0 / 1 Now, Binary value of the given number :-

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1 Course Code : MCS-012 Course Title : Computer Organisation and Assembly Language Programming Assignment Number : MCA(1)/012/Assign/ Maximum Marks : 100 Weightage : 25% Last Dates for Submission : 15th October, 2014 (For July 2014 Session) 15th April, 2015 (For January 2015 Session) Perform the following arithmetic operations :- Using binary signed 2 s complement notation for integers. You may assume that the maximum size of integers is of 9 bits including the sign bit. (Please note that the numbers given here are in decimal notation). i) Add 256 and 206 ii) Subtract 224 from 99 iii) Add 124 and 132 Please indicate the overflow if it occurs. Also write how you identify overflow. i) Add 256 and 206 First, we have to represent the number in binary notation The sign of a binary number is represented by 0 as plus and 1 as minus 0 / 1 Now, Binary value of the given number This number value is of more than 8-bits (i.e.9-bits) in signed 2 s complement notation also the value remains the same. Hence this number cannot fit inside it. To add this numbers we will need one more bit, hence Addition not possible. ii) Subtract 224 from 99 First, we have to represent the number in binary notation The sign of a binary number is represented by 0 as plus and 1 as minus 0 / 1 Now, Binary value of the given number : : In Binary, Subtraction is not done directly it is done by taking a MINUS sign for a positive number. For subtraction changing +224 to -224: : Now, covert it to signed 2 s complement notation:- -99 : :

2 Simple trick to convert any binary value to its signed 2 s complement notation is Check for the firstone (i.e. 1) in the magnitude of the number from Right to Left when you find it, Keep the number unchanged till one (i.e. 1) and remaining number reverse it by changing value from 0 to 1 and vice-verse. -99 : : :- Carry bit Overflow condition occured. The magnitude has been overflowed into carry the given 8-bits are not sufficient for the result of the magnitude. iii) Add 124 and 132 First, we have to represent the number in binary notation The sign of a binary number is represented by 0 as plus and 1 as minus 0 / 1 Now, Binary value of the given number :- +132:- +124:- +132: :- Carry bit Overflow condition occured. The magnitude has been overflowed into sign bit and sign into carry the given 8-bits are not sufficient for the result of the magnitude.

3 (b) Convert the hexadecimal number: ( ABCD01)h into equivalent binary, octal and decimal. The Base Value of Hexadecimal number is 16. The Table represents the value in decimal A 11 B 12 C 13 D 14 E 15 F Any Hexadecimal number can be represented in the following format for converting to Decimal. A* B* C* D* * *16 0 =10* * * * * *16 0 =10* * * * *16 + 1*1 = =

4 = Decimal-to-Binary: Converting can be found on this link Decimal-to-Octal: Converting can be found on this link

5 (c) Convert the following string into equivalent UTF 8 code DNS uses domains like.com,.au etc.. Are these UTF 8 codes same as that used in ASCII? UTF 8 Stands for U Universal Character Set T Transformation F Format 8 8-bits (2 8 =256 characters) ASCII CODE table:- To Convert: DNS uses domains like.com,.au etc. Codes: D=068 N=078 S=083 (space)=032 u=117 s=115 e=101 s=115 (space)=032 d=100 o=111 m=109 a=097 i=105 n=110 s=115 (space)=032 l=108 i=105 k=107 e=101 (space)=032.(dot)=046 c=099 o=111 m=109,(comma)=044 (space)=032.(dot)=046 a=097 u=117 (space)=032 e=101 t=116 c=099.(dot)=046

6 (d) Use a Karnaugh's map to design a odd parity generator circuit for 4 input bits. Before Designing we should know inputs which is 4 That means 2 4 = 16 Combinations of outputs Inputs are represented as ABCD Output is represented by F and some times termed as Function Decimal A B C D F (Function) Before proceeding we should know what odd parity is for ABCD inputs?

7 There are two types of parity even parity and odd parity The parity is 0 or 1 depending upon total numbers of 1 s If count of 1 s is even number then even parity = 0 and odd parity = 1 Similarly If count of 1 s is odd number then even parity = 1 and odd parity = 0 After Finding out odd parity for ABCD inputs We have Table as:- Decimal A B C D F (Function)

8 It s can also be written as F =Σ (0, 3, 5, 6, 9, 10, 12, 15) Only the decimal number where we find 1 s is shown in the Bracket. Final Equation:-

9 (e ) An 8 bit data after transmission is received as Explain how SEC code will detect and correct this problem. SEC means Single Error Correction This if found in Hamming Error Correction Code First find out the number parity bits in SEC Code Formula: 2 i 1 >= N + i where i = number of parity bits in SEC Code N = number of bits in Data Word In this case N=8 i=? So, 2 i 1 >= N + I at i= >= >= 11 (Not True) So, 2 i 1 >= N + I at i= >= >= 12 (True) Condition Satisfied. Coorection bits (parity bits) are 4. Parity bits are placed at i.e respectively P1 P2 D1 P3 D2 D3 D4 P4 D5 D6 D7 D ???? D1 D2 D3 D4 D5 D6 D7 D8 are the data sent and recd.

10 Before sending calculate Parity of Data to be sent. P1 P2 D1 P3 D2 D3 D4 P4 D5 D6 D7 D ?? 1? 0 1 1? P1 =? (Positions) Starting Point will be after Parity position P1 i.e 2 and Take 1 Skip 1till the end of table data. We get It s even parity will be 1 P2 =? (Positions) Starting Point will be after Parity position P2 i.e 3 and Take 2 Skip 2 till the end of table data. We get It s even parity will be 0 P3 =? (Positions) Starting Point will be after Parity position P3 i.e 4 and Take 4 Skip 4 till the end of table data. We get 0111 It s even parity will be 1 P3 =? (Positions) Starting Point will be after Parity position P3 i.e 4 and Take 8 Skip 8 till the end of table data. (Note:- data ends before taking 8 elements.) We get 0011 It s even parity will be 0 We found all parity bits, we will fill in the table:-

11 P1 P2 D1 P3 D2 D3 D4 P4 D5 D6 D7 D Assuming all parity sent correctly The 8-bit Sent Data = The 8-bit Sent Data = Check with parity bits, before that Create a new parity bit with new data P1= D1 D2 D4 D5 D7= even parity will be 1 - correct P2= D1 D3 D4 D6 D7= even parity will be 1 - incorrect P3= D2 D3 D4 D8= even parity will be 0 - incorrect P4= D5 D6 D7 D8= even parity will be 0 - correct After Checking We find that common data bit numbers are D3 & D4 But D4 is also present in P1that means error has been occurred in D3 only. By this we dectect error in D3 and Correct it by replacing it by 0 to 1.