Homework 10

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Homework 10

ECE 460: Communication and Information Theory
Prof. B.-P. Paris
Homework 10
Solution

Problem 7.9

a) The received signal may be expressed as
${ n(t) if s (t) was transmitted r (t) = 0 A + n(t) if s1(t) was transmitted$
Assuming that s(t) has unit energy, then the sampled outputs of the crosscorrelators are
$r = sm + n, m = 0,1$
where s₀ = 0, s₁ = A and the noise term n is a zero-mean Gaussian random variable with variance
$[ ∫ ∫ ] 2 --1- T -1-- T σn = E √T-- 0 n (t)dt√T-- 0 n(τ)dτ ∫ T ∫ T = 1- E [n (t)n (τ )]dtd τ T 0 0 N0- ∫ T ∫ T N0- = 2T 0 0 δ(t - τ)dtdτ = 2$
The probability density function for the sampled output is
$2 f (r|s ) = √-1---e- rN0 0 πN0 1 (r-A√T)2 f (r|s1) = √-----e- N0 πN0$
Since the signals are equally probable, the optimal detector decides in favor of s₀ if $PM (r,s0) = f (r|s0) > f(r|s1) = PM (r,s1)$
otherwise it decides in favor of s₁. The decision rule may be expressed as
$s PM (r,s0) (r-A√T-)2-r2 - (2r-A√T-)A√T 0 ----------= e N0 = e N0 >< 1 PM (r,s1) s 1$
or equivalently
$s1 √-- r >< 1A T 2 s0$
The optimum threshold is 1 2A.
b) The average probability of error is
$1 1 P (e) = -P (e|s0) + -P (e|s1) 2 ∫ 2 ∫ 1 √ -- = 1- ∞ f (r |s )dr + 1- 2A Tf (r |s )dr 2 12A√T-- 0 2 -∞ 1 ∫ ∞ 2 ∫ 1A √T- (r-A√T)2 = 1- √ -√--1---e- rN0dr + 1- 2 √-1---e- --N0---dr 2 12A T πN0 2 - ∞ πN0 ∫ ∫ 1∘ -2- √-- = 1- ∞∘ --- √1--e- x22dx + 1- -2 N0 A T √-1--e- x22 dx 2 12 N2A √T- 2π 2 -∞ 2π [ ∘ 0--- ] 1- -2- √ -- [√ -----] = Q 2 N A T = Q SNR 0$
where $12A2T-- SNR = N0$
Thus, the on-off signaling requires a factor of two more energy to achieve the same probability of error as the antipodal signaling.
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Problem 7.10

Since the rate of transmission is R = 10⁵ bits/sec, the bit interval T_b is 10^-5 sec. The probability of error in a binary PAM system is
$[∘ ---] 2Eb- P (e) = Q N0$
where the bit energy is _b = A²T_b. With P(e) = P₂ = 10^-6, we obtain
$∘ ---- 2 2Eb-= 4.75 =⇒ E = 4.75-N0- = 0.112813 N0 b 2$
Thus
$2 √ --------------- A Tb = 0.112813 =⇒ A = 0.112813 × 105 = 106.21$

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Problem 7.11

a) For a binary PAM system for which the two signals have unequal probability, the optimum detector is
$s1 r > -N√0---ln 1---p-= η < 4 Eb p s2$
The average probability of error is
$P(e) = P(e|s1)P (s1) + P (e|s2)P (s2) = pP∫ (eη |s1) + (1 - p)P(e|s2)∫ ∞ = p f(r|s )dr + (1 - p) f(r|s )dr - ∞ 1 η 1 ∫ η 1 (r-√Eb)2- ∫ ∞ 1 (r+√Eb)2- = p √-----e- N0 dr + (1 - p) √-----e- N0 dr - ∞ ∫ πN0 ∫ η πN0 --1-- η1 - x22 --1-- ∞ - x22- = p√ 2π -∞ e dx + (1 - p)√ 2π η2 e dx$
where $∘ ---- ∘ ---- ∘ ---- ∘ ---- 2Eb- -2- 2Eb- -2- η1 = - N0 + η N0 η2 = N0 + η N0$
Thus,
$[ ∘---- ∘ ---] [∘ ---- ∘ ---] P(e) = pQ 2Eb-- η -2- + (1 - p)Q 2Eb-+ η 2-- N0 N0 N0 N0$

b) If p = 0.3 and _b N₀ = 10, then
$P (e) = 0.3Q [4.3774] + 0.7Q [4.5668 ] = 0.3 × 6.01 × 10-6 + 0.7 × 2.48 × 10-6 = 3.539 × 10 -6$
If the symbols are equiprobable, then $∘ ---- 2Eb- √ --- -6 P (e) = Q [ N0 ] = Q [ 20] = 3.88 × 10$
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Problem 7.12

a) The optimum threshold is given by
$η = -N√0---ln 1---p-= -N√0--ln 2 4 Eb p 4 Eb$

b) The average probability of error is (η = ln 2)
$∫ ∞ 1 -(r+√E-)2∕N P(e) = p(am = - 1) √-----e b 0dr ∫η πN0 +p (a = 1) η √--1--e- (r-√Eb)2∕N0dr m -∞ πN0 ⌊ √ --⌋ ⌊√ -- ⌋ = 2Q ⌈ η∘-+---Eb⌉ + 1-Q ⌈-∘Eb---η⌉ 3 N ∕2 3 N ∕2 ⌊ ∘ --0---- ----0⌋ ⌊ ---- ∘------- ⌋ 2 2N0 ∕Ebln2 ∘ 2E 1 ∘ 2E 2N0 ∕Eb ln 2 = -Q ⌈ -------------+ --b⌉ + --Q ⌈ --b-- ------------⌉ 3 4 N0 3 N0 4$

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