Solutions Communications technology II WS 2009/2010 - Universität ...

SolutionsCommunicationstechnologyIIWS2009/2010Peter KlennerNW1, Room S1260, Tel.: 0421/218-4282E-mail: klenner@ant.uni-bremen.deUniversität Bremen, FB1Institut für Telekommunikation und HochfrequenztechnikArbeitsbereich NachrichtentechnikProf. Dr.-Ing. K. D. KammeyerPostfach 33 04 40D–28334 BremenWWW-Server: http://www.ant.uni-bremen.deVersion 24th February 2010

I WS 2009/2010“Communications technology II” – SolutionsContents1 Entzerrung / Equalization 1Solution to exercise 1(eq03) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Solution to exercise 2(eq08) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2Solution to exercise 3(eq09) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Solution to exercise 4(eq11) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Solution to exercise 5(eq12) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Viterbi 8Solution to exercise 6(vit01) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Solution to exercise 7(vit08) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Solution to exercise 8(vit10) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11Solution to exercise 9(vit14) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Solution to exercise 10(vit15) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Mobile Radio Channel 14Solution to exercise 11(2007-03-mobrad) . . . . . . . . . . . . . . . . . . . . . . . . 14Solution to exercise 12(2007-10-mobrad) . . . . . . . . . . . . . . . . . . . . . . . . 16Solution to exercise 13(2009-10-mobrad) . . . . . . . . . . . . . . . . . . . . . . . . 174 OFDM 19Solution to exercise 14(ofdm03) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19Solution to exercise 15(ofdm04) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Solution to exercise 16(ofdm05) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21Solution to exercise 17(2007-10-6) . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Solution to exercise 18(2009-10-ofdm) . . . . . . . . . . . . . . . . . . . . . . . . . 23

“Communications technology II” – Solutions WS 2009/2010 IIConventionsand Nomenclature• All references to passages in the text (chapter- and page numbers) refer to the book: K.-D. Kammeyer: “Nachrichten”ubertragung”, 2.Edition, B. G. Teubner Stuttgart,1996, ISBN: 3-519-16142-7; 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 thelength of 2T.• The letters f and F represent frequencies (in Hertz), ω and Ω angular frequencies (inrad/s). The following relations are always valid: ω = 2πf resp. Ω = 2πF.• “δ 0 (t)” denotes the continuous(!) Dirac-pulse, whereas “δ(i)” represents the timediscreteimpulse sequence.• So called “ideal” low-, band- and highpassfilter G(jω) have value ’1‘ in the respectivepassing range and value ’0‘ in the stop range.• If a time-discrete data sequence d(i) of rate 1/T stimulates a continues filter withimpulse 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).Abbreviationsi=−∞i=−∞ACF auto-correlation function, -sequence ISI intersymbol-interferenceBW, BB bandwidth; baseband KKF cross-correlation function, -sequenceBP bandpass AF audio frequencyDPCM differential PCM PCM pulse code “modulation”F{·} Fourier-transform PR partial responseH{·} Hilbert-transform S/N=SNR signal-to-noise ratioHP highpass LP lowpassAvailabilityon InternetPDF (or PS) -files of the exercises can be downloaded from:http://www.ant.uni-bremen.de

1 WS 2009/2010 “Communications technology II” – Solutions1 Entzerrung/EqualizationSolution to exercise 1 (eq03):a) DFE Block diagram:y i q ( )y( i)+-2z -1b)c(i) = [ √ 1 1; √ ]2 2y(i) = d(i) ∗ c(i) + n(i)= 1 √2· d(i) + 1 √2· d(i − 1) + n(i)y q (i) = √ 2 · 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) + √ 2 · n(i) − ˆd(i − 1)y q (i) = d(i) + √ 2 · n(i)c)E bN 0= 10 dB = 10γq 2 = 1 S/N-loss caused by noise amplification in task b)2 (√ )P b = 1 2 erfc Eb· γqN 2 = 1 (√ )0 2 erfc 5= 1 2erfc (2.23)= 8 · 10 −4

“Communications technology II” – Solutions WS 2009/2010 2Solution to exercise 2 (eq08):a) Pulse response of the total system:f(t) = g(t) ∗ c(t) ∗ h(t) = g(t) ∗ h(t) ∗ c(t) ( wobei g(t) ∗ h(t) see exercise )1.51.00.52 3sampling returns f 2 (k) = {0, 0.5, 1.5, 1.5, 0.5}b) T/2 - equalization: total pulse response is convolution off 2 (k) and e T/2 : With f 2 (k) ∗e T/2 (k) = the result is0.75 · (0.5 1.5 1.5 0.5+ −0.25 · ( 0.5 1.5 1.5 0.5)0.375 1.125 1.125 0.375(−) 0.125 0.375 0.375 0.125= [0.375 1.0 0.75 0 −0.125 ] Tc) Sampling with even k gives [ 1.0 0 ] T which is an undistorted system.d) Sampling of total pulse response of task a) at symbol clock:f(i) = [ 1.5 0.5 ] TWith f(i) ∗ e(i) = the result is−0.0012 0.0039 0.9987 −0.2997−0.0004 0.0013 0.3329 −0.0999[ −0.0012 0.0035 1.0000 0.0332 −0.0999 ] T

3 WS 2009/2010 “Communications technology II” – SolutionsSolution to exercise 3 (eq09):a) Total pulse response:g(i) = f(i) ∗ e(i) = [1, −α, +α 2 , −α 3 , 0]+[0, α, −α 2 , α 3 , −α 4 ]g(i) = δ(i) − α 4 · δ(i − 4)b)G(z) = 1 − α 4 z −4z 0,ν = |α| · e j· π2 ·ν , ν = 0...3c) ( SN)ISI= 1 α 8Max{∆d(i)} = α 4d) General equalizer of the order n:n∑e(i) = (−1) l α l δ(i − l)l=0f(i) ∗ e(i) = δ(i) − (−1) n · α n · δ(i − n)( ) S 1=NISIα ( 2n)Max{∆d(i)} = α n

“Communications technology II” – Solutions WS 2009/2010 4Solution to exercise 4 (eq11):a)+y(i)d(i)Z -1Z -1b)0.5y(i) = x(i) − 0.5 · ˆd(i − 2) + n(i)= d(i) + 0.5 · d(i − 2) + n(i) − 0.5 · ˆd(i − 2)= d(i) + 0.5 · d(i − 2) + 0.5 · d(i − 2) + n(i)= d(i) + d(i − 2) + n(i)c) .d(i) d(i-2) y(i) error-probability1 1 2+n 01 -1 0+n 1/2-1 1 0+n 1/2-1 -1 -2+n 0P b = 1 4 · (1 2 + 1 2 ) = 1 4The power of the noise has no influence as long as it is so small that}y(i) = 2 + nlead to save decisionsy(i) = −2 + n

Im5 WS 2009/2010 “Communications technology II” – SolutionsSolution to exercise 5 (eq12):(a) Since y(i) = d(i − 1) · 0.5 · e jπ/4 + d(i), the admissible values for y(i) are given byd(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 − jRe(b) The impulse response of the overall system is given by w(i) = ∑ ιh(ι)g(i − ι). Thus,w(0) = h(0)e(0) = 1w(1) = h(1)e(0) + h(0)e(1) = e j5π/4 + e jπ/4 = 0w(2) = h(1)e(1) = (0.5) 2 · e j(5π/4+π/4) = −0.25j.(c) Since z(i) = d(i) ∗ w(i) = d(i − 2) · 0.25 · e j3π/4 + d(i), the admissible values for y(i) aregiven by

“Communications technology II” – Solutions WS 2009/2010 6d(i) d(i − 2) y(i)+1 + j +1 + j +1.25 + j ∗ 0.75+1 + j −1 + j +1.25 + j ∗ 1.25+1 + j −1 − j +0.75 + j ∗ 1.25+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 ∗ 1.25−1 + j −1 − j −1.25 + j ∗ 1.25−1 + j +1 − j −1.25 + j ∗ 0.75−1 − j +1 + j −0.75 − j ∗ 1.25−1 − j −1 + j −0.75 − j ∗ 0.75−1 − j −1 − j −1.25 − j ∗ 0.75−1 − j +1 − j −1.25 − j ∗ 1.25+1 − j +1 + j +1.25 − j ∗ 1.25+1 − j −1 + j +1.25 − j ∗ 0.75+1 − j −1 − j +0.75 − j ∗ 0.75+1 − j +1 − j +0.75 − j ∗ 1.251.510.5imag0−0.5−1−1.5−1.5 −1 −0.5 0 0.5 1 1.5real(d) The squared magnitude frequency response can be calculated by|W(Ω)| 2 = (1 − j0.25e j2Ω )(1 + j0.25e −j2Ω ) (1)= 1 + 0.25 2 − j0.25(e j2Ω − e −j2Ω ) (2)= 1.0625 + 0.5 sin(2Ω) (3)

7 WS 2009/2010 “Communications technology II” – Solutions1.81.61.41.2|W(Ω)| 210.80.60.40.200 pi/4 pi/2 pi 0Ω

“Communications technology II” – Solutions WS 2009/2010 82 ViterbiSolution to exercise 6 (vit01):(a), (b) S0 = {−1 − j}, S1 = {−1 + j}, S2 = {+1 − j}, S3 = {+1 + j}S0S1S3S4d(1)=1+jd(2)=1-jd(3)=-1-jd(4)=-1+jd(5)=-1-jz 0,0 = −2 − 2jz 0,1 = −2z 0,2 = −2jz 0,3 = 0z 1,0 = −2z 1,1 = −2 + 2jz 1,2 = 0z 1,3 = 2jz 2,0 = −2jz 2,1 = 0z 2,2 = 2 − 2jz 2,3 = 2z 3,0 = 0z 3,1 = 2jz 3,2 = 2z 3,3 = 2 + 2jc)λ 1 = |˜z 0,3 − (1.5 + j)| 2 = | + 1 + j − (1.5 + j)| 2 = 0.25λ 2 = |z 3,2 − (2 + 0.5j)| 2 = |2 − (2 + 0.5j)| 2 = 0.25λ 3 = |z 2,0 − (−3j)| 2 = | − 2j − (−3j)| 2 = 1λ 4 = |z 0,1 − (−2)| 2 = | − 2 − (−2)| 2 = 0λ 5 = |z 1,0 − (−2 + j)| 2 = | − 2 − (−2 + j)| 2 = 15∑λ i = 2.5i=1

9 WS 2009/2010 “Communications technology II” – SolutionsSolution to exercise 7 (vit08):a) Trellis element for 2nd-order channel:Trellis diagram for a complete sequence:EinschwingphaseAusschwingphaseS {-1,-1}01620214292828 182721S {-1,1}14171217281718S {1,-1}24162181613S {1,1} 382524924s(i)1.5 -0.5 -1.5 2.5 -2.5 0.5b) Table with the (not standadized) partly path costs:P 1.5 -0.5 -1,5 2.5 -2.5 0.5z 0,−1 = -2.5 16 4 1 25 0 9z 0,1 = -0.5 4 0 1 9z 1,−1 = -1.5 0 16 1 4z 1,1 = 0.5 4 4z 2,−1 = -0.5 0 1 9 4z 2,1 = 1.5 4 9 1z 3,−1 = 0.5 4 4 9z 3,1 = 2.5 16 0

“Communications technology II” – Solutions WS 2009/2010 10Path: see item a)c)d(i) = 1, −1, −1, 1

11 WS 2009/2010 “Communications technology II” – SolutionsSolution to exercise 8 (vit10):(a) channel length=4 ; channel order=3;S 0 = {−1, −1, −1}S 1 = {−1, −1, +1}S 2 = {−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))/2⇒ e = [1 , 0, 1] T⇒ γ 2 min = eH F H Fe = 0.6

“Communications technology II” – Solutions WS 2009/2010 12Solution to exercise 9 (vit14):a) Symbol clock model:( )-10.8 0.6b) Calculate two signal levels exemplarily:w 11 = √ 1][0.8 · (1 + j) + 0.6 · (1 + j)2w 42 = √ 1][0.8 · (1 − j) + 0.6 · (−1 + j)2= √ 1 ] [1.4 + 1.4j 2= √ 1 ] [0.2 − 0.2j 2≈ 1 + jc) ISI (here) leads to a total number of 16 different points in the signal space:QPSK-Signalraum mit ISI≈ 0.14 − 0.14j1w 22w 21 w 12w 11imag →0.50-0.5-1w 23 w 24w 32 w 31w 13 w 14w 42w 43 w 44w 33 w 34w 41-1 -0.5 0 0.5 1real →d) (Solution of the additional exercise)The average signal power ¯σ d 2 has to be calculated from the average quadratic value ofthe 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:[¯σd 2 = 1∣16 ·4 ·0.2 + 0.2j ∣∣∣2∣ ∣ √ + 8 ·1.4 + 0.2j ∣∣∣22∣ √2+ 4 ·∣= 1 8 · [( 0.2 2 + 0.2 2) + 2 · (1.4 2 + 0.2 2) + ( 1.4 2 + 1.4 2)]= 1 8 · [4· 0.2 2 + 4 · 1.4 2] = 1.0∣ ]1.4 + 1.4j ∣∣∣2√2Thus the signal is neither extenuated nor amplified by the channel (“neutral due topower”).

13 WS 2009/2010 “Communications technology II” – SolutionsSolution to exercise 10 (vit15):a) (from p.556) upper path: d(i) = 0 / lower path: d(i) = 1true data sequence (bold): d(i) = 0 1 1 0 0b) estimated sequence (dashed): ˆd(i) = 1 0 1 0 00001101101010101error vector:e = [∆d(i 0 ), . . . , ∆d(i 0 + L f − l − 1] TL f = 4; l = 2; → L f−l−1 = 4 − 2 − 1 = 1An error vector of length 2 results:⇒e = [∆d(i 0 ), ∆d(i 0 + 1)] Te = [−1, 1] Tc) The structure of the AK-matrix for a channel of order 2 is given on p.577. To find thismatrix we have to determine the energy-ACF of the error vector first:r ee (0) = ∑ νr ee (1) = ∑ νe(ν)e(ν + 0) = 2e(ν)e(ν + 1) = −1r ee (λ) = ∑ ν⎡⇒ R E ee =⎢⎣e(ν)e(ν + λ) = 0 f”ur λ ≥ 22 −1 0−1 2 −10 −1 2⎤⎥⎦d) Eigenvalue-equation, cf. Eq.(14.5.48a):det(R E ee − λI) = 0Eq.(14.5.59a): (r ee (0) − λ) 2 − 2|r ee (1)| 2 = 0⇒ λ min = 2 − √ 2S/N-loss: γmin 2 = 2 − √ 2 ˆ= 2.3 dB

“Communications technology II” – Solutions WS 2009/2010 143 MobileRadio ChannelSolution to exercise 11 (2007-03-mobrad):a)f D = v c 0· f 0 · cos (α)f D1 = 185.2Hz f D2 = 0Hz f D3 = −151.7Hzb)10.90.80.7Amplitude0.60.50.40.30.20.10−200 −150 −100 −50 0 50 100 150 200f [Hz]c)Amplitude10.90.80.70.60.50.40.30.20.10t_0 t_1 t_2Delayd)h K (t) =n∑ρ ν · δ (t − τ ν ) = 1 + r 1 · δ (t − τ 1 ) + r 2 · δ (t − τ 2 )ν=0n∑H K (jω) = ρ ν · exp (−jωτ ν ) = 1 + r 1 · exp (−jωτ 1 ) + r 2 · exp (−jωτ 2 )ν=0e) H TP (jω) = H K (j(ω + ω 0 )), |f| < B/2⎧⎨[ ]1 + r1 · e −j(ω+ω 0)τ 1+ r 2 · e −j(ω+ω 0)τ 2für − BH TP (jω) =≤ f ≤ B 2 2⎩0 sonst

15 WS 2009/2010 “Communications technology II” – Solutionsf) Absolute channel transfer function not constant → frequency-selective channel

“Communications technology II” – Solutions WS 2009/2010 16Solution to exercise 12 (2007-10-mobrad):a) BER for BPSK and AWGNP b = 1 2 erfc (√EbN 0)BER for BPSK and AWGN with instantaneous SNR(√ )P b (h) = 1 2 erfc |h| E 2 bN 0Uniform Prob. of occurence: P 1 = P 2 = P 3 = 1/3, E b /N 0 = 7dB = 5¯P b = 1 3 (P b(h 1 ) + P b (h 2 ) + P b (h 3 ))P b (h 1 ) = 1 2 erfc (√P b (h 2 ) = 1 2 erfc (√)|0, 5 · exp(jπ/4)| E 2 b= 1 erfc (1, 12) = 0, 0569N 0 2)|0, 8 · exp(jπ/6)| E 2 b= 1 erfc (1, 789) = 0, 0057N 0 2P b (h 3 ) = 1 (2 erfc |0, 1 + 0, 2j| 2 E )b= 1 erfc (0, 5) = 0, 2398N 0 2¯P b = 0.1008b)¯P b = 0, 6P b (h 1 ) + 0, 3P b (h 2 ) + 0, 1P b (h 3 ) = 0.0598c) Strongest channel coefficient h 2P b,min = P b (h 2 ) = 0, 0057d) data rate R b = 1/T Baud = 1/(50 ns) = 20 Mbit/sTransmitted in 30% of all cases: ¯Rb = 0, 3 · R b = 6 Mbit/s

17 WS 2009/2010 “Communications technology II” – SolutionsSolution to exercise 13 (2009-10-mobrad):a)τ 0 = l 0 /c 0⎛ ⎞h(0) = ρ 0 · exp ⎝−j2π f 0 · τ } {{ } 0⎠f 0 · τ 0 = f 0l 0c 0= 2000 → h(0) = 1l 1f 0 ·τ 1 = f 0 = 666.66... → h(1) = ρ 1 · exp (−j2π (666 + 0.¯6)) = 0.5 · [−0.5 + j · 0.866]c 1= −0.25 + j · 0.433b)|H(jω)| = |1 + 0.5 · exp (−j · 2π · 0.¯6) · exp (−j · ω · ∆τ)|∆τ = τ 1 − τ 0 = l 1c 0− l 0c 0=1400 m3 · 10 8 m/s − 600 m3 · 10 8 m/s = 2.¯6µsMaximum, if exponent multiple of 2πψ 1 − ω max ∆τ = n2π ⇒ f max = −2π · 0.¯6 − n2π2π∆τ=−2π · 0.¯62π · 2.¯6µs − n2.¯6µsf max = −250 kHz + n · 375 kHz, n = 0, ±1, ±2, ±3·Minimum, if exponent odd multiple of πψ 1 − ω min ∆τ = nπ ⇒ f min = ψ 1 − nπ2π∆τ=−2π · 0.¯62π · 2.¯6 µs − n2 · 2.¯6µsf min = −250 kHz + n · 187.5 kHz, n = ±1, ±3, . . .c)f D = f 0 · vc 0· cos (α)f D0 = f Dmax = f 0 · 150Km/h = 138.8889 Hz3 · 10 8 m/s( πf D1 = f Dmax · cos = √4)1 · f Dmax = 98.2093 Hz2r(t) = s(t − τ 0 ) · exp (j2π f Dmax · t) + s(t − τ 1 ) · exp (j2π 0.707 · f Dmax · t)

“Communications technology II” – Solutions WS 2009/2010 182.5125 kHz21.5|H(jω)| 210.5−62.5 kHz0−200 −150 −100 −50 0 50 100 150 200f in kHz

19 WS 2009/2010 “Communications technology II” – Solutions4 OFDMSolution to exercise 14 (ofdm03):a) Kern symbol length: T s = 1∆f = 4 msBandwidth: B = N · ∆f = 512 kHzData rate: R = log 2 (M)·NT s+T g= N T = 20486msb) FFT-Length: N f = 4096= 341 kBit/sN f samples account for the interval with the length of T s since exactly 1 FFT is usedfor the generation of the kern symbol.Sample rate: f A = N fT s= 1024 kHzSamples within guard interval: N g = N f ·T sT g= 2048c)Sender’s power:N 0 = 2 · N02 = 1.2 · 10−4 WsE OFDME b =log 2 (M) · N = 1.4Ws2048 = 6.836 · 10−4 WsE bN 0= 6.836 · 10−4 Ws= 5.7 ≈ 7.55 dB1.2 · 10 −4 Ws(γg 2 = 1 − T )g= 2 T g + T s 3)P b= 1 2 erfc (√EbN 0· γ 2 g= 1 erfc (1.9488)2= 3 · 10 −3P = E OFDMT s + T g= 1.4Ws6ms = 233.3W

“Communications technology II” – Solutions WS 2009/2010 20Solution to exercise 15 (ofdm04):a)a) T = 1 · 1 = Nc 1= 16 · 1= 3, 333 · ∆f u B u 6000 0,8 10−6 sR = N c · ldM 3= 16 · = 14, 4Mbit/sT 3,33·10 −6b) T g = T − T S = Nc( 1 16− 1) = ( 1 − 1) = 0, 66 µsB u 6000 0,8c) N c = R · T 1 = 13, 5 · ldM 106 · 3, 333 · 10 −6 1 = 15 3ÈËÖÖÔÐÑÒØ× ½ºd) h½¼º¼¼ ½ ¾ ¿ Å ½¼½½½¾½¿½½¾À´ÅµÀ´¡¾¡¡¡µ

21 WS 2009/2010 “Communications technology II” – SolutionsSolution to exercise 16 (ofdm05):a)b)c)T G = τ c = 0, 8 µs → T G = 20%T = 10020 · 0, 8 µs = 4 µs → T S = T − T G = 4 − 0, 8 µs = 3, 2 µs∆f = 1T S=1= 312, 5kHz3, 2 · 10 −6 µsSN = 1 − T GT = 1 − 0, 8 = 1 − 0, 2 = 0, 8 = −0.096 dB ≈ −1 dB4B∆f=20 · 106312, 5 · 10 3 = 64d)ld(M) = R · TN= 32 · 106 · 4 · 10 −664= 2 → QPSK

“Communications technology II” – Solutions WS 2009/2010 22Solution to exercise 17 (2007-10-6):a) Advantages and disadvantages to be found in the following table...AdvantageSimple EqualizationFrequency domain adaptationRobust against multipath fadingDisadvantageincreased PAPRGuard lossb)∆f ≈ B N = 15kHz1 1β = =1 + T GTS1 + 16.67µs66.67µs

23 WS 2009/2010 “Communications technology II” – SolutionsSolution to exercise 18 (2009-10-ofdm):a)b)γ 2 =T ( ) ( ) ( )s 1 1 1 1⇒ T g =T s + T g γ − 1 T 2 s =γ − 1 2 ∆f = 0, 794 − 1 110 kHzT g = 25, 9 µsR b = N · log 2(M)⇒ N = ⌈ R b · (T s + T g )⌉ = ⌈629.5⌉ = 630T s + T g log 2 (M)c) N fft = 1024, f a = N · ∆f = 1024 · 10 kHz = 10, 24 MHzd) Maximal date rate if all subcarriers are modulated:R b = N · log 2(M)T s + T g= 1024 · 2125.9 µs= 16, 27 MBit/se)∆n Pi< 1 ⇒ ∆n Pi < T s= 100 = 3, 86T s τ max T g 25, 9