Communication miscellaneous

Communication miscellaneous

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Communication

Communication miscellaneous
  1. Four voice signals, each limited to 4 kHz and sampled by Nyquist rate are converted into binary PCM using 256 quantization levels. The bit transmission rate for the time division multiplexed signal will be—








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    Given N = information bit
    i.e. 2N = 256 or N = 8, n = 4 (no. of channel)
    fs = 2fm = 2 × 4kHz = 8kHz
    Transmission rate = nNfs = 4 x 8 x 8 = 256 kbps

    Correct Option: C

    Given N = information bit
    i.e. 2N = 256 or N = 8, n = 4 (no. of channel)
    fs = 2fm = 2 × 4kHz = 8kHz
    Transmission rate = nNfs = 4 x 8 x 8 = 256 kbps

  1. The Nyquist sampling rate for the signal g(t) = 10 cos (50πt) cos2 (150 πt) where ‘t ’ is in seconds is—








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    Given
    g(t) = 10 cos(50πt) cos2 (150πt)

    = 10 cos(50πt)
    [1 + cos92.150πt]
    2

    = 5 cos(50πt) + 5 cos (50πt) cos (300πt)
    = 10 cos(30πt) +
    5
    		[cos (350πt) + cos (250πt)]
    2

    Hence
    fmmax = 175Hz
    So Nyquist sampling rate
    2fmmax = 2 × 175 = 350 sample per second.

    Correct Option: D

    Given
    g(t) = 10 cos(50πt) cos2 (150πt)

    = 10 cos(50πt)
    [1 + cos92.150πt]
    2

    = 5 cos(50πt) + 5 cos (50πt) cos (300πt)
    = 10 cos(30πt) +
    5
    		[cos (350πt) + cos (250πt)]
    2

    Hence
    fmmax = 175Hz
    So Nyquist sampling rate
    2fmmax = 2 × 175 = 350 sample per second.

  1. The pre-envelop and the envelop of a function f(t) = A cos (ωct)








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    Pre-envelop is given by
    fp(t) = f(t) + jfh(t)
    = A cos ωct + jA cos (ωct – 90°)
    = A cos ωct + jA sin ωct
    and the envelop is the magnitude of fp(t)
    |fpejωct(t)| = |Aejωct = A

    Correct Option: A

    Pre-envelop is given by
    fp(t) = f(t) + jfh(t)
    = A cos ωct + jA cos (ωct – 90°)
    = A cos ωct + jA sin ωct
    and the envelop is the magnitude of fp(t)
    |fpejωct(t)| = |Aejωct = A

  1. An audio signal has a bandwidth of 25 kHz. The maximum value of |m(t )| is 5V. This signal frequency modulates a carrier. If the frequency sensitivity constant of the modulator is 20 kHz/V, then according to Carson’s rule, the bandwidth of the modulator output would be—








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    BW = 2 (β + 1) fm
    where,

    β =
    kfAm
    		=
    20 ×103×5
    		= 4
    fm25 × 103

    BW = 2(4 + 1) 25 kHz = 250 kHz

    Correct Option: B

    BW = 2 (β + 1) fm
    where,

    β =
    kfAm
    		=
    20 ×103×5
    		= 4
    fm25 × 103

    BW = 2(4 + 1) 25 kHz = 250 kHz

  1. For 1 ≤ t ≤ 2 the phase deviation is—








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    Phase deviation,
    Δφ(t)=2πkf ∫t0 m(t)dt
    for
    0 ≤ t ≤ 1, Δφ(t)=2π·25 ∫t0 10tdt = 250πt2
    and
    Δφ(1) = 250π
    Now for 1 ≤ t ≤ 2, Δφ(t) = Δφ (1) + 2π·25 ∫t0 10dt
    = 250π + 500πt
    = 250π (1 + 2t) rad
    Hence alternative (B) is the correct choice.

    Correct Option: B

    Phase deviation,
    Δφ(t)=2πkf ∫t0 m(t)dt
    for
    0 ≤ t ≤ 1, Δφ(t)=2π·25 ∫t0 10tdt = 250πt2
    and
    Δφ(1) = 250π
    Now for 1 ≤ t ≤ 2, Δφ(t) = Δφ (1) + 2π·25 ∫t0 10dt
    = 250π + 500πt
    = 250π (1 + 2t) rad
    Hence alternative (B) is the correct choice.