Chapter 6

Chapter 6 - Chapter 6 1(a For sinc pulse B = 1 2 T s ⇒ T...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

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

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 4-QAM 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(t1-t2); [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....
View Full 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 / 10

Chapter 6 - Chapter 6 1(a For sinc pulse B = 1 2 T s ⇒ T...

This preview shows document pages 1 - 3. Sign up to view the full document.

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