Errata Introduction to Wireless Systems P. Mohana Shankar. Page numbers are shown in blue Corrections are shown in red.

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1 Errata Introduction to Wireless Systems P. Mohana Shankar Page numbers are shown in blue Corrections are shown in red February 25

2 Page 11 below Figure 2.4 The power detected by a typical receiver is shown in Figure 2.5. Page 13 ( ) = ( ) ( ) 2 d d ( d ref P d P d, >, 2.4) r r ref d ref Page 14 Example 2.1; Answer The wavelength at 9 MHz is /9.1 8 = (1/3)m. Expressing the frequency in MHz, eqn. (2.6) can now be expressed as ( ) ( ) L = log f + 2 log d, (2.7) free 1 1 where d must be larger than 1 km. The unit of f is in MHz and d is in kilometers. Page 17 top of the page to account for the terrain. Additional correction factors can be included to account for other factors such as street orientation. These correction factors They take into account the following:

3 Page 15 Figure 2.7 Transmitted power = 1 mw instead of 1 dbm Page 19 Example 2. 3 Find approximate values of the loss parameter, ν, using Hata model for the four geographical regions, namely large city, small-medium city, suburb, and rural area. Answer: Using Figure 2.12, the loss values at a distance of 3 km are db, db, db, and 12.8 db, respectively, for large city, small-medium city, suburb, and rural area. Page 27 Page 28 Page 34 Figure 2.18 X axis label, Envelope a or Power p Sentence below eqn. (2.4) where p is the average power Eqn. (2.4) p out p thr = 1 exp p Frequency-dispersive Behavior of the Channel Page Frequency dispersion vs Time dispersion We have seen that fading can be in the frequency domain or in the time domain. It is possible to treat the fading in wireless communications systems as constituted by independent effects. The channel shows time dispersive behavior when multipath phenomena are present. At the same time, independent of this effect, the channel will also exhibit frequency dispersion if the mobile unit is moving.

4 Page 39 Figure 2.31 Signal bandwidth Time dispersive/ frequency selective Slow and Frequency Selective Fast and Frequency Selective Bc Time selective/ frequency dispersive Slow and Flat Fast and Flat Tc Bit duration

5 Page 4

6 Page 41 Figure K = - db.5.4 K=6 db K=14 db K=18 db f A (a) envelope a

7 Page 41 First sentence on top of the page The Rician probability density function is shown in Figure 2.33 for different values of the parameter K. Page 42 Eqn ( ) f p db 1 = 2πσ 2 db ( p p ) 2 db av exp 2 2σ db Page Summary of Fading The various fading mechanisms and the attenuation described can be summarized in a diagram shown in Fig Note that Rician and Rayleigh arise out of multipath effects and Nakagami can represent them both. This is not shown in the figure. For most of the cases, analyses based on Rayleigh or Rician fading are sufficient to understand the nature of the mobile channel. A number of recent publications (Alhu 1985, Anna 1998, 1999) have suggested the use of Nakagami fading models to provide a generalized view of fading in wireless systems. Page 53 Exercise 2.1 Using MATLAB, generate plots similar to the ones shown in Figure 2.7 to demonstrate the path loss as a function of the loss parameter for distances ranging from 2 Km to 4 Km. Calculate the excess loss (for values of ν >2.) in db.

8 Page 54 Exercise Generate a plot similar to the lognormal fading shown in Figure Page 55 Exercise 2.17 Compare the maximum data transmission capabilities of the two channels characterized by the impulse responses shown below. dbm dbm ns (delay) 4 8 microsec (delay) (a) (b) FIGURE P2.17

9 Page 75 Figure 3.18 M(f)I 2 P(f) f f -R f +R frequency R 2R frequency

10 Page 79 (below eqn. 3.45) The output noise, n(t), has a spectral density, G out (f), given by (See Appendix A.6) Page 8 Eqn. (3.51) Use M in in place of M out ( ) = ( ) ( π ) H f KM f exp j2 ft (3.51) in Page 87 The probability of error can be expressed (Taub 1986) as (Exercise 3.21) 1 E ( ) = erfc N p e 2 2 (3.79) Page 9 1 E ( ) = ( N ) pbpsk e 2 erfc (3.83) The plot of error probability for different values of E/N (signal-to-noise ratio) is shown in Figure Comparing eqn. (3.83) with the bit error rate for ASK systems (eqn. (3.79)), it is seen that the performance of a coherent BPSK is 3dB better than that of coherent ASK. Page 93 Table 3.2 DPSK encoding: d k = b k dk 1

11 Page 94 In Figure 3.32, use x k and y k in place of X k and Y k. Page 98 Data φ n 1 1 π/ π/ π/4 (or 5π/4) 1-1 π/4 (or 7π/4) Table 3.4 Phase encoding in QPSK

12 Page 1 Figure 3.35 In Figure 3.35a, the line at 2 should be broken as shown QPSK (a) amplitude QPSK (shifted by π /4) time (units of T) (b)

13 Page 13 Figure 3.41 In the Figure the line at 1 should be broken as shown. amplitude time(units of T)

14 Page 14 Figure 3.43 {-1 1}

15 Page 116 Figure 3.57 cos[2π(f + f)t FSK T dt + _ Σ z decision 1 if z > (threshold = ) if z < T dt cos[2π(f - f)t Page 124 Detection and Reception of MSK and GMSK Since MSK can be generated starting from OQPSK, the bit error performance of MSK will be identical to that of BPSK, QPSK or OQPSK. However, MSK and GMSK can also be detected using a 1-bit differential detector, 2-bit differential detector or a frequency discriminator. The block diagrams of these receiver structures are shown respectively in Figure 3.69 a, 3.69 b, and 3.69 c. Page 125 Ts = [ ] log M T (3.139) 2 As the number of levels of modulation increases, the minimum power required to maintain a fixed bit error rate decreases increases. This can be seen from the decision boundaries shown in Figure 3.71.

16 Page 129 ( E R N )( B ) C = Blog bits/ s (3.145) = log 1 + ( )( ) N B R E R B 2 (3.146)

17 Page 141 Figure 4.5 Frequency reuse pattern of cells. The alphabets (except D) represent the different channels. D is the distance between cells having the same frequency. D = D 3R= R= 21R R Page 143 Because of this, q =D/R= 3N c is also known as the frequency reuse factor Page 143 TABLE 4.2 i j N c q S/I(dB) Page 143 Page 144 Figure 4.6 Expanded view of the cell structure showing a seven cell reuse pattern. H is the channel reused. Figure 4.7 A three-cell pattern (N c =3) showing six interfering cells D q= = 3Nc = 3 R

18 Page 169 Page 175 p ( γ ) 1 γ = exp, γ (5.5) γ γ Effects of Frequency Selective Fading, Co-Channel Interference As discussed in Chapter 2, frequency selective fading arises when the coherent bandwidth of the channel is less than the message bandwidth. The error rates vary with the form of modulation and demodulation used. The performances of the modems also depend on the ratio of (σ d /T), where σ d is the r.m.s delay spread (eqn. 2.42) and T is the symbol period. The exact equations governing the error probability, taking frequency selective fading into account, are once again very complex, and are beyond the scope of this book. Numerical results are available in a number of research papers (Fung 1986, Guo 199, Liu 1991a,b). Page 176 Figure 5.5 The legends should read db instead of db. Page 189 f M 1 2 M ( γ ) = ( M ) M 1 M γ (5.34) 2 1! γ M The performances of the three signal processing schemes are shown in Figure 5.17, with γ av equal to the signal-to-noise ratio after combining. They are also given in tabular form in Table 5.1. Page 195 ( ) ( ) exp jθ ( ) Hc f = A f f Eqn. 5.47)

19 Page 213 Figure 6.8 two bits a T chip sequence b K T c T c modulo 2 addition of chip and data sequences c

20 Page 214 First Paragraph The PN sequence is unique to each user and is almost orthogonal to the sequences of other users. Thus, the interference from other users in the same band will be much less. The number of orthogonal codes, however, is limited. As the number of users increase, the codes become less and less orthogonal more and more correlated (orthogonality is compromised) and the interference will increase. Page 22 Example 6.1 In a DS-CDMA cell, there are 24 equal power channels that share a common frequency band. The signal is being transmitted on a BPSK format. The data rate is 96 bps. A coherent receiver is used for recovering the data. Assuming the receiver noise to be negligible, calculate the chip rate to maintain a bit error rate of 1e-3. Answer: Assuming that there is no thermal noise, the bit error rate is given by eqn. (6.15) where K is the processing gain and k is the number of channels. BER == 1 3 =.5erfc( z ) where K z = Using the MATLAB function erfinv (.), we can solve for z = erfinv( 1 2* BER) k 1 chip rate We get z = We are given k=24; K=23*4.77= Since K = chip rate = 19.82*96=1.5 Mchip/s. data rate [ ] 2

21 Page 225 Figure 6.21 Block diagram of the transmitter associated with an AMPS system Change to Page 231 Eqns and 6.22 The radio capacity of the system, m, is defined as m = ( t ) ( ) * number of cells( N ) total allocated spectrum B channel bandwidth B c c radio channels/cell. (6.21) Making use of the relationship between q and N c (Chapter 4), q = 3N c, (6.22)

22 Page rd line from the top Let us start by considering a single cell with N (= k of eqn (6.11)) users who share the cell. Page 233 E = R = Bc 1 N (6.27) 1 R N S ( N S ) ( 1) B c where R is the information bit rate and B c is the RF bandwidth. We have divided the signal power (S) in the numerator by the bandwidth (data rate R) of the message data, while the signal power in the numerator has been divided by the bandwidth (B c ) occupied by the interfering signal. (See Section 3.2.7) The quantity (B c /R) is the processing gain K of the CDMA processing, defined in connection with Figure 6.7. Page 233 η = BN c (6.29) Page 234 N ( E ) ( BN ) N = K Nmax + S min ( E ) ( E ) N N s ( E ) ( E ) 1 c min max = 1 + K (6.33) N N min s 1

23 Page 25 g() t = an exp( jn2 π ft) ( A.1.7) n= Page 263 Normal or Gaussian distribution f ( x) 2 x ( x µ ) 2 1 = exp 2, x 2πσ 2σ x Page 266 where Γ is the gamma function w z w e dz z 1 ( ) Γ =

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