Chapter One

Question 1 – Frequencies in the UHF range normally propagate by means of
(a) ground waves
(b) sky waves
(c) surface waves
(d) space waves

Solution – (d) The frequencies in UHF range normally propagate by means of space waves. The high frequency space does not bend with ground but are ideal for frequency modulation.

Question 2 – Which of the following frequencies will be suitable for beyond the horizon communication using sky waves?
(a) \(10\text{ kHz}\)
(b) \(10\text{ MHz}\)
(c) \({1\text{ GHz}\)
(d) \(1000\text{ GHz}\)

Solution – (b) \(10\text{ kHz}\) frequencies cannot be radiated due to large antenna size, \(1\text{ GHz}\) and \(1000\text{ GHz}\) will be penetrated. So, option (b) is correct.

Question 3 – Digital 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)
(b) (ii) and (iii)
(c) (i), (ii) and (iii)
(d) All of (i), (ii), (iii) and (iv)


(c) A digital signal is a discontinuous function of time in contrast to an analog signal. The digital signals can be stored as digital data and cannot be transmitted along the telephone lines. Digital signal cannot utilize decimal signals.

Question 4 – Is 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 \(81\text{ m}\) tall. How much service area can it cover, if the receiving antenna is at the ground level?

Solution – Given, height of antenna \(h = 81\text{ m}\)
Radius of earth \(R = 6.4 \times 1{0^6}\text{ m}\)
No, it is not necessary for line of sight communication, the two antennas may not be at the same height.
Area = \(\pi {d^2}\)
\(\because \) Range \(d = \sqrt {2hR} \)
\(\therefore \) Service Area \( = \pi  \times 2hR = \cfrac{{22}}{7} \times 2 \times 81 \times 6.4 \times {10^6}\)
\( = 3258.5 \times {10^6}{m^2} = 3258.5k{m^2}\)

Question 5 – A carrier wave of peak voltage \(12\text{ V}\) is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of \(75%\)?

Solution – Given, peak voltage \({V_0} = 12\text{ V}\)
Modulation index \(\mu  = 75\%  = \cfrac{{75}}{{100}}\)
We know that
Modulation index \((\mu ) = \cfrac{{{\text{Peak voltage of modulating signal}}\left( {{{\rm{V}}_{\rm{m}}}} \right)}}{{{\text{Peak voltage(}}{{\rm{V}}_0})}}\)
So, peak voltage of modulating signal,
\({ V_m} = \mu \times \) Peak voltage
\( = \cfrac{{75}}{{100}} \times 12 = 9\text{ V}\)

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