Chapter 6

# Chapter 6 - Chapter 6 1(a For sinc pulse B = 1 2Ts Ts = 1...

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Chapter 6 1. (a) For sinc pulse, B = 1 2 T s T s = 1 2 B = 5 × 10 - 5 s (b) SNR = P b N 0 B = 10 Since 4-QAM is multilevel signalling SNR = P b N 0 B = E s N 0 BT s = 2 E s N 0 B ( BT s = 1 2 ) SNR per symbol = E s N 0 = 5 SNR per bit = E b N 0 = 2 . 5 (a symbol has 2 bits in 4QAM) (c) SNR per symbol remains the same as before = E s N 0 = 5 SNR per bit is halved as now there are 4 bits in a symbol E b N 0 = 1 . 25 2. p 0 = 0 . 3 , p 1 = 0 . 7 (a) P e = Pr (0 detected, 1 sent — 1 sent) p (1 sent) + Pr (1 detected, 0 sent — 0 sent) p (0 sent) = 0 . 7 Q d min 2 N 0 + 0 . 3 Q d min 2 N 0 = Q d min 2 N 0 d min = 2 A = Q s 2 A 2 N 0 (b) p ( ˆ m = 0 | m = 1) p ( m = 1) = p ( ˆ m = 1 | m = 0) p ( m = 0) 0 . 7 Q A + a q N 0 2 = 0 . 3 Q A - a q N 0 2 , a > 0 Solving gives us ’a’ for a given A and N 0 (c) p ( ˆ m = 0 | m = 1) p ( m = 1) = p ( ˆ m = 1 | m = 0) p ( m = 0) 0 . 7 Q A q N 0 2 = 0 . 3 Q B q N 0 2 , a > 0 Clearly A > B , for a given A we can find B (d) Take E b N 0 = A 2 N 0 = 10 In part a) P e = 3 . 87 × 10 - 6 In part b) a=0.0203 P e = 3 . 53 × 10 - 6 In part c) B=0.9587 P e = 5 . 42 × 10 - 6 Clearly part (b) is the best way to decode. MATLAB CODE: A = 1; N0 = .1; a = [0:.00001:1]; t1 = .7*Q(A/sqrt(N0/2));

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t2=.3*Q(a/sqrt(N0/2)); diff = abs(t1-t2); [c,d] = min(diff); a(d) c 3. s ( t ) = ± g ( t ) cos 2 πf c t r = ˆ r cos Δ φ where ˆ r is the signal after the sampler if there was no phase offset. Once again, the threshold that minimizes P e is 0 as (cos Δ φ ) acts as a scaling factor for both +1 and -1 levels. P e however increases as numerator is reduced due to multiplication by cos Δ φ P e = Q d min cos Δ φ 2 N 0 4. A 2 c Z T b 0 cos 2 2 πf c tdt = A 2 c Z T b 0 1 + cos 4 πf c t 2 = A 2 c T b 2 + sin(4 πf c T b ) 8 πf c | {z } 0 as f c 1 = A 2 c T b 2 = 1 x ( t ) = 1 + n ( t ) Let prob 1 sent = p 1 and prob 0 sent = p 0 P e = 1 6 [1 .p 1 + 0 .p 0 ] + 2 6 [0 .p 1 + 0 .p 0 ] + 2 6 [0 .p 1 + 0 .p 0 ] + 1 6 [0 .p 1 + 1 .p 0 ] = 1 6 [ p 1 + p 0 ] = 1 6 ( p 1 + p 0 = 1 always ) 5. We will use the approximation P e (average number of nearest neighbors). Q d min 2 N 0 · where number of nearest neighbors = total number of points taht share decision boundary (a) 12 inner points have 5 neighbors 4 outer points have 3 neighbors avg number of neighbors = 4.5 P e = 4 . 5 Q 2 a 2 N 0 · (b) 16QAM, P e = 4 ( 1 - 1 4 ) Q 2 a 2 N 0 · = 3 Q 2 a 2 N 0 · (c) P e 2 × 3+3 × 2 5 Q 2 a 2 N 0 · = 2 . 4 Q 2 a 2 N 0 · (d) P e 1 × 4+4 × 3+4 × 2 9 Q 3 a 2 N 0 · = 2 . 67 Q 3 a 2 N 0 · 6.
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