CSC270 Homework 1 Solutions

Solutions for Homework #1

<limitsmith> Problem #1 was worth 1 point. Problems #2 and 3 were worth 1.5 points each.

Problem #1

Neither assertion is true.

But, if we had substituted U for OR and T for AND in the first assertion, and U for AND and T for OR in the second, both assertions would have been true.

This illustrates the duality of the system we have. if we decide that alpha is 0 and beta 1, then the U operator is an OR, and the T operator is an AND. If we decide the opposite, then U is an AND, and T is an OR.

This explains why de Morgan's laws hold, and why if the NAND is the universal gate, so must the NOR be.

Problem #2

f = a.b + a' . ( b'.c +d' ) obtained with a Karnaugh map.

g is f', so we can bar the expression above and get the result.

Or we can generate a Karnaugh map for g and get g= a'.b + a.d.( c + b' )

Problem #3

Q = a'.b' + b.c

The logic diagram with ands, ors, and nots is obvious once you have Q expressed as above.

Draw the truth table for Q, and put a 1 in rows where a=0 and b=0, and in rows where b=1 and c=1. Then you lift the minterms easily:

Q = Σ( 0, 1, 3, 7 )

</limitsmith>