Question

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.

Answer 1

Question

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