EECE 320 Homework 1, Spring 2007

American University of Beirut Department of Electrical and Computer Engineering EECE 320 – Digital Systems Design Assignment 1 Suggested problems: Solve the drill problems of chapter 2. Problem 1: Complete the following table Decimal Octal Hexad ecimal Unsigned Binary Signed- Magnitude Signed 1’s complement Signed 2’s complement 18.75 22.3 12.3 10010.11 010010.11 001101.11 001110.11 -12.875 -14.7 -C.7 1100.111 11100.111 10011.111 10100.111 63.26 76.32 3E.1A 111110.11010 0111110.11010 0000001.11010 0000010.11010 910.3 1616.3 38E.3 1110001110.0011 01110001110.0011 00001110001.0011 0001110010.0011 202 312 0CA 11001010 011001010 000110101 000110110 -74 -112 -04A 1001010 11001010 10110101 10110110 41 051 029 0101001 00101001 01010110 01010111 -5.375 -05.3 -5.3 0101.011 10101.011 11010.011 -5.3 -5.3 -5.3 101.11 1101.11 1010.11 1011.11 Problem 2: Find the 9’s complement of the decimal number 678 546 900. Find the 16’s complement of the hexadecimal number AB45. _ 9’s complement of 678 546 900: 999999999-678546900 = 321453099. _ 15’s complement of AB45: 54BA (By substracting 15 from each digit). 16’s complement of AB45: 15’s complement+1 = 54BA+1 = 54BB. Problem 3: Perform subtraction directly first then using the 2’s complement of the subtrahend for the following numbers: a) 110010 – 110101 Direct substraction: 110010 - 110101 1 111101 (the following number is in signed 2’s complement) Using 2’s complement: 2’s complement of 110101 is 001010+1 = 001011. Performing addition: 110010 +001011 1 11101 (the following number is in signed 2’s complement) b) 1101 – 101000 Direct substraction: 001101 - 101000 1 100101 (the following number is in signed 2’s complement) Using 2’s complement: 2’s complement of 101000 is 010111+1 = 011000. Performing addition: 001101 +011000 1 100101 (the following number is in signed 2’s complement) Problem 4: Add the following Hexadecimal numbers: 4AB3F7C + 3F6FE4D. 4 (10) (11) 3 (15) 7 (12) + 3 (15) 6 (15) (14) 7 (13) 8 (10) 2 4 (13) (15) 9 Therefore 4AB3F7C + 3F6FE4D = 8A24DF9 Problem 5: Add the following BCD numbers and verify your result by performing the corresponding decimal operation: 1000+0101; 0011+0110; 1001+1001; 0010 1000 + 1000 1000. 1)In Binary: 1000+0101 = 1101. In Decimal: 1000 = 8. 0101 = 5. 1101 = 13. And 8 + 5 = 13. 2) In Binary: 0011+0110 = 1001. In Decimal: 0011 = 3. 0110 = 6. 1001 = 9. And 6 + 3 = 9 . 3) In Binary: 0011+0110 = 1001. In Decimal: 0011 = 3. 0110 = 6. 1001 = 9. And 6 + 3 = 9 4) In Binary: 1001+1001 = 10010. In Decimal: 1001 = 9. 1001 = 9. 10010 = 18. And 9 + 9 = 18. 5) In Binary: 00101000 + 10001000 = 10110000. In Decimal: 00101000 = 40. 10001000 = 136. 10110000 = 176. And 40 + 136 = 176 Problem 6: Using the ASCII table, convert the following message to binary. Good Night! G = 01000111. o = 01101111. d = 01100100. N = 01001110. i = 01101001. g = 01100111. h = 01101000. t = 01110100. Therefore Good Night == 01000111 0110111101101111 01100100 0100111001101001011001110110100001110100