### Communication Systems Class 12th Physics Part Ii CBSE Solution

##### Question 1.Which of the following frequencies will be suitable for beyond-the horizon communication using sky waves?A. 10 kHzB. 10 MHzC. 1 GHzD. 1000 GHzAnswer:10 MHz frequency will be suitable for beyond the horizon communication using skywaves.10 KHz cannot be transmitted because the size of the antenna will be large for this apparently lower frequency.1 and 1000 GHz are very high frequency wave which will penetrate and so cannot be transmitted.Question 2.Frequencies in the UHF range normally propagate by means of:A. Ground waves.B. Sky waves.C. Surface waves.D. Space waves.Answer:The signals having Ultra High Frequency(UHF) can propagate through Space waves. These can neither travel along the trajectory of ground as shown in green nor can propagate by reflection in the ionosphere as shown in figure in red. The UHF signal propagates as shown in Blue in the figure. 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) onlyB. (ii) and (iii) onlyC. (i), (ii) and (iii) but not (iv)D. All of (i), (ii), (iii) and (iv).Answer:Digital signal obviously use the binary signals i.e 0 and 1 only and no other signal other than these. It is also a signal that represents discontinuous values and it cannot utilize decimal values (0 to 9) which are continuous.  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 81m tall. How much service area can it cover if the receiving antenna is at the ground level?Answer:No, it is not necessary for a transmitting antenna and the receiving antenna to be at the same height. Because in this specific case of transmission there is no physical obstruction between the transmitter and the receiver. (b) Height of the given antenna, h = 81 mRadius of earth, R = 6.4 × 106 mFor range, d = (2Rh)1/2, the service area of the antenna is given by the relation:A = π d2= π (2Rh)= 3.14 × 2 × 6.4 × 106 × 81 m2= 3255.55 × 106 m2~ 3256 km2[As 1km = 103 m , so 1km2 = 106 m2]Question 5.A carrier wave of peak voltage 12V 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%?Answer:Amplitude of the carrier wave, Ac = 12 VModulation index, m = 75% = 0.75 (unitless as a ratio)Amplitude of the modulating wave = AmWe know for modulation index:m = Am/Ac⇒ Am = mAc⇒ Am = (0.75 × 12) V = 9VQuestion 6.A modulating signal is a square wave, as shown in Fig. 15.14. The carrier wave is given by c (t) = 2sin(8πt) volts.(i) Sketch the amplitude modulated waveform(ii) What is the modulation index?Answer:It can be observed from the given modulating signal that the amplitude of the modulating signal, Am = 1 VIt is given that the carrier wave c (t) = 2sin(8πt) volts∴ Amplitude of the carrier wave, Ac = 2 VTime period of the modulating signal Tm = 1 sThe angular frequency of the modulating signal is calculated as:(t)m = 2π/Tm= 2π rad s-1 (as Tm = 1)……………….(i)The angular frequency of the carrier signal is calculated as:(t)m = 8π rad s-1 ……….(ii)From equations (i) and (ii), we get:(t)m = 4(t)cThe amplitude modulated waveform of the modulating signal is shown in the following figure. (ii) Modulation index,m = Am/Ac = 1/2 = 0.5Question 7.For 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, μ. What would be the value of μ if the minimum amplitude is zero volt?Answer:Maximum amplitude, Amax = 10 VMinimum amplitude, Amin = 2 VModulation index μ is given by the relation:  ⇒ μ = 8/12∴ μ = 0.67(b)If Amin= 0Then μ = Amax/Amax = 1Question 8.Due 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.Answer:Let ωc and ωs be the respective frequencies of the carrier and signal waves.Signal received at the receiving station, V = V1cos(ωc + ωs)tInstantaneous Voltage of the carrier wave, Vin = Vc cos ωctV.Vin = V1.Vc{cos(ωc + ωs)t. cos ωct}= ( V1.Vc)/2 {2cos(ωc + ωs)t. cosωct}= ( V1.Vc)/2 [cos{(ωc + ωs)t + ωct} + cos {{(ωc + ωs)t - ωct }= (V1.Vc)/2 [cos{(2ωc + ωs)t} + cos ωst ]As the receiving centre allows only high frequency signals to pass through it. Obstructs the low frequency signal ωs Thus, at the receiving station, one can record the modulating signal (V1Vc/2).cosωst , which is the signal frequency.

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