Communication miscellaneous

Communication miscellaneous

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Moderate Questions
Difficult Questions

Communication

Communication miscellaneous
  1. A narrowband FM does not have the following feature—








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    In narrow band FM the lower sideband is negative, i.e. 180° out of phase as compared to the carrier. This makes the amplitude of FM approximately constant. Hence alternative (C) is most suitable choice.

    Correct Option: C

    In narrow band FM the lower sideband is negative, i.e. 180° out of phase as compared to the carrier. This makes the amplitude of FM approximately constant. Hence alternative (C) is most suitable choice.

  1. A non-linear device with a transfer characteristic given by i = (0 + 2Vi + 0·2Vi2) mA is supplied with a carrier of 1V amplitude and a sinusoidal signal of 0.5V amplitude in series. If at the output the frequency component of AM signal is considered, the depth of modulation is—








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    Given transfer characteristic
    i = (10 + 2Vi + 0·2 Vi2) mA …(A)
    Since, according to question both carrier and sinusoidal signals are in series. Let assume Vi is represented as
    Vi = Am cos ωmt + Ac cos ωct
    or
    Vi = 0·5 cos ωmt + cos ωct (˙.˙ Am = 0·5V, Ac = 1V)
    Now, on putting the value of Vi in equation (A) we get amplitude modulated AM signal i.e.,
    i = 10 + 2 (cos ωct + 0·5 cos ωmt) + 0·2 (cos ωct + 0·5 cos ωmt)2
    or
    i = 2 cos ωc t + 0·2 cos ωct + cos ωmt
    (˙.˙ At the output only lower frequency terms are considered and neglecting the higher frequency terms)

    i = 21 +
    0.2
    		cos ωmt cos ωct
    2
    ...(B)
    which is equal to standard AM equation.
    i = Ac (1 + μ cos ωmt) cos ωct …(C)
    So, on comparing equation (B) and (C) we get
    μ =
    0.2
    		 = 0.1
    2

    and% modulation index = 0.1 x 100 = 10%.
    Hence alternative (B) is the correct choice.

    Correct Option: B

    Given transfer characteristic
    i = (10 + 2Vi + 0·2 Vi2) mA …(A)
    Since, according to question both carrier and sinusoidal signals are in series. Let assume Vi is represented as
    Vi = Am cos ωmt + Ac cos ωct
    or
    Vi = 0·5 cos ωmt + cos ωct (˙.˙ Am = 0·5V, Ac = 1V)
    Now, on putting the value of Vi in equation (A) we get amplitude modulated AM signal i.e.,
    i = 10 + 2 (cos ωct + 0·5 cos ωmt) + 0·2 (cos ωct + 0·5 cos ωmt)2
    or
    i = 2 cos ωc t + 0·2 cos ωct + cos ωmt
    (˙.˙ At the output only lower frequency terms are considered and neglecting the higher frequency terms)

    i = 21 +
    0.2
    		cos ωmt cos ωct
    2
    ...(B)
    which is equal to standard AM equation.
    i = Ac (1 + μ cos ωmt) cos ωct …(C)
    So, on comparing equation (B) and (C) we get
    μ =
    0.2
    		 = 0.1
    2

    and% modulation index = 0.1 x 100 = 10%.
    Hence alternative (B) is the correct choice.

  1. A diode detector has a load of 1kΩ shunted by a 10000 pF capacitor. The diode has a forward resistance of 1Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping, with modulating signal frequency of 10kHz will be—








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    We know that

    RC ≤
    1
    		 √
    1 − μ2
    ωmμ2

    given that R = 1 kΩ = 1000Ω
    C = 10000 pf = 10–8 f
    ωm = 2π fm = 2π × 10 × 103 radian/sec
    μ =? (neglecting diode forward resistance i.e.1Ω)
    1000 × 10−8 ≤
    1
    		 √
    1 − μ2
    2π × 10 ×103μ2

    or
    0.628 ≤ √
    1 − μ2
    μ2

    or
    0.394 ≤
    1 − μ2
    μ2

    or
    0.394 μ2 ≤ 1 − μ2
    or
    1.394 μ2 ≤ 1
    or
    μ2 ≤
    1
    1.394

    or
    μ ≤ 0.8469
    or
    μmax ≤ 0.847
    Hence alternative (A) is the correct choice.

    Correct Option: A

    We know that

    RC ≤
    1
    		 √
    1 − μ2
    ωmμ2

    given that R = 1 kΩ = 1000Ω
    C = 10000 pf = 10–8 f
    ωm = 2π fm = 2π × 10 × 103 radian/sec
    μ =? (neglecting diode forward resistance i.e.1Ω)
    1000 × 10−8 ≤
    1
    		 √
    1 − μ2
    2π × 10 ×103μ2

    or
    0.628 ≤ √
    1 − μ2
    μ2

    or
    0.394 ≤
    1 − μ2
    μ2

    or
    0.394 μ2 ≤ 1 − μ2
    or
    1.394 μ2 ≤ 1
    or
    μ2 ≤
    1
    1.394

    or
    μ ≤ 0.8469
    or
    μmax ≤ 0.847
    Hence alternative (A) is the correct choice.

  1. The threshold effect can occur in the system— (i) FM (ii) AM (iii) AM with synchronous detection








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    NA

    Correct Option: C

    NA

  1. Which of the following is the best transmission efficiency?








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    Since threshold effect can occur on those demodulating systems which use envelop detector for demodulating the signal.
    We always use envelop detection for demodulating AM wave, and usually envelop detection for demodulating FM signal, as it is
    low cost process. Hence alternative (C) is the correct choice.

    Correct Option: C

    Since threshold effect can occur on those demodulating systems which use envelop detector for demodulating the signal.
    We always use envelop detection for demodulating AM wave, and usually envelop detection for demodulating FM signal, as it is
    low cost process. Hence alternative (C) is the correct choice.