Solutions Communications Technology II WS 010/011 Yidong Lang, Henning Schepker NW1, Room N350, Tel.: 041/18-6393 E-mail: lang/ schepker@ant.uni-bremen.de Universität Bremen, FB1 Institut für Telekommunikation und Hochfrequenztechnik Arbeitsbereich Nachrichtentechnik Prof. Dr.-Ing. A. Dekorsy Postfach 33 04 40 D 8334 Bremen WWW-Server: http://www.ant.uni-bremen.de Version December 17, 010
I WS 010/011 Communications Technology II Solutions Contents 1 Equalization 1 Solution to exercise 1(eq03).............................. 1 Solution to exercise (eq08).............................. Solution to exercise 3(eq09).............................. 3 Solution to exercise 4(eq11).............................. 4 Solution to exercise 5(eq1).............................. 5 Viterbi 8 Solution to exercise 6(vit01).............................. 8 Solution to exercise 7(vit08).............................. 9 Solution to exercise 8(vit10).............................. 11 Solution to exercise 9(vit14).............................. 1 Solution to exercise 10(vit15).............................. 13 3 Mobile Radio Channel 14 Solution to exercise 11(007-03-mobrad)........................ 14 Solution to exercise 1(007-10-mobrad)........................ 16 Solution to exercise 13(009-10-mobrad)........................ 17 4 OFDM 19 Solution to exercise 14(ofdm03)............................ 19 Solution to exercise 15(ofdm04)............................ 0 Solution to exercise 16(ofdm05)............................ 1 Solution to exercise 17(007-10-6)........................... Solution to exercise 18(009-10-ofdm)......................... 3
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: 3-519-1614-7; 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: 3-519- 06140-6. 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: http://www.ant.uni-bremen.de
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) = 8 10 4
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 ) 1.5 1.0 0.5 3 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 (0.5 1.5 1.5 0.5 + 0.5 ( 0.5 1.5 1.5 0.5) 0.375 1.15 1.15 0.375 ( ) 0.15 0.375 0.375 0.15 = [0.375 1.0 0.75 0 0.15 ] T c) Sampling with even k gives [ 1.0 0 ] T which is an undistorted system. d) Sampling of total impulse response of task a) at symbol clock: f(i) = [ 1.5 0.5 ] T The result is f(i) e(i) = 0.001 0.0039 0.9987 0.997 0.0004 0.0013 0.339 0.0999 [ 0.001 0.0035 1.0000 0.033 0.0999 ] T
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 = 0...3 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
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) + 0.5 d(i ) + n(i) 0.5 ˆd(i ) = d(i) + 0.5 d(i ) + 0.5 d(i ) + n(i) = d(i) + d(i ) + n(i) c). d(i) d(i-) y(i) error-probability 1 1 +n 0 1-1 0+n 1/ -1 1 0+n 1/ -1-1 -+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
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 + 0.5) +1 + j 1 + j +1 0.5 + j +1 + j 1 j +1 + j (1 0.5) +1 + j +1 j +1 + 0.5 + j 1 + j +1 + j 1 + j (1 + 0.5) 1 + j 1 + j 1 (0.5) + j 1 + j 1 j 1 + j (1 0.5) 1 + j +1 j 1 + 0.5 + j 1 j +1 + j 1 + j ( 1 + 0.5) 1 j 1 + j 1 0.5 j 1 j 1 j 1 + j ( 1 0.5) 1 j +1 j 1 + 0.5 j +1 j +1 + j +1 + j ( 1 + 0.5) +1 j 1 + j +1 0.5 j +1 j 1 j +1 + j ( 1 0.5) +1 j +1 j +1 + 0.5 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
Communications Technology II Solutions WS 010/011 6 d(i) d(i ) y(i) +1 + j +1 + j +1.5 + j 0.75 +1 + j 1 + j +1.5 + j 1.5 +1 + j 1 j +0.75 + j 1.5 +1 + j +1 j +0.75 + j 0.75 1 + j +1 + j 0.75 + j 0.75 1 + j 1 + j 0.75 + j 1.5 1 + j 1 j 1.5 + j 1.5 1 + j +1 j 1.5 + j 0.75 1 j +1 + j 0.75 j 1.5 1 j 1 + j 0.75 j 0.75 1 j 1 j 1.5 j 0.75 1 j +1 j 1.5 j 1.5 +1 j +1 + j +1.5 j 1.5 +1 j 1 + j +1.5 j 0.75 +1 j 1 j +0.75 j 0.75 +1 j +1 j +0.75 j 1.5 1.5 1 0.5 imag 0 0.5 1 1.5 1.5 1 0.5 0 0.5 1 1.5 real (d) The squared magnitude frequency response can be calculated by W(Ω) = (1 j0.5e jω )(1 + j0.5e jω ) (1) = 1 + 0.5 j0.5(e jω e jω ) () = 1.065 + 0.5 sin(ω) (3)
7 WS 010/011 Communications Technology II Solutions 1.8 1.6 1.4 1. W(Ω) 1 0.8 0.6 0.4 0. 0 0 pi/4 pi/ pi 0 Ω
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
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} 16 0 1 4 9 8 slow down phase 8 18 7 1 S {-1,1} 1 4 17 1 17 8 17 18 S {1,-1} 4 16 1 8 13 16 S {1,1} 3 s(i) 5 9 8 4 4 1.5-0.5-1.5.5 -.5 0.5 b) Table with the (not standadized) partial path costs: P 1.5-0.5-1,5.5 -.5 0.5 z 0, 1 = -.5 16 4 1 5 0 9 z 0,1 = -0.5 4 0 1 9 z 1, 1 = -1.5 0 16 1 4 z 1,1 = 0.5 4 4 z, 1 = -0.5 0 1 9 4 z,1 = 1.5 4 9 1 z 3, 1 = 0.5 4 4 9 z 3,1 =.5 16 0
Communications Technology II Solutions WS 010/011 10 Path: see item a) c) d(i) = 1, 1, 1, 1
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
Communications Technology II Solutions WS 010/011 1 Solution to exercise 9 (vit14): a) Symbol clock model: ( ) -1 0.8 0.6 b) Calculate two signal levels: w 11 = 1 ] [0.8 (1 + j) + 0.6 (1 + j) w 4 = 1 ] [0.8 (1 j) + 0.6 ( 1 + j) = 1 ] [1.4 + 1.4j = 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 0.14 0.14j 1 w w 1 w 1 w 11 imag 0.5 0-0.5-1 w 3 w 4 w 3 w 31 w 13 w 14 w 4 w 33 w 34 w 43 w 44-1 -0.5 0 0.5 1 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 = 1 16 4 0. + 0.j + 8 1.4 + 0.j + 4 = 1 8 [( 0. + 0. ) + (1.4 + 0. ) + ( 1.4 + 1.4 )] = 1 8 [4 0. + 4 1.4 ] = 1.0 ] 1.4 + 1.4j Thus the signal is neither extenuated nor amplified by the channel ( neutral due to power ).
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) = 0 1 1 0 0 b) estimated sequence (dashed): ˆd(i) = 1 0 1 0 0 00 01 10 11 0 1 0 1 0 1 0 1 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 1 0 1 1 0 1 d) Eigenvalue equation, cf. Eq.(14.5.48a): det(r E ee λi) = 0 Eq.(14.5.59a): (r ee (0) λ) r ee (1) = 0 λ min = S/N-loss: γmin = ˆ=.3 db
Communications Technology II Solutions WS 010/011 14 3 Mobile Radio Channel Solution to exercise 11 (007-03-mobrad): a) f D = v c 0 f 0 cos (α) f D1 = 185.Hz, f D = 0Hz, f D3 = 151.7Hz b) 1 0.9 0.8 0.7 Amplitude 0.6 0.5 0.4 0.3 0. 0.1 0 00 150 100 50 0 50 100 150 00 f [Hz] c) Amplitude 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0.1 0 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
15 WS 010/011 Communications Technology II Solutions f) Absolute channel transfer function not constant frequency-selective channel
Communications Technology II Solutions WS 010/011 16 Solution to exercise 1 (007-10-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, 0569 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 = 0.1008 b) P b = 0, 6P b (h 1 ) + 0, 3P b (h ) + 0, 1P b (h 3 ) = 0.0598 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
17 WS 010/011 Communications Technology II Solutions Solution to exercise 13 (009-10-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 = 666.66... h(1) = ρ 1 exp ( jπ (666 + 0. 6)) = 0.5 [ 0.5 + j 0.866] c 1 = 0.5 + j 0.433 b) H(jω) = 1 + 0.5 exp ( j π 0. 6) exp ( j ω τ) τ = τ 1 τ 0 = l 1 c 0 l 0 c 0 = 1400 m 3 10 8 m/s 600 m 3 10 8 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 187.5 khz, n = ±1, ±3,... c) f D = f 0 v c 0 cos (α) f D0 = f Dmax = f 0 150Km/h = 138.8889 Hz 3 10 8 m/s ( π f D1 = f Dmax cos = 4) 1 f Dmax = 98.093 Hz r(t) = s(t τ 0 ) exp (jπ f Dmax t) + s(t τ 1 ) exp (jπ 0.707 f Dmax t)
Communications Technology II Solutions WS 010/011 18.5 15 khz 1.5 H(jω) 1 0.5 6.5 khz 0 00 150 100 50 0 50 100 150 00 f in khz
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 = 1. 10 4 Ws E OFDM E b = log (M) N = 1.4Ws 048 = 6.836 10 4 Ws E b N 0 = 6.836 10 4 Ws = 5.7 7.55 db 1. 10 4 Ws ( γg = 1 T ) g = T g + T s 3 ) P b = 1 erfc ( Eb N 0 γ g = 1 erfc (1.9488) = 3 10 3 P = E OFDM T s + T g = 1.4Ws 6ms = 33.3W
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 6000 0,8 10 6 s R = N c ldm 3 = 16 = 14, 4Mbit/s T 3,33 10 6 b) T g = T T S = Nc( 1 16 1) = ( 1 1) = 0, 66 µs B u 6000 0,8 c) N c = R T 1 ldm = 13, 5 106 3, 333 10 6 1 3 = 15 d) h ½º ½ ¼º ¼ ¼ ½ ¾ ½¼ ½½ ½¾ ½ ½ ½ Å ¾
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 = 100 0 0, 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 = 0 106 31, 5 10 3 = 64 d) ld(m) = R T N = 3 106 4 10 6 64 = QPSK
Communications Technology II Solutions WS 010/011 Solution to exercise 17 (007-10-6): 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 1 + 16.67µs 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 6 100 = = = 86.4 Mbit/s T S + T G 66.67µs + 16.67µs 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 048 16.67µs 66.67µs = 51 samples
3 WS 010/011 Communications Technology II Solutions Solution to exercise 18 (009-10-ofdm): a) b) γ = T ( ) ( ) ( ) s 1 1 1 1 T g = T s + T g γ 1 T s = γ 1 f = 0, 794 1 1 10 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 = 104 10 khz = 10, 4 MHz d) Maximal data rate, if all subcarriers are modulated: R b = N log (M) T s + T g = 104 15.9 µs = 16, 7 MBit/s e) n Pi < 1 n Pi < T s = 100 = 3, 86 T s τ max T g 5, 9