Spring 2001 CSCI 255 Homework 2 Solution

Problem 1

Add the following six-bit twos-complement number. Which, if any, of the additions result in an overflow?

111000 + 101010100010
010111 + 001111100110

In the first problem, two negative numbers are added and a negative number is the result. This is not an overflow. In the second problem, two postive numbers are added and a negative number is the result. This is an overflow.

Problem 2

Compute the following bit-wise logical operations on six-bit binary numbers.

011011 AND (101000 OR 000001)001001
NOT(001100) AND NOT(011110)100001

Problem 3

Translate the first four characters of your last name into an ASCII hexadecimal string.

Broc42726f6b

Use the table of page 514 to look up the ASCII codes of the four characters.

Problem 4

The following 32-bit strings represent IEEE floating point numbers. Translate the bits into their "normal" floating point representation. This is not an easy problem.

0100000100101000000000000000000010.5
11000100110000000000000000000000-1536

Hopefully, you used the spreadsheet from the January 24 lecture to solve this problem. If not, go look at it now. It explains IEEE floating point better than words.

Problem 5

Draw a circuit that implements the truth table shown below. The "inputs" to the truth table are A, B, and C. The output is Z. Review section 3.3.4 before attempting this problem.

inputoutput
ABCZ
0000
0010
0101
0111
1000
1010
1101
1110

There are three 1's in the output. They are in the rows for A' B C', A' B C, and A B C'. Consequently the Boolean expression is A' B C' + A' B C + A B C'. If you've taken MATH 261, you should be able to determine that this expression can be simplified to A' B + B C'.

Problem 6

Complete a truth table to describe the logic circuit shown below.
Problem 6 logic circuit

inputoutput
ABCX
0000
0010
0100
0111
1000
1011
1100
1111

You can solve the problem in a couple of ways. First, you can just "plug in" all eight input combinations and see what they produce on the output. Or, you can determine what the function is algebraicly, in this case it simplifies to A C + B C, and then complete a truth table. Unless you are particularly good in Boolean algebra, you'll find the first method easier.