Welcome - Jie Bo's Home Page



Chapter 2 – NUMBER SYSTEMS AND CODES2-1.Convert the binary number 101 1011 to its decimal equivalent.a. 5b. 63c. 91d. 92e. 1392-2.What is the weight of the digit ‘3’ in the base-7 number 12345?a. 3b. 7c. 14d. 21e. 492-3.Which of the following has the largest value?a. (101)10b. 10011011)2c. (1111)5d. (9A)16e. (222)82-4.If (321)4 = (57)10, what is the decimal equivalent of (321 000 000)4?a. 57 x 104b. 57 x 106c. 57 x 104d. 57 x 46e. 5742-5.Convert each of the following decimal numbers to binary with at most eight digits in the fractional part, rounded to eight places.a. 2000b. 0.875c. 0.07d. 12.3452-6. Convert each of the decimal numbers in Question 2-5 above to septimal (base 7) with at most four digits in the fractional part, rounded to four places.2-7.Convert each of the decimal numbers in Question 2-5 above to octal (base 8) with at most six digits in the fractional part, rounded to six places.2-8.Convert each of the decimal numbers in Question 2-5 above to hexidecimal (base 16) with at most two digits in the fractional part, rounded to two places.2-9.Which of the following octal values is equivalent to the binary number (110001)2?a. (15)8b. (31)8c. (33)8d. (49)8e. (61)82-10.Convert the binary number (1001101)2 toa. quaternaryb. octalc. decimald. hexadecimal2-11.What is (1011)2 x (101)2?a. (10000)2b. (110111)2c. (111111)2d. (111011)2e. (101101)22-12.Perform the following operations on binary numbersa. (10111110)2 + (10001101)2b. (11010010)2 – (01101101)2c. (11100101)2 – (00101110)22-13.In a 6-bit 2’s complement binary number system, what is the decimal value represented by (100100)2s?a. -4b. 36c. -36 d. -27e. -282-14.In a 6-bit 1’s complement binary number system, what is the decimal value represented by (100100)1s?a. -11b. 43c. -43d. 20e. -202-15.What is the range of values that can be represented in a 5-bit 2’s complement binary system?a. 0 to 31b. -8 to 7c. -8 to 8d. -15 to 15e. -16 to 152-16.In a 4-bit 2’s complement scheme, what is the result of this operation: (1011)2s + (1001)2s?a. 4b. 5c. 20d. -12e. overflow2-17.Assuming a 6-bit 2’s complement system, perform the following subtraction operations:a. (011010)2s – (010000)2sb. (011010)2s– (001101)2sc. (000011)2s – (010000)2s2-18.Assuming a 6-bit 1’s complement system, perform the following subtraction operations:a. (011111)1s – (010101)1sb. (001010)1s – (101101)1sc. (000011)1s – (010000)1s2-20.What does 1 110 1001 represent in this floating-point number scheme: 1-bit sign, 3-bit normalized mantissa, followed by 4-bit 2’s complement exponent?a. 0.125 x 29b. -0.125 x 29c. -0.75 x 2-1d. -0.75 x 2-6e. -0.75 x 2.72-21.What is the decimal value represented by (C3700000)16 in the IEEE Standard 754 single-precision floating point representation?2-22.How to represent (246)10 in the following system/code?a. 10-bit binaryb. BCDc. Excess-3d. 2421 codee. 84-2-1 code2-23.The decimal number 573 is represented as 1111 0110 1011 in an unknown self-complementing code. Find the code for the decimal number 642.2.24.Convert (101011)2 to its corresponding Gray code value.a. (101011)Grayb. (010100)Grayc. (110010)Grayd. (111110)Graye. (43)Gray2-25.Convert (101011)Gray to its corresponding binary value.a. (101011)2b. (010100)2c. (110010)2d. (111110)2e. (010101)2 2-29.According to the fixed-point number format in an 8-bit sign-and-magnitude, fixed at 2, what are the largest and smallest positive values in decimal that can be represented? What are the largest and smallest negative values?2-34.Write the 8-bit sign-and-magnitude, 1’s complement, and the 2’s-complement representations for each of these decimal numbers:8-bit+18+115+79-49-3-100Sign-and-magnitude1’s-complement2’s-complement2-plete the following sequence so that the result is a 3-bit Gray code sequence.011 100 101 0012-40.(a) Convert decimal -125.5 into IEEE Standard 754 single-precision floating point representation, leaving your answer in hexadecimal form.(b) What is the result (in decimal) of interpreting the hexadecimal number 3D40 0000 as an IEEE Standard 754 single-precision floating point number?Answers2-1.(c)2-2.(e)2-3.(c)2-4.(d)2-5.(a) (2000)10 = (1111 1010 000)2(b) (0.875)10 = (0.111)2(c) (0.07)10 = (0.00010010)2(d) (12.345)10 = (1100.01011000)22-6.(a) (2000)10 = (5555)7(b) (0.875)10 = (0.606061)7(c) (0.07)10 = (0.033003)7 / (0.033004)7(d) (12.345)10 = (15.226223)72-7.(a) (2000)10 = (3720)8(b) (0.875)10 = (0.7)8(c) (0.07)10 = (0.0437)8(d) (12.345)10 = (14.2605)82-8.(a) (2000)10 = (7D0)16(b) (0.875)10 = (0.E)16(c) (0.07)10 = (0.12)16(d) (12.345)10 = (C.58)162-9.(e)2-10.(a) (1031)4(b) (115)8(c) (77)10(d) (4D)162-11.(b)2-12.(a) (101001011)2(b) (01100101)2(c) (10110111)22-13.(e)2-14.(d)2-15.(e)2-16.(e)2-17.(a) (001010)2s = (10)10(b) (001101)2s = (13)10(c) (110011)2s = -(13)102-18.(a) (001010)1s = (10)10(b) (011100)1s = (28)10(c) Overflow2-20.(e)2-21.-240 2-22.(a) (0011110110)2(b) (0010 0100 0110)BCD(c) (0101 0111 1001)Excess-3(d) (0010 0100 1100)2421(e) (0110 0100 1010)84-2-12-23.642 = 0100 0000 10012-24.(d)2-25.(c)2-29.-31.752-34.+18+115+79-49-3-100Sign-and-magnitude0001 00100111 00110100 11111011 00011000 00111110 01001’s complement0001 00100111 00110100 11111100 11101111 11001001 10112’s complement0001 00100111 00110100 11111100 11111111 11011001 1100011 111 110 100 101 001 000 0102-39.2-40.(a) C2FB 0000(b) 0.046875 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download