Chapter 7

Chapter 7 - Chapter 7 1. Ps = 103 QPSK, Ps = 2Q( s ) 103 ,...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 7 1. P s = 10 - 3 QPSK, P s = 2 Q ( γ s ) 10 - 3 , γ s γ 0 = 10.8276. P out ( γ 0 ) = M Y i =1 1 - e - γ 0 γ i · γ 1 = 10, γ 2 = 31.6228, γ 3 = 100. M = 1 P out = 1 - e - γ 0 γ 1 · = 0.6613 M = 2 P out = 1 - e - γ 0 γ 1 · ‡ 1 - e - γ 0 γ 2 · = 0.1917 M = 3 P out = 1 - e - γ 0 γ 1 · ‡ 1 - e - γ 0 γ 2 · ‡ 1 - e - γ 0 γ 3 · = 0.0197 2. p γ Σ ( γ ) = M γ £ 1 - e - γ/ γ / M - 1 e - γ/ γ γ = 10 dB = 10 as we increase M, the mass in the pdf keeps on shifting to higher values of γ and so we have higher values of γ and hence lower probability of error. MATLAB CODE gamma = [0:.1:60]; gamma_bar = 10; M = [1 2 4 8 10]; fori=1:length(M) pgamma(i,:) = (M(i)/gamma_bar)*(1-exp(-gamma/gamma_bar)).^. .. (M(i)-1).*(exp(-gamma/gamma_bar)); end 3. P b = Z 0 1 2 e - γ p γ Σ ( γ ) = Z 0 1 2 e - γ M γ h 1 - e - γ/ γ i M - 1 e - γ/ γ = M 2 γ Z 0 e - (1+1 / γ ) γ h 1 - e - γ/ γ i M - 1 = M 2 γ M - 1 X n =0 ± M - 1 n ( - 1) n e - (1+1 / γ ) γ = M 2 M - 1 X n =0 ± M - 1 n ( - 1) n 1 1 + n + γ = desired expression
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
0 10 20 30 40 50 60 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 γ p γ Σ ( ) M = 1 M = 2 M = 4 M = 8 M = 10 Figure 1: Problem 2 4. p γ Σ ( γ ) = Pr { γ 2 < γ τ 1 < γ } γ < γ τ Pr { γ τ γ 1 γ } + Pr { γ 2 < γ τ 1 < γ } γ > γ τ If the distribution is iid this reduces to p γ Σ ( γ ) = P γ 1 ( γ ) P γ 2 ( γ τ ) γ < γ τ Pr { γ τ γ 1 γ } + P γ 1 ( γ ) P γ 2 ( γ τ ) γ > γ τ 5. P b = Z 0 1 2 e - γ p γ Σ ( γ ) p γ Σ ( γ ) = ( ± 1 - e - γ T / γ ) 1 γ e - γ r / γ γ < γ T ± 2 - e - γ T / γ ) 1 γ e - γ r / γ γ > γ T P b = 1 2 γ 1 - e - γ T / γ · Z γ T 0 e - γ/ γ e - γ + 1 2 γ 2 - e - γ r / γ · Z γ T e - γ/ γ e - γ = 1 2( γ + 1) 1 - e - γ T / γ + e - γ T e - γ T / γ · 6. P b P b (10 dB ) P b (20 dB ) no diversity 1 2( γ +1) 0.0455 0.0050 SC(M=2) M 2 M - 1 m =0 ( - 1) m M - 1 m 1+ m + γ 0.0076 9 . 7 × 10 - 5 SSC 1 2( γ +1) ± 1 - e - γ T / γ + e - γ T e - γ T / γ ) 0.0129 2 . 7 × 10 - 4 As SNR increases SSC approaches SC 7. See MATLAB CODE: gammab_dB = [0:.1:20]; gammab = 10.^(gammab_dB/10); M= 2;
Background image of page 2
0 2 4 6 8 10 12 14 16 18 20 10 -7 10 -6 10 -5 10 -4 10 -3 10 -2 10 -1 10 0 γ avg P b,avg (DPSK) M = 2 M = 3 M = 4 Figure 2: Problem 7 for j = 1:length(gammab) Pbs(j) = 0 for m = 0:M-1 f = factorial(M-1)/(factorial(m)*factorial(M-1-m));
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/11/2012 for the course EE 359 taught by Professor Goldsmith during the Fall '08 term at Stanford.

Page1 / 11

Chapter 7 - Chapter 7 1. Ps = 103 QPSK, Ps = 2Q( s ) 103 ,...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online