Bp Lathi Modern Digital And Analog Communications Systems Solutions Manual

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Bp Lathi Modern Digital And Analog Communications Systems Solutions ManualTo browse Academia.edu and the wider internet faster and more securely, please take a few seconds to upgrade your browser. READ PAPER Download pdf. Please choose a different delivery location.Our payment security system encrypts your information during transmission. We don’t share your credit card details with third-party sellers, and we don’t sell your information to others. Please try again.Please try again.Then you can start reading Kindle books on your smartphone, tablet, or computer - no Kindle device required. In order to navigate out of this carousel please use your heading shortcut key to navigate to the next or previous heading. Register a free business account If you are a seller for this product, would you like to suggest updates through seller support ? Amazon calculates a product’s star ratings based on a machine learned model instead of a raw data average. The model takes into account factors including the age of a rating, whether the ratings are from verified purchasers, and factors that establish reviewer trustworthiness. Therefore \n \n Using similar arguments.If g(t) is a n even function of \nt.This is readily seen from the fact that replacing t by - f in the \nfirst component yields the same function. The same maneuver in the serond component yields the negatis.The funct,ion in part (a) is a gate function centered at the \n \n origin and of width 2. The function in part (h) can be expressed as A. For the sinc pulse s inr(g).This multiplyicg iactor \nrepresents a linear phnse spectrkrm -2nd. In this alternate procedure.Hence the filter is unrealizable. Also \n\n \n Figure S3.5-1 \n \n Hence the filter is noncausal and therefore unrealizable. Since h(t) is a Gaussian function delayed by to.This yields \n n frequency of 22.736.4 Hz. \n \n 3.5-4 There is a typo in this example. Hence, the transmission may be considered distortionless. Hence, if g ( t ) is the input. Hence \n \n 28 \n\n \n In evaluating the integral.http://www.agriturismomonteistulargiu.com/userfiles/brawl-busters-ph-manual-patch.xml

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Both the components will be suppressed by the filter.Hence, this system will not do the desired job. \n (c) The readel may verify that the idcntity for cosndct contains a term cosw,t when n is odd.During the negative half cycle.I t alone passes through the bandpass filter \n(centered at 400 kHz). All the other terms arc slippremed.By blocking the dc component A from this \nout put.To show that the energies of r r c ( t ) and t r r h 0 j ale; \nequal.Hence, the filter output no(t) is given \nby \n \n Observe that if both Ai and A are zero.This changes the sound \nof an audio signal slightly.Such distortion may change the quality of speech.In video signals and data transmission.In the present case. The reascjn for \nthis is as follows. When the receiver is tuned to 590 kHz. The sum carrier is suppressed. The sum \n component is suppressed by the IF filter.For the next n \n seconds. Then instantaneously.Doubling the frequency \nof r r r ( t.This spectrum also deca:;s \nrapidly away from the origin.Therefore the Nyquist rate for gi2(t) \nis 400 kHz.Clearly the \nbandwidth of the composite signal is the higher of the two, that is. 100 IIz. The Syquist rate is 200 Hz \n. The bandwidth of the sum is the higher of \n the two. that is. 60 Hz. The Syquist sampling rate is 120 IIa. \n(el \n \n The two signals have bandwidths 25 Hz and 50 Hz respectively. Rom width property of the convolution. It is an even function of t. Consequently, the Xyquist rate is 10 Hz, that is.In the second case. we use \nthe Syquist sampling rate of 10 Hz (Fig. S6.1-4e). The spectrum G(w) consists of b3ck-to-back. Hence, G(y) can be recovered from C ( w ) using an ideal lowpass \nfilter of bandwidth 5 Hz (Fig. S6.1-4f)-The output is 10si1lc2(5rt). Finally, in the last case of oversampling \n(sampling rate 20 Hz).The corresponding amplit.ude response H ( J ) and the ideal amplitude response (lowpgrs) \nrequired for signal reconstruction is rhowa in Fig. S6.1-7c.http://clap-bas.com/userfiles/bravopro-xi-manual.xml \n ( c ) A ininimum of T secs delay is r uired to make I r ( t ) causal (realizable). Such a delay would cause the \nrecotistructed signal in Fi. SB.1-Ta tSbe delayed by 7 usr. \n (d) When thr input to t f e first filter is b ( t ).Hence, each quarter cycle \ncontributes identical energy. Consequently, we can compute tohe power for this signal by averaging its energy \nover a quarter cycle.However, for \nthe bipolar case, the bandwidth nmainr unchanged. There are eight equally likely combinations: 11 1,101, \n110,loO, 01 l,OlO,oOl, and 000. Hence we require only -A M-zvy \n \n Jog2 M \n pulsedsecond. The first ASK signal \noccupies a band fd f 1 MHz, and the second ASK \nsignal occupies a band f,, f 1 MHz.Draw the output of the DSU for this signal. The first 1 is already drawn. \nShow the bit stream which is substituted by the DSU. \n Solution: \n \n 8.1-6 T-1 synchronization at two distant locations is controlled by separate crystal controlled \noscillators which differ in frequency by 125 parts per million. Also, \nlet xo2 and x 4 represent events that the second ship drawn is an aci l l l ta and a PLL, respctively. H a w, we have 43 r 6 combiiatiom in which exactly 5 numbers \nmatch.There arc 4 sub. Hence there are4 r 1287 ways of getting a \nflush. Therefore, \n \n 10.1-13 Sum of 4 can be obtained as (1,3), (2,2) and (3,l). The two dice outcomes are independent. However, it is a unilateral (rectified) version of \nGaussian PDF.For example, I I \n amplitudes of all sample hctions are zero at same \ninstants (one is shown with a dotted tine. Others are not valid PSDs. PSD is always a real, non-negative and even \nfunction ofa). A typical sample \nfunction of this process is shown in Fig. Sl12-10. The signal x(t) changes abNptly in amplitude at random \ninstants. The average number of changes or shifts in mpliides arepper second, and the number of \nchanges are Poisson-distributed. The amplitude aAer a shift is independent of the amplitude prior to the \nshift.http://superbia.lgbt/flotaganis/1647550002 It can be shown that this process is \n \n stationary of order 2. Htnce, p(x;t) can be expressed as p(x). The probability of k shifts \nin r seconds is given by \n \n whm then are on the average Pshifls per second. Moreover, we are given that the \namplitudes before and after a shift an independent. Hence the \nautocorrelation function of R,(r) is the sum of autocorrelation function of each term. They are shown in Fig. S11.5- 1. Also, 2 is the area under. \n \n Fig. S11.5-1 \n\n \n 11.5-2 We follow a procedure similar to that of the solution of Rob. \n1 1.5-1 except tha! the center ikcquencies arc different. Hence we shall choose h(t) to be \n \n p ( 5 - t) rather than 2 p ( q - r). In short, the detector is a phase comparator.The distance bcwen these signal pairs is always 2 f i. Hence, a picture is worth 1000 words \nis very much an underrating or understating the reality. \n \n 15.1-4 (a) Both options are equally likely. This shows that white Guassian noise is the wont possible kind of \nnoise. \n\n \n f \nr Chapter 16 \n i \n \n 16.1-2 (a) There are (7) ways in which j positions can be chosen Imm n. But for a ternen' d e, a digit can \nbe mistaken for two other digits. Ihe other two pattern will not be cbmcted. H' is chosen to have 12 distinct \nrows of four elements with the last 4 rows forming an identity matrix. Then are 12 single enor patterns. Therefore, it can correct all 1-error \npatterns. B i r also yields a remainder I. This means it cannot have a third order factor either. It docs not divide either. Hence this is a single m o r correcting code. \n \n Row 4 is ok. \nRow 3 is left shift of row 4. \nFor row 2, left shift row 3. \nAnd add row 1 to obtain row 2. \nFor row 1, left shift row 2. \nAnd add row 1 to obtain row 1. \n\n \n 16.61 The burst (of length 5 ) detection ability is obvious. The single emr conecting ability can be demonstrated \nas follows. If in any segment of b digits a single error occurs, it will violate the parity in that segment.http://fccsms.com/images/bp-dl750-manual.pdf \nHence we locate the segment whm the aror exists.By checking which bit in the augmented segment violates the parity, we can locate \ntbe wrong bit position exactly. \n \n 16.5-1 The code can conect any 3 bunts of length 10 or less. It can also correct any 3 random mors in each code \n word. \n \n To achieve a value 9.872 x for PE,, we need new value Eb JcAI say El Id.For parts (a) and (b) To explain these results in terms of vector cotacepts we observe from Fig. 2.15. Because of this orthogonality, the length-square of To explain these results in terms of vector concepts we observe from Fig. 2.15 that Ct'e can verify this also nnalytically. Let us first c0mput.e the energies of all the signals If g(t) is a n even function of S2.8-4c shows tlie plot of C, and 8,. This is readily seen from the fact that replacing t by - f in the The same maneuver in the serond component yields the negatis? of To Half wave symmetry. Hence S2.9-2b. By inspection of exponential spectra in Fig. S2.9-la, we obtain (b) The exponential spectra are shown in Fig. S2.9-3. Moreover from ( I ) Therefore from the result in Pmb. The funct,ion in part (a) is a gate function centered at the This is a triangle pulse centered For the sinc pulse s inr(g).Adding pails 7 and 12 in Table 3.1 and then dividing by 2 yields the desired Thcl.efoi e There forc Therefore Fig. S3.3-6. This multiplyicg iactor In this alternate procedure.Using the same argument as for part (b).The two spectra are nono\er lapping because Fig. S3.S-10 This is noncausal. Hence the filter is unrealizable. Also Figure S3.5-1 Since h(t) is a Gaussian function delayed by to. it looks The time delay tolerance rhould be 4 instead of 1. From Exmnple 3.16. The gain variation over the band is only 1.99. Similarly, we find the time delays at the band edges as Hence, the transmission may be considered distortionless. The. We also find. Hence, if g ( t ) is the input.http://andreagarciam.com/wp-content/plugins/formcraft/file-upload/server/content/files/1627f18bc48d62---brother-vx-808-manual-free.pdfHence Hence, the transfer function of the entire system is The renlaining Both the components will be suppressed by the filter. whose Hence, this system will not do the desired job. This is not true During the negative half cycle.Therefore, the diodes act as In the present case. Hence, the output is Fig. 54.2-8 I t alone passes through the bandpass filter All the other terms arc slippremed. The desired output is Hence the output of the lowpass Over a quarter cycle m(t) car1. S4 3-3) Hence. Clearly, if the input is By blocking the dc component A from this The siqnal at point b is Only when t r r ( 1 ) is near its peak, this condition is violated. Hence. From the signal l t l ( t ), we see If wr apply rn h ( t ) at the input of the Hllbert. To show that the energies of r r c ( t ) and t r r h 0 j ale. T h e prodeuct of the incoming signal and the local carrier is cd(t). Hence, the filter output no(t) is given This changes the sound Such distortion may change the quality of speech. but In video signals and data transmission.The two frequencies. In the present case. The reascjn for The outpul i'. The sum carrier is suppressed.For the next n For PSI: This cycles Then instantaneously.The cycle repeats periodically with period lo-' as shoal1 in. Fig. S5.1-2a. For PM. For convenience. we select the Therefore, OOOx sin 2000rt For P M: TO find B P M.Such a pulse and its spectrum are depict.ed in Figs. Doubling the frequency This spectrum also deca:;s This yields Therefore the Nyquist sampling rates Therefore the Nyquist rate for gi2(t) The bandwidth of the first term on the right-hand side is 50 Hz and the second term is 25 Hz. Clearly the The bandwidth of the sum is the higher of The spectrum of the product of two signals is. Rom width property of the convolution.Hence, the Fourier series for the pulse train can be expressed as Proctlcal Filter Consequently, the Xyquist rate is 10 Hz, that is.www.demirdokumservisiankara.com/image/files/bosch-vip-1000-manual.pdfWe present this information in the following Table for three sampling In the second case. we use Hence, G(y) can be recovered from C ( w ) using an ideal lowpass Finally, in the last case of oversampling This acts as the input to The impulse response of the circuit is a rectangular pulse.It sample generates a rectangular pulse at the output.The corresponding amplit.ude response H ( J ) and the ideal amplitude response (lowpgrs) Such a delay would cause the The Nyquist rate is 30 kHz. The SNR for this value of L is - Fig. 56.2-7 Hence, each quarter cycle Consequently, we can compute tohe power for this signal by averaging its energy The power or the mean squared value (energy averaged over N o (In 101)2. The SSR for this value of L is In part (a). the Hence, an increase in the SNR For a bipolar case. we need a The PSDs of the thee ELK.The bandwidths for the three sun, when half-width pulses arc used, are as follows. Clearly, for polar and on-off cases the bandwidth is halved when full-width pulvs are used. However, for The puke shape has only a minor influence in the There are eight equally likely combinations: 11 1,101, The last six combinations have either the first andlor the last bit 0. The fvst two combinations an the only ones which Thus ak are 1 and 1 or - 1 and - I, respectively, yielding Hence the first digit is I. The For on-off case, half the pulses an zero, and for half-width rectangular pulses, the aansmitted power is: Hmce, the bandwidth is reduced by a factor of Hence, we need Hence the average mergy of the M-ary pulse is Hence we require only -A M-zvy Hence, each pulse must The first ASK signal Hence, the Hence, there are a total of 16,800 sampledscc. This combined If yes, why? If no, why not. Solution: No. The bit sequence 001 1 cannot be in an ESF framing bit sequence. The first 1 is already drawn. Show the bit stream which is substituted by the DSU. If the terminal equipment doesn't Circle the correct answer.https://www.airportlimofortlauderdale.net/wp-content/plugins/formcraft/file-upload/server/content/files/1627f18d30ab92---brother-vx707-sewing-machine-manual.pdf a) 5 How long, in terms of TI bits, will a correction signal take to be transmitted to The probability that the numberiappws should be ki not ki. AII these outcomes arc equally likely with H a w, we have 43 r 6 combiiatiom in which exactly 5 numbers For the system in Fig. a. The system fails if the upper and lower branches fail simultaneously.We may consider this system as a cascade of two subsystems xi and x2, whm x, is the parallel combination Let PJ (xi) be the probability of failure ofx. Then Hence, the probability of system not failing Nupbcr of ways of drawing 5 cards of. There arc 4 sub. Hence there are4 r 1287 ways of getting a The two dice outcomes are independent. Let xl be the The probability one such Since the event can occur in 10 ways, its probability is. Hence, the probability of at least three 1's occuring is. Consider case of mor over the first link However, it is a unilateral (rectified) version of. Gaussian PDF. Hence, we can use the expression of Gaussian r.v. with a multiplier of 2. For a gaussian r.v. Taking Fourier transform of both sides, we have Hence, the discriminant of this Hence, x and y are not independent. From Eq. (10.87). we obtain - For example, I I Hence, the Using this result, we obtain Time means of each Jarnple function is different and is not equal to the Others are not valid PSDs. PSD is always a real, non-negative and even Hence, R,(?) is a But because of symmetry of 1 and 0, Now, k m Eq. 1 129, Because of duobinary rule, the neighboring pulses must have the same polarities. Hence, Hence, The amplitude aAer a shift is independent of the amplitude prior to the. It can be shown that this process is The probability of k shifts The probability of no shih is obviously po(r), where Moreover, we are given that the Substituting. Eqs. (3) and (4) in Eq. (2), we get Hence follows the Since the two sources are incoherent, the principle of superposition applies to the PSD. Hence, They are shown in Fig. S11.5- 1.https://www.nationaalgolfcongres.nl/wp-content/plugins/formcraft/file-upload/server/content/files/1627f18eaf0948---brother-vx1435-user-manual.pdf Also, 2 is the area under. S 1 1 5-2. In all the rhne cases, the area under Snc (a) is the. Thus in all 3 cases - - Hence, a reasonable value of time-delay required to make this filter 75 For BSB-SC, Hence, - T h m should be 4x2 in the denomlumr (see below). From Eq. (12.42a) In this case, the demodulator output signal, when passed through the Let w consider. This signal is C I Hence, Hence we shall choose h(t) to be S13.5-2. Also, and Hence Hence, We wish to determine TIIK covarinse matrix is: (0x1 x2 1) and In short, the detector is a phase comparator. It chooses that signal which is Hence Note that The minimum energy equivalent set of orthogonal signal set in Fig. (c) is also given by the set in Fig. (b). The set in Fig. (b) is polar This set is shown b Fig. S14.4-3a. The signal set is now rotated so a to yield a new set shown in Fig. SI4.4-3b. a Hence, the minimum energy signal set is. The minimum energy signals ate The information per picture M e was f a d in Problem 15.1-2 to be 9.96 x 10'bits. Obviously, it is not Hence, a picture is worth 1000 words For ternary coding, we need one dummy message of probability 0. Thus, Hence, it follows I(xIy) 2 I(xlz) 1 M Hence, The signal and noise power S and N respectively an i The maxknum channel capacity over this band ly is given by Using Equation 15.37, we obtain Under this condition, 15.5-5 In this problem, we use the results of Problem 15.5-4. Under the best possible conditions, Thus, we wish to maximize Using Equation 15.37, we have This shows that white Guassian noise is the wont possible kind of But for a ternen' d e, a digit can Hmce the number of possible mrs in j places is Since the Hamming bound is Ford - 0 Such a code can conect up to two errors using majority rule for If a single error occun anywhere in the code word, the parity is violated. Therefore this code can detect a single cmr. Hence O l O l l l O l O. O l l O O l O l 1 One possible is: Ihe other two pattern will not be cbmcted.www.delhigurgaontrophy.com/userfiles/files/compact-4-door-manual.pdfH' is chosen to have 12 distinct Then are 12 single enor patterns. Hence we may be Therefore, it can correct all 1-error Then are 12 single enor patterns. Hence we may be Therefore, it can correct all 1-error A similar procedure is used to fmd the remaining codes (see Table I). This means it cannot have a third order factor either. Hence. It does not divide. So by dividing by. It does not divide. Next try dividing by See table below. O l l O l d O Thw, so far we have T'he 1 is eliminated by adding row 4 to row 2. Hence this is a single m o r correcting code. Row 3 is left shift of row 4. For row 2, left shift row 3. And add row 1 to obtain row 2. For row 1, left shift row 2. And add row 1 to obtain row 1. 16.61 The burst (of length 5 ) detection ability is obvious. The single emr conecting ability can be demonstrated Hence we locate the segment whm the aror exists.It can also correct any 3 random mors in each code. Get started with a FREE account. Let us make life easy on us. Let us be loved ones and lovers. The earth shall be left to no one. ” ? Yunus Emre Istar'?n kaleminden (Turkish Edition ) Get books you want. To add our e-mail address ( ), visit the Personal Document Settings under Preferences tab on Amazon. Some features of WorldCat will not be available.By continuing to use the site, you are agreeing to OCLC’s placement of cookies on your device. Find out more here. Numerous and frequently-updated resource results are available from this WorldCat.org search. OCLC’s WebJunction has pulled together information and resources to assist library staff as they consider how to handle coronavirus issues in their communities.However, formatting rules can vary widely between applications and fields of interest or study. The specific requirements or preferences of your reviewing publisher, classroom teacher, institution or organization should be applied. Please enter recipient e-mail address(es). Please re-enter recipient e-mail address(es). Please enter your name. Please enter the subject. Please enter the message. Author: Ron Taylor; Tao Zhen; B P LathiPlease select Ok if you would like to proceed with this request anyway. All rights reserved. You can easily create a free account. Depression, ADHD, memory loss, agitation: These may seem like inevitable byproducts of modern lives spent multitasking, not getting enough sleep, and operating on digital overload. But while much of the brain’s work still remains a mystery, a growing body of scientific evidence suggests that the food you eat directly affects how well your brain functions. Brain health also pl.The guests range from super celebs (Jamie Foxx, Arnold Schwarzenegger, etc.) and athletes (icons of powerlifting, gymnastics, surfing, etc.) to legendary Special Operations commanders and black-market biochemists. For most of my guests, it’s the first time they. Groups Discussions Quotes Ask the Author New topics include spread-spectrum systems, cellular communication systems, global positioning systems (GPS), and a chapter on emerging digital technologies such as SONET, ISDN and video compression. To see what your friends thought of this book,When an inexpensive radio receiver (which has a poor selectivity in its RF stage band pass filter) is turned to 1500KHZ, the signal is heard loud and clear. This same signal is also heard (not as well) at another dial setting state with reasons at what frequency you will hear this station. The IF frequency is 455KHZ SOLUTION (less) To view it,To view it,To view it,To view it,There are no discussion topics on this book yet. Solutions Manuals are available for thousands of the most popular college and high school textbooks in subjects such as Math, Science ( Physics, Chemistry, Biology ), Engineering ( Mechanical, Electrical, Civil ), Business and more. Understanding Modern Digital and Analog Communication Systems homework has never been easier than with Chegg Study. Unlike static PDF Modern Digital and Analog Communication Systems solution manuals or printed answer keys, our experts show you how to solve each problem step-by-step. No need to wait for office hours or assignments to be graded to find out where you took a wrong turn. You can check your reasoning as you tackle a problem using our interactive solutions viewer. Plus, we regularly update and improve textbook solutions based on student ratings and feedback, so you can be sure you're getting the latest information available. Hit a particularly tricky question. Bookmark it to easily review again before an exam. The best part? 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