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4 problems, 4 pages Exam One Solutions 13 September 2000
Problem 1 (2 parts, 31 points) Boolean Algebra and Standard Forms
Part A (21 points) For each expression, complete the truth table that represents its behavior.
OUTX = A โ
B โ
C + A โ
B โ
C + A โ
B โ
C + A โ
B โ
C
OUTY =
( A + B + C )โ
( A + B + C )โ
( A + B + C )
OUTZ =
A โ B โ C
A B C OUT X A B C OUT Y A B C OUT Z
Part B (10 Points) Transform each of the following Boolean expressions to a form where they can be implemented using switches (i.e., there should be no bars in the expression except for complements of the inputs A, B, C, etc.). The behavior of the expression should remain unchanged.
Out (^) X = A โ
( B +( C โ
D )) A + B โ
( C + D )
Out (^) Y = ( A + B )โ
C +( D + E )โ
F A^ + B + C + D โ
E โ
F
4 problems, 4 pages Exam One Solutions 13 September 2000
Problem 2 (1 part, 15 points) Incomplete Circuit
For the incomplete circuit are shown below. Complete each circuit by adding the needed switching network so the output is pulled high or low for all combinations of inputs (i.e., no floats or shorts). Then write the expression. Assume both the inputs and their compliments are available.
OUTx = ( A โ
B + C )( E + F )+ D โ
G
4 problems, 4 pages Exam One Solutions 13 September 2000
Problem 4 (2 parts, 25 points) Karnaugh Maps
Part A (10 points) Describe the behavior of the following expression by completing the entries in the Karnaugh Map. You only need to put a 1 and 0 in each box. Do not simplify.
Out = A โ
C โ
B + B โ
D + C โ
D + A โ
B โ
D + A โ
B โ
C โ
D
Part B (15 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.
simplified SOP expression (^) A โ
D + B โ
C + A โ
B โ
C + A โ
C โ
D or
A โ
D + B โ
C + B โ
C โ
D + A โ
C โ
D or A โ
D + B โ
C + B โ
C โ
D + A โ
B โ
D
