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JEE Mains Physics · Formula sheet

Communication Systems formulas

7 formulas, 1 reference table and 26 common traps for JEE Mains Physics Communication Systems, grouped by subtopic.

Full notes with worked examples

Communication Systems, Bands and Antenna Size

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Elements of a communication system and the number of channels

Number of channels

N=available bandwidthbandwidth of one channel,f=cλN = \frac{\text{available bandwidth}}{\text{bandwidth of one channel}}, \qquad f = \frac{c}{\lambda}

Antenna length and why a high-frequency carrier is needed

Antenna length and radiated power

lmin⁡=λ4=c4f,P∝(lλ)2l_{\min} = \frac{\lambda}{4} = \frac{c}{4f}, \qquad P \propto \left(\frac{l}{\lambda}\right)^{2}

Frequency bands, propagation modes and atmosphere layers

ServiceFrequency bandModeHow the wave travels
AM broadcast540 to 1600 kHzGround waveAlong the earth's surface, for frequencies up to a few MHz
Short-wave radioA few MHz to about 30 MHzSky waveReflected back to earth by the ionosphere
FM broadcast88 to 108 MHzSpace waveIn a straight line from the transmitting antenna to the receiving one
Television54 to 890 MHzSpace waveIn a straight line; the antenna height sets the range
TV is split into sub-bands (54 to 72, 76 to 88, 174 to 216 and 420 to 890 MHz). A frequency like 64 MHz is TV, not FM.
Satellite uplink5.925 to 6.425 GHzSpace waveStraight up through the ionosphere to the satellite
The uplink is the higher band. The downlink comes back on the lower one.
Satellite downlink3.7 to 4.2 GHzSpace waveFrom the satellite straight down to the earth station
Bands as in NCERT's table. Everything above about 40 MHz travels as a space wave.

Common traps

Find the frequency before taking the percentage

When a source is given by its wavelength, first find f=c/λf = c/\lambda. The percentage is a share of that frequency, not of the wavelength.

A repeater is a receiver and a transmitter

A repeater does not only amplify. It receives the signal, amplifies it and transmits it again, so it combines both ends of the link.

Attenuation and demodulation are easy to swap

Attenuation is the loss of strength in the medium. Demodulation is taking the message back off the carrier at the receiver. Match lists put the two side by side.

kHz for AM, MHz for FM

AM broadcast is 540 to 1600 kHz; FM broadcast is 88 to 108 MHz. An option that gives AM in MHz or FM in kHz is wrong even if the numbers look right.

Uplink is the higher frequency

In satellite links the uplink uses 5.925 to 6.425 GHz and the downlink 3.7 to 4.2 GHz. The two bands are always offered together, so check which one is asked.

Use NCERT's layer labels

NCERT calls the D and E layers part of the stratosphere, F1 part of the mesosphere and F2 part of the thermosphere. Match-list questions follow these labels, so answer by the table.

A quarter, not a half

The minimum antenna length is λ/4\lambda/4. Using λ/2\lambda/2 doubles the answer, and that value is always among the options.

Size the antenna for the carrier

The radiated wave is the carrier with its side bands, all near fcf_c. Using the message frequency fmf_m gives an antenna far too long.

Divide c by the refractive index in a medium

In a dielectric the wave travels at c/εrμrc/\sqrt{\varepsilon_r \mu_r}, so the wavelength is shorter for the same frequency. Using c gives a frequency that is too high.

Largest antenna, lowest frequency

The largest wavelength an antenna can radiate is 4l4l, and the largest wavelength is the lowest frequency. Reading it the other way round inverts the answer.

Line-of-Sight Range of a Tower

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Range of one antenna on a round earth

Line-of-sight range of one antenna

d=2Rh,A=πd2=2πRhd = \sqrt{2Rh}, \qquad A = \pi d^{2} = 2\pi R h

Range between a transmitting and a receiving antenna

Line-of-sight range of two antennas

dM=2RhT+2RhRd_M = \sqrt{2Rh_T} + \sqrt{2Rh_R}

Common traps

Increased by, or increased to

To double a range the height becomes 4h, so it is increased BY 3h. Questions ask both ways, and both values are among the options.

Keep one unit inside the root

Put R and h in metres, or both in kilometres, before multiplying. A tower height in metres with R in kilometres gives a range off by a factor of about 30.

The range grows as the square root

A height 4 times as large gives a range only twice as large, because 4=2\sqrt{4} = 2. Scaling the range by the same factor or percentage as the height is the usual wrong option.

Add the ranges, not the heights

hT+hR\sqrt{h_T} + \sqrt{h_R} is not hT+hR\sqrt{h_T + h_R}. Taking one root of the summed heights gives a range that is too short.

Do not forget the second antenna

When both antennas have a height, both add range. Using only the transmitter's 2RhT\sqrt{2Rh_T} gives an answer that is always one of the options.

Identical towers divide by 8R

For two equal heights, d=22Rhd = 2\sqrt{2Rh}, so h=d2/8Rh = d^{2}/8R. Using d2/2Rd^{2}/2R treats the whole distance as one antenna's range and gives a height 4 times too large.

Modulation Index, Sidebands and Bandwidth

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Amplitude modulation and the modulation index

Modulation index

μ=AmAc,μ≤1\mu = \frac{A_m}{A_c}, \qquad \mu \le 1

Maximum and minimum amplitude of an AM wave

Modulation index from the envelope

μ=Amax⁡−Amin⁡Amax⁡+Amin⁡,Aside band=μAc2=Am2\mu = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}}, \qquad A_{\text{side band}} = \frac{\mu A_c}{2} = \frac{A_m}{2}

Side-band frequencies and the bandwidth of an AM wave

Side bands and bandwidth

fLSB=fc−fm,fUSB=fc+fm,BW=2fmf_{\text{LSB}} = f_c - f_m, \quad f_{\text{USB}} = f_c + f_m, \qquad \text{BW} = 2f_m

Common traps

Message over carrier, not the other way

μ=Am/Ac\mu = A_m/A_c. Writing Ac/AmA_c/A_m gives a value above 1 for any normal AM wave, which should itself be a warning.

Read the amplitude, not the swing

A message drawn between +a and −a has amplitude a, not 2a. Taking the full swing doubles μ\mu.

It is the carrier whose amplitude changes

Statements say the amplitude of the 'modulating' or 'modulated' signal is varied. In AM it is the amplitude of the carrier that is varied, in step with the message.

μ is not the ratio of the largest to the smallest

μ=(Amax⁡−Amin⁡)/(Amax⁡+Amin⁡)\mu = (A_{\max} - A_{\min})/(A_{\max} + A_{\min}). Dividing Amax⁡A_{\max} by Amin⁡A_{\min} gives (1+μ)/(1−μ)(1 + \mu)/(1 - \mu), a different number that the options also carry.

A side band carries half the message amplitude

Each side band has amplitude μAc/2=Am/2\mu A_c/2 = A_m/2. Using μAc\mu A_c gives twice the right value.

Check which ratio is asked

Questions ask for maximum to minimum or minimum to maximum, and as a ratio like 50 : x. Write both amplitudes first, then set them in the order the stem gives.

Convert ω to f first

In sin⁡(ωt)\sin(\omega t) the number in front of t is ω=2πf\omega = 2\pi f. Reading it as f makes every frequency, and the bandwidth, 2π2\pi times too large.

The message frequency is not in the AM wave

An AM wave contains fcf_c and fc±fmf_c \pm f_m only. Listing fmf_m on its own among the frequencies present is wrong.

Bandwidth is twice the message frequency

The bandwidth is 2fm2f_m, set by the message alone. The carrier frequency does not enter it, and fmf_m alone is half the answer.

Each station needs 2fₘ, not fₘ

Dividing the band by fmf_m counts twice as many stations as can really fit. Divide by 2fm2f_m and round down.

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