Baixe Contemporary Communication Systems using Matlab - Proakis and Salehi e outras Notas de estudo em PDF para Engenharia Elétrica, somente na Docsity! RR Ri MAT LAB John G. Proakis Masoud Salehi The PWS BookWare Companion Series 14 CONTEMPORARY COMMUNICATION SYSTEMS | USING MATLAB? | John G. Proakis Masoud Salehi Northeastern University i PWS Publishing Company HP An International Thomson Publishing Company Boston * Albany * Bonn * Cincinnati * London « Madrid + Melbourne * Mexico City * New York Paris » San Francisco « Singapore + Tokyo » Toronto « Washington lou freciy ever highespecd nenwor for ibesc and other goals to be rca necvork estrinistrators. and publishers will neec to communicate fizety and actively asinh each other. We encourage you to parncipute in these exciting developments and become Involver in the BC Secies today, Tf vou save an idea for improving te problem. à demenstration asirs softwara cr multimedia. or an cpporturity to explore, contact us. We thank you eme and al for ym.e continuiag sappert 4 Engincoring Isam Contents BBarerpus com — Acquinitioas Editor ks IG support remote icurnirg activities, In order 2d, educators. students, saliware developers. ness of he BC concept, an exam; SicansGpwscom Assistant Editor Nwilharipwscom — Marketing Manager 1 Signals and Linear Systems 1 LI Preview... 1 ee a cam Produeciom Ei 1.2 Fourier Series a . ne 1 PRockaweilê pus com Production Fiditor 1.2.1 Periudic Sigoals and LTT Systems o TKolivepwscom — Editortal Assistam: H3 Pine: Transforms té 13.1 SamplingTheorem ...... .. 23 1.3.2 Frequency Domain Analysis o! LTI Systems .. o 1.4 Power and Energy o R 1.5 Lownass Equivalent of Bandpass Signals as 2 Random Progosses as 21 Preview AS Generation of Random Variables . . . e as Gaussian and Gauss-Markow Processes .......... st 2.4 Power Spectrum of Random Processes and White Processes s7 2.5 Linear Filtering of Random Processes . . .s 2.6 Lowpass and Bandpass Processes . o 3 Analog Modulation 79 31 Preview RE 79 32 Amplitude Modulation (AM) ” 321 DSB-AM . .. . 30 322 Conventiona AM... close 89 323 S$8-AM a . o 9% 33 Demodulation of AM Signals Cc ON 331 DSB-AM Demodulaton . .. 04 332 SSB-AM Demodulation 106 3.33 Conventional AM Demodulation E 4 Angle Modulation ce 6 4 Analog-to-Digital Conversion Jar 4d Preview... decir 13 42 Measure uf Information 132 vi Lo mu vii 43.2 Pulse-Code Mod atiom 5 Baseband Digitul Transmission Preview 5.2 Binary Signal Transmission sa Optimum Receiver for the AWGN Channel Signal Correinor Matched Filter The Detector Monte Carlo Simulation of a Binary Comtnanicason System . Other Binary Sienal Transmission Meikods Antipodal Signals tor Binary Signal Transmission . On-Olt Signals for Binary Stgnal Transmission... Signal Constellation Diagrams for Binary Signais. . 53 Multiamplitude Sigaal Transmission s21 532 533 sis sas Signal Wavetorms with Four Amplitude Levels Optimum Recesver for the AWGN Channei Signal Correlator The Detector... Signal Wavetorms with Multiple Amplitude Levels . . 5.4 Multidimensional Signals 54.1 542 Multidimensional Orthogona! Signals Brorthogonal Signals é Digital Transmission Through Bandlimited Channels Preview .. The Power Spectrum of a Digital PAM Signal , = Characterization of Banulimited Channels and Channel Distortion Characterization of Intersymbol Interference . Communication System Design for Bandlimited Channels 61 62 6.3 [4 6.5 06 67 651 és.2 653 Linear Equalizers .... 661 Signal Design for Zero [SL ....... Signal Design for Conrolled ISI Precoding for Detection of Partial Response Signals Adaptive Linear Equalicers Nonlinear Fqualizers 7 Digital Transmission via Carrier Modulation Preview 71 72 73 Carrier-Amplizade Modulation . . 724 Carrier-Phase Modulation Demodulaton ot PAM Signals CONTENTS 132 13% 138 146 192 196 200 201 20 22 221 22 22% 2e1 244 2 253 251 265 71 281 281 281 285 287 t i CONTENTS 7.3.1 Phase Demodulation end Detscsion 7.3.2 Differential Phase Moduiatioa and Demogulaticn . Quadrature Amplitude Modularion 74.1 Demedulation and Detection of QAM 74.2. Probability of Error for QAM in ar AWGN Channel Canrier-Frequency Modulation Frequency-Shiit Koying Demodulation and Detecuon of FSK Signals Probabiliny of Brror for Noneatezens Detection of FSK . Inenizaven in Communication Systems Carrier Synchronizanon Clock Synchronization 34 8 Channel Capacity and Coding BL Preview 8.2 Channel Mcdel und Channel Capacity . 82.1 Chanael Capacity 8a Linear Block Codes. . 00... 83.2 Convolutiona! Codes 9 Spread Spectrum Communication Systems 91 Previow . 9.2. Direct-Sequence Spread Specstum Systems 92,1 Signal Demodulation . ........ 922 Probability of Error . 9.23 Two Applications of DS Spread Spectrum Signals 9.3 Generation of PN Sequences . 9.4 Frequency-Hopped Spread Spectrum 9.4.1 Probability of Error (or FH Signals . 942 Use of Signal Diversity to Qvercome Partial- Band Interference . . 291 am 306 308 313 314 316 32 328 326 332 33 363 343 344 355 357 3 Ea! 391 392 395 397 398 a03 409 am am PREFACE There are mamy texcbooks on the imarket today (sat near the basic topics in analog and digital communicatior systems including coding and deced:g algoricams and medulation ané demoduianon iechaques. The socus cf mst of (ese textbuoks is by necessity om the Eeory that underiies Me desen and performance analysis o the various building blocks. e, coders, decoders, modulators, ané demedulaters. that constitute the aasic elements of a communication system. Relarively fem Gt ae teshonks, especially chose iwritien for an- dergeaduars stuécuts, include à namber of applications that serve ta moúvale the students. SCOPE OF THE BOOK “The objecuve oi thus dock is so serve as a eempanior or suaplement, to any cf the com- pesbensive texthonks in communication systems. “The bogk provides a vanety of exercises that may be solved on the computer (generally. à personal vompuses is sufficient) using the popular student edito: of MATLAB. The book is intended to be used primarily dy senir-level undergraduate students and graduate students in electrical ergensering, com puter ergincenng and zomputer science We assume lhat the student (or user) is familiar veth the funidamentals cf MATLAB. Those topigs are not covered, because several tutorial books and manuais on MATLAB are available. By destgn, the treatment cf the various topies is Enef. We provide the motivation ond a short introducuon o each topic, establish the necessary notation. and then illustrate the basic notions hy means of an example. The primary text and the imstructor are expecteé to provide lhe required depth of tre topres treated. Fer example, we introduce the matched filter and the correlator and assert that these devices result im the optimum demodulation cf signals corrupted by asdiave awiute Gaussian noise (AWGN), bat we do not provide à proof ofthis assertion. Such a prowf is generally siven im most textbooks en communication systems. ORGANIZATION OF THE BOOK “The book consists Of nine chapters. The Árst two ctapters on signats and Lincar systems and on random processes treat tha basic background tal is generally cequiteé m che study of communicanon systems. There is one chapter on analog communication techniques, and she remaining five chaptezs are focused on digital communications. Chapter |: Signals and Linear Systems This chapter provides a review of the basic tocis and techniques from linear systems analysis, including both time-domain and frequency -«Juraain chasactecizaticas. Frequency-domain- analysis techniques are emplisized, sines these techniques are most frequently used in the treatment f communication systems x 2 CHAPTER à. SH TALS AND LINEAR SYSTEMS then the output is given by vo= Act pio) dr eo . =a[f mor Sn [ata (123) tm other words, the output is a complex exponential with the same Frequency as the inpuc The (complex) amplitude of the output, however, is the (complex) amplitude of the input amplified by / lee ho gr Note thatthe above quantity isa function ofthe impulse response of the LTI system. h(t), and the frequency of the input signal, fo. Therefore, computing the response of LT systems to exponential inputs is particularly easy. Consequentty, its natutat in incar system analysis 10 look for methods of expancing signals as the sum of complex exponentials. Fourier series and Fourier transforms are tecâniques for expanding signals in terms af complex exponentials, A Fourier series is the orthogonat expansion of periodic signals witit period Ty when the signal set (e/2ru/To JO is employed as the basis for the expansion. With this basis, any periodic signal! x(t) with period Tg can be expressed as sN= L nei TA (12.8) vhere the x's are called the Fourier series coeficients of the signal x(4) and are given by 1 pet, , nel [ elojecitsaito 025 a Here a is an arbitrary constant chosen in such à way that the cormputation of the integral is simplifed. The frequeney fg = 1/75 is catied the fundamental frequency ví the periodic signal, and the frequency f, = n7o às called the nth harmonic. Ja most cases either a = O ore = —Th/2 is à good choice. This type of Fourier series is known as the exponentiu! Fourier series and cane applied to bolhreal-valued and complex-valued signais x (7) as fong as they are períodic. In general, the Fourier series coctficients (xn) are complex numbers even when x(?) is a real-valued signal 1A sutficient condition for the details ee [1] state vF the Fourier series ia that e:+) satusfy the Dirichiet conditivns. Fur 12 Fourier Series Whea xét) is à real veined periodi. sigua, we buve df Dam q =— rare Eta q nl Feam this it is obviors thar “Thus the Fourier series cocificients of a reai-value signal have Hermitian sooemermy: ie. their real part is even and their imuginery part is odd tor. equivalenty. their m even and their phasc is odd) Another form of Fourier series, known as pigememetrio Fourier series, cau ix applied dEly :o real, periadic signals and is obtained by detinng fbu jituce is which, after using Euter's relation em tim os (ame) = ssin mt 612.10) sesults in 2 qem no m= ESA xQ)cos (em) dt 2 quim n bm Tl *(Bsin (15) dr 21 and, therefore, a , . dn = ; -Emeos E) + sin (am) Note that for n = 0 we always have bg = O. so ag = 2x9. By defining +b; (1213) : CHAPTER | SIGNALS AND LINEAR SYSTEMS and using the eeiarion Ts 5º só tising="0 Hitestp -) (1214) q a! Ecustion (52.123 cam be weten in the form ay, É r cur LE + enzos [rr + 6 1.215) ra To pe svbica is the Erg sorm of ihe Founer sentes expansion fer reab and pervodie signais. To generai the Fourice secs coclticieuts rw] for real-valued signals ate related mó, by, Cm, and (4, tbroug -2 mí metal 216) e =ir Pets of [xy] and Lg vecsas avr nf) are called the discrere speciriem ot x(1). The plotof leis usagily calted the ateegrnitide spectrim, art the piotof é xy is referred to as the phase spectrum E egej is real and even af (>) = x(rj—then taking e = —Ty/2, we have E steysin [2 E) de (217) Luas di io which is zezo Sesause the integrand is an odd function of r. Theretore. jor a real and even signal xtr). all aºs aro neal. Tn bis case he trigonometrio Puucier sertes consists of all sosine [unctcns. Simiiarly. if city is real and eddie. if vio) = —x(n)—hen 2 path a al uiipeos (2 =) + (218 is z2rc and all tw' are imaginary. In this case the trigonomei: Huurier series consists vf all sine functions, ENSINA ES Tustrative Problem 1.1 [Fourier series of a rectangular signal train) Let the periodie signal air, with period Ty, be defined by A ch 4 . ; un=an(s > t=&t (1.219) 9. others 13 tomei Seres 5 for ft! = 1572, where ro < To/2. The reciangular si al [ít) :s, as usual, deôned by H=<3 rot (1.220) l nin=44 0. otherwise à plot of x(r) is shawa in Figure LL, Figure 1.1: The signal w(t). m Mustrative Problem 1.1 Assuming 4 = |. To =4 ando = | 1. Determine the Fourier series coetficients of «(t) in exponential and trigonometric form. 2. Plot the discrete spectrum of x17). ED —————— . To derive the Fourier series coefficients in the expansion of x(1), we have ' prints gg [ecra ma] 422) a - 2 1222 5) (222 where sine(s) is defined as Ta) sinctr) = = q2) A plot of the sine iuaction is shown in Figure 5.2 CHAPTER |, SIGNALS AND LINKAR SYSTEMS Pigure 1.2. The sine signal, Obvinusly, all the x,'s are real (since x(t) is real and even), so a = sinc(5) 3! m=0 (1324) c=pe(g) &=02 Note lhat for even nºs. sy = O (With the exception of n = 0, where ap = cg = | and xe = 4). Using lhese coeificients we have & 1 n > ) ane (PY estames do rele e n A at mel) cos (2u15) (1.228) A plot of the Fourier series approximatious to this signal aver ons period fur n = 0,1,3.5,7,9 is shown in Figure 1.3. Note that as 1 increases, the approximation becomes closer ta the original signal x(e) » Note that x, is always real, Therefore. depending on its sign, the phase is either zero or. Tie magnitude of the xs is 3 [sine (2)| . The discrete spectram is shown in Figure Ld. Fourier Series 7 na Figure 1.3: Various Fuurier series approximations for the rectangular pulse in llustrative Problem 1.1 Figuro 1,9: The discrete spectrum of the signal in Hlustrative Problem 11 CHAPIER 1. SIGNALS AND LINEAR SYSTEMS 1.2.1 Períodic Signals and LTI Systems When à periadie segnal tr) is passed througã a boear dime-invaricat (LTD system, as shown m Eipure 7.9. the output sigas! (0) is also periodic, uscally with the same period as the input signal” why), and therefore it has à Foueicr series expansion 1Ê xét) and vit) are expanded es vida Do açenim (1.226) vn = » petit (1227) then ihe relation between tke Fourier series coefficients of x emploving the comvoluion integral 3 and ytr) can be obtained by O sa — esairhar -f DO ane) de (F atue feto de) altas 37 qeitamim (1228) From the above relation we have =ati( o (12.29) dos anti o (UI where Hi f) denotes the transfer function” of the LTI system given as the Fourier cransform vt its impulse response A(t) HO -[ ge TP dr (1.230) Eye say usualty with the same period as che input signal. Can you give an example wbere the period of the ouspur 15 diffecent From the perica of the input? 3 Also imewa as lhe frequeney resperse oF the system, Pintor Series um mos nm um nm om Figure 1.8: Magnitude and phase spectra fur Ulustrative Problem 13. 14 CHAPTER í. SHGNALS AND LINEAR SYSTEMS 4 LTt System st Figure 1.9: Periodic signais thrmgh LTI systems, ANS) Uustrative Problem 1.4 [Filtering af periodic signais] à triangular pulse train (e) wim period To = 2 is defined in one period as t+ -gr<a atn= [== O<tx1 (1230 9. otherwise 1. Determine the Fourier series coefficients of x(º) 2. Plot ihe diserete spectrum of x (4) 3. Assuming that this signal passes through an LTI system whose impulse response is given by 1 Ostel mo = o (1232) O, otherwise plot :he distrete spectrum and the output s(7). Plois of x(1) and A(t) are given in Figure L.IO. Figure 1.10: The input signal and the system impulse response í [2 Fourier Seres 15 ED — 1. Wehave (1.2.34) (12.35) (236) (1237 where we have used the fact Et 4 (7) cenishes cursido the [= 1. 1º isrerval und tha the Fumrier tcansform ot A(t) is siac?: 9. This resul can also be cotained by using the expression for A tr) and Integrating by parts. Obviousty, we have x, = O for all even values of n except fer n = O. 2. A plot of the discrete spectrem aé «(1) :s shown in Figure 1.1]. Tm Figure 1.1: The discrete spectrum af the signal, 3, First we have to derive A(/. tac transfer function of the systerr. Although this can be done analytically, we will adopt a numerical approach, The resulting magnitude 15 CHAPTER |. SIGNALS AND LINCAR SYSTEMS of tie transter funsuon amd also the megusituds of Flini To) — Hín/23 are shown in Figure DI? 1 Tou: jus the diserete spectrum 2: che output we employ the relation The resulting diserote spectrum: of the output is shown in Figure 1.13 The MATLAB seript [or tis problem follows. % MATEAD semper poe Musicativo Prublem 4 Clemire ! echo om r=|-20:1:20], % Emurior corios comgheiemas aj aim vereur “a fsadots; hefzereato.20) 421 BI) zeceçd 2037: & trunsger fenctem Herhihi;6s, jrequeney resuluton Ge Pine enmrcds fistino 1.3 Fourier Transforms The Fourier transforra is she extension of the Founsr series to nonperiodic signals. “Lhe Fuvcier “ranstorm of a signal «(?) that satishies certain conditions. known as Dirichlet's conditions [1], is denoted by X(/) ot equivalentiy, 7 [x(*)) and is defined by Fl = XD =[ acre 413.0 “Tke inverso Founier transform ví X(7) is x(t), given by F xcol==0 =[ XNDerint ar 3 t Í i £3 Fourer Trarméoras ” Frqueneo alema ii [item pm Figure 1.12: The transter fuactioa pf the LTI system and the magnitude of H (5. 2 CHAPTEK 1 SIONALS AND LINEAR SYSTEMS Since the signals are simila: magnitude spectra. Thy the same axas are shown in Pigu mitade spectrum and the two pj LIS and 1.1á, respectively, un R N | | | | i 1 1 | | ! + Freq Figure 1.15: The comenon magnitude spectrum of the signals 1)(7) and 2!) The MATLAB script for thus problem is given below. ln Section 1.3.1 we show how to obtain the Fourier transform of a signal using MATLAB. pt for a cime shift. we would expect them to have the sume % MATLAB acnpt for MMusiranse Problem 5, Choptes 4 dr-007; XMS Therl(at 6 XL id Ts X2a21 dB] =fisenç Kli=k 1/6 X2i=K2yre F=[O Tl ailengria! 1) = 1)]-65/2, ploeit she o PIOUFTS0O-S25: Htstufangle(X HHÇS00-525V I(O0:525) ERS NA angie X2500:52579), =) 4.3. Erurice Transtormas 2 E rsdiams e fem; Figure 1.76: The phase spectra cf the signals x1(t) and a2(1; 1.3.1 Sampling Theorem The sampling iheorem is one of the most important results in signal and system analyais; it forms the basis for the retaçion between continuaus-time signals and disceete-time signals, The sampling theorem says that a bandlimited signal e asignal whose Pourier transform vanishes for 1/1 » W for some W—can be completely described in terms of 1ts sample values taken at intervals 7, as longas 7, < 1/2W. IFhe sampling is done at intervals “= 1/2W. known as the Ayquist interval (or Nuquis: rate), the signal x(1) can be reconstructed from the sample values [e[4] = x(nT7, Ji (as Mn= 5 xnTosine (Wi = nt) (319) E This result is based on the fact that the sampled waveform rs(t) defined as utn= DD xtnTobt nt) 13.20) foral for IF] (13.21) 2» CHAPTER 1. SIGNALS AND LINEAR SYSTEMS so passing itdtrough a lcavmass filter with à bandwidth of W and a gain of 7; in tite passband produce the original sigoa! Figure 11758 a representarian of Fquation (1.2.19) for 7; = 1 and fa[nJl.. [1,1,=1.2,-2,1,23. In other words. wiid xt) = sineçr + 3) + sinetz +2) — simeis + )— 2eame(o) — 2 sineiy — 15 sino! — 2) + 2 sines — 2) Figure 1.17: Representatiun of the sampling theorem. The discrere Fourier transform (DET) of the discrete-time sequence «[n] is expressed as Xutf= PO snes (1329) Comparing Equations (13:22) and (1.3.21) we conclude that XD=THMO tor f<W (1.3.23) which gives the relation betiveen the Fourier transfocm of an analog signal and the discrete Fourier transform of its corresponding sampled signal. 13 Fourtes Transforms 25 Numerical computation of he discrete Fonrier cransform is done via the well-Enown fest Fourier transform “FET) algoritam. In his algorithm à seguence oi length N of samples of the signals x(£) taken at intervals of T, is nser as he representation of the sigaal. The result às a sequence of length M of samples cf Xu) in lhe frequency interval |O, fi], where fe MTE = 2W is the Nyquist frequency. When he sumples are 4 = f,/N apart, the value of A/ gives the frequency resolution af the resulting Fourier transform. In order to improve the frequency resolutioa, we have to increase N, the lengih of the input signal The PET algovithm is eompuationaliy efcient if the ingrh of the inpot sequence, N7, is a power of 2. In many cases if this length is nota power of 2. it is made ta be a power of 2 dy techniques such as sero-pactirg. Note that since the FT algorithen essentially gives the DFT af the sampled sigaal, in order to get he Fourier transform of the analog signal we have to employ Equation (1.3.23). “This means that after computing the EFT. we have co multiply il be Ty, or. equivalently, civic it by 4 in order ta obtain lhe Fourier transform of the original analog signal. The MATLAB funstion ffiseq.em, given below, tas as its input a time sequence rr, the sampling interval 1,, and the required frequency cesolutico df and returns a sequence whose length is à power of 2, the FFT of this sequence M, and the resulung frequency resoiution function [Mm dfi=Msegtmas dj % Mm) = ifrenim sed) 1Mmaf) = ftregom ta) Geesrutes M, the FFT mf be tequence m The sequence as ceru-pudded to meet He required reguency resulinica df as is the sampling incerval, The comepue d? ir the Jirul jrequency resolution, Dugput m ts the cem-puddoa versam of depu mm, M 44 the RET. al=fs/ds; end nZstength(m, ne2º (maximerrpow (ni!) nempowta2)): Mefttmay: m=fm aeros( 1 n=nZi): dt=ts/a. ILUS IRATIVE PROBLEM Mlustrative Problem 1.6 [Analytical und sumerical derivation of the Fourier trans- farm) The signal x(t) is described by 1+2 -2<1<- ao= 1h mistsl (1.3.2) lerg2 o. otherwise and is shown in Figure L.IA. CHAPTER 4. SIENALS AND LINEAR SYSTEMS Figure 1.18: The si al tt) 1. Determine the Faurier transtorm ot x(r) analytically and plot the spectrum of (1). 2. Using MATLAB, determine the Fourier transtorm numericaily ant plot the result. €TD—————— 1. The signal x(t) can he written as norma ( J-am (1325) t31m and therefore X(f)= 4sinci(2f) — sinci(f) (13.26) where we have used, lineurity, scaling, and the fact that the Fourier transfoem vÊ Me) is sinciÇf). Obvicusly the Fourier transform is real. The magnitude-spectrum is shown in Figure 1.19 In arder (o determine the Fourier transform using MATLAB, we first give a rough estimate of the bandwidih of the signal. Since the signal is relatively smocth, its Vianclvsidth is proportionalto the inverse time duration of he signal. The time duration ofthe signal is 4. Tu be ún the safe side we take the bandwidth as ten times Lhe inverso time duration, or BW=lUx1=25 (1327 and therefore the Nyquist frequency is twize the bandwidth and is equal to 5. Hence, the sampling intecval is 7, == 1/f. = 0.2. We consider the signal on the interval (-4,4] and sumpte it at TF, intervals. Wuth this choice, using a simple MATLAB serpt employing the fitseg.m function, we car derive che EET numericatiy. We have shosen the reguired frequency resolution to be 0.01 Hz, so the resulting frequency resolution returned by Iftseq.m is 0.0098 Hz, which mects the requirements of the = TT AT | ! i 1 + i . 1 1 E! i | e | n Poa Í ' ! ! Fregtos Figure 1.19: The magutude-spectrum of 1(t) derived analyically problem. The signal vector x, which has length 41, is zero-padided te à length oi 256 to meet the frequency-tescluton requirement and also to make it a power of 2 for computational efficiency. A plot of the magnitude-speciram of the Fourier itansform is given ia Figure 1.20. The MATLASB script for this problem is given below. % MATLAB senpi for Heero Problem 5. Chupier é ecão ou t=0.2; o ser pusumeçary dE-001; s=ferosti 10) [0:0 2:1,onesç1.9),(1.=Q.2:0) serostt AOJk IXoxdf deriva he FET XI=Xes e acuiing 1550 fl nclengrhço- 19)- 18,2; He 250.001:25]; To frequento cecir far amulvtio approach v=dafsine(2» |), "2—(sincttl)) "2: o esoei Fourier transform, Pause db Prese a key co tee Me plvi ef he Heurier Transform aerivea anabyeculiy es sebplou2,1,1) plextflaba(ç)j: slabel( Frequenoy") ride('Magni tude-pectrum oÉ x1t) derived analyticalIy') pouso dé Press akoy to sec the nlot of tie Fourier ininsjorm derivec! mumeneutty subploi(2.1,2) Va frequentes vector for EFT n CHAPIRR 5 SILINALS AND LINEAR SYSTEMS H27My-2nonest 1.5] U3BAN=ZAZecustd.Sepre32i 1 n82:48)=Beomesti 5 Fur d [XX df Jeitisegixas ar: o specirum vs ho sou E-OraftdetaConghix o -J-Is/2, Mo pequento, vector Xi=ets, o sur % per? à plter sraneter function Hejones(I, cells SdF zeros Length) Becuilit S/N ane 1 cedlit.5 fa) Ye eH; De cmo neem HO: “E oops 0 che filer ; peer? LT! eystem impulso h=izeroaid teil(5/18M Most 182 Leny 2t9) nas À cet fm cenlB rest, 1 cui ren], pecou, Ge cmtpot cj the LTE serem Pause Broca u ey to see apectoum of dir imp pler(ttsmNans De: pace 6 Precios key do sec hr muipue oj the fompass fer coça: Hemgentudhp. pause SE Prestar dev ta nloetl="048.19452) o che omega ef que LF essi 1.4 Power and Energy The energy and the power con:ents ofa real signal xí1). denoted by E x and Px, respectively, are defined as. (4 Px = lim — TT Jr A signal with finite energy is called an energy-type signai. and à signal with positive and fimte power 1s a power-type signal * For instance x(t) = Tltt) is au example of an energy- type signal, whereas x(!) - cost?) 15 an example of a power-type signal. AIL períudic signals? are power-type signafs. The energy spectral densiry of an energy-type signal gives the cistribution cf energy at various irequencies ví Lhe s;pnal and is given by Put) = ICS (142) Therefore, . =[ Bytfrdf 43 ae AThere exist signals that are neither energy Lyye nor pomez type. Qne example of such signals is ate) = ita “The only exeeptioa is those signals rhat are equal to zero 2Jmost exerywhere 14 Power ane Enerey 3 Using the convolution thecrem we have Seis FIRED (144 where Rxir) is the autocorrelarion fimetion Di xir) derincd ss erro [O inair+ ride = meu) HAS) for real-valued signals. For power-type signals we define the time-arerage autocorrelation Suncrior as voçes Ryo = lim — vence —nde (14,6; TT Loro and the pow er apeciral densiry is in general given by Swifi— FlRgíe] CAD The total power is the integral of the power spectral density given by Px f Fo: the special case of a periodie signal x(t) with pericd Ty and Fourier series coefficients xr. lhe power spectral density is given by Sxifidr (1.4.8 sp= Dm (14.9 which means all the power is concentrated at the harmomes of the fundamental frequency, and the power at the ch harmonie (17 To) is [xyl?. ie. the magnitude square af the corre- sponding Fourier series coefficiem When the stgnai (7) passes through a Álter with transfer funciina FH4f5, che output energy spectral density, or power spectral density. is obtained via Brifr= io ES SAN HE Sr) (1.4.10) f we use she diserete-time (sampled) signal, the energy and power 1 Equation (14.4) ia terms of the discrete-sime signal become jons equivalent to Ex-r o ll x (LA; a CHAPTER | SHENALS AND LINEAR SYSTEMS and if the EFT is emploved-- repeated—then . if the length of Lhe sequence is finito and the sequence is so Ex 05 sta N=1 » tm a=b (14.12) The following MATLAB function power.m gives the power content of a signal vector. functica p=powesta) % p= pemero GPCIER Returns the Deer im suqual x psinnem(as" 23 lemgthlaa 1F Xu) is the DFT of the sequence «|n, then the energy spectral density of x(1), the equivalent analog ed by using the Equation (1.3.23) and is given by FIXatS Sr = (14.13) where T, is the sampling interval. The power spectral density of a sequence x(n) is most easily derived by using the MATLAB function spectrum m Blustrative Problem L.$ [Power and power spectrum) The signal x() has a duration of LO and is the sum Of two sinusoidal signals of unst amplitude. one with frequency 47 Hz and the nther with frequency 219 Hz (6) = [UM x 871) + costêm x 2191), 051 <10 = o, otherwise This signal is sampled at a sampling rate of 1000 samples per second. Using MATLAB. find the power content and the power spectral density for this signal. “DD Ting the MATLAB function powerm the power content of the signal is found to be 10003 W. By using spectrumm and specplotam, iwe can plut the power spectral density of the signal, as showa in Figure 1.25. The twin peaks in the power spectrum correspond to the Iyiu frequenties present in ihe signal The MATLAB script for this problem follows. i i 1.5. Lowpass Equivalent of Bandpass Signais 35 Figure 1.25: The power spectral density of the signal consisting of two sinusordal signals at frequencies =47 and f, = 219 Hz. % MATLAB teript for Mlustrucuve Problem 8, Chuprer 1. ts=0.001 1145; t=[0325:10], A=coMepie47 en-cos(ZepivZi8ar); pepowerty: psdaspectrum(s, 1024); pouso % Press a key to ser the pover in she signal ? pouso % Preto a key to see the prver spectrum specplotk pd ts) 1.5 Lowpass Equivalent of Bandpass Signals A bandpass signal is a signal for which all frequency components are located in lhe neigh- borhood of a central frequency fy (and, of course, — fi). In ather words, For a bandpass signal X(f) = 0 for |f£ fo] > W, where W «& fo. A lowpass sipnal is a signal for vrhich the frequency components are located around the zero (regueney; i.e. for fi > W, wehave X(/)= 0. Correspending ta a bandpass signal x(%) we can define the analyric signal (1), whose 3% CHAPTER |. SIGNALS AND LINEAR SYSTEMS Fourier transform is given as Zi) = ua DADO (15) where a (js the unit step function. In lhe time domain shis relation 15 written as ses) + sit (152) where £ (1) denotes the Hilbert iransfarm af x(3) detinedas $t8) = air: inthe frequency domain it is given by Xe = jsgnQNXio (1.5.3) We note that he Elilbert transtorm tuncuion in MATLAB, dengted by hulbort.m, gonerates mplex sequence 2(1). The real part nt (2) is the original sequence, and its imaginary part is the Hilhert transform of the original sequence. The lowpass equivalent of the signal tt), denoted by te), is expressed im terms of xoas amis ater sale (15.4) From this relation we have x() = Rebotne? 58) (O = Imba(r In the frequency domain we have Xi = Ef + fo) = Uolf > MXCS + jo) (15.6) and Xiii XOto fo) + =) as The lowpass equivalent vt a real bandpass signal is, in general, à complex signal. lts real part, denoted by xe(t), is called the in-phase component of x(7) and ils imaginary part is called the quadratsre component cl xtt) and is dennted by x+(7); ie. ati = st) + Past) (1.5.8) In cerms of the in-phase and the quadrature components we have 241) = xots) cos(Zm for) — ate) sin(Zr for) (159) Z(4) = x (ti cos(2M fot) + x. (1) sin(2m for) If we express x:(t) in polar coordinates, we have ato = Vígeoa (510) 3 Í i í i LS Lowpass Eguivalenr of Bancpass Signals 3 shere W(5) and Ott) are called the envelope and the phase of the signal a(t). In terms of these two we have nú) = Ve) custe fo + OC) (51 The envelope and the phase van he expressec as Vo= a) radey (15.12) Oto — arean or, equivalenty, vo (1543 er ftis obvious from the above relations thar she envelope is independent of the choice of fa, whereas the phase depends on this choice We have written some simple MATLAB files to generate the analytic signal, the lowpass representation of a signal, the in-phase and quadrature components, and the envelope and phase. These MATLAB functions are analytic.m, loweq.m, quadcomp.m, and env.phas.m. respectively. À listing of these functions is given Delow funedon zeanalyeio(a) % 2 = umano) GANALHTIC Returns the angíptic «ipnul cnrresponding to vigaat x % zebilbença): > ——————— fimetioa xl=towmeglx,3,10) * af = lowegicasso Returas the lumpuss equivalen: of Me cignet x JO is the center frequency vs de ahe sampling inrval isto(eagitad= 19]. Ibero): xisz sexp(-jsZeginti=t), function [xs xsfequantcorup(a ts, 50 % [zc.xs] = quadcompta.tsj0) COUADCOMP Renems the incphuse and quadramre components uf CHAPTER 1. SIGNALS AND LINEAR SYSTEMS Problems 11 Consider the periudi signal of ustrative Problem 1.1 shown in Figure 1.1, Assuming 4=]1, Ty = 10, and tg = 1, determine and plot the discrete spectrum of the signal. Com- pare your results with those obtained in Mustrarive Problem 1.t and justify the differences. 12 In Tlustrative Problem LL, assuming A = |, Tg = &, and to = 2, determine ané plot the discreze speciram of the signal. Compare your results with those obtained in Nllustrative Problem 1,1 and justify the differences 13 Usingthe mfile fseries.m, determine the Fourier series coefficients of the signal shown in Pigure 1 with 4 = 1 To = 4 andio = À for —24 = 24. Plot the magnitude spectrum of the signal. Now using Equauon (1.2.5) determine the Fourier series coefficients and plot ihe magnitude spectrum. Wihy are the results not exactly the same? 14 Repeat Problem 1.3 with To = 4.6, and compare the results with those abrained by using Equation (1.2 5). Do you observe the same discrepancy between the two results here? why? 15 Using the MATLAB scripi dis.spotm, determine and plot the phase spectram of the períodio signal x(*) with à penod of To = 4.6 and described in the interval [-2.3,2.3] by the relation x(+) = At). Plot the phase spectrum for —24 4 n < 24. Now, analycically determine she Fourter series coefficienis of the signal, and show that alt coefficients are nonnegative real numbers. Does the phase spectrum you plolted earlier agree with this tesult? If not, explain why. 1.6 In Problem 1.5, define xt) = A(t) in the interval [-1.3, 3.3]; the period is still To = 4.6, Determine and plot the mugnitude and phase spectra using the MATLAB script dis.spetm. Note that the signal is the same as the signal m Problem 15. Compare the magnitude and the phase spectra with that obtained in Probiem 1.5. Which one shows a more noticeable difference, the magnitude or the phase spectrum? Why? 1.7 Repeat Hlustrative Problem 1.2 with fa, b] = [-4, 4Janda(1) = cos(rt/8) for kt] < 4 1.8 Repeat [lustrative Problem 1.2 with (a,5) = [-4, 4] and (1) = sintrr/8) for | x 4 and compare your results with those of Problem L.7. 1.9 Numerically determine and piot the magnitude and the phase spectra of a signal x(!) with a period equal to t0-S seconds and defined as —108+0,5, 0<r<5x 107 soy= o otherwise inthe interval 4] < 5 x 1057 Í i i : 1.3. Lunwpass Equivalens of Bandpass Signals 8 LIO À periadic signal x(t) wita period 7h = 6 is defined by x(1) = NGs/3) for |; < 3. This signal passes through an LTI system with an imputse response given by ae = 10 Q<1<4 0, utherwise Numerically determine and plot the discrere spectrum of the output signal. 1.11 Repeat Problem 1.10 with xi) =X far |t <3a0d IL Dstsa nn) = [ É 0, otherwise 112 Verity the convolution thecrera of the Fourier transform for signals x(t) = [Tt") and x() = Att) numerically, onve by determining the convolution directly ane once by sing the Fourier transforms of the Iwo sigoais. 1.13 Plot the magnitude and The phase spectra of a signal given by 1 -Ixix EE d<i lxt<2 O. otherwise (M= LI4 Determine and plot the magnitude spectrum of an even signal x(1) wiich for pasitive values of tis given as :+1, Ostxi z Istx2 n= = La deres 0, otherwise Determine your result both analyticaliy and numerically and compare the results 1-15 The signal described in Problem 1.14 is passcd through an LIT system with an impulse response given by 1, Oxis2 nD=]2, 20123 O, otherwise Determine the miugnitude and the phase spectra of the output signal. 4 CHAPTER 1. SIGNALS AND LINEAR SYSTEMS 1.16 The signal cos(2m x 44) + cos(2r x 2191), Dae a0 “O = fg, otherwise is considered, As in Hinstrative Problem 1.8, assume this signal :s sampled at a rate of 1000 samples/second. Using the MAILAB M-file butterm, design a lowpass Rutterwarth filter ef order 4 with à cutoif Irequeney of OO Ie and pass x(:) through (his filter. Determine ané sketch the output power spectrum and compare it wilh Figure 1.25. Now design à Buterwarth filter Of order 8 th lhe suise cutaff frequency, and determine the output of this filter and plot us puveer spectrum. Compare yowt results in teses two cases. 1.17 Repeat Problem 1.16, but this time design hghoass Butterworth filters with the same orders and the same catoff frequencies, Plot your results und make the vormpurisons. 1.18 Consider the signal cos x 47) = cosf2r x 2191), 0xr< 0 sin= otherwise a. Determine the analytic signal corresponding to this signal e . Determine and plot the Hilbert transtorm of this signal c. Determine and plot the envelope of this signal d. Once assuming fo = 47 and once essuming fo = 219, determine the lowpass equivalent and the in-phase and the quadrature components of this signal. mea qr jr retos mr Chapter 2 Random Processes 21 Preview Te this chapter we iilustrate methods for generatirg random variabtes and samples vf random processes. We begin with the description of a method for generating random variables witha specified prabability distributica function. Then, we consider Gaussian and Gauss-Markow processes and ilfustrate a method for generaling samples of such processes. The third topic hat we consider the caaracterization of a stationary random process by ils atocorrefation in the time domain ané by its power spectrum in the frequency domain. Since linear filters play a very important role in communication systems, we also consider the autocarrelation function and the power spectrum of à linearly filiered random process. The final section of this chapter deals with the characteristics cf lowpass and bandpass random processes. 2.2 Generation of Random Variables Random number generators are often used in practice to simulate the effect oi noiselike signals and other random phenomena that are encountered in the physical werld. Such noise is present in electronic devices and systems and usually limits cur ability to communicate aver large distances and to detect relatively weak signals. By generating such noise um a computer, we are able to study is effecis ihrough simulation of communication systems and to assess the performance of such systems iu the presence nf noise. Most computer software fibraries include a un:form random number generator. Such a random number generator generates a number tetwcen O and | with equal probability; We call the output of the random number generator 3 tandom variable, If 4 denotes such a random variable, its range is the interval O < 4 < |. We know that the numerical output of à digital computer has limited precision, and as a consequente i is impossible to represent the continuum 0€ numbers in the interval 0 < A < 1. However, we may assume that our computer represents each output hy à large number of bits in either fixed poi or Hoating point. Consequentiy, tor al! practical purposes, the number of outputs in the interval O < As 1 is sufficiently large, so lhat we as 46 CHAPTER 2. RANDOM PROCESSES are justibied in assuming that any value in che interval is a possible output from dhe generator. ie uniform probability density function for the random variable A, denoted as jd). is illustrated in Figure 2.1(4), We note that the average value or mean value of A, denated asma isma = 3. The integral of the probability density function, which represents the area under f (4), is called the probabilicy elistriburion function of the random variable 4 and is defined as A FA) -[ Fl) da n For any random variable, this area must always be unity, which is the amaximum value thar can be achicved by a distribution function. Hence, for the uniform random variable A we have ' sm- f flar = 222 and the range Of F(A)ISO 3 F(A) s tor < A < L The probability distribution function is shown in Figure 2.1(b). Ra) FA) dr Figure 2.1; Probability density function f(A) and the probability distribution function F(A) of a uniformly distributed random variable A. [f we wish to generate uniformly distributed noise in an interval (5, b + 1), it can be accomplished simply by using the output À of the random number generator and shifting it by an amount b. Thus a new random variable 8 can be defined às B=A+b 223 vehich now has à mean value ing = 5 + 5. Forexample, ifb = - 1. che random variable & is uniformly distributed in the interval (4, 4), as shown in Figure 2.2(a). Ts probability distribution function F(B) is shown in Figure 2.24b). mandem 22. Generation of Random Variables a à uniformly distributed random variable in the range (0, ) cun be used to generate runelom variables wilh other probability distrisuncn functions. Por example, suppose that We wish 10 generate à random variable C with probability distribution function illustrated in Figure 2.3. RB a ut ' o me o ta) eb) Figure 2.2: Probability density function and the probability distribution function of à zero- mean uniformiy distributed random variable. Figure 2.3: Invere mapping from the uniformty distributed random variable À to the new cundom variable C. Since the range of FIC) is the imterval (0.1), we begin by generating a uniformiy distributed random variable A in the range (0, (). df we set F(C)= A 2.24) C=EA) (2.2.5) CHAPTER 2, RANDOM PROCESSES Property 2: 1f the Gaussian process X(f) is passed througã a linear, timie-invartant (LTT) system, the output of the system is alsu 0 Goussian process, The effect of the system on X(6 is simply iellected by a charge in the mean value and (he covariance of X (1) (STRASS Austrative Problem 2.2 [Generation of samples of a multivariate Gaussian process] Generate samples of a multivariate Gaussian random process Xtr) baving à specified mesn value mr and « covariance Ce First, awe generate a sequente o: » sentistically independent vero mean and unit variance Gaussian andor variables by asins the method cescribed in Section 2.2. Let us denote this sequence cf 2 samples by lhe vector Y = 3a). Secondiy, we faclor Me desired x x 1 covariance matrix €, as 10,5) (23.3) Thea, we define ke linearly teansiormed qn x 1) vecter X us (23.4) “Thus the covanance af X is C=EMX-mMX-mo) = Etc rio) =ciewro” = ciciy 235 The most difficuft step in this process :s the fastocization of be covariance matrix Cy. Let us demonstrate this procedure by means of an example that employs the bivariate Gaussiaa distnbution. Suppose we begin with a pair of statistically independent Gaussian random variables yj and y, which have zero mean and unic variance. We wish to transfarm these into a pair of Gaussian random variadies x1 and 42 witltiean 4a = O and covariance matrix colei, “221 po o; 1 ] 2.3.6 shere 7 and 57 are the variances of x, and 43, respecuvely, and p is the normalized sovariance, defined as ELMO — mo — ma) e aros ai 3. Gaussian and Gsuss- Markov Processes 5 The covarianeo matrix € can be factored as Ce ciicty where Therefore, t [atm -(VI- ly 3-Dyn-tw3 rt The MATLAS scripis for this computation are given below. pr for Aihuctranive Problem 2, Chapter 2 or; 1 sZt2 1 semuliá.gplima Cah. % Comprracinm cf the péf of tal. A Julhous dettac0,3;, al=-Igelta:ã; Nu -Idelta:3 for ieAlengahtx 1d. for j=ttenginta?) Edit /BepiadertOe 1/2)oenplt= 1 72)=(MA ICO RG] maR + EmrrGom + 1 DL CE) AZ) mo: end end: % plotting commund far pa! fitows ameshixL x.) fimesiom [3] ae mulô gp(im,C) % dx = mulrigpím,C) %» MULTIGP generares a muitiuniie Guussian random z process walk mean vector m (coluna vector, aut cuvariance metris C, 54 CHAPTERZ, RANDOM PROCESSES Nelergihtmy: Far dx 1, virgngaoss; vos X=sqamCayem; ES AS GAIN ESSAS AS SS EEESSLG 6 Seção EM Es Figure 2.6: Joint probability density function of x1 and x2. As indicated above, the most difficult step in the computátion is determining (1/72, Given the desired covariance matrix, we may determine the eigenvalues [44,1 < E < 1) and the corresponding eigenvectors (uy, | 5 k < n). Then, he covariance matrix € can be expressed as C= Davi (23.10) ta and since C = Ch2(c1/2y it follows that CR Saiu (2341) Hm 23, Gaussian and Ciuss Markov Processes 55 Definitina: A Markov process Xt4) is é random process whose past has no influence on the future if its present is specified. That is, dt ty > 17-,. then PP sm ds ta) = P[X Co) Sae KG)! 2312 From this definition, it follows ihatifn < 19 << ta, then Xin PER oo < x |xto Definition: A Gauss.Murkoy process X(t) is à Markow process whose probability densicy funetion is Gaussian. The siraplest method for gencrating 4 Markov processis by means ofihe simple recursive formula Ku 6Xa1 + um (23.14 when tum is à sequence of zeru-mean 1i,d. (white) random variables and p is a parameter that determines the degree of correlation between X, and Xy.. Thatis, EAaka = EM) = 69), (23.15) Ciaussian, then the PROBLEM Ifthe seguence (ui ILLUSTRA Hlustrative Problem 2.3 Generate a sequence of :000 tequally spaced) samples of u Gauss-Markov process from the recursive relation An =0.95X, 1 + um. mm 12... 000 (23.16) where Xg = O and (au fis a sequence of zero mean and unit variance i.i.d. Gaussian random variables. Plot the sequence (Xy, 1:77 < 1000) as a function of the time index n and the antocarrelation Ram) = +50 31) Nom DO XnXnom m at where N = LD00. “ED The MATLAB saripis Cor this computation are given below. Figures 2.7 and 2.8 ilustrate the sequence (X,) and the antocorrelatioa function R,(m) CHAPIER 2. RANDOM PROCESSES 169; Chugrer 2 o ana Xe. NJ a queria ess qner(KO, ho, % [XIegues mano, a] GAUSMAR generos ar Guuss-Murkoo proceso «A length M & the cabra proces ds reboa tor he witito Gaussiun & mise mith cer mec ana sn striunce far tn, UWSC) Wok % venercie the mueso puncors end; Eq KM Way de pirer olomeno in she Guuse Manha process M, Xtiy-rhae Xi Ietêrsias So she cemuininy elements end; Figure 2.7: The Gauss-Markov sequence. 24 Power Spectrum of Random Provesses ar Yolte Prncesses s Figure 2.8: The autocorrelation function of the Lauss-Markov process 2.4 Power Spectrum of Random Processes and White Processes A stationary random process X (1) 1s characterized in the frequency domain by its power spectram 84 (f3, which is the Fourier transform of the autocorrelation function R,(1) of the random process. That is. ao san= | êaue imitar es Conversely, the autocorcelation function Re(T) Of a stationary random process X tt) is obtained from the power spectrum S:(/) by means of the inverse Fourier transtorm: i.e, Reto) =f Sape ar (2.4.2) In modeling thermal noise lhat is genecated in electronic devices used in the imple- mentation of communication systems, we often assume that such noise is a white random process. Such a process is defined as follows. Definition: A random process X (7) is called a white process if it has a flat power spectrum, ie, if 8:(f) isa constant for all 7 As indicated above. the importance of white processes stems from the fact that thermal noise can be closely modeled as specirally constant over a wide range of frequencies. Also, CHAPTER 2. RANDOM PROCESSES vogasos 8 B 8 Figure 2.13: Inverse FET ot the power spectrum of the bandlimited random process in Blustrative Problem 2.5 with 32 samples. 5 Lincar Filtering of Random Processos é ni Í | ' 1 al | , Ad, PRAÇA AV V | | ou [ y amb : &—— Figure 2.14: Invecse PET of the power specirum cf the bandlimited random process in Tustrative Problem 2.5 wiih 256 samples. 2.5 Linear Filtering of Random Processes Suppose Mat a stationary random process Xtr) is passed through alincar time-invariant ter that is characterized in the time domain by its impulse response A(f)and in the frequency domain by fis frequency response HD Ê. hate PAS dr (25.1) Tu follows that the output of the linear Álter is the random process rn -f. X(oh(t — nar (252 The mean value of F(r) is a CHAPTER 2 RANDOM PROCESSES Eita (t— dr ms [ med = mato (25,35 oxhere H(0) 15 lhe Irequeney response F(f) of the filter evaluated at / =) The autocarrelanon function of Ftt) is Rato = ElPiryr a + A - f EIXO X a) hide — Dio + T — al dr de =[ fÉ Ras att cet r -addeda sa In the frequency domain, the penser spectrum of (he output process (7) is related 10 the power spectrum of the input process X(r) and the frequency response of the linear filter by the expression 8O= Ê ess This is easily shown by taking the Fourier transform of (2.5.4) ILLUSTRATIVE PROBLEM Ilustrative Problem 2.6 Suppose that a white candom process X (1) with power spectrum 84f) = oral j excites a linear filter with impulse resporse aun= ht CE (1.5.6) 9, Determine the powe: spectrum $4./) af the fihec output “ED ———————— The frequency response of the filter is casily shown ta be 1 =——— (2.5.7 HM TE jaf í ) 2.5, Lincar Fiftcring or Random Processes ss Hence, The graph of Se: 3 is illustrated in Figure 2.15. The MATLAB script fer computing 8,0) for a spesified Sat) and HP) is gnen belos . % MATIAS echo on deita-0.01, 2, ea di Tustcanivo Ponha Figure 2.15: Plor of 8,(./) given by (2.5.8). ILLUSTRATIVE. PROBLEM Rlustrative Problem 2.7 Compute the autocorrelation function Ry(r) corresponding to &y() in the Illustrative Problem 2.6 for the specified 8.) = 1 6 CHAPTER 2. RANDOM PROCESSES “€ED—— Inthis case, we may use the inverse FF algorithm on samples of $,(/) gwen by (2.5.8). Figure 2. 16 illustratos this computation with N = 256 trequency samples and 2 frequeney separátion Af = 0.1. The MATLAB script for this computation is given below % MATIAR seripa for Hlustratve Emislem 7. Chaprer 2 echo on Xap56; o number vf sumpies Haf=0.1; E frequeney cepuranon fOrdeltaF(N/2)sdeitai, —(N/2- 4 adeltaf-deltaf:— celtaf), E cmup the firot dal Bye 1.1 HBupiet)."2k % sumpled! spectrum Rysilisy); % dutacorretaion uj Y So plotumg command falls plotishifitreake Rey: Figure 2.16: Plotof Ry(r) of Ilustrative Problem 2.7, Let us now consider the equivalent discrete-time problem. Suppose that a stationary random process X (1) is sampled and the samples are passed through a discrete-time linear filter syith impulse response k (x). The vutput of the linear filter is given by the convolution sum formula Yi =D adoxa (25.9) th 2.5, Linear Fiitermy 0Y Random Processes g where Xín) = X(3,) are discrete-time values of the input randorm process and Fix) is the output of lhe discrete-time filter. The mean value Dé she outpus process is m.— ElF e) =5 aiiptxio - 6] + =me) hi =m HO) 2510) whero 47 (0) is the frequency response Ft /) of she filtor evaluated at / = O and e HER = SO him eta (2511) amo The autocorrelation functiun of the eutput process is Rom) = ElFin)P (a + m)] e DD uemeettin-txin tm 0] êspico =D Dacia Rom 144) (25.12) test The corresponding expression in the frequency domain is SD = SABE (25.13) where the power spectra are defined as S(D= 5 &uameraia (25.14) and > 3 DD Ryimerrinim (2,3.15) n CHAPTER 2. RANDOM PROCESSES Srj) . i TORA ALA AA ! l DA pa da PD i A | t | | Figure 2.19: Power spesura S.t7) and $y(/) in lustrative Problem 2.9. where X tt) and Xs(1) are called the in-phase and quadratis components ot X 4). The random processes X.(t) and X,(1) are lowpass processes. The following thearem, stated without proof, provides an important relatonshp among X(t), Xctt) and X, (1. Theorem: IE X (3) is a zero-mean stationary random process, the processes X.(1) and Xs (1) axe also zero-mean, jointly stationary processes. In fact, xt can be easily proved (see [1]) that the sucocorrelation functions of Xc(t) and Xs (1) are identical and may be expressed as Rár)= Rári= Retrjcos2a for + Reir)sinZr far (2.6.2) ssbere R$(1) i£ the antocorrelation function of the handpass process X(r) and És (1) is the Hilbert wansform of R+(r), which is defined as os Ra BO) = = Ea 263 x let Also, the cross-vorrelatian function of Xc(r) and Xs(f) is expressed as Resto) = Rm) sin2m for — Rm costa fer (264) Finaliy. the autocorrelation finction of the bandpuss process X (?) is expressed in terms of the autocorrelation function R.(7) and lhe eross-correlation function Kes(T) as RdO = At) cos tr fr — Ran) sinZm for (2.6.5) 2.6. Lowpass aad Dandpass Processes 3 ILLUSTRATIVE PROBLEM Nlustrative Problem 2.10 [Generation ofsamples of a band pass random process) Gen- erate samples ol a bandpass rancom process by frst generating samples of two statistically independent random processes X. ét) and X, tt) and using these to modulute the quadrarure camiers cos 2x fiz and sur 257737, as shown in Figare 2.20. sos 2m ff N do sia Ze was Lowpass | Hut fer | ç Lospass | 48 WON filter en lrfs Figure 2.20: Generation cf a bantipass random process On a digital computer samples of the Jowpass processes X.(t) and Xs(!) are generated by filering two independent white noise processes by twa identical Lompass filters. Thus, «we obtain the samples X.(n) and X,(1). corresponding to the sampled values of Xc(f) and Xs(t). Then, X.(n) modulates the sampled carrier cos 2 for? and sim) modulates the quadrature carrier sir. 27 fan T, where 7 15 the appropriate sampling interval. The MATLAB script for these computatians is given below. Por illussrative purposes, we have selected the lowpass filter 10 have a transter function HC Also, we selected 7 = 1 and fo = 122, The resulung power spectrum ol the bandpass process is shown in Figure 2.21. % MATLAR script for iflusirative Problem 10. Chapter 2 N=100% Se number vy sumpies for ie t2N, LX It XL t=gngauss: [X2tD X2lieilegngaus: end; de standard Guussiun impar moise procerses A-[1 -€.9]; Do Imepese fiber purumeiers a CHAPTER 2. RANDOM PROCESSES 5 eusrier fraquemx sind Zepomfiei! of ihe handpues press ba-specirarm E plain como izabei Abro. autecorr) e folia, Figure 2.21; The power spectrum of the bandpass process in Illustrative Problem 2.10 2.6. Lonvpress and Bandipass Processes B Problems 2.1 Generate a set af 1000 uniform random numbers in the interval (0,1) using the MAT- ZAR function randéL, N) functior.. Plot lhe histogram and the arohabiliy distribution function. for the sequence. The histogram may be determined by quantizing the iuterval am 10 cqual-widih subintervals covering the range [0, 1] and counling the numiers in each subinterval 2.2 Generate a set of 1900 unitorm random numbers in the interval [- using the MATLAB function randil, A). Plotthe hstogram and the probability disuributior functior. far the sequence 2.3 Generate a set of IÇ00 uniform random numbees in the interval [-2, 2] by using the MATLAB functioa randél, 4º). Plot the histogram and Me probability distribution function for the sequence, 2.4 Generate a set of 1000 randum numbers having the linear probability density function U<s<2 O, otherwise foj= Plot the histogramt and the probability distribution function 2.5 Generate a set of IOUO Gaussian random numbers having zero mean and unit variance using the method described in Section 2.2, Plotthe histogram and the probability distribution function for the sequence. In determining the histogram. the range of the random qumbers may be subdivided into subintervals of width o 2/5, beginning with the first interval covering, the range —02/10 « x < 02/10, where o? is the variancs. 2.6 Gencrate à set of 1000 Gaussian random numbers having zero mean and unit vari- ance by using the MATLAB function randn(1, N). Plot the histogram and the probability distribution function for the sequençe, Compare these results with che results obtained in Problem 2.5. 2.7 Generate 1000 pairs ef Gaussian random numbers (x). x2) that have imcan vector m=Ele and covariance matrix Eure EE CHAPTER2. RANDOM PROCESSES à Determine :ke mmeaos of the samples taxis xd. = 1,2... 1000 defined as RR Pi md L e me MS oo de and their covaniance, | me Tomo Dt dinlea a) b. Compare the values cbtained from the samples with the theoretical values. 2.8 Generate à sequence cf 1000 samples vt a Gauss-Markov process desenbed by the recursive relation 1000 Xu = 0X: + Wa n= where Xo = 0,0 = 0.9, and [W,) 15 a sequence of zero-mean and umit variançe iid. Gaussian andom variabies 2.9 Repeat Ilustrative Problem 2.4 with an i.i.d. sequence of zere-mean, unit variancs, Gaussian random variables. 2.10 Repeat Tlustrative Problem 2,5 when the power spectrum of à bandlinited random process is otherwise [= É. If<B sun sis 2.4) Repeat Iustrative Problem 2.6 with a linear filter that has impulse response do fes zo O=ig eo 26. Lowpass and Dandpass Processes ” 2.12 Determine rumerically the antorarrelatica Fencticn of the random aracess atthe output of the lincar filter in Problem 2.! 2.13 Repear Iustrative Problem 2.8 when am DOS nz0 o n<0 2.14 Gencrate an 1. inthe incerval [1 d. sequence Iza | ot N = 1000 uniformly distributed random numbers |. This sequence is passed through linear filter with impulse response mim = TOS nz mi= o. n<0 The recursive equation thar deseribes the output of tus filter as a funcmon of che inputs =005y-ttm n20, a Compute the autocorrelation functions R«(m) and Ryim) of the sequences [xai and [3h] and the corresponding power spectra $e(/) and $y(./) using the relations given in (2.4.6) and (24 7), Compase thus cesuilt for 4,(£) with that obtained in Mustrative Problem 2.8. 2.15 Generate two iid. sequences lua, and ;39,9] 0! N = 1000 uniformiy distributed random numbers in the interval [—3, 4). Each of these sequences is passed through a linear filter svith impulse response whose input-output characteristi is given by the recursive relavion 1 in = qdo + Um 21,20=0 Mm pia nz], xo Thus, we obtain two sequences. [Xcw) and (sa). The output sequence [xcm| modulates the carrier cos(r/2)n, and the output sequence ix,x) modulates the quadrature carrier sin(x/23n. The bandpass signal is formed by combining the modulaicd components as in 26.0 Compute and plot the autocorrelation components R. (m) and sim) for Im] < 10 for the seguences [t.n) and (Xsm]. respecnvely. Compute the autocorrelation funcuon Re(7m) of the handpass signal for |m| < 10. Use the DFT (or the FFT algorithm) to compute the power specira 8.(/). 85(7) and $,(/). Pler these poser spectra and comment on the results. 82 CHAPTER 3. ANALOG MODULATION SNR; i.e. itis given by 5 Pa — | =—— 3.2.8 (5), NoW é ) where Pp is the received power (the power in the modulated signal at lhe receiver), a is the noise power spectral density (assuming white noise), and W is the message bandwidth. ILEUSTRATIVE PROBLEM Illustrative Problem 3.1 [DSB-AM modulation] The message signal m(+) 15 defined as | 0<xi<$% mn=4-2, fera 0, otherwise 2 F This message DSB-AM modulates the carrier c(1) = cos 2yr f.t, and the resulting modulated signal is denoted by u(4). Itis assemed that 19 = 0.15 s and fo = 250 Hz. !. Obtain the expression for u(1). 2. Derive the spectra of m(t) and u(t). 3. Assuming that the message signal is periodic with period Ty = to, determine the power in the modulated signal. 4. If a noise is added to the modulated signal in part (3) such that the resulting SNR is 10 dB, find the noise power. — SHIP 1. m(t) can be written as mt) = (6) — 2 (UE): therefore, t— 0.025 t— 0.075 uct) = [n (o) -21 (os) cos(HMyrt) (3.2.9) 2. Using the standard Fourier transform relation F[M(4)] = sine(t) together with the shifting and the scaling theorems of the Fourier transform, we obtain FIm(tn] = Det ( 1) — 2ãe =iTfnçinç ( 2) = imtuil DÍA sem iinnio = 3 e sinc 3 (1 Ze ) (3.210) Substituting ty = 0.155 gives FUm(t)] = 0.086 205/74 sine(0.05 f) (1 = De Pins ) (3211) 32. Amplitude Modulation (AM) 83 For the modulated signal u(i) we have UI) = 0.025 CSI cinc (005(f - fo) (1 - eU tinti= 10) + 0.025 COINA inc (0.05(f + £)) (1 - Ze ist) Plots of the magnitude of the spectra of the message and the modulated signals are shown in Figure 3.2. Figure 3.2: Magnitude spectra of lhe message and the modulated signals in Ikustrative Problem 3.1. 3. The power in the modulated signal is given by Ai 1 Pa = É Pm = 5Pm where Pm is the power in the message signal; do o doy 5 =— d — |=-= 'm | mi(ndr = a (3 + 3 ) 3 1.666 and 1666 Here P 1010819 (5) = 10 n or Pr = P, = 10P,, which results in 2, = P,/10 = 0,0833. mM CHAPIBR 3. ANALOG MODULATION The MATLAR script for the above problem loliows. de deb o Mestab demenstcanam corpt for DS! Sirtifed er e UI DR MS ad eche an m= 13: 180.691; -250 “t-0.3; t=(0u51] sme-tin= Oo (nei e message signal x, mofomestt dj (3wts)j—2ronEstT Ata zeros) sd, Memyts: Loud -tftsegiuas dry, Valãs: (C estlethseqee so: ctg elengthumi= I9l-f5/2; ságnal pover=possertut 1 lengiiue noise powesesignal poser /ant lim: noise .sld-sqrnoice.pover;; roisenense sul econdn(T engriuu: nor; Rad =ilsegird ct R=Ry(5; paso 5 Press «key do sho the mmcduduced signal pour ErsuRSARRAnARRA AM muitulation The message sena! 25043 amd raro utheruise amet ureto it urettom So sampling interval carrier freqmene SNR ve 4P figurino, sumptina fresuener % dastred jreq verouçiom time vectar iemeur SNR OAB DI; career signal emoelutaneo eiçnal Foucter tranegoem scutina Pourave aransfiem acatina Fourier iruesprems prog, vecior peter de madutmad signal Compute noite power Compute ntise asandard aruunon Generute noso Ada mense te Me modulated tignal apectrum of the sigrulecnse scaiing signaf poser pause Press any key to see u plot nf the message subplo(2:2.1] plonfemt A dengahioo)p alabel(cTêma) til" The message signal”) rasse o Press any key io ser u plot OJ she corner sutolou2,2,2) plegie deagttrto dabel( Tire) inleC' The sexeier") passa % Press any key tu see «plot of the mudalates signai subpler(2,2.3) plot Tlengrtçe) xiabelf" Timo”) tideteuhe modularea sigral”) pruse % Preso any key tm tee a plots ef the mugmitade 0) the messoxe and Me do moudatateu sigral im she fregurency subploda.1,5) plot abstfrsturteMm) xhabel( Frequency") domain 32 Amplitude Modulaios (AM) Bs tutteto Sm es subplori 1 9 Plone abcfis ereção tiley' ssecreim cf ineo , paus P% Pressa kar io see a meio conipto eubplouez 1,13 plot noise lite nor se 5: pads Se Prego der to eee rh madidato auibplat,2.1.29 plottal Elengemoiy liketSigaal ard nesse + eiaholg Dime) prase Prestar hey o tec he mo suboloti2,1,13 pleoEabautfisaniçãos;o tileç: signa. inhate eia! ota! monge de. Pre domo subplur:2.*. 2) almiCabscfisteeR o) titer' Signal a: xtabeltº Freques: ILLUSTRATIVE PROBLEM, Illustrative Problem 3.2 [DSE modulation for an almost band! sage signal m(t) is given by pe fome asda “o vlherwise where 16 = ().1. This message modulates :he carrier cít) = cos(2m fit), where f, = 250 Hz, Determine the medulated signal ut). . Determine the spectra of vm(t) and utr) ta . If he message signal is periodio with period Te imodulated signal 2, s determine the power in the » tf'a Gausstan noise 1$ added to the modulated sigoal such titat the resulting SNR is 10 dB, find the noise power. 86 CHAPTER 3. ANALOG MODILATION sis) = mino sine( 1001) cus(500r), |] < 0.1 a otherwise = sine( LDO TT (51) cos (5006) (213) GL) A plot of the spestra ot m(t) and a(t) is shown in Figure 3.3 As the figure shows, the message signal is an almost bandllimited signal with a band- width of 50 Hz. 2. The power m the modulated signal is half of the power ia the message signal. The power in the message signal is given by fla Pa= 5a f sinettnon de The integral can be computed using MATLAB's quad8.m m-file, which results in Pm = 0.0495 and, hence, P; = 0.0247. 3. Here |Dlogig (5) = 10 = Pa = 0. 1Pe =0.1Py = 000247 The MATLAB seript for the abave problem follows, T—— % dsbêm % Matiuh demonstranon script for DSB-AM modulation. The message signal do is ola) = sínct IDO, echo on n=2; * signal duration 18=0.001; % sumplag inlervul fem250: % carrier frequency q: E SNR in dB (tgarithmic) % sampling frequency % required! jrey. resolution % time veciar k & lincur SNR mesinçi100m1), % lhe mesrage signal ps(Zepiefe ar); he carrier sugeria) e she DSB-AM mudulmed tignai +seqlmuadO): B Fourier transfam % acaling (aid sfitsegins df): % Fourier transform 3.2. Ampliuds Modulaston (AM) -r . nn cof 1) cara axa g SO COD DO SS CN Do MEIO CRE do Eequeney Hz; “19º ê Soo ag z ECA) Frequency Figure 3.3: Spectra of the message and (he modulated signals in Nustrative Problem 3.2. so CHAPTERI, ANALOG MODULATION | ij | ] | e Figure 3.5: The essage and the modulated sigaals in Hlestrative Problem 3.3 4. The poser in lhe mess: e signal can be obtained as 1 nos a Pra —s di+4 dr| = 1.667 15 h bs The power in the cormaiized message signal Bm, is given by pao lo = "É coa67 . 4 4 and the modulation efficiency 15 E Pe, 2x 0.4 no fa S8PAOAG oo TP 14085 x 04167 The power ia the modulased signat is given by (E denotes the expected value) A Paz En +amtnÊ L/ nas =— O3010— 1.7 ato 28) = 0.5088 e Iotogo [1( se )] =10 7 (5) =10 Pa 5 An this case or 32 Amplo Mudalanon AMP s act . | Mis Feepenar uz na no 4 m q! Sig >= SR SA ARO a Emos Figure 3.6: Spectra of the message and she modulaled sigaals ia Mustrative Problem 3.3 a CHAPTER 3. ANALOG MODULATION Substituting q = 02314 and Pg = E A p= =0MIS = 0.5088 ia the above equation yietds In finding the power in the madulated signal, in this problem we could not use the relation because in this problem (4) 18 aot à zero mean signal The MATLAB seripa fo thus prabiesa is given helow. % umum Se Mathab demenstrunon script for DSB-AM imucdolorion. The messuge ssguat Miselpu de te 20/32 Jor 103 eds 2/3 url cem urhemanto echo on do signal duratiun %o dumphing incereut % currier frequeney % SNR im dB ilogarubmic. e omadutamem indes % sampling Pequency % tume vector E % required Jrequency eecotuno ane lina 10"(507/10, % SNR % mersuge tignat m=fones[1 10/(3+18)) —2agnes(1 10 /:3405)) zeros(110/(3+15)+1)) cxcnsiZopieicao. % carrier sigaut me nm/moxçabsimi) e normalized message cignat EMdan df ]=fisegim ts. % Fourier bunsform M/ Es: % eculing FI: Ledlengrhtm) NI -ts/ 2: % fequenoy vecur 143em.n) % mendutased signal [Ucsdfty % Founer transfira Ual/is, o eculioy signal. poseee=pomer(ut Islengu(t)), do peer um medulasça cprat % power in rormalized message pranspoveertani 1 lengthtts))/(mae(abs(m))" 2, “Bapmn)1 va Zeprom). o maduduton effcieney nense power-etaasigmal. power /snr-lim: So munre porver nuise sId=sqrttnoise. peer): de oiee stumluard desamor noisemnoise .sud=randn( lengrhiwj): % generate muiee reutmense; Wo Adil nose to the medulaies ngnai Rad =ifisegte ssa % Fourier arunsjorm ReRyfs; % sculina pausa % Presa a bey ta sãom the modahued signal power signal power pasa Ge Presr u key tu shaw the modulatios efficiency em pause Sb Preso any key to ser a plor of the message subpluit2.2,1) 32, Amplitude Moautarion (AM) 95 alotttn despir axiadD 0.15 2121) xlabelj' “ima +) unter pouso. passe So Preso cm dep mu see cs pos at the cuerier aubplod(2.2,2 plertizt Ulengrnts) aeis(f0 0.15 21 2.º) tide( he carrtar- pause e Preso am ev do eee ue plot mf tl mudistutad signai Eubplox2.2.3% plenti t Ilenggivio) axistiO 2.15 21 2.1) alabelt Time "1 ilet*The modtetazed sigral o) Pause Preso ais be de see cr pluts of the maguntude of she message und she “o modutatei! sugmai in die frequency domain. cubpiaee2 31 plot fast hufceM alapelt Frequentar) lilet:Specrrum of "ha message signal) subplot2. 1.24 piortabstisaitçoo filer" Spectrum cf che modulared signal") Xinbel' Frequency") pause O Press u key do ses « aviso suiple subplot2.:. 1 ploutenoise(tengrhçonh tider'noise sample") alabelt Time”) posse o Prees a key to see the modulaied tignat and noise subplouç2.1,2) ponte t Tfengiot tile/-Signa” and noise'i alabelt Timer) quase 6 Prers a key ta see the mudulated signal und noise iz fhea. domain subdott21. 1) plot abstitshiteetro diet'Signal spectzum") xlabel(' Frequency subpluct.9.2) plor(f abeçFRshivicR o) tile('Sigral and raise spectrun?) atalelt? Fa equency) “The MATLAB m-file am.mode.m given below is a general conventional AM modulator, este) armam fe) RAMMOD iukes signal m sampled at tr and cores % Sire Je as inpus und returns the AM modidated signal, "a is the medatation indes Function u=am.nwdí: % u= % 6 CHAPTER 3. ANALOG MODULATION % une ty e d/28 lengenem; - tre osiZapisfe a); im ve m/mantabetm)): ue(i-arm nome 3.23 SSB-AM SSB-AM is derived from DSB-AM by elimirating one cf the sidebands. Therefore, ir accupies half the bandwidth of DSB-AM. Depending on the sideband that remains, either the upper or the lower sideband), there exist two types of SSB-AM, USSB-AM and LSSB- AM, The time cepresentation For these signals is given by un= dra 5 min) cosa foi ca (o) sinlêm fee) (3.2.21) where the minus siga correspends to USSB-AM and the plus s:gn corresponds to LSSB-AM. Tae signal denoted by di(t) is che Hilbert transform of 2a"), detincd by sk (th = m(r) + 5;+ or, im the Grequeney domain, by ÁTC/) = — sgn( IMC. In other words, the Hilbert transfarm ofa signal represents a 1/2 phase shift in oll signal components. In the Frequency domain, we have p Mir cn Mer+ do A <fI h = (3.2.22) Vussatt) E otherwise (22 and Jimi stat Mif+ oO ais h Urssalf) = . athervise (3.223) Typical plots of thespectra of a message signal and the corresponding USSB-AM modulated sigaal are shown in Figure 3.7. The bandwidth of the SSB signal is hatf the bandwidth ot DSB and conventional AM and so is equal to the bandwidth of the message signal, i.e Bra W (3.2.28) “The power in the SSE signal is given by A? fa= E Pm (3.225) Note that (he power is half of the power in she corresponding DSB-AM signal, because one ofthe sidebands has been removed. On the other hand, since the modutated signal has half the bandwidth of the corresponding DSB-AM signal, the noise power at the front end of the ceceiver is also half of a comparable DSB-AM signal, and therefore the SNR in both systems is the same; i.e, s Pr = E 3.2.26; (o). Now N ? 32. Amplitude Modufation (AM) Figure 3.7: Spectra of the message and (ho USSB-AM signal ILLUSTRATIVE PROBLEM Hlustrative Problem 3.4 (Single-sideband example! The message signal Lo 0<:<% men = 0, oiherwise modulates the carrier c(t) = cost27 fer) using an LSSB-AM scheme. It is assumed that to =0.15sand £ = 250 Hz. 1 Plot the Hilbert transform cf the message signal and the modulated signal s(:) . Find the spectrum of the modulated signal. Assuming the message signal is períodio with period ip, determine the penver in the modulated signal. 4. la noisc 1s added to the modulated signal such that the SNR after dermodulation is 10 dB. determine the power in the noise. — EI — ame 1. The Hilbert transform of the message signal can be computed using the Hilbert transíorm m-tile ot MAILAB, 1.e., hilbertm. H should be noted, however, that this function returas a complex sequence whose real part is the original signal and whose imaginary part is he desited Hilbert transform. Therefore, the Hlilbert transtorm ot the sequence mm is obtained by using ihe cominand imag(hilberi(mn)). 102 CHAPTER 3. ANALOG MODULATION This message DSE-AM modulates the carrier c(t) -« cos 27.1, and the resulting modulated signal is denoted by u(t). lt is essumed that to = 0.15 s and / 1. Obxain the expressian for att) 2. Derive the specira of m(r) and u(r). 3. Demodulate the modulated signal 4(1) and recover m(1). Plot the results 1m the time and frequency domains: Dn 1,2. The firsc wo parts of this problem are the same as the first two parts of the IHluscrative Problem 3.1, and we repeat only those results here; namely, 1— 0.025 t— 0.075 ua) = [n (Se) -20 (55) cos(506r) and 4 2 rim Becotnoas (8) - Rsteçs (0) = Becintois, LN oeritamro = qe! tolsine (E (1-2 ) = 0.08e-008/2/sinc(0.05 1) (1 — 2e-0047/) Therefore, UCP) = 0.025e POUR 2O sine(0,05(f — 250) (1 -20º Srty 250) + 0,025" POSIr(/+25inç(0.05( + 250) (1 = 20 Unts + 250 o . To demodulate, we multiply ur) by cos(27 f.t) 10 obtain the mixer output (1) 0) = ute)cos(2mf.1) 1 = 0.025 1=005)] à (= Jon ES JJ $0me) HT (t- 0.025 OS alm( 005 )an( 0.05 ] If. /1- 0.025 1-0075 ei [o(a (102) cone «whose Fourier transtorm is given by FCF) = 0.025€ POR ginc(0.057) (1 - 2e Mia) u + 0,0125698 In(/=50Dsinç (0.05(/ — 500)) (1 —- 2! tony sem) + 00125000 inc (0.05(f + 500) ( — qerbniay =) 3.3 Demodulation of AM Signais 103 where the first term corresponds to the message signal and lhe last two terms corre spond o the high-Cequency terms actwvice the carrier frequeney. We see that iltering che first term yields the original message signal (up to a groportionaiity constant). À plot of the magmudes of (7) and Y(/) is shown in Figure 2.10. The demo mira Figure 2.10: Spectra of the modulated signal and the mexer output in IMustrative Problem 35. As shown above. the spectrum of the mixer output has a lowpass component that is quile similar to the spectrum of the message signal, except for a factor of 4, and a bandpass component located at &2/, (in this case, 500 Hz). Using à lowpass filter we can simply separate the lospass component om the bandpass component. In order to recover the message signal sm(t). we pass y(r) through a lowpass filler with a bandwidth of 150 Hz. The choice cf the bandwidth of the filtcr is more or less arbitrary here because the message signal is not stricily bandlimited. For a strictly bandhmited message signal, the appropriate choice for the bandwidth of the lowpass filter would be W, the bandwidth of the message signal, Therefore, the ideal lowpass filter employed here has a characteristic 1 fx 150 Hp = - Dto, oihervise A comparison of the spectra of m(t) and the demodulator output is shown in Figure 3.11. and a comparison in the time domain is shown in Figure 3.12. The MATLAB seript for this problem follows. % dsbdemm % Matlab demonsiralon script for DSB-AM demodulation The mesrage signal Misa for O << M/4 2 for 08 «é « 203 uni rem enherniso ectm on 105.15; e signal duremon us CHAPTER 3. ANALOG MODULATION Figure 3.11: Spectra of the message and (he demodulated signals om Tlustrative Problem 35 he sem tais d pol 1 Sd fm Figure 3.12: Message and demoduiator output in Tlustrative Problem 3.5. 33 Demodulation of AM Signais 105 do camping iniorval % carrier frequency So cumpliny frequency % uime vecrar % desired frequercy resolution % message signal mes 0/0 Dets. = Zronesi 1 (Its! east O /6Isdo 3; ceenst2ypimfo my, carrier eigaut u=mte. smoduioted signat meuimy Fourier tramstorm senna Fuaner transfom, seaiing onrier trogisem svaling cce Freq Cf the fer Design che for ts: E cutoft= 150; Recinoff=foor(150/df1) E=jOdflidtiodlengthon = 1H -s, angus paa tiriencentofr Hlengihtfi-m cutolr + enguluf))-2eones 1 n.cutof; DEMEH +Y; * apecirum of che Alter quis demercaltIM(DEM nfs & fer ouipur pause % Press u dy du see the efect ut minor cr subplor(3.1.1) poste sshiitabseMin tidet- Spectrum Df the the Message Signal) xiabel(* Frequency”) subpiorçã. 1,2) plot Efashiiats (UI) tie('Spactrum of the Modulared Signal") alábelk' Frequency") sabplotçã. 1.31 plot shifitabot Pp tibe(' Spectrum cf the Mixer Output") alabelt" Frequency") pause % Press q key to see he efleci of fiteriny cm dhe misor culpas ar subplotç3,1,1) plontE stitabsi cúle(' Speczrum of che Mixer Cutput') alabelt' Frequency") subpÃos(3,9.2) Porte ishificabstHm) Hilet' Lompass Filter Characteristics") Mabel Frequency") subploiçã.1.3) plor(f fishifiçabs DEM) fidel* Spectcum ot Che Demoduiator quiput") alubel(” Prequercy") pause À Press a dey to Compare the spectra of the message an she received signal ar subplot(2.1,1) plortf fshificnbs(M 106 CHAPTER 3. ANALOG MODULATION file( spectrum oi the Macuaye Sicnal') , subplet2,1 pleeti Hieshif cabo DEM Che Demodelarar Grepur) xlabelt' Frequency) pausa G% Prers a tes to see ih subplor(?,1,1) Plot Iilengrhgçio! tuler*The Maesaçe Stgnal! xlnhel Time) sesploti2,1,23 ploctedecnt? lengahityy use" The Demadr ator quepar *) clabeli imo”) ILLUSTRATIVE PROBLEM Ilustrative Problem 3.6 [Effect of phase error on DSB-AM demodulation] In the de- modulation of DSB-AM signals we assumed (hat (he phase of the oca! oscillaor is equal to the phase of the carrier. If that is nor the case. f there exists a phase shift & between the local oscillator and the carrier—how would the demodulation process change? “ED —————— [a this case we have u(1) = A m(r) cost2m ft), and the local oscillator generates a sinu- soidal signal given by cost2r ft + q). Mixing these two signals gives mertuxe une te demudulutor outras sigrals (1) = Amt) cos fer) x cos(2m fit + qd) (333) A, Sem(ncostg) + Sm) costêmja + 4) (0.3.4) As before, there are twa terms present in the mixer output. The bandpass term can be filiered out by a lowpass filter, The lowpass term, 4-mtt) costg?. depends on &, however. The power in the lowpass term is given by A Prcosig (135) Pam = É where Px denotes the power in lhe message signal. We can ses, lherefore, that in this case we can recover the message signal with essentially no distortion. but we will suifer a power lossofcos2g. For p = 7/4 this power loss is 3 dB, and for é = 1/2 nothing is recovered in the demodulation process 3.3.2 SSB-AM Demodulation “The demodulation process of SSB-AM signals is basicaliy the same as the demodulation process for DSB-AM signals, i.e. mixing followed by lowpass fltering. In this case 4 Ae ut) = dem cost fot) = Sto) sunt fer) (313.6) 33. Dermdulmion o! AM Signals 107 where the minus siga corresponds 16 the 1:SSR ad the plus siga corresponds to the LSSR Mixing vir) with the local oscilator omput we obtain Ar - Ar = qmtiteoÊ (Ox fa) 7 (o) sin(22(.0 cosi2m pos) Aeins de Aee . = q mi+4- miticostdz fin) 7 — mis) sind fr) (337 7 4 4 vehich contains bandpass components at 2 f, and à losipass component proportional te the message signal. The lowpass component can be filtered cut using à lowpass filler to tecover the message signal. This process for the USSE-AM case is depicted :n Pigure 3.13 > -— CANA E? 2h 2hrW Figure 3.13: Demodulation of USSB-AM sigrals. — CT Hlustrative Problem 2.7 [LSSB-AM example) in a USSB-AM modulation system, if the message signal is 1 0xi<% mij=(-2, Fera 0. otherwise with to = 0.15 5. and the carrier has a frequency of 250 Hz, find U(f) and Fí compare the demodulated signal with the message signal. “€ED————— The modutated signal and its spectrum are given in Tustrative Problem 3.4, The expression for UC is given by 0.025e POSMD net O5(S — 250)) (1 — Ze tizey UU = | + 0.028 DOME ine(D05(F + 250) (1 — De Ur Wish 0. otherwise ANALOS MODULATION Figare 3.17, A mple enseiope detector. Mathematically, the envelope detcior generates the envelope af the conventional AM signal, wbish is vn = [Lam tn] (339) Because | + imp(t) 2 O, we constude thal ve o Lhemto (3210) uihere ma?) 'S proportonal to the message signal mtt) and 1 eomesponds to the carrier comoneat that can be separated by a de-block circuit. As sera, in Me above procedure there is no need fr knowledge of &, the phase of the carrier signal. That is why such a demudulation scheme is calied nonconerent, or asynckronous, demodulasion. Recall irom Chapter 1 that the envelops ot a bendpass signal cam be expressed as the magnitude of its lowpass equivalens signal. Thus :f utt) is the bandpass signal with central frequency f and the lowpass equivalent to u(t) is denoted by tr(r), then the envelope of uír), denoted by V(), can be expressed as 3311) where uv it) and ue, (t) represent the in-phase and the quadrature components of the bandpass signal u(7). Therefore, in order to obtain the envelope, it is enough to obtain the lowpass equivalent of the handpass signal. The envelope is simply the magnitude of the lovepass mity= (2 0, otherwise moduiates (ho camier ct!) = cos(2m. is assumed that fz = 250 Ha and dg — +) using a conventional AM modulation scheme. It 155. and the modulation index. is a = 0.85 3.3, Demedilation of AM Siginais na Using envelope Estection. demodutate the message sipnal. 3f the message signal is periodic with à perical equal to «9 and É an AWGN pro- 1 added to the modulated signal suzh that the power in the noise process is ume-hundredih Lhe power im the modulated signal, use an envelope demodulator to demodilate the receiver] signai, Compare (has case wilh Lhe case where (here is no noise present Asan Tlustralive Prabiem 2.3, we have —0. ssa cosipr fo UoIs 005 1— U.025 - |! 425 3 = 0.858 1] cosis007!) 1f an envelope detectar is used to Cecmedulate the signal and the carrier component is removed hy à de-block, then the original message m(r) is recoverad. Note that a erucial point in the recovery ofzm(r) is that for atl values of. the expression I+amn(t) is positive; therefore, the envelope of the signal [1 + ama (8) cos(2x /.t), which is Ve) = [14 amyto], is equalto 1 + ama(t) from which m(r) can be recovered easily. A plot of the conventional AM modulated signal and its envelope as detected by the envelope detector are shown in Esgure 3.18. ul Il ) | | | rd | | | Figure 3.18: Conventional AM modulated signal and its envelope. After the envelope detector separates lhe envelope ví lhe moduiated signal, the de component of the sigaral is removed and the signal is scaled to generate the demod- ulator output. Plets of the original message signal and the demadulator output are shown in Figure 3.19. 14 CHAPTER 4 ANALOG MODULATION 1 ad nona | | d | , ! rancor Figure 3.19: The message signal and the demodularsd signal when no noise is present. 2. When noise is present. some distortion due 10 the presence of noise will also be present. Im Figure 3.20 the received signal and its envelope are shown. In Figure 3.21 the message signal and the demodulated signal are compared for this case ppt | fd | | | | Minie ij perotonal Figure 3.20: The received signal and its envelope in the presence of noise. The MATLAB senipt for this problem follows. * um demm %e Matiab demensiration script for envelope detection. The message signal Ris+iforO <r<r0/3-2/%r mr 20/3 and zem othernise, echo on De 15; % signal duration 18=0.00%; % sampling intervat fo=250; % carrier frequeney &3. Demodulation of AM Signals ns pre % mudidanem andor e sampling frequency %o time vector % required frequency vesoluiina de messuge siguul mO-[ones(] «0/(3e1s)), -Dannesi 1 104Qots) 2er0st1 10 /(3ursis 1]; Emcos(Zepim(e ay; % curreer riguat man=m/maxtabstm)) % aúrmahzed message signal [Mana [effisegemas df; % Fourier transform F=[O:aflsdtleilengthtmo— 134 fs/2 o frequency vector u=1tuem.m) sc; % mudulaed signct EU dr =tisegtu sd %o Fourier trunstorm emv=sav.phas(u): % Find he emelupe demi=atenv— fiya: % Remove de und rescule signal. power=poweriu( lengery, % pawer in mudulared signal noise. pomeresigna]. power; 100: * mise power noise std=sqruenaise. power); % neuse suundtard cemanion nove=noise .srdamnda( lengebtu). & penerme nice teuenoise: % Add núise «o the mudulaied rignal [Ra af] jfitsegris di): do Fourier iransform env.rsens. phastr); % envelope, «hem muuse ix present demz=2menv.s— 1a; % Demululate in the presence af noise pama & Pregs any hey ho sec u plet of the message subplotta, 1,15 plot(Lamt 1 tengihteo) axis(fO 0.15 21 2.1 alabeii Time") tine(*The message signal”) pause % Pres any key o see à plos of the menduluted signai subploiça.1,2) plot SC] dengrteiyo) axistlO 0.45 -2.1 2.1) xlabelk*Tima 3 ilet'The modulared signal”) pause % Press a hey to see the envelome vf the maculated signal ar LE CHAPTER 3. ANALOG MODULATION subpiona 1,1 peottaui Ttsagençr aein(O 015 =21 27) xasedt Pinar) he medulatad signal" plotetenv: flengrhuras xlabelt Pimars megulated » contiparo que mens gra) e and me demaitutared signo suoplod(2,12) plstátam E lengiaço axist(O 0.15 —21 21 xlabe:f" Time 1 nil('The message signal + aubplot2, 1.29 dem IC eagihitio Pine") tlet The dec; passe % Pres a key do compare om ie pereonco of noise air subatovt2,t,1) alone dengentado =xisilO 0.15 2124] -e4 signal) he massage signal i subplott2.1.2 porco dem Ilergehutioh xlatelt Time) tidet'Tre demodulated sianal in che presence of noise! In the demodulation process above, we have neglected the effect of the nuise-limiting filter, which is a bandpass filter in the Best stage of any receiver. im practice the received signal rlz) is passed through the noise-limiting filter and then supplied to the enveiope detector. 1n the above example, since the message bandwidth is not ânite, passing r(t) through any bandpass filter will cause distortion on the demodulated message, butit will also decrease the amount of noise in (he demodulator output. Ta Figure 3.22 we have plotted the demodulator cuputs when noise-limiting fiters of different bandwidths are used The case of infinito tanga is equivalent ta the result shown in Figure 3.21 34 Angle Modulation Angie-modulation schemes, which include frequency modulation (FM) and phase modula- tion (PM), belong to the class of nonlinear moculation schemes. “This family of modulaton schemes is charscterized by Lheir high-bandwidih requirements and good performance in the presence of nvise, These schemes can be visualized as modulation techniques úhat trade-off bandwidth for power and. therefore, are used in situations where bandwidth is not the major concern and a high SNR 15 required. Frequency modulation is widely used in 24. Anple Meditar n? A: Vila, o o ! rita Bana Vametocre | N h DA a | | | VM Figure 3.22: Etiectof the bandwidth of the noise-limiting filter on the output of the envelope detector. CHAPTER 3. ANALOO MODULATION and sa frequency mudiaitiom. The message cias! vor UPS 47 4 203 and a uthenvise, signal cturation camping interval carmer jrequency emvelutamem index sampling jrequene Rana * reqicved frequency rerotuizon es1119,/13-189.- Zrones! MACA Ivtsn2erost 1 10/eiss 1; 1-0: ngihio 1 o integre of m ars emtio Dstmi ratipermintets, and Mim lHlj=fisectmas af) % Founer aransjurm s % sculing dl del =tlengetiem— 1] a/2: % frequency vector Jmeos(Bepinfeet+2epiakf int. mb; % mudulated sigra UudflJefiegiutsdo, % Enurier transforma U-tys % sculing muié E Preso amy hey tu see se pior 0) the message and the modulimed tgral sutplo2o 1) otit lemgrhito asisif0 0.15 2.1 2.1) clabelt usa") le Dre message sigaai”] subolon2.1,2) plot ui? lenguiço axufO 045 2.1 24] ctapelk "Time 'y ulet' The nodilates signal > use E Presa any key du se q ploix ef the mugmitade cf the mesmige ana he So mucaluced signal in the fregueney dumain subpn2115 elotctabscaistitiçMos belg Fesuueney") Gdet'Hngri rude spectcwm of the message signal | sibpiot2.1,2) plom abs istutas Dj Hagrirvds-speci Nateil' tregasney") OÉ the modulateú siunal” 34. Angie Modutation 123 ILLUSTRATIVE PROBLEM Mustrutive Problem 3.11 [Frequency modulatian] Let the messsge signal be since T00N, tl otherwise mup= wterc ir = 1, This message n Hz The deriatior constam is &y iulstes the carrier c(r! = cos(2r f.4). where f. = 250 106 1 Plotthe modulated signal ir the time end frequenzy doracim. 2. Coumpare the demodulator amput and the original message signal ED 1 We Best integrare tac message sigaal and then use the se ston ahrt ted, f ntar) jo ind “(?). A plos of a(s) together with the message signa! 1s shown in Figure 3.3 The integral of the message signal is shown in Figure 327 Figure 3.26: The message and the mudulated signals mm A plot of'the modulated signal in the frequency domtuin is shown in Figure 3.28 n To demodulute the FM signal, we first find the phase of the modulated signal u(s) This phase is 27ky 2 m(z) dr, which can be differentiated and divided by 27k; to obtain m(r), Note Matin order to restore the phase and ondo the elf of2x phuse foldings, we employ the unwrapum function of MAIL AB, Plats of (he message signal and the demodulated signal areshown in Figure 5.29. As youcanses, thecemodulated signal is quite similar to the message signal. na CHAPTER 3. ANALOCG MODLLATION do Figure 3.27: integral of the message signal. Figure 3.28: Magnitude-specira of the message and the modulated signal. Dodo Angelo adodrtaçnon 125 A DANE DA Figure 1.20: The ruess signa: and the demodulated signal The MATLAB script for ils problem fo:lows. & fmim E Mankalo desmemenanuom ceripo for frequenci ormdufanim. The message sinal ds mt -sanct dO e cmgoat duruetar Ge sampling enierat % carrier frequency Se SNR q dB dlngarishome! % campling frequemey % requered jrego resolution e teme vertor 00: % devation conse dfx0.25; % required Jrequerey resulunor mesine 100: % the messuxe signal intom(1 md; for i=1lenginto—1 E incegret of m iotom(is Deine m(ipemtizars; end (Mama! sffisegtmas dy: % Furtos srunsform fa; E tcntry 0.4 dE alegam) - Mt ts/2: frequent vector u=cos(Zapisfcet42epirki eim m) & muduluçed sigaut (Uai Miseg(u asdf % Fourier transforma U=U/6s: % realing Cecphaseleeny. phasiu1s.2803 “o demoduiunor ford phase u! a phisunwrap(phase): % restore original phnee Sem A Zopisk)etdifrephitj0s) % demoduiator output, diferente und suute phase pouso % Press uy key to ser v ploi of the message and the madulaied sipacl subploeç2.11) plea(e (1 -lengataçoo) Xlabel(' Time") tilei'The message signal”) 126 CHAPTER 3 ANALOG MODULATION subplo(2,1,2) plouta(tlengthtr))) xlabelt' Time") litie('The modulated signal ') pause % Press any key to see ut plhos of the magnitude vf the message und the , % modulatred signal ir the frequenex domain subplot(2,1,1) plorifabs(Hishift(u1 5) xlabel' Frequency ') . . ude(i'Magnitude-spectrum of che message signa:'l subplot(2,1,2) plottf abs tshifiçU . . ttlei Magnitude-specrrum of the modulated signal ') slahel' Frequency!) , pause % Pres anv kev to see plots of the message and the demecdidator cuiper meth no % nuise subplot2,1,1) plotter length() xlabel(' Time) title(' The message signal '> subplot(2,1,2) plot dem(Tlength(t))) xlabel(' Time) tilet' The demodulated signal') —€ED- A frequençy-modulated signal has constant amplitude. However, in Figure 3.26 the amplitude of the signal =(t) às apparently not constant. Can ycu explain why this happens? 34 Angie Modulaiion 127 Problems 3.1 Signal m(t) is given by Fr, Ol<i<l miti=4-142, l<t<10 e, otherwise in the interval [0, 2). This signal is used to DSB modulate à carrier of frequency 25 Hz and amplitude | to generate the modulated signal utt). Write a MATLAB m-file and use it to do the following. a. Plot the modulated signal. b. Determine the power content of the modulated signal. c. Determine the spectrum of the modulated signal. Determine the power spectral density cf he modulated signal and compare it with the power spectral density of the message signal. 3.2 Repeat Problem 3.1 with t t l 2 IA t, o mit) = | =[+2, EA IA A in the interval (0, 2). What do you think is the difference between this problem and Prob- lem 3.1? 3.3 Repeat Problem 3.1 with sinci(lO), |H<2 n(t) = t) D, otherwise and a carrier with frequency of 100 Hz. 3.4 In Problem 3.1 suppose instead of DSB, a conventional AM scheme is used with a modulation index of a = 0.2. a. Determine and ptoi the spectrum of the modulated signal, b. Change the modulation index from 0.1 t6 0.9 and explain how it affects the spectrum derived in parta. e. Plot the ratio of the power content of the sidebands to the power content of the carrier as a function of the modulation index. CAP ER do MINA Eelao Etiriidhad Eótdo o od dO 42 Measure of Information “The Guipul 0 2m infitrmation source idua, speech, vides, etc. can be modelad as a candom pracess. Foz a discreie-memoryless and stationary random process, which can be thought q as independent drawings SÉ one random variable X, the mformation content, or entropy is defind as. max =—5º píxsiog pix) (2) ER alegro 5% denotes te sourze alphabecand p/2) is le probasility o the leer x. “Che base of he logeritasm is usually chosen to be 2, sehictr sesults in the entropy being expressed in bits. For ke hinary alphabet with probabiliies arc 3 — p, the encopy as denated hoy Hi(p) and is given >y Hip) =(i-phogc — pt (22) A plot of the Dinary ensrepy funcnon is given ix Figure 4.1. Figure 4.1: Plot of te binary entropy uneúon The entropy cf a source provides an essential bouad on the number of bits required to represent a sousce foz (uil recovery. In other words, the average number ul bits per source cusput teguired to encode à source tor error-fee recovery can be mude as close to HI(X) as swe desire, but zannot be less than 462). 4.21 Noiseless Coding Naiseloss coding is the general term for ali schemes Uhat reduce the number of bits required for he representation of a source output for perfect recovery. The noisetess coding theorem. i i clue to Shanaoar states hat for perfect reconsiresnon oi à seurce iLis possible to use a code sesth a rate as close to the entrogy of te source as we desire, “mt it IS nar possible to use & code with a rate less than the scures entropy. In cther words. tor any é > O, we can have a code with rate less than FX) + e, but wwe cannot have a code veta rato less than HCX). vegardless of the complexity ol the encoder and che devoder. Tatece exist various algoriltuns tor noiseless semirce code, Huifmao codiry qd Lempel-Ziv coding are two examples Here we discuss the Huifman coding aleoritam Huffman Coding in Huffman coding we assign longer codewords to the less probable source outputs and shorter codeseords to the more probable ones. To de tis we start by imerging (he two least probahle source cutposs ta generate à new mersed output whose probability is the sum of the corresponding probabilities. This process is repeated until only one xrerged oucput is left. (n tins way we gencrate a sree. Siarung from the rool ef the tres and assigning 0º and t's to any two branches emerging [rom lbe súme node, we geneete the code. ftean be shown that ia this way we generate à code with minimum average length among the class of prefir-free codes The following example shows how to design a Huffman code, ILLUSTRATIVE PROBLEM Iustrative Problem 4.1 [Huffman coding] Design à Hutfman code for a source with alphaber X = (x1.x3.... . 49). and correspondiag probability vector p=(02,015,013,0:2,0:,009,08,0.07,006) Find the average codeword length of the resutung code and compare 1t with the entropy of the source, D—————— We follow the algorithm oultined above to gel (he Lee shown in Figure 4.2 The average codeworá length for this code is L=2x02+3x(0.15401340.12 40.1) 44 x (0094008 4007 +006 =3.1 bits per source output The entropy of the source is given as 9 ROO = — » pilog pu = 3037] bits per source output We observe that L » H(X), as expected, “The MATLAB function entropy.m given Eclow caiculates the entropy of a probability vector p !Prefix-free codes ase condes am selucla iu «usleseuna Sa pref of anathes codenvord, Codewords [ie [o 1010 10LL “Ho um x x x “ CHAPTER4. ANALOG-TO 11 160 no 042 932 mo ou 10 9 lose 0.26 1019 at? 10% 1011 tio 0.13 m tu Figure 4.2: Huffman code tree. “AL CONVERSION Prob. = 1 42 Messyre of Informares us TD Sunerioa h-ensrapyig) % = exoprr. % che probe dE dengaótiadra <9: ns the entrepy funchon eg component (5) ) t abs(sumpo= 1 temor 'NcE a p end h=sum(-p aloga(p): omencs dC qur ada up co 1º) Ihe quantity (423) is called the ejficienos af she Hugfman code. Ohviously, wc always have q < 1. Tn gem. sral, it cur be shown that ihe average codesvord length for any Quifman code satisÃes the inequalities BUD LÊ <HON+L (4.2.4) Tf we design à Huffman code for blocks of length K instead of for single letters, we «will have - I HM) <Ê<HOO+ E (4.25) and, therefore, by increasing X we van get as close to H(X) as we desire. Needess 19 say, increasing K increases the ucmplexity considerahly. Tt should also be noted that the Huffman coding algorithm does not result in à unique code due to the arbitrary way of assigning 0's and 1's to dilferen tree branches. That is why we talk about « Huffman code rather than che Huffman code, The MAILAB function huffman.m, which designs a Huffman code for a discrete- memanyless sauree with probability vector p and returns boththe codesvoras and the average codesward length. is given below. function Fadi=hufiiaantp): GHUFEMAN Mugen conde gerenim % ff = huófmurtp), Hoffman code genenaur % returas h the Husfram code auras and E che % average conlemurd feng far a torce veish & probubitico vector p. 1 lemprhçânec eg, Ea prob. vector, negative compuneac (3) ') df abs(sum(pi—1)10819, na CHAFTER 4. ANALOG-TO-DIGITAL CONVERSION ent Nor a prcd. veczor, cosponents Jo not ade up = for cetiam" —avifindmin-ioi, findtentn= it Jet: ein ineo” etr-ia-tZem Saetn-ita-1) climiZener dr; for jetã= cinnça+ Mono 1QsBandectnmi= astndemin=is:, end ené j==ato= "5 natindiratn= ot a ati fesdeett meitindeoat = D+ 1 inaçme! camino: LUGUI=Iemgrhe Anda hi =32M: «ad tesumto atiy; Nlustrative Problem 42 [Huffman coding] A discrete-memor;ess information source with alphaber «14h and the cortesponding probabilíties p= 10.1.0.3,005,0.09.0.2], 0.25) às to De encoded using Huifman coding 1. Determine the entropy of the source. Find a Huffman cude for the source and determine the efficiency of the Huffman code, ma à, Now desiga a Huffman code for source sequences of length 2 and compare the elficiency of this code with the efficiency of the code derived in part 2. Mereagrire CÊ Br formato 154 1. The entr s derived via he entrcpy.m fuastior and is found to be 2.3548 bits per source symbol y af the sourei Using rhe kofêmaram dircrion ve can design a Huffman code for this soutce The codesvards found 19 be 010, 11, ONO, QUI OG, and 10. The average codeword lengtà for tbis code is found to be 2.38 >inary symbols per souree output. Therefore, 4he eificiencs of this code is 2.3549 — 09895 "238 3. A new source whose outpuis are letter pairs of the auiginal sovrce hrs 36 output letters of the forma feai, €5)]É joy. Since the sourze às memorylesa, the prosability of cach pair is the product of the individual letter probabilities. Thus, in order to obtain the probability vector for the extended source, we must generate a vector with 36 components, each component being the praducz of two probabibiies in the original probability vector p Thiscan he done by emploving lhe MATLAB fun in the form of kraníp, p). The Huftman codewords are given by TIGO0O, OH 70, OTHON, LOTLGON, 1TICO1, 00703. GI TIL, 000. 011010, 00111. 1001, 1100, 11101110,0HLOLI. LHOLLHHO, INIUNTLIT, 10UON, OO1000, LOTIOTO, ONOO, 10140110, JOTIQVO, 101140, 141110, FLIOIO. 1010, LLIOHO, 1OLHIL, VITLO, 0100, GOLO, 1103, 001007, LHLINL, 010), and LOGO and the average code- word length for the extended sauree is 4,7420. The entropy of the extended source is found to be 4.7097. so the efficiency of this Huffman code is 4.7097 4.7220 ma = 0.9932 which shows an improvement compared to the efficiency of the Huffman code de- signed in part 2. ILLUSTRATIVE PROBLEM Nlustrative Problem 4.3 [4 Huffman code with maximum efficiency] Design a Iuff- man code for a source with probability vector 14 256' 256 We use the huffman.m function to determine a Huffman code and the corresponding average codeword length. The resulting codewords are L, 01. 001, 0001, 00001, 000001, O0000B1. 80000009. and 00000001. The average codeword length is 1.9922 binary symbots per sunrce output. 1 we find the entropy of the source using the entropyam function. we see 142 CHAPTER 4. ANALOG-TO-DIGITAL CONVERSION dist-distrevalif! quad (nawtum, a tr ato= Ih told cares] end “DD [QN q foneeian [y.fiaf=aq disigfuníca, bee deita s.tol,pl.p2,93) SUQ DIST Reruons the dirtom df à eniforo quemues aid quarizatten prrnts set to the centroids pr UIQ DASTIFUNECNS,CN DELTA.S.TOLPIP2,P3) funten=sutce density fmetiom givem in um meoite tt ur masa chree parameters, sii.p2.p3. Ebuct=The supor: of the suurce densicy faneiiom, umbro cf ieveis deita-ievei size ne fefimost qusmtasamen segion boundary pi.PZpd=paremeters e che umpui function s=quantsatiom james, distedimercian ol=ihe relativo errar SRAPIFANANAR dé Ceotedeatn=20 emert'Toa many levelo tox this range. p ren end ir asd) n emori The Lafrmost bourdazy too szali.*% retum end df is-(n-2)edelta>c) emori he leftmosr baundary t20 Large end eetum argse( |: for j=fnacgin=? arps=fargs." .p' imtêstriD: end avgs=[args, a(Deb; fer i=2n atit=s+ti- BJedeitas end ata+ti=e; [dissera] use distifunfeo, atol“ args): ILLUSTRATIVE PROBLEM Tlnsrative Problem 4.4 [Determining the centroids] Determine the centroids of the quantization regions for a zero-mean, unit variance Gaussiun distribution. where the bound- aries of the quantization regions are given by (-5, —4, —2,0,1,3,5). The Gaussian distribution is given in the m-file normalm, This distribulionis a foreion of two parameters, the mean and the variance, denoted by m and s (oro), respect ei The support of the Guussian distribution is (—oe, co). but for emploving the reeical rontnes it is enongh to use a range thai is muny times the standard deviation of the dis . sad 43 Quamizaion 142 Forexample. (m 10/35, + 10,/5. can be used. The following rm-file determines the centrords (optima! quantizarion levels) & MATIAS seript ho ihusirative Problem 4, Cheptes 4 10. 5,-4,-20,13.5,19] lengihian— 1 entrei norma? * armha(i+1) 0 091.0,1) for it end This resutts in the foilowng quantization levels: 2168, -2.3706, 0,722 —0,4599, 15101, 3.2827,5 1865) (INES ES Mlustrative Problem 4.5 In [ustrative problem 4.4 determine the mean-square error. EPP—-———— Lettinga = (10, — 0.1, square errar of 0.177 ILEUSTRATIVE PROBLEM Mustra «5, 10) and using imse .distum, we obtain a mean e Problem 4.6 [Uniform quantizer distortion] Determine the mean-square er- sor for à uniform quantizer with :2 quantization levels, each of lengih 1. designed for a aero-mean Gaussiar source with variance of 4, It is assumed that che quantzation regions are symmetric with respect to the mean of the distribution. ET —— >> By the symmetry assumptinn the boundaries of the quantization regions are 0, 1, 2 3, 4. £5, and the quantization regions are (00, 5; 4-5, —4], (4, 3) (3, 2], (=2.=1]. (=1,0), (0.1, (1.2), (2, 31. (3,4), (4,5). and (5, +00). This means that in the ug list funciion we can substitute 6 = 20.0 =20,4=1]n=12,5e-5, tot = 0.001, p; = 0, and pz = 2, Substitutng these values into uy distm we obtain a squared esror distortion ot 0.085] and quantizativa values of +0.4897, + 1.269], 42. 4487. 34286, 54,4089, and +5.6455. The m-file ug.mepnt.m deteciines (he squared error distortion For a symmertio density function when lhe quantization levels are chosen to be the midpoints of (he quantization intervals. Tr this case ihe quantization levels corresponding to the firs: and she last quan- tication regions are chosen 4 be ac distance 4/2 frum: úhe two outermest quantization boundaries. “This means that if the number of quantization levels is even. then the quantiza- tion boundaries are 0, EA. EZA,...,(N/2 — 1)á and the quantization levels are given by =á/2,43A/2....,(N — 134/2. Ifthe number of quantization levels is cdd, then the boundarics are given by £4/2.=34/2...., +(N/2- [A and the quartization levels are eivenby 0, LA. 2A,....(N — VA/2, The mfile uq mdpotm is given below. 14 CHAPTER 4 ANALOG-TO-DIGITAL CONVERSION —€CEE tol.pl.p2,93) Sure quanacer furction disteuq. mdpestfunten h,n ht UU MDENT O Beterns abe dito a with queimtiraticm pres ser to the micpome e DIST=UQMDPNTEFENFON E MDELTA. TOLPL IPS & difenssonree demsigo fometõom gávem cm co da vinte at most iherer parameters, pl % funcao se onsumed o be um evem panction % e sugar of the sonrce eme jumeom & 4 PB p3-purimeters of Me Inpa funcsion 1º 1Bsseceltaaino 1) emeri! Toc cas negin-5 ams atear tn/2- acesa: 2) -delta/2. alorati= Htdelta; Mi 1mati)--dotta/2. nesfuge[" t3e- 6º murmêst(ytido 1 22. unha istteval(ç' quad inesfun.é (1) ario CESAR RE ILLUSTRATIVE PROBLEM Uiustrative Problem 4.7 [Uniform quantizer with levels set to (he midpoinis) Find the distertions den a uniforia quantizer is used to quantize à zero-mean, unit variance Gaus- sian random variable. The number o quanazation levels is 1, and the ength of each quantization region is 1. — ED >>>>>———T la ug.mdpaim we substituto 'normal?? for the density function aame, p| = O und tor the densixy funcuon parameters, 2 = 1 for the number of quantization levels, which chocses pm and A = | for he length of the quantization levels, For the parameter 2The name of tas function should be subsrinuted with "nounual” enciudiag che single quotes, 4? Quamizaiton 145 the support set of he density function, we use lhe value b = 162 = 10. anó we choose tãe tolerance to he N0G:. The resuling cistortior 1s 0,083. Nonuniform Quanticution ln nonuniform quantizatioa the reguirement ihar the quantizatioa regions, except the ficst and the ist, have equal lengchs is relaxed. anc each quantization region can have any lengih. Sinee in this sase aptimization is dore under more relaxed conditions, the result is obviously superior to that of unitarm quantization. The opumanty conditions Fax this case, known as he Lioyd-Mar conditions. can he expressed as (430) From these equatioas we zonclude that (he cptimal quantzaton ievels are lhe centroids Of the quantization regions and lhe optima! koundares hesween «be quantizanon regions are the midpoints between the quantization levels. [a order to obrain “he soluticn 19 she Lloyd-Max equation», we start with a set of quantzanos levels é, From this set we can simply find the set of quartization region Soundaries «. From this set of q,'5 à new set of quantization levels can be obtained. This process is continued anti) he improvement in distortion from ane iteraton to another is noi noticeable This algorithm is guaranteed to sonverge to a local minimum, but in general ihere 15 no guarantee tãat the global minimura can be achieved The provedure of desigaing an optimal quantizer is shown in the m-file fioydmax.m, given below. >» function [ay dist]=lloyuimax(funfen bn el.pl.p2,p3) RLLUDMAX Renurna the dba Li Ma quanticer qa the mesm-squared % qutantzateur estar Jor a symmeimo distribuem % dA LDISTISLLOFDMAMPUNFCS. BN TOLPL,22,P3] à he density famchon queen % car depene oh up tu three % parameters, pt.p2,pà % he cector giving sie Duundartos of the auantizemes regions bb! approxmaias support «f te denerty fm na The number ef quente v=The quanncuten feveis Pi pêpêsParcemrere nf fempên Jolzthe reiative error a regina aee) for jatnnrgin—s args=(arg, pintas eng argolas 146 CHAPTER & ANALOG-PO-DIGIFIAL CONVERSION veevalif* varianceifunfen, -b,b. ral args) escore args) for i-2n ya Nayuyz dist-newdit, [ynewdistlcevall'mse diet !funfer a, cal" args); and ILLUSTRATIVE PROBLEM, lltustrative Problem 4.8 [Lloyd-Max quantizer design] Design 2 10-level Lloyd-Max quantizer for a zero-mean, anit variance Gaussian source. €E-————— Using b = 10,4 = 10. t01= 0.01, py = O, and pa = 1 in lloydmax.m, we obtain the quantization boundaries and quanuzation levels veciors a and y as a 0, =2.16, E 1.51, &0.98, 40.48,0 = &2.52, +1.98, 41.22, 40.72,-024 and the resulting distortioa is 0.02. These values are good approxizmations to the optimal values given in the table by Max [2] 43,2 Pulse-Code Modulation Tn pulse-code modulation an analog signal is first sampled at a rate higher than the Nyquist rate, and then the samples arc quaatized. It is assumed that the analog signal is distributed on an interval denoted by [-Xmax. tras), and che number of quantization levels is large. “Tae quantization levels can be equal ur unequal. In the first case we are dealing with a uniform PCM and in the second case. with a nonunifocm PCM Uniform PCM Inuniform PCM the interval [—xmas. Xmas] Of length Z%mas is divided into N equal subinter- vals, each ot length A = 2xmax/N. EN is large enough, the density function of the input in each subinterval can be assumed to be uniform, resulting in a distortion of D = 42/12. TEN isa power vi 2,or N = 2º, then v bits are required for representation cf cach level. This means chat if the bandwidth of the analog signal is W and if sampling is done at the 43. Quantization 147 Nyquist rate, the required handwidih for transmission of the PUM signal is at least »W (in practice, 1.$v!W is close: to reality). The distortion is gives: by (438) If the power of the analog signal is denoted by X7, the signal-to-quamization-noise ratio (SQNR) is given by (43.9 where X denotes the normalized input defined by sx g= + Xmas The SQNR in decibels is given by SQNRig = 484604 js (43.40) After quantization, the quantized levels are encaded using v bits for each quantized level, The encoding scheme that is usually employed is natura! binary coding (NBC), meaning that the lowest level is mapped into a sequence of all 0's and the highest level is mapped into a sequence of'all 1's. Alt the other levels arc mapped in increasing order of the quantized value. The mefile u.pem.m given below takes as its input à sequence of sampled values and the number of desired quantization levels and Ends the quantized sequence, the encoded sequence, and the resulting SQNR (in deibels) function [sgara.quan, codej=u. pernfa,n) RUPCM Uniform PCM encoding of a sequence. [SONHA QUAN.CODE] = LPCMIAN) a = input sequence n = munber o) quuntizuior levels feveny squr = cruspur SOMR fin dj a quan = quantized ouput bejure encoding code = the encuded cuipar REAR 152 CHAPTER 4, ANALOG-TO-DIGITAL CONVERSION The tiwa desired plois are shown in Figure 4.6. ILLUSTRATIVE PROBLEM Tlustrative Probtem 4.12 Repeat Tnstrativz Proklem 4.1: with number of quantization eveis sel once to “6 and set once to 128. Compare the results. EDP o The cesult fer 16 quanúizatica levels is shown in Figure 4.7, and the result for 128 quantization levels is sheven in Pigure 4.8. Comparing Figures 4.6. 4.7, and 4.8, iris cbxious that che Jarger (ho numbe: oé quantiza- ion levels, the smailer lhe quantizanon error, as expected, Also note that for a large number oé quartizanon levets, the relaúun belwecu the ingur and the quantized values tends to à [oe seit) slope 1 passing through the origin; i.e.. the input and the quantized values become almost equal. Sor à smal; number of quantization levels (16 for instance), this relaton is far from equaliny, às shown in Figure 4.7 Nonuniform PCM ln aomeniform PCM the tapua signal :s fics: passed ihrough a norlincar element to reduce us dynamic range. and che oucput is applied tc à unitorm PCM system. AL the receiving end. tn OUIpUE IS passed through the inverse al tre noalinear element used in the transmitter “The overall esfect is equivalent to u PCM system with nanuniform spacing between levels In gereral, for transmission af speesh signals. the ronlinearitios ihat are employed are either sela or A-law aonlinearities A ylaw nonlinearity is definc by she relanon e go BUT gr) (2311) Togil + 4) where x is the aormatized input (x| £ 15 and já às a parameter that in standard pla aontizsarity is equal to 255. A plot of this nonhnearity [or úfferent values of é is shown in Figure 4.4 The inverse of ju-law nontineavity 1s given by (4.3.12) The two m-files mulawem and iavmulavem given below implement g:-law nonlinearity and tis inverso. “di Suetion [yafeneclaw (saca) SOMULAW melao nontineneity for monuniform PCM, % 7 — MULAWOME) % X* input vector Figure 4.7: Quantization erros for 16 quantizatiun levels. 154 CHAPTER 4 ANALOG-TO-DIGITAL CON Figure 4.8: Quantization error for 128 quantization levels “3, Quantization Figure 4.9: The u-law compander. aemaviabs(ok. logCL+musabs(x 0) log(1+mo)) asignumé. o function xeinvemvlaw(y mu) SeINVMULAR The inverte of muiaw nonbncarioy RASINVMULAWEEMU) Y = Normaiized cumes of ste mula nontinearira K=itTlAmo. (abs(y)= 1), /mU) esigrumivy; The m-file mula pom.m is lhe equivalent of the im-file u.pem.m when using a p-law PCM scheme. This fike is given below. funesion [sque.2 quan, code] =mula. penta. mu) SEMULAPCM mulum PCM encoding vj à vequence ISONRA LUANCODE] = MULA PCAANMU; a = input sequence a = number of quemiza dqnr — utpas SON ton levels fever) am aRaAR 156 CHAPTER 4. ANALOG-TO-DIGITAL CONVERSION % a quam = quersized cuia befere encudding % code = ie encoded crupar Lreratamuml= mulaw(a,rma): ts aqunymula a-quan=maxemumes quam, senr=20elog 1 Gtnoranta) Pnorita—a quam); ILLUSTRATIVE PROBLEM Ilustrative Problem 4.13 [Nonuniform PCM] Generate 2 sequence of random variables ot length 500 according to am A(O, 1) distrikution. Using 16, 64, and 128 quântization Jevels and a qt law noniinearity with é = 255, plot the errar and the inpur-outpu relation for the quantizer in each case. Also determine the SQNR in each case. —E——————————— LeLibe vector a be the vector of length 500 generated according to WO, |), ie., a =: randn(500) Then by using |dista.quan.code] = mula pem(a, 16. 255) we can obtain the quantized sequence and the SQNR for a 16-level quantization. The SQNR will be 13.76dB. For the case of 64 levels we obrain SQNE, = 25.89 dB, and for 128 levels we have SQNR = 31.76 dB. Comparing these results with the uniform PCM, we observe that in all cases the gerormance s inferiur to the uniform PCM. Plots of the input-ouiput reiation for the quantizer and the quantization error are given in Figures 4.10, 4.11, and 4.32. Comparing the ingut-output relation for the uniform and the nonuniform PCM shown in Figures 4.7 and 4.10 clearly shows why the former is called unifarm PCM and the Iatter is called nonuniform PCM. From the above example we see that the performance of the nonuniform PCM, in this case, is not as good as uniform PCM. The reason is that in the abuve example the dynamic range of the input signal is not very large. The next example examines tho case where the performance of the nonuniform PUM is superior to the performance of the uniform PCM. UMA VRATIVE PROBLEM Tlustrative Problem: 4.14 The nonstationary sequence a 0 length 500 consists of two parts. The first 20 samples are generated according to a Gaussian random variable with mean O and variance 400 (9 = 20), and the next 480 samples are drawn according to à Gaussian random variable with mean O and variance 1. This sequence is once quantized using a uniform PCM scheme and once using à nonunitorm PCM scheme. Compare the resulting SQNR in the tiwo cases 4.3. Quantization Figure 4.10: Quantization error and input-output relation for à 16-level gt-lawy PCM. 157 162 CHAPTER <. ANALOG-TO-DIGITAL CONVERSION Letter Prebability 006 0057 DOE 0.0317 0.31 0208 0.01 0.0467 00575 | 0.0008 7 D.0049 vos 00198 0.0574 0.0635 00152 0.0608 0.044 0514 0.0796 0.0228 0.0085 0075 00013 00164 0.0005 Word space 0.189 NI=<|=[ gi<iofntomolsio)z|zicla im] Table 4.1: Probabilities of letters to printed English. 43 Quantization 163 48 For a zero-mean, unit variance Ganssiun source, design optima! nonaniform guantizers with a number of levels Nº = 2.3,4,5,6, 7,8. For cach case determine F7( E), the entropy Of the quantized source, and R, the average codewore length ot a Euffman code designed forthat source. Plos H(Ê), R, and log, N as a function o! N on the same figure. 49 A Laplacian random variable is defined by the probabiltty density funcuion where À > 95 à giver constun. à Verify that the varianco of à Laplacian random variable is equal so 2 db. Assuming À = 1. design umiform quantizers with N - 2,3,4,5,6,7,8 levels for this source. As usuat, take lhe interval of ins —100, 100] where o is the standard deviazion of the sauree, estto €. Plor the entropy cf the quantized source and log, N as funetions of N on the same figure 4.10 Repeat Problem «9, substituting the unstorm quentizer with the cptimal nonuniform quantizer. 41! Design an optmal 8-leve! quantizer for a Lapiacian source and plot the resulting mean-square elistortion as a function of à as 2 changes in the interval 70.1, 5). 412 Design optimal noruniform quantizers with N = 2,3,4,5,6,7,8 for the Laplacian source given in Problem 4.9 with a À = «/2 (note that tbts choice of À results in a zero- mean, unit vartance Laplacian source). Plot the mean-square error as à function of N for this source, Compare these resulis with thosc obtained from quamtizing a zero-mean, unit variance Gaussian squrce. 4.13 The periodic signal x(r) has a period of 2 and in ths interval 10, 2] is detined as Ft) — . =1+2, |si« & Design au B-level uniform PCM quuntizer tor this signal and plot the quantized output of his system. - Plor the quantizatiun error tor this system By calculating the power in the error signal. determine the SQNR for this system in decibels. d. Repeal parts a, h, and e using a t6-level uniform PCM system. vas CIHAPTER 4 ANALOG-TO-DIGITAL CONVERSION 4.14 Generate a Gaussian sequence with mean equai to D and veriance equal to L with 1000 elements. Design + E-level, Ló-leve, 32 level, end 64-levet uniftrm PCM schemes for his sequence and pler che resulting SQNR Gn dezibels) as a fenctior of thz number o x allacated to each source output + » 4.15 Gencrate a 2ero-mean. unis variance Gaussián sequence with à length of 1900 and guamtiee il asing 4 6 trt per symbal uniform PCM scheme. The cesulting 6000 bits are tranoristeé to lhe receiver via a noisy channel. The error probability of the ciannel is denated by p. Plet the overall SQNR in decibels us a funchon cl p fer values of p = 1054,5 x 1053, 1052,5 x :052,0.1.0.2. Fer sunulation of he effect of noise, you can generate binary random sequences witi these probabilities and add them imodulo 23 to he nicaded sequence 4.16 Repear Probiexe <. 3 using a nonunifoor e law PEM with é = 417 Repeat Problem « sing a nonuniform ;a-law PCM ven 4 = 255 4.18 Repeas Problem 4.15 using a noruncfoem ju-Jaw PCM vit qu -- 255. Chapter 5 Baseband Digital Transmission 5.1 Preview a this chapter we consider several buseband digital modulation and dermodu!ation tech- niques far transmitting digital information rrough an additive vinte Gaussian noise chan- nel. We begin with binary pulse modulation and then we intraduce several nonbinary medulation methods. We describe the Optimum receivers for these different signais and consider the evaluation of their performance in terms of the average probabilicy of error. 5.2 Binary Signal Transmission In a binary cominunicativo system, binary data consisting of & sequence of O's and 1's ase transmitted by means of two signal waveforms, say, so(?) and s1(1). Suppose that the data rate is specified as À bits per second. Then. each bit is mapped inta à correspanding signal waveform according to lhe rule d> sl), Ost<Th T+ sto. U<i<% where 7; = 1/R is defined as the bit time interval. We assume that the data bits O and 1 are equally probable, ie., cach occurs with probabihty 4, and arc mutually statistically independent. The channel through hich the sigraf is transmitted is assumed to corrupt the signal by the addition of noise, denated as n(t), which is à sample function of a wixte Gaussian process with power spectrum No/2 wattshertz. Such a channet is called an additive white Gaussian noise (AWGN) channel, Consequently, the received signal waveform is expressed as Fls) = (0) + dt), i=0,1, 0O<r<T 621) 165 166 CHAPTER 5. HASEBAND DIGITAL TRANSMISSION The task of the receiver is to determine whether a Dora L was transmitted afier observing tac received signal +(t) ix the interval U < + £ Th. The receiver is designed 10 mmimze the praoabuliry of error. Such a receiver is called the oprumum receiver 5.2.1 Optimum Receiver for the AWGN Channel Io nezrty all basic digital communication texts, 1t 15 Shown that the optimum resejves for the AWGN chanrel consists of two ku Iding blocks, One matcherl filter. The ather is a detector «ither a ségral correlator ora 5.2.2 Signal Correlator The signal correlator cross-correiates the cece xisted signals sete) and sy(0), as illus computes the two ontputs ved signal r(7) with tãe two possible trans- tec in Figure 5 E That is, (ne signal correlater min = f ristndr o no ritysindr (5.22) o inthe interval O < 4 < Tp, samples the two vutpuis att = T> and feeds the sampled outpus 19 the detector. a Soodr ' so i "9 t Deiector |—» Oueput data sen : KOdr Ao ] Sample —— au Figure 5.1: Cross correlation of the received signal rir) with the two transmitted signals. ILLUSTRATIVE PROBLEM Nustrative Problem 5.L Suppose the signal waveforms sait) and sy(4) are the ones shown in Figure 5.2, and let spt?) be (he transmiticd signal. Then. the receiver signal is &2. Binary Sinal Transmission 167 sun La. 0: Te Í Figure 5.2: Signal maveforms sp(s) and s1€:) fer a binary communication system = mino, Dxiei (5.231 Detesmine the coretator cmpuis at the sampling instanta When the signal ni?) is processed by the two signal curreletors shown in Figure 5.1, the “utputs r9 and 1 at the sampling insiam : = To are mn n=[ Fist dr o -[ oa f nébso(t) dr o y =E tra (52.4 and n E) Hos(ndr U T Tm -[ solte ar+ nús dr o =" 15.25) where ng and nj ate the noisc components at the output of the signal correlatos, m= f n(tsa(r) de ) n n1 =[ (eso de o 1 CHAPTER 5. BASEBAND DIGITAL TRANSVIS ON 5.2.4 The Detector “The detecior observes the correlator or mateked Áltor outpuis 7y and 71 and decides on swltether the transmitted signal wavcform is eilher s0(1) or sit5), wnizh currespond to the trarsmission of either a ora 1. respectively, The aptomem detector is defivecl as the detector that minimizes the probability of error ILLUSTRATIVE PROBLEM Iustrative Problem 5.3 [er us consider the detector tor the signais shown in Bigure 5.2, weiich are equally probable and have equai energies. Tue opimum detector far chese signais compares ra and 7: and decides that a O was “ransmitted wiher r9 > r1 and thata | was transmincd when rm > ro. Determine the probability of error —- ED ——————————— When so(t) is lhe transmitted signal swaveform, the probability cf ror is Pe=Ptri>m= Pla E + ny: Pim ny» Ed (5.2.19 Since ny and no are zero-mean Gaussian random variables, iheir difference x = m — 015 also zero-mean Gaussian. The variance of the random variable x is Elen — ne But E(nin9) = U, because the signal waveforms are orthogonal. Thatis, = E(n7) + Eing) — LElnino) (52.20) PT fo Equnç = f ) sotosdn(Onto ddr No ph ph 5) Í sotr)s(r)Btt— ridrai 2d dh T, E [ ses (5,220 Therefore. (5.2.22) (5223) 52 Binary Signa! Transmission 13 The ratio E /No is calied che signal to-noise ratio (SNRY r Ê I t f | F o 3 o 10 logia ESNo Figure 5.7: Probability bf erzor for orhogenal sipnals “The derivation of the detector performance given above sxas based on the Iransmission of the signal waveform so(i). The reader may verify that the probability of error that 1s obtained when s: ft) is transmitted is identical to that obtained when so(t) is transinitted Because the O's and 1's in the data sequence are equally probable, the average probability of error is that given by (5.2.23). This expression for the probability of error is evaluated by using the MATLAB senpt given below and is plotted in Figure 5.7 as a function of the SNR, where the SNR is displayed on à logarithmic scale (LO log1g E / No). As expected, the probability of errar decreases exponentially as the SNR increases DD ———— % MATLAB seript that generates the probabitins nf ermr versus the signalito-noise ratio drtia] sor final.sor=15, snr.step=0.25: al. sn 30r.step final.sar, lengnhçsnr tm dB, snr=10"(sar.in dR(/103: Peti=Qunet(sgrtsnri: end; semilogy(snt in dB ey; pa CHAPTER 5. BASEBAND DIGITAL TRANSMISSION 5.35 Mente Carlo ulation of a Binary Communication System Morte Carlo computer simulations are asuully performed in practice te estimate the proba- hility of ecear ol a digita! communication system. especially 1n cases where the analysis of lhe dezectar performance is difficult 10 perform. We demonstrate the method for estimating the probability of error fur the binary communication sys:em destribed above. ILLUSTRATIVE PROBLEM Hlustrative Problem 5.4 Usc Monte Carlo simulation ta escimate and plot P, versus SNR for a Sinary communication system lhatemploys correlators er matched filters. “Lhe model vt the system is illustratert in Figare 5.8 | Urifoem candom Giaussian cardom Mumiier gererqtor number generator Output data WE Detector i VE 4 > Gaussian random number generator Figure 5.8; Simulation model tor INusirative Problem 5.4, We simuiale the generation of the random variables 29 and 14, which constitute the input to the detector We begin by gencrating a binary sequence of 0's and 1's that cecur with equal probability and are trutrally statistically independent. To accomplish this task, we usc a random number generator that generates a unifucm random number with a range (0, 15. If the number generated is in the range (0, (1.5), the binary source output is a O, Olherwise, tis a 1. 1a O is generated, then ry -=.E + no and +; =m. If 1 is generated, thenry =ngand | = E pr The additive noise components n9 and ny are generated by means of two Gaussian noise generators. Their means are zero and their variances are o? = E No/2. For conventence, 5.2. Binary Signal Transmission rs we may normalize tas s:vnal energy E to unity (E = 1) and vary 02, Note that the SNR. which is detined as E / Ny. is then equal to 1,292. The detecior Output is compared with the binasy iransmittcê sequence. acd ar error eounter is used to count the number of bit errars Figure 5.9 illustrates the results of th.s siriulatior For the transmission of N=10,000 drts at several different values of SNR. Note che agreement between the simulation results and the theoretical value ví 2, given by (5.2.23). We should also note that a simulation DÊ N=10.000 data his allows us 10 estimate the error probability reliably down to about Pe = 107%, tn other words. swith N=HO.OUO data bits, we should have at east !O exroes for a reliable estimate of 4. MATI.AB seripis for this problem are given below. o MATLAB scope por Mieinunve Problem 4, Chapter £ echa on SNRindB1=0,1.12: SNRimdB2=00 1:12; for i=iilengiiSNRudB!; % sumularea! arror ute smld err -pebvii=smlgPeS48NRingE Lin; end tengihiSNRind2), x PE SNRiadB3isdoag 1934 10), % dhenresicai err rute hem. oem prin Qlunct(sqr(SNRY, end; do pleno command juthos, semilogy(SNRingB 1 sumld.em pr, ep hot semilogy(SNRindB2 aheo. err. pe TN ion plesmidPaSaisur.in dB) du % fpj — amiuPeseiene tn dB) % SMLOPESS fed the probubihiy of error far the given % eme an dB ssprad-tsiso ratio in dB, E SNRecxpisar im 98 elog(10)7103, “o cignal.re-noise rate sgma-E /squiZASNRY Ge rigmia. andar! deveation of noise Ne 1000: % generaston mf dhe binare data ronco for 1=1.N temperand; % 4 uniform random vuriabie over (Qi) if «emp< 0.5) dsuurcoritel e cith prubabalico 192, source outpu? is & else dsgurveii Ke mith probubaisty 1/2, rource encon is 1 end end; % desecnan. und poobubuisty of ervar calcudasion mumotegr-O: foras 16 CHAPTER 5. BASEBAND DIGITAL TRANSMISSION Fe muchos if (dsoueeetid==0), = Bsgnganssisgra): ugausslegrma: Do the source omtpuris O" ouipure else rO=angauss(egona) rl=B+gngnvssisgma); do df she amaro impuros "1 end. %o detector foltoms if drtato deciscO: de deciciom de "07 exe veciset, decision ir ipa ie an error increase the error counter murrotercanumaferr+; end: end p-numoferaN; 2 pemhabitios mf error esiimase 1Dlogig E/Ny Figure 5.9: Error probability from Monte Carlo simulation compared swith theoretical error prohabifity for arthagonal signaling, In figure 5.9, simulation and theuretical results completely agree at ow signal-to-noise ratios, whereas at higher SNRs they agree less. Can you explain why? How should we change the simulation process to result in better agreement at higher signal ta-noise ratos? 52 Binary Signal Transmission 1” 5.2.6 Other Binary Signal Transmission Methods Tae binarv signal transmiss:on method described ahove was based on the use of orthogonid signels. Beltw, we descr.be cwc other methods for transmitting binary information through a communication slannel. Ore method employs antipodsl signais. The other method employs an on-ofi-type signal 5.2.7 Antipodal Signals for Binary Signal [ransmission Two signal waveforms are said to be antipodat if one signal waveform is the negative of the other. For example, one pair ví ancipodal signals is illustrated in Figure 5.:0(a). A second pair ás illustrated in Figure 5.0) tm ut asi i asi A A ar A bo, % D no] o] a Wlio qdo taj A pair ot antipedal signade 1b) Another pair of antipodal signals Figure 5.10: Examples cf antipodal signals. (a) A pair of antipodal signals. (b) Another pair of antipodal signals. Suppase we use antipodal signal waveforms sa(?) = 5(*) and si(1) == —s(f) lo sramsmit binary information, where s(1) is some arbitrary waveform having energy £. The received signal waveform from an AWGN channel may be expressed as ni = Estr) + ati). Os:<5, (5.2.24) The optimum receiver far revovering the binary information employs a single correlator or a single matched filter matched to s(1), followed by a detector, as illustrated in Figure 5.11 Let us suppose that (7) was transmitted, so that the received signal is rin=sto+nt) (5.225) The output of the correlator or maiched filter at the sampling instant =: Ty is r=E+n (5.2.26)