DIGITAL DESIGN THIRD EDITION M MORRIS MANO CHAPTER

DIGITAL DESIGN THIRD EDITION M. MORRIS MANO CHAPTER 1 : BINARY SYSTEMS PROBLEMS

1. 1 -) List the octal and the hexadecimal numbers from 16 to 32. Using A and B for the last two digits, list the numbers from 10 to 26 in base 12. Octal : 16 = 8¹ x 2 + 8º x 0 => (16)10 = (20)8 32 = 8¹ x 4 + 8º x 0 => (32)10 = (40)8 20, 21, 22, 23, 24, 25, 26, 27, 30, 31, 32, 33, 34, 35, 36, 37, 40

Hexadecimal : 16 = 16¹ x 1 + 16º x 0 => (16)10 = (10)16 32 = 16¹ x 2 + 16º x 0 => (32)10 = (20)8 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1 A, 1 B , 1 C, 1 D, 1 E, 1 F, 20 Base-12 : 10 = 12º x A => (10)10 = (A)12 26 = 12¹ x 2 + 12º x 2 => (26)10 = (22)12 A, B, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1 A, 1 B, 20, 21, 22

1. 2 -) What is the exact number of bytes in a system that contains (a) 32 K byte, (b)64 M bytes, and (c)6. 4 G byte ? (a) 32 K byte: 1 K = 2¹º = 1, 024 32 K = 32 x 2¹º = 32 x 1, 024 = 32, 768 32 K byte = 32, 768 byte

(b) 64 M byte: 1 M = 2²º = 1, 048, 576 64 M = 64 x 2²º = 64 x 1, 048, 576 = 67, 108, 864 64 M byte = 67, 108, 864 byte (c) 6. 4 G byte: 1 G = 2³º = 1, 073, 741, 824 6. 4 G = 6. 4 x 2³º = 6. 4 x 1, 073, 741, 824 = 6, 871, 747, 674 6. 4 G byte = 6, 871, 747, 674 byte

1. 3 -) What is the largest binary number that can be expressed with 12 bits? What is the equivalent decimal and hexadecimal ? Binary: (111111)2 Decimal: (111111)2 = 1 x 2º+ 1 x 2¹ + 1 x 2² +…. . + 1 x 2¹¹ + 1 x 2¹² (111111)2 = 4, 095 Hexadecimal: (1111)2 F F F = (FFF)16

1. 4 -) Convert the following numbers with the indicated bases to decimal : (4310)5 , and (198)12. (4310)5 = 0 x 5º + 1 x 5¹ + 3 x 5² + 4 x 5³ = 0 + 5 + 75 + 500 (4310)5 = 580 (198)12 = 8 x 12º + 9 x 12¹ + 1 x 12² = 8 + 108 + 144 (198)12 = 260

1. 5 -) Determine the base of the numbers in each case for the following operations to be correct : (a) 14/2 = 5 ; (b) 54/4 = 13 ; (c) 24+17 = 40. (a) (14)a / (2)a = (5)a Þ (4 x aº + 1 x a¹) / (2 x aº) = 5 x aº Þ (4 + a) / 2 = 5 Þ 4 + a = 10 Þa=6

1. 6 -) The solution to the quadratic equation x² 11 x + 22 = 0 is x=3 and x=6. What is the base of the numbers? x² - 11 x + 22 = (x – 3). (x – 6) x² - 11 x + 22 = x² - (6 + 3)x + (6. 3) Þ (11)a = (6)a + (3)a Þ 1+a=6+3 Þa=8

1. 7 -) Express the following numbers in decimal : (10110. 0101)2 , (16. 5)16 , (26. 24)8. ( 1 0 1 1 0. 0 1 )2 4 3 2 1 0 = 2¹ + 2² + (2^4) +( 2^-2) + (2^-4) -1 -2 -3 -4 (10110. 0101)2 = 2 + 4 + 16 + (1/4) + (1/16) (10110. 0101)2 = 22. 3125

( 1 6. 5 )16 = 6 x 16º + 1 x 16¹ + 5 x (16^-1) 1 0 -1 (16. 5)16 = 6 + 16 + (5/16) (16. 5)16 = 22. 3125 ( 2 6. 2 4 )8 = 6 x 8º + 2 x 8¹ + 2 x (8^-1) + 4 x (8^-2) 1 0 -1 -2 (26. 24)8 = 6 + 16 + (2/8) + (4/64) (26. 24)8 = 22. 3125

1. 8 -) Convert the following binary numbers to hexadecimal and to decimal : (a) 1. 11010 , (b) 1110. Explain why the decimal answer in (b) is 8 times that of (a) ( 1. 1101 0 )2 = ( 1. D )16 = 1 x 16º + D x (16^-1) (b) 1 D 0 0 -1

1. 9 -) Convert the hexadecimal number 68 BE to binary and then from binary convert it to octal. (68 BE) 16 Binary form: (0110 1000 1011 1110)2=(0110100010111110)2 6 8 B E Octal form: (0 110 100 010 111 110)2 0 6 4 2 7 6 =(064276)8

(a) 1. 10 -) Convert the decimal number 345 to binary in two ways : Convert directly to binary; Convert first to hexadecimal, then from hexadecimal to binary. Which method is faster ?

Method 1: Number 345 Divided by Remainder 2 345/2=172 1 172/2=86 0 86 86/2=43 0 43 43/2=21 1 21 21/2=10 10/2=5 0 5 5/2=2 1 2 2/2=1 1 (345)10

Method 2: Number Divided by 16 Remainder 345/16=21 9 21 21/16=1 5 (345)10=(159)16 (1 1001)2

1. 11 -) Do the following conversion problems : (a) Convert decimal 34. 4375 to binary. (b) Calculate the binary equivalent of 1/3 out to 8 places. Then convert from binary to decimal. How close is the result to 1/3 ? (c) Convert the binary result in (b) into hexadecimal. Then convert the result to decimal. Is the answer the same ?

(a) 34. 4375 34 0. 4375 34: 2=17 r=0 0. 4375*2=0. 875 r=0 17: 2=8 r=1 0. 875*2=1. 75 r=1 8: 2=4 r=0 0. 75*2=1. 5 r=1 4: 2=2 r=0 0. 5*2=1. 0 r=1 2: 2=1 r=0 0*2=0 r=0 0. 4375=(0. 01110)2 34=(100010)2 34. 4375=(100010. 01110)2

(b) 1/3=0. 3333… 0. 33333*2=0. 66666 r=0 0. 66666*2=1. 33332 r=1 0. 33332*2=0. 66664 r=0 0. 66664*2=1. 33328 r=1. . . 0. 3333…=(0. 010101…. )= 0+ ¼ + 0 + 1/8 + 0 + 1/32 +… =~0. 33333…

(c) 0. 01010…=0. 0101 (0. 555. . )16=5/16 +5/256 +5/4096 +…=~0. 33203

1. 12 -) Add and multiply the following numbers without converting them to decimal. (a) Binary numbers 1011 and 101. (b) Hexadecimal numbers 2 E and 34. (a) 1011 (11) 101 (5) +_____ 10000(16) 1011(11) 101(5) x_____ 1011 0000 + 1011 _____ 110111 (55)

(b) 2 E (46) 34 (52) +____ 62 (98) 2 E 34 x____ B 8 8 A +____ 958(2392)

1. 13 -) Perform the following division in binary : 1011111 ÷ 101. (1011111)2=95 (101)2=5 95/5=19 (10011)2 1011111 101 10011 000111 101 0101 0000

1. 14 -) Find the 9’s- and the 10’s-complement of the following decimal numbers : (a) 98127634 (b) 72049900 (c) 10000000 (d) 0000. 9’s comlements : (a) 9999 -98127634=01872365 (b) 9999 -72049900=27950099 (c) 9999 -10000000=89999999 (d) 9999 -0000000=9999

10’s complements (a)10000 - 98127634= 01872366 (b)10000 -72049900=27950100 (c)10000 -10000000=90000000 (d)10000 -0000000=0000

1. 15 -) (a) Find the 16’s-complement of AF 3 B. (b) Convert AF 3 B to binary. (c) Find the 2’s-complement of the result in (b) (d) Convert the answer in (c) to hexadecimal and compare with the answer in (a)16^5 -AF 3 B=50 C 5 (b)(AF 3 B)16=1010 1111 0011 1011 (c)1010111100111011 0101000011000101 (d)(d)0101 0000 1100 0101= 50 C 5

1. 16 -) Obtain the 1’s and 2’S complements of the following binary numbers : (a)11101010 (b)01111110 (c)00000001 (d)10000000 (e)0000 1’s complements: (a) 00010101 (b)10000001 (c)11111110 (d)01111111 (e)1111 2’s complement : (a) 00010110 (b)10000010 (c)1111 (d)10000000 (e)0000

1. 17 -) Perform subtraction on the following unsigned numbers using the 2’s-complement of the subtrahend. Where the result shoud be negative, 10’s complement it and affix a minus sign. Verify your answers. (a) 7188 -3049 (b)150 -2100 (c)2997 -7992 (d)1321 -375

(a)7188+6951=4139 answer is correct. (b)150+7900=8050 One carry out so correct answer=-1950 (c)2997+2008=5005 correct answer=-4995 (d)1321+9625=0946 answer is correct. One carry out so

1. 18 -) Perform subtraction on the following unsigned binary numbers using the 2’scomplement of the subtrahend. Where the result should be negative, 2’s complement it and affix a minus sign. (a)11011 -11001 (b)110100 -10101 (c)1011110000 (d)101010 -101011

(a)11011+00111=00010(27 -25=2) (b)110100+01011=011111(52 -21=31) (c)1011+010000=011011 -100101(11 -48=-37) (d)(d)101010+010101=111111 -000001(42 -43=-1)

1. 19 -) The following decimal numbers are shown in sign- magnitude form : +9826 and +801. Convert them to signed 10’s-complement form and perform the following operations : (Note that the sum is +10627 and requires six digits). (a) (+9826)+(+801) (b)(+9826)+(-801) (c)(-9826)+(+801) (d)(-9826)+(-801)

(a)009826+00801=010627 (b)(b)009826+999199=09025 (c)(c)990174+000801=990975 -09025 (d)(d)990174+999199=989373 -10627

1. 20 -) Convert decimal +61 and +27 to binary using the signed-2’s complement representation and enough digits to accomodate the numbers. Then perform the binary equivalent of (+27) + (61) , (- 27) + (+61) and (-27) + (- 61). Convert the answers back to ecimal and verify that they are correct.

+61=0111101 +27=0011011 -61=1000011 -27=1100101 (a)27+(-61)=0011011+1000011=1011110 (b)-27+(+61)=1100101+0111101=0100010 (c)-27+(-61)= 1100101+1000011=0101000(overflow) 11100101+11000011=10101000

1. 21 -) Convert decimal 9126 to both BCD and ASCII codes. For ASCII, an odd parity bit is to be appended at the left. BCD: 1001 0010 0110 ASCII: 10111001 00110010 10110110

1. 22 -) Represent the unsigned decimal numbers 965 and 672 in BCD and then show the steps necessary to form their sum. 965= 1001 0110 0101 672= 0110 0111 0010 +___ ____ 1 0000 1101 0111 +0110 +_________ 0001 0110 0011 0111 (1637)10

1. 23 -) Formulate a weighted binary code for the decimal digits using weights 6, 3, 1, 1.

6 3 1 1 Decimal 0 0 0 0 1 1 2 0 1 0 0 3 0 1 1 0 4(0101) 0 1 1 1 5 1 0 0 0 6 1 0 0 1 7(1010) 1 0 1 1 8 1 1 0 0 9

1. 24 -) Represent decimal number 6027 in (a) BCD, (b) excess-3 code, and (c) 2421 code. (a)6027 BCD : 0110 0000 0010 0111 (b)excess 3: 1001 0011 0101 1010 (c)(c)0110 0000 0010 1101

1. 25 -) Find the 9’s complement of 6027 and express it in 2421 code. Show that the result is the 1’s complement of the answer to (c) in Problem 1. 24. This demonstrates that the 2421 code is self-complementing. 9’s complement of 6027 is 3972 6027 as 2421 code is 0110 0000 0010 1101 3972 as 2421 code is 0011 1101 0010

1. 26 -) Assign a binary code in some orderly manner to the 51 playing cards. Use the minimum number of bits. 2^4 =16 2^5 =32 2^6=64 6 bits are necessary.

1. 27 -) Write the expresion “G. Boole” in ASCII using an eight-bit code. Include the period and the space. Treat the leftmost bit of each character as a parity bit. Each 8 -bit code shouls have even parity. G. B O O L E (01000111)(00101110) (01000010) (01101111) (01101100) (01100101)

1. 28 -) Decode the following ASCII code : 1001010 1100001 1101110 1100101 0100000 1000100 1101111 1100101. Jane Doe

1. 29 -) The following is a string of ASCII characters whose bit patterns have benn converted into hexadecimal for compactness : 4 A EF 68 6 E 20 C 4 EF E 5. Of the 8 bits in each pair of digits, the leftmost is a parity bit. The remaining bits are the ASCII code. 01001010 J 11101111 O 01101000 01101110 00100000 11000100 11101111 11100101 H N ( space) D O E

1. 30 -) How many printing characters are there in ASCII ? How many of them are special characters (not letters or numerals) ? 94 characters 62 of them are numbers and letters. 32 of them are special characters.

1. 31 -) What bit must be complemented to change an ASCII letter from capital to lowercase, and vice versa ? Cevap: Bir ASCII karakteri büyük harften küçük harfe çevirmek için sağdan 6. bit 0 iken 1 yapılır. Küçükten büyüğe çevrilecekse 1 iken 0 yapılır.

1. 32 -) The state of a 12 -bit register is 100010010111. What is its content if it represents (a) three decimal digits in BCD? (b) three decimal digits in the excess-3 code? (c) three decimal digits in 84 -2 -1 code? (d) binary number?

Three Decimal Digits in BCD: 1000 1001 0111 897 Three Decimal Digits in Exces-3 Code: 1000 1001 0111 (8 -3) (9 -3) 564 (7 -3) Three Decimal Digits in the 8 -4 -2 -1 Code: 1000 1001 0111 8 9 897 7 Binary Code: 100010010111 2^11+2^7+2^4+2^2+2+1=2199

1. 33 -) List the ASCII code for the 10 decimal digits with an even parity bit in the leftmos position. 00110000 10110001 10110010 0011 10110100 00110101 00110110111 10111000 00111001

1. 34 -) Assume a 3 -input AND gate with output F and a 3 -input OR gate with output G. Inputs are A, B, and C. Show the signals (by means of a timing diagram) of the outputs F and G as functions of three inputs ABC. Use all possible combinations of ABC.

F: A , BX , CX NOT: X’ler HIGH ya da LOW olabilir , C AX , BX , C