JEE Physics Communication System Notes Set 03

Elements of a Communication System

Every communication system has three parts: a transmitter, a channel (the physical path, either wires or free space) and a receiver. A source of information produces the message; if it is not electrical (for example a voice) a transducer first changes it into an electrical signal.
While travelling, the signal is attenuated and distorted by the channel and noise is added, so the receiver must reconstruct a recognisable copy of the message.
Point-to-point communication links one transmitter to one receiver (telephony). Broadcast communication has one transmitter and very many receivers (radio and television).

Basic Terms

Transducer: converts one form of energy into another (microphone, loudspeaker, photodiode). An amplifier is not a transducer.
Signal: information in electrical form. Analog signals vary continuously (sound, TV picture); digital signals take discrete values, usually the two levels 0 and 1.
Noise: unwanted signal that disturbs transmission. Attenuation: loss of signal strength along the path. Amplification: raising the signal strength using energy from a DC source.
Range: the largest distance over which the signal is received with enough strength. Bandwidth: the span of frequencies occupied by a signal or handled by an equipment.
Modulation: placing the low-frequency message on a high-frequency carrier. Demodulation (detection): recovering the message at the receiver. Repeater: a receiver plus a transmitter that picks up, amplifies and re-sends the signal; a communication satellite is a repeater in space.

Bandwidth of Signals and of Media

Speech needs about 300 Hz to 3100 Hz, so a bandwidth of 2800 Hz suffices for telephony. Music needs about 20 kHz. A video (picture) signal needs about 4.2 MHz, and a complete TV signal is allotted 6 MHz.
A rectangular digital pulse train of fundamental frequency \( f_0 \) is a sum of the harmonics \( f_0, 2f_0, 3f_0,\dots \); an exact copy would need infinite bandwidth, but a few harmonics are enough to recover the information because higher harmonics contribute less.
Media: coaxial cable about 750 MHz bandwidth (used below about 18 GHz); free space for radio waves from a few hundred kHz to a few GHz; optical fibre from about 1 THz to 1000 THz, with usable bandwidth above 100 GHz.

Why Modulation is Needed

Antenna size: an antenna should be at least about \( \lambda/4 \) long. For a 20 kHz signal \( \lambda=15\ \text{km} \), which is impractical, while for 1 MHz \( \lambda=300\ \text{m} \).
Radiated power: for an antenna of length \( l \) the radiated power is proportional to \( (l/\lambda)^2 \), so a low-frequency signal radiates very little power.
Mixing of signals: if many baseband signals were sent together they would overlap and could not be separated. Giving each transmitter its own carrier frequency and band solves this.

Production and Detection of AM

Production: add the message \( A_m\sin\omega_m t \) to the carrier \( A_c\sin\omega_c t \) and pass the sum through a square-law (non-linear) device, \( y=Bx+Cx^2 \). The output contains a dc term and the frequencies \( \omega_m,\ 2\omega_m,\ \omega_c,\ 2\omega_c,\ \omega_c\pm\omega_m \). A band-pass filter centred on \( \omega_c \) keeps only \( \omega_c \) and \( \omega_c\pm\omega_m \), which is the AM wave. A power amplifier then feeds the antenna.
Reception: antenna, amplifier, an intermediate-frequency (IF) stage that lowers the carrier frequency, the detector, and a final amplifier.
Detection: the diode rectifies the AM wave and the RC network charges to the peak (the envelope) and discharges slowly through R. For the envelope to be followed without ripple, \( \frac{1}{f_c}\ll RC\ll\frac{1}{f_m} \). For modulation index m the message frequency must satisfy roughly \( RC\le\frac{\sqrt{1-m^2}}{m\,\omega_m} \), otherwise the capacitor cannot discharge fast enough and the output is clipped.

Power in an AM Wave

Carrier power \( P_c=\frac{A_c^2}{2R} \); total side-band power \( P_{sb}=\frac{m^2}{2}P_c \); total power \( P_t=P_c\left(1+\frac{m^2}{2}\right) \).
In terms of currents, \( I_t=I_c\sqrt{1+\frac{m^2}{2}} \). At m = 1 the total is 1.5 times the carrier power and the side bands share one third of it, so two thirds of the power is wasted in the carrier.
If several tones modulate one carrier the total modulation index is \( m_t=\sqrt{m_1^2+m_2^2+m_3^2+\dots} \).

Worked Examples

Example 1. A carrier of 1 kW is modulated simultaneously by two sine waves with modulation indices 0.3 and 0.4. Find the total modulation index and the total power.
Answer:
Total index \( m_t=\sqrt{0.3^2+0.4^2}=0.5 \).
\( P_t=1000\left(1+\frac{0.25}{2}\right)=1125\ \text{W} \).

Example 2. A TV tower is 81 m high. Find its radio horizon and the area it covers (R = 6400 km).
Answer:
\( d=\sqrt{2Rh}=\sqrt{2\times6.4\times10^6\times81}=32.2\ \text{km} \).
Area \( =\pi d^2=3.26\times10^3\ \text{km}^2 \).