Communication miscellaneous

Communication miscellaneous

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Communication

Communication miscellaneous
  1. An FM signal for broadcast in the 88-108 MHz range, band has a frequency deviation of 15 kHz. Find the% modulation of this signal—








  1. View Hint View Answer Discuss in Forum

    As we know that in band (88–108) MHz, the maximum frequency deviation is ± 75kHz and given signal have frequency deviation of 15 kHz so% modulation given by

    % modulation =
    15kHz
    		×100
    75kHz

    = 20 %

    Correct Option: B

    As we know that in band (88–108) MHz, the maximum frequency deviation is ± 75kHz and given signal have frequency deviation of 15 kHz so% modulation given by

    % modulation =
    15kHz
    		×100
    75kHz

    = 20 %

  1. The angle modulated signal is given by
    s(t) = cos 2π(2 × 106t + 30 sin 150t + 40 cos 150t)
    The maximum frequency deviation in radian/sec is—








  1. View Hint View Answer Discuss in Forum

    Given equation s(t) = cos 2π (2 × 106t + 30 sin 150t + 40 cos 150t) …(1)
    From equation (1)
    θ(t) = 2 × 106t + 30 sin 150t + 40 cos 150t
    or

    dθ(t)
    		 = 2 × 106+ 30.150 cos 150t + 40.150 sin 150t
    dt

    or
    dθ(t)
    		 = 2 × 106 + √ 302 + 402 cos(150t + φ)
    dt

    or
    dθ(t)
    		 = 2 × 106 + 7500 cos (150t + φ)
    dt

    Where,
    φ = tan−1 −
    40
    30

    We know that,
    dθ(t)
    		= ωi
    dt

    2 × 106 = ωc
    Frequency deviation
    ⇒ ωi − ωc = 7500 cos (150t + φ)
    |ωi − ωc|max = 7500 radian/sec.
    Hence alternative (B) is the correct choice.

    Correct Option: B

    Given equation s(t) = cos 2π (2 × 106t + 30 sin 150t + 40 cos 150t) …(1)
    From equation (1)
    θ(t) = 2 × 106t + 30 sin 150t + 40 cos 150t
    or

    dθ(t)
    		 = 2 × 106+ 30.150 cos 150t + 40.150 sin 150t
    dt

    or
    dθ(t)
    		 = 2 × 106 + √ 302 + 402 cos(150t + φ)
    dt

    or
    dθ(t)
    		 = 2 × 106 + 7500 cos (150t + φ)
    dt

    Where,
    φ = tan−1 −
    40
    30

    We know that,
    dθ(t)
    		= ωi
    dt

    2 × 106 = ωc
    Frequency deviation
    ⇒ ωi − ωc = 7500 cos (150t + φ)
    |ωi − ωc|max = 7500 radian/sec.
    Hence alternative (B) is the correct choice.

  1. Multiplying circuit used in FM transmitter—








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    The multiplying circuit used in FM transmitter not only increases the carrier frequency, but also the frequency deviation by same factor.

    Correct Option: C

    The multiplying circuit used in FM transmitter not only increases the carrier frequency, but also the frequency deviation by same factor.

  1. A carrier A cos ωct is modulated by a signal f(t) = 2 cos 104. 2πt + 5 cos 103. 2πt + 3 cos 104. 4πt, then the bandwidth of the FM signal will be—
    (Given that, kf = 15 × 103 Hz per volt)








  1. View Hint View Answer Discuss in Forum

    We know that bandwidth of FM modulated signal is given by

    BW = 2(β + 1)fm =2Δf
    		 1 +
    1
    β

    Here, the maximum frequency component in f (t) is 20 kHz, and the maximum amplitude present in f (t) is 5 volt.
    Therefore the frequency deviation Δf is given by
    Δf = Kf ·Am = 15 × 103 × 5 = 75 kHz
    and the modulation index β is given by
    β =
    Δf
    		 =
    75 kHz
    		 =
    7.5
    		 = 3.75
    fm20kHz2

    Now, BW = 2 (β + 1) fm = 2(3·75 + 1) 20 kHz
    = 9·5 × 20 kHz = 190 kHz
    Hence alternative (B) is the correct choice.

    Correct Option: B

    We know that bandwidth of FM modulated signal is given by

    BW = 2(β + 1)fm =2Δf
    		 1 +
    1
    β

    Here, the maximum frequency component in f (t) is 20 kHz, and the maximum amplitude present in f (t) is 5 volt.
    Therefore the frequency deviation Δf is given by
    Δf = Kf ·Am = 15 × 103 × 5 = 75 kHz
    and the modulation index β is given by
    β =
    Δf
    		 =
    75 kHz
    		 =
    7.5
    		 = 3.75
    fm20kHz2

    Now, BW = 2 (β + 1) fm = 2(3·75 + 1) 20 kHz
    = 9·5 × 20 kHz = 190 kHz
    Hence alternative (B) is the correct choice.

  1. For the data given Q. 52 what will be the new bandwidth, if modulating signal amplitude is doubled?








  1. View Hint View Answer Discuss in Forum

    When the modulating signal amplitude is doubled, the frequency deviation becomes

    Δf' =
    KfA'm
    		=
    Kf'(2Am)
    ωmωm

    = 2Δf = 2 × 75 = 150 kHz
    Now, β′ = 2(Δf ′ + fm) = 2(150 + 10) = 320 kHz

    Correct Option: C

    When the modulating signal amplitude is doubled, the frequency deviation becomes

    Δf' =
    KfA'm
    		=
    Kf'(2Am)
    ωmωm

    = 2Δf = 2 × 75 = 150 kHz
    Now, β′ = 2(Δf ′ + fm) = 2(150 + 10) = 320 kHz