Solution Manual for "Wireless Communications" by A. F. Molisch

Transcription

1 Solution Manual for "Wireless Communications" by A. F. Molisch Peter Almers, Ove Edfors, Fredrik Floren, Anders Johanson, Johan Karedal, Buon Kiong Lau, Andreas F. Molisch, Andre Stranne, Fredrik Tufvesson, Shurjeel Wyne, and Hongyuan Zhang February 5, 006

2 Chapter Diversity. (a) For N r and 0 db we obtain BER 4 : (b) For N r 3 and 0 db we obtain BER approximately 5000 times larger than in (b).. (a) :6 0 7 : The result in (a) is (c) In order to achieve :6 0 7 with a single-antenna system, we would require 4 :6 0 7, so that :6 0 6, which corresponds to 6 db. (b) (c) and for MRC diversity and for MRC diversity The corresponding diversity gains are given by D RSSI,M db (.) D RSSI,M db D MRC,M 0 8 db (.) D MRC,M db D RSSI,M 6 5 db (.3) D RSSI,M3 6 5 db D MRC,M db (.4) D MRC,M db: D RSSI,M db (.5) D RSSI,M3 0 8 db D MRC,M db D RSSI,M db D RSSI,M db (.6) D RSSI,M db D MRC,M db D RSSI,M3 6 5 db It is evident that the diversity gain increases when the speci ed values of BER and outage probability are small.

3 3. Let the average SNR be : (a) (b) M t M t ln ( P out ) : (.7) ln p : (.8) Pout (c) M M 0 db. 6. The scheme in this problem is called the Alamouti scheme (see also Chapter 0). (a) The outputs become r s h + s h + n (.9) r s h s h + n : (b) We compute ^s and ^s ^s h r h r (.0) h (s h + s h + n ) h (s h s h + n ) s jh j + s h h + h n s h h + s jh j h n s jh j + jh j + h n h n s + + h n h n ^s h r + h r h (s h + s h + n ) + h (s h s h + n ) s h h + s jh j + h n + s jh j s h h + h n s + + h n + h n : For the single antenna case, let the transmitted symbol be s, the attenuation h e j ; and the noise n. The SNR then becomes s-a E jhsj E (jnj ) jhj E jsj E s ; (.) N 0 where E s is the signal energy and N 0 is the AWGN variance. For the two transmit antenna case, the SNR becomes t-a E js + j E (jh n h n j ) + E js j ( + ) N + E s : (.) 0 N 0 We see that the scheme can be viewed as an addition of two single antenna SNRs, although a 3 db penalty is paid since the power per antenna must be halved. N 0

4 . The SNR for the k th branch is given by the useful energy divided by the AWGN variance as The total AWGN variance after combining is and the total useful energy after combining is k s k N 0 : (.3) XN r N t N 0 k; (.4) k E t XN r k k s k! : (.5) The SNR after combining thus becomes Cauchy s inequality states that E t (.6) N t PNr k ks k P Nr : N 0 k k! nx a k b k k nx k a k! nx k b k! ; (.7) with equality for a k cb k ; where c is an arbitrary constant. The maximal SNR after combining is therefore given by P Nr P k Nr P k k Nr k k P Nr k ; (.8) N 0 k N k 0 which with k s k gives that XN r k : (.9) The maximal ratio combining rule thus corresponds to adding the branch SNRs. k 3

5 Chapter Channel coding Soft Viterbi decoding (a) Replacing ones and zeros with their antipodal signal constellation points in the trellis stage in Figure B-4.5 a) gives:,,,,,,,,,,,,,,,, A B C D (b) Executing the Viterbi algorithm for all seven trellis stages gives: Received: Keeping only the surviving paths gives the following: 4

6 We should, however, notice that in state B in the second to last trelllis stage there were two equal paths and one was eliminated using the toss of a fair coin. Had the coin given the opposite result, we would have obtained (the equally valid): Neither of the alternatives gave (exactly) the same surviving paths as the hard decoding in Figure B Block codes on fading channels Rewriting the expression against which the BER is proportional, so that we get a ( B ) i -factor in each term, gives NX it+ NX it+ K i + B i K i ( B ) i ( B + ) i + B we can see that the lowest degree ( B )-term has the asymptote K t+ t+ B 7.4 N i (.) N B + B + i t+ (.) for large B. Hence, with error-correcting capability t we achieve diversity order t +. Since t dmin, we have proven that a code with minimum distance dmin achieves a diversity order of dmin +. 5

7 Chapter 3 Speech coding. 6

8 Chapter 4 Equalizers If the channel has a transfer function F (z) + 0:5z, the ZF- lter must have a transfer function E(z) F (z) + 0:5z. (a) The memory of the channel is. (b) The fundamental stage is shown in Figure 4.. (c) The trellis representing the equalization process performed by the Viterbi equalizer is shown in Figure 4.. Tracing the surviving path from the back of the trellis, we end up with the sequence f; ; ; ; g which corresponds to the bit sequence f; ; 0; ; 0g. 7

9 Figure 4.: One trellis stage X Figure 4.: Trellis for the equalization performed by the Viterbi equalizer. 8