Communication Systems
Communication Systems for JEE Main & Advanced
Elements of Communication and Propagation
Communication System Basics and BandwidthTopic 1
Communication is the transfer of information from one point to another. A typical communication system has three main components:
``` Information → [TRANSMITTER] → [CHANNEL] → [RECEIVER] → Information ```
Key elements:
- Transducer: Converts non-electrical signal (sound, image, temperature) into electrical signal. Example: microphone, camera.
- Transmitter: Processes signal (amplifies, modulates) for transmission.
- Channel: The physical medium connecting transmitter and receiver (wire, optical fibre, free space).
- Receiver: Extracts the original signal from the received signal.
- Noise: Unwanted random electrical signal that distorts information.
- Repeater: Amplifies the signal en route to compensate for losses (used in long-distance communication).
Signal Types:
| Type | Description |
|---|---|
| Analog | Continuous variation of voltage with time (e.g., speech, music) |
| Digital | Discrete (binary: $0$ or $1$); used in computers, digital phones |
Bandwidth: Range of frequencies necessary to transmit a signal. Different signals require different bandwidth:
| Signal | Approximate bandwidth |
|---|---|
| Speech (intelligible) | $2.8$ kHz ($300$ Hz – $3.1$ kHz) |
| Music | $20$ kHz ($20$ Hz – $20$ kHz) |
| TV (video + audio) | $6$ MHz |
| Digital data | Depends on bit rate |
Bandwidth of transmission medium:
| Medium | Bandwidth |
|---|---|
| Wire (twisted pair) | $100$ MHz |
| Coaxial cable | $750$ MHz |
| Optical fibre | $\sim$ THz (extremely large) |
| Free space (radio) | All EM frequencies (assigned by regulation) |
Attenuation: Loss of signal strength along the channel. Amplification: Boosting signal strength (compensates attenuation). Range: Maximum distance between transmitter and receiver beyond which signal is too weak to recover.
A communication channel of bandwidth $5$ MHz is used. Find the maximum frequency for transmission if the lower limit is $100$ MHz.
Show solution
Bandwidth = $f_{\max} - f_{\min}$ $f_{\max} = 100 + 5 = 105$ MHz.
Final Answer: $105$ MHz.
An optical-fibre channel allows $10^{13}$ Hz bandwidth. How many telephone-quality voice channels (each $4$ kHz) can it carry simultaneously?
Show solution
$$N = \frac{10^{13}}{4 \times 10^3} = 2.5 \times 10^9 \text{ channels}$$
Final Answer: $2.5 \times 10^9$ channels.
The function of a transducer is to:
Bandwidth of speech for intelligible reproduction:
The medium between transmitter and receiver is called:
A repeater is used to:
Bandwidth of TV signal is approximately:
Propagation of EM WavesTopic 2
EM waves are used for wireless communication. Three modes of propagation are used depending on frequency:
1. Ground Wave (Surface Wave) Propagation:
- Wave travels along Earth's surface
- Used for: AM broadcast ($530$ kHz – $1710$ kHz)
- Range: Few hundred km (high attenuation at higher frequencies)
- Frequency: $< 2$ MHz typically
2. Sky Wave Propagation:
- Wave is reflected by ionosphere back to Earth
- Used for: short-wave broadcasting, amateur radio
- Frequency: $2-30$ MHz (HF)
- Multiple bounces possible — long-distance global coverage
- Limited by ionosphere conditions (day/night, solar activity)
3. Space Wave (Line-of-Sight) Propagation:
- Wave travels directly from transmitter to receiver
- Used for: TV, FM, microwave links, satellite communication, mobile phones
- Frequency: $> 30$ MHz (VHF, UHF, microwave)
- Higher frequencies penetrate ionosphere → useful for satellite communication
Line-of-Sight Distance for Antenna of Height $h_T$: $$d_T = \sqrt{2Rh_T}$$ where $R$ = Earth's radius. For both antennas (transmitter $h_T$, receiver $h_R$): $$d_M = \sqrt{2Rh_T} + \sqrt{2Rh_R}$$
Ionosphere Layers:
| Layer | Height (km) | Density |
|---|---|---|
| D | $65-100$ | Lowest |
| E | $100-150$ | Moderate |
| F1 | $150-250$ | Higher |
| F2 | $250-400$ | Highest |
Critical Frequency: Maximum frequency reflected back at vertical incidence. Above this, wave penetrates the ionosphere.
MUF (Maximum Usable Frequency): Highest frequency that gives reflection from ionosphere for given path.
Satellite Communication: Satellites act as relay stations. Geostationary satellites at $36000$ km appear stationary; three suffice for global coverage. Frequencies used: GHz range (microwaves).
Find line-of-sight distance for a TV antenna at height $80$ m. ($R = 6400$ km)
Show solution
$$d = \sqrt{2Rh} = \sqrt{2 \times 6.4 \times 10^6 \times 80} = \sqrt{1.024 \times 10^9}$$ $$\approx 3.2 \times 10^4 \text{ m} = 32 \text{ km}$$
Final Answer: $d \approx 32$ km.
A transmitter antenna is $200$ m tall, receiver $50$ m. Maximum line-of-sight distance: ($R = 6400$ km)
Show solution
$$d_T = \sqrt{2 \times 6.4 \times 10^6 \times 200} \approx 50.6 \text{ km}$$ $$d_R = \sqrt{2 \times 6.4 \times 10^6 \times 50} \approx 25.3 \text{ km}$$ $$d_M = d_T + d_R \approx 75.9 \text{ km}$$
Final Answer: $\approx 75.9$ km.
AM broadcasts use:
Frequencies above $30$ MHz are best transmitted via:
Ionosphere reflects waves of:
Line-of-sight distance from $h$ m high tower:
Geostationary satellite altitude:
Modulation and Demodulation
Need for Modulation, Amplitude Modulation (AM)Topic 1
Why Modulation? Direct transmission of low-frequency audio (kHz range) faces several problems:
- Antenna size: Effective antenna length $L \geq \lambda/4$. For $f = 1$ kHz: $\lambda = 300$ km, so $L \geq 75$ km — impossibly large. For high $f$ (MHz), antenna is just metres.
- Power radiated: Power radiated by antenna $\propto (l/\lambda)^2$. Higher frequency → much more efficient radiation.
- Mixing of signals: If many transmitters all radiated at audio frequency, signals would overlap. Different carrier frequencies allow many channels simultaneously.
Modulation: Process of superimposing low-frequency (modulating/message) signal on a high-frequency carrier wave.
Three Basic Types:
| Type | What varies with signal | Symbol |
|---|---|---|
| Amplitude Modulation | Amplitude of carrier | AM |
| Frequency Modulation | Frequency of carrier | FM |
| Phase Modulation | Phase of carrier | PM |
Amplitude Modulation (AM):
Carrier: $c(t) = A_c\sin\omega_c t$. Message: $m(t) = A_m\sin\omega_m t$.
AM signal: $c_m(t) = (A_c + A_m\sin\omega_m t)\sin\omega_c t = A_c(1 + \mu\sin\omega_m t)\sin\omega_c t$
where modulation index $\mu = A_m/A_c$; ranges $0 \leq \mu \leq 1$. Over-modulation ($\mu > 1$) causes distortion.
Expanding: $$c_m(t) = A_c\sin\omega_c t + \frac{\mu A_c}{2}\cos(\omega_c - \omega_m)t - \frac{\mu A_c}{2}\cos(\omega_c + \omega_m)t$$
AM signal contains three frequencies:
- Carrier $\omega_c$
- Upper side band (USB) $\omega_c + \omega_m$
- Lower side band (LSB) $\omega_c - \omega_m$
Bandwidth of AM = $2\omega_m$ (or $2f_m$ in Hz).
In terms of voltage amplitudes from waveform: $$\mu = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}}$$
Power in AM:
- Carrier power: $P_c = V_c^2/(2R)$
- Each sideband: $P_{SB} = \mu^2 P_c/4$
- Total power: $P_t = P_c(1 + \mu^2/2)$
- Efficiency: $\eta = \mu^2/(2 + \mu^2)$. For $\mu = 1$: $\eta = 1/3 = 33.3\%$
A carrier of $1$ MHz is amplitude-modulated by audio of $5$ kHz. Find side bands and bandwidth.
Show solution
- USB: $1000 + 5 = 1005$ kHz
- LSB: $1000 - 5 = 995$ kHz
- Bandwidth = $2f_m = 10$ kHz
Final Answer: USB $1005$ kHz, LSB $995$ kHz, BW $= 10$ kHz.
An AM wave has $V_{\max} = 120$ V, $V_{\min} = 40$ V. Find $\mu$.
Show solution
$$\mu = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}} = \frac{120-40}{120+40} = \frac{80}{160} = 0.5$$
Final Answer: $\mu = 0.5$.
Modulation is needed because:
AM signal of carrier $f_c$ and modulating $f_m$ has bandwidth:
Modulation index for AM:
Maximum modulation index without distortion:
The carrier signal in AM has:
Frequency and Phase Modulation, DemodulationTopic 2
Frequency Modulation (FM): Instantaneous frequency of carrier varied in proportion to message amplitude. Carrier amplitude remains constant.
$$f(t) = f_c + k_f m(t)$$
Advantages of FM over AM:
- Better noise immunity (noise mostly affects amplitude, which FM ignores)
- Higher fidelity sound quality
- Used for high-quality audio broadcasting ($88-108$ MHz commercial FM band)
Disadvantages:
- Requires larger bandwidth (typically $200$ kHz vs $10$ kHz for AM)
- More complex circuitry
Carson's Rule (FM bandwidth): $$BW_{FM} = 2(\Delta f + f_m)$$ where $\Delta f$ = peak frequency deviation, $f_m$ = highest modulating frequency.
Phase Modulation (PM): Phase of carrier varied with message: $$\phi(t) = \phi_c + k_p m(t)$$
PM and FM are closely related — differentiating phase gives frequency.
Comparison: AM vs FM
| Feature | AM | FM |
|---|---|---|
| What varies | Amplitude | Frequency |
| Bandwidth | $2f_m$ | $\sim 2(\Delta f + f_m)$ |
| Noise immunity | Poor | Good |
| Power efficiency | Low ($33\%$ max) | High |
| Quality | Medium | High |
| Complexity | Simple | Complex |
| Application | Long-distance, simple radio | High-fidelity radio, TV audio |
Demodulation (Detection): Process of recovering the original message signal from the modulated wave.
AM Demodulation: Uses an envelope detector = diode + RC low-pass filter. The diode rectifies the AM wave (positive envelope), and the RC filter smooths it, leaving the audio envelope.
Time constant $\tau = RC$ should satisfy: $$\frac{1}{f_c} \ll RC \ll \frac{1}{f_m}$$
FM Demodulation: Uses frequency discriminator circuits (slope detector, ratio detector, PLL — phase-locked loop). More complex than AM detection.
Production of AM: Square-law modulator with carrier and message inputs, followed by band-pass filter to keep $\omega_c \pm \omega_m$ components.
Production of FM: Voltage-controlled oscillator (VCO) — frequency depends on input voltage.
An FM signal has carrier $98$ MHz, max deviation $75$ kHz, max audio frequency $15$ kHz. Find bandwidth (Carson).
Show solution
$$BW = 2(\Delta f + f_m) = 2(75 + 15) = 180 \text{ kHz}$$
Final Answer: $180$ kHz.
A carrier of $10$ V is modulated by audio with $\mu = 0.6$. Find total transmitted power if carrier power $= 100$ W.
Show solution
$$P_t = P_c(1 + \mu^2/2) = 100(1 + 0.36/2) = 100 \times 1.18 = 118 \text{ W}$$
Final Answer: $P_t = 118$ W.
In FM, what varies?
Commercial FM uses frequency band:
Bandwidth of AM is:
Envelope detector is used for:
FM has better noise immunity than AM because:
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