Communication is the act of transmission of information.

Transducer : Any device that converts one form of energy into another can be termed as a transducer.

Signal : Information converted in electrical form and suitable for transmission is called a signal.

Noise: Noise is unwanted energy, usually random characters generated by numerous natural or man made events.

Transmitter : It converts the message signal to a form suitable for transmission and then this processed signal is transmitted thru a suitable channel.

Receiver: It is an arrangement in which the transmitted signal from transmitter is picked up at the channel output and then received signal is processed to reproduce the message signal in a suitable form.

ModulationThe process of superimposing a low frequency signal on a high frequency wave, which acts as a carrier wave for long distance transmission is known as modulation

RepeaterIt is combination of receiver and transmitter. It is used to extend the range of the communication.

Amplitude modulationIn this type , the amplitude of the carrier wave is varied according to information signal.Carrier wave: c(t)=Acsin𝜔ct $c(t)={A}_{c}sin{\mathit{\omega}}_{c}t\phantom{\rule{0.22em}{0ex}}$Message signal : m(t)=Amsin𝜔mt $m(t)={A}_{m}sin{\mathit{\omega}}_{m}t\phantom{\rule{0.22em}{0ex}}$ 𝜇=Am

Ac= modulation index $\phantom{\rule{0.22em}{0ex}}\mathit{\mu}=\frac{{A}_{m}}{{A}_{c}}\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}modulation\phantom{\rule{0.22em}{0ex}}index\phantom{\rule{0.22em}{0ex}}$

Frequency ModulationIn this type , the frequency of the carrier wave is varied according to information signal

Space CommunicationThe communication that takes place throught the space surrounding the earth

Bandwidth of signalBandwidth is portion of electromagnetic spectrum occupied by a signal. It is the frequency range over which an information signal is transmitted or over which a receiver or other electronic circuit operates.

The audible range of frequencies extends from 20 Hz to 20 kHz.So to transmit music, an approximate bandwidth of 20 kHz is required .

Video signals for transmission of pictures require about 4.2 MHz of bandwidth. A TV signal contains both voice and picture and is usually allocated 6 MHz of bandwidth for transmission.

Propagation of electromagnetic waves1. Ground Wave2. Sky Wave3. Space Wave

Sky WaveFrequency range from a few MHz up to 30 to 40 MHz, long distance communication can be achieved by ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave propagation and is used by short wave broadcast services. The ionosphere is so called because of the presence of a large number of ions or chargedparticles.

Space WaveA space wave travels in a straight line from transmitting antenna to the receiving antenna. Space waves are used for line-of-sight (LOS) communication as well as satellite communication. At frequencies above 40 MHz, communication is essentially limited to line-of-sight paths. If transmitting antena is at height hT$h}_{T$ then distance to the horizondT=2 R hT$d}_{T}=\sqrt{2\phantom{\rule{0.22em}{0ex}}R\phantom{\rule{0.22em}{0ex}}{h}_{T}$

Electromagnetic waves of frequencies higher than 30 MHz penetrate the ionosphere and escape.

UHF (ultra high frequencies) : 174 MHz to 216 MHz

Amplitude ModulationIn amplitude modulation the amplitude of the carrier is varied in accordance with the information signal.Carrier Wavec(t)=Ac sin 𝜔ct$c(t)={A}_{c}\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{c}t$Messagem(t)=Am sin 𝜔mt$m(t)={A}_{m}\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{m}t$Then modulated signal will becm(t)=(Ac+Am sin 𝜔mt) sin 𝜔ct${c}_{m}(t)=({A}_{c}+{A}_{m}\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{m}t)\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{c}t$cm(t)=Ac(1+Am

Ac sin 𝜔mt) sin 𝜔ct${c}_{m}(t)={A}_{c}(1+\frac{{A}_{m}}{{A}_{c}}\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{m}t)\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{c}t$cm(t)=Ac sin 𝜔ct+𝜇 Ac sin 𝜔mt sin 𝜔ct${c}_{m}(t)={A}_{c}\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{c}t+\mathit{\mu}\phantom{\rule{0.22em}{0ex}}{A}_{c}\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{m}t\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{c}t$Modulation Index𝜇=Am

Ac$\mathit{\mu}=\frac{{A}_{m}}{{A}_{c}}$

cm(t)=Ac sin 𝜔ct+𝜇 Ac

2 cos(𝜔c-𝜔m) t-𝜇 Ac

2 cos(𝜔c+𝜔m) t $c}_{m}(t)={A}_{c}\phantom{\rule{0.22em}{0ex}}sin\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{c}t+\frac{\mathit{\mu}\phantom{\rule{0.22em}{0ex}}{A}_{c}}{2}\phantom{\rule{0.22em}{0ex}}cos({\mathit{\omega}}_{c}-{\mathit{\omega}}_{m})\phantom{\rule{0.22em}{0ex}}t-\frac{\mathit{\mu}\phantom{\rule{0.22em}{0ex}}{A}_{c}}{2}\phantom{\rule{0.22em}{0ex}}cos({\mathit{\omega}}_{c}+{\mathit{\omega}}_{m})\phantom{\rule{0.22em}{0ex}}t\phantom{\rule{0.44em}{0ex}$Lower Side Frequency = (𝜔c-𝜔m)$({\mathit{\omega}}_{c}-{\mathit{\omega}}_{m})$Upper Side Frequency = (𝜔c+𝜔m)$({\mathit{\omega}}_{c}+{\mathit{\omega}}_{m})$

A band pass filter rejects low and high frequencies and allows a band of frequenciesto pass through.

Common Questions

Q.In order to transmit a signal of wavelength 𝜆$\mathit{\lambda}$, thelength of the antenna at the transmitting stationshould at least be equal toa) 𝜆$\mathit{\lambda}$b) 𝜆

2$\frac{\mathit{\lambda}}{2}$c) 𝜆

4$\frac{\mathit{\lambda}}{4}$d) 4𝜆$4\mathit{\lambda}$A.Answer: c

Questions

Q.15.1Which of the following frequencies will be suitable for beyond-the-horizon communication using sky waves?(a) 10 kHz(b) 10 MHz(c) 1 GHz(d) 1000 GHzA.15.1(b) 10 MHz

Q.15.2Frequencies in the UHF range normally propagate by means of:(a) Ground waves.(b) Sky waves.(c) Surface waves.(d) Space waves.A.15.2(d) Space waves.

Q.15.3Digital signals(i) do not provide a continuous set of values,(ii) represent values as discrete steps,(iii) can utilize binary system, and(iv) can utilize decimal as well as binary systems.Which of the above statements are true?(a) (i) and (ii) only(b) (ii) and (iii) only(c) (i), (ii) and (iii) but not (iv)(d) All of (i), (ii), (iii) and (iv).A.15.3Answer (c)Digital signals 1. do not provide a continuous set of values, 2. represent values as discrete steps 3. can utilize binary system. A binary number has only 2 numbers.

Q.15.4Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81m tall. How much service area can it cover if the receiving antenna is at the ground level?A.15.4It is not necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication.If transmitting antena is at height hT$h}_{T$ then distance to the horizondT=2 R hT$d}_{T}=\sqrt{2\phantom{\rule{0.22em}{0ex}}R\phantom{\rule{0.22em}{0ex}}{h}_{T}$So total service areaA=𝜋 d2T=𝜋 2 R hT$A=\mathit{\pi}\phantom{\rule{0.22em}{0ex}}{d}_{T}^{2}=\mathit{\pi}\phantom{\rule{0.22em}{0ex}}2\phantom{\rule{0.22em}{0ex}}R\phantom{\rule{0.22em}{0ex}}{h}_{T}$A=3.14×2×6400000×81$A=3.14\times 2\times 6400000\times 81$A=3255552000 m2$A=3255552000\phantom{\rule{0.22em}{0ex}}{m}^{2}$A=3255.552 km2$A=3255.552\phantom{\rule{0.22em}{0ex}}k{m}^{2}$

Q.15.7For an amplitude modulated wave, the maximum amplitude is found to be 10V while the minimum amplitude is found to be 2V. Determine the modulation index, P. What would be the value of P if the minimum amplitude is zero volt?A.15.7 modulation index = 𝜇=Am

Ac$modulation\phantom{\rule{0.22em}{0ex}}index\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}\mathit{\mu}=\frac{{A}_{m}}{{A}_{c}}$ 𝜇 is kept ⩽1 to avoid distortion$\phantom{\rule{0.22em}{0ex}}\mathit{\mu}\phantom{\rule{0.22em}{0ex}}is\phantom{\rule{0.22em}{0ex}}kept\phantom{\rule{0.22em}{0ex}}\u2a7d1\phantom{\rule{0.22em}{0ex}}to\phantom{\rule{0.22em}{0ex}}avoid\phantom{\rule{0.22em}{0ex}}distortion$So we have considered , Ac>Am$So\phantom{\rule{0.22em}{0ex}}we\phantom{\rule{0.22em}{0ex}}have\phantom{\rule{0.22em}{0ex}}considered\phantom{\rule{0.22em}{0ex}},\phantom{\rule{0.22em}{0ex}}{A}_{c}>{A}_{m}\phantom{\rule{0.22em}{0ex}}$Ac+Am=10 $A}_{c}+{A}_{m}=10\phantom{\rule{0.22em}{0ex}$Ac-Am=2 $A}_{c}-{A}_{m}=2\phantom{\rule{0.22em}{0ex}$⟹2Ac=12 $\u27f92{A}_{c}=12\phantom{\rule{0.22em}{0ex}}$⟹Ac=6 $\u27f9{A}_{c}=6\phantom{\rule{0.22em}{0ex}}$⟹Am=4 $\u27f9{A}_{m}=4\phantom{\rule{0.22em}{0ex}}$ modulation index = 𝜇=Am

Ac=4

6=0.667$\phantom{\rule{0.22em}{0ex}}modulation\phantom{\rule{0.22em}{0ex}}index\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}\mathit{\mu}=\frac{{A}_{m}}{{A}_{c}}=\frac{4}{6}=0.667$if minimum amplitude is 0 thenAc+Am=10 $A}_{c}+{A}_{m}=10\phantom{\rule{0.22em}{0ex}$Ac-Am=0 $A}_{c}-{A}_{m}=0\phantom{\rule{0.22em}{0ex}$⟹2Ac=10 $\u27f92{A}_{c}=10\phantom{\rule{0.22em}{0ex}}$⟹Ac=5 $\u27f9{A}_{c}=5\phantom{\rule{0.22em}{0ex}}$⟹Am=5 $\u27f9{A}_{m}=5\phantom{\rule{0.22em}{0ex}}$ modulation index = 𝜇=Am

Q.15.8Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.A.15.8Received signal is : V=V1 cos(𝜔c+𝜔m) t$V={V}_{1}\phantom{\rule{0.22em}{0ex}}cos({\mathit{\omega}}_{c}+{\mathit{\omega}}_{m})\phantom{\rule{0.22em}{0ex}}t$Carrier signal is : Vin=Vc cos 𝜔c t${V}_{in}={V}_{c}\phantom{\rule{0.22em}{0ex}}cos\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{c}\phantom{\rule{0.22em}{0ex}}t$Multiplying V and Vin$V}_{in$V.Vin=V1 cos(𝜔c+𝜔m)t . Vc cos 𝜔c t $V.{V}_{in}={V}_{1}\phantom{\rule{0.22em}{0ex}}cos({\mathit{\omega}}_{c}+{\mathit{\omega}}_{m})t\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}{V}_{c}\phantom{\rule{0.22em}{0ex}}cos\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{c}\phantom{\rule{0.22em}{0ex}}t\phantom{\rule{0.22em}{0ex}}$V.Vin=V1Vc

2 2 cos(𝜔c+𝜔m)t . Vc cos 𝜔c t $V.{V}_{in}=\frac{{V}_{1}{V}_{c}}{2}\phantom{\rule{0.22em}{0ex}}2\phantom{\rule{0.22em}{0ex}}cos({\mathit{\omega}}_{c}+{\mathit{\omega}}_{m})t\phantom{\rule{0.22em}{0ex}}.\phantom{\rule{0.22em}{0ex}}{V}_{c}\phantom{\rule{0.22em}{0ex}}cos\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{c}\phantom{\rule{0.22em}{0ex}}t\phantom{\rule{0.22em}{0ex}}$V.Vin=V1Vc

2[ cos(2𝜔c+𝜔m)t +cos 𝜔m t]$V.{V}_{in}=\frac{{V}_{1}{V}_{c}}{2}[\phantom{\rule{0.22em}{0ex}}cos(2{\mathit{\omega}}_{c}+{\mathit{\omega}}_{m})t\phantom{\rule{0.22em}{0ex}}+cos\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{m}\phantom{\rule{0.22em}{0ex}}t]\phantom{\rule{0.22em}{0ex}}$This signal is pass thru low pass filter .A low-pass filter is the electric circuit, which passes the low range of frequency signals.So, at the receiving station, we record the modulating signal, V1Vc

2$\frac{{V}_{1}{V}_{c}}{2}$cos 𝜔m t$cos\phantom{\rule{0.22em}{0ex}}{\mathit{\omega}}_{m}\phantom{\rule{0.22em}{0ex}}t$ which is the signal frequency.