Communication miscellaneous

Communication miscellaneous

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Communication

Communication miscellaneous
  1. A 2 MHz carrier is amplitude modulated by a 500 Hz modulation signal to a depth of 60%. If the unmodulated carrier power is 1kW, the power of the modulated signal is—








  1. View Hint View Answer Discuss in Forum

    Given that
    Pc = 1kW
    μ = 60% = 0·6
    Power of modulated signal

    Pt = Pc 1 +
    μ2
    2

    or
    Pt = 1kW 1 +
    0.62
    		= 1.18kW
    2

    Correct Option: C

    Given that
    Pc = 1kW
    μ = 60% = 0·6
    Power of modulated signal

    Pt = Pc 1 +
    μ2
    2

    or
    Pt = 1kW 1 +
    0.62
    		= 1.18kW
    2

  1. Given f(t) = cos ω1t + sin ω2t then fh(t) =?








  1. View Hint View Answer Discuss in Forum

    We know that Hilbert transform [fh (t)] can be obtained by replacing the signal (– 90°) i.e.
    fh (t) = cos (ω1t – 90°) + sin (ω2t – 90°)
    or fh (t) = sin ω1t – cos ω2t
    Hence alternative (A) is the correct choice.

    Correct Option: A

    We know that Hilbert transform [fh (t)] can be obtained by replacing the signal (– 90°) i.e.
    fh (t) = cos (ω1t – 90°) + sin (ω2t – 90°)
    or fh (t) = sin ω1t – cos ω2t
    Hence alternative (A) is the correct choice.

  1. Given f(t) = cos ω1t + sin ω2t then fh(t) =?








  1. View Hint View Answer Discuss in Forum

    We know that Hilbert transform [fh (t)] can be obtained by replacing the signal (– 90°) i.e.
    fh (t) = cos (ω1t – 90°) + sin (ω2t – 90°)
    or fh (t) = sin ω1t – cos ω2t
    Hence alternative (A) is the correct choice.

    Correct Option: A

    We know that Hilbert transform [fh (t)] can be obtained by replacing the signal (– 90°) i.e.
    fh (t) = cos (ω1t – 90°) + sin (ω2t – 90°)
    or fh (t) = sin ω1t – cos ω2t
    Hence alternative (A) is the correct choice.

  1. The allocation of power in each sidebands if percentage of modulation is changed to 80%—








  1. View Hint View Answer Discuss in Forum

    Here given μ = 0·8 and Pc = 1000 W
    so,

    Pt = Pc 1 +
    μ2
    2

    Pt = Pc 1 +
    0.82
    		= 1320 W
    2

    Now,
    Pt = Pc + PUSB + PLSB
    1320 = 1000 + 2PUSB (∴ PUSB = PLSB)
    PUSB =
    320
    		= 160 W
    2

    Correct Option: B

    Here given μ = 0·8 and Pc = 1000 W
    so,

    Pt = Pc 1 +
    μ2
    2

    Pt = Pc 1 +
    0.82
    		= 1320 W
    2

    Now,
    Pt = Pc + PUSB + PLSB
    1320 = 1000 + 2PUSB (∴ PUSB = PLSB)
    PUSB =
    320
    		= 160 W
    2

  1. What is the percentage modulation?








  1. View Hint View Answer Discuss in Forum

    We know that,

    Pt = Pc 1 +
    μ2
    2

    where, Pt= Pc + PUSB = 1000 + 200 + 200 = 1400 W Pc = 1000 W
    Now,
    1400 = 1000 1 +
    μ2
    2

    or  
    μ2
    		= 0.4
    2

    or   μ2 = 0.8
    or   μ = 0.894
    or   μ = 89.4%

    Correct Option: D

    We know that,

    Pt = Pc 1 +
    μ2
    2

    where, Pt= Pc + PUSB = 1000 + 200 + 200 = 1400 W Pc = 1000 W
    Now,
    1400 = 1000 1 +
    μ2
    2

    or  
    μ2
    		= 0.4
    2

    or   μ2 = 0.8
    or   μ = 0.894
    or   μ = 89.4%