CHSE Odisha Elective Computer Science study materials
CHSE Odisha Elective Computer Science study resources

CHSE Odisha Elective Computer Science: Computer Fundamentals and System Organisation

CHSE Odisha Elective Computer Science · Work through your exam, one topic at a time.

All guide sections

Computer Fundamentals and System Organisation

This domain tests whether you can describe a computer as a set of functional units and explain how data and instructions move between them: the block diagram, CPU internals, memory hierarchy, buses and I/O devices.

The block diagram

A computer has five functional units: input, memory, arithmetic logic unit (ALU), control unit (CU) and output. ALU + CU + registers = the CPU. Key registers: program counter (PC) holds the address of the next instruction; instruction register (IR) holds the instruction being decoded; memory address register (MAR) holds the address currently on the address bus; the accumulator holds an ALU operand/result.

Instruction cycle

Fetch → decode → execute → store. In fetch, the PC value goes to the MAR, the instruction is read into the IR, and the PC is incremented.

Memory hierarchy

Registers (fastest, smallest, inside the CPU) → cache → main memory (RAM/ROM) → secondary storage. RAM is volatile and holds running programs; ROM is non-volatile and holds firmware such as the bootstrap loader. Cache exploits locality of reference to cut average access time.

Worked example

A machine has a 20-bit address bus and 1 byte per location. Addressable locations = 2^20 = 1,048,576 = 1M locations, so addressable memory = 1 MB. Compare a 16-bit address bus: 2^16 = 65,536 locations = 64 KB.

Traps

Confusing KB (1024 bytes) with Kb (1024 bits); saying the ALU stores results permanently; placing cache outside the CPU; swapping the roles of PC and MAR; forgetting that ROM is non-volatile.