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Communication Systems

NIOS · Class 12 · Physics

Flashcards for Communication Systems — NIOS Class 12 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.

45 questions30 flashcards5 concepts

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A block diagram illustrating the fundamental components of a communication system: Information Source, Input Transducer, Transmitter, Communication Channel, Receiver, Output Transducer, and Destinatio
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30 Flashcards
Card 1Communication System Basics

State the three essential elements of a communication system.

Answer

1. TRANSMITTER – Converts information into electrical signal and launches it. 2. MEDIUM / CHANNEL – Carries the signal over long distances (e.g., air, cable, fibre). 3. RECEIVER – Intercepts the signa

Card 2Types of Signals

Distinguish between Analogue and Digital signals. Give one example of each.

Answer

ANALOGUE SIGNAL: • Varies continuously with time. • Can take any value between min and max. • Example: Human speech (sound waves). DIGITAL SIGNAL: • Takes only discrete values (ON = 1, OFF = 0). • Re

Card 3Electromagnetic Waves

A radio wave has a frequency of 30 MHz. Calculate its wavelength in free space. (Speed of light c = 3 × 10⁸ m/s)

Answer

Given: ν = 30 MHz = 30 × 10⁶ Hz, c = 3 × 10⁸ m/s Find: Wavelength (λ) Formula: c = νλ → λ = c/ν Step 1: λ = (3 × 10⁸) / (30 × 10⁶) Step 2: λ = (3 × 10⁸) / (3 × 10⁷) Step 3: λ = 10 m Answer: λ = 10

Card 4Bandwidth of Signals

What is signal bandwidth? How is it expressed differently for analogue and digital signals?

Answer

BANDWIDTH = The range of frequencies that a signal occupies. For ANALOGUE signals: • Measured in Hz (hertz) • Represents the frequency range of the signal • Example: Human speech = 200 Hz to 4000 Hz

Card 5Electromagnetic Spectrum

The VHF band covers 30 MHz to 300 MHz. Calculate the wavelength range of VHF waves in vacuum. (c = 3 × 10⁸ m/s)

Answer

Given: ν₁ = 30 MHz = 3 × 10⁷ Hz, ν₂ = 300 MHz = 3 × 10⁸ Hz Find: Wavelength range Formula: λ = c/ν For ν₁ = 30 MHz: λ₁ = (3 × 10⁸) / (3 × 10⁷) = 10 m For ν₂ = 300 MHz: λ₂ = (3 × 10⁸) / (3 × 10⁸) =

Card 6Modulation

Why is modulation necessary for long-distance signal transmission? Give TWO reasons.

Answer

Modulation is needed because low-frequency audio signals CANNOT be transmitted directly over long distances. Reasons: 1. ANTENNA SIZE PROBLEM: • Antenna size must be comparable to wavelength (λ/4). •

Card 7Amplitude Modulation

State the formula for an Amplitude Modulated (AM) wave. Define the modulation index (mₐ).

Answer

AM WAVE FORMULA: v_mod(t) = V_c₀ [1 + mₐ sin(ωₐt)] sin(ωct) Where: • V_c₀ = Amplitude of carrier wave • ωc = Angular frequency of carrier • ωₐ = Angular frequency of audio (modulating) signal • mₐ =

Card 8Amplitude Modulation

An AM wave has a carrier wave of amplitude 5V and the audio signal has amplitude 2V. Calculate the modulation index.

Answer

Given: V_c₀ = 5 V (carrier amplitude), V_a₀ = 2 V (audio amplitude) Find: Modulation index (mₐ) Formula: mₐ = V_a₀ / V_c₀ Step 1: mₐ = 2 / 5 Step 2: mₐ = 0.4 Answer: mₐ = 0.4 (or 40% modulation) S

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Frequently Asked Questions

What are the important topics in Communication Systems for NIOS Class 12 Physics?
Key topics in Communication Systems include Communication Systems – Complete Concept Map, How to Write a Perfect Physics Answer — Step-by-Step, Communication Systems — Complete Chapter Overview. These are the concepts NIOS Class 12 examiners draw on most — study them first, then practise related questions.
How to score full marks in Communication Systems — NIOS Class 12 Physics?
Understand the core concepts first, then work through the 45 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
How many flashcards are available for Communication Systems?
There are 30 flashcards for Communication Systems covering key definitions, formulas, and concepts. Use them daily for 10–15 minutes for best results.

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