فصل اول نمايش اعداد. سيستم نمايش اعداد مبنا :)base( نيازها: مبناي r: ارقام محدود به [1-r,0] (379) 10 محاسبات در هر مبنا تبديل از يک مبنا به مبناي ديگر

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1 فصل اول نمايش اعداد سيستم نمايش اعداد مبنا :)base( نيازها: مبناي r: ارقام محدود به [1-r,0] (379) 10 دسيمال: ( ) 2 باينري: (372) 8 اکتال: (23D9F) 16 هگزادسيمال: محاسبات در هر مبنا تبديل از يک مبنا به مبناي ديگر 2 1

2 (126.53) 10 = 1* * * /26/2012 سيستم نمايش اعداد )دسيمال( اعداد دسيمال: دو بخش صحيح و اعشاري A n-1 A n-2 A 1 A 0. A -1 A -2 A -m+1 A -m که A i عددي بين 0 تا 9 و با وزن 10 i است. 3 سيستم نمايش اعداد )دسيمال( The value of A n-1 A n-2 A 1 A 0. A -1 A -2 A -m+1 A -m is calculated by i=n-1..0 (A i 10 i ) + i=-m..-1 (A i 10 i ) مثال : 5* *

3 سيستم نمايش اعداد )حالت کلي( base r (radix r) N = A n-1 r n-1 + A n-2 r n A 1 r + A 0 + A -1 r -1 + A -2 r A -m r -m Most Significant Digit (MSD) Least Significant Digit (LSD) 5 اعداد باينري )مبناي 2( کامپيوترها داده ها را به صورت رشته اي از بيت ها نمايش مي دهند. بيت: 0 يا 1 مبناي 2: ارقام 0 يا 1 ( ) 2 = (in decimal) = ½ + 0 = (45.5) 10 مثال: 7 3

4 اعداد باينري )مبناي 2( ( ) 2 = (in decimal) = = (9.375) 10 مثال: 8 اعداد باينري ( ) = ( ) B D 9 4

5 توان هاي 2 Memorize at least through اعداد اکتال )مبناي 8( مبناي 8: ارقام 0 تا 7 مثال: (762) 8 = (in decimal) = = (498)

6 اعداد هگزادسيمال )مبناي 16( مبناي 16: ارقام 9, A, B, C, D, E, F, 0, A=10, B=11,, F = 15 مثال: (3FB) 16 = (in decimal) = = (1019) تبديل مبناها هر مبنا )r( دسيمال: آسان )گفته شده( r هر مبناي دسيمال دسيمال باينري اکتال باينري و برعکس هگزادسيمال باينري و برعکس 13 6

7 تبديل دسيمال به هر مبناي r = (?) 16 بخش اعشاري: ضرب متوالي در r خواندن بخش صحيح ها به پايين x 16 = 12.5 int = 12 = C 0.5 x 16 = 8.0 int = 8 Read down = 0.C تبديل دسيمال به هر مبناي r 34, = (?) , ,172 rem rem 12 = C 16 8 rem 7 0 rem 8 بخش صحيح: تقسيم متوالي بر r خواندن باقيمانده ها به باال. Read up 34, = 87C

8 تبديل دسيمال به هر مبناي r = (?) 2 مثالي ديگر 0.1 x 2 = 0.2 int = x 2 = 0.4 int = x 2 = 0.8 int = x 2 = 1.6 int = x 2 = 1.2 int = x 2 = 0.4 int = x 2 = 0.8 int = 0 Read down = اعداد در مبناهاي مختلف Memorize at least Binary and Hex 17 8

9 دسيمال باينري فرض: N يک عدد دسيمال بزرگترين عددي که توان 2 است و با تفريق آن عددي مثبت ( 1 N (حاصل مي شود پيدا کن. يک عدد 1 در MSB قرار بده. مرحلة 1 را با عدد N 1 تکرار کن. در بيت مربوط عدد 1 قرار بده. وقتي اختالف صفر شد توقف کن. 18 باينري به اکتال 8 = 2 3 باينري به اکتال باينري به هگز هر 3 بيت باينري به يک بيت اکتال تبديل مي شود. باينري به هگزادسيمال 16 = 2 4 هر 4 بيت باينري به يک بيت هگزادسيمال تبديل مي شود. 19 9

10 دسيمال باينري مثال: N = (717) = 205 = N = = 77 = N = = 13 = N 3 64 = = 5 = N 4 8 = = 1 = N 5 4 = = 0 = N 6 1 = 2 0 (717) 10 = = ( ) 2 20 Binary Octal ( ) 2 ( ) 2 ( )

11 Binary Hex ( ) 2 ( ) 2 ( 6 A 8. F 5 C ) Octal Hex ازطريق باينري انجام دهيد: Hex Binary Octal Octal Binary Hex 23 11

12 )مثال( تبديل ها جدول را پر کنيد: Decimal Binary Octal Hex ???? ???? 336.5???? F9C7.A 24 اعمال رياضي باينري: جمع Carry قوانين: مانند جمع دسيمال با اين تفاوت که 10 = 1+1 توليد نقلي 0+0 = 0(c0) (sum 0 with carry 0) 0+1 = 1+0 = 1(c0) 1+1 = 0(c1) = 1(c1) Augend Addend Result

13 سرريز )Overflow( اگر تعداد بيت ها = n و حاصل جمع 1+n بيت نياز داشته باشد سرريز 26 اعمال رياضي باينري: تفريق قوانين: 0-0 = 1-1 = 0 (b0) (result 0 with borrow 0) 1-0 = 1 (b0) 0-1 = 1 (b1) Borrow Minuend Subtrahend Result

14 روش انجام محاسبات الگوريتم هاي اعمال رياضي مبناي 10 را به خاطر آوريد. آنها را براي مبناي مورد نظر تعميم دهيد. قانون مبناي مورد نظر را به کار بريد. براي باينري: 10=

15 ضرب باينري Shift-and-add algorithm, as in base 10 M cand M plier Check: 13 * 6 = 78 (1) (2) (3) Sum تمرين 1.7,1.8,1.9 از کتاب.Holdsworth 31 15

16 نمايش اعداد منفي سه روش متداول: اندازه-عالمت magnitude( )Sign مکمل complement( 1 )1 s مکمل complement( 2 )2 s متداولترين: مکمل 2 در بعضي منابع بهجاي مکمل از لغت متمم استفاده شده است. فرض: ماشين با کلمه هاي 4 بيتي: اگر بخواهيم فقط اعداد مثبت را نمايش دهيم 16 عدد مختلف قابل نمايش است ( زا 0 تا.)15 اگر بخواهيم هم اعداد مثبت و هم اعداد منفي را نمايش دهيم بايد يکي از سه روش فوق را بهکار ببريم. 32 روش اندازه-عالمت High order bit is sign: 0 = positive, 1 = negative = = جدول درستی Three low order bits is the magnitude: 0 (000) through 7 (111) Number range for n bits = [-2 n n-1-1] Two representations for 0: instead of 16 numbers, 15 numbers are represented: waste resources Cumbersome addition/subtraction Must compare magnitudes to determine sign of result 33 16

17 مکمل Diminished Radix complement Radix complement 34 روش مکمل 1 اعداد منفی بهصورت مکمل 1 قدرمطلق عدد متناظر ذخیره میشوند. n N = (2-1) - N Example: 1's complement of = = = 0111 Shortcut method: 1000 = -7 in 1's comp. simply compute bit wise complement >

18 روش مکمل = = Some complexities in addition Still two representations of 0! Again, the first bit shows the sign of the number. 36 روش مکمل 2 like 1's comp except shifted one position clockwise Only one representation for One more negative number than positive number -1-8 Again, the first bit shows the sign of the number. نمايش اعداد مثبت: در هر سه سيستم يکسان است = =

19 n N* = 2 - N روش مکمل = Example: Twos complement of 7 sub 7 = = repr. of -7 Example: Twos complement of = sub -7 = = repr. of 7 Shortcut method: Twos complement = bitwise complement > > 1001 (representation of -7) > > 0111 (representation of 7) 38 روش مکمل 2 Here s an easier way to compute the 2 s complement: 1. Leave all least significant 0 s and first 1 unchanged. 2. Replace 0 with 1 and 1 with 0 in all remaining higher significant bits. Examples: complement unchanged complement unchanged N = 1010 N = s complement 2 s complement 39 19

20 جمع و تفريق مکمل 2 جمع: جمع معمولی تفریق: جمع با مکمل دو مفروق منه عدم نیاز به روش جداگانه برای تفریق If )carry-in to sign = carry-out ( then ignore carry (-3) if )carry-in carry-out( then overflow Simpler addition scheme makes twos complement the most common choice for integer number systems within digital systems 40 جمع و تفريق مکمل 2 Why can the carry-out be ignored? -M + N when N > M: n n M* + N = (2 - M) + N = 2 + (N - M) Ignoring carry-out is just like subtracting 2 n n-1 n n -M + (-N) = M* + N* = (2 - M) + (2 - N) n n = 2 - (M + N) + 2 After ignoring the carry, this is just the right twos compl. representation for -(M + N) or (M+N)* 41 20

21 Overflow Conditions سرريز Add two positive numbers to get a negative number or two negative numbers to get a positive number = = Overflow Conditions سرريز Overflow Overflow No overflow No overflow Method 1: Overflow when carry in to sign carry out Method 2: Overflow when sign)a( = sign)b( sign )result( 43 21

22 تمرين 1.11 از کتاب.Holdsworth 44 کدهاي دودويي براي رقم هاي دسيمال ورودي و خروجي سيستمهاي ديجيتال از نوع دسيمال هستند. مثال: ماشين حساب. تبديل اعداد دسيمال به باينري و برعکس زمانبر است

23 Binary-Coded Decimal (BCD) Decimal numbers (0..9) are coded using 4-bit distinct binary words Observe that the codes (decimal ) are NOT represented (invalid BCD codes) 46 Binary-Coded Decimal To code a number with n decimal digits, we need 4n bits in BCD e.g. (365) 10 = ( ) BCD This is different from converting to binary, which is (365) 10 = ( ) 2 Clearly, BCD requires more bits. BUT, it is easier to understand/interpret. Trade off between speed/memory

24 BCD Addition Case 1: Case 2: (0) 0110 (0) (0) 1011 (1) 1 WRONG! Case 3: (1) 0001 (1) 7 Note that for cases 2 and 3, adding a factor of 6 (0110) gives us the correct result. 48 BCD Addition (cont.) BCD addition is therefore performed as follows 1) Add the two BCD digits together using normal binary addition 2) Check if correction is needed a) 4-bit sum is in range of 1010 to 1111 b) carry out of MSB = 1 3) If correction is required, add 0110 to 4-bit sum to get the correct result; BCD carry out =

25 BCD Negative Number Representation BCD 9 s complement invert each BCD digit (0 9, 1 8, 2 7,3 6, 7 2, 8 1, 9 0) BCD 10 s complement -N 10 n - N; 9 s complement BCD Addition (cont.) Example: Add 448 and 489 in BCD (448 in BCD) (489 in BCD) (greater than 9, add 6) (carry 1 into middle digit) 1101 (greater than 9, add 6) (carry 1 into leftmost digit) (BCD coding of ) 51 25

26 تمرين: چرا جمع با Excess-3 مانند BCD ولي هر رقم 3+ self-comlpement code )مکمل هر رقم = مکمل 9 آن( برابري تعداد صفرها و يک ها

27 ASCII character code We also need to represent letters and other symbols alphanumeric codes ASCII = American Standard Code for Information Interchange. Also known as Western European It contains 128 characters: 94 printable ( 26 upper case and 26 lower case letters, 10 digits, 32 special symbols) 34 non-printable (for control functions) Uses 7-bit binary codes to represent each of the 128 characters 54 ASCII Table Null Space Bell BkSpc Tab Line Fd Escape Crg Ret 55 27

28 Unicode Established standard (16-bit alphanumeric code) for international character sets Since it is 16-bit, it has 65,536 codes Represented by 4 Hex digits ASCII is between B Unicode Table

29 Unicode 062B 1579 ث 062C 1580 ج 062D 1581 ح 062E 1582 خ س ش ص ض 063B 1595 ػ 063C 1596 ػ 063D 1597 ػ 063E 1598 ػ ك ل م ن 064B C D E B C D E ٣ ٤ ٥ ٦ 066B C D E 1646 ٮ ٳ ٵ ٶ 067B 1659 ٻ 067C 1660 ټ 067D 1661 ٽ 067E 1662 پ ڃ ڄ څ چ 068B 1675 ڋ 068C 1676 ڌ 068D 1677 ڍ 068E 1678 ڎ 59 ASCII Parity Bit Parity coding is used to detect errors in data communication and processing An 8 th bit is added to the 7-bit ASCII code Even (Odd) parity: set the parity bit so as to make the # of 1 s in the 8-bit code even (odd) 60 29

30 ASCII Parity Bit (cont.) For example: Make the 7-bit code an 8-bit even parity code Make the 7-bit code an 8-bit odd parity code Error Checking: Both even and odd parity codes can detect an odd number of error. An even number of errors goes undetected. 61 Gray Codes Gray codes are minimum change codes From one numeric representation to the next, only one bit changes Applications: جدول کارنو. در سايت درس آدرس ويکي پدياي اين کد گذاشته شده است 62 30

31 Gray Codes (cont.) Binary Gray Binary Gray Binary Gray توجه شود که تعداد بيتهای اين کد مانند کد باينری متغير است و به عدد بستگی دارد. يکی از مزايا: جلوگيری از ورود شوک به سيستم 63 How to construct gray codes? غیربازگشتی.1 2. Recursively A 1-bit Gray Code has 2 code words, 0, 1 The first 2 n code words of an (n+1)-bit Gray code equal the code words of an n-bit Gray Code, written in order with a leading 0 appended. The last 2 n code words equal the code words of an n-bit Gray Code, but written in reverse order with a leading 1 appended

32 Binary to Gray Code Conversion (BC) (GC) MSB does not change as a result of conversion Start with MSB of binary number and add it to neighboring binary bit to get the next Gray code bit Repeat for subsequent Gray coded bits 65 Gray to Binary Code Conversion (GC) (BC) MSB does not change as a result of conversion Start with MSB of binary number and add it to the second MSB of the Gray code to get the next binary bit Repeat for subsequent binary coded bits 66 32

33 تمرين 1.12,1.16,1.17 از کتاب.Holdsworth 67 33

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