JEE Main & Advanced

Communication Systems

Communication Systems for JEE Main & Advanced

1
Module 1

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:

TypeDescription
AnalogContinuous variation of voltage with time (e.g., speech, music)
DigitalDiscrete (binary: $0$ or $1$); used in computers, digital phones

Bandwidth: Range of frequencies necessary to transmit a signal. Different signals require different bandwidth:

SignalApproximate bandwidth
Speech (intelligible)$2.8$ kHz ($300$ Hz – $3.1$ kHz)
Music$20$ kHz ($20$ Hz – $20$ kHz)
TV (video + audio)$6$ MHz
Digital dataDepends on bit rate

Bandwidth of transmission medium:

MediumBandwidth
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.

Worked Examples
1

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.

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

The function of a transducer is to:

Q2.

Bandwidth of speech for intelligible reproduction:

Q3.

The medium between transmitter and receiver is called:

Q4.

A repeater is used to:

Q5.

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:

LayerHeight (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).

Worked Examples
1

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.

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

AM broadcasts use:

Q2.

Frequencies above $30$ MHz are best transmitted via:

Q3.

Ionosphere reflects waves of:

Q4.

Line-of-sight distance from $h$ m high tower:

Q5.

Geostationary satellite altitude:

2
Module 2

Modulation and Demodulation

Need for Modulation, Amplitude Modulation (AM)Topic 1

Why Modulation? Direct transmission of low-frequency audio (kHz range) faces several problems:

  1. 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.
  2. Power radiated: Power radiated by antenna $\propto (l/\lambda)^2$. Higher frequency → much more efficient radiation.
  3. 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:

TypeWhat varies with signalSymbol
Amplitude ModulationAmplitude of carrierAM
Frequency ModulationFrequency of carrierFM
Phase ModulationPhase of carrierPM

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 $\mu = 1$: $\eta = 1/3 = 33.3\%$
Worked Examples
1

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.

2

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: $\mu = 0.5$.

✎ Self-Check — 5 questions0 / 5
Q1.

Modulation is needed because:

Q2.

AM signal of carrier $f_c$ and modulating $f_m$ has bandwidth:

Q3.

Modulation index for AM:

Q4.

Maximum modulation index without distortion:

Q5.

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

FeatureAMFM
What variesAmplitudeFrequency
Bandwidth$2f_m$$\sim 2(\Delta f + f_m)$
Noise immunityPoorGood
Power efficiencyLow ($33\%$ max)High
QualityMediumHigh
ComplexitySimpleComplex
ApplicationLong-distance, simple radioHigh-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.

Worked Examples
1

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.

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

In FM, what varies?

Q2.

Commercial FM uses frequency band:

Q3.

Bandwidth of AM is:

Q4.

Envelope detector is used for:

Q5.

FM has better noise immunity than AM because:

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