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• 1. DIGITAL DESIGN THIRD EDITION M. MORRIS MANO CHAPTER 1 : BINARY SYSTEMS PROBLEMS
• 2. 1.1-) List the octal and the hexadecimal numbersfrom 16 to 32. Using A and B for the last twodigits, list the numbers from 10 to 26 in base 12 .Octal : 16 = 8&#xB9; x 2 + 8&#xBA; x 0 =&gt; (16)10 = (20)8 32 = 8&#xB9; x 4 + 8&#xBA; x 0 =&gt; (32)10 = (40)8 20, 21, 22, 23, 24, 25, 26, 27, 30, 31, 32, 33, 34, 35, 36, 37, 40
• 3. Hexadecimal : 16 = 16&#xB9; x 1 + 16&#xBA; x 0 =&gt; (16)10 = (10)16 32 = 16&#xB9; x 2 + 16&#xBA; x 0 =&gt; (32)10 = (20)8 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B , 1C, 1D, 1E, 1F, 20Base-12 : 10 = 12&#xBA; x A =&gt; (10)10 = (A)12 26 = 12&#xB9; x 2 + 12&#xBA; x 2 =&gt; (26)10 = (22)12 A, B, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 20, 21, 22
• 4. 1.2-) What is the exact number of bytes in asystem that contains (a) 32K byte, (b)64M bytes,and (c)6.4G byte ?(a) 32K byte: 1K = 2&#xB9;&#xBA; = 1,024 32K = 32 x 2&#xB9;&#xBA; = 32 x 1,024 = 32,768 32K byte = 32,768 byte
• 5. (b) 64M byte: 1M = 2&#xB2;&#xBA; = 1,048,576 64M = 64 x 2&#xB2;&#xBA; = 64 x 1,048,576 = 67,108,864 64M byte = 67,108,864 byte(c) 6.4G byte: 1G = 2&#xB3;&#xBA; = 1,073,741,824 6.4G = 6.4 x 2&#xB3;&#xBA; = 6.4 x 1,073,741,824 = 6,871,747,674 6.4G byte = 6,871,747,674 byte
• 6. 1.3-) What is the largest binary number that canbe expressed with 12 bits? What is the equivalentdecimal and hexadecimal ?Binary:(111111111111)2Decimal:(111111111111)2 = 1x 2&#xBA;+ 1 x 2&#xB9; + 1 x 2&#xB2; +&#x2026;..+ 1 x 2&#xB9;&#xB9; + 1 x 2&#xB9;&#xB2;(111111111111)2 = 4,095Hexadecimal:(1111 1111 1111)2 = (FFF)16 F F F
• 7. 1.4-) Convert the following numbers with theindicated bases to decimal : (4310)5 , and(198)12 .(4310)5 = 0 x 5&#xBA; + 1 x 5&#xB9; + 3 x 5&#xB2; + 4 x 5&#xB3; = 0 + 5 + 75 + 500(4310)5 = 580(198)12 = 8 x 12&#xBA; + 9 x 12&#xB9; + 1 x 12&#xB2; = 8 + 108 + 144(198)12 = 260
• 8. 1.5-) Determine the base of the numbers in eachcase 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 &#x21D2; (4 x a&#xBA; + 1 x a&#xB9;) / (2 x a&#xBA;) = 5 x a&#xBA; &#x21D2; (4 + a) / 2 = 5 &#x21D2; 4 + a = 10 &#x21D2;a=6
• 9. (b) (54)b / (4)b = (13)b &#x21D2; (4 x b&#xBA; + 5 x b&#xB9;) / (4 x b&#xBA;) = 3 x b&#xBA; + 1 x b&#xB9; &#x21D2; (4 + 5b) / 4 = 3 + b &#x21D2; 4 + 5b = 12 + 4b &#x21D2;b=8(c) (24)c + (17)c = (40)c &#x21D2; (4 x c&#xBA; + 2 x c&#xB9;) + (7 x c&#xBA; + 1 x c&#xB9;) = 4 x c&#xB9; &#x21D2; 4 + 2c + 7 + c = 4c &#x21D2; c = 11
• 10. 1.6-) The solution to the quadratic equation x&#xB2; -11x + 22 = 0 is x=3 and x=6. What is the baseof the numbers? x&#xB2; - 11x + 22 = (x &#x2013; 3) . (x &#x2013; 6) x&#xB2; - 11x + 22 = x&#xB2; - (6 + 3)x + (6.3) &#x21D2; (11)a = (6)a + (3)a &#x21D2;1+a=6+3 &#x21D2;a=8
• 11. 1.7-) Express the following numbers in decimal :(10110.0101)2 , (16.5)16 , (26.24)8 .( 1 0 1 1 0 . 0 1 0 1 )2 = 2&#xB9; + 2&#xB2; + (2^4) +( 2^-2) + (2^-4) 4 3 2 1 0 -1 -2 -3 -4(10110.0101)2 = 2 + 4 + 16 + (1/4) + (1/16)(10110.0101)2 = 22.3125
• 12. ( 1 6 . 5 )16 = 6 x 16&#xBA; + 1 x 16&#xB9; + 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 8&#xBA; + 2 x 8&#xB9; + 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
• 13. 1.8-) Convert the following binary numbers tohexadecimal and to decimal : (a) 1.11010 , (b)1110.10 . Explain why the decimal answer in (b)is 8 times that of (a) . (a) ( 1 . 1101 0 )2 = ( 1 . D )16 = 1 x 16&#xBA; + D x (16^-1) 1 D 0 0 -1
• 14. 1.9-) Convert the hexadecimal number 68BE tobinary and then from binary convert it to octal . (68BE) 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
• 15. (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 ?
• 16. Method 1: Number Divided by 2 Remainder(345)10 345 345/2=172 1 172 172/2=86 0 86 86/2=43 0 43 43/2=21 1 21 21/2=10 1 10 10/2=5 0 5 5/2=2 1 2 2/2=1 1
• 17. Method 2: Divided by Number Remainder 16 345 345/16=21 9 21 21/16=1 5 (345)10=(159)16 (1 101 1001)2
• 18. 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 ?
• 19. (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
• 20. (b) 1/3=0.3333&#x2026; 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&#x2026;=(0.010101&#x2026;.)= 0+ &#xBC; + 0 + 1/8 + 0 + 1/32 +&#x2026; =~0.33333&#x2026;
• 21. (c) 0.010101010&#x2026;=0.0101 0101 0101(0.555..)16=5/16 +5/256 +5/4096 +&#x2026;=~0.33203
• 22. 1.12-) Add and multiply the following numberswithoutconverting them to decimal.(a) Binary numbers 1011 and 101 .(b) Hexadecimal numbers 2E and 34 . (a) 1011 (11) 1011(11) 101 (5) 101(5) +__________ x_____ 10000(16) 1011 0000 + 1011 _________ 110111 (55)
• 23. (b) 2E (46) 2E 34 (52) 34 +____ x____ 62 (98) B8 8A +____ 958(2392)
• 24. 1.13-) Perform the following division in binary :1011111 &#xF7; 101 . (1011111)2=95 (101)2=5 95/5=19&#xF0E0; (10011)2 1011111 101 101 10011 000111 101 0101 101 0000
• 25. 1.14-) Find the 9&#x2019;s- and the 10&#x2019;s-complement ofthe following decimal numbers :(a) 98127634 (b) 72049900 (c) 10000000 (d)00000000 . 9&#x2019;s comlements : (a) 99999999-98127634=01872365 (b) 99999999-72049900=27950099 (c) 99999999-10000000=89999999 (d) 99999999-0000000=99999999
• 26. 10&#x2019;s complements(a)100000000- 98127634= 01872366(b)100000000-72049900=27950100(c)100000000-10000000=90000000(d)100000000-0000000=00000000
• 27. 1.15-) (a) Find the 16&#x2019;s-complement of AF3B . (b) Convert AF3B to binary . (c) Find the 2&#x2019;s-complement of the result in (b) (d) Convert the answer in (c) to hexadecimal and compare with the answer in (a) (a)16^5-AF3B=50C5 (b)(AF3B)16=1010 1111 0011 1011 (c)1010111100111011 0101000011000101 (d)0101 0000 1100 0101= 50C5
• 28. 1.16-) Obtain the 1&#x2019;s and 2&#x2019;S complements of thefollowing binary numbers :(a)11101010 (b)01111110 (c)00000001(d)10000000 (e)000000001&#x2019;s complements:(a) 00010101 (b)10000001 (c)11111110 (d)01111111(e)111111112&#x2019;s complement :(a) 00010110 (b)10000010 (c)11111111 (d)10000000(e)00000000
• 29. 1.17-) Perform subtraction on the followingunsigned numbers using the 2&#x2019;s-complement ofthe subtrahend. Where the result shoud benegative, 10&#x2019;s complement it and affix a minussign. Verify your answers .(a) 7188-3049 (b)150-2100 (c)2997-7992(d)1321-375
• 31. 1.18-) Perform subtraction on the followingunsigned binary numbers using the 2&#x2019;s-complement of the subtrahend. Where the resultshould be negative, 2&#x2019;s complement it and affix aminus sign .(a)11011-11001 (b)110100-10101 (c)1011-110000 (d)101010-101011
• 32. (a)11011+00111=00010(27-25=2)(b)110100+01011=011111(52-21=31)(c)1011+010000=011011 &#xF0E0;-100101(11-48=-37)(d)101010+010101=111111&#xF0E0;-000001(42-43=-1)
• 33. 1.19-) The following decimal numbers are shownin sign- magnitude form : +9826 and +801.Convert them to signed 10&#x2019;s-complement formand perform the following operations : (Note thatthe sum is +10627 and requires six digits).(a) (+9826)+(+801) (b)(+9826)+(-801)(c)(-9826)+(+801) (d)(-9826)+(-801)
• 34. (a)009826+00801=010627(b)009826+999199=09025(c)990174+000801=990975&#xF0E0; -09025(d)990174+999199=989373&#xF0E0; -10627
• 35. 1.20-) Convert decimal +61 and +27 to binaryusing the signed-2&#x2019;s complement representationand 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 thatthey are correct .
• 36. +61=0111101 -61=1000011+27=0011011 -27=1100101(a)27+(-61)=0011011+1000011=1011110(b)-27+(+61)=1100101+0111101=0100010(c)-27+(-61)= 1100101+1000011=0101000(overflow)&#xF0E0;11100101+11000011=10101000
• 37. 1.21-) Convert decimal 9126 to both BCD andASCII codes. For ASCII, an odd parity bit is to beappended at the left . BCD: 1001 0001 0010 0110ASCII: 10111001 00110001 00110010 10110110
• 38. 1.22-) Represent the unsigned decimal numbers965 and 672 in BCD and then show the stepsnecessary to form their sum . 965= 1001 0110 0101 672= 0110 0111 0010 +___ ___ ____ 1 0000 1101 0111 +0110 +0110 +_________________ 0001 0110 0011 0111 &#xF0E0; (1637)10
• 39. 1.23-) Formulate a weighted binary code for thedecimal digits using weights 6, 3, 1, 1 .
• 40. 6 3 1 1 Decimal0 0 0 0 00 0 0 1 10 0 1 1 20 1 0 0 30 1 1 0 4(0101)0 1 1 1 51 0 0 0 61 0 0 1 7(1010)1 0 1 1 81 1 0 0 9
• 41. 1.24-) Represent decimal number 6027 in(a) BCD, (b) excess-3 code, and (c)2421 code .(a)6027&#xF0E0; BCD : 0110 0000 0010 0111(b)excess3: 1001 0011 0101 1010(c)(c)0110 0000 0010 1101
• 42. 1.25-) Find the 9&#x2019;s complement of 6027 and express it in 2421 code. Show that the result is the 1&#x2019;s complement of the answer to (c) in Problem 1.24 . This demonstrates that the 2421 code is self-complementing .9&#x2019;s complement of 6027 is 39726027 as 2421 code is &#xF0E0; 0110 0000 0010 11013972 as 2421 code is &#xF0E0;0011 1111 1101 0010
• 43. 1.26-) Assign a binary code in some orderlymanner to the 51 playing cards. Use theminimum number of bits. 2^4 =16 2^5 =32 2^6=64 &#xF0E0; 6 bits are necessary.
• 44. 1.27-) Write the expresion &#x201C;G. Boole&#x201D; 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) (01101111) (01101100) (01100101)
• 45. 1.28-) Decode the following ASCII code : 100101011000011101110 1100101 0100000 1000100 11011111100101 . Jane Doe
• 46. 1.29-) The following is a string of ASCII characters whose bit patterns have benn converted into hexadecimal for compactness : 4A EF 68 6E 20 C4 EF E5 . Of the 8 bits in each pair of digits, the leftmost is a parity bit. The remaining bits are the ASCII code.01001010 11101111 01101000 01101110 00100000 11000100 11101111 11100101 J O H N (space) D O E
• 47. 1.30-) How many printing characters are there inASCII ?How many of them are special characters (notletters or numerals) ? 94 characters 62 of them are numbers and letters. 32 of them are special characters.
• 48. 1.31-) What bit must be complemented to changean ASCII letter from capital to lowercase, andvice versa ?Cevap: Bir ASCII karakteri b&#xFC;y&#xFC;k harften k&#xFC;&#xE7;&#xFC;k harfe&#xE7;evirmek i&#xE7;in sa&#x11F;dan 6. bit 0 iken 1 yap&#x131;l&#x131;r. K&#xFC;&#xE7;&#xFC;ktenb&#xFC;y&#xFC;&#x11F;e &#xE7;evrilecekse 1 iken 0 yap&#x131;l&#x131;r.
• 49. 1.32-) The state of a 12-bit register is100010010111 . What is its content if itrepresents(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?
• 50. Three Decimal Digits in BCD: 1000 1001 0111 897Three Decimal Digits in Exces-3 Code: 1000 1001 0111 564 (8-3) (9-3) (7-3)Three Decimal Digits in the 8-4-2-1 Code: 1000 1001 0111 897 8 9 7Binary Code: 100010010111 2^11+2^7+2^4+2^2+2+1=2199
• 51. 1.33-) List the ASCII code for the 10 decimaldigits with an even parity bit in the leftmosposition. 00110000 10110001 10110010 00110011 10110100 00110101 00110110 10110111 10111000 00111001
• 52. 1.34-) Assume a 3-input AND gate with output Fand a 3-input OR gate with output G. Inputs areA, B, and C . Show the signals (by means of atiming diagram) of the outputs F and G asfunctions of three inputs ABC. Use all possiblecombinations of ABC.
• 53. F: &#xF0E8;A ,B ,C F: &#xF0E8;A , BX , CX &#xF0E8; AX ,B , CX &#xF0E8; AX , BX ,CNOT: X&#x2019;ler HIGH ya da LOW olabilir