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Unformatted text preview: 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 B = 10 Since 4QAM is multilevel signalling SNR = P b N B = E s N BT s = 2 E s N B ( ∵ BT s = 1 2 ) ∴ SNR per symbol = E s N = 5 SNR per bit = E b N = 2 . 5 (a symbol has 2 bits in 4QAM) (c) SNR per symbol remains the same as before = E s N = 5 SNR per bit is halved as now there are 4 bits in a symbol E b N = 1 . 25 2. p = 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 . 3 Q d min √ 2 N ¶ = Q d min √ 2 N ¶ d min = 2 A = Q s 2 A 2 N (b) p ( ˆ m = 0  m = 1) p ( m = 1) = p ( ˆ m = 1  m = 0) p ( m = 0) . 7 Q A + a q N 2 = 0 . 3 Q A a q N 2 ,a > Solving gives us ’a’ for a given A and N (c) p ( ˆ m = 0  m = 1) p ( m = 1) = p ( ˆ m = 1  m = 0) p ( m = 0) . 7 Q A q N 2 = 0 . 3 Q B q N 2 ,a > Clearly A > B , for a given A we can find B (d) Take E b N = A 2 N = 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)); t2=.3*Q(a/sqrt(N0/2)); diff = abs(t1t2); [c,d] = min(diff); a(d) c 3. s ( t ) = ± g ( t )cos2 πf c t r = ˆ r cosΔ 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 (cosΔ φ ) acts as a scaling factor for both +1 and 1 levels. P e however increases as numerator is reduced due to multiplication by cosΔ φ P e = Q d min cosΔ φ √ 2 N ¶ 4. A 2 c Z T b cos 2 2 πf c tdt = A 2 c Z T b 1 + cos4 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 P e = 1 6 [1 .p 1 + 0 .p ] + 2 6 [0 .p 1 + 0 .p ] + 2 6 [0 .p 1 + 0 .p ] + 1 6 [0 .p 1 + 1 .p ] = 1 6 [ p 1 + p ] = 1 6 ( ∵ p 1 + p = 1 always ) 5. We will use the approximation P e ∼ (average number of nearest neighbors). Q ‡ d min √ 2 N · 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 · (b) 16QAM, P e = 4 ( 1 1 4 ) Q ‡ 2 a √ 2 N · = 3 Q ‡ 2 a √ 2 N · (c) P e ∼ 2 × 3+3 × 2 5 Q ‡ 2 a √ 2 N · = 2 . 4 Q ‡ 2 a √ 2 N · (d) P e ∼ 1 × 4+4 × 3+4 × 2 9 Q ‡ 3 a √ 2 N · = 2 . 67 Q ‡ 3 a √ 2 N · 6....
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This note was uploaded on 01/11/2012 for the course EE 359 taught by Professor Goldsmith during the Fall '08 term at Stanford.
 Fall '08
 Goldsmith

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