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4 problems, 5 pages Exam Two Solutions 18 October 2000
Problem 1 (2 parts, 22 points) Arithmetic
Part A (10 points) The adder below adds two four bit numbers A and B and produces a four bit result S. Add extra digital logic to support subtraction as well as addition. Label inputs X 3 , X 2 ,
X 1 , X 0 , Y 3 , Y 2 , Y 1 , Y 0 , ADD / SUB and outputs Z 3 , Z 2 , Z 1 , Z 0. Do not consider error determination here.
Part B (12 points) Consider a function that determines whether an error occurred when adding or subtracting two two’s complement numbers. Suppose XMSB and YMSB are the most significant bits of the two numbers being processed and ZMSB is the most significant bit of the result. Complete the truth table below to indicate an error (Err) when one has occurred.
Addition (X+Y=Z) Subtraction (X-Y=Z) XMSB YMSB ZMSB Err^ XMSB YMSB ZMSB Err 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 1 0 1 1 1 0 0 0 0 1 1 0 0 1 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 1 1 1 1 1 0 1 1 1 0
4 problems, 5 pages Exam Two Solutions 18 October 2000
Problem 2 (3 parts, 28 points) Transparent Latch
Part A (10 points) For the circuit below, complete the truth table describing its behavior. Indicate the cases where a meta-stable state can result.
A B Out
0 0 1 meta 1 0 0 0 1 1 1 1 Qo
Part B (12 points) Now implement a transparent latch using this circuit plus additional basic gates (AND, OR, NAND, NOR, and NOT). Label inputs IN and EN. Label output OUT. Do not attempt to employ mixed logic notation.
Part C (6 points) How many transistors are employed in the transparent latch in Part B? How many transistors are employed in the widely used transparent latch implementation using only inverters and pass gates?
transistors used in part B 18
transistors used in inverter/pass gate version of transparent latch 10
4 problems, 5 pages Exam Two Solutions 18 October 2000
Problem 4 (3 parts, 30 points) Numbers and Karnaugh Maps
Part A (12 points) For the following behavior (in map format), derive a simplified products of sums expression using a Karnaugh Map. Circle and list the prime implicants, indicating which are essential. Then write the simplified POS expression.
simplified POS expression (^) ( A + B )⋅( C + D )
Part B (9 points) Convert some binary values (and powers of two) into decimal notation:
binary notation decimal notation
1001.101 8+1+.5+.125 = 9.
111101010 256+128+64+32+8+2=
225 32 million
Part C (9 points) Convert the following octal values into hexadecimal notation:
octal notation hexadecimal notation
100 10000002 = 40
450 1001010002 = 128
123.45 1010011.1001012 = 53.
