Your SlideShare is downloading. ×
0
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Bolum1cozumler
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Bolum1cozumler

2,807

Published on

sdf

sdf

Published in: Education
0 Comments
4 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
2,807
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
55
Comments
0
Likes
4
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide
  • 00
  • 11
  • Transcript

    • 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 = 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
    • 3. 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, 1A, 1B , 1C, 1D, 1E, 1F, 20Base-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, 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 = 2¹º = 1,024 32K = 32 x 2¹º = 32 x 1,024 = 32,768 32K byte = 32,768 byte
    • 5. (b) 64M byte: 1M = 2²º = 1,048,576 64M = 64 x 2²º = 64 x 1,048,576 = 67,108,864 64M byte = 67,108,864 byte(c) 6.4G byte: 1G = 2³º = 1,073,741,824 6.4G = 6.4 x 2³º = 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 2º+ 1 x 2¹ + 1 x 2² +…..+ 1 x 2¹¹ + 1 x 2¹²(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 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
    • 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 ⇒ (4 x aº + 1 x a¹) / (2 x aº) = 5 x aº ⇒ (4 + a) / 2 = 5 ⇒ 4 + a = 10 ⇒a=6
    • 9. (b) (54)b / (4)b = (13)b ⇒ (4 x bº + 5 x b¹) / (4 x bº) = 3 x bº + 1 x b¹ ⇒ (4 + 5b) / 4 = 3 + b ⇒ 4 + 5b = 12 + 4b ⇒b=8(c) (24)c + (17)c = (40)c ⇒ (4 x cº + 2 x c¹) + (7 x cº + 1 x c¹) = 4 x c¹ ⇒ 4 + 2c + 7 + c = 4c ⇒ c = 11
    • 10. 1.6-) The solution to the quadratic equation x² -11x + 22 = 0 is x=3 and x=6. What is the baseof the numbers? x² - 11x + 22 = (x – 3) . (x – 6) x² - 11x + 22 = x² - (6 + 3)x + (6.3) ⇒ (11)a = (6)a + (3)a ⇒1+a=6+3 ⇒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 = 2¹ + 2² + (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 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
    • 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 16º + 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… 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…
    • 21. (c) 0.010101010…=0.0101 0101 0101(0.555..)16=5/16 +5/256 +5/4096 +…=~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 ÷ 101 . (1011111)2=95 (101)2=5 95/5=19 (10011)2 1011111 101 101 10011 000111 101 0101 101 0000
    • 25. 1.14-) Find the 9’s- and the 10’s-complement ofthe following decimal numbers :(a) 98127634 (b) 72049900 (c) 10000000 (d)00000000 . 9’s comlements : (a) 99999999-98127634=01872365 (b) 99999999-72049900=27950099 (c) 99999999-10000000=89999999 (d) 99999999-0000000=99999999
    • 26. 10’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’s-complement of AF3B . (b) Convert AF3B 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) (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’s and 2’S complements of thefollowing binary numbers :(a)11101010 (b)01111110 (c)00000001(d)10000000 (e)000000001’s complements:(a) 00010101 (b)10000001 (c)11111110 (d)01111111(e)111111112’s complement :(a) 00010110 (b)10000010 (c)11111111 (d)10000000(e)00000000
    • 29. 1.17-) Perform subtraction on the followingunsigned numbers using the 2’s-complement ofthe subtrahend. Where the result shoud benegative, 10’s complement it and affix a minussign. Verify your answers .(a) 7188-3049 (b)150-2100 (c)2997-7992(d)1321-375
    • 30. (a)7188+6951=4139 One carry out soanswer is correct.(b)150+7900=8050 correct answer=-1950(c)2997+2008=5005 correct answer=-4995(d)1321+9625=0946 One carry out soanswer is correct.
    • 31. 1.18-) Perform subtraction on the followingunsigned binary numbers using the 2’s-complement of the subtrahend. Where the resultshould be negative, 2’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 -100101(11-48=-37)(d)101010+010101=111111-000001(42-43=-1)
    • 33. 1.19-) The following decimal numbers are shownin sign- magnitude form : +9826 and +801.Convert them to signed 10’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 -09025(d)990174+999199=989373 -10627
    • 35. 1.20-) Convert decimal +61 and +27 to binaryusing the signed-2’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)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  (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 BCD : 0110 0000 0010 0111(b)excess3: 1001 0011 0101 1010(c)(c)0110 0000 0010 1101
    • 42. 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 39726027 as 2421 code is  0110 0000 0010 11013972 as 2421 code is 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  6 bits are necessary.
    • 44. 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) (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 büyük harften küçük harfeçevirmek için sağdan 6. bit 0 iken 1 yapılır. Küçüktenbüyüğe çevrilecekse 1 iken 0 yapılır.
    • 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: A ,B ,C F: A , BX , CX  AX ,B , CX  AX , BX ,CNOT: X’ler HIGH ya da LOW olabilir

    ×