{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Hw8SolnBinder_1

# Hw8SolnBinder_1 - l ECE 5670 Digital Communications ECE...

This preview shows pages 1–7. Sign up to view the full content.

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

View Full Document

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: l. ECE 5670 : Digital Communications ECE department, Cornell University, Spring 2011 Homework 8 Solutions Instructor: Salman Avestimehr Ofﬁce 3225 Rhodes Hall (a) Since we have a two tap channel (L = 2), the minimum cyclic preﬁx is L —- 1 = 1. (b) We know that ~ N°_1 27m! in = h ea: — 1 n g l P ( No ) ( ) Hence with ho = 1, in = 0.25 and NC = 2, £0 = be + h1 = 1.25 (2) ill = ho + h1€\$p(—7l') (3) = 0.75 (4) (0) Figures sh0w the plot of |hk |2 at points ’33:” for different values of N6. (C!) Figure below shows the plot of |H(f)|2 Vs. frequency. As we increase Na, we are sampling IH ( f ) |2 with higher resolution. hi? w. muting, t8 L4 0.6 0.4 02 Figure 1: Plot of lhk|2 Vs ’33: with N: = 32. Fm. ......... ooooooooooooo m? w. It'W'dNg Flgure 4: Plot of I'flkl2 Vs ’33” with NC = 256. Figure 5: Plot of lH ( f ) |2 Vs frequency. 2. We know that ~ Ive—1 27ml ht = 1; hit-EXP (4 N1 ) (5) With he 21,111 = %, we have ~ 1 3 ho — l + 5 — E (6) ~ 1 1‘11 = 1+ Eexp (—111) (7) ._ 1 (8 _ 2. ) The optimal power allocation is given by 1 02 )+ P’“ = — — t 9 k. (A lhklz ( ) A is so chosen that Pg‘ + Pf = P Substituting for it)“ we have 1 402 + P* = — 10 0 (A 9 ) ( ) 1 + P; = (A—40’2) . (11) (as) If 4 32 P g 402 — 902 = 9 02, (12) then 1 4 = P 2 1 A + 90 ( 3) and P; = P (14) Pf = 0 (15) o On the other hand, if 32 P > 9 0'2, (16) then 1 1 32 2 2 _ = _ _ __ 7 /\ 2 (P 9 or ) +40 (1 ) P 20 2 = — — 13 2+9cr () In this case 13* = — — —— 0 2 + 9 0' 90' (19) P 16 2 — §+§a (20) * __ P 16 2 P1 — 5*30 (2].) (b) When P<< a2, (22) then 32 P s 302, (23) and the ﬁrst case happens, i.e., P6“ — P (24) p; = 0. (25) The idea is that when transmit power is scarce, we should invest it “wisely” (i.e., use all of it only on the better of the two sub—channels). (c) When P > .72, (26) then we have P 16 3 >> 302. (27) In this case, the difference in power allocation for both the channels is negligible and both the channels get half the power of g. 3. The received sequence is y[0] = 1.1, y[1] = 2.9, y[2] = 0.5, y[3] = —3.6, y[4} = 2.6, y[5] = —1.2, y[6] = —l.9 and y[7] = 2.3. The trellis diagram is shown below. m=1 m=2 m=3 m=4 m=5 m=6 m=7 (1,1) 0—t0——*O—>O—>O—’O—>O (-1.1) O O (1,-1) O (-1,-1) At stage 1 there are four possible paths, (1,1),(1,—1),(—1,1),(—1,—1), with costs C(1,1) = (1.5 x 1 — 1.1)2 + (1.5 x 1 + 0.6 x1— 2.9)2 = 0.80 C(1,—1)=[1.5 x 1 — 1.1)2 + (—1.5 x 1+ 0.6 x 1 — 2.9)2 = 14.60 0(—1, 1) = (—1.5 x 1 — 1.1)2 + (1.5 x 1 — 0.6 x 1 — 2.9)2 = 10.76 0(—1,—1) = (1.5 x 1 — 1.1)2 + (1.5 X 1 + 0.6 x 1— 2.9)? = 31.76 At stage 2 we pick the least costly two paths entering each vertex in the trellis at m = 2. C(1,1,1) = C(1, 1) + (1.5 x 1 + 0.6 x 1 — 0.5)2 = 0.80 + 2.56 = 3.36 C(1, 1, +1) = 0(1, 1) + (—1.5 x 1+ 0.6 x 1 — 0.5)2 = 0.80 +1.96 = 2.76 C(1, —1, 1) = C(1,—1)+ (1.5 x 1— 0.6 x 1 — 0.5)2 = 14.60 + 0.16 = 14.76 C(1,—1,—1) = C(1, —1)+ [—1.5 x 1— 0.6 x 1 — 0.5)2 = 14.60 + 6.76 = 21.36 As all four paths have £[0] = 1 in common, we can decode this symbol immediately. At stage three we have 0(1,1,1,1) = 3.36 + (1.5 + 0.6 + 3.6)2 = 35.85 C(l, 1,1, —1) = 8.36 + (+1.5 + 0.6 + 3.6)2 = 10.65 C(1, 1, —1, 1) = 276+ (1.5 — 0.6 +3.6)2 = 23.01 ’ C(1,1,—1,—1) = 2.76 + (—1.5 — 0.6 + 3.6)2 = 5.01 As all four paths have the same second symbol, we can immediately decode £[1] = 1. At stage four we have C(1,1,—1, 1,1) = 23.01 + (1.5 + 0.6 — 2.6)2 = 23.26 0(1, 1, —1, 1, —1) = 2301+ (—1.5 + 0.6 — 2.6)2 = 35.26 0(1, 1, +1, —1, 1) = 5.01 + (1.5 — 0.6 — 2.6)2 = 7.90 C(1,1,—1,—1,—~1) = 5.01 +(-1.5 — 0.6 — 2.6)2 = 27.10 Thus we can immediately decode :E[2} = —1. At stage ﬁve we have 0(1,1,—1,—1,1,1)= 7.90 + (1.5 + 0.6 + 1.2)2 =18.79 C(1,1,—1,—1,1,—1)= 7.90 + (—1.5 +0.6 +1.2)2 = 7.99 0(1, 1, —1, —1, —1, 1) = 27.10 + (1.5 — 0.6 +1.2)2 = 31.51 C(1,1,—1,—1,—1,—1) = 27.10 + (—1.5 — 0.6 + 1.2)2 = 27.91 Thus £[3] = —1. At stage six we compute C(1,1,—1,—1,1,1,1)=18.79 + (1.5 + 0.6 +1.9)2 = 34.79 0(1,1,—1,—1,1,1,—1)= 18.79 + (—1.5 + 0.6 + 1.9)2 = 19.79 0(1, 1, —1, —1, 1, —1, 1) = 7.99 + (1.5 — 0.6 + 1.9)2 = 15.83 C(1, 1, +1, +1, 1, —1, +1) = 7.99 + (—1.5 — 0.6 + 1.9)2 = 8.03 6 Thus §3[4] = 1. At stage seven we compute C(1,1,—1,—1,1,—1,1,1) = 15.83 + (1.5 + 0.6 — 2.3)2 = 15.87 0(1,1, —1,—1, 1, —1,1,—1) 2 15.83 + (—1.5 + 0.6 — 2.3)2 = 26.07 C(1,1,—1,—1,1,~1,—1,1) = 8.03 + (1.5 — 0.6 — 2.3)2 = 9.99 C(1,1,—1,—1,1,—1,—1,—1) = 8.03 + (—1.5 — 0.6 — 2.3)2 = 27.39 The most probable path, and hence the path the Viterbi algorithm outputs, is then £[0] = l, 2f:[1] = 1, 652] = —1,:ﬁ[3]= —1,§:[4] = 1, 55[5] = —1, £[6] = —1,5.r:-[7] = 1. Comparing this to the transmitted sequence we have an error for x[2], but all other symbols are decoded correctly. ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

Hw8SolnBinder_1 - l ECE 5670 Digital Communications ECE...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online