Solutions Communications Technology II WS 2010/2011

Transcription

1 Solutions Communications Technology II WS 010/011 Yidong Lang, Henning Schepker NW1, Room N350, Tel.: 041/ lang/ Universität Bremen, FB1 Institut für Telekommunikation und Hochfrequenztechnik Arbeitsbereich Nachrichtentechnik Prof. Dr.-Ing. A. Dekorsy Postfach D 8334 Bremen WWW-Server: Version December 17, 010

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3 I WS 010/011 Communications Technology II Solutions Contents 1 Equalization 1 Solution to exercise 1(eq03) Solution to exercise (eq08) Solution to exercise 3(eq09) Solution to exercise 4(eq11) Solution to exercise 5(eq1) Viterbi 8 Solution to exercise 6(vit01) Solution to exercise 7(vit08) Solution to exercise 8(vit10) Solution to exercise 9(vit14) Solution to exercise 10(vit15) Mobile Radio Channel 14 Solution to exercise 11( mobrad) Solution to exercise 1( mobrad) Solution to exercise 13( mobrad) OFDM 19 Solution to exercise 14(ofdm03) Solution to exercise 15(ofdm04) Solution to exercise 16(ofdm05) Solution to exercise 17( ) Solution to exercise 18( ofdm)

4 Communications Technology II Solutions WS 010/011 II Conventions and Nomenclature All references to passages in the text (chapter- and page numbers) refer to the book: K.-D. Kammeyer: Nachrichtenübertragung,.Edition, B. G. Teubner Stuttgart, 1996, ISBN: ; References of equations of type (1.1.1) refer to the book, too, whereas these of type (1) refer to the solutions of the exercises. The functions rect ( ) and tri( ) are defined analogous to: N. Fliege: Systemtheorie, 1.Edition, B. G. Teubner Stuttgart, 1991, ISBN: Thus rect (t/t) has the temporal expanse T, whereas tri(t/t) is not zero for the length of T. The letters f and F represent frequencies (in Hertz), ω and Ω angular frequencies (in rad/s). The following relations are always valid: ω = πf resp. Ω = πf. δ 0 (t) denotes the continuous(!) Dirac-pulse, whereas δ(i) represents the timediscrete impulse sequence. So called ideal low-, band- and highpassfilter G(jω) have value 1 in the respective passing range and value 0 in the stop range. If a time-discrete data sequence d(i) of rate 1/T stimulates a continues filter with impulse response g(t), it has to be interpreted as [ ] x(t) = T d(i) δ 0 (t it) g(t) = T d(i) g(t it). Abbreviations i= i= ACF auto-correlation function, sequence ISI inter-symbol interference BW, BB bandwidth, baseband KKF cross-correlation function, sequence BP bandpass AF audio frequency DPCM differential PCM PCM pulse-code modulation F{ } Fourier transform PR partial response H{ } Hilbert transform S/N = SNR signal-to-noise ratio HP highpass LP lowpass Availability on Internet PDF (or PS) -files of the exercises can be downloaded from:

5 1 WS 010/011 Communications Technology II Solutions 1 Equalization Solution to exercise 1 (eq03): a) DFE block diagram: y i q ( ) y( i) + - z -1 b) c(i) = [ 1 1 ; ] y(i) = d(i) c(i) + n(i) = 1 d(i) + 1 d(i 1) + n(i) y q (i) = y(i) ˆd(i 1) no wrong decisions: ˆd(i 1) = d(i 1) The DFE amplifies the noise by 3dB. = d(i) + d(i 1) + n(i) ˆd(i 1) y q (i) = d(i) + n(i) c) E b N 0 = 10 db = 10 γq = 1 S/N-loss caused by noise amplification in task b) ( ) P b = 1 erfc Eb γq N = 1 ( ) 0 erfc 5 = 1 erfc (.3) =

6 Communications Technology II Solutions WS 010/011 Solution to exercise (eq08): a) Impulse response of the total system: f(t) = g(t) c(t) h(t) = g(t) h(t) c(t) ( where g(t) h(t) is given exercise ) sampling returns f (k) = {0.5, 1.5, 1.5, 0.5} b) T/ - equalization: total impulse response is convolution off (k) and e T/ : With f (k) e T/ (k) = the result is 0.75 ( ( ) ( ) = [ ] T c) Sampling with even k gives [ ] T which is an undistorted system. d) Sampling of total impulse response of task a) at symbol clock: f(i) = [ ] T The result is f(i) e(i) = [ ] T

7 3 WS 010/011 Communications Technology II Solutions Solution to exercise 3 (eq09): a) Total impulse response: g(i) = f(i) e(i) = [1, α, +α, α 3, 0] +[0, α, α, α 3, α 4 ] g(i) = δ(i) α 4 δ(i 4) b) G(z) = 1 α 4 z 4 z 0,ν = α e j π ν, ν = c) ( ) S = 1 N ISI α 8 d) General equalizer of the order n: e(i) = Max{ d(i)} = α 4 n 1 ( 1) l α l δ(i l) l=0 f(i) e(i) = δ(i) ( 1) n α n δ(i n) ( ) S = 1 N ISI α n Max{ d(i)} = α n

8 Communications Technology II Solutions WS 010/011 4 Solution to exercise 4 (eq11): a) + y(i) d(i) Z -1 Z -1 b) 0.5 y(i) = x(i) 0.5 ˆd(i ) = d(i) d(i ) + n(i) 0.5 ˆd(i ) = d(i) d(i ) d(i ) + n(i) = d(i) + d(i ) + n(i) c). d(i) d(i-) y(i) error-probability 1 1 +n n 1/ n 1/ n 0 P b = 1 4 (1 + 1 ) = 1 4 The power of the noise has no influence as long as it is so small that } y(i) = + n lead to safe decisions y(i) = + n

9 Im 5 WS 010/011 Communications Technology II Solutions Solution to exercise 5 (eq1): (a) Since y(i) = d(i 1) 0.5 e jπ/4 + d(i), the admissible values for y(i) are given by d(i) d(i 1) y(i) +1 + j +1 + j +1 + j ( ) +1 + j 1 + j j +1 + j 1 j +1 + j (1 0.5) +1 + j +1 j j 1 + j +1 + j 1 + j ( ) 1 + j 1 + j 1 (0.5) + j 1 + j 1 j 1 + j (1 0.5) 1 + j +1 j j 1 j +1 + j 1 + j ( ) 1 j 1 + j j 1 j 1 j 1 + j ( 1 0.5) 1 j +1 j j +1 j +1 + j +1 + j ( ) +1 j 1 + j j +1 j 1 j +1 + j ( 1 0.5) +1 j +1 j j Re (b) The impulse response of the overall system is given by w(i) = ι h(ι)g(i ι). Thus, w(0) = h(0)e(0) = 1 w(1) = h(1)e(0) + h(0)e(1) = e j5π/4 + e jπ/4 = 0 w() = h(1)e(1) = (0.5) e j(5π/4+π/4) = 0.5j. (c) Since z(i) = d(i) w(i) = d(i ) 0.5 e j3π/ + d(i), the admissible values for y(i) are given by

10 Communications Technology II Solutions WS 010/011 6 d(i) d(i ) y(i) +1 + j +1 + j j j 1 + j j j 1 j j j +1 j j j +1 + j j j 1 + j j j 1 j j j +1 j j j +1 + j 0.75 j j 1 + j 0.75 j j 1 j 1.5 j j +1 j 1.5 j j +1 + j +1.5 j j 1 + j +1.5 j j 1 j j j +1 j j imag real (d) The squared magnitude frequency response can be calculated by W(Ω) = (1 j0.5e jω )(1 + j0.5e jω ) (1) = j0.5(e jω e jω ) () = sin(ω) (3)

11 7 WS 010/011 Communications Technology II Solutions W(Ω) pi/4 pi/ pi 0 Ω

12 Communications Technology II Solutions WS 010/011 8 Viterbi Solution to exercise 6 (vit01): (a), (b) S0 = { 1 j}, S1 = { 1 + j}, S = {+1 j}, S3 = {+1 + j} S0 S1 S3 S4 d(1)=1+j d()=1-j d(3)=-1-j d(4)=-1+j d(5)=-1-j z 0,0 = j z 0,1 = z 0, = j z 0,3 = 0 z 1,0 = z 1,1 = + j z 1, = 0 z 1,3 = j z,0 = j z,1 = 0 z, = j z,3 = z 3,0 = 0 z 3,1 = j z 3, = z 3,3 = + j c) λ 1 = z 0,3 (0.5 + j) = 0 (0.5 + j) = 1.5 λ = z 3, ( + 0.5j) = ( + 0.5j) = 0.5 λ 3 = z,0 ( 3j) = j ( 3j) = 1 λ 4 = z 0,1 ( ) = ( ) = 0 λ 5 = z 1,0 ( + j) = ( + j) = 1 5 λ i = 3.5 i=1

13 9 WS 010/011 Communications Technology II Solutions Solution to exercise 7 (vit08): a) Trellis element for nd order channel: Trellis diagram for a complete sequence: ramp-up phase S 0{-1,-1} slow down phase S {-1,1} S {1,-1} S {1,1} 3 s(i) b) Table with the (not standadized) partial path costs: P , z 0, 1 = z 0,1 = z 1, 1 = z 1,1 = z, 1 = z,1 = z 3, 1 = z 3,1 =

14 Communications Technology II Solutions WS 010/ Path: see item a) c) d(i) = 1, 1, 1, 1

15 11 WS 010/011 Communications Technology II Solutions Solution to exercise 8 (vit10): (a) channel length=4 ; channel order=3; S 0 = { 1, 1, 1} S 1 = {+1, 1, 1} S = { 1, +1, 1} S 3 = {+1, +1, 1} S 4 = { 1, 1, +1} S 5 = {+1, 1, +1} S 6 = { 1, +1, +1} S 7 = {+1, +1, +1} (b) d(i) = {+1, 1, +1, 1, +1, +1, 1, +1, 1, 1, 1} (c) f(i) = {+1, 1, 1, 1, 1, +1, 1, +1, 1, 1, 1} error vector: e(i) = (d(i) f(i))/ e = [1, 0, 1] T γ min = eh F H Fe = 0.6

16 Communications Technology II Solutions WS 010/011 1 Solution to exercise 9 (vit14): a) Symbol clock model: ( ) b) Calculate two signal levels: w 11 = 1 ] [0.8 (1 + j) (1 + j) w 4 = 1 ] [0.8 (1 j) ( 1 + j) = 1 ] [ j = 1 ] [0. 0.j 1 + j c) ISI (here) leads to a total number of 16 different points in the signal space: QPSK symbol space with ISI j 1 w w 1 w 1 w 11 imag w 3 w 4 w 3 w 31 w 13 w 14 w 4 w 33 w 34 w 43 w real w 41 d) (Solution of the additional exercise) The average signal power σ d has to be calculated from the average quadratic value of the signal space points. This calulation is trivial for the symbol alphabet at the input: The symbol s average power is 1, since all symbols also have the absolute value of 1. Average signal power at the channel s output: [ σ d = j j + 4 = 1 8 [( ) + ( ) + ( )] = 1 8 [ ] = 1.0 ] j Thus the signal is neither extenuated nor amplified by the channel ( neutral due to power ).

17 13 WS 010/011 Communications Technology II Solutions Solution to exercise 10 (vit15): a) (from p.556) upper path: d(i) = 0 / lower path: d(i) = 1 true data sequence (bold): d(i) = b) estimated sequence (dashed): ˆd(i) = error vector: e = [ d(i 0 ),..., d(i 0 + L f l 1)] T L f = 4; l = ; L f l 1 = 4 1 = 1 An error vector of length results: e = [ d(i 0 ), d(i 0 + 1)] T e = [ 1, 1] T c) The structure of the AC matrix for a channel of order is given on p.577. To find this matrix we have to determine the energy ACF of the error vector first: r ee (0) = ν r ee (1) = ν e(ν)e(ν + 0) = e(ν)e(ν + 1) = 1 r ee (λ) = ν R E ee = e(ν)e(ν + λ) = 0 forλ d) Eigenvalue equation, cf. Eq.( a): det(r E ee λi) = 0 Eq.( a): (r ee (0) λ) r ee (1) = 0 λ min = S/N-loss: γmin = ˆ=.3 db

18 Communications Technology II Solutions WS 010/ Mobile Radio Channel Solution to exercise 11 ( mobrad): a) f D = v c 0 f 0 cos (α) f D1 = 185.Hz, f D = 0Hz, f D3 = 151.7Hz b) Amplitude f [Hz] c) Amplitude t_0 t_1 t_ Delay d) h K (t) = n ρ ν δ (t τ ν ) = 1 + r 1 δ (t τ 1 ) + r δ (t τ ) ν=0 n H K (jω) = ρ ν exp ( jωτ ν ) = 1 + r 1 exp ( jωτ 1 ) + r exp ( jωτ ) ν=0 e) H TP (jω) = H K (j(ω + ω 0 )), f < B/ [ ] 1 + r1 e j(ω+ω 0)τ 1 + r e j(ω+ω 0)τ for B H TP (jω) = f B 0 otherwise

19 15 WS 010/011 Communications Technology II Solutions f) Absolute channel transfer function not constant frequency-selective channel

20 Communications Technology II Solutions WS 010/ Solution to exercise 1 ( mobrad): a) BER for BPSK and AWGN ( ) P b = 1 erfc Eb N 0 BER for BPSK and AWGN with instantaneous SNR ( ) P b (h) = 1 erfc h E b N 0 Uniform Prob. of occurence: P 1 = P = P 3 = 1/3, E b /N 0 = 7dB = 5 P b = 1 3 (P b(h 1 ) + P b (h ) + P b (h 3 )) P b (h 1 ) = 1 erfc ( P b (h ) = 1 erfc ( 0, 5 exp(jπ/4) E b N 0 ) ) = 1 erfc (1, 1) = 0, , 8 exp(jπ/6) E b = 1 erfc (1, 789) = 0, 0057 N 0 P b (h 3 ) = 1 ( erfc 0, 1 + 0, j E ) b = 1 erfc (0, 5) = 0, 398 N 0 P b = b) P b = 0, 6P b (h 1 ) + 0, 3P b (h ) + 0, 1P b (h 3 ) = c) Strongest channel coefficient h P b,min = P b (h ) = 0, 0057 d) data rate R b = 1/T Baud = 1/(50 ns) = 0 Mbit/s Transmitted in 30% of all cases: Rb = 0, 3 R b = 6 Mbit/s

21 17 WS 010/011 Communications Technology II Solutions Solution to exercise 13 ( mobrad): a) τ 0 = l 0 /c 0 h(0) = ρ 0 exp jπ f 0 τ }{{} 0 f 0 τ 0 = f 0 l 0 c 0 = 000 h(0) = 1 l 1 f 0 τ 1 = f 0 = h(1) = ρ 1 exp ( jπ ( )) = 0.5 [ j 0.866] c 1 = j b) H(jω) = exp ( j π 0. 6) exp ( j ω τ) τ = τ 1 τ 0 = l 1 c 0 l 0 c 0 = 1400 m m/s 600 m m/s =. 6µs Maximum, if exponent is a multiple of π ψ 1 ω max τ = nπ f max = π 0. 6 nπ π τ = π 0. 6 π. 6µs n. 6 µs f max = 50 khz + n 375 khz, n = 0, ±1, ±, ±3 Minimum, if exponent is an odd multiple of π ψ 1 ω min τ = nπ f min = ψ 1 nπ π τ = π 0. 6 π. 6 µs n. 6µs f min = 50 khz + n khz, n = ±1, ±3,... c) f D = f 0 v c 0 cos (α) f D0 = f Dmax = f 0 150Km/h = Hz m/s ( π f D1 = f Dmax cos = 4) 1 f Dmax = Hz r(t) = s(t τ 0 ) exp (jπ f Dmax t) + s(t τ 1 ) exp (jπ f Dmax t)

22 Communications Technology II Solutions WS 010/ khz 1.5 H(jω) khz f in khz

23 19 WS 010/011 Communications Technology II Solutions 4 OFDM Solution to exercise 14 (ofdm03): a) Kernel symbol length: T s = 1 f = 4 ms Bandwidth: B = N f = 51 khz Data rate: R = log (M) N T s+t g = N T = 048 6ms b) FFT length: N f = 4096 = 341 kbit/s N f samples account for the interval with the length of T s since exactly 1 FFT is used for the generation of the kernel symbol. Sample rate: f A = N f T s = 104 khz Samples within guard interval: N g = N f T s T g = 048 c) Transmitter power: N 0 = N0 = Ws E OFDM E b = log (M) N = 1.4Ws 048 = Ws E b N 0 = Ws = db Ws ( γg = 1 T ) g = T g + T s 3 ) P b = 1 erfc ( Eb N 0 γ g = 1 erfc (1.9488) = P = E OFDM T s + T g = 1.4Ws 6ms = 33.3W

24 À ŵ À ¾ µ Communications Technology II Solutions WS 010/011 0 Solution to exercise 15 (ofdm04): a) T = 1 1 = Nc 1 = 16 1 = 3, 333 f u B u , s R = N c ldm 3 = 16 = 14, 4Mbit/s T 3, b) T g = T T S = Nc( ) = ( 1 1) = 0, 66 µs B u ,8 c) N c = R T 1 ldm = 13, , = 15 d) h ½º ½ ¼º ¼ ¼ ½ ¾ ½¼ ½½ ½¾ ½ ½ ½ Å ¾

25 1 WS 010/011 Communications Technology II Solutions Solution to exercise 16 (ofdm05): a) b) c) T G = τ c = 0, 8 µs T G = 0% T = , 8 µs = 4 µs T S = T T G = 4 µs 0, 8 µs = 3, µs f = 1 T S = 1 = 31, 5kHz 3, 10 6 µs S N = 1 T G T = 1 0, 8 = 1 0, = 0, 8 = 0.96 db 1 db 4 B f = , = 64 d) ld(m) = R T N = = QPSK

26 Communications Technology II Solutions WS 010/011 Solution to exercise 17 ( ): a) Advantages and disadvantages to be found in the following table... Advantage Simple Equalization Frequency domain adaptation Robust against multipath fading Disadvantage increased PAPR Guard loss b) f B N = 15kHz 1 1 β = = 1 + T G TS µs 66.67µs <= T G = 16.67µs τ c = 0.8 c) N off R b = N B max 18MHz = 048 = 848 subcarriers f 15kHz ld (M) N = = = 86.4 Mbit/s T S + T G 66.67µs µs d) FFT length N f = 048 f A N g = N f = 048 T S 66.67µs = N f T G T S = = 30.7 MHz µs 66.67µs = 51 samples

27 3 WS 010/011 Communications Technology II Solutions Solution to exercise 18 ( ofdm): a) b) γ = T ( ) ( ) ( ) s T g = T s + T g γ 1 T s = γ 1 f = 0, khz T g = 5, 9 µs R b = N log (M) N = R b (T s + T g ) = 69.5 = 630 T s + T g log (M) c) N fft = 104, f a = N f = khz = 10, 4 MHz d) Maximal data rate, if all subcarriers are modulated: R b = N log (M) T s + T g = µs = 16, 7 MBit/s e) n Pi < 1 n Pi < T s = 100 = 3, 86 T s τ max T g 5, 9