Solutions Communications Technology II WS 2011/2012

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“Communications Technology II” – Solutions WS 2011/2012 IIConventionsandNomenclature• All references to passages in the text (chapter- and page numbers) refer to the book:K.-D. Kammeyer: “Nachrichtenübertragung”, 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 references 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 highpass filter 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 inter-symbol interferenceBW, BB bandwidth, baseband KKF cross-correlation function, sequenceBP bandpass AF audio frequencyDPCM differential PCM PCM pulse-code modulationF{·} Fourier transform PR partial responseH{·} Hilbert transform S/N = SNR signal-to-noise ratioHP highpass LP lowpassAvailabilityonInternetPDF-files of the exercises can be downloaded from:http://www.ant.uni-bremen.de
1 WS 2011/2012 “Communications Technology II” – Solutions1 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= 10dB = 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 2 erfc(2.23)= 8·10 −4
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