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# Communication Systems I ECE 5660

Utah State University

GPA 3.77

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This 18 page Class Notes was uploaded by Nelle Braun DDS on Wednesday October 28, 2015. The Class Notes belongs to ECE 5660 at Utah State University taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/230403/ece-5660-utah-state-university in ELECTRICAL AND COMPUTER ENGINEERING at Utah State University.

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Date Created: 10/28/15

ECE 5660 Utah State Continuoustime Sampling Spring 2006 University Discretetime Sampling 1 Continuoustime Sampling Let Mt be a continuoustime signal that is to be sampled at a rate f5 lT samplessecond to produce the discretetime sequence x7 xtltnT 96nT The sampling process may be modeled by modulating the signal to be sampled iiei xt in this case with a train of Dirac impulses followed by an operation that integrates the impulses to extract the areas This illustrated in Figure l 95170 pa i 6t7nT n7ocgt Figure 1 Model of the process of sampling a continuoustime signal to produce a discretetime sequence De ne the impulse train by pt i 6t7nT The modulated signal is mama i menT i mwtenT i xnT6t7nTi Note that the desired sample values xnT weight the delta functions and may therefore be extracted by integration Integrating over the interval 2T 7 2T produces the ith sample as follows zT zT 00 mach Z xnT6tenTdt Te Teg 00 zT Z xnT 6t7nTdt eyee 1T7 V 1 239n 61771 0 2393 n Z xnT6ln The two signal processing operations involved in sampling a continuoustime signal ie those illustrated in Figure 1 can be analyzed in the frequency domain Let the continuoustime Fourier transforms CTFT of 95t and xpt be Xw and Xpw respectively and let the discretetime Fourier transform of 957 be Xemi Suppose 95t has the Fourier transform shown in Figure 2 l l l l l i r U 67r 47r 27r 0 27r T T T T HI 4 lt72 e Figure 2 Fourier transform Xw of ECE 5660 Utah State Continuoustime Sampling Spring 2006 University Discretetime Sampling The CTFT ofpt is pt lt gt Pww i0 6w7kw w 2 7ri s 700 s s T This is illustrated in Figure 3 for frequencies from 7 671 to GT However keep in mind that PW is periodic with period 2Tquot and therefore extends from 700 to 00 2TW 7 2 26W Mw 2 21 2 21 2 21 21 T T T T T T T PM l l l l l l l w 6 4 2 0 2 4 e T T T T T T Figure 3 Fourier transform PW ofpti The modulation property of the Fourier transform leads to Xltgt71XltgtPltgtilixlt k 1 p w 7 27quot w w 7 T 700 w as Graphically this says that Xw is simply reproduced at integer multiples of the sample frequency as 2 quot T i The amplitude of each replica is scaled by The CTFT of xpt is illustrated in Figure 4 w 7 Xw XW l l l l l l l T T T T T T T X170 l l l l l l l W 61 41 21 0 2 4 e T T T T T T Figure 4 The Fourier transform Xpw of xpt consists of scaled replicas of Xw appearing at multiples of the sample rate as The relationship between CTFT of xpt and the DTFT of gen can be derived as follows First note that the DTFT of gen is My 2 WW 2 n7ocgt ECE 5660 Utah State Continuoustime Sampling Spring 2006 University Discretetime Sampling Now consider the CTFT of xpt Xpw xpte jmdt Z 6t7nTe jmdt oo Z xnTe jWTn n7ocgt i0 xne jWTn 3 Comparing 2 and 3 indicates that v 9 W Xp T 4 Now substituting from 1 connects Xem to Xw as follows v 1 Q 27r J 7 Xe Tkng T Tk Graphically this says that Xw is simply reproduced at integer multiples of the sample frequency as scaled by and then the frequencies are multiplied by T ie 9 UT The DTFT of 957 is illustrated in Figure 5 Xejinl XW Xejquot l l l l l l l T T T T T T Xem l l l l l l i r 9 76 747139 27r 0 27r 47r 67r Figure 5 Fourier transform Xem of 957 For convenience Fourier transforms of all signals involved in the sampling process ie those shown in Figure 1 are shown together in Figure 6 Note that the highest frequence cum in Xw gets mapped to 0m me through the sampling process In drawing Figure 6 the sample frequency was implicitly assumed to satisfy 5 gt 2w If this were not the case the replicas of Xw in Xpw would overlap and aliasing would result ECE 5660 Utah State Continuoustime Sampling Spring 2006 University Discretetime Sampling 1 m X 1 T T T T T T T r W 7 6 7 7 4 7 7 2 quot 0 2 7 4 quot 6 quot T T T T T T 2 7 2 quot 2 7 2 quot 2 7 2 quot 2 quot T T T T T T T PM T T T T T T T 7 w 7 6 7 47w 7 27w 0 2 4 e T T T T T T l l l l l l l T T T T T T T X2200 T T T T T T T W 7 61 7 47w 7 27w 0 2 4 e T T T T T T l l l l l l l T T T T T T T X em T T T T T T T 9 76w 74w 27r 0 27r 47r 67r 0m me Figure 6 Fourier transforms of all signals involved in the continuoustime sampling process shown in Figure 1 ECE 5660 Utah State Spring 2006 Continuoustime Sampling University Discretetime Sampling 2 Continuoustime Reconstruction Reconstructing xt from samples 957 performs the operations illustrated in Figure 7 3 Se uence to xplt can q LPF m mpulse tr n Figure 7 Model of the process of reconstructing a continuoustime signal from a discretetime sequence First a weighted impulse train is created 7 i0 9576t7nT n 951 1n the frequency domain this corresponds to X1200 XWWT 5 which is the reverse of Graphically this says that the frequencies in Xem are divided by T ie 40 The conversion from Xem to Xpw is illustrated in Figure 1 1 1 1 1 1 1 Xem i i i i i i 9 76 747139 27r 0 27r 47r 67r 1 1 1 1 1 1 1 X2700 l l l l l l l r W 76 74 quot 72 0 2 quot 4 quot 6 quot T T T T T T Figure 8 The DTFT of 957 and CTFT of xpt which are related by Xpw XejWT through a scaling of the frequency axis ie 40 A low pass lter with impulse response Mt is used to remove the unwanted frequency components in xpti Suppose an ideal low pass lter is used Figure 9 shows the low pass lter frequency response The cutoff frequency is half the sample rate 2 1 1n the time domain the action of the lter Mt may be viewed as interpolating between the values of xpt at the sample times t nT 1 HM 2 0 i i 761 741 i i gt w 1 27r 47r 67r T T T T Figure 9 Frequency response of the low pass lter that removes the unwanted components of Xp Filtering xpt by Mt leads to xt with the Fourier transform given by W Xpw M lt f 0 M gt 75 T Substituting from 5 leads to wk WW M T 0 lwlgti ECE 5660 Utah State Continuoustime Sampling Spring 2006 University Discretetime Sampling The Fourier transform of the reconstructed continuoustime signal is shown in Figure 10 along with Fourier transform of the other signals involved in the reconstruction process ie the signals shown in Figure 7 Xem 0 HI 4 lt72 l 6 l 6 l Hw w HI 4 I 67f 47f 27f 1 0 27f 47f 67f TT TT TT T T T T 1 XW l l l l l l i r U 6 4 2 0 21 4 6 T T T T T T 9 W T Figure 10 Fourier transforms of the signals involved in the reconstruction process shown in Figure 7 ECE 5660 Utah State Continuoustime Sampling Spring 2006 University Discretetime Sampling 3 Discretetime Sampling Down Sampling Let yon be a discretetime sequence that is to be down sampled by a factor of N to produce the sequence y ani The down sampling process is modeled by modulation with a train of Kronecker impulses followed by an operation that removes the zeros introduced by modulationi This is illustrated in Figure 11 00 107 Z SnilcN k7ocgt Figure 11 Model of the process of down sampling a discretetime sequence De ne the impulse train by 00 1 E 57k IcN k7ocgt This is a sequence of zeros except for every Nth sample which is one The modulated signal is 00 00 00 1171 In E SnilcN Z xn nikN Z xlcN nikN k7ocgt 1c Note that the desired down sampled sequence ka weights the nonzero samples of wni The down sampled sequence is produced by removing the N 7 1 zeros in every N samples that are generated by modulation with p yk wkN The two signal processing operations involved in down sampling can be analyzed in the frequency domain Let the DTFTs of x wn y be Xejn Wham Yem respectively The DTFT of pn is 00 7 27r 00 N71 pw 1 osmgwami 71195 as g 13 00697 ms 7 27 as N The DTFTs of gen and pn are illustrated in Figure 12 Note that in the plot of Pem N 3 was used 1 1 1 X in 5 l l l l l l l r 9 73 727139 77139 0 7quot 27r 37r P 1390 2wquot 2wquot 2wquot 2wquot 2wquot 2wquot 2wquot 2wquot 2wquot e N3 l l l l l l l l l 9 73 727139 77139 0 7quot 27r 37r Figure 12 The DTFTs of gen and p The modulation property of the DTFT involves circular convolution IIIem Xem 93PM ECE 5660 Utah State Continuoustime Sampling Spring 2006 University Discretetime Sampling Applying this property leads to 1 N71 W m X 9495 6 e gt N e gt lt gt Graphically this says that Xem is simply reproduced at integer multiples of the sample frequency 95 2 quot The amplitude of each replica is scaled by These operations are illustrated in Figure 13 for the case N 3 IIIem f t t t t t t t N 3 i i i i i i 9 73 727139 77139 0 7quot 27r 37r Figure 13 The DTFT of 11 contains N replicas of the DTFT of 957 which occur at integer multiples of the the sample frequency 95 Wquot The relationship between the DTFT of w and yn can be derived as follows W Zykeimk k2 2 wkNeij k 1c Z w e jn N change of variables in summation 7LkN 00 Z w e mnN because 11 has N 7 1 zeros between each nonzero sample n7oo WW39QN 7 Finally substituting from 6 leads to Z 71 my X ejm kQWVN s 0 a H Graphically 7 says that Yejn can be thought of as a frequency scaled version of Wem Equation 8 says that Yem can be generated by reproducing Xem at integer multiples of the sample frequency 95 scaling the amplitude by and then multiplying all frequencies N iiei 0N The DTFT of y is illustrated in Figure 14 along with the other signals involved in the down sampling process Notice that the highest frequency 90 in 957 is mapped to 91 QON in y through the down sampling process In drawing Figure 14 the sample frequency 95 was implicitly assumed to satisfy 95 gt 290 If this were not the case the replicas of Xem in Wejn would overlap and aliasing would result ECE 5660 Utah State Continuoustime Sampling Spring 2006 University Discretetime Sampling 1 1 9 1 X em 1 1 1 1 1 1 1 r 9 737139 727139 77139 0 7quot 27139 37139 1 271 271 271 271 271 271 271 271 271 PM W W W W W W W W W N 3 1 l l l 1 l l l 1 l l l 1 17 737139 727139 77139 0 7quot 27139 37139 WM t N 1 1 1 1 1 1 1 Q 737139 727139 77139 0 7quot 27139 37139 i L L N N N Yem 1 1 1 1 1 1 1 Q 737139 727139 77139 0 7quot 27139 37139 01 QON Figure 14 The DTFTS of the signals involved in the clown sampling processi ECE 5660 Utah State Spring 2006 Continuoustime Sampling University Discretetime Sampling 4 Discretetime Reconstruction Up Sampling Up sampling to obtain y from yon reverses the operations in down sampling as illustrated in Figure 15 Figure 15 Model of the up sampling process First wn is obtained by inserting N 7 1 zeros in between each sample of xki yonN n is a multiple of N w n 0 otherwise In the frequency domain this corresponds to IIIem Xw Ny 9 Graphically this says that the frequencies in Xejn are divided by N ie 0 A This creates N replicas of Xejni The desired replica is obtained by low pass ltering The transition band edges are Qpass and Qsmp 7 The DTFT of the lter output y is N V 1399 Q 1 We we gt m lt 92 I 0 gt 32 Ni Substituting from 9 leads to We XMW M lt f 0 M gt 75 The signals and systems involved in the up sampling process are illustrated in Figure 16 for the case N 3 Notice that the highest frequency 90 in gen gets mapped to 91 l in the up sampled signal yni ECE 5660 Utah State Continuoustime Sampling Spring 2006 University Discretetime Sampling 1 1 9 1 X em 1 1 1 1 1 1 1 r 9 737139 727139 77139 0 7quot 27139 37139 WM 1 1 1 1 1 1 1 1 1 N 1 1 1 1 1 1 1 r 9 737139 727139 77139 0 7quot 27139 37139 1 1 1 H 519 9 1 1 1 1 1 1 1 1 1 737139 727139 77139 0 7quot 27139 37139 1 1 1 Yem 1 1 1 1 1 1 1 Q 737139 727139 77139 0 7quot 27139 37139 91 To Figure 16 The DTFTS of the signals involved in up sampling ECE 5660 Spring 2006 Utah State University Sampling and Aliasing 1 Analog and DiscreteTime Sinusoids Properties of Analog and DiscreteTime Sinusoids xatAcosQtt9 7ltxgtlttltltxgt Q 27rF radianssecond F cyclessecond or Hertz Periodicity xat is periodic for 700 lt 9 lt 0 Uniqueness cos91t cosQQt for all t if and only if 91 92 Frequency range Unlimited frequency range lncreasing 9 increases the oscillations xnAcoswn9 7271012 w 27rf w radianssample yclessample Periodicity is periodic if and only if f rational Uniqueness cosw1n cosw2n for all n if and only if lt02 lt01 27rk Frequency range Limited frequency range Highest rate of oscillation is w in or f i De nition The Fundamental period is the smallest integer N that satis es xn N Example Calculate the fundamental period of cos Engn Solution Express f as a ratio of relatively prime integers 12 1 175 N73 1 Example Calculate the fundamental period of cos Engn Solution Principle Small changes Example Uniqueness of sinusoids Conclusion In general discretetime Sinusoids cos27rf0 kn cosw0 27rkn Example Limited frequency range Consider just slightly larger than 7r 13 N 36 2 f 36 7 in the frequency can lead to large changes in the fundamental period 4 27r ngt 3 cos lt 1 cos 27r1 1 cos 27m 27r n 1 cos lt27r ngt are not unique cos27rf0n for all integers k n cosw0n for all integers k n a discretetime sinusoid With frequency w 7r e that is cos7r en cos7r en 7 27m 9 cos77r en 10 cos77r 7 en 11 cos7r 7 en 12 ECE 5660 Utah State Spring 2006 University Sampling and Ahasmg Conclusion In discretetime a cosinusoid with frequency 7re is equal to a cosinusoid with frequency 7r7 6 Therefore 7r e is a lower frequency than 7r Now let 5 A 0 and we see that 7r is the highest frequency in discretetime Since 40 7r radians per sample is the same as f cycles per sample we have the bounds 1 5 13 1 77rltwlt7r figfg 2 Sampling So far we have only compared the properties of analog and discretetime sinusoids Now let s consider sampling analog sinusoids to create discretetime sinusoids Let s assume we use uniform sampling which means that xat when t Tsn where Ts is called the sample period Note that is only de ned for n in the integers The sample frequency or sample rate is de ned to be F5 Consider sampling a sinusoidal signal xat cos27rFt 14 zaTsn cos27rFTsn 15 F cos lt27rEngt 16 The discretetime frequency of the sampled sinusoid is f The highest frequency represented in discretetime is cycles per sample Hence if we want to avoid aliasing we must have assuming F gt 0 F 1 F 1 7r F s O F 17 f Fslt2 lt2 2T5 or lt7rs Ts In general an analog sinusoid with frequency F will alias to a discretetime sinusoid with frequency f given by F F 1 7 18 f F5 F5 1 2T 3 Reconstruction Sampling Theorem If the highest frequency contained in an analog signal xat is Fm and the signal is sampled at a rate F5 Tis gt 2me then xat can be exactly recovered from its sample values using the interpolation function sin27rFst gm 2W 19gt Thus xat may be expressed as wt 2 xanTsgtinTs 20 n7ocgt Proof Assume that F5 gt 2Fm u This means that Xam 0 for m gt T1 21 The samples xanTs can be extracted by modulating xat by the periodic impulse train pt given by W Z w 7 nTs i Pm 2T1 Z 6 lt9 7 22 5 k7oo s n7ocgt Let xdt xatpt be the output of the modulator 13 i xanTs6t7nTs LXMQ Ti i X m 2 23 5 k7oo 5 n7ocgt ECE 5660 Utah State Spring 2006 University Sampling and Aliasmg Note that xdt is zero except at the sampling instants nTs The km sample value can be extracted by integrating xdt from k 7 Ts to k Ts Notice that the spectrum of X0109 is the sum of shifted replicas of Xa 1f xat is bandlimited as in 21 then these shifted replicas do not overlap and xat can be recovered by an ideal lowpass lter This is the approach that we take next However if xat violates 21 by having signal content above T15 then these high frequency regions overlap when XajQ is shifted and summed to produce X0309 This overlapping in the frequencydomain is called aliasing Aliasing is an irreversible eliect De ne an interpolation lter gt by 7 sin27rFst j 7 Ts for lt Tl gt 7 27rFst lt gt GOQ 7 0 fer 2 Tl 24 Due to 24 and 21 we have XaUQ GUdeUQ 25 In the timedomain xat is given by the following convolution W 7 mm 7 rgtdT 26gt 1 2 xanTs6r 7 nTsgt TM 27 Z xanTs 100 67 7 nTsgt 7 Td7 28 Z xanTsgt 7 nTs 29 n7oo 4 Simulation of Continuoustime Sinusoidal Signals in Discrete time Using Matlab 41 Time Domain Simulation Suppose you want to simulate the analog signal Ct cosQCt 95 cos27rFCt 95 for over the time interval 0 S t S 2 in Matlab where FC 10 Hz and 9C radians According to the sampling theorem Ct can be represented by discretetime samples provided the sample rate F5 is more than twice the largest frequency in Ct The highest frequency and only frequency in Ct is FC 10 Hz Therefore the sample frequency must be greater than 20 Hz to avoid aliasing 1f visualization or demonstration is the purpose of the simulation it is a good idea to choose the sample rate 10 or more times faster than is required by the sampling theorem On the other hand if representation by discretetime samples is all that is required then sampling at 21 Hz will do Let s suppose that we choose F5 5000 Hz as the sample rate In Matlab we will work with the discretetime sequence c7 CnT instead of Ct where T Before generating the sequence cn we must generate the time vector tn nT to represent time over the interval 0 3 seconds The following Matlab commands will accomplish the construction of cw fs 5000 Z Sample frequency T 1fs Z Sample period tn 0T3 39Z Time vector with samples spaced T seconds appart fc 10 Z Frequency of the sinusoid theta pi3 39Z Phase of the sinusoid cn cos2pifctn theta Z Construct the sinusoid plottncn Z Plot the sinusoid The nal command plots the sinusoid Here is what the plot looks like ECE 5660 Utah State Spring 2006 University and 1 i i i i i 05 n 39U E i 0 E u 05 c 1 0 05 25 3 1 1 5 2 time seconds You can zoom in and use the time axis to measure the period of the sinusoidi Invert the period measure ment and verify that the frequency really is 10 Hz You can also measure the frequency of the sinusoid by looking for a peak of the spectrum of the signali 42 Viewing the Spectrum of the Sampled Signal Matlab s fft function may be used to compute the discrete Fourier transform DFT of the sampled signali Recall that discrete frequencies occupy the interval 7 The correspondence between discretetime f A i f and quot V f A i F is given by the formula F f F 5 If we want frequency components in the sampled signal to appear in a plot of the spectrum at their true continuous time frequencies assuming no aliasing we have to generate a frequency vector that covers the range 7 Fs Fs with steps of F5 N which is the sample frequency times the FFT bin widthi Plotting the spectrum can be accomplished in Matlab using the following commands N 2 14 39Z FFT size f 0N 1N 05fs 39Z The frequency vector for plotting C fftshiftltfftcnN 39Z Compute the FFT and rearrange the output plotf10log10absC 39Z Plot the magnitude of the spectrum on a X log scale Here is what the spectrum looks like Note that the xaxis has the correct frequency labeling and scaling m 10 i i i 39m on 2000 1000 0 1000 2000 ECE 5660 Utah State Spring 2006 University Sampling and Ahasmg It is difficult to see the details of the spectrum around 10 Hz The plot below shows the details around 10 Hz Circles have been added around the data points which are connected by straight lines by the plot functions 10 log C 15 20 10 frequency Hz It appears that there is a peak in the spectrum at 10 Hz However zooming in to the interval 8 12 Hz as shown in the gure below reveals that the peak is not actually at 10 Hz This is due to the spectral leakage77 phenomenon frequency Hz If the frequency of the continuoustime signal is changed to 9765625 Hertz which is a FFT bin center frequency mapped back to a continuous time frequency then the plot looks like this ECE 5660 Utah State Spring 2006 University and 30 20 Q 107 r U 2 9 0 7 10 20 quot 2000 1000 0 1000 2000 frequency Hz Note that except for the two peaks which are unresolvable because they are at low frequency the spectrum is effectively zero less than 100 dB Here is what the picture looks like zoomed to the frequency range 8 to 12 10 log C 10 frequency Hz Next we explain how to construct sinusoids with frequencies which are DFT bin center frequencies 43 Constructing Bin Center Discretetime Sinusoids Spectral leakage due to processing nite length data records srnears the spectral lines of pure tones across several bins around the tonal frequency Sinusoids with frequencies which are FFT bin centers are not srneared however The FFT bin center frequencies are 0 W cycles per sample The fundamental range77 can be shifted by to the following set of frequencies 7 7 7 i Using the correspondence between continuous and discrete f A i the co 1 quot quot tirne f that are not srneared are given by 1 k Fbin center F5NFs k01N71 30 ECE 5660 Spring 2006 Utah State University Sampling and Aliasing Exercises 1 9quotquot What is the frequency in Hertz of the sinusoidal signal xt sin 197131 If this signal is sampled at a rate of 3 samples per second What is the sampled signal and What is the perceived frequency What is the fundamental period in samples of cos Simulate the continuoustime signal Ct cos27rFCt 95 in Matlab using a sample rate F5 15 Hz Let 1 27r3 radians Let N 212 be the size of the FFT that Will be used for analyzing the sampled signal a Choose FC so that the sampled sinusoid c7 CnT has a discrete time frequency that is an FFT bin center frequency Fa 0 is not allowed Remember that this is a demonstration So we want F5 to be much greater than Fa What value of FC did you choose What FFT bin center frequency did you choose b How many seconds of data should you generate in Matlab so that you Will have exactly N samples of the sinusoid c Generate the sinusoid in Matlab and plot it The xaxis should be properly scaled so that the units are in secon s d Using the fft function compute the spectrum and plot it The xaxis in this plot should be properly scaled so that the units are in Hertz Does your plot appear as predicted

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