Is Communication Systems part of the EAPCET Physics syllabus?

Yes. Communication Systems is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Physics questions are there in EAPCET?

40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.

What mistakes do students make in Communication Systems?

Mixing kilometres and metres inside √(2hR) — put both h and R in metres, then convert the answer. Using √(hR) instead of √(2hR) for antenna range. For two antennas, adding their heights before taking the root — the range is √(2h_TR) + √(2h_RR). Allowing an AM modulation index greater than 1 — over-modulation distorts the signal, so m should be at most 1.

EAPCET Physics · Intermediate second year

Communication Systems

Communication Systems covers the elements of a communication system, modulation, propagation of signals and antenna range.

Subject: Physics
Taught in: Intermediate second year
Physics in the paper: 40 of 160 questions

Key formulas and results

A communication system has a transmitter, a channel and a receiver
Modulation index for AM: $m = A_{m}/A_{c}$
Range of a transmitting antenna of height $h$ : $d = \sqrt{2hR}$
Ground wave, sky wave (ionospheric reflection) and space wave (line of sight) propagation
AM wave limits: $A_{\max} = A_{c} + A_{m}$ and $A_{\min} = A_{c} - A_{m}$ , so $m = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}}$
Side band frequencies: $f_{c} + f_{m}$ and $f_{c} - f_{m}$ ; bandwidth of an AM signal $= 2f_{m}$
Antenna height needed for range $d$ : $h = \frac{d^{2}}{2R}$ ; area covered $= \pi d^{2} = 2\pi Rh$
Minimum practical antenna length $\approx \lambda/4$

What this chapter covers

Elements of a communication system
Bandwidth
Amplitude and frequency modulation
Propagation of electromagnetic waves
Range of an antenna

How it is usually asked

Modulation index
Antenna range
Modes of propagation

Understanding Communication Systems

Every communication system has three parts: a transmitter that converts the message into a signal, a channel that carries it, and a receiver that recovers the message. Low-frequency audio signals cannot be radiated efficiently by antennas of practical size, so they are carried on a high-frequency carrier wave by modulation. In amplitude modulation the carrier’s amplitude follows the message; in frequency modulation its frequency does.

For AM, the modulation index m = A_m/A_c should be less than 1 to avoid distortion, and the modulated wave contains the carrier plus sidebands at f_c ± f_m, so the bandwidth is 2f_m. Signals travel as ground waves at low frequencies, as sky waves reflected by the ionosphere up to a few tens of MHz, and as line-of-sight space waves at higher frequencies.

Because the earth is curved, a space wave from a tower antenna of height h reaches only up to a distance √(2hR) along the ground, which is why television and mobile towers are built tall. EAPCET questions usually ask for the modulation index, sideband frequencies and bandwidth, antenna range, and which propagation mode suits a given frequency band.

Worked example

EAPCET-style question

A transmitting antenna is 20 m high. Taking the radius of the earth as $6.4 \times 10^{6}\ \text{m}$ , the maximum distance up to which its signal can be received directly on the ground (line of sight) is

(1)16 km
(2)11.3 km
(3)32 km
(4)256 km

Solution

Range $d = \sqrt{2hR}$ .
$d = \sqrt{2 \times 20 \times 6.4 \times 10^{6}} = \sqrt{2.56 \times 10^{8}}\ \text{m}$ .
$d = 1.6 \times 10^{4}\ \text{m} = 16\ \text{km}$ .

Answer: 16 km

Shortcuts and results to remember

A repeater is a receiver and a transmitter combined; it picks up a signal, amplifies it and retransmits it to extend range.
Attenuation is the loss of signal strength while it travels through a medium; amplification increases the amplitude.
A transducer converts one form of energy into another, such as a microphone turning sound into an electrical signal.
Power radiated by a linear antenna of length $l$ is proportional to $(l/\lambda)^{2}$ , which is why high-frequency carriers are used.

Common mistakes to avoid

Mixing kilometres and metres inside √(2hR) — put both h and R in metres, then convert the answer.
Using √(hR) instead of √(2hR) for antenna range.
For two antennas, adding their heights before taking the root — the range is √(2h_TR) + √(2h_RR).
Allowing an AM modulation index greater than 1 — over-modulation distorts the signal, so m should be at most 1.

How to prepare it

A short, formula-light chapter — a quick revision gives reliable marks.

Once the formulas feel familiar, test them the way the exam does: practise Communication Systems questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

Semiconductor ElectronicsIntermediate second year

See every chapter in the EAPCET Physics syllabus.

Practise Communication Systems chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Communication Systems part of the EAPCET Physics syllabus?: Yes. Communication Systems is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Communication Systems for EAPCET?: A short, formula-light chapter — a quick revision gives reliable marks.
How many Physics questions are there in EAPCET?: 40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Communication Systems?: Mixing kilometres and metres inside √(2hR) — put both h and R in metres, then convert the answer. Using √(hR) instead of √(2hR) for antenna range. For two antennas, adding their heights before taking the root — the range is √(2h_TR) + √(2h_RR). Allowing an AM modulation index greater than 1 — over-modulation distorts the signal, so m should be at most 1.