Question

ASAP Computer Architecture and Organization CSC 264 all questions

Question 1.

1. Write a note on general-purpose and special-purpose computers.
2. Briefly explain logic gates and various gates used in digital logic design.
3. Explain how data is represented in a digital computer and state how the signed binary numbers are represented.
4. Using the Hamming code algorithm (7, 4), convert a data message (0110) using 7bit.
   • Find the number of parity bits needed
   • Evaluate the values of parity bits
   • Final message bits with parity bits
   • Inject an error (o or 1) at position 4 and identify the error position

Answer 1

General purpose computers are designed with programs such that they can solve various types of problems. It takes less time to solve even the complex data. They have large memory to store data and so versatile that the most of the bussiness companies and commercial sites use general purpose computers these days.

Special purpose computers are specially designed and programmed in a way that it can solve only a speccific type of problem. These are mostly used as training simulators. It has many features of the general purpose computers but are made only for a specific purpose.

Logic gate is an electronic device which takes one or two binary inputs and give one output on the basis of a certain logic.

different types of logic gates are:

AND gate: works on the boolean expression Q=A.B working table: A B Q 0 0 0 0 1 0 1 0 0 1 1 1

OR gate: works on the boolean expression Q=A+B working table: A B Q 0 0 0 0 1 1 1 0 1 1 1 1

NOT gate: works on the boolean expression Q=~A (takes one input and gives opposite of it) working table: A B    0 1 1 0

Hex buffer : works on the boolean expression Q=A (takes one input and gives same output) working table: A B    0 0 1 1

NAND gate: works on the boolean expression Q=~(A.B) {AND+NOT} working table: A B Q 0 0 1 0 1 1 1 0 1 1 1 0

NOR gate: works on the boolean expression Q=~(A+B) {NOT+OR}   working table: A B Q 0 0 1 0 1 0 1 0 0 1 1 0

XOR gate: works on the boolean expression Q=A⊕B   {A OR B but not both} working table: A B Q 0 0 0 0 1 1 1 0 1 1 1 0

Ex-NOR gate: works on the boolean expression Q=~(A⊕B) {A AND B both same} working table: A B Q 0 0 1 0 1 0 1 0 0 1 1 1

In a digital computers any form of data is represented in binary form where the high level is 1 and low level is 0. In signed binary numbers, the most significant bit(MSB) is used as a signed bit, if it is 0 then it is a positive number and if it is 1 then it is a negative number.

Using hamming code algorithm (7,4) i.e 7 bit and 4 message bits and considering even parity check so the parity and message bits are shown as: P1 P2 m3 P4 m5 m6 m7 putting the given data in the message bits: 0 1 1 0 1st parity check: (P1, m3, m5, m7)= P1, 0, 1, 0 therefore, to make it even parity i.e with even numbers of 1s, P1 must be 1. 2nd parity check: (P2, m3, m6, m7)= P2, 0, 1, 0 therefore, to make it even parity i.e with even numbers of 1s, P2 must be 1. 3rd parity check: (P4, m5, m6, m7)= P4, 1, 1, 0 therefore, to make it even parity i.e with even numbers of 1s, P4 must be 0. So the converted data will be: 1 1 0 0 1 1 0

3 parity bits are needed in a 7 bit code where two of them are 1 and one is 0.

After injecting error at position 4 we get: 1 1 0 1 1 1 0

for detection of error: 1st parity check (P1, m3, m5, m7): 1, 0, 1, 0 =even 2nd parity check (P2, m3, m6, m7): 1, 0, 1, 0 =even 3rd parity check (P4, m5, m6, m7): 1, 1, 1, 0= odd (ERROR) P1,P2,P4 being 111. The error is in 7th position as decimal value for 111 is 7.

so corrected data will be obtained by changing the 7th bit value i.e 1 1 0 1 1 1 1