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20 June, 10:49

1. A sine wave is to be used for two different signaling schemes: a. PSK b. QPSK The duration of a signal element is 10-5 s. If the received signal is of the following form: s (t) = 0.005 sin (2 106 t +  ) volts And if the measured noise power at the receiver is 2.5 x 10-8 watts, determine the Eb/N0 (in dB) for each case.

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  1. 20 June, 10:59
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    A sine wave is to be used for two different signaling schemes: a. PSK b. QPSK The duration of a signal element is 10^-5 s. If the received signal is of the following form: s (t) = 0.005sin (2π10^6t + θ) And if the measured noise power at the receiver is 2.5 x 10-8 watts, determine the Eb/N0 (in dB) for each case.

    Answer:

    a. 27dB

    b. 24dB

    Explanation:

    a.

    For case 1;

    Given

    Amplitude = A = 0.005

    Let

    t1 = Signal element period

    t2 = Bit period

    Such that t1 = t2 = 10^-5 sec

    Noise Power = 2.5 * 10^-8 W

    Calculating N°

    N° = N * t1 = 2.5 * 10^-8 * 10-5 =

    N° = 2.5E-13

    Calculating Eb

    Eb = P * t2

    Where P = 25E-6 1/t1 ∫s² (t) {t1,0}

    P = 1/10^-5 * 25E-6 ∫ sin² (2π10^6t + θ) dt {10^-5,0}

    Sin²θ = ½ (1 - cos2θ)

    P = 10^5 * 25E-6 ∫½ (1 - cos (4π10^6t + 2θ)) dt {10^-5,0}

    P = 10^5 * ½ * 25E-6∫ (1 - cos (4π10^6t + 2θ)) dt {10^-5,0}

    P = 10^5 * ½ * 25E-6 * t {10^-5,0}

    P = 10^5 * ½ * 25E-6 * (10^-5 - 0)

    P = 10^5 * ½ * 25E-6 * 10^-5

    P = ½ * 25E-6

    Eb = P * t2

    Eb = ½ * 25E-6 * 10^-5

    Eb = 12.5E-11

    Eb/N° = 12.5E-11/2.5E-13

    Eb/N° = 500

    Convert to dB

    = 10Log (500)

    = 27dB

    b.

    For case 2;

    Amplitude = A = 0.005

    Let

    t1 = Signal element period

    t2 = Bit period

    Such that t2 = ½t1 = 10^-5 sec

    Eb = P * ½t1

    N° = 2.5 E-8 * t1

    Eb/N° = 25E-6 * ½t1/2.5E-8 * t1

    Eb/N° = 250

    Convert to dB

    10Log (250) = 24dB
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