Digital Communication Systems: Handwritten Exercises with Solution, Exercises of Digital Communication Systems

Download Digital Communication Systems: Handwritten Exercises with Solution and more Exercises Digital Communication Systems in PDF only on Docsity! hype 7 Gl () C= W hog. (i+ sue) C= 3000 Leg, (1001) 2 30 000 bik fy. (2) 2 = j+sNR SVR= 27-1 = Qtr? = 2.03 = 3,08 dB ~| = $3,45= 17,248 192%C/g,00 | (O swR= 2 1,92 ce W log. Cit SMR) = Be00 fog (") = 3000 * 3.46 = !0, 380 bE fa, “s zt 4 (H) uf toa 4+ ¢ og, e+ the 16 44 pe Oe B® 7, H = Z fe toy- C4: ) = 15-(0.06) bog, (Ki) + 26S (0.00035 lo, ar = 3653+ LIGS = = £ dg tee zy I = 4,196 Ree = 4.148 x 1000 = 4798 bebe /a. 45 (a) He = BR hen OH) 6 (qe gat) = FM fader Ha, x Beminde [tte 4 Rae = bela /etmondt * 32x2«/0° amine fre. = 2.56 xo* Ate fate () Haw = GFxi6 (ime Lag, (6*64)) * 10 bE [aly Rar = Ha * 32x 2810alemedily = 6.44n/0" lols /a 9-2  With 4 Mis dale Pll, moe Gf fir ce Pte foeturtr- bo eednc/ Ant it fe getsumted ha? thes 6 Quaclatle pn he bancluntl buley “a tae f- PK bn ly, Ait rele By 7 ecrtaston fanolartlX (47 Cf, Cypaes GY ineuseed Cefn) for a 0 LEE. 9-6  7.9 fn EIRP ys t & EIRP Gp > a> 2 2+ =f“ N Riel. ? ® ” RELL Eh )ap = ZIRP +6. (R th +L,+L,) all en AR INTELSAT We, tak, = 176 dB By = Ace Dscs 1: pscs Tr: Eum, = 20.648 5 Ky = 0.002 ELg, = 2148; Ry = 1 44 GAPSAT: ELA, = 3.648, Ry = 0,00! Raw ¥ 6 tyrewsay Te 4 SEVERELY BW LiMiTED psc IC : @ 8 UnueD SEE Fe, OD * oar one Oneee 8. D3iar bore O0O7E 0.0834 6, 900g ® Dses aed tL © garsar Pewen mire SEVERELY B.000NS Pow LiMiTED 9-7 =0 mR a, 8 ~ as % — ‘ & e = ~ mo (97) Coen) Goan) - 23S x IO” 9-10 3°” ahle for 712 Barc ti -. Q fe Z GA9 He (bang Wh (5 ry GAY 0b bit, ane 9-11 Pitnemsllid po 9b00 bee 4 bets fogmtel Wilf. the (122 G2) BCH cole the 2400 will be bupandid ayoo «22 ~ 3313 (wheed fuse he 3400 Jeg usta tnd ) The enete provides thy follerng tren efit anes ; fin -(@) (tara) (1-4 az x10"*) re! = 103% ° R= 1-G-& ye? = BI x10 Wheih rl the ape. Y fe < tot Tnbobensing in Aregucree i hanoll. 9-12 For the (127, 64) BCH code, we note that the rate of the code is k/n=64/127~1/2. Hence, repeating the above computation yields approximately the same channel bit-error probability ( p, =1.18x10-?) as before. However, in this case of entering p, on the abscissa of Figure 6.21, we are using the transfer function of the (127, 64) code, which yields a decoded bit-error probability that meets the required P, 10-7. We now consider the final candidate, the (127, 36) BCH code, and Tepeat the same computations shown above. (Remember to use the given transfer-function intercepts to guide you in making graphical estimates.) The computations now yield p,~3.8x10%. When entering this point on the abscissa of Figure 6.21, and examining the transfer function of the (127, 36) BCH code, we find that P, >10-’. Hence, of the three candidate codes, only one, the (127, 64) BCH code meets the requirements. (b) From Figure 6.21 it should be clear that the (127, 36) BCH code is the most capable of the group. Even its label states that t= 15, meaning that within a block of 127 bits, this code can comnect any combination of 15 or fewer errors. A natural initial guess might be to choose the (127, 36) BCH code-—but that would have been incorrect. The reason that the more capable (127, 36) code does not meet our requirements is related to the fact that in a real-time communications system, there are two mechanisms at work: 1) more redundancy makes for more powerful codes and, 2) more redundancy results in Jess energy per channel bit. As the rate of a code decreases (from 1 to 0), there will be an improvement in Ps due to mechanism 1. But eventually, mechanism 2 results in more errors out of the demodulator and outweighs the benefits of mechanism 1, (See Section 9.7.7.2.) (c) Using Equations (9.25) and (9.27) we compute the uncoded received E,/No corresponding to p,=10-7, as follows: 9-15  2h, a (H 20) Ny wn( 5) log, M log, Af 1.5x10°7 ala} 3827) 5 |: Q(x) where Q(x) = seo 2/2) E, Ey [E) (1 By trial and error: x = 5. 13, Thus, = N, =89.85 and W Fe #| (3 0 a E. and thus, E: W we, =29.2=14.8 dB. Therefore, the coding gain from Equation (9.32) is: G(dB) = 14.8 - 10 = 4.8 dB. 9.14 4ad A § T, =T,+(LF-1)290 = 290+(20~-1)290 = 5800K (or37.6dB-K). Since (Z,/N,), = Mx(£,/No) for (E,/No ) with the parameters given and the basic link margin ___EIRP G/T (ETN naga R kLs Ly = 5.2 dB (or 3.31). Since the channel is bandlimited, we choose MPSK modulation. To meet the bandwidth requirement of 3000 Hz, i and at the same time conserve power, we choose the smallest M/-ary i value for MPSK, which is 16-PSK. This modulation (with perfect filtering) will require a symbol rate (and transmission bandwidth of i Rilog, M = 9600/4 = 2400 Hz, We also need to select a BCH code | Space loss is Z, = a i =3.94x10" (or 156 dB). System temperature rgd? aNd margin is 0 dB (or 1), we solve equation M = which yields the value of (£,/N,) 9-16 |  such that the 2400 Hz modulation bandwidth is not expanded beyond 3000 Hz. Hence, this places a restriction on the code redundancy n/k, and for a (127, k) code, the smallest value allowable for kis dictated by the fact that 127/A must not exceed 3000/2400. Therefore, & must be larger than 102, and our choice from Table 9.2 to provide the most redundancy (and still meet our bandwidth constraint) is the (127, 106) triple error correcting code. The chosen 16-PSK modulation dictates that the available (&, / Ny), will yield an £,/Nq=(log, M)(k/n)(E,/Nq)=4x(106/127)»3,31=11.05. We use this value for finding symbol-error probability, P;, in the following relationship: P,(0)=20| ae ea where Q( ) is the complementary error function. With the computed value of £,/N,=11.05, we find that Py = 0.359. Since Gray coding is called for, the probability of channel bit error is p, =a 0.09. 2 Now, we can calculate the decoded bit-error probability, Pp, using the approximation of Equation (6,46). Note, that if p, is small (or E,/N, reasonably large) then the first two terms in the summation of Equation (6.46) are usually adequate because of rapid convergence. However, in this problem, convergence occurs after about 25 terms, yielding a decoded Pz = 0.09 (There is no coding gain!) If only the first two terms in the sunmmation of Equation (6.46) had been used, the result would give the erroneous appearance of acceptable error performance. 9-17 2 NO) a ty, im Cn") aM~ = Avtrage porte for MK EO ee (e) for « fod & ama As Enns, oe SA ee Oe tg fronelin of MT Hyena the cane Y GAA] avenge frmeay wees te A ntar fumelion R _ 28.8 kbits/s : W 3439 Ha 8-4 bits/s/Hz S) ER C E, R| (b) C=Whog (1+) =Wt0g see © Liog [r | ) 2 N ON 0 | Ltt W NW E, C) w(20" 1) g Ha: CL ne abe \ =) LetR=C: Then, 2 iff | and a xt 10 342920738294 ae =l0 By trial and error, C ~ 20,300 bits/s, weer) soo —1}=40.1 (or = 16 4B) a 3 N, Cc 28, tynatittelerrn zh 4 ton -234] Arey Argmel powen for the Halt apoare eo. part & her anplilede OS ngs” ft b har ancplide fasts se peer 2.57 pre © a> Prewege Pree» Boss Lae) $ (ue) -F) SD The _ conetllalicn Nepures ar, ancl Tun orbhorpnel . tlh 2 Am pltictin Oud [sar Vb (PRee, BL Mienerbita bmp llirtes wth lawn tg gn 0 paacec. the, 9-21  9.20 For this problem, the circuit, trellis, and Ungerboeck partitioning diagrams are seen in Figures 9.29, 9.30, and 9.32, respectively. The error-event path with the minimum distance is seen in Figure 9.24 as the darkened path Jabeled with waveform numbers 2, 1, 2. The minimum distance-squared ay =36as is shown in Equation (9.61). Note that in this example, the parallel paths do not characterize the etror-event with the minimum distance. Using Equation (9.63), the average power for the signal waveforms is S,,=21. In this example, the reference waveform set was chosen such that @?.. =4, and the 1? +167 5 =128.5. average power for this reference set is Si, = Coding gain is computed using Equation (9.62) as follows: d2/S, ( 36/21 \ G(AB) = 10 log, oe | = 10 logy oes jr i74ap ref ¢ a \ . Of course this answer seems to violate the Shannon prediction of coding gain. Would anyone use such a paradoxical reference set as the one given here? Absolutely not. However, sometimes the choice of a reference set involves judgement. Generally, any reasonable choice for a reference set yields similar coding gains. But, this problem purposely starts with a very unreasonable reference choice to emphasize that the resulting 17.4 dB coding gain teflects two different nechanisms: 1) the improvement due to coding, and 2) the improvement due to the better signal-waveform set (compared. to the sub-optimum reference set). Because trellis-coded modulation involves coding in conjunction with modulation, the “so called” coding gain can be made to appear arbitrarily large by simply starting with a reference set that is sufficiently poor. 9-25 9.21 We draw the trellis diagram, and we assign waveforms to trellis branches according to the Ungerboeck assignment rules. We then label each state according to the waveforms assigned to the branches that emanate (top to bottom) from that state. For example, the first state and its branches are labeled as follows: STATE D 0426 4 2 6 After drawing and labeling three sections of the cight-state machine, a methodical search for the error-event path with the minimum distance yields the darkened path shown below. We find d ; using Figure 9.22 as dP =d? +d? +d? =2+0.585+2=4.585 i For the 4-PSK reference set, we find d?,,=2 from Figure 9.23. Te 9-26 1e5 1537 4062 5173 2604 3715 F351 s a The asymptotic coding gain relative to a 4-PSK reference set is found. using Equation (9.62) as follows: ft 21S, G(4B)=10 log,, (Z a | a? Sa, ref8¥ The average power is unity for both the signal waveform set and the reference waveform set. Therefore, G(dB)=10 lob | =3.6 dB. 9-27