Computer Fundamentals

Computer evolution is divided into five generations based on the core switching technology:

1st Gen (1940-1956): Vacuum tubes. Examples: ENIAC, UNIVAC, EDVAC. Bulky, high heat, machine language.
2nd Gen (1956-1963): Transistors. Smaller, faster, less heat. Assembly + early high-level languages (FORTRAN, COBOL).
3rd Gen (1964-1971): Integrated Circuits (ICs). Keyboards/monitors, OS introduced (IBM 360).
4th Gen (1971-present): Microprocessors (VLSI). Intel 4004 was the first. PCs, GUIs, networks.
5th Gen (present-future): Artificial Intelligence, ULSI, parallel processing, quantum/robotics.

Memory aid: 'Very Tiny Insects May Argue' = Vacuum tube, Transistor, IC, Microprocessor, AI. Bank PO exams frequently ask which technology defines which generation.

By DATA HANDLING: Analog (measures continuous data, e.g. speedometer), Digital (discrete 0/1 data — most computers), Hybrid (both, e.g. ECG/hospital ICU machines, petrol pumps).

By SIZE/POWER (largest to smallest): Supercomputer > Mainframe > Minicomputer > Microcomputer.

• Supercomputer: fastest, used for weather forecasting, molecular modelling. India's PARAM (C-DAC), CRAY series.
• Mainframe: large organisations, banks, handle thousands of users simultaneously.
• Minicomputer (midrange): mid-sized firms.
• Microcomputer: PCs, laptops, tablets, smartphones — single microprocessor.

By PURPOSE: General-purpose vs Special-purpose. Memory tip for size order: 'Some Men Make Money' (Super, Mainframe, Mini, Micro).

Q: A computer uses VLSI technology and supports GUI-based operating systems with internet connectivity. Which generation and what does VLSI mean?

Step 1: VLSI = Very Large Scale Integration, packing thousands of transistors on one chip — this is the microprocessor era.
Step 2: Microprocessor = 4th Generation (1971-present).
Step 3: GUI + internet confirms 4th Gen (5th Gen is defined by AI/ULSI, not just GUI).

Answer: Fourth Generation, VLSI = Very Large Scale Integration.

Speed trick: Map keyword-to-generation instantly — Vacuum tube=1, Transistor=2, IC=3, Microprocessor/VLSI=4, AI/ULSI=5. The word in the question is your shortcut.

The Central Processing Unit (CPU) is the brain of the computer, comprising three core parts:

1. ALU (Arithmetic Logic Unit): performs all arithmetic (+, -, x, /) and logical (AND, OR, comparison) operations.
2. CU (Control Unit): directs and coordinates all operations; fetches, decodes and executes instructions but does NOT process data itself.
3. Registers/MU: small high-speed temporary storage inside the CPU (e.g. Accumulator, PC, MAR, MBR).

Von Neumann architecture (stored-program concept) keeps both instructions and data in the SAME memory. The Machine Cycle = Fetch -> Decode -> Execute -> Store (FDES). CPU speed is measured in Hertz (GHz). Memory aid: ALU calculates, CU controls, Registers remember.

The MOTHERBOARD is the main printed circuit board connecting all components. Key elements:

• BIOS (Basic Input Output System): firmware on a ROM chip that boots the system (POST — Power On Self Test).
• CMOS battery: keeps BIOS settings and the real-time clock alive.
• Chipset: Northbridge (fast — CPU, RAM, graphics) and Southbridge (slower — I/O, USB).

BUSES carry data: Data bus (carries data), Address bus (carries memory addresses — its width sets max addressable memory), Control bus (carries control signals). Collectively the System Bus.

PORTS: USB (universal), HDMI/VGA (display), Ethernet/RJ-45 (network), PS/2 (old keyboard/mouse). Expansion slots: PCI, PCIe, AGP (old graphics).

Q: If a CPU has a 16-bit address bus, what is the maximum memory it can directly address?

Formula: Addressable memory = 2^(address bus width) locations.
Step 1: Width = 16 bits, so 2^16 = 65,536 locations = 64 KB (since 2^10 = 1 KB, 2^16 = 2^6 x 2^10 = 64 KB).

Answer: 64 KB.

Quick reference for the exam: 2^10=1K, 2^20=1M, 2^30=1G. A 20-bit bus = 2^20 = 1 MB; a 32-bit bus = 2^32 = 4 GB. Speed trick: subtract 10 from the exponent per K-step. 2^16 -> (16-10)=6, so 2^6=64 K = 64 KB. This 'subtract 10' trick saves time in DI-style memory questions.

PRIMARY (main) memory is directly accessible by the CPU:

• RAM (Random Access Memory): volatile (loses data on power-off), read/write, holds running programs. Types: DRAM (needs refreshing, used as main memory) and SRAM (faster, no refresh, used in cache).
• ROM (Read Only Memory): non-volatile, stores permanent boot firmware. Types: PROM (programmable once), EPROM (erased by UV light), EEPROM (electrically erasable — e.g. BIOS flash).
• CACHE: very fast SRAM between CPU and RAM; levels L1 (fastest, smallest, on-core) < L2 < L3 (largest, shared).

Speed hierarchy (fastest to slowest): Registers > Cache > RAM > SSD > HDD > Optical/Tape. Cost and speed rise as you go up; capacity rises as you go down. Memory aid: 'Volatile = Vanishes' (RAM vanishes without power).

Memory is measured in bits and bytes. 1 Byte = 8 bits. The ascending order:
Bit < Nibble (4 bits) < Byte (8 bits) < KB < MB < GB < TB < PB < EB < ZB < YB.

1 KB = 1024 bytes (2^10)
1 MB = 1024 KB (2^20)
1 GB = 1024 MB (2^30)
1 TB = 1024 GB (2^40)
1 PB = 1024 TB (2^50)

Memory aid for the ladder: 'Kilo Mega Giga Tera Peta Exa Zetta Yotta' = 'Kind Men Give Tea Para Even Zebra Yearly'. Note: in marketing, manufacturers often use powers of 10 (1 KB = 1000 bytes), but for exam binary calculations use 1024. A nibble = half a byte = 4 bits is a favourite trick question.

Q: How many songs of 4 MB each can be stored on a 2 GB pen drive (use 1 GB = 1024 MB)?

Step 1: Convert 2 GB to MB: 2 × 1024 = 2048 MB.
Step 2: Divide by file size: 2048 ÷ 4 = 512 songs.

Answer: 512 songs.

Variation trick: If asked in powers of 2, note 2 GB = 2^11 MB and 4 MB = 2^2 MB, so 2^11 ÷ 2^2 = 2^9 = 512. Working in exponents avoids long division. Bank PO often disguises memory conversions as DI/arithmetic word problems — recognising '÷ file size' is the key shortcut.

INPUT devices feed data INTO the computer: keyboard, mouse, scanner, joystick, light pen, microphone, webcam, barcode reader, MICR (Magnetic Ink Character Recognition — used on bank CHEQUES), OMR (Optical Mark Reader — exam OMR sheets), OCR (Optical Character Recognition).

OUTPUT devices send data OUT to the user: monitor (VDU), printer, speaker, projector, plotter.

BOTH input and output (I/O): touchscreen, modem, network card, headset, hard disk (storage I/O).

Printers: Impact (Dot-Matrix — strikes ribbon, noisy) vs Non-impact (Inkjet — sprays ink; Laser — uses toner, fastest, page-at-a-time). Banking relevance: MICR encodes the cheque number band, OMR reads filled bubbles. Memory aid: if it 'gives info TO you', it is OUTPUT.

Computers use four number systems:

• Binary (base 2): digits 0-1 — the machine's native language.
• Octal (base 8): digits 0-7.
• Decimal (base 10): digits 0-9 — human everyday numbers.
• Hexadecimal (base 16): digits 0-9 and A-F (A=10, B=11, C=12, D=13, E=14, F=15).

CONVERSIONS:

• Decimal to Binary: divide by 2 repeatedly, read remainders bottom-up.
• Binary to Decimal: multiply each bit by its place value (powers of 2) and sum.
• Binary↔Octal: group bits in 3s. Binary↔Hex: group bits in 4s.

Memory aid: Hex 'A through F' = 10 through 15. 1 hex digit = 4 bits (a nibble); 1 octal digit = 3 bits. This grouping trick makes conversions instant.

Q: Convert binary 1101 to decimal.

Method (place values, right to left = 2^0,2^1,2^2,2^3):
1101 = (1×2^3) + (1×2^2) + (0×2^1) + (1×2^0)
= 8 + 4 + 0 + 1 = 13.

Answer: 13.

Reverse check (decimal 13 to binary): 13÷2=6 r1, 6÷2=3 r0, 3÷2=1 r1, 1÷2=0 r1 -> read remainders bottom-up = 1101. Matches.

Speed trick for binary→decimal: write the place values 8 4 2 1 above the bits and just ADD the values where a 1 appears (8+4+1=13). For binary→hex, split into nibbles: 1101 = D directly. Knowing 8-4-2-1 by heart converts any 4-bit number in seconds.