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
