HOMEWORK # 2 SOLUTIO


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1 HOMEWORK # 2 SOLUTIO Problem 1 (2 points) a. There are 313 characters in the Tamil language. If every character is to be encoded into a unique bit pattern, what is the minimum number of bits required to do this? 8 bits can used to encode 2 8 = 256 characters and 9 bits can be used to encode 2 9 = 512 characters. So, we would need 9 bits. b. How many more characters can be accommodated in the language without requiring additional bits for each character? = 199 Problem 2 (4 points) Convert the following 2's complement binary numbers to decimal numbers. a First bit is 1. So it is a ve number. 2 s complement of 1010 = = So the answer is 6. b This is a +ve number since it starts with 0 Answer is 2. c This is a ve number since it starts with 1. Its 2 s complement is = So the answer is 1 d This is a +ve number since it starts with 0. The answer is 31. Problem 3 (4 points) a. What is the largest positive number one can represent in a 16bit 2's complement code? Write your result in binary and decimal binary and = decimal
2 b. What is the greatest magnitude negative number one can represent in a 16bit 2's complement code? Write your result in binary and decimal binary and = decimal c. What is the largest positive number one can represent in a 16bit signed magnitude code? Write your result in binary and decimal binary and = decimal d. What is the greatest magnitude negative number one can represent in a 16bit signed magnitude code? Write your result in binary and decimal binary and (2 15 1) = decimal Problem 4 (2 points) What are the 8bit patterns used to represent each of the characters in the string "This Is Easy!"? (Only represent the characters between the quotation marks.) Character Hex (from ASCII table) Binary equivalent T h i s Space I s Space E a s y ! Problem 5 (4 points) Convert the following decimal numbers to 8bit 2's complement binary numbers. If there is problem while doing this, describe it. a
3 b. 64 c Does not fit in an 8bit signed number d Problem 6 (4 points) The following binary numbers are 4bit 2's complement binary numbers. Which of the following operations generate overflow? Justify your answers by translating the operands and results into decimal. a No overflow Answer is 0100 binary = 4 decimal [7 + (3)] b Overflow Answer is 0111 binary = 7 decimal. But actual answer is 9 [(7) + (2)]
4 c No overflow Answer is 1000 binary = 8 decimal [(1) + (7)] d Overflow Answer is 1000 binary = 8 decimal. But actual answer is 8 [3 + 5] Problem 7 (2 points) A computer programmer wrote a program that adds two numbers. The programmer ran the program and observed that when 5 is added to 8, the result is the character m. Explain why this program is behaving erroneously. The error that is occurring here is that 5 and 8 are being interpreted as characters 5 and 8 respectively. As a result, the addition that is taking place is not 5 + 8; rather, it is If we look up values in the ASCII table, 5 is 0x35 and 8 is 0x38. 0x35 + 0x38 = 0x6d, which is the ASCII value for m.
5 Problem 8 (2 points) Compute the following: a. OT(1011) OR (1011) NOT(1011) = 0100 Answer = (0100) OR (1011) = 1111 b. OT(1001 A D (0100 OR 0110)) 0100 OR 0110 = AND 0110 = 0000 Answer = NOT(0000) = 1111 Problem 9 (4 points) Write the decimal equivalents for these IEEE floating point numbers. a Sign bit is 0 (+ve). Exponent = 127. Fraction = 1* *22 = 0.75 Answer = (+) 1.fraction * 2exponent 127 = 1.75 * 2 0 = 1.75 b Sign bit is 1 (ve). Exponent = 125. Fraction = 1*21 = 0.5 Answer = () 1.fraction * 2exponent 127 = * 22 =
6 Problem 10 (2 points) Given a black box which takes n bits as input and produces one bit for output, what is the maximum number of unique functions that the black box can implement? (Hint: Try to visualize a truth table for a single function of n bits. Determine how many rows such a truth table has. Then determine how many combinations are possible with the number of rows that you just found) Consider a single function that this black box implements. If there are n binary inputs, the truth table contains 2 n rows. Now, each of these rows in the truth table can be filled with 0 or 1. The number of ways in which we can fill in these rows (using 0 and 1) gives us the number of unique functions. Since each of the rows can be filled in using 2 possible values and since the number of rows is 2 n, the number of ways = 2 power (2 n ).
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