chap_3-vectorist

# chap_3-vectorist - 3-1.— “1.53;” ma“ I-S'Tg I LIT...

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Unformatted text preview: 3-1 ..—._._..-.-.. - “1.53;” ma“; --------------- I-S'Tg, I LIT; L‘s?” {{3ﬂta)#* 2/ (60? Wait +m‘7rftf)&f 1 4.17; :0 "- M (at) 30=Trf M @= 45% #4“ =§£=o fAC'DAm‘d if (34%(31—9 en'ffabﬁ + if‘M-aﬁﬂ? .mﬁfatlr *0 gm (a) £=£ aim—IL (by: (ATP-"7L [My ”mg: a = 0 d. 1310/ 1% 4+3 mew/AWN! ﬂax—055+ 7[y[ﬂ)o£z‘/[A)pg)dz‘ i» rJ: +[-’*‘ﬂ.+ [raw—f [111469 4’43)#:F/ZA)/—a)aa 7g%ﬂ)# I 2. ﬂ ., 1. [email protected])/'ﬂ)‘d +_/ f’UK—‘vdf = Ail—v9 -/-}+A =0 f 7/2 (a 9466') ad =/ famamﬁmmm = 2A“, ”2&1, -,_-. (0/ Waoas 2,44% [M]: I 2AM+2A1= 4H1- 7; {.1 W) Z/HI=/ r: W MW 924,66) (7’); (.5) ,l/a-'"' t _,_ 1 '1. t “kip—7. la 17- F;— 3’3 0 do ____= _ eéﬁﬂqez'ﬁ)ﬂ_ +fE—t1'ﬂeﬁdf =0 ° 1': ‘° ._ “Met-1493)“ 70.51248 3W6 =0 -40 ‘L' o 9 _ "3&9” et'AEB + [F8 t-l- £3 2 D T H50 3 a l”f_—[—l+ﬁ]= Z’iﬁco 3 3 3 “246.— =2 3 A =5 : QO) MW 8;) WWW? IE: a/s—g'z 72M K=74 s. 104.3M/d,_ 3-5 ¢£H|ﬂﬂll W Fidg, (2,5,) M -' H. a. +4 “9502 ' ”KT—"”543” 0:1 H1 9' PCS.) H. 0-...1— > a {3(a) __ @- f1 qt — qt. £3: PC-S'IB '— KO Y = 0' o 3 0 1T £33? (’5'? = - 0.0171 W T 3—6 5.. Eva-LE“ fag/J/‘nggz'ng/ﬁ MEN #4 £52 =125‘Mfﬁy (5) NEW (276) ~—.-. JEOJFVDRS [,37smtbf OH“) ”rm”? 1/“ =1 0.1 3? mm EM (270 L; __L.;_. ’ :mm V‘ 7. 1400 —-{(900 _ of. (600 a ’ EL VW ,4 6,; dz W ‘3?” Wm” #5.. KW? 60 59000 W/xb (*1) 64M W ) W (z 7; Lam-9.2L (1+?) K5 ) Wu P = gamW’ = ;(l+l> mac . 3-9 3—10 3-11 3.13 Signaling with R2 pulses represents an example of orthogonal signaling. Therefore, for coherent detection, we can use Equation (3.71) as E _ AZT a—lel 10—3 =Q[.l°£\}OT]=Q(x) Using Table B.l to ﬁnd x, yields x=3.l. Thus, 0.01T 2><10‘8 23.1, T=19.2,us, and R=52,083bits/S 3.14 Signaling with NRZ pulses represents an example of antipodal signaling. Therefore, for coherent detection, we can use Equation (3.70) as PB=Q _ 2A2 . 1°3=Qllﬂli§eﬁml=wl Using Table B.l to ﬁnd x, yields x=3l Thus, FA (11”0 5? 00.0). = 3-1, A? = 0.268. Thus, if there were no signal power loss, the minimum power needed would be approximately 260 mW. With a 3-dB loss, 538 mW are needed. 3-12 3.1 5 The power spectral density for a random bipolar (antipodal) sequence in Equation (1.38) is expressed in the form of , 2 TS[51:}§TS ] , where TS is the symbol duration. The total area 3 under the Spectral plot is found by integrating as follows: 2 2 GO ‘ T on ' “4ng df = 222 i [—y—f—l a Letx = 7973, then df =cit/7tTS , and the area is: 2 2T5 J‘m[SiDX] dx : EE : 0 TITS x it The two-sided Nyquist minimum bandwidth extends from 1 l 1 Thus, the one-sided (baseband) bandwidth is 2LT— = 555—. S The sketch below illustrates the rectangular construction having the same area as the signal power spectral density. The width (bandwidth) of this rectangle is RS (two-sided) and Rﬂ (one— sided), which is the same as the Nyquist minimum bandwidth for ideal—shaped bipolar pulses. 3.16 The output of an MF is a time series, such as seen in Figure 3.7b (e.g., a succession of increasing positive-and negative c0rrelations to an input sine wave). Such an MP output sequence can be equated to several correlators operating at different starting points of the input time series. Unlike an MF, a correlator only computes an output Once per Symbol time. A bank of N t 6 correlators is shown in Figure l, where the reference signal for the ﬁrst one is 31(3) , and the reference for each of the others is a symbol-time-offset copy, 510‘ * kT ) of the ﬁrst reference. It is convenient to refer to the reference signals as templates. Since the correlator emulates a matched ﬁlter, the "matching" is often provided by choosing each of the S, (I) templates to be a square-root Nyquist shaped pulse, and thus the overall system transfer function being the product of two root- raised cosine functions, is a raised cosine function. Figure 2 is a pictorial of the 6 shaped-pulse templates, each one occupying 6 symbol times, and each one offset from its staggered neighbor (above and below) by exactly one symbol time. Each of the template signals will be orthogonal to one another, provided that the time offset is chosen to be an integer number of symbols. Each correlator performs product-integration of the received pulse sequence, r(z‘), by using its respective template. The time- shifted templates account for the staggered time over which each cerrelator operates. That is, the ﬁrst correlator processes the r0) waveform over the time intervals 0 to 6, then 6 to 12, and so forth. The second correlator operates over the intervals 1 to 7, then 7 to 13, and so forth. The sixth correlator operates over the intervals 5 to 11, then 11 to 17, and so forth. In Figure 1, following the bank of correlators is a commutating switch connecting the correlator outputs to a sampling switch. Startup consists of loading the correlators with 6-symbol durations of the received waveform, after which the commutating switch simply "sits" on the Output of each correlator for one symbol duration before moving on to the next correlator. Even though a correlator only produces an output 3-14 at the end of a symbol time, the commutating switch acts to form a time-series from the outputs of the staggered correlators. The output of the commutating switch is a discrete approximation of the demodulated raised-cosine (smeared) analog waveform seen in Figure 3.23b. This output is now ready for sampling and detection. The comrmtating switch itself can be implemented to act as the sampling switch. Recall that the beneﬁcial attribute of a matched ﬁlter or correlator is that it gathers the signal energy that is matched to its template, yielding some peak amplitude at the end of a symbol time. Each correlator, operating On the "smeared" signal, gathers the energy that matches its template over 6-symbol times, and when sampled at the appropriate time, produces an output ready to be detected. REFERENCE 5, (r) SIGNAL CDMMUTES ONE POSITION EACH SYMBOL TIME SAMPLE AND : '3” a . . or : s, (t -s:r) : at 117 O ¢ J- ‘ 53" Figure 1 3.16 (cont’d) Figure 2 16 3 3.16 (cont’d) For this example, Figure 3 shows the signal into the staggered correlators. We see 6 successive Views of the smeared signal appearing as “snapshots” through a sliding window (6-symbol times in duration). 2r o _2 .. 21. 2 4 £3 10 12 0 g.— ‘29 2 45—— s s to .112 2 0 E j -2 l ____4 g l [ # D l I Figure 3. Time intervals processed by successive correlators 3-17 3.16 (cont’d) For this example, Figure 4 shows the output of each successive correlator. We see 6 successive results from each windowed signal in Figure 3 that has here been product-integrated with each of the staggered templates..Note that the signal values at the successive sampling times 6, 7, , 11 correspond to the PAM signal values that had been sent. Figure 4. Outputs of successive correlators 3-18 3.17 The overall (channel and system) impulse response is Mr) 2 5(t)+oc8(t—T). We need a compensating (equalizing) ﬁlter with impulse response 60‘) that forces h(t)*c(t)= 60‘) and zero everywhere else (zero-forcing ﬁlter). The impulse response of the equalizing ﬁlter can have the following form: C(t) = 005(1) + 0,5(1— T) + (225(15— 21‘) + c350 —3T) + . -- where {ck} are the weights or ﬁlter values at times k = 0, l, 2, 3, After equalizing, the system output is obtained by cenvolving the overall impulse response with the ﬁlter impulse response, as follows: I10) * C(r) : 605(3) + 6,50: - T) + 6250? — 2T) + 6350‘ — 31") + - .. +OLCOSO‘~T)+(1615(t+—2T)+a625(t-3T)+--- We solve for the {ck} weights recursively, forcing the output to be equal to 1 at time t= 0, and to be 0 elsewhere. At t = Contribution to on ut r 0 C“ - T [ Cl+0tCO Therefore, the ﬁlter impulse response is: c(r) = 5(2‘) -—o.5(t — T) + 0850— 27‘) —-a’5(t — 3T) And the output is: r(r)=h(t)*c(r)=1x5(t)+0x6(tLZT)+0xB(t-3T)-0t“ x5(t—4T) =5(t)—a‘5(t-4T) The ﬁlter can be designed as a tapped delay line. The longer it is (more taps), the more 181 terms can be forced to zero. If or = 1/2, then the 4-tap ﬁlter described above has an impulse response represented by a 1 lus three Os, and the resulting 181 has a magnitude of (1/2) = 1/256. Further ISI suppression can be accomplished with a longer ﬁlter. 0 9‘1. 7‘, IX. 49, MM“ ' Mu W 44 wziEW (3.871,) . 115:4 71% is a": warm W' e“ 0.91:1. 0.1.371, -o.7.|07 o C. = -O:3071 9. 9'3»? 0, \$513 i 0.1.101 -o. 3019‘ 0,31“ 0 3-20 ’11! WW {50.)} W M M ,f‘ {4% [may £5!an 3-253, Fan/ﬁe {300} = 0-14'5) M67? 0.0000) ) Lbaog manna! —o, 5’91) 0, “4.3 WM 1 = 0.180"; M “6? ISI 5‘“- “TJFSI .. 0.42?! 3-21 3.19 Channel response: [0.01 0.02 -0.03 0.10 1.00 0.20 -0.10 0.05 0.02] Matrix description of problem 0.01 0 0 0 0 0 0 0 0 0 0.02 0.01 0 0 0 0 0 0 0 CO 0 0.03 0.02 0.01 0 0 0 0 0 0 01 0 0.10 0.03 002 0.01 0 0 0 0 0 02 0 1.00 0.10 0.03 0.02 0.01 0 0 0 0 C3 0 0.20 1.00 0.10 0.03 0.02 0.01 0 0 0 C4 0 0.10 0.20 1.00 0.10 0.03 0.02 0.01 0 0 C5 0 0.05 0.10 0.20 1.00 0.10 0.03 0.02 0.01 0 0'5 0 0.02 0.05 0.10 0.20 1.00 0.10 0.03 0.02 0.01 07 : 1 0 0.02 0.05 0.10 0.20 1.00 0.10 0.03 0.02 ‘33 0 0 0 0.02 0.05 0.10 0.20 1.00 0.10 0.03 0 0 0 0 0.02 0.05 0.10 0.20 1.00 0.10 0 0 0 0 0 .002 0.05 0.10 0.20 1.00 0 0 0 0 0 0 0.02 0.05 0.10 0.20. 0 0 0 0 0 0 0 0.02 0.05 0.10 0 0 0 o 0 0 0 0 0.02 0.05 0 0 0 0 0 0 o 0 0 0.02 0 Equivalent form xc = z , -1 Form of Solutlon c=(x7x) x77. Output of equalized channel: 3011111011 -—0.0000 —0.0003 0.0001 0.0003 -—0.0000 00 0.0033 0.0000 “0.0000 0.0024 0.9953 0.0111 3; 'glgggg -0.0947 0.0202 —0.0061 0.0063 —0.0024 03 __ 0.1232 —0.0011 0.0001 04 — 1.0521 , . . . . 05 0.2225 Pnor to equalization, the maXImum smgle ISI 2? 23223 magnitude was 0.2, and the sum of all the ISI C8 0.0039 magnitude contributions was 0.530. After equalization, the maximum single ISI magnitude is 0.0947 and the sum of all the ISI magnitude contributions is 0.1450. 3-22 3.20 G1 02 Signals at points A, B, C, and D have units of volts (which. characterizes most any signal—processing device). If the transfer function or gain of the multiplier is G1, then its unit are: r(t) x s,(r) >< G1 = volts (point C) Thus, volts x volts x GI = volts Units of G1 = 1/volts If the gain of the integrator is 62, then its units are: volts (point C) integrated over T seconds x G; = volts (point D) Thus, volt-seconds x G; = volts Units of G2 = l/seconds Therefore, the overall gain or transfer function of the product integrator is 1/volt-seconds. We thus can view the overall transformation as an input energy (volt-squared seconds) times a gain factor of 1/volt—seconds yielding volts/volt-squared-seconds (i.e., an output voltage proportional to energy). 3—23 ...
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• Spring '08
• Nagaraj,SV

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