Which of the following frequencies will be suitable for beyond-th
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    Which of the following frequencies will be suitable for beyond-the-horizon communication using sky waves?

    • 10 kHZ

    • 10 MHz

    • 1 GHz

    • 1000 GHz

    A.

    10 kHZ

    10 kHz cannot be radiated because the size of antenna is large. 1 GHz and 1000 GHz will penetrate.
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    Frequencies in the UHF range normally propagate by means of
    • Ground waves.
    • Sky waves.
    • Surface waves.
    • Space waves.

    D.

    Space waves.Reason: Propagation of waves in ultra high frequency range 0.3 - 3 GHz is possible by space waves. Because these waves can reflect from ionosphere.
    1402 Views
    Answer
    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?
    • (i) and (ii) only
    • (ii) and (iii) only
    • (i), (ii) and (iii) but not (iv)
    • All of (i), (ii), (iii) and (iv).

    C.

    (i), (ii) and (iii) but not (iv)(c) Reason: Digital signals connot utilise decimal system which represents a continuous set of values.
    961 Views
    Answer
    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 m tall. How much service area can it cover if the receiving antenna is at the ground level?
    No. Height of receiving antenna should be LOS of transmitting antenna.
    straight d subscript straight T superscript 2 space equals space left parenthesis straight R plus straight h right parenthesis squared minus straight R squared space space space space space space equals space straight R squared open square brackets open parentheses 1 plus straight h over straight R close parentheses squared minus 1 close square brackets space space space space space space equals space straight R squared open square brackets 1 plus fraction numerator 2 straight h over denominator straight R end fraction minus 1 close square brackets space space space space space space equals straight R squared. fraction numerator 2 straight h over denominator straight R end fraction straight d subscript straight T superscript 2 space equals space 2 straight h space. space straight R

    No. Height of receiving antenna should be LOS of transmitting antenna
    ∴   Service area will be
    straight A space equals space straight pi left parenthesis straight d subscript straight T right parenthesis squared space space space space equals space 22 over 7 cross times left parenthesis 2 hR right parenthesis space space space space equals space 22 over 7 cross times 2 cross times 81 cross times 6.4 cross times 10 to the power of 6 space space space space space equals space 3258 space km squared.
     
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    Answer
    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%?
    Peak voltage, Ac = 12 V 

    Modulation index = 75 % = 0.75  
     
                       μ = 0.75 = AmAc 

    Hence,       Am = 0.75 Ac 

    ⇒     Am = 0.75 × 12 V = 9 V is the peak voltage. 
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    Answer
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