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2's Complemet-Basic Unified Engineering-Assignment Solution, Exercises of Engineering

Jiwaji UniversityEngineering

This is solution to assignment of Basic Unified Engineering course. It was submitted to Prof. Yasaar Verma at Jiwaji University. It includes: 2's, Complement, Notation, Convert, Base, Compute, Binary, Addition, Bit, String, Overflow

Typology: Exercises

2011/2012

Uploaded on 07/22/2012

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C5 Solutions
1. Convert the following base 10 numbers into 8-bit 2’s complement notation
0, -1, -12
To Compute 0
0 = 00000000
To Compute –1
Step 1. Convert 1 to binary
00000001
Step 2. Flip the bits
11111110
Step3. Add 1
11111111
Therefore –1 = 11111111
To Compute –12
Step 1. Convert 12 to binary
00001100
Step 2. Flip the bits
11110011
Step3. Add 1
11110100
Therefore –12 = 11110100
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C5 Solutions

  1. Convert the following base 10 numbers into 8-bit 2’s complement notation 0, -1, -

To Compute 0

0 = 00000000

To Compute –

Step 1. Convert 1 to binary 00000001

Step 2. Flip the bits 11111110

Step3. Add 1 11111111

Therefore –1 = 11111111

To Compute –

Step 1. Convert 12 to binary 00001100

Step 2. Flip the bits 11110011

Step3. Add 1 11110100

Therefore –12 = 11110100

  1. Perform each of the following additions assuming that the bit strings represent values in 2’s complement notation. Identify the cases in which the answer is incorrect because of overflow.

Answer = 11110 Overflow = 0 ∴ Answer is correct

01111

  • 10001 100000

Answer = 00000 Overflow = 1 ∴ Answer is incorrect

Answer = 11000 Overflow = 0 ∴ Answer is correct

  1. -- convert to modular type
  2. Place_Holder := Byte'Val(Integer'Pos(Number_To_Convert));
  4. -- flip the bits
  5. Place_Holder := Place_Holder xor 2#11111111#;
  6. -- add 1
  7. Place_Holder := Place_Holder + 2#1#;
  8. -- reconvert to integer
  9. Number_To_Convert := Integer'Val(Byte'Pos(Place_Holder));
  11. end if;
  13. -- decimal to binary conversion
  14. -- fill in the bit pattern from left to right
  15. loop
  16. exit when Count = 0;
  17. -- if the remainder is non-zero, the bit is set to 1
  18. -- else the bit is 0
  19. if (Number_To_Convert mod 2) = 1 then
  20. Binary_Number(Count) :='1';
  21. else
  22. Binary_Number(Count) :='0';
  23. end if;
  25. Count := Count -1;
  26. Number_To_Convert := Number_To_Convert/2;
  28. end loop;
  30. Put(Binary_Number);
  32. end Decimal_To_Binary;

67 lines: No errors

C

  1. How many bits do you need to represent a number in excess-16 format? What is the excess-16 representation of 12?

16 = 2^4 = 2 N-1^ ⇒ N =5.

Five bits are needed to represent the number in excess-16 format.

Step 1. Add 16 to the number 16+12=

Step 2. Convert to binary 12 in excess-16 = 11100

  1. Convert 29/8 into binary 8-bit floating-point representation.

Step1. Set the sign bit to zero since number is positive

Step2. Convert the number into binary representation 29/8 = 3 + 5/ = 011.

Step 3. Normalize the binary representation

0.11101 * 2^2

Step 4.Convert the exponent into excess-

2 = 110

Step 5. Fill in the mantissa

Therefore 29/8 = 01101110

  1. Sketch the basic von Neumann architecture and describe each component in a few lines.

The von Neumann architecture describes a computer with four main sections: the Arithmetic and Logic Unit (ALU) the control unit (CU) the memory the input and output devices (collectively termed I/O) These parts are interconnected by a bundle of wires, a Bus.

C

  1. Write an algorithm to implement the subtraction operation for two positive integers in assembly language. 1. Let the numbers be A, B and the operation be A-B 2. Convert A into binary 3. Convert B into binary 4. Compute 2’s complement of B i. Invert the bits in B using B xor 11111111 ii. Add 1 to B 5. Add A and the 2’s complement of B.
  2. Implement your algorithm in the assembly language describe in the machine language handout. Test your implementation using the SimpleSim simulator.

; Program name : Subtraction using add only ;Programmer : Joe B ;Last Modified : Sep 16 2003

load R1,1 ;1 added for computing 2's complement load R2,FFh ;mask for flipping the bits load R3,first_number ; load R4, second_number ; xor R5, R4,R2 ; flip the 0's and 1's in the second number addi R5,R5,R1 ; add 1 to the flipped bits to get the 2's complement addi R5,R5,R3 ; add the numbers to obtain A - B halt

first_number: db 8 ;A in A-B second_number: db 5 ;B in A-B