Communication miscellaneous

Communication miscellaneous

Easy Questions
Moderate Questions
Difficult Questions

Communication

Communication miscellaneous
  1. The carrier swing—








  1. View Hint View Answer Discuss in Forum

    Carrier swing = 107.218 – 107.196 = 0.22 MHz

    Correct Option: A

    Carrier swing = 107.218 – 107.196 = 0.22 MHz

  1. Which of the following AM techniques provide the advantages of greater signal power and reduction of bandwidth?








  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: B

    NA

  1. The sum of two signals e1 = 3 sin (4π × 103t) and e1= 5 sin(4π × 256t) is sampled at 1024 Hz. The sampled signal is passed through a low pass filter with cut off at 2048 Hz. The output of the filter will contain components at—








  1. View Hint View Answer Discuss in Forum

    Given that
    e1= 3 sin(4π × 103t), fm1 = 2 × 103 Hz
    e2 = 5 sin(4 π × 256t), fm2 = 256 Hz
    sampling frequency, fs = 1024 Hz
    cut-off frequency of LPF, ωc = 2048 Hz
    Since for proper reconstruction of signal fs ≥ 2fm
    For fm1, fs less than 2fm1 so it will pass through the LPF,
    while for fm2, fs > 2fm2 [i.e. 1024 > (2 × 256)]
    So the output of the filter will contain 256 Hz component only.

    Correct Option: C

    Given that
    e1= 3 sin(4π × 103t), fm1 = 2 × 103 Hz
    e2 = 5 sin(4 π × 256t), fm2 = 256 Hz
    sampling frequency, fs = 1024 Hz
    cut-off frequency of LPF, ωc = 2048 Hz
    Since for proper reconstruction of signal fs ≥ 2fm
    For fm1, fs less than 2fm1 so it will pass through the LPF,
    while for fm2, fs > 2fm2 [i.e. 1024 > (2 × 256)]
    So the output of the filter will contain 256 Hz component only.

  1. In a broadcast transmitter, the RF output is represented as
    e(t) = 50 [1 + 0·89 cos 5000 t + 0·30 sin 9000t] cos (6 × 106t) volt
    What are the sidebands of the signals in radians?








  1. View Hint View Answer Discuss in Forum

    The given equation. e(t) = 50[1 + 0·89 cos 5000 t + 0·30 sin 9000t] cos (6 × 106t) volt
    Let 5000 = ωm1 or ωm1 = 0·005 MHz
    9000 = ωm2 or ωm2 = 0·009 MHz
    6 × 106 = ωc1 or ωc1 = 6 MHz
    e(t) = 50[cos ωc1 t + 0·89 cos ωm1t. cos ωc1t + 0·30 sin ωm2t]
    Sidebands of the signals are at
    ωc1 ± ωm1 and ωc1 ± ωm2, frequency
    i.e. (6 ± 0·005) MHz and (6 ± 0·009) MHz
    6·005 × 106, 5·995 × 106, 6·009 × 106, 5·991 × 106
    Hence alternative (B) is the correct choice.

    Correct Option: B

    The given equation. e(t) = 50[1 + 0·89 cos 5000 t + 0·30 sin 9000t] cos (6 × 106t) volt
    Let 5000 = ωm1 or ωm1 = 0·005 MHz
    9000 = ωm2 or ωm2 = 0·009 MHz
    6 × 106 = ωc1 or ωc1 = 6 MHz
    e(t) = 50[cos ωc1 t + 0·89 cos ωm1t. cos ωc1t + 0·30 sin ωm2t]
    Sidebands of the signals are at
    ωc1 ± ωm1 and ωc1 ± ωm2, frequency
    i.e. (6 ± 0·005) MHz and (6 ± 0·009) MHz
    6·005 × 106, 5·995 × 106, 6·009 × 106, 5·991 × 106
    Hence alternative (B) is the correct choice.

  1. Which one of the following is represented by
    V(t) = 5[cos (106 πt) – sin (103 πt) × sin (106 πt)]








  1. View Hint View Answer Discuss in Forum

    NA

    Correct Option: D

    NA