a minimum amount of propagation delay. F = D(A( B C + B) + C A )
Expressions in POS or SOP forms translate to gate implementations with the minimum amount of delay because they have only two levels. Note, the expression does not need to be simplified but it does need to be in POS or SOP form.
F = D(A B C + AB + C A ) = A B CD + ABD + A CD = CD + ABD
D
A
B
C
F
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 = A B + B D + B C
G :
AB
@CD
@ @@
00
A
A
B
B
B
}
}
C C
︷ ︸︸ ︷︷ ︸︸ ︷
D D D
︸ ︷︷ ︸︸ ︷︷ ︸︸ ︷︷ ︸
G = C D + A C + A B D + AB C D
